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Which sentence has the same meaning? "it was nice of you to hel-Turito
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Which sentence has the same meaning?
"It was nice of you to helps tie the boy's shoe”
A. It was kind of you to help tie the boy's shoe
It was mean of you to help tie the boy's shoe
The correct answer is: It was kind of you to help tie the boy's shoe
Correct answer - a) It was kind of you to help tie the boy's shoe
Explanation - kind is a synonym of nice.
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Isometry is a geometric transformation that preserves the distances between points.
In simpler terms, if two points have a certain distance between them before the transformation, that distance remains unchanged afterward.
This means that every isometry transforms a shape into a congruent shape.
For example, the first triangle is rotated 90° clockwise and shifted to the right. The distances between points A, B, and C remain the same.
Therefore, rotation and translation are both examples of isometric transformations.
Additionally, both triangles are congruent because their side lengths stay the same.
The term 'isometric' is often used interchangeably with 'congruent'. The word has Greek origins, where 'isos' means 'same' and 'metros' means 'measure'. For example, the two triangles mentioned
earlier are congruent because their sides and angles remain congruent even after rotation and translation.
Transformations that result in an isometry are called isometric transformations or rigid transformations (or rigid motions) because they do not alter the shape or size of objects.
Shapes that remain unchanged under isometry are called isometric figures, as they retain the same measurements.
Types of Isometries
The main types of isometric transformations are:
Compositions of Isometries
When you apply two or more isometric transformations in a specific sequence to a geometric figure, the final result can be described as a composition of isometries.
For example, you can translate a triangle 10 cm along a horizontal line (translation) and then rotate it 180° (rotation).
It's important to note that the composition of two isometries is still an isometry.
Since each isometry individually preserves distances and shapes, a composition of isometries will also retain these properties.
In other words, a figure subjected to a composition of isometries maintains its original shape and size.
Note: However, the composition of isometries is not always commutative. This means the order in which you perform the geometric transformations can affect the final result. For example, a rotation
followed by a translation yields a different result than a translation followed by a rotation.
Isometric compositions can have specific names depending on the transformations combined. For example:
• Glide Reflection
The combination of a reflection and a translation is known as a "glide reflection".
• Rototranslation
The combination of a rotation and a translation is referred to as a "rototranslation".
Invariants in Isometry
In an isometry, certain characteristics of a shape may change, such as its position, while others remain unchanged.
The features that do not change are known as "invariants" in isometry.
The main invariant properties during an isometry are:
• Distance between points
In isometries, the distance between points remains unchanged. This means that, for example, if the distance between two points A and B is 5 cm in the original figure, it will remain the same in
the resulting figure after an isometric transformation.
• Length of segments
The length of segments does not change between the original figure and its image. For instance, the segments AB, BC, and AB have the same length in both triangles.
• Angle measures
The measure of the angles remains constant after an isometry. For example, the angles α, β, and γ of the triangle maintain the same measure after the isometric transformation.
• Area of the figure
The total area of the figure does not change. For example, the area of the triangle remains the same after the isometric transformation.
• Shape and size
In an isometry, the shape and size of the geometric figure do not change. These properties are invariant.
• Alignment of points
If two or more points are aligned in the original figure, they will also be aligned in its image. For example, points A and B are aligned in both the original triangle and the resulting triangle
after the transformation.
• Parallelism of lines
Lines that are parallel in the original figure remain parallel in the image.
• Perpendicularity of lines
Lines that are perpendicular in the original figure maintain this characteristic in their image as well. For example, segments AB and BC are perpendicular in both the first triangle (original)
and the second triangle (isometric image).
Additional Observations
Here are some additional observations and side notes on isometries:
• Isometries are congruences
Isometries are geometric transformations that produce congruent figures because they preserve segment lengths and angle measures. As a result, the transformed figures are point-for-point
superimposable onto the original figures after some rigid motions, fulfilling the definition of congruence in Euclidean geometry.
• Isometries are a specific type of similarity
Isometries are similarities with a similarity ratio of 1. They satisfy all the properties of similarities:
□ they preserve the parallelism between segments,
□ corresponding angles are congruent, and
□ corresponding segments are congruent.
Therefore, in an isometry, the shape of the figure does not change, which is an invariant property (as in similarities), because the angles remain congruent. Additionally, the corresponding
segments have the same length, meaning they are proportional with a ratio of 1.
And so on. | {"url":"https://www.andreaminini.net/math/isometry","timestamp":"2024-11-01T23:36:38Z","content_type":"text/html","content_length":"18096","record_id":"<urn:uuid:dd369f96-3c43-4a15-99b8-f3b630242721>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00401.warc.gz"} |
A "New Set of Numbers" (An Introduction of Integers)
Barbara Barlow Caldwell School
8546 S. Cregier
Chicago IL 60617
Students of the sixth grade will be able to add and subtract positive and
negative integers. Incidentally, negative numbers were invented by the Chinese.
Materials Needed:
Group of fifteen students
Numbers with positive signs (written on construction paper)
Numbers with negative signs (written on construction paper)
A large zero (written on construction paper)
Two paper bags (one labeled positive and one labeled negative) with
directions inside which tell the students what to do on the number line. Some
examples are:
Take 3 negative steps Add negative 3
Take 5 positive steps Subtract positive 4
Review of number line on chalkboard:
Number line will be made, using students as the numbers. Students to the
right of the zero will be called positive integers and all students to the left
of the zero will be called negative integers. The distance from the zero to the
next point (going in either direction) will be called the "unit distance." Each
student will tell how many units they are from zero. The students will now
receive their signed numbers written on construction paper.
The "additive inverse" concept will now be introduced and explained. The
students will then find their additive inverse.
Students will move a given number of units (to the left or to the right) to
represent the process of adding and subtracting integers. They will then name
their new value.
Students will now pull directions from a bag and each take turns moving in
the given direction; Ex: Take 2 positive steps.
Students will be given a sheet of problems to practice adding and
subtracting integers.
Performance Assessment:
Students will solve a puzzle by adding and subtracting integers.
After presenting this lesson, students will have a greater understanding of
integers and the concept of positive and negative numbers. They will also know
how to add and subtract integers.
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A Bayesian updating framework for calibrating the hydrological parameters of road networks using taxi GPS data
Articles | Volume 27, issue 20
© Author(s) 2023. This work is distributed under the Creative Commons Attribution 4.0 License.
A Bayesian updating framework for calibrating the hydrological parameters of road networks using taxi GPS data
Hydrological parameters should pass through a careful calibration procedure before being used in a hydrological model that aids decision making. However, significant difficulty is encountered when
applying existing calibration methods to regions in which runoff data are inadequate. To achieve accurate hydrological calibration for ungauged road networks, we propose a Bayesian updating framework
that calibrates hydrological parameters based on taxi GPS data. Hydrological parameters were calibrated by adjusting their values such that the runoff generated by acceptable parameter sets
corresponded to the road disruption periods during which no taxi points are observed. The proposed method was validated on 10 flood-prone roads in Shenzhen and the results revealed that the trends of
runoff could be correctly predicted for 8 of 10 roads. This study demonstrates that the integration of hydrological models and taxi GPS data can provide viable alternative measures for model
calibration to derive actionable insights for flood hazard mitigation.
Received: 05 Jan 2023 – Discussion started: 24 Jan 2023 – Revised: 20 Jun 2023 – Accepted: 11 Sep 2023 – Published: 27 Oct 2023
In the context of climate change and increased urbanization, flooding poses far-reaching threats to urban road networks of coastal metropolises (Balistrocchi et al., 2020). In Australia,
approximately 53% of flood-related drowning deaths were the result of vehicles driving into flood waters between 2004 and 2014. Additionally, indirect losses caused by flooding such as canceled
commutes, mandatory detours, and travel time delays often outweigh direct losses (Kasmalkar et al., 2020). Quantifying the impact of flood exposure requires the prediction of surface runoff over
roads and road disruptions induced by runoff, which are critical for the implementation of flood mitigation, traffic resilience improvement, and early warning systems.
Public concerns regarding road flooding hazards have created pressure to develop fine-grained and accurate models for hydrological simulation. Hydrological modeling is based on a relatively
well-established theory that can provide approximations of real-world hydrological systems and has been widely used in many road-related studies (Versini et al., 2010; Yin et al., 2016;
Safaei-Moghadam et al., 2023). Because hydrological modeling is subject to uncertainty that arises from the oversimplified reflection of hydrological systems, initial and boundary conditions, and
lack of true knowledge, parameters for hydrological models must be carefully calibrated prior to their application to practical problems, so that models can closely match the historical trends (Gupta
et al., 1998). As uncalibrated models are indefensible and sterile, very few models documented in the literature have been applied without a calibration procedure (Beven, 2012).
Over the past four decades, numerous studies have been conducted on the development of calibration methods. Methodologies for model calibration range from simple trial-and-error methods that adjust
one parameter value in each iteration until the differences between predicted and observed values are satisfactory to Bayesian updating frameworks that reject the concept of a single correct
solution. To a great extent, the success of model calibration is dominated by the availability of field-observed runoff data. However, runoff data are generally only gathered at a few sites, and some
cities never measure runoff data in built-up regions (Gebremedhin et al., 2020). Although runoff data can be effectively collected by administration departments in some cities, these cities are not
always motivated to share these data with the public. For example, among China's top 10 largest cities,^1 only Shenzhen has shared runoff-related data on an open data platform. For model calibration
at the road scale, runoff data are even more difficult to acquire because road networks are far denser than river networks and flood gauges are only installed in a few flood-prone roads based on
their high measurement cost, leaving most roads ungauged. As pointed out by Beven (2012, p. 55), “the ungauged catchment problem is one of the real challenges for hydrological modelers”.
This lack of hydrological data has prompted researchers to seek additional data sources to support flood-related decision making. Based on the advancement of mobile telecommunication technologies,
big data are emerging as alternative sources of information for coping with flood risks (Paul et al., 2018; Li et al., 2018; Gebremedhin et al., 2020). Citizens can voluntarily or passively act as
human sensors to generate georeferenced data to improve flood monitoring. Many studies have leveraged crowdsourced social media data (Brouwer et al., 2017; Sadler et al., 2018; Zahura et al., 2020),
mobile phone data (Yabe et al., 2018; Balistrocchi et al., 2020), and taxi GPS data (She et al., 2019; Kong et al., 2022). However, most previous works have concentrated on using big data either for
flood mapping or mining spatiotemporal patterns (Restrepo-Estrada et al., 2018), and parameter calibration for ungauged roads based on big data remains a problem.
This study extends our previous study (Kong et al., 2022) by going a step further than simply recognizing flooded roads. We propose a calibration method for road-related hydrological parameters based
on taxi GPS data. Many studies have shown that vehicle-related information during rainfall, including vehicle volume, speed, and trajectory information, is useful for flooded road detection (Zhang et
al., 2019; Qi et al., 2020; Yao et al., 2020). When a road segment is inundated by heavy rainfall, the vehicle volume may exhibit a sharp or gradual drop depending on the intensity of the rainfall
event. Conversely, an abnormal drop in vehicle volume during the rainfall may imply that a road has experienced rainfall-induced inundation. This motivates us to use traffic-related data sources to
calibrate hydrological parameters. In this study, we developed a transformation process that converts rainfall time series data into a time series of probabilities that no taxis will drive on a road
(“no-taxi-passing probability” hereafter) for a given hydrological parameter set. We then assigned a probability to every parameter set by integrating the no-taxi-passing probability with observed
taxi GPS data. We outlined a generalized taxi-data-driven calibration framework and implemented a framework with specific hydrological and transportation models.
2.1Bayesian updating procedure
Observed data are not always as informative as expected and may be inconsistent with other data sources. Hydrologists typically adopt the Bayesian framework to update hydrological parameters, which
provides a generalized formalism that integrates prior probability representing prior knowledge with likelihood that reflects how accurately a model can reproduce observations to form a posterior
probability. Suppose we have several versions of a hydrological model, each with different sets of parameters. Then, the purpose of the Bayesian updating procedure adopted in this study is to assign
a posterior probability to every hydrological parameter set as new taxi data become available.
Two components are critical for this Bayesian updating procedure: one is the prior probability and the other is the likelihood function. Regarding the prior probability, for their famous calibration
model called generalized likelihood uncertainty estimation, Beven and Binley (1992) stated that all parameter combinations are considered equally probable before additional information is introduced.
After the first update, the prior probability of each updating iteration can be replaced by the posterior probability of the latest updating iteration. Likelihood, which is a measurement of how well
a given model conforms to the observed taxi behavior, is not as easy to compute as the prior probability because the parameter set to be estimated is hydrology related, whereas the observed evidence
is taxi related. Therefore, we must determine how to construct a taxi-based proxy whose probability is equal to the associated hydrological parameter and construct a function enabling the
transformation from hydrological parameters to taxi-related proxies.
The proxy selected in this study was the time series of the no-taxi-passing probability. Figure 1 presents a generalized procedure for converting a rainfall time series into a time series of
no-taxi-passing probabilities for each hydrological parameter. This procedure consists of three steps. First, a hydrological model is used to convert a rainfall time series into a hydrograph. Second,
a runoff-disruption function that relates runoff to the probability that a road is blocked is used to transform the hydrograph into a time series of road disruption probabilities. Third, the taxi
arrival rate is combined with the time series of road disruption probabilities to derive a time series of no-taxi-passing probabilities. The hydrological model and taxi arrival rate are considered to
be unique for every road and are invariable within a short period, whereas the runoff-disruption function is identical for all roads.
Integrating this three-step process with the Bayesian equation enables us to compute the posterior probability of a parameter set based on taxi data. For a specific road, suppose there are N
hydrological parameter sets to be estimated. Because the runoff-disruption function and taxi arrival rate are assumed to be fixed for the road, we can construct a composite function converting the i
th parameter set, which is denoted as θ^(i), into a time series of no-taxi-passing probabilities, which is denoted as Ω^(i). Therefore, the probability of θ^(i) being optimal is equal to the
probability of Ω^(i) being true, which can be expressed as follows:
$\begin{array}{}\text{(1)}& P\left({\mathbit{\theta }}^{\left(i\right)}\right)=P\left({\mathbf{\Omega }}^{\left(i\right)}\right),\end{array}$
where P(θ^(i)) and P(Ω^(i)) are the prior probabilities of θ^(i) and Ω^(i), respectively. As taxi observations become available, P(θ^(i)) (or P(Ω^(i))) can be updated using the Bayes theorem as
$\begin{array}{}\text{(2)}& P\left({\mathbit{\theta }}^{\left(i\right)}\mathrm{|}\mathbf{X}\right)=P\left({\mathbf{\Omega }}^{\left(i\right)}\mathrm{|}\mathbf{X}\right)\propto P\left({\mathbit{\theta
}}^{\left(i\right)}\right)L\left(\mathbf{X}\mathrm{|}{\mathbit{\theta }}^{\left(i\right)}\right),\end{array}$
where X is the taxi observation, and P(θ^(i)|X) and P(Ω^(i)|X) are the posterior probabilities of θ^(i) and Ω^(i), respectively, conditional on the taxi observation. L(X|θ^(i)) is the likelihood of
X given θ^(i). The optimal parameter set is that which yields the Ω^(i) that most closely fits the observed taxi data.
Solving Eq. (2) involves the calculation of P(θ^(i)) and L(X|θ^(i)). The derivation of P(θ^(i)) depends on prior knowledge regarding the parameter distribution, which is typically obtained using
traditional hydrological methods. However, this prerequisite knowledge may not always be readily accessible based on limited data availability. In such cases, Beven and Binley (1992) suggested that
any parameter set combination could be considered to be equally likely. This implies that the parameter set is drawn from a uniform distribution as follows:
$\begin{array}{}\text{(3)}& P\left({\mathbit{\theta }}^{\left(i\right)}\right)=\mathrm{1}/N.\end{array}$
In this study, we compared the effects of two types of prior parameter distributions, namely a uniform distribution and a distribution derived from digital elevation model (DEM) data, on the
resulting posterior distributions.
L(X|θ^(i)), which is a likelihood function, describes the joint probability of the observed taxi data X as a function of the chosen θ^(i). Consider a rainfall event that is divided into T 5min
intervals. From the taxi data, we can obtain a sequence of taxi-related observations, which are denoted as $\mathbf{X}=\mathit{\left\{}{x}_{\mathrm{1}},{x}_{\mathrm{2}},\mathrm{\dots },{x}_{T}\mathit
{\right\}}$, where x[t]=1 if the observed road has at least one taxi pass during the tth interval, and x[t]=0 otherwise. ${\mathbf{\Omega }}^{\left(i\right)}=\mathit{\left\{}{\mathit{\omega }}_{\
mathrm{1}}^{\left(i\right)},{\mathit{\omega }}_{\mathrm{2}}^{\left(i\right)},\mathrm{\dots },{\mathit{\omega }}_{T}^{\left(i\right)}\mathit{\right\}}$ is also a T-dimensional vector, where ${\mathit
{\omega }}_{t}^{\left(i\right)}$ is the no-taxi-passing probability in the tth interval with θ^(i) as the parameter set. Note that Ω^(i) is only determined by the chosen hydrological parameter and
rainfall time series, and is not measured from observed data. Considering that the arrival of taxis is independent of time, L(X|θ^(i)) can be formulated as
$\begin{array}{}\text{(4)}& L\left(\mathbf{X}\mathrm{|}{\mathbit{\theta }}^{\left(i\right)}\right)=L\left(\mathbf{X}\mathrm{|}{\mathbf{\Omega }}^{\left(i\right)}\right)=\prod _{t=\mathrm{1}}^{T}\left
(\mathrm{1}-{\mathit{\omega }}_{t}^{\left(i\right)}{\right)}^{{x}_{t}}\left({\mathit{\omega }}_{t}^{\left(i\right)}{\right)}^{\mathrm{1}-{x}_{t}}.\end{array}$
By substituting Eq. (4) into Eq. (2), the following equation can be obtained:
$\begin{array}{}\text{(5)}& P\left({\mathbit{\theta }}^{\left(i\right)}\mathrm{|}\mathbf{X}\right)\propto P\left({\mathbit{\theta }}^{\left(i\right)}\right)\prod _{t=\mathrm{1}}^{T}\left(\mathrm{1}-
{\mathit{\omega }}_{t}^{\left(i\right)}{\right)}^{{x}_{t}}\left({\mathit{\omega }}_{t}^{\left(i\right)}{\right)}^{\mathrm{1}-{x}_{t}}.\end{array}$
Equation (5) is the proposed Bayesian updating model for calibrating hydrological parameters based on taxi data, where X can be directly measured and ${\mathit{\omega }}_{t}^{\left(i\right)}$ is
calculated through the three-step process illustrated in Fig. 1, which will be discussed in detail in the following section. Having selected an updating model, the optimal parameter for one period of
observations may not be optimal for another period. Because the model may have continuing inputs of new taxi observations, the posterior probability for θ^(i) should be updated as new evidence
becomes available. For the second update, the posterior probability from the first observation becomes the prior probability for the second observation and the posterior probability for θ^(i) is
recursively updated as
$\begin{array}{}\text{(6)}& P\left({\mathbit{\theta }}^{\left(i\right)}\mathrm{|}{\mathbf{X}}_{\mathrm{2}}\right)\propto L\left({\mathbf{X}}_{\mathrm{2}}\mathrm{|}{\mathbit{\theta }}^{\left(i\right)}
\right)P\left({\mathbit{\theta }}^{\left(i\right)}\mathrm{|}{\mathbf{X}}_{\mathrm{1}}\right),\end{array}$
where X[1] and X[2] are the first and the second taxi observations.
2.2Instantiation of the three-step procedure
Section 2.1 presented a generalized three-step procedure for converting a rainfall time series into a time series of no-taxi-passing probabilities. In this section, we specialize this process by
integrating existing theories with our model. The three conceptualized steps illustrated in Fig. 1 were replaced with three concrete submodels. First, a Soil Conservation Service (SCS) unit
hydrograph was used to convert rainfall excess into a hydrograph of the target road. Second, an empirical runoff-disruption function based on data extracted from various experimental, observational,
and modeling studies was applied to convert the hydrograph into a time series of road disruption probabilities. Third, a Poisson distribution representing the distribution of taxi arrival rate was
combined with the road disruption probability time series to derive a no-taxi-passing probability time series.
2.2.1Step 1: converting rainfall into runoff based on the SCS unit hydrograph
Not all rainfall produces runoff because soil storage can absorb a certain amount of rain. However, in urbanized areas, only a small proportion of rainfall infiltrates the soil or is retained on the
land surface, leaving most rain to flow across urban surfaces and become direct runoff. The rainfall that becomes direct runoff is referred to as rainfall excess. The Natural Resources Conservation
Service (NRCS)^2 developed a method to estimate rainfall excess based on soil types and land uses using the following curve number equation:
where P[e] is the accumulated rainfall excess in centimeters, P[a] is the accumulated rainfall in centimeters, and S is the potential retention after runoff begins, which is defined as a function of
the curve number, that is,
$\begin{array}{}\text{(8)}& S=\mathrm{2.54}×\left(\mathrm{1000}/\mathrm{CN}-\mathrm{10}\right),\end{array}$
where CN is the curve number. For urban and residential land, the curve number varies from 40 to 95 depending on the impervious area (Natural Resources Conservation Service, 2010a). Because prior
knowledge on the CN is unavailable, it was considered as a calibrated parameter in this study.
The rainfall excess derived using Eq. (7) was inputted into the unit hydrograph to derive the runoff. The unit hydrograph is a commonly used rainfall-runoff model that converts rainfall excess into a
temporal distribution of direct runoff. First proposed by Sherman (1932), the unit hydrograph is defined as the hydrograph resulting from one unit of rainfall excess distributed uniformly over a
catchment area. It assumes that rainfall is uniform over the catchment area and that runoff increases linearly with rainfall excess. Although these assumptions cannot be perfectly satisfied under
most conditions, the results obtained from the unit hydrograph are generally acceptable for most practical cases. The model, originally designed for larger watersheds, has been found to be applicable
to some catchment areas less than 5000m^2 in size (Chow et al., 1988).
The unit hydrograph is only applicable to watershed areas where runoff data are measured. The paucity of runoff data motivated the development of the synthetic unit hydrograph (SUH) concept. The term
“synthetic” in SUH refers to a unit hydrograph derived from watershed characteristics, rather than empirical rainfall-runoff relationships. In this study, we utilized the SCS unit hydrograph, which
is a dimensionless SUH proposed by the NRCS. For the dimensionless SUH, the discharge (i.e., y axis) is expressed as the ratio of discharge q to peak discharge q[p] and the time (i.e., x axis) is
expressed as the ratio of time t to peak time t[p]. Therefore, the SCS unit hydrograph is not exactly an SUH itself, but it is a useful tool for constructing an SUH.
The shape of an SCS unit hydrograph is entirely determined by the peak rate factor. A standard value of 2.08 for the peak rate factor is recommended and commonly used by the NRCS (Fig. 2). To
construct an SUH from an SCS unit hydrograph, the x axis of the SCS unit hydrograph is multiplied by t[p] and the y axis is multiplied by q[p]. The values of q[p] and t[p] are functions of the
catchment area and time of concentration as follows:
$\begin{array}{}\text{(9)}& {t}_{\mathrm{p}}=\mathrm{0.6}{t}_{\mathrm{C}}+D/\mathrm{2}\text{(10)}& {q}_{\mathrm{p}}=\mathrm{2.08}A/{t}_{\mathrm{p}},\end{array}$
where t[C] is the time of concentration in hours, A is the catchment area in square kilometers, and D is the duration of unit rainfall excess in hours, which was set to one-twelfth of an hour (i.e.,
5min) in this study. Notably, the catchment area and time of concentration are required to construct an SUH, and they are the other two hydrological parameters that should be calibrated based on
taxi data. Although numerous tools and theories have been developed for estimating catchment area and time of concentration, these two parameters are still prone to significant errors, particularly
in urban areas, because of challenges in accurately delineating urban catchments (Huang and Jin, 2019; Li et al., 2020). Urban catchment delineation is more complex than natural catchment
delineation. Urban catchments have spatially heterogeneous surface cover types, which change with city development and construction, and modify runoff parameters (Goodwin et al., 2009). High
densities of residential and commercial buildings obstruct flow paths and alter flow directions of storm water runoff, complicating rainfall-runoff and overland flow processes in urban areas (Ji and
Qiuwen, 2015). Additionally, accurate urban catchment delineation necessitates high-resolution DEMs, which are not always available. In many Chinese cities, high-resolution DEMs are considered
confidential data and are generally inaccessible to non-governmental organizations. Based on these challenges, deriving accurate catchment area and time of concentration data in urban areas is
difficult in Shenzhen.
For the sake of simplicity, the peak rate factor was not calibrated and was at 2.08, although some studies have indicated that it has a wide range from 0.43 for steep terrain to 2.58 for very flat
terrain (Chow et al., 1988). After t[C] and A are chosen, an SUH can be constructed and used to convert rainfall excess into runoff by applying the discrete convolution equation. The detailed
computation process of the discrete convolution equation can be found in most hydrological textbooks (e.g., Chow et al., 1988) and will not be discussed here. The workflow in Fig. 3 illustrates the
transformation of rainfall time series data into a hydrograph for every parameter set.
2.2.2Step 2: derivation of the road disruption probability using the runoff-disruption function
The goal of Step 2 is to convert the hydrograph generated in Step 1 into a time series of road disruption probabilities, or more specifically, the probability that a taxi driver chooses to turn their
car when arriving at a flooded road. Most models in the literature assume that a road is either open or closed, which does not correspond to the empirical evidence that many drivers may take risks to
drive on inundated roads. To transition from a binary view of a flooded road being considered “open” or “closed,” Pregnolato et al. (2017) proposed the use of a curve that relates the depth of
floodwater to a reduction in vehicle speed to indicate the probability of road disruption. This idea was soon adopted by Contreras-Jara et al. (2018) and Nieto et al. (2021).
A driver will turn around when they believe that the flow rate is too risky for their vehicle configuration. From this perspective, the road disruption probability is equal to the probability that
vehicle performance is less than the flow rate perceived by a driver. However, it is difficult to quantify the factors that influence the willingness of people to drive through a flooded roadway, and
impossible to obtain the precise knowledge regarding all taxi-flood intersections. Alternatively, to ensure vehicle stability in flood flows, guidelines are typically recommended based on the
limiting value of depth times velocity. Many researchers have conducted laboratory testing on the stability of different types of vehicle models exposed to different combinations of depth and
velocity (Merz and Thieken, 2009; Shah et al., 2018). As suggested by Pregnolato et al. (2017), we constructed our runoff-disruption function by integrating data from the literature and authoritative
guidelines. In this study, the road disruption probability was defined as the probability that the product of flow velocity and flow depth was greater than the stability limits extracted from the
literature, which are listed in Table 1 and plotted in Fig. 4. The expression of the fitting curve is
$\begin{array}{}\text{(11)}& y=\left[\mathrm{1}+\mathrm{exp}\left(-\mathrm{16.6}\left(x-\mathrm{0.48}{\right)}^{\mathrm{2}}\right){\right]}^{-\mathrm{1}},\end{array}$
where x is the product of flow velocity and flow depth, and y is the disruption probability. According to Eq. (11), a road has a disruption probability of 50% when the product of flow velocity and
flow depth is 0.47m^2s^−1 and is totally disrupted when the product is greater than 0.80m^2s^−1. By applying the fitting curve, we can easily convert the flood runoff into the disruption
probability as follows:
$\begin{array}{}\text{(12)}& P\left(\mathrm{Disrupt}{\right)}_{t}^{\left(i\right)}=\left[\mathrm{1}+\mathrm{exp}\left(-\mathrm{16.6}\left({q}_{t}^{\left(i\right)}/W-\mathrm{0.48}{\right)}^{\mathrm
where $P\left(\mathrm{Disrupt}{\right)}_{t}^{\left(i\right)}$ and ${q}_{t}^{\left(i\right)}$ are the road disruption probability and discharge in the tth interval derived from the hydrological model
with the parameter set θ^(i), respectively, and W is the road width.
2.2.3Step 3: derivation of the time series of no-taxi-passing probabilities
A road is considered to have no taxis passing in a fixed time interval if the road has no taxis arriving or if every taxi that arrives at the road turns around. Therefore, the no-taxi-passing
probability can be calculated using the following equation:
$\begin{array}{}\text{(13)}& {\mathit{\omega }}_{t}^{\left(i\right)}=\sum _{n=\mathrm{0}}^{\mathrm{\infty }}P\left(\mathrm{Arrival}\mathit{_}\mathrm{taxi}=n{\right)}_{t}×\left(P\left(\mathrm{Disrupt}
where ${\mathit{\omega }}_{t}^{\left(i\right)}$ is the no-taxi-passing probability in the tth interval and P(Arrival_taxi=n)[t] is the probability that n taxis arrive at the road segment during the t
th interval. Equation (13) indicates that if every taxi that arrives at the road segment makes a turn because of the flooded roadway, then the taxi volume on the road will be zero. In this study, P
(Arrival_taxi=n)[t] was assumed to follow the Poisson distribution,
$\begin{array}{}\text{(14)}& P\left(\mathrm{Arrival}\mathit{_}\mathrm{taxi}=n{\right)}_{t}={e}^{-{\mathit{\lambda }}_{t}}{\mathit{\lambda }}_{\mathrm{t}}^{n}/n\mathrm{!},\end{array}$
where λ[t] is the average number of taxis arriving at the road during the tth interval. By substituting Eq. (14) in Eq. (13), we obtain
$\begin{array}{}\text{(15)}& {\mathit{\omega }}_{t}^{\left(i\right)}=\sum _{n=\mathrm{0}}^{\mathrm{\infty }}\left({\mathrm{e}}^{-{\mathit{\lambda }}_{\mathrm{t}}}{\mathit{\lambda }}_{\mathrm{t}}^{n}/
By applying ${e}^{x}={\sum }_{n=\mathrm{0}}^{\mathrm{\infty }}{x}^{n}/n\mathrm{!}$, Eq. (15) can be converted into
$\begin{array}{}\text{(16)}& \begin{array}{rl}{\mathit{\omega }}_{t}^{\left(i\right)}& ={\mathrm{e}}^{-{\mathit{\lambda }}_{t}}\sum _{n=\mathrm{0}}^{\mathrm{\infty }}\left(P\left(\mathrm{Disrupt}{\
right)}_{t}^{\left(i\right)}{\mathit{\lambda }}_{\mathrm{t}}{\right)}^{n}/n\mathrm{!}\\ & =\mathrm{exp}\left({\mathit{\lambda }}_{\mathrm{t}}\left(P\left(\mathrm{Disrupt}{\right)}_{t}^{\left(i\
Equation (16) indicates that ${\mathit{\omega }}_{t}^{\left(i\right)}$ is entirely determined by λ[t] and $P\left(\mathrm{Disrupt}{\right)}_{t}^{\left(i\right)}$. Because $P\left(\mathrm{Disrupt}{\
right)}_{t}^{\left(i\right)}$ is obtained from Step 2, what is left to determine is the value of λ[t]. The value of λ[t] fluctuates according to the time of day, indicating higher taxi volume during
congested periods and lower volume during non-congested periods. Therefore, we calculate λ[t] by averaging the taxi volume during the tth interval to account for time-of-day variations. It should be
noted that as the intensity of rain increases, experienced taxi drivers will avoid flood-prone roads in advance, meaning that strictly speaking, λ[t] is a decreasing function of rainfall intensity.
However, fitting the rainfall–λ[t] curve requires many taxi GPS trajectories to inspect the route choices of taxi drivers under heavy rain, which is outside the scope of this study. Therefore, we
assumed that λ[t] was independent of rainfall.
Table 2 lists all the submodels and parameters used in the three-step process. The core principle of the three-step process was to calculate the time series of no-taxi-passing probabilities, Ω^(i),
given each parameter set θ^(i). Because the best choice of model is often data specific, it is likely that the model combination proposed in this study is not optimal for other scenarios. To apply
the proposed calibration method in practice, the submodels for the three-step process must be specified according to the available data, prior knowledge, and accuracy requirements.
The method outlined above was tested on the intersection of Xinzhou Road and Hongli Road in Shenzhen, which is recognized as a flood-prone point by the Water Authority of Shenzhen Municipality.
Recall that the parameters to be calibrated are the curve number CN, catchment area A, and time of concentration t[C]. The parameter spaces for CN, A, and t[C] are determined by DEMs and other prior
knowledge, which will be discussed in Sect. 4. Table 3 presents the details of the parameter sets to be calibrated, which form 8×20×30=4800 possible combinations. For ease of exposition, we
assume that all parameters are uniformly distributed.
Taxi GPS data collected during two storm events that occurred on 9 and 23 May 2015 were used to calibrate the parameter sets for the target intersection. Rainfall time series data and taxi
observations during these two storms are presented in Fig. 5. Each taxi observation contains two time series: the time series of taxi volumes at 5min intervals and the time series of road statuses
at 5min intervals. These series were derived from the taxi volumes with a value of one if the taxi volume was greater than zero and a value of zero if the taxi volume was zero.
Given the rainfall on 9 May 2015, we must calculate the time series of no-taxi-passing probabilities for each parameter combination. Because there are 4800 parameter sets, we can derive 4800 possible
time series of no-taxi-passing probabilities. For simplicity, we only present the 3120th parameter set (i.e., CN=65, A=0.2km^2, and t[C]=2.75h) as an example to demonstrate the working of the
proposed method. According to the three-step process, the first step is to convert the original rainfall into rainfall excess using the curve number method given CN=65 (Fig. 6a). Then, we
calculated the peak discharge q[p] and peak time t[p] using Eqs. (9) and (10):
$\begin{array}{ll}& {t}_{\mathrm{p}}=\mathrm{0.6}×\mathrm{2.75}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{12}}\approx \mathrm{1.69}\phantom{\rule{0.125em}{0ex}}\mathrm{h}\\ & {q}_{\mathrm{p}}=\mathrm
{2.08}×\frac{\mathrm{0.2}}{\mathrm{1.69}}\approx \mathrm{0.24}\phantom{\rule{0.125em}{0ex}}{\mathrm{m}}^{\mathrm{3}}{\mathrm{s}}^{-\mathrm{1}}.\end{array}$
The SUH was derived through multiplication by t[p] on the x axis and q[p] on the y axis of the standard SCS unit hydrograph (Fig. 6b). Next, the rainfall excess presented in Fig. 6a was combined with
the derived SUH to obtain a hydrograph through convolution (Fig. 6c).
In the second step, the runoff was transformed into a time series of road disruption probabilities based on the runoff-disruption function (Fig. 6d). The runoff-disruption function takes the product
of water depth and velocity (in units of m^2s^−1) as inputs. Therefore, the original runoff (in units of m^3s^−1) derived in the first step should be divided by the road width before inputting it
into the runoff-disruption function.
In the third step, the time series of road disruption probabilities (Fig. 6e) was converted to no-taxi-passing probabilities using Eq. (16) (Fig. 6f). The average number of taxis during the flooding
period is presented in Fig. 6f, and the derived time series of no-taxi-passing probabilities is presented in Fig. 6g.
After the time series of no-taxi-passing probabilities for every parameter set were derived, the degree of belief that a given parameter set is optimal was calculated by integrating it with the taxi
observations on 9 May 2015. According to Eq. (5), the posterior probability of the 3120th parameter set is calculated as
$\begin{array}{ll}P\left({\mathbit{\theta }}^{\left(\mathrm{3120}\right)}\mathrm{|}\mathbf{X}\right)& \propto L\left(\mathbf{X}\mathrm{|}{\mathbit{\theta }}^{\left(\mathrm{3120}\right)}\right)P\left
({\mathbit{\theta }}^{\left(\mathrm{3120}\right)}\right)=\mathrm{1}/\mathrm{4800}\\ & ×\prod _{t=\mathrm{1}}^{T}\left(\mathrm{1}-{\mathit{\omega }}_{t}^{\left(\mathrm{3120}\right)}{\right)}^{{x}_{t}}
\left({\mathit{\omega }}_{t}^{\left(\mathrm{3120}\right)}{\right)}^{\mathrm{1}-{x}_{t}},\end{array}$
where L(θ^(3120)|X) is the posterior distribution of probabilities that the 3120th parameter set is optimal conditional on X, which represents the taxi observations on 9 May 2015 presented in
Fig. 5c. P(θ^(3120)) is the prior probability of the 3120th parameter set being optimal and its value is $\mathrm{1}/\mathrm{4800}$ because there are 4800 possible combinations.
By following this process, we can calculate the posterior probabilities for every parameter set. Additionally, the posterior probability distribution of a parameter set can be updated using the taxi
observations and rainfall data on 23 May 2015 as
$P\left({\mathbit{\theta }}^{\left(i\right)}\mathrm{|}{\mathbf{X}}_{\mathrm{2}}\right)\propto L\left({\mathbf{X}}_{\mathrm{2}}\mathrm{|}{\mathbit{\theta }}^{\left(i\right)}\right)P\left({\mathbit{\
theta }}^{\left(i\right)}\mathrm{|}{\mathbf{X}}_{\mathrm{1}}\right),$
where P(θ^(i)|X[1]) is the original posterior probability distribution calibrated based on the storm on 9 May 2015 and P(θ^(i)|X[2]) is the updated posterior distribution after the data of the storm
from 23 May 2015 are added. Figure 7 illustrates the evolution of the probability distribution with the availability of additional taxi data. The first row in Fig. 7 represents the prior joint
distribution of hydrological parameter sets, and the second and third rows represent the posterior distribution after each round of updating. The posterior distribution dominates the uniform prior
distribution after the first update, and the distribution is refined slightly after the second update.
4.1Data description
The proposed method was validated on flood-prone roads located in Shenzhen, China, which is a coastal city frequently hit by extreme storms during summer. To the best of our knowledge, Shenzhen is
the only city that has shared runoff-related data with the public in China. Three data sources, namely taxi GPS data, rainfall data, and authoritative water level data, were used to validate our
parameter calibration method. Hydrological parameters were calibrated using the first two data sources and the water level data acted as the ground truth to validate the proposed method. Taxi GPS
data were anonymized and aggregated in 5min intervals. Rainfall data, which were also collected in 5min intervals, were measured at 115 gauging stations citywide and mapped to the road network
throughout Shenzhen using the ordinary Kriging spatial interpolation algorithm. The water level data were only measured at certain flood-prone points with a dynamic sampling interval ranging from
5min when the weather was rainy to 1h when the weather was clear. The proposed calibration method was validated by analyzing the hydrographs derived from the calibrated hydrological models against
the authoritative water levels for 10 selected roads. Detailed information on the three data sources is provided in Table 4.
^1 The complete taxi GPS data and rainfall data are not openly accessible owing to the requirements of data policy. To validate our research findings, we have uploaded the necessary data to Zenodo
(Kong, 2022). ^2 Openly available at: https://opendata.sz.gov.cn/data/dataSet/toDataDetails/29200_01403147 (last access: 6 September 2022).
The two storm events on 9 and 23 May 2015 were treated as calibration events, and a storm on 11 June 2019 was retained for testing. Clearly, there is a 4year gap between the calibration data and
validation data based on data availability. The hydrological environments of flood-prone roads may have changed during these years, which could render the parameters calibrated based on data from
2015 inaccurate for analysis in 2019. To reduce the validation error caused by this time gap, the roads to be validated should have been vulnerable to flooding in both 2015 and 2019 so that the
hydrological parameters of these roads would have a higher chance of remaining unchanged. Therefore, a total of 10 flood-prone roads that were labeled as such in both the List of 2015 Flood-prone
Roads in Shenzhen (Water Authority of Shenzhen Municipality, 2015) and the List of 2019 Flood-prone Roads in Shenzhen (Water Authority of Shenzhen Municipality, 2019) were carefully selected
(Fig. 8).
4.2Prior distributions of calibrated parameters
We introduced two types of prior distributions to demonstrate the effects of prior distributions on calibrated parameters. The first prior distribution was determined based on prior knowledge and
DEMs from Shenzhen, which were obtained from ASTER GDEM V3, which is a product of NASA and Japan's Ministry of Economy, Trade, and Industry (METI) (Ministry of Economy, Trade, and Industry (METI) of
Japan and the United States National Aeronautics and Space Administration (NASA), 2023). This global DEM covers the entire land surface of the earth with a 30m resolution, exhibiting notable
improvements in horizontal and vertical accuracy while reducing anomalies compared with previous versions. We inputted the DEMs from Shenzhen into the hydrological software PCSWMM to delineate
catchments and calculate the catchment area. Subsequently, we computed the time of concentration using the watershed lag method (Natural Resources Conservation Service, 2010b). As suggested by Zhang
and Huang (2018), we used the average curve number for Shenzhen in 2015, which was assessed to be 60, as the estimated curve number for each road under validation.
We then constructed a discretized parameter space for the three parameters for each road as follows: for the curve number, we examined eight possible values centered on 60 with steps of five. For the
catchment area, we considered 20 possible values centered on the estimated value with steps of 0.01km^2. For the time of concentration, we considered 30 possible values centered on the estimated
value with steps of 5min. After constructing the parameter space for the parameters, we assigned a triangular prior distribution to each, which assumed the maximum probability at the estimated value
and linearly decreased to zero at the parameter space boundaries, as depicted in Fig. 9.
The second prior distribution assumed that the three parameters all follow uniform distributions. The parameter spaces for the second prior distribution were the same as those for the first. As a
result, the joint probability of each parameter set was equal to $\left(\mathrm{1}/\mathrm{20}\right)×\left(\mathrm{1}/\mathrm{30}\right)×\left(\mathrm{1}/\mathrm{8}\right)$. To facilitate
comparisons, we present the detailed information on the two types of prior distributions in Table 5.
4.3Posterior distributions after calibration
We first calibrated the parameters based on the prior distributions calculated according to the DEMs and other prior knowledge. The resulting posterior distributions are presented in Fig. 10. Each
row in Fig. 10 represents a different road, and each column represents a curve number. Each subplot presents the joint probability distribution of the catchment area and time of concentration for a
given curve number. The color intensity in Fig. 10 represents the magnitude of the probabilities. Following two iterations of updating, the posterior probability distributions for both the catchment
area and time of concentration converge around the optimal parameter sets for most flood-prone roads. This demonstrates that incorporating taxi observations significantly reduces the uncertainty
associated with catchment area and time of concentration. The probability typically achieves its maximum value when the curve number is either 55 or 60. Furthermore, each subplot contains a salient
cluster with higher probability than other regions, suggesting that there may be multiple acceptable parameter sets.
Furthermore, the optimal catchment area under a given curve number decreases as the curve number increases, whereas the optimal time of concentration under a given curve number increases with the
curve number. This is logical, because a higher curve number corresponds to increased rainfall excess under identical rainfall conditions, requiring a reduction in catchment area to maintain the
runoff that best aligns with the taxi observations. Similarly, an increase in the time of concentration diminishes the peak runoff produced by the additional runoff generated by a higher curve
number, thereby preserving the optimal runoff status.
We also present the marginal distributions of the three parameters for 10 roads before and after calibration in Fig. 11. In Fig. 11, the marginal posterior distributions of the curve number appear
relatively similar to the marginal prior distributions. It seems that the proposed method employing taxi data provides limited information regarding the distribution of curve numbers compared with
the catchment area and time of concentration. This outcome may be a result of the range and discretization granularity of the parameter spaces. Catchment area and time of concentration encompass 20
and 30 possible values, respectively, whereas the curve number has only 8 potential values. The smaller parameter space of the curve number reduces the search space, and its impact on the
no-taxi-passing probability is comparatively lower than that of the catchment area and time of concentration.
For example, for road ID=6, the optimal parameter set consists of a catchment area of 0.19km^2, time of concentration of 0.9h, and curve number of 55. To investigate the effects of these
parameters on the hydrograph and time series of no-taxi-passing probabilities, we held two parameters constant at their optimal values and observed the impact of changing the third parameter. Our
findings are presented in Fig. 12. One can see that when the catchment area varies from 0.04 to 0.23km^2 the maximum no-taxi-passing probability increases from 20% to 100% and the duration for
which the no-taxi-passing probability exceeds 0.5 increases from 0.0 to 1.3h. Similarly, when the time of concentration fluctuates from 0.1 to 1.9h, the peak time of the no-taxi-passing probability
varies from 0.5 to 1.8h. In contrast, when the curve number varies from 40 to 75, the maximum no-taxi-passing probability is fixed at 100%, the duration for which the no-taxi-passing probability
exceeds 0.5 extends from 1.1 to 1.3h, and the peak time of the no-taxi-passing probability remains fixed at 1.1h. These small fluctuations in the time series of no-taxi-passing probabilities are
representative of why the distribution of curve numbers remains relatively stable after calibration compared with the catchment area and time of concentration.
The posterior distributions calibrated based on the uniform prior distribution are presented in Fig. 13. When comparing two posterior distributions derived from two prior distributions, it is clear
that the posterior distributions of the catchment area and time of concentration are very similar, indicating that the impact of prior distributions on these parameters rapidly diminishes after
taxi-related knowledge is added. As stated by Beven and Binley (1992, p. 286), “as soon as information is added in terms of comparisons between observed and predicted responses then, if this
information has value, the distribution of calculated likelihood values should dominate the uniform prior distribution when uncertainty estimates are recalculated”.
4.4Validation results
After the parameter sets were calibrated, they were combined with an SCS unit hydrograph to construct an SUH, which was combined with the rainfall data from 11 June 2019 to produce the predicted
hydrograph. Because the posterior probability associated with each parameter set can be regarded as a fuzzy measure reflecting the degree of belief that the parameter set is true, the weighted runoff
values for each parameter set were summed to calculate the final predicted runoff:
$\begin{array}{}\text{(17)}& Q=\sum _{i=\mathrm{1}}^{N}P\left({\mathit{\theta }}^{\left(i\right)}\mathrm{|}X\right){Q}^{\left(i\right)}.\end{array}$
Here, Q is the final predicted runoff, Q^(i) is the simulated runoff derived from the ith parameter set, and P(θ^(i)|X) is the posterior probability of the ith parameter set, which acts as a weight.
The output of the calibrated hydrological model is runoff (with units of m^3s^−1), whereas the validation data are water level data (with units of m). Because the calibration data and validation
data came from multiple sources and have different units, conventional error-based statistics, such as the mean absolute error, were not suitable for this study. The discharge of a stream is rarely
measured directly. Instead, streamflow is typically determined by converting measured water depth (i.e., water stage) into discharge through a rating curve, which provides a functional relationship
between the water stage and discharge at a specified point (Le Coz et al., 2014). Inspired by the application of the rating curve, we validated our method by estimating the goodness of fit between
the water level which was measured in the field and the corresponding runoff which was predicted based on the proposed calibration method. A higher goodness of fit indicates synchronous trends
between the runoff and water level, which indirectly demonstrates the feasibility of the proposed method.
Because the posterior distributions derived from the two types of prior distributions were very similar, we only considered the posterior distribution calibrated based on prior distributions derived
from DEMs and other prior knowledge for validation. Comparisons between the observed water depth and simulated runoff for 10 selected roads are presented in Fig. 14, and corresponding scatter plots
are presented in Fig. 15. We use the Pearson correlation coefficient, which measures the linear correlation between two variables, as a goodness of fit indicator. One can see that 8 of 10 roads are
characterized by significant positive Pearson coefficients, indicating that the runoff and water have similar and consistent variation trends.
It is noteworthy that goodness of fit simply describes the degree of correlation between the observed and simulated data, and may contain validation bias. As suggested by Legates and McCabe (1999),
correlation-based statistics are insensitive to additive and proportional differences between simulations and observations. Therefore, the fitting of a rating curve is only a partial validation and
the usefulness of the proposed calibration method requires further analysis.
Four main points about the proposed calibration method are worthy of further discussion. The first is that although the presented validation results support the use of taxi GPS data to calibrate
hydrological parameters for poorly gauged road networks, the proposed method is more applicable to roads that are frequently visited by taxis. Uncertainty increases as the taxi volume on a road
decreases. A road is considered to be passable when at least one taxi GPS point is observed during a time interval, but we cannot assert that a road is disrupted when the taxi volume is zero. When a
road with frequent taxi traffic is observed with no taxi GPS points during a storm, it is highly probable that the road is disrupted by flooding, which provides relatively reliable information for
parameter calibration. Conversely, when a road with little taxi traffic has no taxi points during a storm, there is a relatively high likelihood that the road remains passable and is simply
exhibiting its typical trend of no taxis. Therefore, the proposed calibration method becomes relatively unreliable when a no-taxi-passing period is no longer a good proxy for the disruption period on
a road with sparse taxi data. To compensate for a shortage of taxi GPS data, additional data sources, such as ride-hailing data and bus data, should be incorporated in future work.
Second, the disruption of one road may cause cascading failures, where the disruption is rapidly propagated from the inundated road to adjacent non-inundated roads under the constraints of road
connectivity. For a road that is disrupted, but not inundated by a storm, the implementation of the proposed calibration method may be subject to structural errors. Consider two connected roads
called Road 1 and Road 2 that are both disrupted during a storm and have taxi volumes of zero (Fig. 16). In this case, Road 1 is disrupted by the flooding, whereas Road 2 is only disrupted because it
is connected to Road 1. If taxi data are the only data used for calibration, then the posterior distributions of the hydrological parameters for Road 1 and Road 2 will be identical after calibration,
because the sequences of taxi volume are identical for both roads. However, we know that the hydrological parameters of these two roads are not the same, because only one road is flooded. Just like
we cannot simply treat the no-taxi-passing period as the disruption period, we cannot confuse the disruption period with the flooded period. In future work, an algorithm that facilitates
distinguishing the flooding-induced disruption from connectivity-induced disruption should be developed.
Third, the proposed three-step process, which consists of an SCS unit hydrograph, empirical runoff-disruption function, and Poisson distribution, is a realization of the generalized framework
presented in Fig. 1. The submodels used in the three-step process can be flexibly replaced with other submodels according to complexity requirements and data availability. For example, an alternative
to the SCS unit hydrograph is the distributed hydrological model. Unlike the SCS unit hydrograph, the distributed hydrological model partitions a watershed into physically homogeneous units and
captures the complex spatial variation induced by human activity in high resolution, which may be more applicable to urbanized environments such as road networks. However, considering that some
critical data, such as road drainage data and land use data, are missing, as well as the extreme computational cost associated with the distributed hydrological model, we did not adopt this model in
this study. Another assumption we made in this study is that the number of taxis arriving at a road follows a Poisson distribution. By conducting the chi-square goodness of fit test, we found that
the frequency distribution of taxi volumes adheres to a Poisson distribution for more than 50% of 5min intervals for 7 of the 10 roads presented in Fig. 8, indicating that the Poisson model appears
to be a suitable assumption. However, this hypothesis may not be universally applicable, particularly in different urban contexts, where alternate distributions, such as the Weibull distribution, may
provide a more accurate representation.
Fourth, it is imperative to acknowledge that the parameter values in this study were discretized, although hydrological model parameters are inherently continuous. This discretization approach could
result in the omission of optimal solutions, particularly when hydrological models exhibit sensitivity to these parameters. It is important to note that discretization is neither a requisite nor a
recommended strategy. Future research should address the optimization or posterior inference problem in a continuous parameter space based on established methods such as the Monte Carlo algorithm.
An urban flooding model requires various types of data for calibration. In this study, we proposed a Bayesian calibration framework for the hydrological parameters of a road network based on taxi GPS
data. A three-step procedure consisting of a rainfall-runoff model, runoff-disruption function, and no-taxi-passing probability model enabled us to transform a given rainfall time series into a time
series of no-taxi-passing probabilities for each parameter set, which is key to taxi-data-driven model calibration. The calculated no-taxi-passing probabilities, which acted as a proxy for the
associated hydrological parameter sets, were compared with observed taxi data based on the Bayes equation to assess the posterior probability distributions of the hydrological parameter sets. Three
parameters, namely the curve number, catchment area, and time of concentration, were calibrated. The proposed calibration method was instantiated by combining classical hydrological models with
traffic flow models and was validated on 10 flood-prone roads in Shenzhen. The validation results indicate that the trends of runoff could be correctly predicted for eight roads, which demonstrates
the potential of calibrating hydrological parameters based on taxi GPS data.
This study highlights the potential of integrating transportation-related data with hydrological theory for the transportation resilience improvement and flood risk management of road networks. We
hope that our study can provide a flexible calibration framework for countries that have little runoff data but rich taxi data. We acknowledge that the application of the proposed method is currently
limited by the heterogeneous spatial distributions of taxis citywide and the cascading effects of road inundation, but we expect this to change with the increasing availability of vehicle data and
continuous optimization of modeling approaches.
JY conceptualized the article and collected field data. XK designed the methodology and was responsible for code compilation. KX plotted the figures and revised the manuscript. BD managed the
implementation of research activities. SJ discussed results and contributed to method validation. XK wrote the final version of the article with contributions from all co-authors.
The contact author has declared that none of the authors has any competing interests.
Publisher’s note: Copernicus Publications remains neutral with regard to jurisdictional claims made in the text, published maps, institutional affiliations, or any other geographical representation
in this paper. While Copernicus Publications makes every effort to include appropriate place names, the final responsibility lies with the authors.
Shan Jiang thanks for the support by Tufts University.
This research has been supported by the National Key Research and Development Program of China (grant no. 2022YFC3303100).
This paper was edited by Yue-Ping Xu and reviewed by Jeffrey M. Sadler and two anonymous referees.
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97–110, https://doi.org/10.1080/15230406.2016.1271356, 2018.
Martínez-Gomariz, E., Gómez, M., Russo, B., and Djordjeviæ, S.: A new experiments-based methodology to define the stability threshold for any vehicle exposed to flooding, Urban Water J., 14, 930–939,
https://doi.org/10.1080/1573062X.2017.1301501, 2017.
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[data set], https://asterweb.jpl.nasa.gov/gdem.asp, last access: 5 May 2023.
Moore, K. A. and Power, R. K.: Safe buffer distances for offstream earth dams, Australas. J. Water Resour., 6, 1–15, https://doi.org/10.1080/13241583.2002.11465206, 2002.
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21422 (last access: 14 October 2023), 2007.
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Ranked by the resident population in 2021.
The NRCS was originally called the US Soil Conservation Service. | {"url":"https://hess.copernicus.org/articles/27/3803/2023/","timestamp":"2024-11-06T15:48:27Z","content_type":"text/html","content_length":"308845","record_id":"<urn:uuid:8b5571aa-3930-43b8-8eab-fadcf73fdae9>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00844.warc.gz"} |
A Free Body Diagram Problem - Physics 1 Free Body Diagrams
A Free Body Diagram Problem
As shown in the figure, two blocks with masses m and M (M>m) are pushed by a force F in both Case I and Case II. The surface is horizontal and frictionless. Let R1 be the force that m exerts on M in
Case I and RII be the force that m exerts on M in Case II. Which of the following statements is true?
SOLUTION MISSING: Unfortunately the author of this youtube video removed their content. You may be able to find a similar problem by checking the other problems in this subject. If you want to
contribute, leave a comment with the link to your solution.
Posted by Rick Weaver a year ago
Related Problems
Two boxes with masses 17 kg and 15 kg are connected by a light string that passes over a frictionless pulley of negligible mass as shown in the figure below. The surfaces of the planes are
frictionless. The blocks are released from rest. T1 and T2 are the tensions in the strings. Which of the following statements is correct?
An Object with mass m and initial velocity v is brought to rest by a constant force F acting for a time t and through a distance d. Possible expressions for the magnitude of the force F are: i. $\
frac{mv^2}{2d}$ ii. $\frac{2md}{t^2}$ iii. $\frac{mv}{t}$
ii only
iii only
i and ii only
ii and iii only
i, ii, and iii
A toy car of mass 6 kg moving in a straight path, experiences a net force given by the function F = -3t. At time t=0, the car has a velocity of 4 m/s in the positive direction and is located +8 m
from the origin. The car will come instantaneously to rest at time t equal to
2/3 s
sqrt( 4/3 ) s
sqrt( 8/3 ) s
sqrt( 8) s
4 s
A block of mass M1 on a horizontal table is connected to a hanging block of mass M2 by a string that passes over a pulley, as shown in the figure below. The acceleration of the blocks is 0.6g. Assume
that friction and the mass of the string are negligible. The tension T in the string is | {"url":"https://www.practiceproblems.org/problem/Physics_1/Free_Body_Diagrams/A_Free_Body_Diagram_Problem","timestamp":"2024-11-11T11:50:48Z","content_type":"text/html","content_length":"42968","record_id":"<urn:uuid:971ab2fe-9568-4c9b-8a74-95a139b0f7a1>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00599.warc.gz"} |
Total Turfing
Calculator - Measuring Guide - Bark and Mulch.
Rectangle & Square
What units are you measuring in:
Select depth type:
Finding the volume in a rectangle is simple.
Multiply the short side by the long side then multiply the depth.
No fancy tricks, just length times width times depth times factor.
Its the same with the square area, multiply the two sides then the factor.
Rectangle = Length x Width x Depth
Square = Length x Width x Depth
The factor used in the calculations is 1.3 and all values are rounded up to the nearest full bag.
Length = Width = Depth =
Total = Bulk Bags Required
What units are you measuring in:
Select depth type:
If the circle is 20 feet across take half of that (10) and multiply it by its self then multiply it by the depth, ensuring that you do it in the same units.
Circle's Volume = (3.14 x Radius²) x depth The factor used in the calculations is 1.3 and all values are rounded up to the nearest full bag.
Radius = Depth =
Total = Bulk Bags Required
What units are you measuring in:
Select depth type:
Right-Angled Triangles are kind of like rectangles, taking calculations form a right angle multiply a x b then divide by 1.3 then multiply by the depth.
This only works with RIGHT-ANGLED triangles.
That means that one corner has to be square (90º).
Triangle's Volume = ((a x b) ÷ 2) x depth.
To convert to the number of bags multiply by the factor. The factor used in the calculations is 1.3 and all values are rounded up to the nearest full bag.
A = B = Depth =
Total = Bulk Bags Required
If your triangle isn't right-angled you need to find the area of an irregular triangle and cut it into two pieces.
Start at the corner opposite the longest side.
Go straight towards the long side, making two right-angled triangles.
Now find the area for each one of the little triangles.
Add them together to get the total area of the triangle.
Irregular Shapes
If your area is of an unusual shape, draw a sketch of it and then by using the four shapes above you can break it up into smaller shapes and calculate from there!
If you are unsure, please don't hesitate to contact us for advise Tel: 01294 822005. | {"url":"https://totalturfing.co.uk/calculate-area-bark_mulch.asp","timestamp":"2024-11-10T08:46:48Z","content_type":"text/html","content_length":"25877","record_id":"<urn:uuid:bdaba0ae-37d2-4f91-83cb-1dc97ec6e526>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00531.warc.gz"} |
Our tools for solving, counting and sampling
Development, Research, SAT, ToolsANF, CNF, model counting, samplingmsoos
This post is just a bit of a recap of what we have developed over the years as part of our toolset of SAT solvers, counters, and samplers. Many of these tools depend on each other, and have taken
greatly from other tools, papers, and ideas. These dependencies are too long to list here, but the list is long, probably starting somewhere around the Greek period, and goes all the way to recent
work such as SharpSAT-td or B+E. My personal work stretches back to the beginning of CryptoMiniSat in 2009, and the last addition to our list is Pepin.
Firstly when I say “we” I loosely refer to the work of my colleagues and myself, often but not always part of the research group lead by Prof Kuldeep Meel. Secondly, almost all these tools depend on
CryptoMiniSat, a SAT solver that I have been writing since around 2009. This is because most of these tools use DIMACS CNF as the input format and/or make use of a SAT solver, and CryptoMiniSat is
excellent at reading, transforming , and solving CNFs. Thirdly, many of these tools have python interface, some connected to PySAT. Finally, all these tools are maintained by me personally, and all
have a static Linux executable as part of their release, but many have a MacOS binary, and some even a Windows binary. All of them build with open source toolchains using open source libraries, and
all of them are either MIT licensed or GPL licensed. There are no stale issues in their respective GitHub repositories, and most of them are fuzzed.
CryptoMiniSat (research paper) our SAT solver that can solve and pre- and inprocess CNFs. It is currently approx 30k+ lines of code, with a large amount of codebase dedicated to CNF transformations,
which are also called “inprocessing” steps. These transformations are accessible to the outside via an API that many of the other tools take advantage of. CryptoMiniSat used to be a state-of-the-art
SAT solver, and while it’s not too shabby even now, it hasn’t had the chance to shine at a SAT competition since 2020, when it came 3rd place. It’s hard to keep SAT solver competitive, there are many
aspects to such an endeavor, but mostly it’s energy and time, some of which I have lately redirected into other projects, see below. Nevertheless, it’s a cornerstone of many of our tools, and e.g.
large portions of ApproxMC and Arjun are in fact implemented in CryptoMiniSat, so that improvement in one tool can benefit all other tools.
Arjun (research paper) is our tool to make CNFs easier to count with ApproxMC, our approximate counter. Arjun takes a CNF with or without a projection set, and computes a small projection set for it.
What this means is that if say the question was: “How many solutions does this CNF has if we only count solutions to be distinct over variables v4, v5, and v6?”, Arjun can compute that in fact it’s
sufficient to e.g. compute the solutions over variables v4 and v5, and that will be the same as the solutions over v4, v5, and v6. This can make a huge difference for large CNFs where e.g. the
original projection set can be 100k variables, but Arjun can compute a projection set sometimes as small as a few hundred. Hence, Arjun is used as a preprocessor for our model counters ApproxMC and
ApproxMC (research paper) is our probabilistically approximate model counter for CNFs. This means that when e.g. ApproxMC gives a result, it gives it in a form of “The model count is between 0.9*M
and 1.1*M, with a probability of 99%, and with a probability of 1%, it can be any value”. Which is very often enough for most cases of counting, and is much easier to compute than an exact count. It
counts by basically halfing the solution space K times and then counts the remaining number of solutions. Then, the count is estimated to be 2^(how many times we halved)*(how many solutions
remained). This halfing is done using XOR constraints, which CryptoMiniSat is very efficient at. In fact, no other state-of-the-art SAT solver can currently perform XOR reasoning other than
UniGen (research paper) is an approximate probabilistic uniform sample generator for CNFs. Basically, it generates samples that are probabilistically approximately uniform. This can be hepful for
example if you want to generate test cases for a problem, and you need the samples to be almost uniform. It uses ApproxMC to first count and then the same idea as ApproxMC to sample: add as many XORs
as needed to half the solution space, and then take K random elements from the remaining (small) set of solutions. These will be the samples returned. Notice that UniGen depends on ApproxMC for
counting, Arjun for projection minimization, and CryptoMiniSat for the heavy-lifting of solution/UNSAT finding.
GANAK (research paper, binary) is our probabilistic exact model counter. In other words, it returns a solution such as “This CNF has 847365 solutions, with a probability of 99.99%, and with 0.01%
probability, any other value”. GANAK is based on SharpSAT and some parts of SharpSAT-td and GPMC. In its currently released form, it is in its infancy, and while usable, it needs e.g. Arjun to be ran
on the CNF before, and while competitive, its ease-of-use could be improved. Vast improvements are in the works, though, and hopefully things will be better for the next Model Counting Competition.
CMSGen (research paper) is our fast, weighted, uniform-like sampler, which means it tries to give uniform samples the best it can, but it provides no guarantees for its correctness. While it provides
no guarantees, it is surprisingly good at generating uniform samples. While these samples cannot be trusted in scenarios where the samples must be uniform, they are very effective in scenarios where
a less-than-uniform sample will only degrade the performance of a system. For example, they are great at refining machine learning models, where the samples are taken uniformly at random from the
area of input where the ML model performs poorly, to further train (i.e. refine) the model on inputs where it is performing poorly. Here, if the sample is not uniform, it will only slow down the
learning, but not make it incorrect. However, generating provably uniform samples in such scenarios may be prohibitively expensive. CMSGen is derived from CryptoMiniSat, but does not import it as a
Bosphorus (research paper) is our ANF solver, where ANF stands for Algebraic Normal Form. It’s a format used widely in cryptography to describe constraints over a finite field via multivariate
polynomials over a the field of GF(2). Essentially, it’s equations such as “a XOR b XOR (b AND c) XOR true = false” where a,b,c are booleans. These allow some problems to be expressed in a very
compact way and solving them can often be tantamount to breaking a cryptographic primitive such as a symmetric cipher. Bosphorus takes such a set of polynomials as input and either tries to simplify
them via a set of inprocessing steps and SAT solving, and/or tries to solve them via translation to a SAT problem. It can output an equivalent CNF, too, that can e.g. be counted via GANAK, which will
give the count of solutions to the original ANF. In this sense, Bosphorus is a bridge from ANF into our set of CNF tools above, allowing cryptographers to make use of the wide array of tools we have
developed for solving, counting, and sampling CNFs.
Pepin (research paper) is our probabilistically approximate DNF counter. DNF is basically the reverse of CNF — it’s trivial to ascertain if there is a solution, but it’s very hard to know if all
solutions are present. However, it is actually extremely fast to probabilistically approximate how many solutions a DNF has. Pepin does exactly that. It’s one of the very few tools we have that
doesn’t depend on CryptoMiniSat, as it deals with DNFs, and not CNFs. It basically blows all other such approximate counters out of the water, and of course its speed is basically incomparable to
that of exact counters. If you need to count a DNF formula, and you don’t need an exact result, Pepin is a great tool of choice.
My personal philosophy has been that if a tool is not easily accessible (e.g. having to email the authors) and has no support, it essentially doesn’t exist. Hence, I try my best to keep the tools I
feel responsible for accessible and well-supported. In fact, this runs so deep, that e.g. CryptoMiniSat uses the symmetry breaking tool BreakID, and so I made that tool into a robust library, which
is now being packaged by Fedora, because it’s needed by CryptoMiniSat. In other words, I am pulling other people’s tools into the “maintained and supported” list of projects that I work with, because
I want to make use of them (e.g. BreakID now builds on Linux, MacOS, and Windows). I did the same with e.g. the Louvain Community library, which had a few oddities/issues I wanted to fix.
Another oddity of mine is that I try my best to make our tools make sense to the user, work as intended, give meaningful (error) messages, and good help pages. For example, none of the tools I
develop call subprocesses that make it hard to stop a computation, and none use a random number seed that can lead to reproducibility issues. While I am aware that working tools are sometimes less
respected than a highly cited research paper, and so in some sense I am investing my time in a slightly suboptimal way, I still feel obliged to make sure the tax money spent on my academic salary
gives something tangible back to the people who pay for it. | {"url":"https://www.msoos.org/2024/02/our-tools-for-solving-counting-and-sampling/","timestamp":"2024-11-10T05:53:04Z","content_type":"text/html","content_length":"57214","record_id":"<urn:uuid:b0c84b68-aa4c-4183-88b6-8a6d65ceeb55>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00817.warc.gz"} |
Calculating Standard Deviation On Excel - SurveyPoint
Excel goes far beyond just rows and columns. It can be a great platform to store your data and keep things organized. The platform also helps you take a break from tedious manual work and apply
built-in formulas to make work easy and fast. And, if you are still struggling with calculating a standard deviation of the data you collected, we are here to do the math for you.
Wait, but what is the standard deviation?
Read the blog to explore what it is, the formula and some examples.
What does Standard Deviation mean?
This technique is a statistical measure that allows you to calculate the dataset’s dispersion relative to the mean. In short, it is calculated as the square root of the spread of numbers in your
By using this technique, you can calculate the variation between different data points. It can also be used to compare the variation among the data points with the same standard but a diverse data
Let’s all this understand with an example.
Here are two data sets.
• 15, 15, 15, 14, 16
• 2, 7, 14, 22, 30
These data sets may look the same, but the last one is more evenly spread.
Technically speaking, if the data points are situated further from the mean, the deviation within that set will be higher. In other words, the more space between the data points, the greater the
This technique is usually used in financing, where it is often applied to the annual return rate of an investment. A higher deviation would mean a more significant price range. Volatile stocks
usually have a higher variation, while a company with a good reputation has a lower standard deviation.
Calculating Standard Deviation In Excel
Because standard deviation is used to measure market volatility, it is often used to strategize investments and trading. Several portfolio managers and traders use this technique to calculate the
degree of risk involved in a particular investment.
Now, you might think it’s tricky to calculate. But excel can help you with that. Here’s how to calculate deviation using excel.
To get started, there are six formulas you can use to determine the standard deviation.
To calculate sample deviation, you can use formulas like STDEV.S, STDEV and STDEVA. However, if you need to calculate the deviation for an entire population, you can use formulas like STDEV.P, STDEVP
and STDEVPA.
In this case, the term population implies that you are considering entire datasets present in your data. If using the total population is impossible, you can use sample formulas. Technically, you can
use the sample formula to calculate a sample deviation and then infer it from the entire population.
Here’s an explanation of the three sample standard deviation formulas:
• STDEV.S: You can use the formula if your data set is numeric, as it will ignore all logical values.
• STDEVA: This formula is used when the data set includes both numeric values and logical values. In the case of logic, “False” is taken as 0 and “True” is taken as 1.
• STDEV: Compatible with older versions of Excel, this formula does the job same as STDEV.S, but on Excel, that is, 2007 or prior.
Using STDEV.S In Excel
Again, STDEV.S only considers numerical values and disregards textual and logical values.
Excel’s STDEV.S function has the following syntax: STDEV.S(number1,[number 2],…
The first argument in the formula must be the number 1. The first number represents the sample’s initial element. In this case, you may use a range name, an array, or a display reference instead of
arguments separated by commas.
Formula number 2 is an optional argument. It can refer to an array, a named range, a data point, or an array reference. It is possible to use up to 254 additional arguments.
Example Of STDEV.S
Consider a dataset with a variety of weights taken from a population sample. The formula, when used with the numbers in column A, will result in the following: =STDEV.S(A2:A10).
Excel will then deliver the average and standard deviation of the applied data. The majority of the group would fall within the weight range of 150-2 or 150+2 if the average is 150 and the standard
deviation is 2.
When a formula is an input, Excel’s Formula bar will also display it. The equal sign must always be used when entering a formula.
Let Us Do The Maths for You
Basic Statistics + Comparative Analysis = Tangible Solutions
Seeing a difference between two numbers is easy, but determining whether that difference is statistically significant takes a little more effort. Especially if your question has several possible
answers or you’re comparing findings from different groups of respondents, the process can be tricky.
Don’t refrain from investing in the proper technologies to alleviate the burden of manual analysis. Take the hassle out of the equation and let SurveyPoint do all the heavy lifting for you.
Heena Shah – Content Writer at Sambodhi | {"url":"https://surveypoint.ai/blog/2022/12/17/calculating-standard-deviation-on-excel/","timestamp":"2024-11-02T02:15:13Z","content_type":"text/html","content_length":"154239","record_id":"<urn:uuid:3e644aa7-1047-4687-8405-ebf0dcaf2123>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00302.warc.gz"} |
Sparse PCA through Low-rank Approximations
Sparse PCA through Low-rank Approximations
Proceedings of the 30th International Conference on Machine Learning, PMLR 28(3):747-755, 2013.
We introduce a novel algorithm that computes the k-sparse principal component of a positive semidefinite matrix A. Our algorithm is combinatorial and operates by examining a discrete set of special
vectors lying in a low-dimensional eigen-subspace of A. We obtain provable approximation guarantees that depend on the spectral profile of the matrix: the faster the eigenvalue decay, the better the
quality of our approximation. For example, if the eigenvalues of A follow a power-law decay, we obtain a polynomial-time approximation algorithm for any desired accuracy. We implement our algorithm
and test it on multiple artificial and real data sets. Due to a feature elimination step, it is possible to perform sparse PCA on data sets consisting of millions of entries in a few minutes. Our
experimental evaluation shows that our scheme is nearly optimal while finding very sparse vectors. We compare to the prior state of the art and show that our scheme matches or outperforms previous
algorithms in all tested data sets.
Cite this Paper
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Now we are going ahead to start new topic i.e. derivation of flexure formula or bending equation for pure bending in the strength of material with the help of this post.
Let us go ahead step by step for easy understanding, however if there is any issue we can discuss it in comment box which is provided below this post.
Let us consider one structural member such as beam with rectangular cross section, we can select any type of cross section for beam but we have considered here that following beam has rectangular
cross section.Â
First of all we will find here the expression for bending stress in a layer of the beam subjected to pure bending and aftre that we will understand the concept of moment of resistance and once we
will have these two information, we can easily secure the bending equation or flexure formula for beams.Â
So let us first find out the expression for bending stress acting on a layer of the beam subjected to pure bending.
Bending stress
Let us assume that following beam PQ is horizontal and supported at its two extreme ends i.e. at end P and at end Q, therefore we can say that we have considered here the condition of simply
supported beam.
Once load W will be applied over the simply supported horizontal beam PQ as displayed above, beam PQ will be bending in the form of a curve and we have tried to show the condition of bending of beam
PQ due to load W in the above figure.
Now let us consider one small portion of the beam PQ, which is subjected to a simple bending, as displayed here in following figure. Let us consider two sections AB and CD as shown in following
Now we have following information from the above figure.
AB and CD: Two vertical sections in a portion of the considered beam
N.A: Neutral axis which is displayed in above figure
EF: Layer at neutral axis
dx = Length of the beam between sections AB and CD
Let us consider one layer GH at a distance y below the neutral layer EF. We can see here that length of the neutral layer and length of the layer GH will be equal and it will be dx.
Original length of the neutral layer EF = Original length of the layer GH = dx
Now we will analyze here the condition of assumed portion of the beam and section of the beam after bending action and we have displayed here in following figure.Â
As we can see here that portion of the beam will be bent in the form of a curve due to bending action and hence we will have following information from above figure.
Section AB and CD will be now section A'B' and C'D'
Similarly, layer GH will be now G'H' and we can see here that length of layer GH will be increased now and it will be now G'H'
Neutral layer EF will be now E'F', but as we have discussed during studying of the variousÂ
assumptions made in theory of simple bending, length of the neutral layer EF will not be changed.
Length of neutral layer EF = E'F' = dx
A'B' and C'D' are meeting with each other at center O as displayed in above figure
Radius of neutral layer E'F' is R as displayed in above figure
Angle made by A'B' and C'D' at center O is θ as displayed in above figure
Distance of the layer G'H' from neutral layer E'F' is y as displayed in above figure
Length of the neutral layer E'F' = R x θ
Original length of the layer GH = Length of the neutral layer EF = Length of the neutral layer E'F' = R x θ
Length of the layer G'H' = (R + y) x θ
As we have discussed above that length of the layer GH will be increased due to bending action of the beam and therefore we can write here the following equation to secure the value of change in
length of the layer GH due to bending action of the beam.
Change in length of the layer GH = Length of the layer G'H'- original length of the layer GH
Change in length of the layer GH = (R + y) x θ - R x θ
Change in length of the layer GH = y x θ
Strain in the length of the layer GH = Change in length of the layer GH/ Original length of the layer GH
Strain in the length of the layer GH = y x θ/ R x θ
Strain in the length of the layer GH = y/R
As we can see here that strain will be directionally proportional to the distance y i.e. distance of the layer from neutral layer or neutral axis and therefore as we will go towards bottom side layer
of the beam or towards top side layer of the beam, there will be more strain in the layer of the beam.
At neutral axis, value of y will be zero and hence there will be no strain in the layer of the beam at neutral axis.
Let us recall the concept of Hook’s Law
According to Hook’s Law, within elastic limit, stress applied over an elastic material will be directionally proportional to the strain produced due to external loading and mathematically we can
write above law as mentioned here.
Where E is the Young’s Modulus of elasticity of the material
Let us consider the above equation and putting the value of strain secure above, we will have following equation as mentioned here.
We can conclude from above equation that stress acting on layer of the beam will be directionally proportional to the distance y of the layer from the neutral axis.
Moment of resistance
As we have discussed that when a beam will be subjected with a pure bending, layers above the neutral axis will be subjected with compressive stresses and layers below the neutral axis will be
subjected with tensile stresses.
Therefore, there will be force acting on the layers of the beams due to these stresses and hence there will be moment of these forces about the neutral axis too.
Total moment of these forces about the neutral axis for a section will be termed as moment of resistance of that section.
As we have already assumed that we are working here with a beam having rectangular cross-section and let us consider the cross-section of the beam as displayed here in following figure.
Let us assume one strip of thickness dy and area dA at a distance y from the neutral axis as displayed in above figure.
Let us determine the force acting on the layer due to bending stress and we will have following equation
dF = σ x dA
Let us determine the moment of this layer about the neutral axis, dM as mentioned here
Total moment of the forces on the section of the beam around the neutral axis, also termed as moment of resistance, could be secured by integrating the above equation and we will have
Let us consider the above equation of moment of resistance and equation that we have secured for bending stress in case of bending action; we will have following equation which is termed as bending
equation or flexural formula of bending equation.
We will discuss another topic in the category of strength of material in our next post.
Strength of material, By R. K. Bansal
Image Courtesy: Google
Also read
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Introduction to Commutative Algebra
by Thomas J. Haines
Publisher: University of Maryland 2024
Number of pages: 93
Notes for an introductory course on commutative algebra. Algebraic geometry uses commutative algebraic as its 'local machinery'. The goal of these lecture notes is to study commutative algebra and
some topics in algebraic geometry in a parallel manner.
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Network design principle for robust oscillatory behaviors with respect to biological noise
Oscillatory behaviors, which are ubiquitous in transcriptional regulatory networks, are often subject to inevitable biological noise. Thus, a natural question is how transcriptional regulatory
networks can robustly achieve accurate oscillation in the presence of biological noise. Here, we search all two- and three-node transcriptional regulatory network topologies for those robustly
capable of accurate oscillation against the parameter variability (extrinsic noise) or stochasticity of chemical reactions (intrinsic noise). We find that, no matter what source of the noise is
applied, the topologies containing the repressilator with positive autoregulation show higher robustness of accurate oscillation than those containing the activator-inhibitor oscillator, and
additional positive autoregulation enhances the robustness against noise. Nevertheless, the attenuation of different sources of noise is governed by distinct mechanisms: the parameter variability is
buffered by the long period, while the stochasticity of chemical reactions is filtered by the high amplitude. Furthermore, we analyze the noise of a synthetic human nuclear factor κB (NF-κB)
signaling network by varying three different topologies and verify that the addition of a repressilator to the activator-inhibitor oscillator, which leads to the emergence of high-robustness
motif—the repressilator with positive autoregulation—improves the oscillation accuracy in comparison to the topology with only an activator-inhibitor oscillator. These design principles may be
applicable to other oscillatory circuits.
The authors study the important problem of how to achieve accurate oscillation robustly in biological networks where noise level may be high. The authors adopted a comprehensive approach and study
how different network configurations affect oscillation. This work makes an important contribution to the field as it offers the first comprehensive survey of networks motifs capable of oscillation,
with further characterization of their robustness.
Oscillatory behaviors have been observed in a broad range of biological processes, such as cell cycle (Ferrell et al., 2011; Tyson, 1991), circadian rhythms (Partch et al., 2014), and mitotic wave in
Drosophila embryo (Deneke et al., 2016). Oscillatory features, including period and amplitude, can encode functional information, which plays an essential role in coordinating gene regulation (Cai et
al., 2008) or transmitting distinct stimuli (Hao and O’Shea, 2012; Heltberg et al., 2019). In past decades, negative feedback, time delay, and nonlinearity have been identified as key mechanisms for
biochemical oscillation (Novák and Tyson, 2008), following which researchers artificially synthesized biochemical networks capable of oscillation (Atkinson et al., 2003; Chen et al., 2015; Elowitz
and Leibler, 2000; Potvin-Trottier et al., 2016; Stricker et al., 2008; Tigges et al., 2010; Zhang et al., 2017). Repressilator (Elowitz and Leibler, 2000) and activator-inhibitor oscillator (
Atkinson et al., 2003) are the most famous of these synthetic oscillators.
While many synthetic biological circuits can oscillate, their dynamics are typically irregular, owing to ubiquitous biological noise such as fluctuations in the microenvironment and inherent
stochasticity of chemical reactions (Elowitz et al., 2002; Li et al., 2009; Potvin-Trottier et al., 2016; Raser and O’Shea, 2004; Swain et al., 2002; Yu et al., 2018). Thus, a natural question is how
the biological systems achieve accurate oscillation in the presence of noise. Previous studies revealed that many kinetic parameters can influence the robustness of the biological oscillators, such
as the system size and degree of cooperativity of reactions (Gonze et al., 2002a), timescale of the promoter interaction (Forger and Peskin, 2005), repressor degradation rate (Potvin-Trottier et al.,
2016), free energy cost measured by ATP/ADP ratios (Cao et al., 2015; Fei et al., 2018; Qin et al., 2021), and kinetic parameter-determined oscillation mechanisms (i.e., limit cycle or force driving)
(Monti et al., 2018). Moreover, growing evidence suggests the existence of the relationship between network configurations and noise buffering capabilities for biochemical oscillators. For example,
in a synthetic microbial consortium oscillator composed of two different types of bacteria, adding negative autoregulation to the negative feedback loop increases the parameter space to oscillate
persistently in the face of noise (Chen et al., 2015); an additional positive feedback loop in the biochemical oscillator consisting of the negative feedback loop can decrease the coefficient of
variation (CV) of period when considering the stochasticity of reactions (Mather et al., 2009) and possess nearly constant period when varying the synthesis rate (Stricker et al., 2008).
Instead of exploring mechanisms to achieve accurate oscillation case by case, we try to understand the general network design principles of accurate oscillation using the bottom-up approach (Ma et
al., 2009; Qiao et al., 2019) and discover the specific network topologies that can oscillate and attenuate noise simultaneously. Here, we systematically explore the relationship between the network
topology and robustness to different sources of noise in both two- and three-node networks. We first perform an exhausting search of two- and three-node network topologies to identify those capable
of oscillation in the absence of noise, and then investigate the abilities of those oscillatory topologies to achieve accurate oscillation in the presence of different sources of noise. Two different
sources are considered: parameters are perturbed by noise terms whose magnitudes are proportional to parameters (i.e., extrinsic noise); chemical reactions induce stochasticity due to a small copy
number of proteins (i.e., intrinsic noise). We classify all oscillatory topologies according to what core motifs they include, and then compare the ability to execute accurate oscillation in the
presence of noise among different categories. Two categories whose core motifs include a repressilator with a positive feedback perform better than others. Importantly, the existence of positive
autoregulation always enhances the performance. While these results hold regardless of what source of noise exists, mechanisms to attenuate different sources of noise are distinct: long period
buffers the extrinsic noise, and high amplitude attenuates the intrinsic noise. Moreover, we experimentally validate that adding a repressilator to the activator-inhibitor topology in synthetic NF-κB
signaling circuits can improve the performance to buffer noise, indicating the important role of the repressilator with a positive autoregulation in filtering noise.
Searching for topologies robustly executing accurate oscillation
Index for measuring the oscillation accuracy
To measure the accuracy of the oscillatory behavior, we use the dimensionless correlation time, which is the correlation time $τ$ divided by the period $T$. The correlation time $τ$ describes how
fast the autocorrelation function $C(t)$ exponentially decays. To be specific, for a noisy dynamic trajectory of the oscillator, $C(t)$ displays a damped oscillation (Figure 1A):
$C\left(t\right)\equiv \frac{{〈\left(x\left(t+s\right)-〈x〉\right)\left(x\left(s\right)-〈x〉\right)〉}_{s}}{〈{x}^{2}〉-{〈x〉}^{2}}=\mathrm{exp}\left(-t/\tau \right)×\mathrm{c}\mathrm{o}\mathrm
{s}\left(2\pi t/T\right)$
Searching all possible two-node and three-node network topologies for oscillatory topologies with high accuracy in the presence of noise.
where $⋯s$ is defined by $f(s)s=limS→∞1S ∫0Sf(s)ds$, and $⋯$ is the ensemble average; $T$ is the period (time needed from one peak to the next peak). If fluctuations of the noisy trajectory are
small, the autocorrelation decays slowly, leading to a large value of $τ$. The correlation time $τ$ has the same unit as that of the period, so $τ/T$ is dimensionless. Therefore, instead of using $τ$
, we utilize $τ/T$ to measure the accuracy, which is equal to the quantity that previous work has used (Cao et al., 2015) except a constant factor.
We limit ourselves to network topologies with two or three nodes (Figure 1B) and search for topologies capable of accurate oscillation using a bottom-up concept. While the signaling pathway of the
oscillator in nature is complex, the core motif executing functions may be simple (Lim et al., 2013; Ma et al., 2009; Novák and Tyson, 2008; Qiao et al., 2019), and thus two- or three-node networks
might be enough to capture key features. Besides, the number of all two- and three-node network topologies is 3^9 because there are nine links in total and each link has three options: activation,
inhibition, or does not exist; however, by excluding topologies with isolated nodes or symmetrical to existing topologies, the number of possible two- and three-node network topologies is reduced
from 3^9 to 1955. Here, two typical oscillatory topologies are shown in Figure 1C: the activator-inhibitor and repressilator topologies. For the activator-inhibitor topology, the activator (node A)
has a positive autoregulation and positively regulates the inhibitor (node B), but is negatively regulated by the inhibitor; for the repressilator topology, each node acts as a repressor to inhibit
its next node, thus constituting a cyclic negative-feedback loop.
To model two- and three-node network topologies, we use transcriptional regulations to describe interactions among nodes (Figure 1D; see ‘Methods’). In a transcriptional regulatory network, nodes and
links represent genes’ products and transcriptional regulations, respectively; genes’ products work as transcription factors to interact with the regulatory sequence of other genes and activate or
inhibit the transcription, regulating the production rates of other genes’ products, that is, other nodes. Moreover, when multiple transcription factors regulate the same gene simultaneously, the
competitive inhibition logic is adopted: those transcription factors compete for the same binding sites. Thus, the transcriptional activity of a gene depends on the relative weights of transcription
factors activating this gene and those inhibiting this gene. Figure 1D illustrates the ordinary differential equation describing dynamics of node A when node A not only activates itself but also is
inhibited by node B. In this equation, the variable $A$ represents the concentration of the product of gene A; $kbasal$ is the basal production rate (much smaller than other terms); $vAA$ is the
maximum production rate caused by product A; $KAA$ and $KBA$ are binding affinities of products $A$ and $B$ to gene A, respectively; $rA$ is the degradation rate; $n$ is the Hill coefficient; and the
production rate is determined by relative weights of $A$ and $B$.
Based on the above deterministic model, we develop stochastic models to describe the oscillatory behavior in the presence of noise. According to the source of noise, the biological noise can be
decomposed into extrinsic and intrinsic components. On the one hand, we model the extrinsic noise as the variability of parameters including the maximum production rate $v$ and the degradation rate
$r$ (Figure 1E; see ‘Methods’): each of these parameters is added by a noise term with zero mean, and the standard deviation of the noise term is proportional to the value of the kinetic parameter.
On the other hand, the intrinsic noise, generated by the stochasticity of discrete chemical reactions, is modeled by directly simulating the dynamics of molecular numbers rather than concentrations.
To this end, we introduce the cell volume $V$, and naturally the molecular number of each node is the product of the cell volume $V$ and the concentration. As reactions progress, the molecular
numbers would randomly increase or decrease by one at some time point (Figure 1E; see ‘Methods’), and the waiting time of the increase and decrease obeys exponential distributions with parameters
determined by the production and decay rates in the deterministic model, respectively. This stochastic process can be exactly solved by the Gillespie algorithm, which has been widely used in previous
studies (Liu et al., 2020; Thattai and van Oudenaarden, 2001; Veliz-Cuba et al., 2015; Zhao et al., 2021); however, the computation cost is high, and thus we use chemical Langevin equations as
approximations to reduce the cost (Gillespie, 2000). Although the biological noise in nature usually has the extrinsic and intrinsic components simultaneously, we only consider the case where only
one source of noise exists for simplicity, that is, only extrinsic noise exists or only intrinsic noise exists.
Procedures to search for network topologies robustly executing accurate oscillation
To search for two- and three-node network topologies that can robustly achieve accurate oscillation (i.e., high dimensionless correlation time $τ/T$), two steps are performed (Figure 1F): the first
step is to identify topologies capable of oscillation in the whole network topology space (the upper panel in Figure 1F); the second step is to use the 90-percentile value of $τ/T$ to quantify the
robustness of each oscillatory network topology to achieve accurate oscillation (the lower panel in Figure 1F). For a given topology, the 90-percentile value of $τ/T$ is defined as the value of $τ/T$
below which 90% of $τ/T$’s fall when 1000 parameter sets are randomly assigned. We refer the reader to ‘Methods’ for details, and here we only show major procedures. In the first step (the upper
panel in Figure 1F), to obtain oscillatory network topologies in the whole network topology space, we randomly assign 10,000 parameter sets for each network topology and simulate the deterministic
dynamics. The oscillatory network topology is chosen by the following two criteria: the network topology without repressilator is regarded as an oscillatory network topology if at least 80 parameter
sets are capable of oscillation; the network topology with repressilator is defined as an oscillatory network topology if at least 10 parameter sets achieve oscillation. In this way, we finally
obtain 474 oscillatory network topologies, and nearly 35% of them are with the repressilator. If we used the threshold of 80 for all network topologies, oscillatory network topologies with
repressilator only occupy 20% of all oscillatory network topologies, which may lose the generality of conclusions about the repressilator. In the second step (the lower panel in Figure 1F), for each
of these 474 oscillatory network topologies, we sample 1000 parameter sets capable of oscillation in the absence of noise and calculate the 90-percentile value of $τ/T$ in the presence of extrinsic
noise or intrinsic noise. This value measures the robustness of the given topology against noise: the higher the value is, the larger probability to achieve accurate oscillation the topology has.
The robustness of accurate oscillation against extrinsic noise for different network topologies
Classification of all 474 oscillatory network topologies
We start by classifying all 474 oscillatory network topologies according to five types of core motifs. These five types of core motifs are as follows: the first core motif (shown in brown in Figure
2A) is composed of the repressilator and a positive autoregulation, but the node with the positive autoregulation is not allowed to have a positive incoming link; the second core motif (shown in
orange in Figure 2A) is similar to the first core motif except that the positive incoming link to the positive autoregulated node is required; the third type of core motifs include the
activator-inhibitor topology and its two variants (shown in green in Figure 2A); the fourth and fifth core motifs are the repressilator and delayed negative feedback (Figure 2A), respectively. Based
on the identification of five types of core motifs, we define C1 category as the network topologies that contain only the first type of core motif, and so do the C2 category, C3 category, C4
category, and C5 category. The above five categories constitute near 59% of all 474 oscillatory network topologies, while the rest of topologies are those containing at least two of these five types
of core motifs. Note that these oscillatory network topologies all have a negative feedback structure, which is consistent with previous studies (Glass and Pasternack, 1978; Novák and Tyson, 2008).
Figure 2
with 2 supplements
see all
The relationship between network topology and robustness to extrinsic noise.
The topologies containing repressilator with positive autoregulation perform better than those containing the activator-inhibitor topology when facing extrinsic noise, and the positive autoregulation
enhances the robustness against extrinsic noise
Next we compare the robustness of accurate oscillation against extrinsic noise among above C1, C2, …, and C5 categories. Figure 2B shows the violin plots of 90 percentiles of $τ/T$ in the presence of
extrinsic noise for the five categories, where each violin corresponds to one category. By applying one-tailed Wilcoxon rank-sum tests to adjacent two categories, we find that 90 percentiles of $τ/T$
for C1 category are significantly larger than those for C2 category, and this relation also holds between C2 and C3 categories, between C3 and C4 categories, and between C4 and C5 categories. These
findings indicate that the order of these five categories according to the robustness of accurate oscillation to extrinsic noise is C1 > C2 > C3 > C4 > C5. The facts that C1 > C3 and C2 > C3
demonstrate that topologies containing the repressilator with positive autoregulation achieve higher robustness against extrinsic noise than those containing the activator-inhibitor topology.
Besides, core motifs in both C1 and C2 categories have an extra positive autoregulation in comparison to the core motif (the repressilator) in C4 category, suggesting that the higher robustness of C1
and C2 categories than C4 category against extrinsic noise is due to the effect of positive autoregulation in improving the robustness to extrinsic noise. This effect is also validated by the
comparison of the robustness between C3 and C5 categories.
The above analyses focused on the oscillatory network topologies in C1, C2, …, and C5 categories, which account for nearly 59% of all oscillatory network topologies. To perform a complete research,
we also investigate the robustness to extrinsic noise for the remaining 41% oscillatory network topologies. These topologies all contain at least two types of core motifs and can be classified into
seven categories, based on what core motifs the topology has. Then we compare each of them with its ‘component’ category (i.e., C1, C2, …, or C5 categories). For example, C13 category is composed of
topologies that contain both the first and third types of core motifs, and its two ‘component’ categories are defined as C1 and C3 categories. The comparison is made in Figure 2C, where each group of
violin plots separated by dashed lines represents the 90 percentiles of $τ/T$ (in the presence of extrinsic noise) for the category with combined core motifs and its ‘component’ categories. It can be
seen that the category with combined core motifs usually shows intermediate robustness among its ‘component’ categories. That is to say, if a network topology has low robustness against extrinsic
noise, adding a high-robustness core motif usually improves the robustness, but the combined topology cannot outperform the added high-robustness core motif.
Topologies with long period achieve high robustness against extrinsic noise
The above analyses indicate that network topologies differ widely in their robustness to achieve accurate oscillation, then we ask what mechanisms cause these differences. Note that how the system
responds to the noise is often linked to the deterministic features (Monti et al., 2018; Paulsson, 2004; Wang et al., 2010). For example, Monti et al. found that the circuit’s ability to sense time
under input noise becomes worse when this circuit’s deterministic behavior cannot generate the limit cycle; Wang et al. adopted a similar form of noise and demonstrated the importance of signed
activation time, a quantity calculated based on deterministic behavior, on the noise attenuation; by using an $Ω$-expansion to approximate the birth-and-death Markov process, Paulsson obtained the
variance of the protein in gene networks and found that it is related to the network’s elasticity, which is calculated from the deterministic model. Based on these observations, we explore two
important characteristics for the oscillator: period and amplitude. Instead of focusing on a specific oscillatory network topology, we consider all 474 oscillatory network topologies and study what
period and amplitude each topology prefers. To be precise, for each network topology, we calculate the distributions of period and amplitude from 1000 randomly sampled oscillation parameter sets and
approximate mean values of period and amplitude by $Topt$ and $Aopt$ , respectively. Here, we refer to the amplitude as the maximal peak value among nodes A–C; $Topt$ (or $Aopt$) is defined as the
expectation of the best-fit exponential distribution of 1000 periods (or 1000 amplitudes) (Figure 2D). Therefore, the topology with large $Topt$ tends to oscillate with long period, and the topology
with large $Aopt$ usually indicates an oscillation with high amplitude. Note that these two quantities are calculated in the noise-free system, and thus are not affected by the amplitude of the noise
source or the type of noise.
To investigate the role of above two quantities $Topt$ and $Aopt$ in the robustness of accurate oscillation against extrinsic noise, we calculate Spearman coefficients between these two quantities
and 90 percentiles of $τ/T$ for all 474 oscillatory network topologies (Figure 2E). In Figure 2E, each dot represents an oscillatory network topology, with the x-axis representing the ranking
according to $Topt$ (left panel) or $Aopt$ (right panel). The Spearman coefficient between $Topt$ and 90 percentile of $τ/T$ for all 474 oscillatory network topologies is 0.94, which is larger than
that between $Aopt$ and 90 percentile of $τ/T$ (0.88). This result not only holds for all 474 oscillatory network topologies, but also holds within each of C1, C2, …, and C5 categories (Figure
2—figure supplement 1). These findings indicate that the robustness to extrinsic noise is more highly correlated with long period rather than high amplitude.
Since the long period benefits the robustness to extrinsic noise, then we ask how network topologies affect the period and whether those topologies with long period indeed show high robustness to
extrinsic noise. To answer these questions, we analyze $Topt$ for C1, C2, …, and C5 categories (Figure 2F). The ranking of these five categories according to $Topt$ is C1 > C2 ≈ C3 > C4 > C5, which
is obtained by the one-tailed Wilcoxon rank-sum tests for each adjacent two categories. This ranking is almost the same as that according to the robustness of accurate oscillation to extrinsic noise
(C1 > C2 > C3 > C4 > C5) except rankings for C2 and C3 categories, suggesting that the topology with long period usually leads to high robustness of accurate oscillation to extrinsic noise (Figure 2G
). The only inconsistency is that C2 and C3 categories differ in the robustness but have no significant difference in the probability to achieve long period. That is to say, C2 category might show
better robustness to extrinsic noise than C3 category though they have the same period. Figure 2H shows typical dynamics for five different topologies when extrinsic noise exists. Those topologies
from the top panel to the bottom panel belong to categories C1 to C5, respectively. Their dynamics have almost the same amplitude, but the period, as well as the autocorrelation, decreases when
categories vary from C1 to C5. These findings suggest that topologies with prolonged period tend to have good performance to filter extrinsic noise, and this correlation is less likely due to that
they have different amplitude.
The robustness of accurate oscillation against intrinsic noise for different network topologies
In the presence of only intrinsic noise, the repressilator with positive autoregulation is still better than the activator-inhibitor, and the advantage of positive autoregulation still holds
Unlike considering the robustness of accurate oscillation against parameter variability in the previous section, we next study the case where only intrinsic noise exists. With the same oscillatory
network topology categories present in Figure 2A, 90 percentiles of the dimensionless correlation time ($τ/T$) in the presence of only intrinsic noise also show a roughly similar trend from C1 to C5
categories (Figure 3A) except that C2 category exhibits the same robustness with C1 category to intrinsic noise while C2 category shows lower robustness than C1 in the presence of extrinsic noise.
Moreover, the higher robustness of C1 and C2 categories compared with C3 category validates the better performance of the repressilator with positive autoregulation than the activator-inhibitor
topology; the improvement of robustness from C4 category to C1 (or C2) category indicates the effect of positive autoregulation on the robustness to intrinsic noise; the comparison between the
robustness of C5 and C3 categories also implies the advantage of the positive autoregulation. These findings are consistent with those when noise is only originated from the extrinsic noise. Another
consistency is that the hybrid of core motifs imparts an intermediate robustness not only in the presence of the extrinsic noise (Figure 2C) but also in the presence of the intrinsic noise (Figure 3B
Figure 3
with 4 supplements
see all
The relationship between network topology and robustness to intrinsic noise.
Topologies with the high amplitude enable high robustness against intrinsic noise
Similar to the analysis of robustness to extrinsic noise, we try to answer whether the period or amplitude is highly correlated with the robustness to intrinsic noise. In the presence of only
intrinsic noise, the Spearman correlation coefficient of 90 percentiles of $τ/T$ and $Topt$ for all 474 oscillatory network topologies is 0.72, which is smaller than that of 90 percentiles of $τ/T$
and $Aopt$ (0.81) (Figure 3C). These findings suggest that unlike the case of extrinsic noise where the robustness is more strongly correlated with period, the robustness of accurate oscillation
against intrinsic noise is more highly correlated with amplitude. In other words, the topology with the high amplitude has a larger probability to achieve high robustness against intrinsic noise than
that with long period. However, it should be noted that the correlation coefficient between 90 percentiles of $τ/T$ and $Aopt$ for all oscillatory network topologies is not very close to 1 (the right
panel in Figure 3C), and it is also much smaller than 1 even in each category (Figure 3—figure supplement 1), implying that the relation between the amplitude and robustness to intrinsic noise is not
very strong, and that some topologies with small amplitude may perform better than those with high amplitude. Therefore, there might exist other mechanisms to attenuate intrinsic noise.
Furthermore, by applying Wilcoxon rank-sum tests to amplitude average ($Aopt$) for neighboring two categories, we find that the ranking of five network topology categories according to amplitude is
C1 > C2≈ C3 > C4 > C5 (Figure 3D). This ranking is almost the same as that according to the robustness to intrinsic noise (C1 ≈ C2 > C3 > C4 > C5) (Figure 3E), implying that the amplitude might link
the topology category and the robustness to intrinsic noise. The only exception is C2 category: because of the fact that C1 > C2 ≈ C3 according to $Aopt$ and the fact that the amplitude strongly
correlates with the robustness, the C2 category is supposed to show the same robustness to intrinsic noise as C3 category and exhibit lower robustness than C1 category; however, the robustness to
intrinsic noise for C2 category is actually at the same level of C1 category, further demonstrating that the high amplitude is not the only mechanism to enhance the robustness to intrinsic noise (
Figure 3—figure supplement 1). Figure 3F shows typical dynamics in the presence of intrinsic noise whose topologies belong to distinct categories. Those dynamics exhibit near period, but their
amplitudes and autocorrelations decrease from category C1 to category C5, which supplies a possibility to enhance the robustness to intrinsic noise through varying topologies while maintaining
Simulations using the Gillespie algorithm lead to similar conclusions
The above analyses are based on simulations for chemical Langevin equations, which can only give approximate solutions of the dynamical behavior in the presence of intrinsic noise. To test whether
this approximation is feasible, we use the Gillespie algorithm to exactly solve the stochastic dynamical behavior when facing intrinsic noise, and then conduct similar analyses (Figure 3—figure
supplement 2) as the previous section has done. According to the robustness rankings for C1–C5 categories, the repressilator with positive autoregulation performs better than the activator-inhibitor,
and the topologies with positive autoregulation are better than that without positive autoregulation. Besides, the robustness is more correlated with the mean amplitude rather than the mean period,
and the order of the five categories sorted by the mean amplitude is almost the same as that sorted by robustness, indicating the bridge role of amplitude to link topologies and the robustness to
intrinsic noise. These results are consistent with the conclusions based on chemical Langevin equations. We also find that the Gillespie algorithm leads to higher dimensionless correlation times than
chemical Langevin equations since the maximal correlation in Figure 3A is near 6 and that in Figure 3—figure supplement 2A is 40. However, this difference does not indicate that chemical Langevin
equations are bad approximations: when the system behaves normal noise filtering capability, these two methods give similar dimensionless correlation times (Figure 3—figure supplement 3A); when the
system buffer noise perfectly, dimensionless correlation times calculated through these two methods differ a lot, but autocorrelation functions remain similar (Figure 3—figure supplement 3B), which
indicates that chemical Langevin equations still capture the system’s ability to buffer noise. The reason why large and extremely large dimensionless correlation times result in almost same
correlations might be that doubling long correlation time cannot increase autocorrelation efficiently due to the property of the exponential function.
Relations between period/amplitude and oscillation accuracy against noise are validated by analytical approaches
The above simulations revealed relations between two important features (i.e., period and amplitude) of the oscillator and the oscillator’s robustness to noise. However, these results just showed the
correlation rather than the causal relationship. Besides, because the period and amplitude are usually positively correlated (Figure 4—figure supplement 1), it is hard to control one feature and
analyze the effect of the other feature. Fortunately, these two problems can be solved by introducing the timescale or rescaling parameters. In this way, we can change one feature while maintaining
the other feature, and then analytically derive causal relations between period or amplitude and the oscillation accuracy. We will illustrate these methods and corresponding results below.
To study the relation between period and oscillation accuracy against noise, we maintain the amplitude and tune the period through changing the factor $M$ on the right-hand side in ordinary
differential equations (Figure 4A), and then analyze the phase noise through the analytical approach proposed by Demir et al., 2000. Varying $M$ can be regarded as the rescaling of time $t$, so the
period is changed while maintaining the amplitude, and thus we can focus on the effect of period on the oscillation accuracy. In order to analyze the system with variable $M$, we first summarize
Demir et al.’s work. They carried out nonlinear perturbation analysis for oscillators and obtained an exact equation for phase deviation. We only summarize the main results below. The dynamics of a
perturbed oscillator can be described as a set of differential equations:
$\stackrel{˙}{x}=f\left(x\right)+B\left(x\right)\xi \left(t\right)$
Effects of tuning period or amplitude on oscillation accuracies.
where $x∈R3$ , $f(⋅):R3→R3$ , $B(⋅):R3→R3×3$ and $ξ(t)∈R3$ is random noise. The unperturbed system $x˙=fx$ has a periodic solution $xst$ (with period $T$). It can be proved that the variance of the
phase deviation $σ2(t)$ satisfies $σ2(t)=ct$, and $c$ is as follows:
(1) $\begin{array}{c}c=\frac{1}{T}{\int }_{0}^{T}{v}_{1}^{T}\left({t}^{`}\right)B\left({x}_{s}\left({t}^{`}\right)\right){B}^{T}\left({x}_{s}\left({t}^{`}\right)\right){v}_{1}\left({t}^{`}\right)d{t}
where $v1Tt$ is the first row of the matrix $V(t)$. Here, the first column of $V-1(t)$ is $x˙st$ , and $V-1(t)diag[μ1,μ2,μ3]Vt$ is the state transition matrix for $w˙=Atw(t)$ where $μi$ ’s are
Floquet exponents and $At=∂f(x)∂x|x=xst$ (see ‘Methods’ for details). The Equation 01 gives an analytic expression describing the phase noise, so we use the dimensionless $c$, that is, $c/T$, to
measure the oscillation accuracy instead of $τ/T$. To calculate $c/T$ for the systems with $M$, we use $T$ to denote the period for the system without $M$, and then the period for this new system is
$MT$. Besides, $v1t$ becomes $Mv1tM$ (see ‘Methods’ for details). For the noise term, we merge $1M$ with kinetic parameters $vij,δi,ri$ , that is, these parameters become $1M$ of original values, and
then we model the extrinsic and intrinsic noise as that in Figure 1E. Therefore, $Bx$ becomes $Bx1M$ when facing extrinsic noise as the magnitude of noise source is proportional to the kinetic
parameters. However, in the presence of only intrinsic noise, $Bx$ becomes $Bx1M$ because the noise term in the chemical Langevin equation is usually the square root of reaction rates. Then we can
calculate the ratio of the slope of the variance of the phase noise to the period ($c/T$) using the Equation 01 (see ‘Methods’ for details). We find that the $c/T$ in the presence of only extrinsic
noise is proportional to the $1/M$, and that in the presence of only intrinsic noise is not affected by $M$. Note that the smaller the $c/T$ is, the more accurate the oscillation is. Thus, large $M$
enhances the oscillation accuracy against extrinsic noise, which is also numerically validated by the trend of dimensionless correlation times for the system with different $M$ (right lower panel in
Figure 4A). Since large $M$ leads to long period but has no influence on amplitude, the prolonged period might be the reason for high oscillation accuracy in the presence of extrinsic noise.
For the relation between amplitude and oscillation accuracy against noise, we keep the period and tune the amplitude through the rescaling parameter $N$ and then analyze the rescaled system. For a
fixed topology with a set of oscillation kinetic parameters, we replace the variables $A$, $B$, and $C$ with $A~/N$, $B~/N$, and $C~/N$, respectively (Figure 4B). This rescaling makes amplitudes of
$A~$$N$ times as high as that of $A$, and so do $B~$ and $C~$ . However, this rescaling has no influence on the period, so we can focus on the role of amplitude in the oscillation accuracy. The
system with rescaled variables $A~$ , $B~$ , and $C~$ shows unchanged oscillation accuracy against extrinsic noise with varied $N$, but the oscillation accuracy against intrinsic noise increases with
increased $N$ (see ‘Methods’). Taken together, large $N$ not only increases the amplitude but also improves the oscillation accuracy to intrinsic noise while maintaining the period. These results are
consistent with numerical simulations for tendencies of period, amplitude, and dimensionless correlation times (lower panel in Figure 4B). These results indicate that the improvement of the
oscillation accuracy to intrinsic noise may due to the high amplitude rather than period.
Analyses of synthetic NF-κB signaling circuits demonstrate the improvement of the oscillation accuracy when adding a repressilator topology to the activator-inhibitor
In previous sections, we have used two- and three-node networks to approximate biological systems and focused on the noise coming from the variability in kinetic parameters or chemical reactions.
Though biological systems are more complex than two- or three-node networks and face noise from various sources besides the above noises, the investigation of a specific biological system— a
synthetic NF-κB signaling circuit is consistent with the theoretical results in previous sections. As described in our previous work, we implement the design of negative feedback-only circuit 1 (
Figure 5A) by integrating the synthetic RelA-IκBα signaling circuit into the yeast MAPK pathway. The nuclear-to-cytoplasmic RelA oscillations can be triggered by inducing the degradation of IκBα
through the activation of yeast MAP kinase Fus3, and we can monitor these single-cell oscillations for up to 10 hr. Base on this simple circuit, we then modify its structure by adding extra
regulations. One modification is adding constantly expressed IκBα. This copy of IκBα also inhibits RelA and is inhibited by Fus3, so it provides another pathway from Fus3 to activate RelA (the orange
link in circuit 2 in Figure 5A). Another modification is adding a yeast MAPK phosphatase Msg5 (the orange link in circuit 3 in Figure 5A), which is activated by RelA and can dephosphorylate Fus3. In
circuit 3, Msg5, RelA, and IκBα form a repressilator topology. To obtain the single-cell time trajectories for these three circuits, we employ time-lapse microscopy to track the RelA nuclear
localization dynamics for over 10 hr. The period lengths are determined as the time intervals between the successive peaks of these trajectories, and then we calculate the CV of those period lengths
for over 50 cells. We find that the circuit 2 shows similar CV of period as that for the circuit 1, but that circuit 3 exhibits lower CV of period than circuit 1 (Figure 5B). These results suggest
that the additional repressilator topology indeed facilitates the noise buffering capability for the activator-inhibitor topology.
Experimental evidence of the improvement of oscillation accuracy when adding a repressilator topology to the activator-inhibitor.
Such improvement of the oscillation accuracy when adding a repressilator topology to the activator-inhibitor in the synthetic NF-κB circuit can also be validated using the mathematical models as
described in Figure 1. We use nodes A, B, and C to denote IκBα, RelA, and Fus3, respectively, and thus circuits 1 and 3 in Figure 5A are networks shown in upper and lower panels in Figure 5C,
respectively. These two networks are interconvertible by tuning $KCA$ (the binding affinity of protein C to gene A): (near) zero $1/KCA$ indicates the little effect of protein C to protein A, and
thus protein A is not affected by protein C, leading to the activator-inhibitor; non-zero $1/KCA$ implies the existence of the inhibition from protein C to protein A, resulting in the network with an
activator-inhibitor and a repressilator. For a given oscillation parameter set for the activator-inhibitor, we first set $1/KCA$ to be (near) zero and calculate period, amplitude, and dimensionless
correlation time in the presence of extrinsic or intrinsic noise (the first points in Figure 5D–G); then we increase $1/KCA$ , and calculate the corresponding quantities (i.e., period, amplitude, and
dimensionless correlation time, as shown from the second points in Figure 5D–G). We can find that the period, amplitude, and dimensionless correlation time for the activator-inhibitor are usually
smaller than those with an additional repressilator, and this gap enlarges with increased $1/KCA$ , that is, the increased strength of the negative regulation from the additional node C to the
inhibitor node A (Figure 5C). Therefore, we demonstrate that adding the repressilator to the activator-inhibitor enhances the oscillation accuracy. This is consistent with that C1 and C2 categories
exhibit higher robustness than C3 categories. Moreover, the prolonged period and increased amplitude, which are also observed in C1 and C2 categories, may be the reason for such enhancement (Figure
5D and E).
It remains the major challenge in biology to understand how living systems perform complex behaviors accurately in the presence of inevitable noise. Instead of studying biological networks case by
case, we try to answer whether there exist general network design principles for living systems to execute biological functions by using a bottom-up strategy (Lim et al., 2013; Zhang and Tang, 2019).
Here, we systematically explore the network design principles for accurate oscillation in both two- and three-node networks. We identify several key motifs that have distinct robustness to noise. The
motif —— a repressilator with positive autoregulation behaves better than other motifs present in Figure 2A in most cases, especially the activator-inhibitor oscillator; the additional positive
autoregulation can improve the robustness. These results are consistent in spite of sources of noise. However, different sources of noise utilize distinct mechanisms to filter noise: the variability
of parameters, a type of extrinsic noise, is largely filtered through long period, and the intrinsic noise is buffered by high amplitude.
Interestingly, investigations of three engineered NF-κB signaling circuits partly validate our simulation results. For the negative-feedback loop circuit, if the additional new regulations form a
repressilator, low variance of period will occur, but if no repressilator emerges, the variance of period shows no significant change. These findings show the advantage of the repressilator against
While modifying network topology and changing regulation strength for a fixed topology are both options to improve the robustness of accurate oscillation, each network’s robustness is an indicator of
the probability of this network topology achieving accurate oscillation with varied regulatory strengths (Figure 2—figure supplement 2, Figure 3—figure supplement 4): the network topology with high
robustness tends to show high dimensionless autocorrelation time when varying regulatory strengths, that is, accurate oscillation (first 10 bars in Figure 2—figure supplement 2 and Figure 3—figure
supplement 4); the network topology with low robustness displays a bad performance of oscillation accuracy in the whole parameter space (last 10 bars in Figure 2—figure supplement 2 and Figure
3—figure supplement 4). Besides, our work also suggests that tuning network topology is more efficient than changing regulatory strength. This is based on the observations that network topologies
with low robustness (last 10 bars in Figure 2—figure supplement 2 and Figure 3—figure supplement 4) cannot have a high oscillation accuracy even when searching all kinetic parameter space, but
changing topologies may increase the probability of high oscillation accuracy. So we suggest that a feasible way to improve the oscillation accuracy in synthetic networks is to first modify the
topology to avoid low-robustness ones and then tune the regulation strength, as illustrated in Figure 5C.
Mechanisms to buffer different sources of noise in the oscillator can be dramatically different. On the one hand, long period is able to attenuate extrinsic noise, which is also called the
time-averaging strategy. This strategy has been widely studied in nonoscillatory networks, such as circuits that are sensitive to the stimulus (Hornung and Barkai, 2008), circuits with
‘‘switch-like’’ behaviors (Wang et al., 2010), and adaptive circuits (Nie et al., 2020; Sartori and Tu, 2011). For these nonoscillatory circuits, fluctuations in output have been proven to be related
to some key timescales, and long timescales often result in the output with small variance. On the other hand, the intrinsic noise is hard to be attenuated through time averaging, such as the
adaptive incoherent feed-forward loop (Sartori and Tu, 2011). Actually, the right way to buffer intrinsic noise in biological oscillators was found to depend on levels of molecules. For example, the
importance of protein numbers has also been demonstrated in the work of Potvin-Trottier et al., 2016. They found that increasing the peak and bottom values decreases the CV in the decay phase of the
oscillator. Based on these results, it is suggestive that the network topology with long period and high amplitude may enable good robustness to both extrinsic and intrinsic noise. Interestingly, it
is usually not hard to obtain long period and high amplitude simultaneously since the long period tends to allow the protein number to climb to a high level (Figure 4—figure supplement 1).
Our work only focused on the effects of biological noise on oscillation accuracy, neglecting other dynamic changes caused by noise. These dynamics may include the loss of multistability and
occurrence of oscillation. Specifically, the way to model the noise may cause the loss of multistability (Duncan et al., 2015; Vellela and Qian, 2009); the presence of noise can produce oscillation
even when the corresponding deterministic model cannot oscillate, which has been validated in the toggle-switch system and excitable system (Lindner et al., 2004; Terebus et al., 2019; Zaks et al.,
2005). The possible reason might be the noise-induced transition between different states. Since our work only studied network topologies whose deterministic model can generate oscillation, we did
not count the topologies that cannot oscillate in the deterministic model but begin to oscillate in the stochastic model. Due to the popularity of such topologies, how these topologies buffer noise
will be of interest and may lead to the discoveries of new principles.
In this work, the extrinsic noise is assumed only from fluctuations in kinetic parameters, and its magnitude linearly depends on the level of the parameter. Except this type of extrinsic noise, cells
also face the random partitioning that occurs during cell division, noisy stimulus, and so on (Monti et al., 2018; Veliz-Cuba et al., 2015). Since two different types of noise studied in this work
require different mechanisms to buffer, other sources of noise may also need new mechanisms to filter. Thus, some unknown principles need to be further revealed and incorporated into network design
as the increasingly improved complexity and multiple sources of noise.
Another limitation of our work is that we did not decompose the reactions in the deterministic model into detailed elementary reaction steps when using the Gillespie algorithm. The advantage of
simulating nonelementary reactions with Hill-type rate functions is the low computation cost, and in some biological networks, it leads to consistent results with the model composed of all elementary
reactions (Gonze et al., 2002b; Kim et al., 2014; Sanft et al., 2011). However, this approach may not be always accurate, depending on the timescale separation of reactions (Kim et al., 2014; Sanft
et al., 2011); for example, the Hill-type reaction rate is based on the quasi-steady-state approximation, which does not hold when binding/unbinding of TF to the promoter is slow or comparable to the
timescales of protein production or decay (Choi et al., 2008). Furthermore, this method neglects detailed reaction in gene regulatory networks, and thus fails to study the roles of these reactions in
stochasticity. These detailed reactions include the binding and unbinding of the transcription factor to the promoter, dimerization of transcription factors, transcription, and translation (Cao et
al., 2018; Shahrezaei and Swain, 2008; Terebus et al., 2019). We anticipate the need for a more detailed model where every reaction of Hill-type form is decomposed into the elementary reactions. The
recent development about stochastic algorithms with fast computation makes it feasible to simulate such detailed model for all two- and three-node network topologies, for example, algorithms focusing
on solving the chemical master equations (Cao et al., 2010; Cao et al., 2016; Munsky and Khammash, 2006; Terebus et al., 2021) and variants of the Gillespie algorithms that directly simulate the
temporal dynamics (Gillespie and Petzold, 2003). Besides, the construction of probability surfaces through these algorithms may shed light on new principles for accurate oscillation.
To model two-node and three-node network topologies, we use transcriptional regulatory networks and assume competitive inhibition among multiple transcription factors. The competitive inhibition
means that multiple transcription factors compete for the same binding sites if they regulate one gene simultaneously (Shi et al., 2017). So the gene expression depends on the relative weight of
transcription factors inhibiting this gene and that activating this gene. The following set of ordinary equations is used to describe the deterministic dynamics of a three-node transcriptional
regulatory network:
(2) $\left\{\begin{array}{l}\frac{dA}{dt}={k}_{basal}+\frac{\sum _{i}{v}_{iA}{\left(\frac{{x}_{i}}{{K}_{iA}}\right)}^{3}+{\delta }_{A}}{1+\sum _{i}{\left(\frac{{x}_{i}}{{K}_{iA}}\right)}^{3}+\sum _
{i}{\left(\frac{{y}_{i}}{{K}_{iA}}\right)}^{3}}-{r}_{A}A\\ \frac{dB}{dt}={k}_{basal}+\frac{\sum _{i}{v}_{iB}{\left(\frac{{x}_{i}}{{K}_{iB}}\right)}^{3}+{\delta }_{B}}{1+\sum _{i}{\left(\frac{{x}_{i}}
{{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{y}_{i}}{{K}_{iB}}\right)}^{3}}-{r}_{B}B\\ \frac{dC}{dt}={k}_{basal}+\frac{\sum _{i}{v}_{iC}{\left(\frac{{x}_{i}}{{K}_{iC}}\right)}^{3}+{\delta }_{C}}{1+\
sum _{i}{\left(\frac{{x}_{i}}{{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{y}_{i}}{{K}_{iC}}\right)}^{3}}-{r}_{C}C\end{array}$
where $A$, $B$, and $C$ are the concentrations of proteins A, B, and C. $xi=A, B$ or $C$ in each equation denotes the concentration of protein activating the gene, and $yi=A,B$ or $C$ the
concentration of protein inhibiting the gene. The production rate constant $vij$ represents the maximal production rate of protein $i$ regulated by protein $j$, with $Kij$ binding affinity. If there
exist proteins activating gene i, $δi$ is zero; if no protein activates gene i, $δi$ is non-zero and represents the production rate caused by other proteins. $kbasal$ is the basal production rate.
The equations for the two-node transcriptional regulatory network can be obtained in a similar way.
To provide a better explanation about the nonlinear reaction term in above equations, we took the following case as an example: protein A (i.e., TF) binds to gene B to inhibit the gene expression,
and protein B binds to the same site in gene B to activate the gene expression. We assumed that (1) there are three binding sites in gene B, which once protein A (or B) binds to, then B (or A)
cannot. The elementary reactions are described as follows:
$A+{G}_{B}⇌{G}_{B}\bullet A,A+{G}_{B}\bullet A⇌{G}_{B}\bullet {A}_{2},A+{G}_{B}\bullet {A}_{2}⇌{G}_{B}\bullet {A}_{3},$
$B+{G}_{B}⇌{G}_{B}\bullet B,B+{G}_{B}\bullet B⇌{G}_{B}\bullet {B}_{2},B+{G}_{B}\bullet {B}_{2}⇌{G}_{B}\bullet {B}_{3},$
where $GB, A, B$ denote gene B, protein A, and protein B, respectively. The dissociation rates ($kreverse/kforward$) for these six reactions are $K1, K2, ⋯, K6$ . Therefore, the fraction of the gene
B at the state $GB∙B3$ in equilibrium is given by
Furthermore, we assumed that (2) $K6≪K5,K4$ and $K3≪K2,K1$ , so that this fraction can be rewritten as
We further assumed (3) that only gene B staying at the state $GB∙B3$ can lead to transcription and subsequent translation for protein B, and (4) that the binding/unbinding of TFs to a gene can
achieve a rapid equilibrium as TF levels change, and thus the production rate of protein B is modeled as
where $vBB$ is the maximal production rate when gene B is bound with three protein B. This form is the same as those in (1) if $K1K2K3$ and $K4K5K6$ are substituted by $KAB3$ and $KBB3$ ,
Stochastic model in the presence of extrinsic noise
To generate extrinsic noise, we perturb each kinetic parameter $vij,δi,ri$ by multiplying the sum of 1 and an independent temporal noise term, and obtain a new system described by the following
stochastic differential equations:
(3) $\left\{\begin{array}{l}\frac{dA}{dt}={k}_{basal}+\frac{\sum _{i}{v}_{iA}\left(1+\epsilon {\eta }_{iA}\right){\left(\frac{{x}_{i}}{{K}_{iA}}\right)}^{3}+{\delta }_{A}\left(1+\epsilon {\xi }_{A}\
right)}{1+\sum _{i}{\left(\frac{{x}_{i}}{{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\frac{{y}_{i}}{{K}_{iA}}\right)}^{3}}-{r}_{A}A\left(1+\epsilon {\zeta }_{A}\right)\\ \frac{dB}{dt}={k}_{basal}+\frac{\
sum _{i}{v}_{iB}\left(1+\epsilon {\eta }_{iB}\right){\left(\frac{{x}_{i}}{{K}_{iB}}\right)}^{3}+{\delta }_{B}\left(1+\epsilon {\xi }_{B}\right)}{1+\sum _{i}{\left(\frac{{x}_{i}}{{K}_{iB}}\right)}^{3}
+\sum _{i}{\left(\frac{{y}_{i}}{{K}_{iB}}\right)}^{3}}-{r}_{B}B\left(1+\epsilon {\zeta }_{B}\right)\\ \frac{dC}{dt}={k}_{basal}+\frac{\sum _{i}{v}_{iC}\left(1+\epsilon {\eta }_{iC}\right){\left(\frac
{{x}_{i}}{{K}_{iC}}\right)}^{3}+{\delta }_{C}\left(1+\epsilon {\xi }_{C}\right)}{1+\sum _{i}{\left(\frac{{x}_{i}}{{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{y}_{i}}{{K}_{iC}}\right)}^{3}}-{r}_{C}C\
left(1+\epsilon {\zeta }_{C}\right)\end{array}$
Here, the control parameter $ε$ indicates magnitude of perturbation of kinetic parameters, and large $ε$ represents big perturbation of kinetic parameters. $ηij$ , $ξi$, $ζi$ are independent noise
terms and all modeled by the Ornstein–Uhlenbeck process:
(4) ${\tau }^{noise}dz=-zdt+\mathrm{\sigma }d{W}_{t}$
where $Wt$ is standard Wiener processes. This equation implies that $z(t)$ has zero mean and variance $σ22τnoise$ .
Stochastic model in the presence of intrinsic noise
To induce intrinsic noise, we replace the concentration of protein with the number of protein by introducing the cell volume $V$ and assume that production events and degradation events occur
independently and randomly. To be precise, we use $XA, XB, XC$ to denote numbers of proteins A, B, C, respectively, and then we replace $A, B,C$ in Equation 2 by $XA/V, XB/V, XC/V$, respectively.
Therefore, the dynamics of protein numbers $XA, XB, XC$ are described by the following reactions:
${X}_{A}\underset{{X}_{A}\stackrel{{r}_{A}{X}_{A}}{\to }{X}_{A}-1}{\overset{V{k}_{basal}+V\left(\sum _{i}{v}_{iA}{\left(\frac{{X}_{i}}{{K}_{iA}}\right)}^{3}+{\delta }_{A}{V}^{3}\right)/\left({V}^{3}+
\sum _{i}{\left(\frac{{X}_{i}}{{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{i}}{{K}_{iA}}\right)}^{3}\right)}{\to }}{X}_{A}+1$
${X}_{B}\underset{{X}_{B}\stackrel{{r}_{B}{X}_{B}}{\to }{X}_{B}-1}{\overset{V{k}_{basal}+V\left(\sum _{i}{v}_{iB}{\left(\frac{{X}_{i}}{{K}_{iB}}\right)}^{3}+{\delta }_{B}{V}^{3}\right)/\left({V}^{3}+
\sum _{i}{\left(\frac{{X}_{i}}{{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{i}}{{K}_{iB}}\right)}^{3}\right)}{\to }}{X}_{B}+1$
${X}_{C}\underset{{X}_{C}\stackrel{{r}_{C}{X}_{C}}{\to }{X}_{C}-1}{\overset{V{k}_{basal}+V\left(\sum _{i}{v}_{iC}{\left(\frac{{X}_{i}}{{K}_{iC}}\right)}^{3}+{\delta }_{C}{V}^{3}\right)/\left({V}^{3}+
\sum _{i}{\left(\frac{{X}_{i}}{{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{i}}{{K}_{iC}}\right)}^{3}\right)}{\to }}{X}_{C}+1$
We used the following two algorithms to simulate the above system.
We used the standard Gillespie algorithm to simulate the system. There are six reactions in total (as shown above), and the propensity functions are the reaction rates listed above the arrow. Note
that we did not decompose the reactions with the Hill function rate into the elementary reactions; the reaction rate with the Hill function type has also been applied to other discrete stochastic
models (Gonze and Goldbeter, 2006; Gonze et al., 2002b; Veliz-Cuba et al., 2015; Wang et al., 2019; Zhao et al., 2021) and proven to be an accurate approximation for the model composed of all
elementary reactions under certain circumstances (Kim et al., 2014; Sanft et al., 2011). In our simulations, each of the six reactions occurs with the random waiting time, which obeys an exponential
distribution with mean of the inverse of the propensity function. For example, the reaction of decreasing protein A by 1 has the propensity function r[A]X[A], and increasing protein A by 1
corresponds to $kbasel+V(ΣiviA(XiKiA)3+δAV3)/(V3+Σi(XiKiA)3+Σi(YiKiA)3)$. Once we get one trajectory, we can calculate the autocorrelation time. See Figure 3—figure supplement 2 and Figure 3—figure
supplement 3 for simulation results.
We also used the Langevin equation, a good approximation of this system under certain conditions (Gillespie, 2000), to model the system. The corresponding chemical Langevin equations are as follows:
$\left\{\begin{array}{l}d{X}_{A}=\left(V{k}_{basal}+V\frac{\sum _{i}{v}_{iA}{\left(\frac{{X}_{i}}{{K}_{iA}}\right)}^{3}+{\delta }_{A}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{i}}{{K}_{iA}}\right)}^
{3}+\sum _{i}{\left(\frac{{Y}_{i}}{{K}_{iA}}\right)}^{3}}-{r}_{A}{X}_{A}\right)dt+\sqrt{V{k}_{basal}+V\frac{\sum _{i}{v}_{iA}{\left(\frac{{X}_{i}}{{K}_{iA}}\right)}^{3}+{\delta }_{A}{V}^{3}}{{V}^{3}+
\sum _{i}{\left(\frac{{X}_{i}}{{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{i}}{{K}_{iA}}\right)}^{3}}+{r}_{A}{X}_{A}}d{W}_{t}^{A}\\ d{X}_{B}=\left(V{k}_{basal}+V\frac{\sum _{i}{v}_{iB}{\left(\
frac{{X}_{i}}{{K}_{iB}}\right)}^{3}+{\delta }_{B}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{i}}{{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{i}}{{K}_{iB}}\right)}^{3}}-{r}_{B}{X}_{B}\right)dt+\
sqrt{V{k}_{basal}+V\frac{\sum _{i}{v}_{iB}{\left(\frac{{X}_{i}}{{K}_{iB}}\right)}^{3}+{\delta }_{B}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{i}}{{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{i}}
{{K}_{iB}}\right)}^{3}}+{r}_{B}{X}_{B}}d{W}_{t}^{B}\\ d{X}_{C}=\left(V{k}_{basal}+V\frac{\sum _{i}{v}_{iC}{\left(\frac{{X}_{i}}{{K}_{iC}}\right)}^{3}+{\delta }_{C}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\
frac{{X}_{i}}{{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{i}}{{K}_{iC}}\right)}^{3}}-{r}_{C}{X}_{C}\right)dt+\sqrt{V{k}_{basal}+V\frac{\sum _{i}{v}_{iC}{\left(\frac{{X}_{i}}{{K}_{iC}}\right)}^
{3}+{\delta }_{C}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{i}}{{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{i}}{{K}_{iC}}\right)}^{3}}+{r}_{C}{X}_{C}}d{W}_{t}^{C}\end{array}$
where $Xi$ is the number of protein $i$, and $Wti$ is the standard Wiener process. The control parameter $V$ reflects the magnitude of stochasticity of biological reactions. The big $V$ indicates
small degree of stochasticity of biological reactions. See the next section for settings in the numerical simulation.
Numerical simulations for deterministic models were carried out in MATLAB (see https://github.com/LingxiaQiao/oscillation, (copy archived at swh:1:rev:72a2d3d1146b14e7988c1cc06208fe1252e9a6f5; Qiao,
2022) for MATLAB scripts). We use the solver ode15s to simulate the dynamics. Simulations for stochastic models were also implemented in MATLAB. In the presence of extrinsic noise, we used the
Milstein scheme (Kloeden and Platen, 2013) to numerically solve the noise term $ηij,ξj,ζj$ and used the Euler scheme to solve the dynamics of proteins’ concentrations. To be specific, the noise term
$z$ ($z=ηij,ξj,ζj$) at $n+1$ time step is determined by the following manner ($τnoise=1$):
${z}^{\left(n+1\right)}={z}^{\left(n\right)}-{z}^{\left(n\right)}∆t+\sigma {\delta W}_{n}+\frac{1}{2}{\sigma }^{2}\left[{\left({\delta W}_{n}\right)}^{2}-∆t\right]$
where $∆t$ is the time step, and $δWn$ obeys the normal distribution with mean zero and variance $∆t$. Then, the protein’s concentration is solved by the Euler scheme (taking A as an example):
${\left[A\right]}^{n+1}={\left[A\right]}^{n}+dt\left({k}_{basal}+\frac{{\sum }_{i}{v}_{iA}\left(1+\epsilon {\eta }_{iA}^{\left(n\right)}\right){\left(\frac{{x}_{i}^{\left(n\right)}}{{K}_{iA}}\right)}
^{3}+{\delta }_{A}\left(1+\epsilon {\xi }_{A}^{\left(n\right)}\right)}{1+{\sum }_{i}{\left(\frac{{x}_{i}^{\left(n\right)}}{{K}_{iA}}\right)}^{3}+{\sum }_{i}{\left(\frac{{y}_{i}^{\left(n\right)}}{{K}_
{iA}}\right)}^{3}}-{r}_{A}\left(1+\epsilon {\zeta }_{A}^{\left(n\right)}\right){A}^{\left(n\right)}\right).$
In the presence of intrinsic noise, we also used the Milstein scheme to numerically solve the dynamics of proteins’ copy numbers. Taking $XA$ as an example, its value at $n+1$ time step is as
${X}_{A}^{\left(n+1\right)}={X}_{A}^{\left(n\right)}+\left(V{k}_{basal}+V\frac{{\sum }_{i}{v}_{iA}{\left(\frac{{X}_{i}^{\left(n\right)}}{{K}_{iA}}\right)}^{3}+{\delta }_{A}{V}^{3}}{{V}^{3}+{\sum }_
{i}{\left(\frac{{X}_{i}^{\left(n\right)}}{{K}_{iA}}\right)}^{3}+{\sum }_{i}{\left(\frac{{Y}_{i}^{\left(n\right)}}{{K}_{iA}}\right)}^{3}}-{r}_{A}{X}_{A}^{\left(n\right)}\right)∆t+{\sigma }_{A}{\delta
W}_{A}^{\left(n\right)}+\frac{1}{2}{\sigma }_{A}^{2}\left[{\left({\delta W}_{A}^{\left(n\right)}\right)}^{2}-∆t\right],$
${\sigma }_{A}=\sqrt{V{k}_{basal}+V\frac{{\sum }_{i}{v}_{iA}{\left(\frac{{X}_{i}^{\left(n\right)}}{{K}_{iA}}\right)}^{3}+{\delta }_{A}{V}^{3}}{{V}^{3}+{\sum }_{i}{\left(\frac{{X}_{i}^{\left(n\
right)}}{{K}_{iA}}\right)}^{3}+{\sum }_{i}{\left(\frac{{Y}_{i}^{\left(n\right)}}{{K}_{iA}}\right)}^{3}}+{r}_{A}{X}_{A}^{\left(n\right)}}$
Searching for topologies capable of accurate oscillation
There are two steps for searching for topologies robustly capable of accurate oscillation. The first step is to select network topologies that are able to robustly oscillate among all two- and
three-node network topologies. For each topology, 10,000 sets of kinetic parameters are assigned randomly, with ranges shown in Supplementary file 1a; for each parameter set, the initial state of the
protein concentration is set to be 0, and we use ode15s in MATLAB to simulate the deterministic dynamics in the time interval [0, 1000]. The dynamics is regarded as oscillation if the following two
requirements are satisfied: every protein concentration cannot maintain unchanged in the time interval [700, 1000]; peaks in the time interval [700, 1000] cannot differ a lot. The first requirement
excludes the dynamics reaching the steady state, and the second the damping oscillator. We record the number of oscillatory dynamics for each topology, and then regard the topology with this number
larger than 80 as the oscillatory topology. But for the topology with the repressilator, if the number of oscillatory dynamics exceeds 10, we still regard this topology as the oscillatory topology.
This loose threshold ensures enough oscillatory topologies with the repressilator. In this way, we get 474 oscillatory topologies.
The second step is to explore the robustness of accurate oscillation for the above 474 oscillatory topologies. For each oscillatory topology, we sample enough parameter sets until there are 1000
parameter sets capable of oscillation. For each of these 1000 parameter sets, we record the period T and the amplitude (the maximal peak value among all protein concentrations) from deterministic
behavior; next we simulate the stochastic behavior in the time interval [0, 100T]. In the presence of only extrinsic noise, the initial concentration is set to be the state when the concentration B
reaches the highest value in a period, but in the presence of intrinsic noise, the initial concentration is converted to the copy number by multiplying the concentration with the cell volume $V$. We
use schemes mentioned in the previous sections to numerically solve the stochastic dynamics, with the time step in Supplementary file 1a. For a given noisy trajectory, the dimensionless
autocorrelation time $τ/T$ is $-1/logc$ , where $c$ is the autocorrelation coefficient at T. Since there are two or three trajectories each of which corresponds to a type of protein, so there are
two or three dimensionless autocorrelation times, and we use the largest one as the final dimensionless autocorrelation time. Finally, we use the 90 percentile of dimensionless autocorrelation time
to measure the robustness of this topology against noise. The 90 percentile is averaged over five repeated simulations.
Analytical results for the relation between robustness and period when tuning the timescale M
Phase noise in Demir et al.’s study
In this section, we briefly summarize Demir et al., 2000 study about the phase noise. We consider the dynamics described by the following equations:
(5) $\stackrel{˙}{x}=f\left(x\right)+B\left(x\right)\xi \left(t\right)$
where $x∈R3$ , $f(⋅):R3→R3$ , $B(⋅):R3→R3×3$ and $ξ(t)∈R3$ is random noise. Note that the noise amplitude $B$ is only related to $x$, which is not affected by the time $t$. The unperturbed system $x˙
=fx$ has a periodic solution $xst$ (with period $T$). Linearizing the noise-perturbed system around $xst$ gives the following system:
$\stackrel{˙}{w}=A\left(t\right)w\left(t\right)+B\left({x}_{s}\left(t\right)\right)\xi \left(t\right)$
where $wt=xt- xst$ , $At=∂f(x)∂x|x=xst$ is $T$–periodic. From Floquet theory, the state transition matrix $Φ(t,s)$ for $w˙=Atw(t)$ is given by
(6) $\mathrm{\Phi }\left(t,s\right)=U\left(t\right)\mathrm{exp}\left(D\left(t-s\right)\right)V\left(s\right)={\int }_{i=1}^{3}{u}_{i}\left(t\right)exp\left({\mu }_{i}\left(t-s\right)\right){v}_{i}^
where $U(t)$ is $T$-periodic nonsingular matrix with columns denoted by $uit$ , and $Vt$ with rows denoted by $viT(t)$ is $U-1(t)$. The $μi$ ’s are Floquet exponents. We can further choose $μ1=0$ and
$u1t=x˙s(t)$. Then corresponding $v1t$ will play an important role in calculating the phase noise.
From the nonlinear perturbation analysis, $zt=xst+α(t)+y(t)$ solves Equation 5 for a small $y(t)$. The $α(t)$ and $y(t)$ are called as phase noise and deviation noise, respectively. It can be proved
that the variance of the phase noise $α(t)$ increases linearly with time $t$, that is,
(7) $\mathrm{V}\mathrm{a}\mathrm{r}\left(\alpha \left(t\right)\right)=ct$
(8) $c=\frac{1}{T}{\int }_{0}^{T}{v}_{1}^{T}\left({t}^{`}\right)B\left({x}_{s}\left({t}^{`}\right)\right){B}^{T}\left({x}_{s}\left({t}^{`}\right)\right){v}_{1}\left({t}^{`}\right)d{t}^{`}.$
Since $c$ has the same unit as $T$, we divide $c$ by $T$ to ensure a dimensionless index when measuring the phase noise.
The vector $v1t$ when tuning $M$
We first consider the oscillator governed by the following equation:
(9) $\stackrel{˙}{x}=f\left(x\right)$
We still use notations in the previous section to denote the quantities for this system. For example, the solution $xst$ , the period $T$, the Jacobi $At$ of $fx$ at the solution $xst$ . We assume
the state transition matrix $Φt,s=UtexpDt-sVs$ , which satisfies the first Floquet exponent is 0 and the first column of $Ut$ is the time derivative of $xs(t)$ . Then the first row of $Vt$ is
denoted as $v1Tt$ , which can be used to calculate the variance of phase noise.
Next, we explore how $v1t$ changes when the right-hand term is divided by the timescale $M$. By this way, we obtain
(10) $\begin{array}{c}\stackrel{˙}{x}=f\left(x\right)\frac{1}{M}\end{array}$
It is easy to verify that the system governed by Equation 10 has a periodic solution $xst/M$ with period $MT$. The linearization of this system gives
where $wt=xt- xst/M$ . According to the definition of $Φt,s$ , $Φt,s$ satisfies
$\frac{\mathrm{d}\mathrm{\Phi }\left(t,s\right)}{\mathrm{d}t}=A\left(t\right)\mathrm{\Phi }\left(t,s\right),\mathrm{}\mathrm{}\mathrm{\Phi }\left(s,s\right)=I.$
So $ΦtM,sM$ satisfies
$\frac{\mathrm{d}\mathrm{\Phi }\left(\frac{t}{M},\frac{s}{M}\right)}{\mathrm{d}t}=\frac{1}{M}A\left(\frac{t}{M}\right)\mathrm{\Phi }\left(\frac{t}{M},\frac{s}{M}\right),\mathrm{}\mathrm{}\mathrm{\Phi
Therefore, $ΦtM,sM$ is the state transition matrix for $w˙=1MAtMwt$ . Since $Φt,s=UtexpDt-sVs$ , we can obtain
$\mathrm{\Phi }\left(\frac{\mathrm{t}}{\mathrm{M}},\frac{\mathrm{s}}{\mathrm{M}}\right)=\frac{1}{\mathrm{M}}\mathrm{U}\left(\frac{\mathrm{t}}{\mathrm{M}}\right)\mathrm{exp}\left(\frac{\mathrm{D}\left
where the first term $1M$ in the right-hand term is to ensure the first column of $1MUtM$ is the time derivative of $xstM$ , that is, $1Mx˙stM$ . Thus, the first row of $MVtM$ is $Mv1TtM$ , which can
be used to calculate the variance of phase noise.
Oscillation accuracy against extrinsic noise when tuning the time scale
For the system governed by Equation 2, we add $1M$ to the right-hand side and perturb the $vij, δi,ri$ to introduce the extrinsic noise, thus leading to the following equations:
$\left\{\begin{array}{l}\frac{dA}{dt}=\left({k}_{basal}+\frac{\sum _{i}{v}_{iA}{\left(\frac{{x}_{i}}{{K}_{iA}}\right)}^{3}+{\delta }_{A}}{1+\sum _{i}{\left(\frac{{x}_{i}}{{K}_{iA}}\right)}^{3}+\sum _
{i}{\left(\frac{{y}_{i}}{{K}_{iA}}\right)}^{3}}-{r}_{A}A\right)\frac{1}{M}+\epsilon \frac{1}{M}\left(\frac{\sum _{i}{v}_{iA}{\eta }_{iA}{\left(\frac{{x}_{i}}{{K}_{iA}}\right)}^{3}+{\delta }_{A}{\xi }
_{A}}{1+\sum _{i}{\left(\frac{{x}_{i}}{{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\frac{{y}_{i}}{{K}_{iA}}\right)}^{3}}-{r}_{A}A{\zeta }_{A}\right)\\ \frac{dB}{dt}=\left({k}_{basal}+\frac{\sum _{i}{v}_
{iB}{\left(\frac{{x}_{i}}{{K}_{iB}}\right)}^{3}+{\delta }_{B}}{1+\sum _{i}{\left(\frac{{x}_{i}}{{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{y}_{i}}{{K}_{iB}}\right)}^{3}}-{r}_{B}B\right)\frac{1}{M}+
\epsilon \frac{1}{M}\left(\frac{\sum _{i}{v}_{iB}{\eta }_{iB}{\left(\frac{{x}_{i}}{{K}_{iB}}\right)}^{3}+{\delta }_{B}{\xi }_{B}}{1+\sum _{i}{\left(\frac{{x}_{i}}{{K}_{iB}}\right)}^{3}+\sum _{i}{\
left(\frac{{y}_{i}}{{K}_{iB}}\right)}^{3}}-{r}_{B}B{\zeta }_{B}\right)\\ \frac{dC}{dt}=\left({k}_{basal}+\frac{\sum _{i}{v}_{iC}{\left(\frac{{x}_{i}}{{K}_{iC}}\right)}^{3}+{\delta }_{C}}{1+\sum _{i}
{\left(\frac{{x}_{i}}{{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{y}_{i}}{{K}_{iC}}\right)}^{3}}-{r}_{C}C\right)\frac{1}{M}+\epsilon \frac{1}{M}\left(\frac{\sum _{i}{v}_{iC}{\eta }_{iC}{\left(\frac
{{x}_{i}}{{K}_{iC}}\right)}^{3}+{\delta }_{C}{\xi }_{C}}{1+\sum _{i}{\left(\frac{{x}_{i}}{{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{y}_{i}}{{K}_{iC}}\right)}^{3}}-{r}_{C}C{\zeta }_{C}\right)\end
For simplicity, we still use $x$ and $fx$ to denote $A,B,C$ and the terms in the first brackets in the right-hand terms, respectively. Thus, the above equations can be rewritten as
$\stackrel{˙}{x}=f\left(x\right)\frac{1}{M}+\frac{1}{M}{B}_{ex}\left(x\right)\Pi \left(t\right)$
where $Π(t)$=$(ηAA,ηBA,ηCA, ξA,ζA,ηAB,ηBB,ηCB, ξB,ζB,ηAC,ηBC,ηCC, ξC,ζC)$, and $Bex(x)$ is the matrix whose elements are coefficients of the random noise when $M=1$. Recall that the ‘ $v1t$ ’ in
Equation 8 is $Mv1tM$ for the system $x˙=fx1M$ , so the slope of variance of phase noise over period for the system with timescale $M$ is
$\frac{1}{MT}\frac{1}{MT}{\int }_{0}^{MT}M{v}_{1}^{T}\left(\frac{{t}^{`}}{\mathrm{M}}\right)\frac{1}{M}{B}_{ex}\left({x}_{s}\left({t}^{`}/M\right)\right)\frac{1}{M}{B}_{ex}^{T}\left({x}_{s}\left({t}^
By replacing $t`$ with $t``M$ and using $t`$ to denote $t``$ , we obtain
$\frac{1}{M{T}^{2}}{\int }_{0}^{T}{v}_{1}^{T}\left({t}^{`}\right)B\left({x}_{s}\left({t}^{`}\right)\right){B}^{T}\left({x}_{s}\left({t}^{`}\right)\right){v}_{1}\left({t}^{`}\right)d{t}^{`}$
So it can be concluded that large $M$ causes small normalized phase noise in the presence of extrinsic noise. As large $M$ also leads to long period but has no effect on the amplitude, the long
period might be the reason for high oscillation accuracy in the presence of extrinsic noise.
Oscillation accuracy against intrinsic noise when tuning the timescale
Similarly, for the system governed by Equation 2, we add $1M$ to the right-hand side and introduce the cell volume $V$ to incorporate the intrinsic noise, thus leading to the following chemical
Langevin equations:
$\left\{\begin{array}{l}d{X}_{A}=\left(V{k}_{basal}+V\frac{\sum _{i}{v}_{iA}{\left(\frac{{X}_{i}}{{K}_{iA}}\right)}^{3}+{\delta }_{A}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{i}}{{K}_{iA}}\right)}^
{3}+\sum _{i}{\left(\frac{{Y}_{i}}{{K}_{iA}}\right)}^{3}}-{r}_{A}{X}_{A}\right)\frac{1}{M}dt+\sqrt{\frac{1}{M}}\sqrt{V{k}_{basal}+V\frac{\sum _{i}{v}_{iA}{\left(\frac{{X}_{i}}{{K}_{iA}}\right)}^{3}+
{\delta }_{A}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{i}}{{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{i}}{{K}_{iA}}\right)}^{3}}+{r}_{A}{X}_{A}}d{W}_{t}^{A}\\ d{X}_{B}=\left(V{k}_{basal}+V\
frac{\sum _{i}{v}_{iB}{\left(\frac{{X}_{i}}{{K}_{iB}}\right)}^{3}+{\delta }_{B}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{i}}{{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{i}}{{K}_{iB}}\right)}^
{3}}-{r}_{B}{X}_{B}\right)\frac{1}{M}dt+\sqrt{\frac{1}{M}}\sqrt{V{k}_{basal}+V\frac{\sum _{i}{v}_{iB}{\left(\frac{{X}_{i}}{{K}_{iB}}\right)}^{3}+{\delta }_{B}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac
{{X}_{i}}{{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{i}}{{K}_{iB}}\right)}^{3}}+{r}_{B}{X}_{B}}d{W}_{t}^{B}\\ d{X}_{C}=\left(V{k}_{basal}+V\frac{\sum _{i}{v}_{iC}{\left(\frac{{X}_{i}}{{K}_{iC}}
\right)}^{3}+{\delta }_{C}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{i}}{{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{i}}{{K}_{iC}}\right)}^{3}}-{r}_{C}{X}_{C}\right)\frac{1}{M}dt+\sqrt{\frac{1}
{M}}\sqrt{V{k}_{basal}+V\frac{\sum _{i}{v}_{iC}{\left(\frac{{X}_{i}}{{K}_{iC}}\right)}^{3}+{\delta }_{C}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{i}}{{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}
We use $X$ and $fX$ to denote $XA, XB, XC$ and the terms in the first brackets in the right-hand terms, respectively. Thus, the above equations can be rewritten as
$\stackrel{˙}{X}=f\left(X\right)\frac{1}{M}+\sqrt{\frac{1}{M}}{B}_{in}\left(X\right)\Lambda \left(t\right)$
where $Λ(t)$ is $(dWA/dt,dWB/dt,dWC/dt)$, and $Bin(X)$ is the matrix whose elements are coefficients of the random noise when $M=1$. If we use $v1int$ to represent the ‘ $v1t$ ’ for the system $X˙=
fX$ , the ‘ $v1t$ ’ for the system $X˙=fX1M$ is $Mv1intM$ . So the slope of variance of phase noise over period for the system with cell volume $V$ is
$\frac{1}{MT}\frac{1}{MT}{\int }_{0}^{MT}M{\left({v}_{1}^{in}\left(\frac{{t}^{`}}{\mathrm{M}}\right)\right)}^{T}\sqrt{\frac{1}{M}}{B}_{in}\left({x}_{s}\left({t}^{`}/M\right)\right)\sqrt{\frac{1}{M}}
By replacing $t`$ with $t``M$ and using $t`$ to denote $t``$ , we obtain
$\frac{1}{{T}^{2}}{\int }_{0}^{T}{\left({v}_{1}^{in}\left({t}^{`}\right)\right)}^{T}B\left({x}_{s}\left({t}^{`}\right)\right){B}^{T}\left({x}_{s}\left({t}^{`}\right)\right){v}_{1}^{in}\left({t}^{`}\
It can be seen that $M$ has no effect on the normalized slope of variance of phase noise, so the long period might not influence the noise in proteins when facing intrinsic noise.
Analytical results for the relation between robustness and amplitude when tuning the rescaling parameter $N$
Deterministic model with rescaled variables
To analyze the relation between amplitude and oscillation accuracy against noise, we replace $(A, B,C)$ in Equation 2 with $A~/N, B~/N,C~/N$ , which allows us to tune the amplitude by varying $N$.
After this rescaling, we obtain the following equations for $A~$ , $B~$ and $C~$:
(11) $\left\{\begin{array}{l}\frac{d\stackrel{\sim }{A}}{dt}=N{k}_{basal}+\frac{\sum _{i}N{v}_{iA}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iA}}\right)}^{3}+N{\delta }_{A}}{1+\sum _{i}{\left(\
frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\frac{{\stackrel{\sim }{y}}_{i}}{N{K}_{iA}}\right)}^{3}}-{r}_{A}\stackrel{\sim }{A}\\ \frac{d\stackrel{\sim }{B}}{dt}=N{k}_
{basal}+\frac{\sum _{i}N{v}_{iB}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iB}}\right)}^{3}+N{\delta }_{B}}{1+\sum _{i}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iB}}\right)}^{3}+\sum _{i}{\
left(\frac{{\stackrel{\sim }{y}}_{i}}{N{K}_{iB}}\right)}^{3}}-{r}_{B}\stackrel{\sim }{B}\\ \frac{d\stackrel{\sim }{C}}{dt}=N{k}_{basal}+\frac{\sum _{i}N{v}_{iC}{\left(\frac{{\stackrel{\sim }{x}}_{i}}
{N{K}_{iC}}\right)}^{3}+N{\delta }_{C}}{1+\sum _{i}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{\stackrel{\sim }{y}}_{i}}{N{K}_{iC}}\right)}^{3}}-{r}_{C}\
stackrel{\sim }{C}\end{array}$
where $x~i=xiN, y~i=yiN$ . If $N=1$, this equation is the same as Equation 2, so $A~$ , $B~$, and $C~$ show same amplitudes with $A, B$, and $C$, respectively. Nevertheless, if $N≠1$, the amplitude
of $A~$ , $B~$, or $C~$ is $N$ times as high as that of $A, B$, or $C$. Note that $N$ has no effect on the period.
Oscillation accuracy against extrinsic noise when tuning the rescaling parameter
In the system governed by Equation 11, we assume that $N$ causes binding affinities, $vij$ ’s, and $δi$ ’s to $N$ times their original values, but the $ri$ ’s remain unchanged. Next, we consider the
system described by Equation 11 in the presence of only extrinsic noise. We perturb each kinetic parameter $vij,δi,ri$ by the same method as mentioned in the section ‘Mathematical modeling’ and
obtain a new system described by the following equations:
$\left\{\begin{array}{l}\frac{d\stackrel{\sim }{A}}{dt}=N{k}_{basal}+\frac{\sum _{i}N{v}_{iA}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iA}}\right)}^{3}+N{\delta }_{A}}{1+\sum _{i}{\left(\frac{{\
stackrel{\sim }{x}}_{i}}{N{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\frac{{\stackrel{\sim }{y}}_{i}}{N{K}_{iA}}\right)}^{3}}-{r}_{A}\stackrel{\sim }{A}+\epsilon \left(\frac{\sum _{i}N{v}_{iA}{\eta }_{iA}
{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iA}}\right)}^{3}+N{\delta }_{A}{\xi }_{A}}{1+\sum _{i}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\frac{{\stackrel{\
sim }{y}}_{i}}{N{K}_{iA}}\right)}^{3}}-{r}_{A}\stackrel{\sim }{A}{\zeta }_{A}\right)\\ \frac{d\stackrel{\sim }{B}}{dt}=N{k}_{basal}+\frac{\sum _{i}N{v}_{iB}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N
{K}_{iB}}\right)}^{3}+N{\delta }_{B}}{1+\sum _{i}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{\stackrel{\sim }{y}}_{i}}{N{K}_{iB}}\right)}^{3}}-{r}_{B}\
stackrel{\sim }{B}+\epsilon \left(\frac{\sum _{i}N{v}_{iB}{\eta }_{iA}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iB}}\right)}^{3}+N{\delta }_{B}{\xi }_{A}}{1+\sum _{i}{\left(\frac{{\stackrel{\sim
}{x}}_{i}}{N{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{\stackrel{\sim }{y}}_{i}}{N{K}_{iB}}\right)}^{3}}-{r}_{B}\stackrel{\sim }{B}{\zeta }_{A}\right)\\ \frac{d\stackrel{\sim }{C}}{dt}=N{k}_{basal}
+\frac{\sum _{i}N{v}_{iC}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iC}}\right)}^{3}+N{\delta }_{C}}{1+\sum _{i}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac
{{\stackrel{\sim }{y}}_{i}}{N{K}_{iC}}\right)}^{3}}-{r}_{C}\stackrel{\sim }{C}+\epsilon \left(\frac{\sum _{i}N{v}_{iC}{\eta }_{iA}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iC}}\right)}^{3}+N{\
delta }_{C}{\xi }_{A}}{1+\sum _{i}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{\stackrel{\sim }{y}}_{i}}{N{K}_{iC}}\right)}^{3}}-{r}_{C}\stackrel{\sim }{C}{\
zeta }_{A}\right)\end{array}$
where $ηij$ , $ξi$, $ζi$ are independent noise terms and all modeled by Equation 4 By multiplying $1/N$ to both sides of above equations, we get
$\left\{\begin{array}{l}\frac{1}{N}\frac{d\stackrel{\sim }{A}}{dt}={k}_{basal}+\frac{\sum _{i}{v}_{iA}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iA}}\right)}^{3}+{\delta }_{A}}{1+\sum _{i}{\left(\
frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\frac{{\stackrel{\sim }{y}}_{i}}{N{K}_{iA}}\right)}^{3}}-{r}_{A}\frac{1}{N}\stackrel{\sim }{A}+\epsilon \left(\frac{\sum _{i}{v}
_{iA}{\eta }_{iA}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iA}}\right)}^{3}+{\delta }_{A}{\xi }_{A}}{1+\sum _{i}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\
frac{{\stackrel{\sim }{y}}_{i}}{N{K}_{iA}}\right)}^{3}}-{r}_{A}\frac{1}{N}\stackrel{\sim }{A}{\zeta }_{A}\right)\\ \frac{1}{N}\frac{d\stackrel{\sim }{B}}{dt}={k}_{basal}+\frac{\sum _{i}{v}_{iB}{\left
(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iB}}\right)}^{3}+{\delta }_{B}}{1+\sum _{i}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{\stackrel{\sim }{y}}_{i}}{N{K}
_{iB}}\right)}^{3}}-{r}_{B}\frac{1}{N}\stackrel{\sim }{B}+\epsilon \left(\frac{\sum _{i}{v}_{iB}{\eta }_{iA}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iB}}\right)}^{3}+{\delta }_{B}{\xi }_{A}}{1+\
sum _{i}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{\stackrel{\sim }{y}}_{i}}{N{K}_{iB}}\right)}^{3}}-{r}_{B}\frac{1}{N}\stackrel{\sim }{B}{\zeta }_{A}\
right)\\ \frac{1}{N}\frac{d\stackrel{\sim }{C}}{dt}={k}_{basal}+\frac{\sum _{i}{v}_{iC}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iC}}\right)}^{3}+{\delta }_{C}}{1+\sum _{i}{\left(\frac{{\stackrel
{\sim }{x}}_{i}}{N{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{\stackrel{\sim }{y}}_{i}}{N{K}_{iC}}\right)}^{3}}-{r}_{C}\frac{1}{N}\stackrel{\sim }{C}+\epsilon \left(\frac{\sum _{i}{v}_{iC}{\eta }_
{iA}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iC}}\right)}^{3}+{\delta }_{C}{\xi }_{A}}{1+\sum _{i}{\left(\frac{{\stackrel{\sim }{x}}_{i}}{N{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{\stackrel
{\sim }{y}}_{i}}{N{K}_{iC}}\right)}^{3}}-{r}_{C}\frac{1}{N}\stackrel{\sim }{C}{\zeta }_{A}\right)\end{array}$
Let $A~~=A~N,$$B~~=B~N$ , and $B~~=B~N$ , the set of equations for $A~~$ , $B~~$ , and $C~~$ is the same as that in Equation 3, in which $N$ do not appear. So the dynamics of $A~~$ , $B~~$ , or $C~~$
will not change with varied $N$, thus leading to the same accuracy of oscillation when varying $N$. Based on $A~~N=A~,$$B~~N=B~$, and $C~~N=C~$ and the fact that the rescaling has no effect on the
correlation function, $A~$ shows the same accuracy of oscillation as $A~~$ , and so does $B~$ or $C~$ . Therefore, in the system for $A~$ , $B~$, and $C~$ , its oscillation accuracy remains the same
with varied $N$. Since $N$ influences the amplitude while maintaining the period, the change in the amplitude will not affect the oscillation noise against extrinsic noise.
Oscillation accuracy against intrinsic noise when tuning the rescaling parameter
The dynamics of the system described by Equation 11. in the presence of only intrinsic noise is governed by
$\left\{\begin{array}{l}d{X}_{\stackrel{\sim }{A}}=\left(VN{k}_{basal}+V\frac{\sum _{i}N{v}_{iA}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iA}}\right)}^{3}+N{\delta }_{A}{V}^{3}}{{V}^{3}+\sum _{i}
{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{i}}}{N{K}_{iA}}\right)}^{3}}-{r}_{A}{X}_{\stackrel{\sim }{A}}\right)dt\text{ }\text{ }\text{
}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{
}\\ +\sqrt{VN{k}_{basal}+V\frac{\sum _{i}N{v}_{iA}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iA}}\right)}^{3}+N{\delta }_{A}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_
{iA}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{i}}}{N{K}_{iA}}\right)}^{3}}+{r}_{A}{X}_{\stackrel{\sim }{A}}}d{W}_{t}^{A}\\ d{X}_{\stackrel{\sim }{B}}=\left(VN{k}_{basal}+V\frac{\sum _
{i}N{v}_{iB}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iB}}\right)}^{3}+N{\delta }_{B}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac
{{Y}_{\stackrel{\sim }{i}}}{N{K}_{iB}}\right)}^{3}}-{r}_{B}{X}_{\stackrel{\sim }{B}}\right)dt\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\
text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\\ +\sqrt{VN{k}_{basal}+V\frac{\sum _{i}N{v}_{iB}{\left(\frac{{X}_{\stackrel{\
sim }{i}}}{N{K}_{iB}}\right)}^{3}+N{\delta }_{B}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{i}}}{N{K}_{iB}}\
right)}^{3}}+{r}_{B}{X}_{\stackrel{\sim }{B}}}d{W}_{t}^{B}\\ d{X}_{\stackrel{\sim }{C}}=\left(VN{k}_{basal}+V\frac{\sum _{i}N{v}_{iC}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iC}}\right)}^{3}+N{\
delta }_{C}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{i}}}{N{K}_{iC}}\right)}^{3}}-{r}_{C}{X}_{\stackrel{\sim
}{C}}\right)dt\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\
text{ }\text{ }\text{ }\text{ }\text{ }\\ +\sqrt{VN{k}_{basal}+V\frac{\sum _{i}N{v}_{iC}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iC}}\right)}^{3}+N{\delta }_{C}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\
frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{i}}}{N{K}_{iC}}\right)}^{3}}+{r}_{C}{X}_{\stackrel{\sim }{C}}}d{W}_{t}^{C}\end{array}$
where $XA~=VA~$ , $XB~=VB~$, and $XC~=VC~$ , and $V$ is the cell volume. By multiplying $1/N$ to both sides of above equations, we get
$\left\{\begin{array}{l}\frac{1}{N}d{X}_{\stackrel{\sim }{A}}=\left(V{k}_{basal}+V\frac{\sum _{i}{v}_{iA}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iA}}\right)}^{3}+{\delta }_{A}{V}^{3}}{{V}^{3}+\
sum _{i}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{i}}}{N{K}_{iA}}\right)}^{3}}-{r}_{A}\frac{1}{N}{X}_{\stackrel{\sim }{A}}\right)dt\
text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }
\text{ }\text{ }\text{ }\text{ }\text{ }\\ \text{ }+\sqrt{\frac{1}{N}}\sqrt{V{k}_{basal}+V\frac{\sum _{i}{v}_{iA}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iA}}\right)}^{3}+{\delta }_{A}{V}^{3}}
{{V}^{3}+\sum _{i}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{i}}}{N{K}_{iA}}\right)}^{3}}+{r}_{A}\frac{1}{N}{X}_{\stackrel{\sim }{A}}}d
{W}_{t}^{A}\\ \frac{1}{N}d{X}_{\stackrel{\sim }{B}}=\left(V{k}_{basal}+V\frac{\sum _{i}{v}_{iB}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iB}}\right)}^{3}+{\delta }_{B}{V}^{3}}{{V}^{3}+\sum _{i}{\
left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{i}}}{N{K}_{iB}}\right)}^{3}}-{r}_{B}\frac{1}{N}{X}_{\stackrel{\sim }{B}}\right)dt\text{ }\text
{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\
text{ }\text{ }\text{ }\text{ }\\ +\sqrt{\frac{1}{N}}\sqrt{V{k}_{basal}+V\frac{\sum _{i}{v}_{iB}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iB}}\right)}^{3}+{\delta }_{B}{V}^{3}}{{V}^{3}+\sum _{i}
{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{i}}}{N{K}_{iB}}\right)}^{3}}+{r}_{B}\frac{1}{N}{X}_{\stackrel{\sim }{B}}}d{W}_{t}^{B}\\ \
frac{1}{N}d{X}_{\stackrel{\sim }{C}}=\left(V{k}_{basal}+V\frac{\sum _{i}{v}_{iC}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iC}}\right)}^{3}+{\delta }_{C}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_
{\stackrel{\sim }{i}}}{N{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{i}}}{N{K}_{iC}}\right)}^{3}}-{r}_{C}\frac{1}{N}{X}_{\stackrel{\sim }{C}}\right)dt\text{ }\text{ }\text{ }\
text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }
\text{ }\text{ }\\ +\sqrt{\frac{1}{N}}\sqrt{V{k}_{basal}+V\frac{\sum _{i}{v}_{iC}{\left(\frac{{X}_{\stackrel{\sim }{i}}}{N{K}_{iC}}\right)}^{3}+{\delta }_{C}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}
_{\stackrel{\sim }{i}}}{N{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{i}}}{N{K}_{iC}}\right)}^{3}}+{r}_{C}\frac{1}{N}{X}_{\stackrel{\sim }{C}}}d{W}_{t}^{C}\end{array}$
Let $XA~~=XA~N,$$XB~~=XB~N,$ and $XC~~=XC~N$ , equations for $XA~~$ , $XB~~$ , and $XC~~$ are
$\left\{\begin{array}{l}d{X}_{\stackrel{\sim }{\stackrel{\sim }{A}}}=\left(V{k}_{basal}+V\frac{\sum _{i}{v}_{iA}{\left(\frac{{X}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iA}}\right)}^{3}+{\delta
}_{A}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iA}}\right)}^
{3}}-{r}_{A}{X}_{\stackrel{\sim }{\stackrel{\sim }{A}}}\right)dt\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{
}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\\ \text{ }+\sqrt{\frac{1}{N}}\sqrt{V{k}_{basal}+V\frac{\sum _{i}{v}_{iA}{\left(\frac{{X}_{\
stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iA}}\right)}^{3}+{\delta }_{A}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iA}}\right)}^{3}+\sum _{i}{\left(\
frac{{Y}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iA}}\right)}^{3}}+{r}_{A}{X}_{\stackrel{\sim }{\stackrel{\sim }{A}}}}d{W}_{t}^{A}\\ d{X}_{\stackrel{\sim }{\stackrel{\sim }{B}}}=\left(V{k}_
{basal}+V\frac{\sum _{i}{v}_{iB}{\left(\frac{{X}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iB}}\right)}^{3}+{\delta }_{B}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{\stackrel{\sim }{\stackrel{\
sim }{i}}}}{{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iB}}\right)}^{3}}-{r}_{B}{X}_{\stackrel{\sim }{\stackrel{\sim }{B}}}\right)dt\text{ }\text{ }
\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{
}\text{ }\text{ }\text{ }\\ +\sqrt{\frac{1}{N}}\sqrt{V{k}_{basal}+V\frac{\sum _{i}{v}_{iB}{\left(\frac{{X}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iB}}\right)}^{3}+{\delta }_{B}{V}^{3}}{{V}^{3}
+\sum _{i}{\left(\frac{{X}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iB}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iB}}\right)}^{3}}+{r}_{B}{X}_{\
stackrel{\sim }{\stackrel{\sim }{B}}}}d{W}_{t}^{B}\\ d{X}_{\stackrel{\sim }{\stackrel{\sim }{C}}}=\left(V{k}_{basal}+V\frac{\sum _{i}{v}_{iC}{\left(\frac{{X}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}
{{K}_{iC}}\right)}^{3}+{\delta }_{C}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iC}}\right)}^{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{\stackrel{\
sim }{i}}}}{{K}_{iC}}\right)}^{3}}-{r}_{C}{X}_{\stackrel{\sim }{\stackrel{\sim }{C}}}\right)dt\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\
text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\\ +\sqrt{\frac{1}{N}}\sqrt{V{k}_{basal}+V\frac{\sum _{i}{v}_
{iC}{\left(\frac{{X}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iC}}\right)}^{3}+{\delta }_{C}{V}^{3}}{{V}^{3}+\sum _{i}{\left(\frac{{X}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iC}}\right)}^
{3}+\sum _{i}{\left(\frac{{Y}_{\stackrel{\sim }{\stackrel{\sim }{i}}}}{{K}_{iC}}\right)}^{3}}+{r}_{C}{X}_{\stackrel{\sim }{\stackrel{\sim }{C}}}}d{W}_{t}^{C}\end{array}$
In above equations, $N$ only negatively affects the magnitude of noise term, so the oscillation accuracies of $XA~~$ , $XB~~$ , and $XC~~$ increased with increased $N$. Thus, the oscillation
accuracies of $XA~$ , $XB~$ , and $XC~$ also increased with increased $N$ because the correlation function is not affected by the rescaling operation. Besides, large $N$ increase the amplitude while
maintaining the period. Taken together, the high amplitude may enhance the oscillation noise against intrinsic noise.
The current manuscript is a computational study. Modelling code and NFKB data for plotting are uploaded to GitHub at https://github.com/LingxiaQiao/oscillation, (copy archived at
27. Book
Numerical Solution of Stochastic Differential Equations
Berlin: Springer-Verlag.
Article and author information
Author details
National Key Research and Development Program of China (2021YFF1200500)
National Natural Science Foundation of China (12050002)
National Key Basic Research Program of China (2018YFA0902800)
National Natural Science Foundation of China (31622022)
The funders had no role in study design, data collection and interpretation, or the decision to submit the work for publication.
LZ was partly supported by the National Key Research and Development Program of China 2021YFF1200500 and National Natural Science Foundation of China No. 12050002. PW was partly supported by the
National Key Basic Research Program of China 2018YFA0902800 and the National Natural Science Foundation of China 31622022.
© 2022, Qiao, Zhang, Zhao et al.
This article is distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use and redistribution provided that the original author and source are credited.
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1. Lingxia Qiao
2. Zhi-Bo Zhang
3. Wei Zhao
4. Ping Wei
5. Lei Zhang
Network design principle for robust oscillatory behaviors with respect to biological noise
eLife 11:e76188.
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polarized system. | {"url":"https://elifesciences.org/articles/76188","timestamp":"2024-11-04T12:36:50Z","content_type":"text/html","content_length":"612591","record_id":"<urn:uuid:d4917465-cb6b-4f98-8986-0200d70f754f>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00669.warc.gz"} |
Enzyme with Const() on a vector throws an error
Hello all,
I would like to autodiff a function where some vector-valued arguments are kept constant. I thought to use Enzyme with the Const() functionality but run into issues when applying Const() to vectors.
The following is a minimal working example that reproduces the problem. I use
Enzyme v0.13.12 with julia 1.11
using Enzyme
# this will work
function f(x::Array{Float64}, c::Vector{Float64})
y = (x[1]-c[1]) * (x[1]-c[1]) + (x[2]-c[2]) * (x[2]-c[2])
return y
# this won't work
function h(x::Array{Float64}, c::Vector{Float64})
y = sum( (x-c).^2 )
return y
# this will work
function h2(x::Array{Float64}, c1::Float64, c2::Float64)
c = [c1, c2]
y = sum( (x-c).^2 )
return y
The three functions compute the squared norm between the two vectors x and c. For example
x = [4.0, 3.0];
c = [2.0, 1.0];
f(x, c)
f(x, c) == h(x, c) == h2(x, c[1], c[2]) # returns true
Now, autodiff on f and h2 works
dx = [0.0, 0.0]
autodiff(Reverse, f, Active, Duplicated(x, dx), Const(c));
2*(x-c) == dx # true
dx = [0.0, 0.0]
autodiff(Reverse, h2, Active, Duplicated(x, dx), Const(c[1]), Const(c[2]));
2*(x-c) == dx # true
However, for h I get a Constant memory is stored (or returned) to a differentiable variable. error
dx = [0.0, 0.0]
autodiff(Reverse, h, Active, Duplicated(x, dx), Const(c));
I am new to Julia and its AD system and puzzled by this error. It seems to me that Const(c) did not work when c is a vector? What would I need to change to make it work? Manually expanding the vector
c into scalars won’t be an option for me.
Many thanks for your help.
I can’t reproduce this on Julia 1.10.6 with Enzyme v0.13.12. Maybe you’re using an older version of Enzyme?
Enzyme on julia 1.11 has some issues still use it on Julia 1.10
1 Like
This seems to be indeed a Julia version issue. The posted code works with Julia 1.10.6, Enzyme v0.13.12 but not with Julia 1.11 (again with Enzyme v0.13.12).
The issue seems to be with the julia version (v 1.11.1) and not the Enzyme version; I was using the same as you. Things indeed work with Julia v 1.10.6. Many thanks for the response.
Yeah Enzyme still has a few things to work out for v1.11 | {"url":"https://discourse.julialang.org/t/enzyme-with-const-on-a-vector-throws-an-error/121927","timestamp":"2024-11-05T04:38:26Z","content_type":"text/html","content_length":"29685","record_id":"<urn:uuid:2184a400-ce6e-4353-8305-bcd55da18625>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00759.warc.gz"} |
How do you simplify (2yz)(-3y^3)?
| HIX Tutor
How do you simplify #(2yz)(-3y^3)#?
Answer 1
#(2yz)(-3y^3)# means #(2)xxyxxz xx (-3)xxy^3#
Regrouping: #(2)xx(-3) xx y xx y^3 xx z#
#= (-6)xxy^4xxz# or #-6y^4z#
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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• Readily available 24/7 | {"url":"https://tutor.hix.ai/question/how-do-you-simplify-2yz-3y-3-8f9af9538e","timestamp":"2024-11-05T22:58:22Z","content_type":"text/html","content_length":"565902","record_id":"<urn:uuid:762e4cf6-ab70-470e-9b59-fdc2caa5388b>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00089.warc.gz"} |
Writing a tiny tRPC client | tRPC
Writing a tiny tRPC client
ยท 12 min read
Ever wondered how tRPC works? Maybe you want to start contributing to the project but you're frightened by the internals? The aim of this post is to familiarize you with the internals of tRPC by
writing a minimal client that covers the big parts of how tRPC works.
It's recommended that you understand some of the core concepts in TypeScript such as generics, conditional types, the extends keyword and recursion. If you're not familiar with these, I recommend
going through Matt Pocock's Beginner TypeScript tutorial to get familiar with these concepts before reading on.
Let's assume we have a simple tRPC router with three procedures that looks like this:
type Post = { id: string; title: string };
const posts: Post[] = [];
const appRouter = router({
post: router({
byId: publicProcedure
.input(z.object({ id: z.string() }))
.query(({ input }) => {
const post = posts.find((p) => p.id === input.id);
if (!post) throw new TRPCError({ code: "NOT_FOUND" });
return post;
byTitle: publicProcedure
.input(z.object({ title: z.string() }))
.query(({ input }) => {
const post = posts.find((p) => p.title === input.title);
if (!post) throw new TRPCError({ code: "NOT_FOUND" });
return post;
create: publicProcedure
.input(z.object({ title: z.string() }))
.mutation(({ input }) => {
const post = { id: uuid(), ...input };
return post;
type Post = { id: string; title: string };
const posts: Post[] = [];
const appRouter = router({
post: router({
byId: publicProcedure
.input(z.object({ id: z.string() }))
.query(({ input }) => {
const post = posts.find((p) => p.id === input.id);
if (!post) throw new TRPCError({ code: "NOT_FOUND" });
return post;
byTitle: publicProcedure
.input(z.object({ title: z.string() }))
.query(({ input }) => {
const post = posts.find((p) => p.title === input.title);
if (!post) throw new TRPCError({ code: "NOT_FOUND" });
return post;
create: publicProcedure
.input(z.object({ title: z.string() }))
.mutation(({ input }) => {
const post = { id: uuid(), ...input };
return post;
The goal of our client is to mimic this object structure on our client so that we can call procedures like:
const post1 = await client.post.byId.query({ id: '123' });
const post2 = await client.post.byTitle.query({ title: 'Hello world' });
const newPost = await client.post.create.mutate({ title: 'Foo' });
const post1 = await client.post.byId.query({ id: '123' });
const post2 = await client.post.byTitle.query({ title: 'Hello world' });
const newPost = await client.post.create.mutate({ title: 'Foo' });
To do this, tRPC uses a combination of Proxy-objects and some TypeScript magic to augment the object structure with the .query and .mutate methods on them - meaning we actually LIE to you about what
you're doing (more on that later) in order to provide an excellent developer experience!
On a high level, what we want to do is to map post.byId.query() to a GET request to our server, and post.create.mutate() to a POST request, and the types should all be propagated from back to front.
So, how do we do this?
Implementing a tiny tRPC clientโ
๐ ง โ โ ๏ธ The TypeScript magicโ
Let's start with the fun TypeScript magic to unlock the awesome autocompletion and typesafety we all know and love from using tRPC.
We'll need to use recursive types so that we can infer arbitrary deep router structures. Also, we know that we want our procedures post.byId and post.create to have the .query and .mutate methods on
them respectively - in tRPC, we call this decorating the procedures. In @trpc/server, we have some inference helpers that will infer the input and output types of our procedures with these resolved
methods, which we'll use to infer the types for these functions, so let's write some code!
Let's consider what we want to achieve to provide autocompletion on paths as well as inference of the procedures input and output types:
• If we're on a router, we want to be able to access it's sub-routers and procedures. (we'll get to this in a little bit)
• If we're on a query procedure, we want to be able to call .query on it.
• If we're on a mutation procedure, we want to be able to call .mutate on it.
• If we're trying to access anything else, we want to get a type error indicating that procedure doesn't exist on the backend.
So let's create a type that will do this for us:
type DecorateProcedure<TProcedure> = TProcedure extends AnyTRPCQueryProcedure
? {
query: Resolver<TProcedure>;
: TProcedure extends AnyTRPCMutationProcedure
? {
mutate: Resolver<TProcedure>;
: never;
type DecorateProcedure<TProcedure> = TProcedure extends AnyTRPCQueryProcedure
? {
query: Resolver<TProcedure>;
: TProcedure extends AnyTRPCMutationProcedure
? {
mutate: Resolver<TProcedure>;
: never;
We'll use some of tRPC's built-in inference helpers to infer the input and output types of our procedures to define the Resolver type.
import type {
} from '@trpc/server';
type Resolver<TProcedure extends AnyTRPCProcedure> = (
input: inferProcedureInput<TProcedure>,
) => Promise<inferProcedureOutput<TProcedure>>;
import type {
} from '@trpc/server';
type Resolver<TProcedure extends AnyTRPCProcedure> = (
input: inferProcedureInput<TProcedure>,
) => Promise<inferProcedureOutput<TProcedure>>;
Let's try this out on our post.byId procedure:
type PostById = Resolver<AppRouter['post']['byId']>;
type PostById = (input: {
id: string;
}) => Promise<Post>
type PostById = Resolver<AppRouter['post']['byId']>;
type PostById = (input: {
id: string;
}) => Promise<Post>
Nice, that's what we expected - we can now call .query on our procedure and get the correct input and output types inferred!
Finally, we'll create a type that will recursively traverse the router and decorate all procedures along the way:
import type { TRPCRouterRecord } from "@trpc/server";
import type { AnyTRPCRouter } from "@trpc/server";
type DecorateRouterRecord<TRecord extends TRPCRouterRecord> = {
[TKey in keyof TRecord]: TRecord[TKey] extends infer $Value
? $Value extends TRPCRouterRecord
? DecorateRouterRecord<$Value>
: $Value extends AnyTRPCProcedure
? DecorateProcedure<$Value>
: never
: never;
import type { TRPCRouterRecord } from "@trpc/server";
import type { AnyTRPCRouter } from "@trpc/server";
type DecorateRouterRecord<TRecord extends TRPCRouterRecord> = {
[TKey in keyof TRecord]: TRecord[TKey] extends infer $Value
? $Value extends TRPCRouterRecord
? DecorateRouterRecord<$Value>
: $Value extends AnyTRPCProcedure
? DecorateProcedure<$Value>
: never
: never;
Let's digest this type a bit:
1. We pass a TRPCRouterRecord to the type as a generic, which is a type containing all the procedures and sub-routers that exists on a tRPC router.
2. We iterate over the keys of the record, which are the procedure or router names, and do the following:
□ If the key maps to a router, we recursively call the type on that router's procedure record, which will decorate all the procedures in that router. This will provide autocompletion as we
traverse the path.
□ If the key maps to a procedure, we decorate the procedure using the DecorateProcedure type we created earlier.
□ If the key doesn't map to a procedure or router, we assign the never type which is like saying "this key doesn't exist" which will cause a type error if we try to access it.
๐ คฏ The Proxy remappingโ
Now that we got all the types setup, we need to actually implement the functionality which will augment the server's router definition on the client so we can invoke the procedures like normal
We'll first create a helper function for creating recursive proxies - createRecursiveProxy:
This is almost the exact implementation used in production, with the exception that we aren't handling some edge cases. See for yourself!
interface ProxyCallbackOptions {
path: readonly string[];
args: readonly unknown[];
type ProxyCallback = (opts: ProxyCallbackOptions) => unknown;
function createRecursiveProxy(callback: ProxyCallback, path: readonly string[]) {
const proxy: unknown = new Proxy(
() => {
// dummy no-op function since we don't have any
// client-side target we want to remap to
get(_obj, key) {
if (typeof key !== 'string') return undefined;
// Recursively compose the full path until a function is invoked
return createRecursiveProxy(callback, [...path, key]);
apply(_1, _2, args) {
// Call the callback function with the entire path we
// recursively created and forward the arguments
return callback({
return proxy;
interface ProxyCallbackOptions {
path: readonly string[];
args: readonly unknown[];
type ProxyCallback = (opts: ProxyCallbackOptions) => unknown;
function createRecursiveProxy(callback: ProxyCallback, path: readonly string[]) {
const proxy: unknown = new Proxy(
() => {
// dummy no-op function since we don't have any
// client-side target we want to remap to
get(_obj, key) {
if (typeof key !== 'string') return undefined;
// Recursively compose the full path until a function is invoked
return createRecursiveProxy(callback, [...path, key]);
apply(_1, _2, args) {
// Call the callback function with the entire path we
// recursively created and forward the arguments
return callback({
return proxy;
This looks a bit magical, what does this do?
• The get method handles property accesses such as post.byId. The key is the property name we're accessing, so when we type post our key will be post, and when we type post.byId our key will be
byId. The recursive proxy combines all of these keys into a final path, e.g. ["post", "byId", "query"], that we can use to determine the URL we want to send a request to.
• The apply method is called when we invoke a function on the proxy, such as .query(args). The args is the arguments we pass to the function, so when we call post.byId.query(args) our args will be
our input, which we'll provide as query parameters or request body depending on the type of procedure. The createRecursiveProxy takes in a callback function which we'll map the apply to with the
path and args.
Below is a visual representation of how the proxy works on the call trpc.post.byId.query({ id: 1 }):
๐ งฉ Putting it all togetherโ
Now that we have this helper and know what it does, let's use it to create our client. We'll provide the createRecursiveProxy a callback that will take the path and args and request the server using
fetch. We'll need to add a generic to the function that will accept any tRPC router type (AnyTRPCRouter), and then we'll cast the return type to the DecorateRouterRecord type we created earlier:
import { TRPCResponse } from '@trpc/server/rpc';
export const createTinyRPCClient = <TRouter extends AnyTRPCRouter>(
baseUrl: string,
) =>
createRecursiveProxy(async (opts) => {
const path = [...opts.path]; // e.g. ["post", "byId", "query"]
const method = path.pop()! as 'query' | 'mutate';
const dotPath = path.join('.'); // "post.byId" - this is the path procedures have on the backend
let uri = `${baseUrl}/${dotPath}`;
const [input] = opts.args;
const stringifiedInput = input !== undefined && JSON.stringify(input);
let body: undefined | string = undefined;
if (stringifiedInput !== false) {
if (method === 'query') {
uri += `?input=${encodeURIComponent(stringifiedInput)}`;
} else {
body = stringifiedInput;
const json: TRPCResponse = await fetch(uri, {
method: method === 'query' ? 'GET' : 'POST',
headers: {
'Content-Type': 'application/json',
}).then((res) => res.json());
if ('error' in json) {
throw new Error(`Error: ${json.error.message}`);
// No error - all good. Return the data.
return json.result.data;
}, []) as DecorateRouterRecord<TRouter['_def']['record']>;
// ^? provide empty array as path to begin with
import { TRPCResponse } from '@trpc/server/rpc';
export const createTinyRPCClient = <TRouter extends AnyTRPCRouter>(
baseUrl: string,
) =>
createRecursiveProxy(async (opts) => {
const path = [...opts.path]; // e.g. ["post", "byId", "query"]
const method = path.pop()! as 'query' | 'mutate';
const dotPath = path.join('.'); // "post.byId" - this is the path procedures have on the backend
let uri = `${baseUrl}/${dotPath}`;
const [input] = opts.args;
const stringifiedInput = input !== undefined && JSON.stringify(input);
let body: undefined | string = undefined;
if (stringifiedInput !== false) {
if (method === 'query') {
uri += `?input=${encodeURIComponent(stringifiedInput)}`;
} else {
body = stringifiedInput;
const json: TRPCResponse = await fetch(uri, {
method: method === 'query' ? 'GET' : 'POST',
headers: {
'Content-Type': 'application/json',
}).then((res) => res.json());
if ('error' in json) {
throw new Error(`Error: ${json.error.message}`);
// No error - all good. Return the data.
return json.result.data;
}, []) as DecorateRouterRecord<TRouter['_def']['record']>;
// ^? provide empty array as path to begin with
Most notably here is that our path is .-separated instead of /. This allows us to have a single API handler on the server which will process all requests, and not one for each procedure. If you're
using a framework with filebased routing such as Next.js, you might recognize the catchall /api/trpc/[trpc].ts file which will match all procedure paths.
We also have a TRPCResponse type annotation on the fetch-request. This determines the JSONRPC-compliant response format that the server responds with. You can read more on that here. TL;DR, we get
back either a result or an error object, which we can use to determine if the request was successful or not and do appropriate error handling if something went wrong.
And that's it! This is all the code you'll need to call your tRPC procedures on your client as if they were local functions. On the surface, it looks like we're just calling the publicProcedure.query
/ mutation's resolver function via normal property accesses, but we're actually crossing a network boundary so we can use server-side libraries such as Prisma without leaking database credentials.
Trying it out!โ
Now, create the client and provide it your server's url and you'll get full autocompletion and type safety when you call your procedures!
const url = 'http://localhost:3000/api/trpc';
const client = createTinyRPCClient<AppRouter>(url);
// ๐ ง โ โ ๏ธ magic autocompletion
• byId
• byTitle
// ๐ fully typesafe
const post = await client.post.byId.query({ id: '123' });
const post: {
id: string;
title: string;
const url = 'http://localhost:3000/api/trpc';
const client = createTinyRPCClient<AppRouter>(url);
// ๐ ง โ โ ๏ธ magic autocompletion
• byId
• byTitle
// ๐ fully typesafe
const post = await client.post.byId.query({ id: '123' });
const post: {
id: string;
title: string;
The full code for the client can be found here, and tests showing the usage here.
I hope you enjoyed this article and learned something about how tRPC works. You should probably not use this to in favor of @trpc/client which is only a couple of KBs bigger - it comes with a lot
more flexibility than what we're showcasing here:
• Query options for abort signals, ssr etc...
• Links
• Procedure batching
• WebSockets / subscriptions
• Nice error handling
• Data transformers
• Edge cases handling like when we don't get a tRPC-compliant response
We also didn't cover much of the server-side of things today, maybe we'll cover that in a future article. If you have any questions, feel free to bug me on Twitter. | {"url":"https://trpc.io/blog/tinyrpc-client","timestamp":"2024-11-03T16:11:01Z","content_type":"text/html","content_length":"180552","record_id":"<urn:uuid:11dcfdb0-fd33-4975-a5ae-c18b7650168a>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00692.warc.gz"} |
Should the design matrix for removeBatchEffect include an intercept?
Last seen 8 weeks ago
I am removing a batch effect from my data using limma's removeBatchEffect. The variable of interest should not be removed, hence I include it in the design matrix. Should the design matrix include an
intercept term? Does it matter? | {"url":"https://support.bioconductor.org/p/128434/","timestamp":"2024-11-09T19:37:37Z","content_type":"text/html","content_length":"17705","record_id":"<urn:uuid:61a458ea-2ae6-4589-8674-ab6b145b73d8>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00619.warc.gz"} |
Chinese-American Mathematician Thomas Hou Finds Singularity Invalidating Euler's Equation - PandailyChinese-American Mathematician Thomas Hou Finds Singularity Invalidating Euler’s Equation
Chinese-American Mathematician Thomas Hou Finds Singularity Invalidating Euler’s Equation
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The famous Euler equations describe the flow of an ideal, incompressible fluid: a fluid with no viscosity or internal friction, that cannot be forced into a smaller volume. Almost all nonlinear fluid
equations are derived from these. However, there are still many unknowns about Euler equations. Mathematicians have long suspected the existence of initial conditions that cause the equations to
fail, but they cannot prove this.
But recently, a preprint submitted by Chinese-American mathematician Thomas Yizhao Hou and his former graduate student and Chinese mathematician Jiajie Chen proved that a particular version of the
Euler equations does indeed sometimes fail. This proof marks a major breakthrough – although it does not completely solve the problem of the more general version of the equations, it brings people
hope that a more general breakthrough can ultimately be achieved.
Jiajie Chen, a mathematician at New York University (Source: Jiajie Chen)
As early as 2013, Thomas Hou and Guo Luo, who now works at Hang Seng University of Hong Kong, put forward a hypothesis: Euler equations will lead to a singularity. They developed a computer
simulation of a fluid in a cylinder whose top half swirled clockwise while its bottom half swirled counterclockwise.
(Source: Merrill Sherman/Quanta Magazine)
In the movement of these two opposite currents, other complicated situations occur, such as the flow of water circulating up and down. At the point where they meet, the vorticity of the fluid (a
hydrodynamic concept describing the rotation of the fluid) grows at an extremely fast rate, and it seems that it will “blow up” at any time.
However, their research at that time could only be said to be enlightening for the existence of singularities, and there was no real evidence. This is because it is impossible for a computer to
calculate infinity. It can calculate an approximation that is very close to the singularity, but it is not the exact one.
In fact, when detected by more powerful computational methods, the obvious singularities have disappeared. For this reason, Charlie Fefferman, a mathematician at Princeton University, commented on
past research on this matter, claiming that these problems are so delicate that the road is littered with the wreckage of previous simulations.
SEE ALSO: Mathematician Yitang Zhang Talks about Landau-Siegel Zeros Conjecture and Analytic Number Theory
In 2022, Thomas Hou and his former graduate student Jiajie Chen successfully proved the existence of the nearby singularity. They first carefully analyzed the 2013 study and found that the
approximate solution seemed to have a special structure. As time goes by, the solutions of these equations will show a so-called “self-similar pattern,” and its shape will later look like its
previous one, but it will be rescaled in a specific way.
Therefore, they thought that there was no need to study the singularity itself. On the contrary, they could pay attention to an earlier time point to study it indirectly. By amplifying this part of
the solution at the correct rate (which is determined by the self-similar structure of the solution), they can simulate what will happen later. What they need to do next is to prove that there is an
exact solution near the singularity.
Thomas Hou, Charles Lee Powell Professor of Applied and Computational Mathematics at the California Institute of Technology, specializes in numerical analysis and mathematical analysis. Born in
Guangdong Province, China in 1962, he studied at South China University of Technology as an undergraduate and obtained his bachelor’s degree in 1982. His doctoral career was completed at the
University of California, Los Angeles. From 1989 to 1993, he taught at the Courant Institute of Mathematical Sciences at New York University. He has been teaching at the California Institute of
Technology since 1993. In 2011, he was elected as a Fellow of the American Academy of Art and Sciences.
The other author of the study, Jiajie Chen, graduated from the School of Mathematical Sciences at China’s Peking University and is currently a mathematician at New York University. During his
postgraduate years, he proved that various fluid equations can “blow up.”
Sign up today for 5 free articles monthly! | {"url":"https://wordp-appli-oeiffwjv3h0b-1837223528.ap-south-1.elb.amazonaws.com/chinese-american-mathematician-thomas-hou-finds-singularity-invalidating-eulers-equation/","timestamp":"2024-11-11T11:50:43Z","content_type":"text/html","content_length":"73644","record_id":"<urn:uuid:009a6119-ebc2-4cb3-bde1-f36e67422bb2>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00407.warc.gz"} |
power functions
Power functions in Excel are a type of mathematical formula that allow users to perform calculations involving exponents, or powers. They are commonly used in financial analysis, statistical
analysis, and various other applications where the need arises to calculate values raised to certain powers.
The most commonly used power function in Excel is the “POWER” function, which is used to calculate a base number raised to a certain power. The syntax for the POWER function is as follows:
=POWER(base, exponent)
For example, if you wanted to calculate 2 raised to the power of 3, you would enter the following formula into a cell:
This would return the result 8, as 2 raised to the power of 3 is equal to 8.
Another commonly used power function in Excel is the “SQRT” function, which is used to calculate the square root of a number. The syntax for the SQRT function is as follows:
For example, if you wanted to calculate the square root of 25, you would enter the following formula into a cell:
This would return the result 5, as the square root of 25 is equal to 5.
Power functions can also be used to calculate other types of roots, such as cube roots and fourth roots. To calculate these roots, you can use the “NTHROOT” function, which has the following syntax:
=NTHROOT(number, root)
For example, if you wanted to calculate the fourth root of 256, you would enter the following formula into a cell:
This would return the result 4, as the fourth root of 256 is equal to 4.
Power functions can also be used to calculate exponential growth or decay. To do this, you can use the “EXP” function, which has the following syntax:
This function calculates the value of “e” raised to the power of the number entered. For example, if you wanted to calculate the value of “e” raised to the power of 2, you would enter the following
formula into a cell:
This would return the result 7.38905609893065, as the value of “e” raised to the power of 2 is equal to 7.38905609893065.
In addition to the POWER, SQRT, NTHROOT, and EXP functions, there are several other power functions available in Excel that can be used for various purposes. These include the “POWER” function, which
calculates the result of a number raised to a fractional power, and the “ROOT” function, which calculates the nth root of a number.
Power functions can be used in a wide variety of applications in Excel, and are particularly useful for performing financial analysis. For example, if you are analyzing the performance of a company’s
stock, you can use power functions to calculate the rate of return on the company’s investments. | {"url":"https://excelguru.pk/power-function/","timestamp":"2024-11-06T08:10:18Z","content_type":"text/html","content_length":"64023","record_id":"<urn:uuid:6f439800-ae85-4d98-98bb-d0f467f92b36>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00105.warc.gz"} |
The POISSON.DIST function calculates the Poisson distribution, which is a statistical distribution that shows how many times an event is likely to occur within a specified period of time, given the
average rate of occurrence. This function is commonly used in probability theory to model the number of times an event occurs in a fixed interval of time or space. The resulting value represents the
probability of the event occurring a certain number of times. If the cumulative argument is set to TRUE (or omitted), the function returns the cumulative distribution function; if set to FALSE, it
returns the probability density function.
Use the POISSON.DIST formula with the syntax shown below, it has 2 required parameters and 1 optional parameter:
=POISSON.DIST(x, mean, [cumulative])
1. x (required):
The number of events that you want to find the probability for.
2. mean (required):
The average rate of occurrence of the event per interval.
3. cumulative (optional):
An optional boolean value that determines the form of the function. If set to TRUE or omitted, the function returns the cumulative distribution function; if set to FALSE, it returns the
probability density function.
Here are a few example use cases that explain how to use the POISSON.DIST formula in Google Sheets.
Calculating the probability of a certain number of events occurring
Suppose you want to find the probability that a certain number of customers will visit your store during a given hour, knowing that the average rate of customer arrivals is 10 per hour. You can use
the POISSON.DIST function to calculate this probability.
Modeling the number of defects in a manufacturing process
A manufacturer produces items with a known average rate of defects per unit. By applying the Poisson distribution, the manufacturer can estimate the likelihood of a certain number of defects in a
given number of units produced.
Predicting website traffic
A website owner can use the Poisson distribution to predict the number of visitors to their site in a given time period based on historical data on visitor arrivals.
Common Mistakes
POISSON.DIST not working? Here are some common mistakes people make when using the POISSON.DIST Google Sheets Formula:
Misunderstanding the mean parameter
One common mistake is to use the actual number of events instead of the expected number of events as the mean parameter. The mean parameter represents the average number of events over a specified
period of time, not the actual number of events in a single sample.
Forgetting to specify the cumulative parameter
Another common mistake is to omit the cumulative parameter when calculating the cumulative distribution function. If the cumulative parameter is not specified, the function will return the
probability density function.
Related Formulas
The following functions are similar to POISSON.DIST or are often used with it in a formula:
• POISSON
The POISSON function returns the Poisson distribution probability density function, which is used to show the probability of a certain number of events occurring in a fixed interval of time or
space. It takes in the values for the number of events (x), the mean (mean), and a boolean for whether or not to return the cumulative distribution (cumulative).
• BINOM.DIST
The BINOM.DIST function returns the probability of a certain number of successes in a fixed number of trials given a probability of success in each trial. It is most commonly used in statistical
analysis and hypothesis testing. The function can calculate either the probability mass function (PMF) or the cumulative distribution function (CDF) depending on the value of the cumulative
• NORM.DIST
The NORM.DIST formula is a statistical function that returns the normal distribution of a specified variable. It is used to determine the probability of a random variable falling within a
specified range of values. This function is commonly used in finance and scientific research.
• TDIST
The TDIST function calculates the probability associated with a Student's t-Test. It returns the probability that the difference between two data sets is greater than or equal to a certain value.
This function is commonly used in hypothesis testing.
• CHISQ.DIST
The CHISQ.DIST function calculates the probability density function or the cumulative distribution function of a chi-squared distribution. This function is commonly used in hypothesis testing to
determine the significance of the difference between expected and observed values. The output of this function can be used to make decisions about the null hypothesis.
Learn More
You can learn more about the POISSON.DIST Google Sheets function on Google Support. | {"url":"https://checksheet.app/google-sheets-formulas/poisson.dist/","timestamp":"2024-11-10T05:25:25Z","content_type":"text/html","content_length":"47962","record_id":"<urn:uuid:d0eedf50-f1fb-45b9-8473-054b8976d41b>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00164.warc.gz"} |
Physics Reduced Syllabus - Physics Syllabus SimplifiedPhysics Reduced Syllabus - Physics Syllabus Simplified
Physics Reduced Syllabus: A Comprehensive Overview
The reduction of the physics syllabus in recent years has been a significant development in educational circles. This adjustment aims to alleviate the burden on students, allowing them to focus on
core concepts and applications. In this article, we will delve into the specifics of the reduced syllabus, exploring the key topics that have been retained and those that have been removed.
Key Topics Retained in the Reduced Syllabus
While some portions of the physics syllabus have been eliminated, several fundamental topics remain essential for students to grasp. These include:
• Units and Measurements: Understanding units and their conversions is crucial for accurate scientific calculations.
• Kinematics: The study of motion, including velocity, acceleration, and displacement, forms the basis of many physics concepts.
• Laws of Motion: Newton's laws of motion provide a framework for understanding the relationship between force and motion.
• Work, Energy, and Power: These concepts are fundamental to understanding the transfer and transformation of energy.
• Rotational Motion: The study of objects rotating around an axis, including angular velocity, torque, and angular momentum.
• Properties of Solids and Liquids: Understanding the properties of matter, such as elasticity, viscosity, and surface tension, is essential in various fields.
• Thermodynamics: The study of heat and energy transfer, including the laws of thermodynamics and applications like heat engines and refrigerators.
• Oscillations and Waves: This topic explores periodic motion, including simple harmonic motion, wave propagation, and resonance.
• Electrostatics: The study of electric charges at rest, including electric fields, potential, and capacitance.
• Current Electricity: The study of electric charges in motion, including Ohm's law, circuits, and electrical instruments.
Topics Removed from the Reduced Syllabus
To accommodate the reduction in syllabus content, certain topics have been removed. These include:
• Lubrication from the Laws of Motion unit.
• Rolling Motion from the Rotational Motion unit.
• Poisson's Ratio, Elastic Energy, Reynolds Number, Qualitative Ideas of Black Body Radiation, Greenhouse Effect, and Newton's Law of Cooling from the Properties of Solids and Liquids unit.
• Kinetic Theory of Gases from the Thermodynamics unit.
• Free, Forced, and Damped Oscillations from the Oscillations and Waves unit.
• Free and Bound Charges Inside a Conductor and Van de Graaff Generator from the Electrostatics unit.
• Carbon Resistor, Color Combination, and Potentiometer from the Current Electricity unit.
Rationale for Syllabus Reduction
The decision to reduce the physics syllabus is based on several factors:
• Alleviation of Student Burden: The reduction aims to lighten the workload on students, allowing them to focus on core concepts and applications.
• Improved Learning Outcomes: By focusing on essential topics, students can develop a deeper understanding of physics principles.
• Alignment with Contemporary Needs: The reduced syllabus reflects the evolving landscape of physics and its applications.
1. Why was the physics syllabus reduced?
The reduction was implemented to alleviate student workload and ensure a focus on core concepts.
2. Which topics have been completely removed?
Topics such as lubrication, rolling motion, and specific aspects of thermodynamics and electrostatics have been removed.
3. Will the reduced syllabus affect future career prospects?
The core concepts retained in the syllabus provide a solid foundation for pursuing careers in physics, engineering, and related fields.
4. Are there any resources available to help students with the reduced syllabus?
Many textbooks, online resources, and tutoring services can assist students in understanding the reduced syllabus.
5. Will the reduced syllabus be revised in the future?
The syllabus may be subject to future revisions based on evolving educational needs and advancements in the field of physics.
The reduction of the physics syllabus represents a significant change in educational practices. By focusing on essential topics, students can develop a strong understanding of physics principles and
apply them to real-world problems. While certain topics have been removed, the core concepts retained in the syllabus provide a valuable foundation for further study and exploration. | {"url":"https://www.vhtc.org/2024/10/physics-reduced-syllabus.html","timestamp":"2024-11-10T12:25:58Z","content_type":"application/xhtml+xml","content_length":"250502","record_id":"<urn:uuid:2f16ad52-b2c9-468f-960c-d446ed77323f>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00249.warc.gz"} |
The Block Library - Indicator Blocks - Arrow Algo
Trading Indicators #
Unlock a spectrum of technical insights with Indicator Blocks, offering an extensive range of indicators, from straightforward moving averages to intricate calculations. This section empowers users
to integrate these indicators seamlessly into their strategies, enhancing their ability to discern market trends and pinpoint potential entry or exit points with precision.
To create a new Indicator block, double click in the blueprint screen, or click the + button in the menu. In the search bar, type: “indicator/” followed by the abbreviations below, or simply type the
abbreviation below.
Remember, Different strategies require different types of indicators. The best trading indicators complement one another, without duplicating information.
Some of our favourites #
Indicator: ticker
Last Signal Profit: last_profit #
Last signal profit facilitates the calculation of profit since the last trading signal, providing users with insights into the performance of their strategy over a specific time frame. Useful for
creating stop triggers.
Candlestick Pattern Detection: candlestick #
Candlestick Pattern Detection aids in detecting Candlestick Patterns, such as bull, bear, hammers, shooting stars, and more, assisting users in identifying key market trends and potential reversal
signals based on candlestick formations.
Relative Strength Index: rsi #
Relative Strength Index measures the speed and rate of change in price movements within the market; it oscillates between zero and 100. It provides insights into whether an asset is overbought or
oversold, helping users identify potential trend reversals and market conditions.
Simple Moving Average: sma #
Simple Moving Average calculates the average price of an asset over a specified number of periods, providing a smooth trend line. It is useful for identifying general market direction.
Exponential Moving Average: ema #
Exponential Moving Average is similar but gives more weight to recent prices, making it more responsive to short-term price changes. It is beneficial for capturing more immediate market trends.
SuperTrend: SuperTrend #
Supertrend calculates the Supertrend value based on the market’s price and volatility, helping users determine the current trend’s direction.
SuperTrendKO: SuperTrendKO #
SuperTrendKO Similar to Supertrend, SupertrendKO is a modified version that factors in market noise and aims to provide more accurate trend signals by minimising false positives.
The Full Trading Indicator List #
Vector Absolute Value: abs #
Vector Absolute Value calculates the absolute value of each element in an array.
Vector Arccosine: acos #
Vector Arccosine calculates the Trigonometric arccosine of each element in an array.
Accumulation Distribution Line: ad #
Accumulation Distribution Line determines the trend of a stock, using the relation between the volume flow and the stock’s price.
Add: add #
Add adds two arrays together.
Accumulation Distribution Oscillator: adosc #
Accumulation Distribution Oscillator is calculated by taking an exponential moving average of short periods of accumulation distribution line subtracted from an exponential moving average of long
periods of accumulation distribution line.
Average Directional Movement Index: adx #
Average Directional Movement Index shows the strength of a trend through a value in a range of 0 to 100.
Average Directional Movement Index Rating: adxr #
Average Directional Movement Index Rating is the same as the average directional movement index but is smoother. This indicator gets less affected than adx from the fast short-term market
Awesome Oscillator: ao #
Awesome Oscillator measures the momentum of the market.
Absolute Price Oscillator: apo #
Absolute Price Oscillator is the difference between the short-period exponential moving average and the long-period exponential moving average.
Aroon: aroon #
Aroon comprises two indicators: Aroon-Up and Aroon-Down. Aroon can identify the beginning of a trend, its strength, and any changes.
Aroon Oscillator: aroonosc #
Aroon Oscillator is the difference between Aroon-Up and Aroon-Down indicators, and the output would be a value between 0 and 100.
Vector Arcsine: asin #
Vector Arcsine calculates the trigonometric arcsine of each element in an array.
Vector Arctangent: atan #
Vector Arctangent calculates the trigonometric arctangent of each element in an array.
Average True Range: atr #
Average True Range measures market volatility over a stock’s price range for a specified period.
Average Price: avgprice #
Average Price shows the mean of open, high, low, and close prices of a stock.
Bollinger Bands: bbands #
Bollinger Bands contains the upper, middle, and lower bands. The middle one is a moving average indicator, and the upper and lower bands are on the sides of the middle one. The value of the standard
deviations determines the distance between the middle band and the upper and lower ones.
Balance of Power: bop #
Balance of Power evaluates the strength of buyers and sellers in the market.
Candlestick Pattern Detection: candlestick #
Candlestick Pattern Detection aids in detecting Candlestick Patterns, such as bull, bear, hammers, shooting stars, and more, assisting users in identifying key market trends and potential reversal
signals based on candlestick formations.
Commodity Channel Index: cci #
Commodity Channel Index would be high when prices are far above the average and would be low when prices are far below it. So cci can identify overbought and oversold areas of price action. Besides
that, it gets used to discover reversals and divergences.
Vector Ceiling: ceil #
Vector Ceiling shows the smallest integer from the elements of an array.
Chande Momentum Oscillator: cmo #
Chande Momentum Oscillator calculates the price of momentum on bullish or/and bearish days. In other words, it computes the difference between the sum of higher closes and the sum of lower closes,
dividing by the sum of all price movements.
Vector Cosine: cos #
Vector Cosine calculates the trigonometric cosine of each element in an array.
Vector Hyperbolic Cosine: cosh #
Vector Hyperbolic Cosine calculates the trigonometric hyperbolic cosine of each element in an array.
Cross Any: crossany #
Crossany continuously detects whether the inputs are crossing each other.
Cross Over: crossover #
Crossover continuously detects whether the first input is crossing over the other one. It means, against the crossany indicator, the only situation that matters is when the first input would place
above the other one.
Chaikins Volatility: cvi #
Chaikins Volatility calculates the difference between the high and low prices for each period.
Decay: decay #
Decay saves an array of recent signals. It is a useful indicator, especially in machine learning algorithms.
Double Exponential Moving Average: dema #
Double Exponential Moving Average is the same as the exponential moving average, but due to allocating more weight to recent data points, delivers fewer lag data.
Directional Indicator: di #
Directional Indicator comprises positive directional indicator and negative directional indicator lines that show the price trend movement. Crossing these two lines propagates the buy and sell
signals; If the positive line crosses up through the negative one, it is a Buy signal, and vice versa.
Vector Division: div #
Vector Division divides the provided inputs.
Directional Movement: dm #
Directional Movement draws positive directional movement and negative directional movement lines. They get calculated using the prior high and low prices.
Detrended Price Oscillator: dpo #
Detrended Price Oscillator removes price trends to make it easier to identify peaks and troughs. Thus, estimating the cycle lengths using the indicator is much simpler.
Directional Movement Index: dx #
Directional Movement Index, which is also referred to as dmi, contains two directional movement lines and the average directional movement index indicator.
Exponential Decay: edecay #
Exponential Decay is almost the same as decay but faster for the same period.
Exponential Moving Average: ema #
Exponential Moving Average shows the direction of the price changes over a period. EMA is like a Simple Moving Average, but where the SMA directly calculates the average price values, EMA applies
more weight to the recent prices.
Ease of Movement: emv #
Ease of Movement investigates the relationship between price fluctuations and trading volume.
Vector Exponential: exp #
Vector Exponential returns the exponential for each number in the input arary. That is, it calculates Euler’s constant, e, raised to the power of each input element.
Fisher Transform: fisher #
Fisher Transform aims to enhance the predictability of turning points in a price series by making prices more normally distributed. This transformation makes it easier to identify extreme values and
potential reversals
The Vector Floor: floor #
The Vector Floor of a value is the largest integer less than or equal to it.
Forecast Oscillator: fosc #
Forecast Oscillator predicts the upcoming stock’s price by monitoring the difference between the current stock’s price and a linear regression price resulting from the Time Series Forecast function.
Hull Moving Average: hma #
Hull Moving Average is an improved moving average that removes the lags (and thus is super fast) and is smoother than the other traditional moving average indicators.
Kaufman Adaptive Moving Average: kama #
Kaufman Adaptive Moving Average reduces false signals by eliminating short-term price fluctuations. In other words, kama removes the market noises, so if the market volatility is low, it will heel
the current market price.
Klinger Volume Oscillator: kvo #
Klinger Volume Oscillator forecasts market reversals by comparing the volume to the price.
Lag Block: lag #
Lag block delays the input data by a specified amount. For example, with a lag of 1 on 15-minute candles, it outputs data from the previous candle. This is useful for comparing current values with
past ones.
Laguerre Filter: laguerrefilter #
Laguerre filter is used to smooth price data and identify trends. It applies a Laguerre filter algorithm to market data, reducing noise and providing a clearer representation of the underlying trend.
Last signal profit: last_profit #
Last signal profit facilitates the calculation of profit since the last trading signal, providing users with insights into the performance of their strategy over a specific time frame.
Linear Regression: linreg #
Linear Regression plots the ending values of linear regression lines for a specific number of bars.
Linear Regression Intercept: linregintercept #
Linear Regression Intercept returns the height of the linear regression line for the first input bar in the moving period.
Linear Regression Slope: linregslope #
Linear Regression Slope determines the direction of trend strength. The indicator determines the slope for each bar using the current bar and the n-1 previous bars where n is the period specified by
the trader.
Vector Natural Log: ln #
Vector Natural Log calculates the natural logarithm for each element in an input array.
Vector Base-10 Log: log10 #
Vector Base-10 Log calculates the base-10 logarithm for each element in an input array.
Moving Average Convergence Divergence: macd #
Moving Average Convergence Divergence determines the direction of the stock price. Consider not using this indicator for detecting trend reversals since it can detect them only after they happen. It
is not usually used to identify overbought or oversold conditions as well.
Market Facilitation Index: marketfi #
Market Facilitation Index measures the trend strength and predicts the starting of a trend when it is about to occur. It calculates the price movement per volume unit.
Mass Index detects: mass #
Mass Index detects helps traders identify potential trend reversals by measuring the expansion and contraction of the trading range (the difference between the high and low prices) over a specified
period using Exponential Moving Averages.
Maximum In Period: max #
Maximum In Period returns the maximum value in the last n bars.
Mean Deviation Over Period: md #
Mean Deviation Over Period computes the absolute mean deviation over a period.
Median Price: medprice #
Median Price computes the mean of the high and low prices for a bar.
The Money Flow Index: mfi #
The Money Flow Index measures the trading pressure by monitoring both the price and volume and returns a value between 0 and 100.
Minimum In Period: min #
Minimum In Period returns the minimum value in the last n bars.
Momentum: mom #
Momentum computes the change between the current price and the price of the n-th bar from the last.
Mesa Sine Wave: msw #
Mesa Sine Wave detects whether the market is in a cycle mode or a trend mode.
Vector Multiplication: mul #
Vector Multiplication takes two input arrays and multiplies them.
Normalized Average True Range: natr #
Normalized Average True Range is a normalized version of the average true range and gets calculated with the following formula: NATR = (ATR / Close) * 100.
Negative Volume Index: nvi #
Negative Volume Index is a cumulative indicator and is sensitive to the market volume. It argued that high market volume is because of uninformative traders, so it doesn’t care about the high-volume
days. On low-volume days, informed traders are more active, and therefore nvi indicator gets affected by them; the nvi value will rise on positive price changes and will fall on negative price
On Balance Volume: obv #
On Balance Volume is a cumulative indicator that calculates buying and selling pressures. It increases on up days and decreases on down days.
Percentage Price Oscillator: ppo #
Percentage Price Oscillator calculates the difference between two exponential moving averages with different periods divided by the longer one.
Predict: predict #
Predict block aims to leverage machine learning to predict the next price movement. Care: This block is still in the process of learning and has not been fully trained. Until such time, it may not
produce accurate predictions.
Parabolic SAR: psar #
Parabolic SAR helps to figure out stop points and potential reversals in trends. Indeed SAR stands for stop and reverse, which describes its application nicely.
Positive Volume Index: pvi #
Positive Volume Index is the same as Negative Volume Index nvi – and often gets used in conjunction with it – but is sensitive to high-volume days.
Qstick: qstick #
Qstick as a momentum indicator applies a simple moving average on the difference between the stock close and open prices.
Rate of Change: roc #
Rate of Change computes the percentage change between the current price and the price n periods ago.
Rate of Change Ratio: rocr #
Rate of Change Ratio computes the change between the current price and the price n periods ago.
Vector Round: round #
Vector Round returns the closest integer for each element in an array.
Relative Strength Index: rsi #
Relative Strength Index measures the speed and rate of change in price movements within the market; it oscillates between zero and 100.
Vector Sine: sin #
Vector Sine computes the Trigonometric sine of each element in an array.
Vector Hyperbolic Sine: sinh #
Vector Hyperbolic Sine computes the Trigonometric hyperbolic sine of each element in an array.
Simple Moving Average: sma #
Simple Moving Average shows the direction of the price changes over a period by calculating the average price value.
Vector Square Root: sqrt #
Vector Square Root computes the square root of each element in an array.
Standard Deviation Over Period: stddev #
Standard Deviation Over Period measures the difference between the current price and the average price over a period.
Standard Error Over Period: stderr #
Standard Error Over Period shows how different the population mean is from the sample mean.
Stochastic Oscillator: stoch #
Stochastic Oscillator compares the last close price to the highest and lowest prices over a period and ranges from zero to 100.
Stochastic RSI: stochrsi #
Stochastic RSI is a combination of two indicators: stoch and rsi. Actually, it’s applying a stoch indicator on a rsi indicator, which means it’s a measure of rsi relative to its high/low range over a
Stock to flow: stocktoflow #
Stock to flow assesses the scarcity of a particular asset, often applied to cryptocurrencies like Bitcoin. It compares the existing stock (current supply) to the flow (new production), offering
insights into the asset’s potential value and market dynamics.
Vector Subtraction: sub #
Vector Subtraction returns the subtraction of the two inputs (a – b).
Super Trend: SuperTrend #
Super Trend calculates the Super Trend value based on the market’s price and volatility, helping users determine the current trend’s direction.
Super Trend KO: SuperTrendKO #
Super Trend KO Similar to Super Trend, Super Trend KO is a modified version that factors in market noise and aims to provide more accurate trend signals by minimizing false positives.
Sum Over Period: sum #
Sum Over Period returns the sum of the last n bars.
Vector Tangent: tan #
Vector Tangent calculates the Trigonometric tangent of each element in an array.
Vector Hyperbolic Tangent: tanh #
Vector Hyperbolic Tangent calculates the Trigonometric hyperbolic tangent of each element in an array.
Triple Exponential Moving Average: tema #
Triple Exponential Moving Average is a high-speed moving average with smoother data. It reduces the lags by placing more weight on the recent data and thus is more appropriate for short-term trading.
Vector Degree Conversion: todeg #
Vector Degree Conversion converts an array of radians into an array of degrees.
Vector Radian Conversion: torad #
Vector Radian Conversion converts an array of degrees into an array of radians.
True Range: tr #
True Range returns the greater value of:
The daySubtraction of the high and low prices of the same day.
• Day’s high minus day’s low
• The absolute value of the day’s high minus the previous day’s close
• The absolute value of the day’s low minus the previous day’s close
Triangular Moving Average: trima #
Triangular Moving Average is the same as Simple Moving Average, sma, but it’s averaged twice; In other words, trima is a sma that applies to another sma. This approach leads to a smoother line that
places more weight on the middle bars.
Triple Exponential Moving Average: trix #
Triple Exponential Moving Average shows the percentage change of a triple-smoothed ema (applying an ema three times).
Vector Truncate: trunc #
Vector Truncate returns only the integer part of a number for each element in an array.
Time Series Forecast: tsf #
Time Series Forecast, as expected from the name, predicts future trends based on past data. It is more sensitive to sudden price changes compared to the moving average indicators.
Typical Price: typprice #
Typical Price computes the arithmetic mean of the high, low, and close prices.
Ultimate Oscillator: ultosc #
Ultimate Oscillator measures buying pressure by considering three different time frames. These periods (7, 14, 28) describe short, medium, and long-term market trends.
Variance Over Period: var #
Variance Over Period measures the variation by calculating the average of squared deviations from the mean.
Vertical Horizontal Filter: vhf #
Vertical Horizontal Filter monitors the price movements and indicates the prices phase, that they are in the trading or the congestion phase.
Variable Index Dynamic Average: vidya #
Variable Index Dynamic Average calculates an ema with a dynamic period depending on the market volatility.
Annualized Historical Volatility: volatility #
Annualized Historical Volatility measures the deviation of the annual average stock price over a period.
Volume Oscillator: vosc #
Volume Oscillator calculates the difference between a fast volume moving average and a slow volume moving average. Monitoring volume changes in this manner has more technical importance than
monitoring volume itself.
Volume Weighted Moving Average: vwma #
Volume Weighted Moving Average is just like most moving average indicators but considers the market volume in its calculations. It actually gives more weight to the high-volume prices than the
low-volume prices.
Williams Accumulation/Distribution: wad #
Williams Accumulation/Distribution is the accumulated sum of accumulation and distribution price changes. Accumulation and distribution describe a market controlled by buyers and sellers,
respectively. Indeed, the wad indicator measures the positive and negative market pressures.
Weighted Close Price: wcprice #
Weighted Close Price is simply the average of high, low, and doubled closing prices.
Wilder’s Smoothing: wilders #
Wilder's Smoothing is the same as ema, but wilder’s smoothing uses a different smoothing factor, which leads to a slower response to price changes.
Williams %R: willr #
Williams %R identifies overbought and oversold markets by comparing the position of the most recent closing price to the highest and lowest prices over a period.
Weighted Moving Average: wma #
Weighted Moving Average is the same as sma, but puts more weight on the recent data. This way, it responds faster to price changes and will stay closer to the market price.
Zero-Lag Exponential Moving Average: zlema #
Zero-Lag Exponential Moving Average follows the same goal as dema and tema. It eliminates the lags to improve the speed and track the price more closely. | {"url":"https://arrowalgo.com/docs/indicator-blocks/","timestamp":"2024-11-03T09:31:59Z","content_type":"text/html","content_length":"205204","record_id":"<urn:uuid:10c17877-b2c8-4d91-8cc7-163833b074c3>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00159.warc.gz"} |
Linear-time implementation of static SPQR-trees.
The class StaticSPQRTree maintains the arrangement of the triconnected components of a biconnected multi-graph G [Hopcroft, Tarjan 1973] as a so-called SPQR tree T [Fi Battista, Tamassia, 1996]. We
call G the original graph of T. The class StaticSPQRTree supports only the statical construction of an SPQR-tree for a given graph G, dynamic updates are not supported.
Each node of the tree has an associated type (represented by SPQRTree::NodeType), which is either SNode, PNode, or RNode, and a skeleton (represented by the class StaticSkeleton). The skeletons of
the nodes of T are in one-to-one correspondence to the triconnected components of G, i.e., S-nodes correspond to polygons, P-nodes to bonds, and R-nodes to triconnected graphs.
In our representation of SPQR-trees, Q-nodes are omitted. Instead, the skeleton S of a node v in T contains two types of edges: real edges, which correspond to edges in G, and virtual edges, which
correspond to edges in T having v as an endpoint. There is a special edge er in G at which T is rooted, i.e., the root node of T is the node whose skeleton contains the real edge corresponding to er.
The reference edge of the skeleton of the root node is er, the reference edge of the skeleton S of a non-root node v is the virtual edge in S that corresponds to the tree edge (parent(v),v).
Definition at line 73 of file StaticSPQRTree.h. | {"url":"https://ogdf.netlify.app/classogdf_1_1_static_s_p_q_r_tree","timestamp":"2024-11-12T00:29:10Z","content_type":"application/xhtml+xml","content_length":"98177","record_id":"<urn:uuid:2a779521-7a2c-4512-83dc-9087f5e2d259>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00093.warc.gz"} |
Steel Tank Heat Loss
Calculate the heat loss from a steel tank.
General Info
of lights
Light Heat (Btu/h)
of people
People Heat (Btu/h)
Total Light Heat (Btu/h)
Total People Heat (Btu/h)
Tank Info
Radius (ft)
Height (ft)
Heat Loss (Bth/ft^2h)
of Units
Motor (kW)
Tank Surface Area (ft^2)
Heat Loss (Btu/h)
Motor Heat (Btu/h)
Total Heat (Btu/h)
Calculation Reference
Heat Loss
Heat Flow from Steel Tanks
Thermal Analysis
To calculate the heat loss from a steel tank, you will need the tank's dimensions (radius and height), heat loss rate per unit surface area, and any additional heat sources such as lights and people.
Here's how you can perform the calculations:
1. Calculate the tank surface area: The surface area of a cylindrical tank can be calculated using the formula:
Surface Area = 2πrh + πr^2
Where π is the mathematical constant pi (approximately 3.14159), r is the radius of the tank, and h is the height of the tank.
2. Calculate the heat loss from the tank: Multiply the tank surface area by the heat loss rate per unit surface area to obtain the heat loss in Btu/h. The heat loss rate per unit surface area can be
provided as a value in Btu/(ft^2·h) or W/m^2.
Heat Loss = Surface Area × Heat Loss Rate
3. Calculate the heat contribution from lights and people: If there are additional heat sources, such as lights or people, calculate their respective heat contributions in Btu/h. Multiply the number
of lights by the heat generated by each light in Btu/h, and multiply the number of people by the heat generated by each person in Btu/h.
Total Light Heat = Number of Lights × Light Heat Total People Heat = Number of People × People Heat
4. Calculate the total heat loss: Sum up the heat loss from the tank, the heat contribution from lights, and the heat contribution from people to obtain the total heat loss in Btu/h.
Total Heat Loss = Heat Loss + Total Light Heat + Total People Heat
Note: Ensure that all the units used in the calculations are consistent (e.g., radius and height in the same unit, heat loss rate and surface area in the same unit).
Calculation Preview
Full download access to any calculation is available to users with a paid or awarded subscription (XLC Pro).
Subscriptions are free to contributors to the site, alternatively they can be purchased.
Click here for information on subscriptions
11 years ago
Your debut calculation - I have extended your XLC Pro subscription by 3 months by way of thanks. | {"url":"https://www.excelcalcs.com/calcs/repository/Heat/Combined/Steel-Tank-Heat-Loss/","timestamp":"2024-11-07T22:09:00Z","content_type":"text/html","content_length":"28821","record_id":"<urn:uuid:5c3f7637-ba39-4ef8-8db8-2bc561204bfe>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00739.warc.gz"} |
MarkLogic 9 Product Documentation
MarkLogic 9 Product Documentation
p1 as cts.point,
p2 as cts.point,
[options as String[]]
) as Number
Returns the true bearing in radians of the path from the first point to the second. An error is raised if the two points are the same.
p1 The first point.
p2 The second point.
Options for the operation. The default is ().
Options include:
Use the given coordinate system. Valid values are:
The WGS84 coordinate system.
The WGS84 coordinate system at double precision.
The ETRS89 coordinate system.
options etrs89/double
The ETRS89 coordinate system at double precision.
The raw (unmapped) coordinate system.
The raw coordinate system at double precision.
Use the coordinate system at the given precision. Allowed values: float and double.
Unit of measure of the tolerance value. Valid values are miles (default), km, feet, meters.
Tolerance is the largest allowable variation in geometry calculations. If the distance between two points is less than tolerance, then the two points are considered equal. For the raw
coordinate system, use the units of the coordinates. For geographic coordinate systems, use the units specified by the units option.
Usage Notes
The value of the precision option takes precedence over that implied by the governing coordinate system name, including the value of the coordinate-system option. For example, if the governing
coordinate system is "wgs84/double" and the precision option is "float", then the operation uses single precision.
Tolerance reflects how accurate you believe the data is. Computing a bearing between two points that are effectively equal within the limits of data accuracy is likely to produce useless results: The
tolerance parameter can be used to force an error in this situation. Effective tolerance may be limited by the limits of precision.
See Also
const sf = cts.point(37, -122);
const ny = cts.point(40, -73);
geo.bearing(sf, ny);
=> 1.2212785952625
Stack Overflow: Get the most useful answers to questions from the MarkLogic community, or ask your own question. | {"url":"http://docs.marklogic.com/9.0/geo.bearing","timestamp":"2024-11-02T17:45:34Z","content_type":"application/xhtml+xml","content_length":"32390","record_id":"<urn:uuid:78e60a53-b235-4f49-a8a2-6aacecd5373f>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00100.warc.gz"} |
(technical) Post-treatment bias can be anti-conservative!
A little rant on the sad state of knowledge about post-treatment bias: For some reason I still see a lot of people using control strategies (typically, regression) that use post-treatment outcomes
that are intermediate between the treatment and endpoint outcome of interest. I have heard people who do so say that this is somehow necessary to show that the “effects” that they estimate in the
reduced form regression of treatment on endpoint outcome are not spurious. Of course this is incorrect. To show the relationship “goes away” after controlling for the intermediate outcome does not
indicate that the effect is spurious. It could just as well be that the treatment affects the endpoint outcome mostly through the intermediate outcome.
I have also heard people say that controlling for intermediate, post-treatment outcomes is somehow “conservative” because controlling for the post-treatment outcome “will only take away from the
association” between the treatment and the outcome. Of course, this is also incorrect. Controlling for a post-treatment variable can easily be anti-conservative, producing a coefficient on the
treatment that is substantially larger than the actual treatment effect. This happens when the intermediate outcome exhibits a “suppression” effect, for example, when the treatment has a negative
association with the intermediate outcome, but the intermediate outcome then positively affects the endpoint outcome. Here is a straightforward demonstration (done in R):
N <- 200
z <- rbinom(N,1,.5)
ed <- rnorm(N)
d <- -z + ed
ey <- rnorm(N)
y <- z + d + ey
print(coef(summary(lm(y~z))), digits=2)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.0049 0.14 0.035 0.97
z -0.1109 0.20 -0.555 0.58
print(coef(summary(lm(y~z+d))), digits=2)
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.078 0.093 -0.84 4.0e-01
z 1.034 0.149 6.95 5.3e-11
d 1.046 0.064 16.23 3.6e-38
In the example above, z is the treatment variable, and y is the endpoint outcome, while d is an intermediate outcome. (The data generating process resembles a binomial assignment experiment.) The
causal effect of z is properly estimated in the first regression. The effect is indistinguishable from 0. The problems that arise when controlling for a post-treatment intermediate outcome are shown
in the second regression. Now the coefficient on z is 1 with a very low p-value!
A question I received offline was along the lines of “what if you control for the post-treatment variable and your effect estimate doesn’t change. Surely this strengthens the case that what you’ve
found is not spurious.” I don’t think that is correct. The case for having a well identified effect estimate is based only on having properly addressed pre-treatment confounding. To show that a
post-treatment variable does not alter the estimate has no bearing on whether this has been achieved or not. Thus, the post-treatment conditioning is pretty much useless for demonstrating that a
causal relation is not spurious.
The one case where post-treatment conditioning provides some causal content is in the case of mediation. But there, exclusion restriction or effect-homogeneity assumptions have to hold, otherwise the
mediation analysis may produce misleading results. On these points, I suggest looking at this very clear paper by Green, Ha, and Bullock (ungated preprint). A more elaborate paper (though not quite
as intuitive in its presentation) is this one by Imai, Keele, Tingley, and Yamamoto (working paper). | {"url":"https://cyrussamii.com/?p=730","timestamp":"2024-11-13T19:34:27Z","content_type":"text/html","content_length":"82328","record_id":"<urn:uuid:eb33206a-cd3f-4338-8798-a18d34c54802>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00355.warc.gz"} |
The Evil GM - Simple Grappling system? (Edition Neutral)
I was messing around with the idea of a quick and dirty grapple system for my games. I wanted something simple enough it can be resolved in a matter of a roll or two but not so simple one side over
powers the other.
So I decided to take the method of rolling by taking the attackers d20 roll and adding any Strength bonuses vs. the defenders d20 roll +Strength bonuses as the base.
Then I decided that depending on their size you’d also roll an additional die such as:
• ·        Small – d4
• ·        Medium – d6
• ·        Large – d8
After you roll this secondary die, you minus it off your opponents result total
Example: John is playing a Halfling who gets a d4 secondary die to his roll, while Tim is playing a human who gets a d6 as his secondary die.
• ·        John rolls a 15 + 2 (str mod) = 17
• ·        John rolls d4 and gets a 3
• ·        Tim rolls a 18 +1 (str mod) = 19
• ·        Tim rolls d6 and gets a 2
-     -  John’s new total is 15 (17 total minus Tim’s d6 roll of 2)
-     - Tim’s new total is 16 (19 total minus John’s d4 roll of 3)
Tim is the winner of the grapple.
Now people are most likely saying, well what about if its small vs. Large? I decided that when its small vs. Large the secondary die totals are swapped, giving a small the d8 and the large a d4.
Of course this method is based on humanoids only and should be modified when using other sizes.
Just some thoughts. | {"url":"https://www.theevildm.com/p/the-evil-gm-simple-grappling-system-edition-neutral","timestamp":"2024-11-05T06:19:32Z","content_type":"text/html","content_length":"161112","record_id":"<urn:uuid:7fab3875-f56b-4152-96cf-50c68985048a>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00032.warc.gz"} |
How do you simplify (x-2)/(x^2+3x-10)? | HIX Tutor
How do you simplify #(x-2)/(x^2+3x-10)#?
Answer 1
at first do factorization of #x^2+3x-10#, #x^2+5x-2x-10# #x(x+5)-2(x+5)# #(x+5)(x-2)#
now put #(x+5)(x-2)# at the given equation, #cancel(x-2)/((x+5)cancel((x-2)) ) #
we finally get, #1/(x+5)#
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Answer 2
To simplify the expression (x-2)/(x^2+3x-10), we can factor the denominator and then cancel out any common factors. The denominator can be factored as (x+5)(x-2). Therefore, the expression simplifies
to 1/(x+5).
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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Pebble guided optimal treasure hunt in anonymous graphs
We study the problem of treasure hunt in a graph by a mobile agent. The nodes in the graph are anonymous and the edges at any node v of degree deg(v) are labeled arbitrarily as 0,1,…,deg(v)−1. A
mobile agent, starting from a node, must find a stationary object, called treasure that is located on an unknown node at a distance D from its initial position. The agent finds the treasure when it
reaches the node where the treasure is present. The time of treasure hunt is defined as the number of edges the agent visits before it finds the treasure. The agent does not have any prior knowledge
about the graph or the position of the treasure. An Oracle, that knows the graph, the initial position of the agent, and the position of the treasure, places some pebbles on the nodes, at most one
per node, of the graph to guide the agent towards the treasure. We target to answer the question: what is the fastest possible treasure hunt algorithm regardless of the number of pebbles are placed?
We show an algorithm that uses O(DlogΔ) pebbles to find the treasure in a graph G in time O(DlogΔ), where Δ is the maximum degree of a node in G and D is the distance from the initial position of
the agent to the treasure. We show a matching lower bound of Ω(DlogΔ) on time of the treasure hunt using any number of pebbles.
• Anonymous graph
• Deterministic algorithms
• Mobile agent
• Pebbles
• Treasure hunt
ASJC Scopus subject areas
• Theoretical Computer Science
• General Computer Science
Dive into the research topics of 'Pebble guided optimal treasure hunt in anonymous graphs'. Together they form a unique fingerprint. | {"url":"https://cris.bgu.ac.il/en/publications/pebble-guided-optimal-treasure-hunt-in-anonymous-graphs","timestamp":"2024-11-04T17:13:32Z","content_type":"text/html","content_length":"56997","record_id":"<urn:uuid:26a84685-cd33-4612-9e64-30ebe6029021>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00263.warc.gz"} |
What is the bond polarity of the water molecule? | Socratic
What is the bond polarity of the water molecule?
1 Answer
The polarity of water can be calculated by finding the sum of the two dipole moments of both $O - H$ bonds.
For ionic compounds, the dipole moment could be calculated by:
$\mu = Q \times r$
where, $\mu$ is the dipole moment,
$Q$ is the coulomb charge $Q = 1.60 \times {10}^{- 19} C$,
and $r$ is the bond length or the distance between two ions.
For covalent compounds, the expression becomes:
$\mu = \delta \times r$
where, $\delta$ is the partial charge on atoms.
For water, the partial charges are distributed as follows:
It is more complicated to calculate the partial charge on each atom, that is why I will skip this part.
The dipole moment of the $O - H$ bond is $\mu = 1.5 D$, where $D$ is the Debye unit where, $1 D = 3.34 \times {10}^{- 30} C \cdot m$.
So the net dipole moment of water could be calculated by summing the two dipole moments of both $O - H$ bonds
${\mu}_{\text{total}} = 2 \times 1.5 D \times \cos \left(\frac{104.5}{2}\right) = 1.84 D$
Note that ${104.5}^{\circ}$ is the bonds angle in water.
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Toward a 12-second Block Time | Ethereum Foundation Blog
One of the annoyances of the blockchain as a decentralized platform is the sheer length of delay before a transaction gets finalized. One confirmation in the Bitcoin network takes ten minutes on
average, but in reality due to statistical effects when one sends a transaction one can only expect a confirmation within ten minutes 63.2% of the time; 36.8% of the time it will take longer than ten
minutes, 13.5% of the time longer than twenty minutes and 0.25% of the time longer than an hour. Because of fine technical points involving Finney attacks and sub-50% double spends, for many use
cases even one confirmation is not enough; gambling sites and exchanges often need to wait for three to six blocks to appear, often taking over an hour, before a deposit is confirmed. In the time
before a transaction gets into a block, security is close to zero; although many miners refuse to forward along transactions that conflict with transactions that had already been sent earlier, there
is no economic necessity for them to do so (in fact quite the contrary), and some don't, so reversing an unconfirmed transaction is possible with about a 10-20% success rate.
In many cases, this is fine; if you pay for a laptop online, and then manage to yank back the funds five minutes later, the merchant can simply cancel the shipping; online subscription services work
the same way. However, in the context of some in-person purchases and digital goods purchases, it is highly inconvenient. In the case of Ethereum, the inconvenience is greater; we are trying to be
not just a currency, but rather a generalized platform for decentralized applications, and especially in the context of non-financial apps people tend to expect a much more rapid response time. Thus,
for our purposes, having a blockchain that is faster than 10 minutes is critical. However, the question is, how low can we go, and if we go too low does that destabilize anything?
Overview of Mining
First off, let us have a quick overview of how mining works. The Bitcoin blockchain is a series of blocks, with each one pointing to (ie. containing the hash of) the previous. Each miner in the
network attempts to produce blocks by first grabbing up the necessary data (previous block, transactions, time, etc), building up the block header, and then continually changing a value called the
nonce until the nonce satisfies a function called a "proof of work condition" (or "mining algorithm"). This algorithm is random and usually fails; on average, in Bitcoin the network needs to
collectively make about 10^20 attempts before a valid block is found. Once some random miner finds a block that is valid (ie. it points to a valid previous block, its transactions and metadata are
valid, and its nonce satisfies the PoW condition), then that block is broadcast to the network and the cycle begins again. As a reward, the miner of that block gets some quantity of coins (25 BTC in
Bitcoin) as a reward.
The "score" of a block is defined in a simplified model as the number of blocks in the chain going back from it all the way to the genesis (formally, it's the total mining difficulty, so if the
difficulty of the proof of work condition increases blocks created under this new more stringent condition count for more). The block that has the highest score is taken to be "truth". A subtle, but
important, point is that in this model the incentive for miners is always to mine on the block with the highest score, because the block with the highest score is what users ultimately care about,
and there are never any factors that make a lower-score block better. If we fool around with the scoring model, then if we are not careful this might change; but more on this later.
We can model this kind of network thus:
However, the problems arise when we take into account the fact that network propagation is not instant. According to a 2013 paper from Decker and Wattenhofer in Zurich, once a miner produces a block
on average it takes 6.5 seconds for the block to reach 50% of nodes, 40 seconds for it to reach 95% of nodes and the mean delay is 12.6 seconds. Thus, a more accurate model might be:
This gives rise to the following problem: if, at time T = 500, miner M mines a block B' on top of B (where "on top of" is understood to mean "pointing to as the previous block in the chain"), then
miner N might not hear about the block until time T = 510, so until T = 510 miner N will still be mining on B. If miner B finds a block in that interval, then the rest of the network will reject
miner B's block because they already saw miner M's block which has an equal score:
Stales, Efficiency and Centralization
So what's wrong with this? Actually, two things. First, it weakens the absolute strength of the network against attacks. At a block time of 600 seconds, as in Bitcoin, this is not an issue; 12
seconds is a very small amount of time, and Decker and Wattenhofer estimate the total stale rate as being around 1.7%. Hence, an attacker does not actually need 50.001% of the network in order to
launch a 51% attack; if the attacker is a single node, they would only need 0.983 / 1 + 0.983 = 49.5%. We can estimate this via a mathematical formula: if transit time is 12 seconds, then after a
block is produced the network will be producing stales for 12 seconds before the block propagates, so we can assume an average of 12 / 600 = 0.02 stales per valid block or a stale rate of 1.97%. At
60 seconds per block, however, we get 12 / 60 = 0.2 stales per valid block or a stale rate of 16.67%. At 12 seconds per block, we get 12 / 12 = 1 stale per valid block, or a stale rate of 50%. Thus,
we can see the network get substantially weaker against attacks.
However, there is also another negative consequence of stale rates. One of the more pressing issues in the mining ecosystem is the problem of mining centralization. Currently, most of the Bitcoin
network is split up into a small number of "mining pools", centralized constructions where miners share resources in order to receive a more even reward, and the largest of these pools has for months
been bouncing between 33% and 51% of network hashpower. In the future, even individual miners may prove threatening; right now 25% of all new bitcoin mining devices are coming out of a single factory
in Shenzhen, and if the pessimistic version of my economic analysis proves correct that may eventually morph into 25% of all Bitcoin miners being in a single factory in Shenzhen.
So how do stale rates affect centralization? The answer is a clever one. Suppose that you have a network with 7000 pools with 0.01% hashpower, and one pool with 30% hashpower. 70% of the time, the
last block is produced by one of these miners, and the network hears about it in 12 seconds, and things are somewhat inefficient but nevertheless fair. 30% of the time, however, it is the 30%
hashpower mining pool that produced the last block; thus, it "hears" about the block instantly and has a 0% stale rate, whereas everyone else still has their full stale rate.
Because our model is still pretty simple, we can still do some math on an approximation in closed form. Assuming a 12 second transit time and a 60-second block time, we have a stale rate of 16.67% as
described above. The 30% mining pool will have a 0% stale rate 30% of the time, so its efficiency multiplier will be 0.833 * 0.7 + 1 * 0.3 = 0.8831, whereas everyone else will have an efficiency
multiplier of 0.833; that's a 5.7% efficiency gain which is pretty economically significant especially for mining pools where the difference in fees is only a few percent either way. Thus, if we want
a 60 second block time, we need a better strategy.
The beginnings of a better approach come from a paper entitled "Fast Money Grows on Trees, not Chains", published by Aviv Zohar and Yonatan Sompolinsky in December 2013. The idea is that even though
stale blocks are not currently counted as part of the total weight of the chain, they could be; hence they propose a blockchain scoring system which takes stale blocks into account even if they are
not part of the main chain. As a result, even if the main chain is only 50% efficient or even 5% efficient, an attacker attempting to pull off a 51% attack would still need to overcome the weight of
the entire network. This, theoretically, solves the efficiency issue all the way down to 1-second block times. However, there is a problem: the protocol, as described, only includes stales in the
scoring of a blockchain; it does not assign the stales a block reward. Hence, it does nothing to solve the centralization problem; in fact, with a 1-second block time the most likely scenario
involves the 30% mining pool simply producing every block. Of course, the 30% mining pool producing every block on the main chain is fine, but only if the blocks off chain are also fairly rewarded,
so the 30% mining pool still collects not much more than 30% of the revenue. But for that rewarding stales will be required.
Now, we can't reward all stales always and forever; that would be a bookkeeping nightmare (the algorithm would need to check very diligently that a newly included uncle had never been included
before, so we would need an "uncle tree" in each block alongside the transaction tree and state tree) and more importantly it would make double-spends cost-free. Thus, let us construct our first
protocol, single-level GHOST, which does the minimal thing and takes uncles only up to one level (this is the algorithm used in Ethereum up to now):
1. Every block must point to a parent (ie. previous block), and can also include zero or more uncles. An "uncle" is defined as a block with a valid header (the block itself need not be valid, since
we only care about its proof-of-work) which is the child of the parent of the parent of the block but not the parent (ie. the standard definition of "uncle" from genealogy that you learned at age
2. A block on the main chain gets a reward of 1. When a block includes an uncle, the uncle gets a reward of 7/8 and the block including the uncle gets a reward of 1/16.
3. The score of a block is zero for the genesis block, otherwise the score of the parent plus the difficulty of the block multiplied by one plus the number of included uncles.
Thus, in the graphical blockchain example given above, we'll instead have something like this:
Here, the math gets more complex, so we'll make some intuitive arguments and then take the lazy approach and simulate the whole thing. The basic intuitive argument is this: in the basic mining
protocol, for the reasons we described above, the stale rate is roughly t/(T+t) where t is the transit time and T is the block interval, because t/T of the time miners are mining on old data. With
single-level GHOST, the failure condition changes from mining one stale to mining two stales in a row (since uncles can get included but relatives with a divergence of 2 or higher cannot), so the
stale rate should be (t/T)^2, ie. about 2.7% instead of 16.7%. Now, let's use a Python script to test that theory:
### PRINTING RESULTS ###
1 1.0
10 10.2268527074
25 25.3904084273
5 4.93500893242
15 14.5675475882
Total blocks produced: 16687
Total blocks in chain: 16350
Efficiency: 0.979804638341
Average uncles: 0.1584242596
Length of chain: 14114
Block time: 70.8516366728
The results can be parsed as follows. The top five numbers are a centralization indicator; here, we see that a miner with 25% hashpower gets 25.39x as much reward as a miner with 1% hashpower. The
efficiency is 0.9798 meaning that 2.02% of all blocks are not included at all, and there are 0.158 uncles per block; hence, our intuitions about a ~16% stale rate without uncle inclusion and 2.7%
with uncle inclusion are confirmed almost exactly. Note that the actual block time is 70.85s because even though there is a valid proof of work solution every 60s, 2% of them are lost and 14% of them
make it into only the next block as an uncle, not into the main chain.
Now, there is a problem here. The original authors of the GHOST paper did not include uncle/stale rewards, and although I believe it is a good idea to deviate from their prescription for the reasons
I described above, they did not do so for a reason: it makes the economic analysis more uncomfortable. Specifically, when only the main chain gets rewarded there is an unambiguous argument why it's
always worth it to mine on the head and not some previous block, namely the fact that the only thing that conceivably differentiates any two blocks is their score and higher score is obviously better
than lower score, but once uncle rewards are introduced there are other factors that make things somewhat tricky.
Specifically, suppose that the main chain has its last block M (score 502) with parent L (score 501) with parent K (score 500). Also suppose that K has two stale children, both of which were produced
after M so there was no chance for them to be included in M as uncles. If you mine on M, you would produce a block with score 502 + 1 = 503 and reward 1, but if you mine on L you would be able to
include K's children and get a block with score 501 + 1 + 2 = 504 and reward 1 + 0.0625 * 2 = 1.125.
Additionally, there is a selfish-mining-esque attack against single-level GHOST. The argument is as follows: if a mining pool with 25% hashpower were not to include any other blocks, then in the
short term it would hurt itself because it would no longer receive the 1/16x nephew reward but it would hurt others more. Because in the long-term mining is a zero-sum game since the block time
rebalances to keep issuance constant, this means that not including uncles might actually be a dominant strategy, so centralization concerns are not entirely gone (specifically, they still remain 30%
of the time). Additionally, if we decide to crank up the speed further, say to a 12 second target block time, single-level is just not good enough. Here's a result with those statistics:
### PRINTING RESULTS ###
1 1.0
10 10.4567533177
15 16.3077390517
5 5.0859101624
25 29.6409432377
Total blocks produced: 83315
Total blocks in chain: 66866
Efficiency: 0.802568565084
Average uncles: 0.491246459555
Length of chain: 44839
Block time: 22.3020138719
18% centralization gain. Thus, we need a new strategy.
A New Strategy
The first idea I tried about one week ago was requiring every block to have five uncles; this would in a sense decentralize the production of each block further, ensuring that no miner had a clear
advantage in making the next block. Since the math for that is pretty hopelessly intractable (well, if you try hard at it for months maybe you could come up with something involving nested Poisson
processes and combinatorical generating functions, but I'd rather not), here's the sim script. Note that there are actually two ways you can do the algorithm: require the parent to be the lowest-hash
child of the grandparent, or require the parent to be the highest-score child of the grandparent. The first way (to do this yourself, modify line 56 to if newblock["id"] > self.blocks[self.head]
["id"]:, we get this:
### PRINTING RESULTS ###
1 1.0
10 9.59485744106
25 24.366668248
5 4.82484937616
15 14.0160823568
Total blocks produced: 8033
Total blocks in chain: 2312
Efficiency: 0.287812772314
Average uncles: 385.333333333
Length of chain: 6
Block time: 13333.3333333
Ooooops! Well, let's try the highest-score model:
### PRINTING RESULTS ###
1 1.0
10 9.76531271652
15 14.1038046954
5 5.00654546181
25 23.9234131003
Total blocks produced: 7989
Total blocks in chain: 6543
Efficiency: 0.819001126549
Average uncles: 9.06232686981
Length of chain: 722
Block time: 110.8033241
So here we have a very counterintuitive result: the 25% hashpower mining pool gets only 24x as much as a 1% hashpower pool. Economic sublinearity is a cryptoeconomic holy grail, but unfortunately it
is also somewhat of a perpetual motion machine; unless you rely on some specific thing that people have a certain amount of (eg. home heating demand, unused CPU power), there is no way to get around
the fact even if you come up with some clever sublinear concoction an entity with 25x as much power going in will at the very least be able to pretend to be 25 separate entities and thus claim a 1x
reward. Thus, we have an unambiguous (okay, fine, 99 point something percent confidence) empirical proof that the 25x miners are acting suboptimally, meaning that the optimal strategy in this
environment is not to always mine the block with the highest score.
The reasoning here is this: if you mine on a block that has the highest score, then there is some chance that someone else will discover a new uncle one level back, and then mine a block on top of
that, creating a new block at the same level as your block but with a slightly higher score and leaving you in the dust. However, if you try to be one of those uncles, then the highest-score block at
the next level will certainly want to include you, so you will get the uncle reward. The presence of one non-standard strategy strongly suggests the existence of other, and more exploitative,
non-standard strategies, so we're not going this route. However, I chose to include it in the blog post to show an example of what the dangers are.
So what is the best way forward? As it turns out, it's pretty simple. Go back to single level GHOST, but allow uncles to come from up to 5 blocks back. Hence, the child of a parent of a parent
(hereinafter, -2,+1-ancestor) is a valid uncle, a -3,+1-ancestor is a valid uncle, as is a -4,+1-ancestor and a -5,+1-ancestor, but a -6,+1-ancestor or a -4,+2-ancestor (ie. c(c(P(P(P(P(head))))))
where no simplification is possible) is not. Additionally, we increase the uncle reward to 15/16, and cut the nephew reward to 1/32. First, let's make sure that it works under standard strategies. In
the GHOST sim script, set UNCLE_DEPTH to 4, POW_SOLUTION_TIME to 12, TRANSIT_TIME to 12, UNCLE_REWARD_COEFF to 15/16 and NEPHEW_REWARD_COEFF to 1/32 and see what happens:
### PRINTING RESULTS ###
1 1.0
10 10.1329810896
25 25.6107014231
5 4.96386947539
15 15.0251826297
Total blocks produced: 83426
Total blocks in chain: 77306
Efficiency: 0.926641574569
Average uncles: 0.693116362601
Length of chain: 45659
Block time: 21.901487111
Completely reasonable all around, although note that the actual block time is 21s due to inefficiency and uncles rather than the 12s we targeted. Now, let's try a few more trials for enlightenment
and fun:
• UNCLE_REWARD_COEFF = 0.998, NEPHEW_REWARD_COEFF = 0.001 lead to the 25% mining pool getting a roughly 25.3x return, and setting UNCLE_REWARD_COEFF = 7/8 and NEPHEW_REWARD_COEFF = 1/16 leads to
the 25% mining pool getting a 26.26% return. Obviously setting the UNCLE_REWARD_COEFF all the way to zero would negate the benefit completely, so it's good to have it be as close to one as
possible, but if it's too close to one than there's no incentive to include uncles. UNCLE_REWARD_COEFF = 15/16 seems to be a fair middle ground, giving the 25% miner a 2.5% centralization
• Allowing uncles going back 50 blocks, surprisingly, has fairly little substantial efficiency gain. The reason is that the dominant weakness of -5,+1 GHOST is the +1, not the -5, ie. stale c(c(P(P
(..P(head)..)))) blocks are the problem. As far as centralization goes, with 0.998/0.001 rewards it knocks the 25% mining pool's reward down to essentially 25.0x. With 15/16 and 1/32 rewards
there is no substantial gain over the -4,+1 approach.
• Allowing -4,+3 children increases efficiency to effectively 100%, and cuts centralization to near-zero assuming 0.998/0.001 rewards and has negligible benefit assuming 15/16 and 1/32 rewards.
• If we reduce the target block time to 3 seconds, efficiency goes down to 66% and the 25% miner gets a 31.5x return (ie. 26% centralization gain). If we couple this with a -50,+1 rule, the effect
is negligible (25% -> 31.3x), but if we use a -4,+3 rule efficiency goes up to 83% and the 25% miner only gets a 27.5x return (the way to add this to the sim script is to add after line 65 for c2
in self.children.get(c, {}): u[c2] = True for a -n,+2 rule and then similarly nest down one level further for -n,+3). Additionally, the actual block time in all three of these scenarios is around
10 seconds.
• If we reduce the target block time to 6 seconds, then we get an actual block time of 15 seconds and the efficiency is 82% and the 25% miner gets 26.8x even without improvements.
Now, let's look at the other two risks of limited GHOST that we discussed above: the non-head dominant strategy and the selfish-mining attack. Note that there are actually two non-head strategies:
try to take more uncles, and try to be an uncle. Trying to take more uncles was useful in the -2,+1 case, and trying to be an uncle was useful in the cas of my abortive mandatory-5-uncles idea.
Trying to be an uncle is not really useful when multiple uncles are not required, since the reason why that alternative strategy worked in the mandatory-5-uncle case is that a new block is useless
for further mining without siblings. Thus, the only potentially problematic strategy is trying to include uncles. In the one-block case, it was a problem, but here is it not because most uncles that
can be included after n blocks can also be included after n+1 blocks, so the practical extent to which it will matter is limited.
The selfish-mining attack also no longer works for a similar reason. If you fail to include uncles, then the guy after you will. There are four chances for an uncle to get in, so not including uncles
is a 4-party prisoner's dilemma between anonymous players - a game that is doomed to end badly for everyone involved (except of course the uncles themselves). There is also one last concern with this
strategy: we saw that rewarding all uncles makes 51% attacks cost-free, so are they cost-free here? Beyond one block, the answer is no; although the first block of an attempted fork will get in as an
uncle and receive its 15/16x reward, the second and third and all subsequent ones will not, so starting from two confirmations attacks still cost miners almost as much as they did before.
Twelve seconds, really?
The most surprising finding about Decker and Wattenhofer's finding is the sheer length of time that blocks take to propagate - an amazingly slow 12 seconds. In Decker and Wattenhofer's analysis, the
12 second delay is actually mostly because of the need to download and verify the blocks themselves; ie. the algorithm that Bitcoin clients follow is:
def on_receive_block(b):
if not verify_pow_and_header(b):
if not verify_transactions(b):
However, Decker and Wattenhofer did propose a superior strategy which looks something like this:
def on_receive_header(h):
if not verify_pow_and_header(h):
ask_for_full_block(h, callback)
def callback(b):
if not verify_transactions(b):
This allows all of the steps to happen in parallel; headers can get broadcasted first, then blocks, and the verifications do not need to all be done in series. Although Decker and Wattenhofer do not
provide their own estimate, intuitively this seems like it may speed up propagation by 25-50%. The algorithm is still non-exploitable because in order to produce an invalid block that passes the
first check a miner would still need to produce a valid proof of work, so there is nothing that the miner could gain. Another point that the paper makes is that the transit time is, beyond a certain
point, proportional to block size; hence, cutting block size by 50% will also cut transit time to something like 25-40%; the nonscaling portion of the transit time is something like 2s. Hence, a
3-second target block time (and 5s actual block time) may be quite viable. As usual, we'll be more conservative at first and not take things that far, but a block time of 12s does nevertheless seem
to be very much achievable. | {"url":"https://blog.ethereum.org/2014/07/11/toward-a-12-second-block-time?ref=fin.plaid.com","timestamp":"2024-11-13T19:51:45Z","content_type":"text/html","content_length":"231930","record_id":"<urn:uuid:2c986fa9-7851-41c7-a45d-58431b76a091>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00272.warc.gz"} |
All About calculate cement sand quantity in 1:6 mortar
Cement and sand are essential components in any construction project, and their proper quantity is crucial for a strong and durable structure. In a 1:6 mortar mixture, the ratio of cement to sand is
1 part cement to 6 parts sand, making it a commonly used ratio in various construction applications. However, calculating the exact quantity of cement and sand required for a 1:6 mortar can be a
daunting task, especially for inexperienced individuals. In this article, we will delve into the details of how to accurately calculate the cement and sand quantity needed for a 1:6 mortar, ensuring
a successful construction project.
How to calculate cement sand quantity in 1:6 mortar?
When it comes to construction, the right quantity of materials is crucial in ensuring the strength and quality of a structure. One of the most commonly used materials in construction is cement sand
mortar, which is a mixture of cement, sand, and water. Knowing how to calculate the quantity of cement and sand in a mortar mix is essential for any civil engineer. Here’s a step-by-step guide on how
to do it for a 1:6 mortar mix.
Step 1: Determine the required volume of mortar
The first step in calculating the quantity of cement and sand in a 1:6 mortar mix is to determine the required volume of mortar. This volume will depend on the area to be covered and the thickness of
the mortar layer. For example, if you have a wall with an area of 10 square meters and a thickness of 0.02 meters, the required volume of mortar would be 0.2 cubic meters (10 sqm x 0.02m).
Step 2: Calculate the dry volume of mortar
Next, you need to calculate the dry volume of the mortar, which is the volume of the mixture before adding water. To do this, you need to add the volume of cement and sand together. The ratio of
cement and sand in a 1:6 mortar mix is 1 part cement to 6 parts sand. So, for every 1 cubic meter of mortar, you will need 1/7 cubic meters of cement and 6/7 cubic meters of sand. Therefore, the dry
volume of mortar would be 0.2 cubic meters x (1+6) = 1.2 cubic meters.
Step 3: Calculate the quantity of cement
Now, using the dry volume of mortar, you can calculate the quantity of cement required for the mix. The standard density of cement is 1440 kg/m3, which means one cubic meter of cement weighs 1440
kilograms. So, to convert the dry volume of mortar into kilograms, you need to multiply the dry volume by the density of cement. In this case, it would be 1.2 cubic meters x 1440 kg/m3 = 1728 kg.
Step 4: Calculate the quantity of sand
Next, you need to calculate the quantity of sand required for the mortar mix. The density of sand is 1600 kg/m3, which means one cubic meter of sand weighs 1600 kilograms. So, to convert the dry
volume of mortar into kilograms, you need to multiply the dry volume by the density of sand. In this case, it would be 1.2 cubic meters x 1600 kg/m3 = 1920 kg.
Step 5: Convert the quantity of sand and cement into bags
Finally, you need to convert the quantity of sand and cement into bags for better understanding. A standard cement bag weighs 50 kilograms, while a standard sand bag weighs 40 kilograms. So, for the
1:6 mortar mix with a required volume of 0.2 cubic meters, the quantity of cement would be 1728 kg/ 50 kg per bag = 34.56 bags, and the quantity of sand would be 1920 kg/ 40 kg per bag = 48 bags.
In conclusion, calculating the quantity of cement and sand in a 1:6 mortar mix is a crucial step in construction. By following the steps mentioned above, civil engineers and construction
The quantity of cement required for 1m3 of cement mortar (1:6)
Cement mortar is a mixture of cement, sand, and water used in various construction projects, such as brick masonry, plastering, and pointing. The ratio of cement to sand in a mortar mix is known as
the cement mortar ratio. In this article, we will discuss the quantity of cement required for 1m3 of cement mortar with a ratio of 1:6.
Before we dive into the calculation, it is essential to understand the meaning of a 1:6 mortar mix ratio. This ratio represents that for every one part of cement, six parts of sand are used in the
mix. This ratio ensures a strong and durable mortar that is suitable for general construction purposes.
To determine the quantity of cement required for 1m3 of cement mortar, we need to follow these steps:
Step 1: Determine the volume of the mortar mix
The volume of the mortar mix is calculated by multiplying the length, width, and height of the area where the mortar will be applied. For example, if the dimension of the area is 1m x 1m x 1m, then
the volume will be 1m3.
Step 2: Calculate the dry volume of mortar mix
Since the mortar mix contains both solid and voids, we need to determine the dry volume to avoid any wastage of materials. The dry volume of the mortar mix can be calculated by multiplying the wet
volume by a factor of 1.35.
Dry volume = 1m3 x 1.35 = 1.35m3
Step 3: Calculate the quantity of cement
To calculate the quantity of cement required, we need to first convert the dry volume of the mortar mix to a wet volume. This can be done by dividing the dry volume by the sum of the cement ratio.
Wet volume = 1.35m3 / (1+6) = 0.225m3
Next, we need to multiply the wet volume by the cement ratio (1). Therefore, the quantity of cement required for 1m3 of cement mortar will be:
Quantity of cement = 0.225m3 x 1 = 0.225m3
Step 4: Convert to bags
Cement is commonly sold in bags, so we need to convert the quantity of cement into the number of bags. A common cement bag has a volume of 0.0347m3. Therefore, to obtain the number of bags, we divide
the quantity of cement by the volume of one bag.
Number of bags = 0.225m3 / 0.0347m3 = 6.49 bags
Hence, for a 1m3 volume of mortar with a 1:6 cement ratio, 6.49 bags of cement will be required.
In conclusion, the quantity of cement required for 1m3 of cement mortar (1:6) is 6.49 bags. It is important to note that this calculation is based on a perfect mortar mix without any wastage. In
practical applications, a certain amount of wastage should be considered, and the quantity of cement may vary accordingly.
The quantity of sand required for 1m3 of cement mortar (1:6)
Cement mortar is a mixture of cement, sand, and water used in construction for binding building blocks together. It is an essential component in construction and is used for various applications such
as brickwork, plastering, and masonry repairs. The ratio of cement to sand in mortar is known as the mix or building ratio. The most commonly used building ratio is 1:6, which means one part of
cement is mixed with six parts of sand.
When it comes to calculating the quantity of sand required for 1m3 (cubic meter) of cement mortar at a 1:6 ratio, there are a few factors that need to be considered:
1. Mixing Ratio: As mentioned earlier, the mixing ratio for 1:6 mortar is one part of cement to six parts of sand. This means that for every 1 kg of cement, 6 kg of sand will be required.
2. Volume of Cement: To determine the volume of cement required for 1m3 of mortar, we need to first calculate the volume of the mix. The total volume of the mix is calculated by multiplying the
length, width, and height of the area where the mortar will be used. For example, if the area to be covered is 10m x 10m x 10m = 100m3. As the mixing ratio is 1:6, the volume of cement required will
be 1/7 x 100 = 14.3m3.
3. Volume of Sand: To determine the volume of sand required, we need to first calculate the volume of the mix. In this case, it will be 6/7 x 100 = 85.7m3. This means that 85.7m3 of sand will be
required for every 14.3m3 of cement.
4. Density of Sand: The density of sand can vary depending on its size, moisture content, and compaction. Generally, the density of sand used in construction ranges from 1600-1825 kg/m3. Taking the
average density as 1713 kg/m3, we can calculate the mass of sand required as 85.7 x 1713 = 146575 kg.
Therefore, the quantity of sand required for 1m3 of cement mortar (1:6) is 146575 kg or 146.6 tonnes. This amount of sand may seem significant, but it is important to use the correct amount to ensure
the strength and durability of the mortar.
In conclusion, the quantity of sand required for 1m3 of cement mortar (1:6) is calculated based on the mixing ratio, volume of cement, volume of sand, and the density of sand. It is essential to
accurately calculate and use the correct amount of sand to achieve the desired strength and quality of the mortar in construction projects.
The quantity of sand & cement required for 1m3 of cement mortar (1:6)
Cement mortar is a mixture of cement, sand, and water that is commonly used in construction for bonding bricks and other types of masonry units. The strength and quality of cement mortar depend
greatly on the ratio of cement and sand used in its preparation. In this article, we will discuss the calculation of the quantity of sand and cement required for 1m3 of cement mortar with a ratio of
The 1:6 ratio of cement mortar means that it contains one part of cement and six parts of sand by volume. This ratio is generally used for non-load bearing walls. The following are the steps to
calculate the quantity of sand and cement required for 1m3 of cement mortar.
Step 1: Determine the dry volume of mortar
The first step is to determine the dry volume of mortar required for 1m3. The dry volume is the volume of the mortar without considering the voids between the particles of sand and cement. It can be
calculated by multiplying the total volume with the ratio of cement and sand.
Dry volume = 1m3 x (1+6) = 7m3
Step 2: Calculate the materials’ quantity
The quantity of cement and sand can be calculated by using the following formula:
Cement quantity = (Dry volume of mortar x Cement ratio) / (Sum of the ratio)
Sand quantity = (Dry volume of mortar x Sand ratio) / (Sum of the ratio)
Substituting the values in the above formula, we get:
Cement quantity = (7m3 x 1) / (1+6) = 1m3
Sand quantity = (7m3 x 6) / (1+6) = 6m3
Step 3: Convert to weight
Cement and sand are usually measured by weight, so the next step is to convert the quantity of cement and sand into weight. The density of cement is 1440 kg/m3 and the density of sand is 1600 kg/m3.
Therefore, the weight of cement required for 1m3 of mortar = 1m3 x 1440 kg/m3 = 1440 kg, and the weight of sand required for 1m3 of mortar = 6m3 x 1600 kg/m3 = 9600 kg.
Step 4: Determine the number of bags
The final step is to determine the number of bags of cement and volume of sand required based on the average weight of a bag of cement and the density of sand.
One bag of cement weighs 50 kg, so the number of bags of cement required for 1m3 of mortar is 1440 kg/50 kg = 28.8 bags. As sand is measured in cubic meters, the volume of sand required for 1m3 of
mortar is 9600 kg/1600 kg/m3 = 6 m3.
In conclusion, for the preparation of 1m3 of cement mortar with a mix ratio of 1:6, we require 28.8 bags of cement and 6m3 of sand. It is important to note that the actual quantity may vary slightly
depending on factors such as the quality of sand and the water-cement ratio. Accurate measurement and proportioning of materials are crucial for achieving a strong and durable cement mortar mix.
The quantity of cement required for 1m3 of cement mortar (1:5)
Cement mortar is a mixture of cement, sand, and water used in various construction activities such as bricklaying, plastering, and masonry works. It provides strength and acts as a binding agent
between bricks or blocks. The proportion of each ingredient is crucial in determining the strength and workability of the mortar.
In the construction industry, the most commonly used ratio for cement mortar is 1:5, which means one part of cement is mixed with five parts of sand. This ratio is generally used for non-load bearing
walls and plastering works.
To calculate the quantity of cement required for 1m3 of cement mortar (1:5), we need to know the total volume of the mortar and the proportion of cement in the mix.
Step 1: Calculate the volume of cement mortar
To calculate the volume of cement mortar, we need to know the thickness of the mortar. For example, if we consider a wall with dimensions 6m x 3m x 0.1m, where 0.1m is the thickness of the mortar,
then the volume of mortar will be:
Volume = 6m x 3m x 0.1m = 1.8m3
Step 2: Calculate the proportion of cement in the mix
As mentioned earlier, the ratio of cement and sand in 1:5 mortar mix is 1:5. Therefore, the proportion of cement in the mix is:
1 / (1+5) = 1/6
Step 3: Calculate the quantity of cement required
To calculate the quantity of cement required, we need to multiply the volume of mortar with the proportion of cement in the mix.
Quantity of cement = 1.8m3 x 1/6 = 0.3m3
Step 4: Convert the volume into bags of cement
Cement is sold in bags, and the standard bag size is 50kg. To convert the volume of cement into bags, we need to use the following formula:
Number of bags = (Quantity of cement x 1440) / 50
Where 1440 is the density of cement in kg/m3.
Substituting the value of the quantity of cement, we get:
Number of bags = (0.3 x 1440) / 50 = 8.64 bags
Therefore, 8.64 bags of cement will be required for 1m3 of 1:5 cement mortar.
Note: It is essential to consider a wastage factor of 5-10% while calculating the quantity of cement to account for spillage and uneven mixing.
In conclusion, the quantity of cement required for 1m3 of 1:5 cement mortar is 8.64 bags, assuming a wastage factor of 5-10%. It is crucial to follow the specified ratio and accurately measure the
ingredients to ensure the quality and strength of the mortar.
The quantity of sand required for 1m3 of cement mortar (1:5)
Cement mortar is a mixture of cement, sand, and water, used in various construction activities such as masonry work, plastering, and flooring. The ratio of cement to sand in cement mortar is
generally referred to as the proportion, for example, 1:5 would mean 1 part of cement and 5 parts of sand. One of the most common mix proportions used in construction is 1:5, which means for 1 cubic
meter of cement mortar, 1 cubic meter of cement and 5 cubic meters of sand are needed.
The quantity of sand required for 1m3 of cement mortar (1:5) can be calculated by first determining the volume of sand required. This can be done by multiplying the ratio of sand to cement (5) by the
volume of cement in the mix (1m3).
Volume of sand required = (5/1) x 1m3 = 5m3
Therefore, for 1m3 of cement mortar (1:5), 5m3 of sand is needed.
It is important to note that the quantity of sand required for cement mortar may vary slightly depending on factors such as the type of sand, moisture content, and compaction. So it is always
recommended to conduct a small test batch before starting a large construction project.
Besides the quantity, the type of sand used also plays a crucial role in the strength and quality of cement mortar. The sand should be clean, free from impurities, and have well-graded particles.
Sharp sand or angular sand is preferred over rounded sand as it provides better bonding and reduces shrinkage.
Moreover, the sand should be properly graded to ensure a good mix. The ideal particle size for sand is between 0.15mm to 4.75mm. If the sand particles are too fine, it can lead to a weaker mortar
mix, while too coarse sand can result in a rough surface finish.
It is also essential to have the right amount of water in the mix. Too little water can result in a dry and crumbly mix, while too much water can weaken the final product. The ratio of water to
cement should be carefully controlled, and the mix should be thoroughly mixed and continuously checked for consistency.
In conclusion, for 1m3 of cement mortar with a ratio of 1:5, 5m3 of sand is required. The quantity of sand can vary depending on various factors, but it is crucial to use clean, well-graded sand and
to control the water-cement ratio to ensure a strong and durable cement mortar mix.
The quantity of cement required for 1m3 of cement mortar (1:4)
Cement mortar is a commonly used construction material that is composed of cement, sand, and water. It is used as a binding material to hold bricks or blocks together in masonry work. The ratio of
cement to sand in cement mortar is expressed as 1:4, meaning that for every part of cement, four parts of sand are used.
If you are a civil engineer working on a construction project, it is essential to know the quantity of cement required for different tasks. One of the most commonly asked questions is the amount of
cement required for 1m3 of cement mortar (1:4). The answer to this question is crucial as it affects the structural integrity, strength, and durability of the building.
So, how much cement is required for 1m3 of cement mortar (1:4)? Let’s find out.
Firstly, we need to understand the meaning of 1m3. 1m3 is a unit of measurement for volume, and it represents one cubic meter. This means that 1m3 of cement mortar (1:4) will cover an area of one
cubic meter.
To calculate the quantity of cement required, we need to know the density of cement and the volume of the mixture.
The density of cement is around 1440 kg/m3. This means that 1440 kilograms of cement are required to fill one cubic meter. However, we need to account for the voids between particles. So, the actual
volume of cement with voids is approximately 30% more. Therefore, the volume of cement required for 1m3 of cement mortar (1:4) is:
= 1m3 x 1.30
= 1.3m3
Next, we need to determine the proportion of cement in the mixture. In the case of 1:4 ratio, it means that one part is cement, and four parts are sand. So, for 1.3m3 of cement mortar, the quantity
of cement required is:
= 1.3m3 x 1/5 (1 part cement out of 5 parts total)
= 0.26m3
Now, we need to convert the volume of cement into kilograms. As mentioned earlier, the density of cement is 1440 kg/m3. Therefore, the quantity of cement required for 1m3 of cement mortar (1:4) is:
= 0.26m3 x 1440 kg/m3
= 374.4 kg
Hence, 374.4 kilograms of cement is required for 1m3 of cement mortar (1:4).
In conclusion, the quantity of cement required for 1m3 of cement mortar (1:4) is 374.4 kilograms. Keep in mind that this is an estimate, and the actual amount may vary depending on the quality of
materials used and other factors such as wastage. It is always recommended to conduct a trial mix to determine the exact quantity of material required before starting any construction work.
The quantity of sand required for 1m3 of cement mortar (1:4)
Cement mortar is a mixture of cement, sand, and water used for various construction purposes such as masonry, plastering, and flooring. The ratio of cement to sand in mortar is usually expressed in
terms of parts, with the standard ratio being 1:4. This means that for every part of cement, four parts of sand are used. So, for 1m3 of cement mortar (with a ratio of 1:4), the quantity of sand
required would be:
1m3 x 4 parts of sand = 4m3 of sand
In other words, 4 cubic meters of sand are needed to make 1m3 of cement mortar with a ratio of 1:4.
It is important to note that the quantity of sand required may vary slightly depending on the quality and density of the sand used. Also, some adjustments may need to be made based on the thickness
of the mortar layer being applied.
To get a better understanding, let’s break down the components of the mixture and their roles in the final quantity of sand required for 1m3 of cement mortar (1:4):
1. Cement: Cement is the binding agent in mortar and holds all the other ingredients together. In a ratio of 1:4, one part of cement (usually measured in kilogram weight) is added with four parts of
sand (measured in cubic meters) to create the perfect mixture.
2. Sand: Sand is the main aggregate used in mortar and it provides bulk to the mixture. It also helps in filling the gaps between cement particles, creating a strong bond and increasing the
workability of the mortar.
The type of sand used in mortar is also important. It is recommended to use fine sand (also known as masonry sand) as it produces a smoother and more workable mortar compared to coarse sand. Also,
make sure to use clean and well-graded sand to ensure the quality of the mortar.
3. Water: Water is the liquid component that binds all the other ingredients together and helps in the hydration process of the cement. The amount of water used should be carefully measured to
achieve the desired consistency of the mortar.
In conclusion, for 1m3 of cement mortar (1:4), the quantity of sand required is 4 cubic meters. Proper measurement of the ingredients and using good quality materials is important in achieving
high-quality and durable mortar for construction purposes. Any deviations in the ratio or quality of materials can affect the strength and workability of the mortar, leading to potential structural
issues in the future.
The quantity of cement required for 1m3 of cement mortar (1:3)
Cement mortar is a common building material used in construction to bond bricks, stones, and other materials together. It is composed of cement, sand, and water in different proportions depending on
the desired strength and workability. As a civil engineer, it is crucial to understand the quantity of each component required for a certain volume of mortar to ensure proper mixing and efficient use
of materials.
In general, the most commonly used ratio for cement mortar is 1:3, which means that for every part of cement, three parts of sand are used. This ratio is often used for general masonry and plastering
work. The following is the calculation of the quantity of cement required for 1m3 (cubic meter) of cement mortar in the ratio of 1:3.
Step 1: Determine the volume of mortar required
To calculate the volume of mortar required, we first need to know the volume of 1m3 of mortar. The volume of cement mortar is calculated by multiplying the area of the surface to be plastered or
masonry work (in square meters) by the thickness of the mortar layer (in meters).
For example, if we consider a wall with an area of 10 square meters and a thickness of 0.01 meters, the volume of mortar required would be 10 x 0.01 = 0.1m3.
Step 2: Calculate the volume of cement
To find the volume of cement required, we need to use the same ratio of 1:3. This means that for every 1 part of cement, we need 3 parts of sand.
So, for 1m3 of cement mortar, we need 1/4 (1+3) = 0.25m3 of cement.
Step 3: Convert volume to weight
Cement is usually sold in bags, and the quantity is measured in kilograms. To determine the weight of cement required, we need to multiply the volume of cement by its density, which is typically
around 1440 kg/m3.
Therefore, the weight of cement required for 1m3 of mortar in the ratio of 1:3 would be 0.25 x 1440 = 360 kg.
In conclusion, for 1m3 of cement mortar in the ratio of 1:3, we need 360 kg of cement. It is important to note that the amount of water required will also vary depending on factors like humidity,
temperature, and the type of sand used. It is recommended to follow the manufacturer’s instructions for the best results. Proper calculation of the quantity of cement and other materials needed for
construction not only ensures cost-effectiveness but also helps in maintaining the quality of the structure. As a civil engineer, it is essential to have a deep understanding of these calculations
and their application in construction projects.
The quantity of sand required for 1m3 of cement mortar (1:3)
In construction projects, mortar is an essential material used to bind together bricks, stones, and other building materials. It is composed of cement, sand, and water in varying proportions,
depending on the intended use. The ratio of cement and sand in mortar is typically denoted as 1:X, where X represents the amount of sand required per unit of cement. In this article, we will focus on
the quantity of sand required for 1m3 of cement mortar with a ratio of 1:3.
Before we dive into the calculations, it is important to understand the role of sand in mortar. Sand is used as a fine aggregate in mortar to fill the voids between the coarse aggregates (such as
gravel or crushed stone). It provides the necessary workability, cohesiveness, and strength to the mortar. Hence, it is crucial to use the right amount of sand for a well-performing and durable
cement mortar.
To determine the quantity of sand required for 1m3 of cement mortar with a ratio of 1:3, we can follow these steps:
Step 1: Calculate the volume of mortar
The first step is to calculate the volume of mortar required for 1m3 of cement mortar. It can be done by multiplying the volume of the wall or surface to be plastered by the thickness of the mortar
layer. For example, if the wall size is 5m x 3m and the thickness of the mortar layer is 1cm (0.01m), then the volume of mortar required will be 5m x 3m x 0.01m = 0.15m3.
Step 2: Calculate the volume of cement
Next, we need to calculate the volume of cement required for 1m3 of cement mortar with a ratio of 1:3. As per the ratio, for every unit of cement, 3 units of sand are needed. Therefore, the volume of
cement for 1m3 of cement mortar will be 1/3 = 0.33m3.
Step 3: Calculate the volume of sand
Now, we can calculate the volume of sand required for 1m3 of cement mortar by subtracting the volume of cement (0.33m3) from the total volume of mortar (0.15m3). Thus, the volume of sand required
will be 0.15m3 – 0.33m3 = 0.12m3.
Step 4: Convert the volume to weight
Lastly, we need to convert the volume of sand to weight, as sand is typically measured in kilograms. To do this, we need to know the bulk density of the sand. Assuming the bulk density of sand to be
1600 kg/m3, the weight of sand required for 1m3 of cement mortar will be 0.12m3 x 1600 kg/m3 = 192 kg.
Therefore, the quantity of sand required for 1m3 of cement mortar (1:3) is 0.12m3 or 192kg. It is important to note that this is an estimate and the actual quantity of sand used may vary slightly
based on factors such as the moisture content of the sand and the mixing techniques. It is always recommended to do a small test batch to determine the exact amount of sand required before starting a
large project.
In conclusion, for 1m3 of cement mortar
The quantity of sand & cement required for 1m3 of cement mortar (1:5)
Cement mortar is a fundamental element in construction, widely used for various applications such as brickwork, plastering, and masonry. It is made by mixing cement, sand, and water in specific
proportions. This mixture acts as a bonding agent, holding the construction materials together and providing strength to the structure. As a civil engineer, it is essential to have a good
understanding of the quantity of materials required for various types of cement mortars. In this article, we will discuss the quantity of sand and cement required for 1m3 of cement mortar with a
ratio of 1:5.
What is 1:5 cement mortar?
The ratio 1:5 in cement mortar means that there is one part of cement and five parts of sand by volume. This ratio is commonly used for internal plastering, brickwork, and masonry works. The strength
of the mortar depends on the proportion of the cement and sand used. A 1:5 mortar mixture provides a good balance between workability and strength, making it suitable for most general construction
Quantity of sand required for 1m3 of cement mortar (1:5):
To determine the quantity of sand required for 1m3 of cement mortar, we first need to calculate the volume of sand. The volume of sand is the total volume of the mixture minus the volume of cement.
Cement volume = (1/6) x 1m3 = 0.167m3
Sand volume = 1m3 – 0.167m3 = 0.833m3
Therefore, for a 1:5 cement mortar mixture, we require 0.833m3 of sand.
Quantity of cement required for 1m3 of cement mortar (1:5):
To calculate the quantity of cement required for 1m3 of cement mortar, we need to multiply the volume of cement by the mix ratio (1/6). This will give us the volume of cement in cubic meters.
Cement volume = (1/6) x 1m3 = 0.167m3
We know that the density of cement is 1440 kg/m3. Therefore, the mass of cement required for 1m3 of cement mortar would be:
Mass of cement = 1440 kg/m3 x 0.167m3 = 241.92 kg
Hence, for a 1:5 cement mortar mixture, we would need 241.92 kg of cement.
In conclusion, for 1m3 of cement mortar with a mix ratio of 1:5, we would require 0.833m3 of sand and 241.92 kg of cement. It is essential to note that these quantities may vary slightly depending on
the moisture content of the sand and the compaction of the mixture. It is always recommended to conduct a small trial batch before starting a construction project to get an accurate idea of the
materials’ required quantity. As a civil engineer, it is essential to have an in-depth understanding of material quantities to ensure cost-effective and efficient construction.
The quantity of sand & cement required for 1m3 of cement mortar (1:4)
Cement mortar is a mixture of cement, sand, and water that is commonly used in construction for various purposes such as bonding bricks or covering surfaces. The ratio of cement to sand in cement
mortar is crucial, as it determines the strength and durability of the structure. In this essay, we will discuss the quantity of sand and cement required for 1m3 of cement mortar with a ratio of 1:4.
Before we calculate the quantity of sand and cement needed for 1m3 of cement mortar (1:4), we must understand the term “1m3.” 1m3 is the volume of 1 cubic meter, which is equal to 1000 liters. In
simpler terms, 1m3 of cement mortar is equivalent to 10 full-sized buckets of mortar.
The first step in determining the quantity of sand and cement for 1m3 of cement mortar (1:4) is to calculate the dry volume of mortar. This volume accounts for the shrinkage that occurs when sand,
cement, and water are mixed together. The dry volume of mortar is calculated by multiplying the wet volume with a factor of 1.35.
Dry Volume of Mortar = Wet Volume of Mortar x 1.35
Now, let us assume the wet volume of mortar is 1m3.
Therefore, Dry Volume of Mortar = 1m3 x 1.35
= 1.35m3
Next, we need to calculate the quantity of cement required for 1m3 of cement mortar (1:4). The ratio of 1:4 indicates that one part of cement is mixed with four parts of sand. In other words, for
every one unit of cement, we need four units of sand.
Quantity of Cement = (Dry Volume of Mortar / Total Ratio) x Cement Ratio
= (1.35m3 / 1+4) x 1
= 0.27m3
We know that the density of cement is about 1440 kg/m3. Therefore, the weight of 0.27m3 of cement is calculated as:
Weight of Cement = 0.27m3 x 1440kg/m3
= 388.8 kg
Lastly, we can determine the quantity of sand required for 1m3 of cement mortar (1:4). Using the ratio, we already know that one unit of cement requires four units of sand. So, the total quantity of
sand needed is calculated as:
Quantity of Sand = Quantity of Cement x 4
= 388.8 kg x 4
= 1555.2 kg or 1.5552 tonnes
In conclusion, for 1m3 of cement mortar with a ratio of 1:4, we need 388.8 kg of cement and 1.5552 tonnes of sand. It is essential to note that the quantities may vary slightly depending on the
properties of the materials used. It is always advisable to consult a professional engineer or refer to a construction material handbook for accurate quantities. The correct ratio of cement to sand
is crucial for the structural integrity of the building, so it is essential to follow the ratio and measure the quantities accurately.
The quantity of sand & cement required for 1m3 of cement mortar (1:3)
Cement mortar is a mixture of cement and sand that is commonly used in various construction and repair works. It is known for its high strength, durability, and ability to bond different materials
together. However, in order to achieve these desired qualities, it is crucial to have the right proportion of cement and sand in the mixture.
In this article, we will discuss the quantity of sand and cement required for 1m3 of cement mortar with a ratio of 1:3, which means one part of cement and three parts of sand.
For a typical 1m3 volume of cement mortar, the required materials include 1440 kg of cement and 360 kg of sand. This makes the total weight of the mortar mixture to be 1800 kg.
Now, let’s break down the calculation to understand how to arrive at these quantities.
1. Calculation for Cement Quantity:
Since the ratio of cement to sand is 1:3, the weight of the cement can be calculated as:
Weight of cement = (volume of cement mortar * ratio of cement) * density of cement
= (1m3 * 1) * 1440 kg/m3
= 1440 kg
2. Calculation for Sand Quantity:
The weight of sand can be calculated as:
Weight of sand = (volume of cement mortar * ratio of sand) * density of sand
= (1m3 * 3) * 1600 kg/m3
= 360 kg
Note: The density of sand may vary depending on the type and source of sand. The value used here is an estimated average.
3. Total Weight of Cement Mortar:
To obtain the total weight of cement mortar, we simply need to sum up the weight of cement and sand:
Total weight of cement mortar = weight of cement + weight of sand
= 1440 kg + 360 kg
= 1800 kg
Therefore, to prepare 1m3 of cement mortar with a ratio of 1:3, you will need 1440 kg of cement and 360 kg of sand.
It is also worth noting that the above calculation is based on the assumption of using loose and dry materials. In practical situations, slight variations in the quantities may occur due to factors
such as moisture content in the materials and the method of mixing.
In conclusion, the correct proportion of sand and cement is crucial in achieving the desired strength and durability of cement mortar. It is important to follow the recommended ratio and accurately
measure the required quantities to ensure the quality and stability of the construction work.
RR Masonry (Random rubble masonry)
RR masonry, or Random Rubble masonry, is a type of construction method commonly used in civil engineering. It involves using undressed and irregularly-shaped stones to build walls, façades, and other
structures. This type of masonry dates back to ancient times and is still used in various parts of the world today.
The term “random rubble” refers to the fact that the stones used in this type of masonry are not cut or shaped in any particular way. They are typically collected from natural sources such as
quarries, riverbeds, or fields. These stones vary in size, shape, and color, giving the final structure a unique and rugged appearance.
Construction of RR masonry typically involves a team of masons who carefully select and arrange the stones by hand. The stones are placed on a bed of mortar and are often packed with smaller pieces
of stone or rubble to fill any gaps. The masons use their skills and knowledge to create a stable and aesthetically pleasing structure that will support the loads placed upon it.
One of the most significant advantages of RR masonry is its cost-effectiveness. Since the stones used are found in their natural state, they do not require any cutting or shaping, reducing the
overall construction costs. This type of masonry is also durable and can withstand harsh weather conditions, making it suitable for use in various climates.
Apart from its cost-effectiveness and durability, RR masonry also offers excellent thermal insulation. The gaps and voids between the stones allow for proper ventilation, minimizing the transfer of
heat between the interior and exterior of the structure. This makes RR masonry a popular choice for buildings in hot and humid climates.
However, one of the main challenges of using RR masonry is its lack of uniformity. The stones used are not of the same shape or size, making it challenging to achieve a consistent appearance
throughout the structure. This can affect the structural integrity of the building and may require frequent maintenance.
To address this issue, various techniques are used to strengthen RR masonry walls. These include the use of quoins or cornerstones, which are larger, well-shaped stones placed at the corners of the
wall for added stability. The use of tie stones or bonding stones is also common, which act as anchors between the inner and outer surfaces of the wall.
In conclusion, RR masonry is a traditional yet practical construction method widely used in civil engineering. Its unique appearance, cost-effectiveness, and thermal insulation properties make it a
preferred choice for many architects and engineers. However, careful planning and construction techniques must be employed to ensure the structural stability of the final structure.
RR masonry 1:5 ratio cement and sand calculation
RR masonry is a type of brickwork commonly used in construction projects. It consists of bricks laid in a specific pattern with a mortar mixture of cement and sand. The 1:5 ratio of cement and sand
means one part cement to five parts sand by volume.
Calculating the amount of cement and sand required for RR masonry is crucial for estimating project costs and ensuring there is enough material available. Below is a step-by-step guide on how to
calculate the amount of cement and sand needed for 1 square meter of RR masonry with a 1:5 ratio.
Step 1: Calculate the Volume of Mortar
First, determine the volume of the RR masonry by multiplying the length, width, and thickness of the wall in meters. For example, if the wall is 4 meters long, has a width of 0.5 meters, and a
thickness of 0.2 meters, the volume would be:
4m x 0.5m x 0.2m = 0.4 cubic meters
Step 2: Find the Volume Proportion of Cement and Sand
Using the 1:5 ratio, the volume proportion of cement and sand is:
Cement = (1/6) x 0.4 = 0.067 cubic meters
Sand = (5/6) x 0.4 = 0.333 cubic meters
Step 3: Convert the Volume of Cement and Sand to Mass
To convert the volume to mass, we need to multiply the volume by the dry density of the respective material. The dry density of cement is typically around 1440 kg/m3, and the dry density of sand is
around 1600 kg/m3. Therefore, the mass of cement and sand are:
Mass of cement = 0.067 cubic meters x 1440 kg/m3 = 96.5 kg
Mass of sand = 0.333 cubic meters x 1600 kg/m3 = 532.8 kg
Step 4: Adjust for Wastage and Shrinkage
It is essential to account for wastage and shrinkage in the material calculation. It is recommended to add an extra 10% of the total amount to account for these factors. Therefore, the final
quantities of cement and sand required for 1 square meter of RR masonry would be:
Cement = 96.5 kg + (0.1 x 96.5 kg) = 106.15 kg
Sand = 532.8 kg + (0.1 x 532.8 kg) = 586.08 kg
In summary, the 1:5 ratio of cement and sand for RR masonry requires 106.15 kg of cement and 586.08 kg of sand for every 1 square meter of a wall with a thickness of 0.2 meters. It is important to
note that these calculations may vary depending on the specific project and should be re-evaluated for accuracy. Additionally, it is recommended to use a mortar mix calculator for precise
calculations. Properly calculating the amount of materials required for RR masonry will result in a strong and durable wall.
RR masonry 1:4 ratio cement and sand calculation
RR masonry, also known as random rubble masonry, is a type of brickwork construction method in which stones of varying sizes and shapes are used to create a strong and durable structure. This type of
masonry is often used for non-load bearing walls and is known for its rustic and irregular appearance.
When constructing RR masonry with a 1:4 ratio of cement and sand, it is important to accurately calculate the amount of materials needed to ensure the strength and stability of the wall. The
following are the steps for calculating the quantity of cement and sand required for RR masonry 1:4 ratio.
Step 1: Determine the volume of the wall
To calculate the volume of the wall, multiply the length, height, and thickness of the wall. For example, if the wall is 10 feet long, 8 feet high, and 1 foot thick, the volume would be 10 x 8 x 1 =
80 cubic feet.
Step 2: Calculate the volume of cement
The amount of cement needed for the wall can be calculated by using the volume of the wall and the ratio of cement to sand. In this case, the ratio is 1:4, meaning that for every 1 part of cement, 4
parts of sand are used. Therefore, the volume of cement required would be 1/5th of the total volume (1+4).
Volume of cement = (1/5) x 80 = 16 cubic feet
Step 3: Calculate the volume of sand
To calculate the volume of sand, multiply the volume of cement by 4.
Volume of sand = 4 x 16 = 64 cubic feet
Step 4: Convert volume to weight
Cement and sand are typically sold by weight, so it is important to convert the volume to weight. The specific gravity of cement is approximately 3.15 and the specific gravity of sand is
approximately 2.65.
Weight of cement = 16 x 3.15 = 50.4 lbs
Weight of sand = 64 x 2.65 = 169.6 lbs
Step 5: Add 10% extra
It is recommended to add an extra 10% of materials to account for wastage and unevenness in the stones. Therefore, the final quantities of materials needed for RR masonry 1:4 ratio would be 55.4 lbs
of cement and 186.6 lbs of sand.
In conclusion, proper calculation of materials is crucial for the construction of RR masonry with a 1:4 ratio of cement and sand to ensure a strong and stable structure. It is also important to note
that the density and specific gravity of cement and sand may vary, so it is always best to double-check the calculations with the specific materials being used.
What is the quantity of stone required for 10 m3 of rubble masonry?
Rubble masonry is a type of masonry construction that uses rough, irregularly-shaped stones set in mortar to build walls. This type of construction is commonly used for foundations, retaining walls,
and boundary walls due to its strength and durability.
When constructing rubble masonry walls, it is important to calculate the quantity of stone required to ensure the structural integrity of the wall. The quantity of stone required depends on the
volume of the wall, which is measured in cubic meters (m3).
In order to determine the quantity of stone required for 10 m3 of rubble masonry, the following factors need to be considered:
1. Size of the stones:
The size of the stones used in rubble masonry can vary greatly, from small pebbles to large rocks. The size of the stones will affect the volume required to fill 1 m3 of wall. Smaller stones will
require more pieces to fill a cubic meter, while larger stones will require fewer pieces.
2. Type of stone:
The type of stone used in rubble masonry can also impact the quantity required. Some types of stones, such as granite and basalt, are denser and heavier than others and will require fewer pieces to
fill the same volume.
3. Thickness of the wall:
The thickness of the wall will also play a role in determining the quantity of stone required. A thicker wall will require more stone to fill 1 m3 compared to a thinner wall.
4. Mortar joints:
Mortar is used to fill the gaps between the stones in rubble masonry. The thickness of the mortar joints will also affect the quantity of stone required. A thin joint will require more stones to fill
1 m3 compared to a thicker joint.
Therefore, to determine the quantity of stone required for 10 m3 of rubble masonry, the first step is to estimate the volume of stone required for 1 m3 of wall. This can be done by using the average
size of the stones, the type of stone, and the thickness of the wall and joints. The quantity for 1 m3 of wall can then be multiplied by 10 to get the total quantity required for 10 m3 of wall.
In general, it is recommended to have approximately 30% more stone on hand than the calculated quantity to account for any cutting or shaping of the stones during construction.
In conclusion, the exact quantity of stone required for 10 m3 of rubble masonry will vary depending on the factors mentioned above. It is important to accurately estimate the volume of stone required
to ensure a strong and stable wall. Therefore, it is always recommended to consult with a professional engineer or mason for an accurate calculation.
In conclusion, calculating the quantity of cement and sand needed for 1:6 mortar is a crucial step in any construction project. By following the simple formula and understanding the factors that can
affect the quantity, one can accurately determine the amount of materials needed for their project. Additionally, it is important to consider the quality of the materials and to always have a slight
excess to account for any discrepancies. This knowledge will not only save time and money, but also ensure that the final result is strong and durable. With practice and careful measurement, anyone
can become proficient in calculating cement and sand quantity for 1:6 mortar. So, go ahead and use this knowledge to successfully complete your next construction project.
Leave a Comment | {"url":"https://civilstep.com/all-about-calculate-cement-sand-quantity-in-16-mortar-2/","timestamp":"2024-11-09T03:58:16Z","content_type":"text/html","content_length":"243596","record_id":"<urn:uuid:9a12ecdb-a3f8-4d48-801f-0b15961eb6c3>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00099.warc.gz"} |
Need help to programm j≠i
Hello all,
I want to programm a formulation: sum(t, sum(j∈N & j≠i, P(i,j,t))). The code is as follows:
i / 1,2,3 /
t / 1*24 /;
sum((t,j)$(t,j<>i), p(i,j,t))
However, the programm repor some errors:
148 Dimension different - The symbol is referenced with more/less indices as declared
Could someone can help me to correct it.
Your code snippet is obviously incomplete and if executed, it does not result in the error you describe.
• it does not declare p
• the sum() must be assigned to some other symbol
• the sum does not control i
• not sure what the t in the dollar condition is supposed to do
• j<>i is no valid GAMS syntax (if j and i are sets)
Maybe you want to do something like:
set i / 1,2,3 /
t / 1*24 /;
parameter p(i,j,t);
scalar x;
p(i,j,t) = uniformint(1,10);
x = sum((i,j,t)$(not sameas(i,j)), p(i,j,t));
display p,x;
I hope this helps!
Hi Fred,
Thank you for your prompt reply. It solves my problem perfectly.
Just as you said, I misinterpret the GAMS syntax 'j ≠i ’ as 'j<>i '.
Again, thank you! | {"url":"https://forum.gams.com/t/need-help-to-programm-j-i/2700","timestamp":"2024-11-07T17:15:36Z","content_type":"text/html","content_length":"17400","record_id":"<urn:uuid:e942cac7-8b31-463e-b3c0-d55f0947a890>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00367.warc.gz"} |
Math 1060Q — Precalculus (Fall 2020) | Math Courses
Math 1060Q — Precalculus (Fall 2020)
Course Coordinator (Storrs): Maksym Derevyagin
Description: Precalculus is a preparation for calculus which includes a thorough review of algebra. Emphasis will be on functions and their applications; in particular, polynomials, rational
functions, exponentials, logarithms, and trigonometric functions.
Prerequisites: A qualifying score of 17 on the mathematics placement exam (MPE), unless you began attending UConn prior to Fall 2016 (in which case it is still recommended). Students who fail to
achieve this minimum score are required to spend time on the preparatory and learning modules before re-taking the MPE or register for a lower level Mathematics course. Not open for credit to
students who have passed MATH 1120, 1125Q, or 1131Q. Students may not receive credit for this course and MATH 1040Q.
Course Materials:
Precalculus 9ed, by Larson (Required textbook). You can buy the textbook from the UConn Bookstore. You will need a WebAssign access code to access your homework assignments. When you buy the textbook
from the bookstore or the web site above, the WebAssign code will come bundled with the textbook. The unbundled version of the book (that is, the book without a WebAssign access code) can be obtained
in many places, but the cost of buying the unbundled text and the WebAssign code separately will almost certainly be significantly greater.
There are two ways to purchase the text and the WebAssign access code:
1. BEST VALUE: Get the text and WebAssign access code bundled together at the UConn Bookstore. Together, the book and online homework code (with e-book included) will cost $80.
2. NOT Recommended: Get the text separately from anywhere (or don’t get it at all, if you want to use the e-book exclusively), and buy the WebAssign access code when you access your homework
through HuskyCT. Using this option, the WebAssign code alone (including the e-book) will cost $100 and your code (and e-book) will last only one semester. The first option lets you use your access
code for the life of the edition of the textbook.
There is only ONE way to register for and access WebAssign. Once the semester officially begins, simply log into your HuskyCT account (lms.uconn.edu), navigate to the page for Math 1060Q, then
follow the link on the left hand side for WebAssign Homework. When logging into WebAssign (through HuskyCT), do not use Internet Explorer or Safari. Use Firefox or Chrome.
You will have two weeks of free access to WebAssign, which includes the e-book, so you can get started right away in case you need some time to arrange to buy textbook with the access code.
Calculators: The use of calculators IS NOT permitted on exams or quizzes.
Homework and WebAssign:
WebAssign Homework: To access the WebAssign homework you will have to go through HuksyCT single sign-on. In the left sidebar of your Math 1060Q HuskyCT page, you will find a link to do your homework
using WebAssign. There will usually be 2-3 homework assignments per week (one for each textbook section covered). Each assignment will be made available on WebAssign several days before the section
is covered in class. The due date for each assignment will generally be two or three days after the material is covered in class.
You will get five attempts for each question that is not multiple choice; the exact number of attempts for multiple choice questions will depend on the number of choices. This means True/False
questions and questions with two answer choices have only one attempt – choose carefully! After each attempt, you will be told whether your answer is correct or not. If you are not able to get the
correct answer after your initial attempts, we recommend that before your final attempt, you seek help from your instructor, the Q-Center, a tutor, or another student.
When accessing your online homework, use Firefox or Chrome as your browser; there are problems that can occur if you use Internet Explorer or Safari. See the document here for tips on using
WebAssign, including entering answers and finding useful settings.
Your lowest WebAssign score will be dropped at the end of the semester.
Written Homework: Throughout the semester, you will have written textbook assignments and worksheets due in class. You can find the assignments under the “Learning Activities” link above, and the due
dates are listed in the course outline. There will be no late work accepted except in extenuating circumstances with proper documentation.
Your lowest written assignment score will be dropped at the end of the semester.
Quizzes and Exams:
Quizzes will be video proctored by your instructor and will be given approximately weekly, beginning with the second week of classes. There are no make-up quizzes – if you miss a quiz, then you will
receive a score of zero unless you have proper documentation of an extenuating circumstance. Your lowest quiz score will be dropped at the end of the semester.
The midterm exams will be video proctored by your instructor and will be held during class periods on October 6 and November 10; the 2-hour common final exam will be held during finals week, December
14-20. More information will listed under exam info as the dates approach. Make-up exams are not provided, except in extenuating circumstances. You should put the exam dates into your calendar and
plan to attend. If you are unable to make it to an exam or quiz for any reason you must notify your instructor as early as possible. All approved makeup exams and quizzes must be completed within one
week of the original quiz/exam date.
Per University policy, all requests to reschedule or make up the final exam must be submitted to the Dean of Students for approval. Please note that vacations, previously purchased tickets or
reservations, and social events are not viable excuses for missing a final exam. If you think that your situation warrants permission to reschedule, please contact the Dean of Students Office with
any questions. Thank you in advance for your cooperation.
Calculators: The use of calculators IS NOT permitted on exams or quizzes.
WebAssign Homework 10%
Quizzes 15%
Written Worksheets and Homework 10%
Exam 1: (Tuesday, October 6) 20%
Exam 2: (Tuesday, November 10) 20%
Final Exam: (Exact time and date provided by University) 25%
Software/Technical Requirements:
The software/technical requirements for this course include:
• Computer with microphone, speaker, and camera for live video/audio interaction
• HuskyCT/Blackboard
• WebEx
• WebAssign
• Apps and cameras for taking pictures of your work and uploading them to HuskyCT in the PDF format
• Dedicated access to high-speed internet with a minimum speed of 1.5 Mbps (4 Mbps or higher is recommended).
NOTE: This course has NOT been designed for use with mobile devices.
Some Tips:
1. If you’ve taken precalculus before, be warned — this course is harder. We will likely cover more material, and it will be more in-depth, than what you’ve done before.
2. Don’t miss class! Each day builds on the previous days, so if you miss class, you get behind very quickly. If you do get sick or have to miss class, talk to your classmates and instructor to
catch up before the next class. The outline for the course that is available using the link above will provide you with information about the topics to be covered in lecture.
3. Watch videos, do worksheets (see the Learning Activities tab of this page), and use the preparatory and learning modules for the MPE (they are free to use via HuskyCT). The videos and worksheets
cover some of the most difficult and/or critical concepts.
4. Seek help early if you think you may need it! Some great resources for help are your instructor’s office hours, the Q-Center, a tutor, and other students.
Academic Integrity and Honesty:
This course expects all students to act in accordance with the Guidelines for Academic Integrity at the University of Connecticut. In mathematics, this means that all work that you turn in should be
written up independently by you, in your own words, and should represent your honest understanding of the material. On exams and quizzes, it should be noted in particular that this means you must not
consult any sources or materials: neighbors’ papers, calculators, and any notes, books, or electronic devices are off-limits. If you have questions about academic integrity or intellectual property,
you should consult with your instructor. Additionally, consult UConn’s guidelines for academic integrity.
Students with Disabilities:
The University of Connecticut is committed to protecting the rights of individuals with disabilities and assuring that the learning environment is accessible. If you anticipate or experience
physical or academic barriers based on disability or pregnancy, please let me know immediately so that we can discuss options. Students who require accommodations should contact the Center for
Students with Disabilities, Wilbur Cross Building Room 204, (860) 486-2020, or http://csd.uconn.edu/. | {"url":"https://courses.math.uconn.edu/fall2020/math-1060-2/","timestamp":"2024-11-08T09:32:16Z","content_type":"text/html","content_length":"62955","record_id":"<urn:uuid:a1d6ca78-9920-4894-8869-ff2b69a62616>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00718.warc.gz"} |
BISE Faisalabad Board 9th Class Math Paper 2023
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papers from the past 5 years on any topic of your choice through Jano.Com.Pk. It is one of the best platforms, where you can easily find all possible ways to prepare for the exam and other past
papers, such as the past papers of the Lahore International Committee on Mathematics, etc. In these ways, you can even easily get full marks in the exam. There is a multiple-choice question, most of
which is repeat in the annual exam, most of which found in the online math MCQ test. Lahore Board of Directors 2nd grade Mathematics past papers is that you will have the opportunity to get the
highest score by remembering the calculations for each question.
Download In Pdf 9th Class Math past Papers Faisalabad Board
9th Class Math Past Papers 2023 BISE Faisalabad Board Download – View
Mathematics Past Papers Faisalabad Board 9th Class
Lahore Board under the BISE Board is considered the largest education committee in the province and the country, as thousands of students appear under this committee every year. Therefore, 10th grade
students related to the Lahore Committee and regions related to the Committee can easily obtain past papers in Mathematics on the Jano.Com.Pk website. Past papers on Mathematics topics in category 10
are available on the Jano.Com.Pk website. These past papers are very helpful to students because they can prepare and practice their writing skills and knowledge by using these papers. BISE (Board of
Intermediate and Secondary Education) conducts examinations for the 10th grade in March and April each year. The examination is conducted under an authority.
Exam Pattern for BISE Faisalabad Board 9th Class Math Paper 2023
Before diving into the preparation tips, it is important to understand the exam pattern for the BISE Faisalabad Board 9th Class Math Paper 2023. The exam will consist of two parts: objective and
When will the BISE Faisalabad Board 9th Class Math Paper 2023 be conducted?
The BISE Faisalabad Board 9th Class Math Paper 2023 will be conducted in the month of March/April.
What is the duration of the BISE Faisalabad Board 9th Class Math Paper 2023?
The total time duration for the BISE Faisalabad Board 9th Class Math Paper 2023 will be 3 hours.
What is the passing marks for the BISE Faisalabad Board 9th Class Math Paper 2023?
The passing marks for the BISE Faisalabad Board 9th class math paper is 33%. | {"url":"https://jano.com.pk/bise-faisalabad-board-9th-class-math-paper-2023","timestamp":"2024-11-13T10:58:19Z","content_type":"text/html","content_length":"53892","record_id":"<urn:uuid:a1511448-00fa-4650-8a3b-00223395b144>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00819.warc.gz"} |
Constrained min/max value of an array
Another puzzle courtesy the Daily Coding Problem email list.
Daily Coding Problem: Problem #1578 [Hard]
This problem was asked by Facebook.
Given an array of numbers of length N, find both the minimum and maximum using less than 2 * (N - 2) comparisons.
Github repo for my solution.
Development journey
These interview problems often have a catch for length zero or one inputs. This one seems to have that catch:
N 2(N - 2)
0 -4
1 -2
What is a program going to do, give back some comparisons for 0-length arrays? For single element arrays, min and max values are just that of the only element, but a program still has to make a
comparison of the array length to 1.
For arrays of length 2, zero comparisons doesn’t make sense. At least 1 comparison has to be done to check if array[0] is less than array[1]. Should the interview candidate point this out?
I’ll assume arrays of more than 2 elements.
The obvious algorithm sets variables minimum and maximum to the smallest number and largest number representable by whatever type the variables possess, then plows through all of the array one
element at a time comparing variables minimum and maximum to each element. That would require 2N comparisons, one to check if array element is less than current value of minimum one to check if array
element is greater than current value of maximum.
The first change to make to get to the goal is to observe that you can set minimum and maximum to the value of the first element of the array instead of smallest and largest representable values. You
can start comparing with the second element of the array. You’ve avoided 2 comparisons, so exactly 2(N - 1) comparisons.
The second change to make to get to the goal is to observe that variables min and max always contain values that at least equate. If some array value compares less than min value, it won’t compare
greater than max value. The program can skip some comparisons: if an array value is less than min, the program does not have to compare that array value to max.
The problem is not all arrangements of values in the array cause the skip to happen. To make less than 2(N-2) comparisons, at least 2 array values have to compare less than min.
Third change: eliminate 1 comparison by checking 1st and 2nd array values, and setting min and max appropriately, then iterating through array values starting with the third element. The program
makes one comparison to set min and max, but it skips two comparisons, so 2(N - 2) + 1 comparisons total.
Here’s where my ideas ran out, and I googled for an answer. I didn’t figure it out on my own, I guess I would not get a job at Facebook.
I did try one of the less abstruse solutions from a stackexchange. My implementation of that algorithm does indeed use 3N/2 - 2 comparisons.
The stackexchange algorithm steps through the array by 2, but uses an index and index+1 values, which usually causes developers to fall prey to a very common off-by-one bug. When the array is an odd
length, stepping through the array by 2 leaves one unexamined element at the end of the array. As in most cases, this involves some unavoidable code repetition to examine that final value. Any
maintainer will have to figure that out, and make changes in two places.
Interview analysis
This strikes me as a dumb interview question: it won’t let an interviewer see much of a candidate’s coding style or ability. There’s a fairly simple flow-of-control for the obvious, 2N comparison
program, and none of the “optimizations” change it much. The interviewer won’t see much programming.
If the candidate realizes that one comparison gets made every time through a for-loop, the candidate may spend time trying to puzzle out a recursive solution, to avoid the comparisons made in the
for-loop test. Candidates might decide they’re spending N (or so) comparisons on the for-loop, then trying to figure out a probably impossible N-comparison method of setting min and max variables.
Some bright spark candidate might think of a way to set min and max with an esoteric combination of bitwise and arithmetic operators, but without an explicit less-than operator. This subverts the
interviewer’s ability to judge the candidate’s programming ability.
Beyond that, even the 3N/2-2 comparison solution is still O(N). Any of these “optimizations” affect run time only very slightly, and only in some arrangements of array values.
If you’re interested in your developers creating “business logic” as quickly as possible, you really don’t want them to do this kind of “optimization”. If you’re interested in human-readable programs
that don’t create cognitive load on the readers, and cause fewer bugs on modification, you don’t want this kind of “optimization”.
This is not a good question for an interview. | {"url":"https://bruceediger.com/posts/minmaxarray/","timestamp":"2024-11-06T09:23:32Z","content_type":"text/html","content_length":"13167","record_id":"<urn:uuid:0ce7656c-4f3b-43bb-9957-81899f6b9d7b>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00671.warc.gz"} |
CSCI 5521 Homework 0 Linear regression solution
Problem 1: Linear regression learns a linear function of feature variables X to fit the
responses y. In this problem, you will derive the closed-form solution for linear regression
1. The standard linear regression can be formulated as solving a least square problem
||Xw − y||2
where X ∈ R
n×m (n ≥ m) represents the feature matrix, y ∈ R
represents the
response vector and w ∈ R
is the vector variable of the linear coefficients. This is
a convex objective function of w. Derive the optimal w by setting the derivative of the
function wrt w to zero to minimize the objective function.
2. In practice, a L2-norm regularizer is often introduced with the least squares, called
Ridge Regression, to overcome ill-posed problems where the hessian matrix is not
positive definite. The objective function of ridge regression is defined as
||Xw − y||2 + λ||w||2
where λ > 0. This objective function is strictly convex. Derive the solution of the
ridge regression problem to find the optimal w.
Problem 2: Consider a coin with probability of heads equal to Pr(H) = p and probability
of tails Pr(T) = 1 − p. You toss it 5 times and get outcomes H,H,T,T,H.
1. What is the probability of observing the sequence H,H,T,T,H in five tosses. Also give
the formula for the natural logarithm of this probability. Your formulas should be a
function of p.
2. You have a box containing exactly 2 coins, one fair with p = 1/2 and one biased with
p = 2/3. You choose one of these two coins at random with equal probability, toss it
5 times and get the outcome H,H,T,T,H.
(a) Give the joint probability that the coin chosen was the fair coin (p = 1/2) and
the outcome was H,H,T,T,H.
(b) Give the joint probability that the coin chosen was the biased coin (p = 2/3) and
the outcome was H,H,T,T,H.
Instructors: Rui Kuang (kuang@umn.edu). TAs: Tianci Song (song0309@umn.edu)
3. What should the bias p = Pr(H) be to maximize the probability of observing H,H,T,T,H,
and what is the corresponding probability of observing H,H,T,T,H (i.e., what is the
maximum likelihood estimate for p), assuming p were unknown? Show the derivation.
Hint: maximize the log of the function.
Problem 3: Below is the pseudo-code of perceptron algorithm for binary classification,
1 w = w0.
2 Do Iterate until convergence
3 For each sample (x
, rt
4 If(< w, xt > rt ≤ 0)
5 w = w + r
1. Implement the perceptron algorithm and test it on the provided data. To begin, load
the file data1.mat into Python. X ∈ R
is the feature matrix of 40 samples in 2
dimensions and y ∈ R
is the label vector (+1/−1). Initialize w to be vector [1, −1].
Visualize all the samples (use different colors for different classes) and plot the decision
boundary defined by the initial w.
Now, run your perceptron algorithm on the given data. How many iterations does
it take to converge? Plot the decision boundary defined by the w returned by the
perceptron program.
Hint: To load data in MATLAB format, you may consider to use the function loadmat,
which is included in the io module of scipy package. To visualize the samples you
could use the function scatter(), which is included in the pyplot module of matplotlib
package. Plotting the boundary is equivalent to plot the line w
T x = 0. Therefore, you
could first generate a vector a to be your x-axis, then compute the y-axis vector b as
b = −
w(2) . Once the plot is generated you could use xlim() and ylim() functions in
pyplot module of matplotlib package to make sure your axes are in a right range.
When you are done your plots will look like the following figures:
2. In previous question the samples from the two classes are linearly separable. Now let’s
look at a linearly non-separable case. Load the file data2.mat into Python and run
your perceptron algorithm with w = [1, −1]. Can the perceptron algorithm converge?
Explain why. To improve the algorithm, we can introduce a ”soft” linear classifier to
tolerate errors. It turns out we can solve the following LP problem:
subject to r
T x
) ≥ 1 − ξ
t ≥ 0.
Here ξ
is the error which needs to be minimized. The function linprog() from optimize
module of scipy package can be used for the problem. Now, run the following Python
code to solve the LP problem on data2.mat.
import numpy a s np
from s ci p y . o p timi z e import l i n p r o g
m, n = np . shape (X)
X = np . h s ta c k ( (X, np . one s ( (m, 1 ) ) ) )
n = n + 1
f = np . append ( np . z e r o s ( n ) , np . one s (m) )
A1 = np . h s ta c k ( (X∗np . t i l e ( y , ( n , 1 ) ) . T, np . eye (m) ) )
A2 = np . h s ta c k ( ( np . z e r o s ( (m, n ) ) , np . eye (m) ) )
A = −np . v s ta c k ( ( A1, A2 ) )
b = np . append(−np . one s (m) , np . z e r o s (m) )
x = l i n p r o g ( f ,A, b )
w = x [ ’ x ’ ] [ 0 : n ]
Apply this algorithm to data2.mat, visualize the data and plot the boundary by the
w returned by LP.
• Things to submit:
1. hw0 sol.pdf: a document contains all the derivations of Problem 1&2 and the
three plots asked by Problem 3.
2. MyPerceptron.py: a Python function defined as MyPerceptron(X, y, w0) with w
and step returned, where X is the feature matrix, y is a label vector and w0 is
the initialization of the parameter w. In the output, w is the parameter found
by perceptron and step represents the number of steps the algorithm takes to
converge. The function should also display the plot of samples and boundary.
Note that only numpy pacakge and functions mentioned above are allowed to use
in this assignment.
3. Zip all the files into a single zipped file and name it as your name.
• Submit: All material must be submitted electronically via Canvas. This homework
will not be graded but required as a proof of satisfying the prerequisites for taking the | {"url":"https://jarviscodinghub.com/assignment/csci-5521-homework-0-linear-regression-solution/","timestamp":"2024-11-03T18:15:52Z","content_type":"text/html","content_length":"111123","record_id":"<urn:uuid:cd678d6b-bd4b-4396-bd78-2549e19aafa8>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00803.warc.gz"} |
Power Factor Correction - Polytechnic Hub
Capacitive Power Factor correction is applied to circuits which include induction motors as a means of reducing the inductive component of the current and thereby reduce the losses in the supply.
There should be no effect on the operation of the motor itself.
An induction motor draws current from the supply, that is made up of resistive components and inductive components.
The resistive components are:
1. Load current
2. Loss current
The inductive components are:
1. Le akage reactance
2. Magnetizing current
The current due to the leakage reactance is dependant on the total current drawn by the motor, but the magnetizing current is independent of the load on the motor.
The magnetizing current will typically be between 20% and 60% of the rated full load current of the motor. The magnetizing current is the current that establishes the flux in the iron and is very
necessary if the motor is going to operate.
The magnetizing current does not actually contribute to the actual work output of the motor. It is the catalyst that allows the motor to work properly. The magnetizing current and the leakage
reactance can be considered passenger components of current that will not affect the power drawn by the motor, but will contribute to the power dissipated in the supply and distribution system.
Take for example a motor with a current draw of 100 Amps and a power factor of 0.75 The resistive component of the current is 75 Amps and this is what the KWh meter measures. The higher current will
result in an increase in the distribution losses of (100 x 100) /(75 x 75) = 1.777 or a 78% increase in the supply losses.
In the interest of reducing the losses in the distribution system, power factor correction is added to neutralize a portion of the magnetizing current of the motor.
Typically, the corrected power factor will be 0.92 – 0.95 Some power retailers offer incentives for operating with a power factor of better than 0.9, while others penalize consumers with a poor power
factor. There are many ways that this is mete red, but the net result is that in order to reduce wasted energy in the distribution system, the consumer will be encouraged to apply power factor
Power factor correction is achieved by the addition of capacitors in parallel with the connected motor circuits and can be applied at the starter, or applied at the switchboard or distribution panel.
The resulting capacitive current is leading current and is used to cancel the lagging inductive current flowing from the supply.
Methods of Power Factor Correction
Capacitors connected at each starter and controlled by each starter is known as “Static Power Factor Correction” while capacitors connected at a distribution board and controlled independently from
the individual starters is known as “Bulk Correction”.
Bulk Correction
The Power factor of the total current supplied to the distribution board is monitored by a controller which then switches capacitor banks In a fashion to maintain a power factor better than a preset
limit. (Typically 0.95)
Ideally, the power factor should be as close to unity (Power factor of “1”) as possible. There is no problem with bulk correction operating at unity.
Static Correction
As a large proportion of the inductive or lagging current on the supply is due to the magnetizing current of induction motors, it is easy to correct each individual motor by connecting the correction
capacitors to the motor starters.
With static correction, it is important that the capacitive current is less than the inductive magnetizing current of the induction motor.
In many installations employing static power factor correction, the correction capacitors are connected directly in parallel with the motor windings. When the motor is off-line, the capacitors are
also off-line. When the motor is connected to the supply, the capacitors are also connected providing correction at all times that the motor is connected to the supply. This removes the requirement
for any expensive power factor monitoring and control equipment.
In this situation, the capacitors remain connected to the motor terminals as the motor slows down. An induction motor, while connected to the supply, is driven by a rotating magnetic field in the
stator which induces current into the rotor.
When the motor is disconnected from the supply, there is for a period of time, a magnetic field associated with the rotor. As the motor decelerates, it generates voltage out its terminals at a
frequency which is related to it’s speed.
The capacitors connected across the motor terminals, form a resonant circuit with the motor inductance.
If the motor is critically corrected, (corrected to a power factor of 1.0) the inductive reactance equals the capacitive reactance at the line frequency and therefore the resonant frequency is equal
to the line frequency.
If the motor is over corrected, the resonant frequency will be below the line frequency.
If the frequency of the voltage generated by the decelerating motor passes throu gh the resonant frequency of the corrected motor, there will be high currents and voltages around the motor/capacitor
circuit. This can result in sever damage to the capacitors and motor. It is imperative that motors are never over corrected or critically corrected when static correction is employed.
Static power factor correction should provide capacitive current equal to 80% of the magnetizing current, which is essentially the open shaft current of the motor.
The magnetizing current for induction motors can vary considerably. Typically, magnetizing currents for large two pole machines can be as low as 20% of the rated current of the motor while smaller
low speed motors can have a magnetizing current as high as 60% of the rated full load current of the motor.
It is not practical to use a “Standard table” for the correction of induction motors giving optimum correction on all motors. Tables result in under correction on most motors but can result in over
correction in some cases. Where the open shaft current can not be measured, and the magnetizing current is not quoted, an approximate level for the maximum correction that can be applied can be
calculated from the half load characteristics of the motor.
It is dangerous to base correction on the full load characteristics of the motor as in some cases, motors can exhibit a high leakage reactance and correction to 0.95 at full load will result in over
correction under no load, or disconnected conditions.
Static correction is commonly applied by using one contactor to control both the motor and the capacitors. It is better practice to use two contactors, one for the motor and one for the capacitors.
Where one contactor is employed, it should be up sized for the capacitive load. The use of a second contactor eliminates the problems of resonance between the motor and the capacitors. | {"url":"https://www.polytechnichub.com/power-factor-correction/","timestamp":"2024-11-14T08:44:29Z","content_type":"text/html","content_length":"52906","record_id":"<urn:uuid:0a214987-59ee-49b9-b15f-5d055cafa597>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00325.warc.gz"} |
Re: st: tricks to speed up -xtmelogit-
Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: st: tricks to speed up -xtmelogit-
From Jeph Herrin <[email protected]>
To [email protected]
Subject Re: st: tricks to speed up -xtmelogit-
Date Wed, 22 Dec 2010 08:33:15 -0500
There's not really a way to reduce the variables; I took several
thousands of medical diagnosis and procedure codes, classified them
into 500 related groups (according to a published scheme) and
then reduced the 500 to 100 by running 500 linear probability
models and keeping those with the biggest abs(t-value). (100
was arbitrary but turns out that only 1 procedure group and 1
diagnostic group had P > 0.05 in the final model.) The other
variables are categories of age, sex, length of stay, number
of admissions, etc. Anyway, if I drop any of the x's, I would hvae
to re-estimate the "main" model, which means 3 weeks wasted :)
In the end, I have decided to look at only one additional model
and to give it 3 weeks.
On 12/21/2010 3:28 PM, Sergiy Radyakin wrote:
Hi, Jeph,
very interesting problem. Are the 150 variables related? E.g. are these 150 a
single group of dummies? Or are they all independent: height/age/gender?
With 6mln observations there is some chance you will have some duplicates,
which may give you a possibility to reduce your sample a bit (just adjust the
Given the rareness of your outcome taking a simple subsample may yield just
a few positives in the subsample. May I suggest also to consider taking all
positives and a random subsample of negatives, estimate the candidate and
then run the full sample on that?
Finally, this command is not in the MP report, but have you investigated how
does it perform as N(CPU) grows?
Best regards, Sergiy
On Tue, Dec 21, 2010 at 2:15 PM, Jeph Herrin<[email protected]> wrote:
I am trying to estimate a series of models using 6 million observations;
the observations are nested within 3000 groups, and the dichotomous
outcome is somewhat rare, occurring in about 0.5% of observations.
There are about 150 independent variables, and so my basic model looks
like this:
. xtmelogit Y x1-x150 || group:
This took approximately 3 weeks to converge on a high end machine
(3.2GHz, Intel Core i7, 24GB RAM). I saved the estimation result
. est save main
but now would like to estimate some related models of the form
. xtmelogit Y x1-x150 z1 z2 || group:
and would like to think I can shave some considerable time off the
estimation using the prior information available. I tried
. est use main
. matrix b = e(b)
. xtmelogit Y x1-x150 z1 z2 || group:, from(b) refineopts(iterate(0))
but this gave me an error that the likelihood was flat and nothing
proceed. So I've thought of some other approaches, but am not sure what
I expect to be most efficient, and would prefer not to spend weeks
figuring it out.
One idea was to use a sample, estimate the big model, and then use
that as a starting point:
. est use main
. matrix b = e(b)
. gen byte sample = (uniform()*1000)<1
. xtmelogit Y x1-x150 z1 z2 if sample || group:, from(b)
. matrix b = e(b)
. xtmelogit Y x1-x150 z1 z2 || group:, from(b) refineopts(iterate(0))
Another was to first use Laplace iteration, and start with that result:
. est use main
. matrix b = e(b)
. xtmelogit Y x1-x150 z1 z2 if sample || group:, from(b) laplace
. matrix b = e(b)
. xtmelogit Y x1-x150 z1 z2 || group:, from(b) refineopts(iterate(0))
I'd appreciate any insight into which of these approaches might shave
a meaningful amount of time off of getting the final estimates, or if
there is another that I could try.
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"https://www.stata.com/statalist/archive/2010-12/msg00862.html","timestamp":"2024-11-12T18:55:24Z","content_type":"text/html","content_length":"13858","record_id":"<urn:uuid:af4f6550-6e95-48e0-aa10-d0fe9a652fc4>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00081.warc.gz"} |
xintsession – Interactive computing sessions (fractions, floating points, polynomials)
This package provides support for interactive computing sessions with etex (or pdftex) executed on the command line, on the basis of the xintexpr and polexpr packages.
Once xintsession is loaded, ε-TeX becomes an interactive computing software capable of executing arbitrary precision calculations, or exact calculations with arbitrarily big fractions. It can also
manipulate polynomials as algebraic entities.
Numerical variables and functions can be defined during the session, and each evaluation result is stored in automatically labeled variables. A file is automatically created storing inputs and
Sources /macros/plain/contrib/xintsession
Version 0.4alpha 2021-11-01
Licenses The LaTeX Project Public License 1.3c
Copyright 2021 Jean-François Burnol
Maintainer Jean-François Burnol
Contained in TeXLive as xintsession
MiKTeX as xintsession
Topics Maths
Download the contents of this package in one zip archive (9.5k).
Community Comments
Maybe you are interested in the following packages as well. | {"url":"https://www.ctan.org/pkg/xintsession","timestamp":"2024-11-02T04:48:26Z","content_type":"text/html","content_length":"17101","record_id":"<urn:uuid:d0610ac9-300c-469e-991a-e15fa4970f42>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00264.warc.gz"} |
Area or Grid Model Worksheets
Area models have evolved from arrays and have become increasingly popular in recent times as they can be used in conjunction with algebra tiles.
Arrays are predominantly used at KS1 and 2 when introducing multiplication, finding fact families, linking multiplication and division, and working with factors. Extend beyond this and they can be
developed to provide an area model for multiplication of large integers or with the addition of algebraic terms.
Both arrays and area models illustrate the distributive and commutative properties of multiplication. Area models are especially effective in promoting a deeper conceptual understanding, fitting well
within the ‘representation and structure’ aspect of maths mastery. By allowing students to visualise abstract concepts, area models help in building fluency and problem-solving skills, essential
components of a strong mastery approach.
Here at Cazoom, we have developed resources using the grid method for multiplication at KS2 and 3, including multiplication of fractions. In addition to this, worksheets combining the area model and
algebra tiles are used to expand single, double and even triple brackets, factorise into single brackets, factorise quadratic expressions, and complete the square.
We’ve included a printable page of algebra tiles for classroom or home use. Explore our growing collection of resources, suitable for KS3 and KS4 learners, and discover new ways to integrate area
models into your lessons or tutorials.
Printable PDF Area Model Worksheets with Answers
Area models are a powerful tool for deepening students’ understanding of multiplication and algebraic concepts! Cazoom Maths offers a wide variety of printable PDF worksheets that utilise area
models, ideal for KS3 to KS4 learners. These resources help students visualise and explore multiplication, algebraic terms, and more, reinforcing their understanding of the distributive and
commutative properties.
Our worksheets guide students through the progression from basic arrays to more advanced area models, supporting the development of fluency and problem-solving skills. Each worksheet includes
separate answers and is available in an easy-to-download PDF format, perfect for both classroom and home learning.
Area models are a visual representation used to illustrate multiplication and algebraic expressions. Here’s why they’re effective:
• Visual Learning: They break down complex calculations into manageable visual chunks, making abstract ideas more accessible.
• Versatile Applications: Area models can be used for multiplying large numbers, expanding brackets, factorising expressions, and more.
• Concrete to Abstract: They offer a bridge from concrete arrays in KS1 and 2 to more abstract algebraic concepts in KS3 and 4.
How are Area Models Used in Learning?
• Understanding basic multiplication through arrays in KS1 and 2.
• Applying area models to large number multiplication and fractions.
• Combining area models with algebra tiles to expand brackets, factorise expressions, and complete the square.
Importance of Area Models
Area models are crucial for promoting a deeper conceptual understanding in maths. Here’s why they are beneficial:
• They fit seamlessly into the ‘representation and structure’ aspect of maths mastery.
• Area models provide a clear visual breakdown of problems, making complex concepts more accessible.
• They help students develop a solid grasp of multiplication and algebra, essential for mastering higher-level maths.
Applications of Area Models in Real Life
Area models go beyond the classroom, providing skills that can be applied in everyday situations:
• Construction and Design: Area calculations are vital in construction projects and interior design.
• Budgeting: Area models can help in visualising and planning budgets, breaking them into manageable parts.
• Problem Solving: Understanding the structure of problems enhances logical thinking and decision-making. | {"url":"https://www.cazoommaths.com/maths-worksheets/area-or-grid-model-worksheets/","timestamp":"2024-11-04T16:58:59Z","content_type":"text/html","content_length":"479391","record_id":"<urn:uuid:5c0401e5-8d84-453b-abd8-4269d541ed72>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00112.warc.gz"} |
QuantMath Alternatives - Rust Computation | LibHunt
Financial maths library for risk-neutral pricing and risk.
Programming language: Rust
License: MIT License
QuantMath alternatives and similar packages
Based on the "Computation" category.
Alternatively, view QuantMath alternatives based on common mentions on social networks and blogs.
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Add another 'Computation' Package
Financial maths library for risk-neutral pricing and risk
Some quant math libraries are really just a collection of pricing formulae. This hopes to be that (see the math module) but also much more. This library is intended to be plugged into the risk and
pricing infrastructure in an investment bank or hedge fund. This does not preclude the use of the library for academic work, but will introduce a level of real-world messiness that is often missing
from academia.
Lifecycle and Flows
QuantMath is responsible for managing the lifecycle of financial instruments (see the instrument module). As products make payments or as dividends go ex, this results in the instrument splitting
into multiple flows. Nothing ever disappears. The SecDB library at Goldman Sachs is famous for taking this philosophy to extremes, but QuantMath is at least capable of the same level of discipline.
It is vital to correctly model when flows go ex -- when they cease to be part of the value of this instrument, and are owned by the counterparty (even if not yet settled).
Most investment banks skirt around the issues of settlement. What is the value of an equity? Is it the spot price, or is it the spot price discounted from the date when the payment for the equity
would actually be received (e.g. T+2). QuantMath intends to allow rigour in settlement, in the premium payment, the payoff, and in the underlying hedges.
Risk and Scenarios
QuantMath is designed to make it easy to reuse calculations that have not changed as a result of a risk bump. For example, if you have an exotic product with multiple equity underlyings, bumping one
of those underlyings only results in the affected Monte-Carlo paths being reevaluated. In my experience this is a vital optimisation, and QuantMath makes it possible from the lowest levels such as
bootstrapping dividend curves, to the highest, such as reuse of Longstaff-Schwarz optimisation.
Recursive Instruments
It is common to build instruments recursively. A basket contains composite or quanto underliers, then a dynamic index maintains the basket -- finally an exotic product is written with the dynamic
index as an underlying. The library must therefore manage this sort of recursive product, whether valuing in Monte-Carlo, analytically or via a finite difference engine.
Simplicity, Orthogonality and Encapsulation
The library must be easy for quants to work in and for IT systems to work with. Adding a new risk, instrument or model should normally mean changes to only one file (and maybe a list of files in
mod.rs). The interface to IT should be data-driven, so IT do not need to rebuild every time an instrument or model is added. Models, instruments and risks should be orthogonal, so any can be used
with any (subject to sensible mathematical restrictions). If things go wrong, it should be easy to debug just QuantMath, without having to debug the containing IT system. This means that QuantMath
should be runnable purely from serialised state, such as JSON files.
The Architecture
The library has a strict hierarchy of modules. Ideally there should be no backward dependencies, such that the library could be split into a separate crate for each module. If you are looking at the
library for the first time, it may be best to start from the top level (Facade). Starting at the top level, the modules are:
This is the interface that IT systems talk to. It is data-driven, so adding a new product or model should not affect the IT systems at all.
A pricer evaluates an instrument given market data and a choice of model. We currently have two pricers: Monte-Carlo, which evaluates instruments by averaging across many random paths; self-pricer,
which relies on instruments knowing how to price themselves. I hope to add at least one finite difference backward-induction engine.
Models of how we expect prices to change in the future. All are stochastic, but some have stochastic volatility or rates. Examples of models are BGM (Brace Gatarek Musiela), Black, Heston.
Defines how market data can be bumped, and manages the dependencies when this happens. This contain definitions of risk reports, such as Delta, Gamma, Vega and Volga for all underliers matching some
Defines financial products, indices, assets and currencies. Anything that has a price. Some instruments know how to price themselves (basically, any instrument where the price is well-defined and not
model-dependent -- remember this module is lower than models). Some instruments know how to price themselves in a Monte-Carlo framework, given paths of their underliers.
The input market data; vol surfaces, dividends, spot prices, yield curves etc. Also defines bumps to these data items. Most risks are calculated by bumping these inputs.
Low level mathematical formulae, from the Black-Scholes formula to interpolation and quadrature. Where possible, we use functionality from well-established crates in Rust, such as ndarray and statrs,
so this is mainly quant-specific maths.
Dates are very important for financial maths software. We use explicit dates everywhere rather than year-fractions, which is essential for handling settlement correctly. This module also handles date
arithmetic, such as date rules and day counts.
Very low-level functionality, such as the definition of the Error struct, and required extensions to serde, such as dedup (deduplication of nodes in directed acyclic graph) and factories (using
tagged_serde to handle polymorphic nodes in serialization and deserialization). | {"url":"https://rust.libhunt.com/quantmath-alternatives","timestamp":"2024-11-07T20:19:56Z","content_type":"text/html","content_length":"59404","record_id":"<urn:uuid:1fcb33b9-a677-4783-b9e0-28d8b448de8e>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00878.warc.gz"} |
Difference between convex and non-convex cost function; what does it mean when a cost function is non-convex?
A convex function: given any two points on the curve there will be no intersection with any other points, for non convex function there will be at least one intersection.
In terms of cost function with a convex type you are always guaranteed to have a global minimum, whilst for a non convex only local minima. | {"url":"https://discuss.boardinfinity.com/t/difference-between-convex-and-non-convex-cost-function-what-does-it-mean-when-a-cost-function-is-non-convex/2858","timestamp":"2024-11-07T13:17:21Z","content_type":"text/html","content_length":"16862","record_id":"<urn:uuid:ac104cb1-471e-4c87-85a3-32b72f7a7dfd>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00808.warc.gz"} |
enVision Math Common Core Grade 2 Answer Key Topic 12 Measuring Length
Practice with the help of enVision Math Common Core Grade 2 Answer Key Topic 12 Measuring Length regularly and improve your accuracy in solving questions.
enVision Math Common Core 2nd Grade Answers Key Topic 12 Measuring Length
Essential Question:
What are ways to measure length?
enVision STEM Project: Growing and Measuring
Find Out Grow bean plants. Give them numbers. Put some in sunlight. Put some in a dark place. Water some of the plants. Do not water some of the plants. See how the plants in each group grow.
Journal: Make a Book Show what you learn in a book. In your book, also:
• Tell if plants need sunlight and water to grow.
• Find plants to measure. Draw pictures of the plants. Tell how tall each plant is.
Review What You Know
Question 1.
Draw a line under the bat to show its length.
Question 2.
School is getting out. Circle a.m. or p.m.
4 pm.
The time shown on the clock is 4 pm.
Question 3.
Draw clock hands to show quarter past 10.
The clock shows 10:25.
Estimating and Measuring Length
Question 4.
Use snap cubes.
Estimate the length.
about ________ cubes
Measure the length.
about ________ cubes
The estimated length is about 2 cubes.
The measuring length is about 2 cubes.
The estimated length of the above image is two cubes and the length of the measurement is two cubes.
Question 5.
Use snap cubes.
Estimate the length.
about ________ cubes
Measure the length.
about ________ cubes
The estimated length is about 3 cubes.
The measuring length is about 3 cubes.
The estimated length of the above image is three cubes and the length of the measurement is three cubes.
Skip Counting
Question 6.
Write the missing numbers.
5, 10, _____, 20, _____
210, 220, _____, 240
400, ____, 600, 700
5, 10, 15, 20, 25.
210, 220, 230, 240.
400, 500, 600, 700.
The difference between the first number and the second number is 10 – 5 which is 5, so the missing numbers are 15,25.
5, 10, 15, 20, 25.
The difference between the second number and the first number is 220 – 210 which is 10, so the missing number is 230.
210, 220, 230, 240.
The difference between the last number and the last before the number is 700 – 600 which is 100, so the missing numbers are 400, 500, 600, 700.
Pick a Project
PROJECT 12A
How are measurements used to design clothing?
Project: Measure Feet and Create Sock Designs
Here, we will place a tape measure on the floor and we will take a position back of the heel at the zero marks on the tape. Then we will measure to the longest toe. And with the foot circumstances,
we will make sure the sock wearer’s foot is placed on the ground. So we can get an accurate measurement.
PROJECT 12B
What units should you use to measure longer distances?
Project: Compare the Measurements of Sports Fields
We will use kilometers to measure the longer distances.
PROJECT 12C
What can help you remember different measurement facts?
Project: Create a Booklet of Measurement Rhymes
PROJECT 12D
How long or how tall are some animals and insects?
Project: Make a Poster of Snake Lengths
Lesson 12.1 Estimating Length
Solve & Share
Your thumb is about 1 inch long. Use your thumb to find three objects that are each about 1 inch long. Draw the objects. From your elbow to your fingers is about I foot long. Use this part of your
arm to find three objects that are each about I foot long. Draw the objects.
I can … estimate the length of an object by relating the length of the object to a measurement I know.
Visual Learning Bridge
Convince Me! Is your height closer to 4 feet or 4 yards? How do you know?
Yes, my height is closer to 4 feet or 4 yards.
Given that some objects and their estimated lengths. So, here we have 3 + 3 + 3 + 3 which is 12 feet. As 1 yard is 3 feet, so 12 feet is 4 yards. So my height is closer to 4 feet or 4 yards.
Guided Practice
Write the name and length of an object whose length you know. Then use that object to help you estimate the length of the object shown.
Pencil is 6 inch long.
Book is 1 foot long.
Desk is 2 feet long.
The pencil is about 6 inch long and my book is 1 foot long and my desk is about 2 feet long.
Independent Practice
Write the name and length of an object whose length you know. te Then use that object to help you estimate the length of the object shown.
My thumb is 1 inch long.
My hand is about 6 inches long.
My book is 1 foot long.
My chair is about 3 feet high.
My thumb is 1 inch long.
My hand is about 6 inches long.
My book is 1 foot long.
My chair is about 3 feet high.
Question 5.
Higher Order Thinking Would you estimate the distance from your classroom to the principal’s office in inches, feet, or yards? How many units? Explain.
Given that the estimated distance from your classroom to the principal’s office is a gaint step is about a yard. So we would use yards because it covers more length at one time.
Problem Solving
Solve each problem.
Question 6.
Vocabulary Complete each sentence using one of the words below.
An ___________ measurement is a good guess.
The height of a kitchen window is about 1 ________.
A small paper clip is about 1 _______ long.
An estimated measurement is a good guess.
The height of a kitchen window is about 1 yard.
A small paper clip is about 1 inch long.
Question 7.
Reasoning Joy and Kyle estimate the height of their classroom. Joy estimates the height to be 10 feet. Kyle estimates the height to be 10 yards. Who has the better estimate? Explain.
Given that Joy and Kyle estimate the height of their classroom and Joy estimates the height to be 10 feet and Kyle estimates the height to be 10 yards. So Joy has the better estimation as we know
that 1 yard is 3 feet and 10 yards is 30 feet which is big.
Question 8.
Higher Order Thinking A city wants to build a bridge over a river. Should they plan out an exact length of the bridge, or is an estimated length good enough? Explain.
Here, we need an exact length because then cars or people would tall so if it was too short
Question 9.
Assessment Practice Draw a line from each estimate to a matching object.
The book is about 1 foot and the stamp is about 1 inch and the umbrella is about 1 yard.
In the above image, we can see that the book is about 1 foot and the stamp is about 1 inch and the umbrella is about 1 yard.
Lesson 12.2 Measure with Inches
Solve & Share
The orange square is I inch long. How can you use I inch squares to find the length of the line in inches? Measure the line and explain.
I can … estimate measures and use a ruler to measure length and height to the nearest inch.
The line is about _____ inches long.
The line is about 5 inches long.
Visual Learning Bridge
Convince Me! Use a ruler to measure. What classroom objects are about 12 inches long?
The classroom objects which are about 12 inches long is books, plates, sink, Chromebook and tissue box.
Guided Practice
Estimate the height or length of each real object. Then use a ruler to measure to the nearest inch.
Question 1.
Question 2.
The estimated length of the pencil case is about 8 inches and the measured value is about 7 inches.
Independent Practice
Estimate the height or length of each real object. Then use a ruler to measure. Compare your estimate and measurement.
Question 3.
The estimated width of the bag is about 6 inches and the measured value is about 12 inches.
Question 4.
The estimated length of the paintbrush is about 5 inches and the measured value is about 8 inches.
Question 5.
The estimated height of a cup is about 4 inches and the measured value is about 6 inches.
Question 6.
The estimated length of a crayon box is about 4 inches and the measured value is about 5 inches.
Higher Order Thinking Think about how to use a ruler to solve each problem.
Question 7.
Jason measures an object. The object is just shorter than the halfway mark between 8 and 9 on his inch ruler. How long is the object?
about _______ inches
The length of the object is about 8 inches.
Given that Jason measures an object and the object is just shorter than the halfway mark between 8 and 9 on his inch ruler, so the length of the object is about 8 inches.
Question 8.
Gina measures an object. The object is just longer than the halfway mark between 9 and 10 on her inch ruler. How long is the object?
about _______ inches
The length of the object is about 10 inches.
Given that Gina measures an object and the object is just longer than the halfway mark between 9 and 10 on her inch ruler. So the length of the object is about 10 inches.
Problem Solving
Solve each problem.
Question 9.
Explain Pam says that each cherry is about 1 inch wide. Is she correct? Explain.
Yes, she is correct.
As Pam says that each cherry is about 1 inch wide, yes she is correct. Because it is longer than the halfway mark.
Question 10.
Vocabulary Find an object in the classroom that measures about 6 inches. Write a sentence to describe the object. Use these words.
Estimated: 8 inches.
Actual: 10 inches.
The estimated length of the chrome book is about 8 inches and its actual length of the chrome book is about 10 inches.
Question 11.
Higher Order Thinking Explain how to use an inch ruler to measure the length of an object.
I would hold up the ruler next to the object to measure.
Question 12.
Assessment Practice Use a ruler. About how many inches long are the two stamps together?
A. 4 inches
B. 3 inches
C. 2 inches
D. 1 inch
2 inches.
The length of the two stamps together is 2 inches.
Lesson 12.3 Inches, Feet, and Yards
Solve & Share
Which objects in the classroom are about 1 inch, about I foot, and about 1 yard long? Show these objects below.
I can … estimate measures and use tools to measure the length and height of objects to the nearest inch, foot, and yard.
about 1 inch
about 1 foot
about 1 yard
Visual Learning Bridge
Convince Me! Would you measure the length of a school building in inches or yards? Why?
The length of the school building is measured in yards because they are large units.
Guided Practice
Match each object with a reasonable estimate of its length.
Independent Practice
Estimate the length of each real object. Choose a ruler, yardstick, or measuring tape to measure. Write the tool you used.
Question 4.
3 feet.
The length of the pencil is about 3 feet.
Question 5.
6 feet.
The length of the table is about 6 feet.
Question 6.
11 feet.
The length of the door is about 11 feet.
Question 7.
Higher Order Thinking Explain how you could use a foot ruler to measure the length of a room in feet.
We will start at one end and then we will make a mark for one foot. Then we will move the ruler to the mark and measure a second foot then we will keep going to the end of the room.
Problem Solving
Solve each problem.
Question 8.
Generalize Circle the real object that is about 4 feet in length.
The real object that is about 4 feet in length is a cycle.
Question 9.
Number Sense Explain how to use a yardstick to measure the length of an object. Estimate the length of your classroom in yards. Then measure.
A yardstick is used to measure larger objects. A yardstick also has inches on it.
Question 10.
Higher Order Thinking Find an object in the classroom that you estimate measures about 2 feet. Draw the object.
What tool would you use to measure it? Explain why you chose the tool you did.
The object in the classroom is bookshelves and I would use a book to measure because they are 1 foot.
Question 11.
Assessment Practice Jon sets two of the same real objects next to each other. Together, they have a length of about 4 feet. Which is the object Jon uses?
In the above objects which are about 4 feet is table.
Lesson 12.4 Measure Length Using Different Customary Units
Solve & Share
Choose an object. Measure your object in feet. Then measure it in inches. Do you need more units of feet or inches to measure your object? Why?
I can … estimate and measure the length and height of objects in inches, feet, and yards.
about ______ feet long
about ________ inches long
about 1 foot long,
about 12 inches long.
Here, my laptop is about 1 foot long and it is about 12 inches long. I need more units of inches because it took 12 of them.
Visual Learning Bridge
Convince Me! Juan measures the height of a wall in his room. He lines the wall with one-foot rulers. He could find this height with yardsticks. Would Juan need more rulers or yardsticks? Explain.
Guided Practice
Measure each real object using different units. Circle the unit you use more of to measure each object.
Question 1.
about ______ feet
about _______ yards
I use more units of:
about 3 feet.
about 1 yard.
The measurement of the object in the above image is about 3 feet and about 1 yard.
Question 2.
about ______ feet
about ______ inches
I use more units of:
about 3 feet.
about 36 inches.
The measurement of the object in the above image is about 3 feet and about 36 inches.
Independent Practice
Measure each real object using different units. Circle the unit you use fewer of to measure each object.
Question 3.
about _______ inches
about _______ feet
I use fewer units of:
about 3 feet.
about 36 inches.
The measurement of the object in the above image is about 3 feet and about 36 inches.
Question 4.
about _______ feet
about _______ yards
I use fewer units of:
about 6 feet.
about 2 yards.
The measurement of the object in the above image is about 6 feet and about 2 yards.
Question 5.
about _______ feet
about _______ yards
I use fewer units of:
about 3 feet.
about 1 yard.
The measurement of the object in the above image is about 3 feet and about 1 yard.
Number Sense Circle the best estimate for the length of each object.
Question 6.
About how long is a key?
2 inches
2 feet
2 yards
Which tool would you use to measure the length of a key?
And the length is 2 inches.
The tool would we use to measure the length of a key is a Ruler and the length of a key is 2 inches.
Question 7.
About how long is a suitcase?
2 inches
2 feet
2 yards
Which tool would you use to measure the length of a suitcase?
And the length is 2 feet.
The tool would we use to measure the length of a suitcase is a Ruler and the length of a key is 2 feet
Problem Solving
Solve each problem.
Question 8.
Use Tools Measure the length of an object in your classroom using two different units.
Object: ___________
about ________ about _________
Which unit did you use more of?
Circle which tool you used.
measuring tape
Question 9.
Higher Order Thinking Andrew wants to measure the length of a football field. Should he use feet or yards to measure it? Which tool should he use? Explain.
Andrew uses yards.
Here, Andrew wants to measure the length of a football field, so he uses yards to measure. Because the football field is very long. So he uses yards.
Question 10.
Assessment Practice Which unit would you need the fewest of to measure the length of the table?
A. inches
B. feet
C. yards
D. all the same
The table is measured in yards.
The fewest of to measure the length of the table is yards, as the length of the table will be one yard.
Question 11.
Assessment Practice Which is the best estimate for the length of a vegetable garden?
A. about 5 inches
B. about I foot
C. about 5 yards
D. about 20 inches
About 5 yards.
The best estimate for the length of a vegetable garden is about 5 yards.
Lesson 12.5 Measure with Centimeters
Solve & Share
The green cube is I centimeter long. How can you use I centimeter cubes to find the length of the line in centimeters? Measure the line and explain.
I can … estimate measures and use a ruler to measure length and height to the nearest centimeter.
The line is about _______ centimeters long.
The line is about 7 centimeters long.
The line is about 7centimeters long.
Visual Learning Bridge
Convince Mel Explain how you know the length of the paper clip above is about 3 centimeters long.
We can see where it is measured on the ruler.
Guided Practice
Estimate the height or length of each real object. Use a ruler to measure to the nearest centimeter.
Question 1.
The estimated length is 15 centimeters and the actual measure is 18 centimeters.
The estimated length of the stapler is about 15 centimeters and the actual measurement is 18 centimeters.
Question 2.
The estimated length is 22 centimeters and the actual measure is 25 centimeters.
The estimated length of the stapler is about 22 centimeters and the actual measurement is 25 centimeters.
Independent Practice
Estimate the width, height, or length of each real object. Then use a ruler to measure. Compare your estimate and measurement.
Question 3.
width of a shoelace
The estimated width is 3 centimeters and the actual measure is 2 centimeters.
The estimated width of the shoelace is about 3 centimeters and the actual measurement is 2 centimeters.
Question 4.
width of a chair
The estimated width is 25 centimeters and the actual measure is 35 centimeters.
The estimated width of the shoelace is about 25 centimeters and the actual measurement is 35 centimeters.
Question 5.
length of a pencil
The estimated length is 15 centimeters and the actual measure is 12 centimeters.
The estimated length of a pencil is about 15 centimeters and the actual measurement is 12 centimeters.
Question 6.
height of scissors
The estimated height is 10 centimeters and the actual measure is 13 centimeters.
The estimated height of the scissors is about 10 centimeters and the actual measurement is 13 centimeters.
Higher Order Thinking Explain whether each estimate is reasonable or not.
Question 7.
Josh estimated that the length of his reading book is about 6 centimeters.
No, Josh estimated length is not correct.
No, Josh estimated length is not correct. Because 6 centimeters is too small.
Question 8.
Shae estimated that the height of her desk is about 10 centimeters.
No, Shae estimated height is not correct.
No, Josh estimated height is not correct. Because 10 centimeters is too small.
Problem Solving
Solve each problem.
Question 9.
Vocabulary Find an object that is about 10 centimeters long. Write a sentence to describe your object using these words.
Here, we have taken the eraser as an object which is about 10 centimeters long. The estimated length of the eraser is about 12 centimeters.
Question 10.
Look for Patterns Nick wants to put another pen end to end with this one. About how long would the two pens be together?
about _________ centimeters
14 centimeters.
The length of the two pens be together would be 7 + 7 which is 14 centimeters.
Question 11.
Higher Order Thinking Paul says that a toothbrush is about 19 centimeters long. Sarah says it is about 50 centimeters long. Who is correct? Explain.
Paul is correct.
Paul is correct, we know that by measuring. So Sarah is wrong.
Question 12.
Assessment Practice Mary measures the length of her eraser to the nearest centimeter. What is the length of her eraser to the nearest centimeter?
__________ centimeters
4 centimeters.
The length of her eraser to the nearest centimeter is centimeters.
Lesson 12.6 Centimeters and Meters
Solve & Share
Which objects in the classroom are about 3 centimeters long? Which objects are about I meter long?
Show these objects below.
I can … estimate measures and use a ruler, a meter stick, or a tape measure to measure length and height to the nearest centimeter or meter.
about 3 centimeters
about 1 meter
Sharpener lid, pencil, and a marker cap.
The objects that are about 3 centimeters long are a sharpener lid, pencil, and a marker cap.
Visual Learning Bridge
Convince Me! Would you measure the length of a house in centimeters or meters? Why?
The length of the house is measured in meters because meters are a bigger unit of measurement.
Guided Practice
Match each object with a reasonable estimate of its length.
Independent Practice
Estimate the length or height of each real object shown. Then choose a tool and measure. Compare your estimate and measurement.
Question 5.
The estimated length is 5 cm and the measured value is 6 cm.
The estimated length of the above object is about 5 cm and the measured value is about 6 cm. We have used a ruler to measure the object.
Question 6.
The estimated height is 1 m and the measured value is 1 m.
The estimated height of the above object is about 1 m and the measured value is about 1 m. We have used a meterstick to measure the object.
Question 7.
The estimated length is 8 cm and the measured value is 12 cm.
The estimated length of the above object is about 8 cm and the measured value is about 12 cm. We have used a ruler to measure the object.
Question 8.
The estimated height is 1 m and the measured value is 1 m.
The estimated height of the above object is about 1 m and the measured value is about 1 m. We have used a meterstick to measure the object.
Question 9.
Tom uses a meter stick to measure the length of a fence. He places the meter stick 5 times on the fence to measure from one end to the other end. How long is the fence?
________ meters
5 meters.
Here, Tom uses a meter stick to measure the length of a fence and he places the meter stick 5 times on the fence to measure from one end to the other end, so the length of the fence is 5 meters.
Question 10.
Higher Order Thinking Debbie says that her doll is about 30 meters long. Do you think this is a good estimate? Why or why not?
No, Debbie’s estimation is incorrect.
No, Debbie’s estimation is incorrect because it is too big.
Problem Solving
Solve each problem.
Question 11.
Be Precise Choose an object to measure. Use metric units. Estimate first and then measure.
Draw the object and write your estimate and measurement. Was your estimate reasonable?
Question 12.
Circle the real object that would be about 2 meters long.
Object 1 is about 2 meters longer than object 2.
Here, Object 1 is about 2 meters longer than object 2.
Question 13.
Higher Order Thinking Each side of a place-value cube is I centimeter long. Use a place-value cube to draw a 5-centimeter ruler.
Here, we have placed I centimeter long 5 cubes.
Question 14.
Assessment Practice Choose an appropriate tool. Measure each line. Which lines are at least 6 centimeters long? Choose all that apply.
Option A and Option B are at least 6 centimeters long.
Lesson 12.7 Measure Length Using Different Metric Units
Solve & Share
Measure this pencil in inches. Then measure it again in centimeters. Which measurement has more units?
I can … measure the length and height of objects using different metric units.
about _________ inches
about _________ centimeters
Which has more units? ___________
about 6 inches,
about 15 centimeters.
Centimeter has more units.
The length of the pencil in inches is about 6 inches and in centimeters, it is about 15 centimeters. So, here centimeter has more units.
Visual Learning Bridge
Convince Me! Tina measures the length of a room with centimeter rulers. She could find this length with meter sticks. Would Tina need fewer rulers or meter sticks? Explain.
Here, centimeters are more because they are smaller.
Guided Practice
Measure each real object using different units. Circle the unit you use more of to measure each object.
Question 1.
about _________ centimeters
about _________ meters
I use more units of:
about 200 centimeters,
about 2 meters.
Centimeters are more.
The above image is about 200 centimeters and about 2 meters. Here, centimeters are more.
Question 2.
about _________ centimeters
about _________ meters
I use more units of:
about 300 centimeters,
about 3 meters.
Centimeters are more.
The above image is about 300 centimeters and about 3 meters. Here, centimeters are more.
Independent Practice
Measure each real object using different units. Circle the unit you use fewer of to measure each object.
Question 3.
about _________ meters
about _________ centimeters
I use fewer units of:
about 200 centimeters,
about 2 meters.
Meters are fewer.
The above image is about 200 centimeters and about 2 meters. Here, meters are fewer.
Question 4.
about _________ meters
about _________ centimeters
I use fewer units of:
about 300 centimeters,
about 3 meters.
Meters are fewer.
The above image is about 300 centimeters and about 3 meters. Here, meters are fewer.
Question 5.
Higher Order Thinking Jay measured the height of his bedroom in both centimeters and meters. Did he use fewer units of centimeters or meters? Explain.
Here, Jay used fewer meters because they are larger units of measurement.
Problem Solving
Solve each problem.
Question 6.
Explain If you had to measure the length of the hallway outside of your classroom, would you use centimeters or meters? Explain.
We would use meters because they are larger.
Question 7.
Higher Order Thinking A meter stick is about 39 inches long. Is a meter longer or shorter than a yard? Explain.
Given that a meter stick is about 39 inches long. Now we will check that meter is longer or shorter than a yard. As we know that 1 foot is 12 inches and for 3 feet it will be 12 × 3 which is 36
inches. And here meter stick is 39 inches, so the meter is longer.
Question 8.
Assessment Practice Estimate the length of a baseball bat in centimeters and meters.
_________ centimeters
________ meters
Which number must be greater? Explain.
Question 9.
Assessment Practice Tina measures the length of a jump rope using different units. How will her measurements compare?
Lesson 12.8 Compare Lengths
Solve & Share
Circle two paths. Estimate which one is longer. How can you check if your estimate is correct?
I can … tell how much longer one object is than another.
Estimate: The _________ path is longer.
Measure: The __________ path is longer.
Estimate: The blue path is longer.
Measure: The blue path is longer.
Here, the blue path is longer.
Visual Learning Bridge
Convince Me! How can you find the length of a path that is not straight?
Here, we will move the ruler and then we need to add both parts.
Guided Practice
Estimate the length of each path. Then use a centimeter ruler to measure each path.
Question 1.
Question 2.
Path B
Estimate: about _______ cm
Measure: about ________ cm
Estimate: about 7 cm.
Measure: about 6 cm.
The estimated length of path B is 7 cm and the measured value is about 6 cm.
Question 3.
Which path is longer?
Path A.
From the above paths, Path A is longer than Path B.
Question 4.
How much longer?
about _________ cm longer
About 4 cm longer.
In the above, we can see that Path A measured value is about 10 cm and Path B measured value is about 6 cm. So the Path A is 10 – 6 which is 4 cm longer than Path B.
Independent Practice
Estimate the length of each path. Then use a centimeter ruler to measure each path. Compare your estimate and measurement.
Question 5.
Estimate: about ________ centimeters
Measure: about __________ centimeters
Estimate: about 10 cm.
Measure: about 11 cm.
The estimated length of path C is 10 cm and the measured value is about 11 cm.
Question 6.
Path D
Estimate: about ________ centimeters
Measure: about __________ centimeters
Estimate: about 8 cm.
Measure: about 7 cm.
The estimated length of path D is 8 cm and the measured value is about 7 cm.
Question 7.
Which path is longer?
Path C.
From the above paths, Path C is longer than Path D.
Question 8.
How much longer?
about ________ centimeters longer
About 4 cm longer.
In the above, we can see that Path C measured value is about 11 cm and Path D measured value is about 7 cm. So the Path A is 11 – 7 which is 4 cm longer than Path D.
Higher Order Thinking Think about the length of each object. Circle the best estimate of its length.
Question 9.
a key
about l cm
about 6 cm
about 20 cm
About 6 cm.
Here, the key is about 6 cm.
Question 10.
a pen
about 2 cm
about 4 cm
about 15 cm
Use your estimates to complete:
A pen is about ________ cm than a ________.
A pen is about 9 cm than a key.
In the above, we can see that the key is about 6 cm long and the pen is about 15 cm long. So a pen is 15 – 6 which is 9 cm longer than a key.
Problem Solving
Solve each problem.
Question 11.
Explain A path has two parts. The total length of the path is 12 cm. If one part is 8 cm, how long is the other part? Explain.
________ centimeters
About 4 cm long.
Given that the total length of the path is 12 cm and one part is 8 cm, so the other part is 12 – 8 which is 4 cm long.
Question 12.
Higher Order Thinking Draw a path with two parts. Measure the length to the nearest centimeter. Write an equation to show the length of your path.
The equation is 7 cm + 3 cm = 10 cm.
Here, we have taken a path with two parts Path A and Path B with 7 cm and 3 cm. So the equation will be 7 + 3 = 10 cm.
Question 13.
Higher Order Thinking Beth drew a picture of a bike path. Use tools. Measure the length of the path below. Write the total length.
about __________ centimeters
The length is about 12 cm.
The length of Beth’s path is 5 + 2 + 5 which is 12 cm.
Question 14.
Assessment Practice Measure each path in centimeters.
How much longer is Path A than Path B? Show your work.
Path A is 2 cm longer than Path B.
In the above image, the length of Path A is 14 cm and the length of Path B is 12 cm. So Path A is 14 – 12 = 2 cm longer than Path B.
Lesson 12.9 Problem Solving
Solve & Share
Zeke measures the snake and says it is about 4 inches long. Jay says it is about 5 inches long. Who measures the snake more precisely? Measure and explain.
I can … choose tools, units, and methods that help me be precise when I measure.
Length: __________
The length of the snake is 5 inches long.
Here, Jay measures it more precisely because I measured to check his answer. The length of the snake is 5 inches long.
Thinking Habits
Which unit of measure will I use? Is my work precise?
Visual Learning Bridge
Convince Me! How does using a string help Anna use precision to measure the worm?
Here, the string helps to measure the curves.
Guided Practice
Question 1.
Bev measures the crayon and says it is 5 centimeters long. Is her work precise? Explain.
Yes, Bev’s work is precise.
Yes, Bev’s work is precise. Because we can see the measurement.
Independent Practice
Solve each problem.
Question 2.
Steve uses centimeter cubes. He says the pencil is 9 centimeters long. Is his work precise? Explain.
No, Steve’s work is not precise.
No, Steve’s work is not precise. Because he left the spaces between the centimeter cubes.
Question 3.
Use a centimeter ruler to measure the pencil yourself. How long is the pencil? Explain what you did to make sure your work is precise.
The pencil is 11 cm long.
It is 11 cm long and here we have used a ruler to measure.
Question 4.
Find the difference in the lengths of the paths in inches. Is your answer precise? Explain.
The difference between the two paths is 3.
The length of Path A is 3 inches and the length of Path B is 6 inches. So the difference between the two paths is 6 – 3 which is 3.
Problem Solving
Performance Task
Shoestring Katie lost a shoestring. The shoestring has the same length as the shoestring at the right. What is the length of the shoestring Katie lost?
Question 5.
Make Sense Estimate the length of the shoestring in the picture. Explain how your estimate helps you measure.
The length of the shoelace is 15 cm.
The estimation of each part of the shoelace is to be 5 cm. So the total will be 15 cm.
Question 6.
Use Tools What tools can you use to measure the shoestring? Explain.
Here, we can use a string to measure the shoestring.
Question 7.
Be Precise Measure the shoestring. Tell why your work is precise.
Here, we have measured 30 cm with the string and then we measure the string on the ruler.
Topic 12 Fluency Practice Activity
Find a Match
Find a partner. Point to a clue. Read the clue.
Look below the clues to find a match. Write the clue letter in the box next to the match. Find a match for every clue.
I can … add and subtract within 100.
A. The sum is between 47 and 53.
B. The difference equals 56 – 20.
C. The sum equals 100.
D. The difference equals 79 – 27.
E. The difference is between 25 and 35.
F. The sum equals 41 + 56.
G. The difference is less than 20.
H. The sum equals 26 + 19.
A. The sum is between 47 and 53 is 100 which is 56 + 44 is 100.
B. The difference equals 56 – 20 = 36 which is 60 – 24 is 36.
C. The sum equals 100 which is 56 + 44 is 100.
D. The difference equals 79 – 27 is 52 which is 81 – 29 is 52.
E. The difference is between 25 and 35 is 10.
F. The sum equals 41 + 56 is 97 which is 32 + 65 is 97.
G. The difference is less than 20 is 47 – 31 is 16.
H. The sum equals 26 + 19 is 45 which is 34 + 11 is 45.
Topic 12 Vocabulary Review
Understand Vocabulary
Word List
• centimeter (cm)
• estimate
• foot (ft)
• height
• inch (in.)
• length
• meter (m)
• nearest centimeter
• nearest inch
• yard (yd)
Question 1.
Circle the unit that has the greatest length.
Question 2.
Circle the unit that has the shortest length.
Question 3.
Cross out the unit you would NOT use to measure the length of a book.
Question 4.
Cross out the unit you would NOT use to measure the height of a house.
Estimate the length of each item.
Question 5.
Question 6.
paper clip
Question 7.
school desk
Use Vocabulary in Writing
Question 8.
Use words to tell how to find the height of a table. Use terms from the Word List.
Topic 12 Reteaching
Set A
There are 12 inches in I foot. There are 3 feet in I yard. You can use lengths of objects you know to estimate lengths of other objects.
Estimate the lengths of two classroom objects in feet. Name each object and write your estimate.
Question 1.
Object: __________
about ____ feet
Object: Window
About 6 feet.
The estimated value of the window is 6 feet and the sink is 2 feet.
Question 2.
Object: ____________
about ________ feet
Object: Sink
About 2 feet.
The estimated value of the sink is 2 feet.
Set B
You can measure the length of an object to the nearest inch.
The string is longer than halfway between 1 and 2.
So, use the greater number.
The string is about 2 inches.
Find objects like the ones shown. Use a ruler to measure their lengths.
Question 3.
about _______ inches
About 5 inches.
The length of the object is about 5 inches.
Question 4.
about _______ inches
About 3 inches.
The length of the object is about 3 inches.
Set C
The measure of the height of a window takes more units of feet than yards.
Measure the real object in inches and feet. Circle the unit you needed more of.
Question 5.
about ________ feet
about ______ yard
Feets are more.
The measurement of the object is about 3 feet and about 1 yard. Here, feets are more.
Set D
You can measure the length of an object to the nearest centimeter.
The paper clip is less than halfway between 3 and 4.
So, use the lesser number.
The paper clip is about 3 cm.
Find real objects like the ones shown. Use a ruler to measure their lengths.
Question 6.
about _________ cm
About 16 cm.
The measurement of the object is about 16 cm.
Question 7.
about _________ cm
About 7 cm.
The measurement of the object is about 7 cm.
Set E
Reteaching Continued
There are 100 centimeters in 1 meter.
Circle the picture of the real object that is about each length or height.
Question 8.
about 1 centimeter
The height of the pen is about 1 centimeter.
The height of the pen is about 1 centimeter.
Question 9.
about 1 meter
The length of the belt is about 1 meter.
The length of the belt is about 1 meter.
Set F
The measure of the height of this cart takes fewer units of meters than centimeters.
about 93 centimeters
about 1 meters
Measure the real object in centimeters and meters. Circle the unit you needed fewer of.
Question 10.
about ________ centimeters
about _______ meters
About 200 centimeters,
About 2 meters.
The length of the door is about 200 centimeters and about 2 meters. So it is fewer in meters.
Set G
Which path is longer? How much longer? Measure each part. Then add the lengths.
Subtract the lengths to compare.
6 – 5 = 1
The purple path is about 1 centimeter longer.
Use a centimeter ruler. Measure each path.
Question 11.
Red path: ________ centimeters
9 centimeters.
The red path is 9 centimeters.
Question 12.
Blue path: ________centimeters
7 centimeters.
The blue path is 7 centimeters.
Question 13.
Which path is longer?
How much longer is it?
about _________ centimeters longer
Red path.
The longer path is the red path and 9 – 7 = 2 centimeters longer.
Set H
Thinking Habits
Attend to Precision Which unit of measure will I use? Is my work precise?
Question 14.
Measure the length of the bottom of this page in feet and in inches.
about _______ ft
about ________ in.
about 1 ft
about 10 in.
The length of the bottom of this page in feet is about 1 foot and in inches, it is about 10 inches.
Question 15.
Which measure in Item 14 is more precise? Explain.
Here, 10 inches is more precise because it is the exact measurement.
Topic 12 Assessment Practice
Question 1.
Estimate. About how tall is the flower?
A. about 5 cm
B. about 10 cm
C. about 15 cm
D. about 1 meter
The flower is about 10 cm.
Question 2.
Draw a line from each estimate to a matching object.
about 1 yard
about 1 foot
about 1 inch
The shoe is about 1 foot.
The coin is about 1 inch.
The ribbon is about 1 yard.
Question 3.
Which units would you need the fewest of to measure the height of a fence?
A. inches
B. feet
C. yards
D. all the same
The fewest of to measure the height of a fence is yards.
Question 4.
Dan measures the width of a window with answer precise? Explain.
No, Dan’s answer was not precise.
No, Dan’s answer was not precise. Because he didn’t mention the measurements.
Question 5.
Use a ruler to measure each line to the nearest centimeter. Which are about 3 centimeters long? Choose all that apply.
Option 1.
The first option is about 3 centimeters long.
Question 6.
Use a ruler to measure the length of the pencil in inches. Which is the correct measurement?
A. 2 inches
B. 3 inches
C. 4 inches
D. 5 inches
The pencil is 3 inches long.
Question 7.
Circle the unit you need fewer of to measure the length of a kitchen.
Circle the unit you need fewer of to measure the length of a table.
Fewer meters.
Fewer yards.
The length of the kitchen is fewer meters.
The length of the table os fewer yards.
Question 8.
Use a ruler. Measure each path to the nearest inch.
Which path is longer? ________
How much longer? about ________ longer
Path B.
In the above image path B is longer and 4 – 3 = 1 inch longer.
Question 9.
Use a ruler. Measure the length of the marker to the nearest centimeter. How long is the marker?
A. 6 centimeters
B. 9 centimeters
C. 12 centimeters
D. 15 centimeters
9 centimeters.
The length of the marker to the nearest centimeter is 9 centimeters.
Question 10.
A path has two parts. The total length of the path is 15 cm. One part of the path is 9 cm long. How long is the other part?
A. 24 cm
B. 15 cm
C. 9 cm
D. 6 cm
Given that total length is 15 cm and one part is 9 cm long, so the length of the other part is 15 – 9 which is 6 cm.
Question 11.
Use a ruler. Measure each path to the nearest centimeter.
Which path is longer? ________
How much longer?
about _________ longer.
PAth A, 2 cm longer.
The longer path is PAth A and 8 – 6 which is 2 cm longer.
Question 12.
Measure the gray line with tools you need to be precise. Choose all the measurements that are precise.
☐ 4 centimeters
☐ 4 inches
☐ 10
☐ 10 centimeters
☐ 4
4 inches,
10 centimeters.
The gray line with tools we need to be precise the measurements is 4 inches and 10 centimeters.
Question 13.
Juan uses different units to measure a jump rope. Compare the measurements. Choose all that apply.
☐ more inches than feet
☐ fewer centimeters than meters
☐ fewer inches than feet
☐ more yards than feet
☐ more centimeters than meters
☐ fewer yards than feet
Option A and Option E.
The measurements that apply are more inches than feet and more centimeters than meters.
Question 14. What is the length of the crayon to the nearest centimeter? What would be the combined length of two crayons?
The crayon is _________ centimeters.
Two crayons would measure ________ centimeters.
The crayon is 4 centimeters.
Two crayons would measure 8 centimeters.
The length of the crayon to the nearest centimeter is 4 cm and the measurement of the two caryons will be 4 + 4 which is 8 cm.
Question 15.
Kim’s softball bat is 1 yard long. She uses 3 bats to measure the length of the classroom whiteboard. About how long is the whiteboard?
A. 3 inches
B. 3 feet
C. 1 yard
D. 3 yards
3 yards.
The length of the whiteboard is 3 yards.
Question 16.
Kevin measured the length of a car in inches and in feet. Why is the number of feet less than the number of inches?
Feets are larger than inches.
Here, feets are larger than inches so we need fewer of them.
Topic 12 Performance Task
Happy Hiking! The Torres family loves to hike. They use this map to plan their hiking trip.
Question 1.
Use a centimeter ruler.
Find the total length of the triangle hiking path shown on the map.
about ________ centimeters
Explain how you found the length.
Question 2.
Debbie Torres uses a backpack for hiking. She wants to measure its width. She wants to be precise. Should she use inches, feet, or yards? Explain your answer.
Question 3.
Daniel Torres estimates the height of his water bottle. Is his estimate reasonable? Explain.
Question 5.
On the hike, the Torres family sees a caterpillar. Use the picture below to answer the questions.
Part A
To be precise, which unit would you choose to measure the length of the caterpillar? Explain.
The unit we will choose to measure the length of the caterpillar is centimeter.
The unit we will choose to measure the length of the caterpillar is centimeter, because if we choose a unit that is too small then we will get a larger number that is harder to keep track.
Part B
Estimate and then measure the length of the caterpillar. Then explain how you measured.
Estimate: ___________
Measurement: ___________
Estimate: 5 centimeters.
Measurement: 4 centimeters.
The lenth of the caterpillar is 4 centimeters and it was measured by using ruler.
Question 4.
Maria Torres says that it would take more units of yards than feet to measure the height of the tower. Do you agree? Circle Yes or No. Explain.
No, Maria Torres is not correct.
No, Maria Torres is not correct. As the height of the tower is measured in feet.
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MCQ ON CONNECTIVE TISSUES - Biologysir
MCQ ON CONNECTIVE TISSUES OF STRUCTURAL ORGANISATION IN ANIMALS class 11 for NEET | CONNECTIVE TISSUES OF STRUCTURAL ORGANISATION IN ANIMALS class 11 MCQ | CONNECTIVE TISSUES OF STRUCTURAL
ORGANISATION IN ANIMALS with Answer | Check the below NCERT MCQ question for class 11 Biology chapter 7 based on CONNECTIVE TISSUES OF STRUCTURAL ORGANISATION IN ANIMALS with Answers.
MCQ on CONNECTIVE TISSUES OF STRUCTURAL ORGANISATION IN ANIMALS class 11 Biology with answers were prepared based on the latest pattern.We have provided class 11 Biology MCQs questions on CONNECTIVE
TISSUES OF STRUCTURAL ORGANISATION IN ANIMALS with Answers to help students understand the concept very well.
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MCQ on CONNECTIVE TISSUES OF STRUCTURAL ORGANISATION IN ANIMALS is useful for NEET / CSIR / UGC / CBSE / ICSE / AIIMS / EXAM / AFMC EXAM / STATE LEVEL MEDICAL EXAM 2022-23
The human body is composed of billions of cells to perform various functions. Connective tissues are most abundant and widely distributed in the body of complex animals.They are named connective
tissue because of their special functions of Linking and supporting others tissue organ of the body. They range from soft connective tissues to support types , which includes cartilage , bone ,
adipose and blood.
MCQ ON CONNECTIVE TISSUES OF STRUCTURAL ORGANISATION IN ANIMALS class 11 for NEET
1. The mast cells secrete
(a) haemoglobin
(b) heparin
(c) myoglobin
(d) histamine
Ans (d) histamine
2. Areolar connective tissue joins
(a) bones with bones
(b) fat body with muscles
(c) integument with muscles
(d) bones with muscles
Ans. (c) integument with muscles
3. Which of the following is a transparent tissue ?
(a) tendon
(b) hyaline cartilage
(c) fibrous cartilage
(d) all of these
Ans. (b) hyaline cartilage
4. Tendons and ligaments are specialized types of
(a) nervous system
(b) muscular tissue
(c) epithelial tissue
(d) fibrous connective tissue
Ans.(d) fibrous connective tissue
5. Which of the following is secreted by mast cells?
(a) histamine
(b) heparin
(c) serotonin
(d) all the above
Ans.(d) all the above
6. Tendons are specialized connective tissue made of
(a) bone with bones
(b) bones with muscles
(c) both a and b
(d) none of the above
Ans.(b) bones with muscles
7. Grave yard of RBCs
(a) gall bladder
(b) kidney
(c) spleen
(d) liver
Ans.(c) spleen
8. The Areolar tissue present beneath
(a) skin
(b) bones
(c) cartilage
(d) blood
Ans.(a) skin
9. The adipose tissue is specialized for storing
(a) carbohydrates
(b) fats
(c) protein
(d) all the above
Ans. (b) fats
10. Ligaments is dense connective tissue which attach
(a) bone to bone
(b) bone to muscles
(c) muscles to muscles
(d) muscles to bones
Ans. (a) bones to bones
11. Loose connective tissue has cells and fibres loosely arranged in a semi fluid ground substances . They are found in
(a) Areolar tissue
(b) adipose tissue
(c) both a and b
(d) tendons
Ans.(c) both a and b
12. Blood is
(a) epithelium tissue
(b) connective tissue
(c) muscular tissue
(d) all the above
Ans.(b) connective tissue
13. The cells of cartilage is
(a) ostein
(b) chondrocytes
(c) heamoglobin
(d) all the above
Ans. (b) chondrocytes
14. Tissue present in the tip of nose , outer ear joints, limbs and hands in adults.
(a) bones
(b) cartilage
(c) Areolar tissue
(d) hyaline
Ans.(b) cartilage
15. The bones cells are.
(a) chondrocytes
(b) myoglobin
(c) adipose tissue
(d) osteocytes
Ans. (d) osteocytes
16. The blood is a fluid connective tissue containing
(a) plasma and platelets
(b) RBCs
(c) WBCs
(d) all the above
Ans.(d) all the above
17. The intercellular materials of cartilage is
(a) solid
(b) pliable
(c) resist compression
(d) all the above
Ans.(d) all the above
18. The bone cells osteocytes are present in the space called
(a) lacunae
(b) tendons
(c) ligaments
(d) all the above
And.(a) lacunae
19. The connective tissue include
(a) bone
(b) adipose
(c) cartilage
(d) all the above
Ans. (d) all the above
20. Volkman’s canals occur in
(a) bones
(b) cartilage
(c) liver
(d) internal ear
Ans.(a) bones
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Classification of Overhead Transmission Lines
The important considerations in the design and operation of a transmission line are the determination of voltage drop, line losses and efficiency of transmission. These values are greatly influenced
by the line constants R, L and C of the transmission line. For instance the voltage drop in the line depends upon the values of above three line constants. Similarly, the resistance of transmission
line conductors is the most important cause of power loss in the line and determines the transmission efficiency. In this chapter, we shall develop formulas by which we can calculate voltage
regulation, line losses and efficiency of transmission lines. These formulas are important for two principal reasons. Firstly, they provide an opportunity to understand the effects of the parameters
of the line on bus voltages and the flow of power. Secondly, they help in developing an overall understanding of what is occuring on electric power system.
A transmission line has *three constants R, L and C distributed uniformly along the whole length of the line. The resistance and inductance form the series impedance. The capacitance existing between
conductors for 1-phase line or from a conductor to neutral for a 3-phase line forms a shunt path throughout the length of the line. Therefore, capacitance effects introduce complications in
transmission line calculations. Depending upon the manner in which capacitance is taken into account, the overhead transmission lines are classified as :
( i) Short transmission lines. When the length of an overhead transmission line is upto about 50 km and the line voltage is comparatively low (< 20 kV), it is usually considered as a short
transmission line. Due to smaller length and lower voltage, the capacitance effects are small and hence can be neglected. Therefore, while studying the performance of a short transmission line, only
resistance and inductance of the line are taken into account.
( ii) Medium transmission lines. When the length of an overhead transmission line is about 50-150 km and the line voltage is moderatly high (>20 kV < 100 kV), it is considered as a medium
transmission line. Due to sufficient length and voltage of the line, the capacitance effects are taken into account. For purposes of calculations, the distributed capacitance of the line is divided
and lumped in the form of condensers shunted across the line at one or more points.
( iii) Long transmission lines. When the length of an overhead transmission line is more than 150 km and line voltage is very high (> 100 kV), it is considered as a long transmission line. For the
treatment of such a line, the line constants are considered uniformly distributed over the whole length of the line and rigorous methods are employed for solution.
It may be emphasised here that exact solution of any tranmission line must consider the fact that the constants of the line are not lumped but are distributed unfiormly throughout the length of the
However, reasonable accuracy can be obtained by considering these constants as lumped for short and medium transmission lines.
Important Terms
While studying the performance of a transmission line, it is desirable to determine its voltage regulation and transmission efficiency. We shall explain these two terms in turn.
( i) Voltage regulation. When a transmission line is carrying current, there is a voltage drop in the line due to resistance and inductance of the line. The result is that receiving end voltage ( VR)
of the line is generally less than the sending end voltage ( VS ). This voltage drop ( Vs −V R) in the line is expressed as a percentage of receiving end voltage V and is called voltage regulation.
The difference in voltage at the receiving end of a transmission line **between conditions of no load and full load is called voltage regulationand is expressed as a percentage of the receiving end
( ii) Transmission efficiency. The power obtained at the receiving end of a transmission line is generally less than the sending end power due to losses in the line resistance.
The ratio of receiving end power to the sending end power of a transmission line is known as the transmission efficiency of the line | {"url":"https://www.brainkart.com/article/Classification-of-Overhead-Transmission-Lines_12367/","timestamp":"2024-11-11T00:27:32Z","content_type":"text/html","content_length":"36089","record_id":"<urn:uuid:3a52c368-a204-4616-ab8c-a6054ca74ad3>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00125.warc.gz"} |
ENTROPY - 2x4 Solutions
by Ester Mittermeier | 8. Feb, 2024 | PS IMAGO PRO, PS QUAESTIO PRO | 0 comments
By NATALIA GOLONKA (Predictive Solutions)
Entropy is a measure of disorder or uncertainty in a probability distribution.The concept was first introduced in 1854 by the physicist Rudolf Clausius, dealing with thermodynamic issues, and in this
sense the definition of entropy refers (in a nutshell) to the course of spontaneous processes. Today, the concept of entropy also has its application in statistics. This is because it allows us to
determine a kind of heterogeneity in a set of data.
When analysing qualitative variables, the measure of entropy gives us information about how much variability (and therefore ‘disorder’) is introduced by the individual variables. We get an indication
of the degree of randomness of individual qualitative variables. Entropy was introduced into the field of statistics on the basis of information theory, and one of its most commonly used measures is
the “Shannon entropy”.
If the entropy score is 0, it means that the variable takes on only one value. Such variables are called constants: they do not allow for any additional information.
The higher the entropy score, the greater the variety of categories that the variable takes. The result will depend on the number of unique categories of the analysed variable and the frequency of
their occurrence with respect to all observations. The value of entropy, on the other hand, does not depend on the size of the Dataset: if a variable has four categories, representing respectively
40%, 30%, 20% and 10% of the distribution, the entropy will have the same value, regardless of whether we have 10 or 10000 observations.
The resulting entropy can also be presented as a percentage compared to its maximum value. Reaching the largest possible value (100% of the maximum entropy value) tells us that all values of the
variable are equally likely. Such a case is when, in a given set, each observation has a different score (category) of the analysed variable, or all its categories are otherwise equal.
As the entropy value in itself is not easily interpretable, it is useful to present this result in percentage form. In so doing we then know that the maximum value is 100% and this is our reference
point for interpreting the result obtained.
Table 1. Entropy results for the analysed variable
Let’s look at the example shown in Table 1 where the entropy for the analysed variable of college degree completed is 1.28. This result is not close to 0, so there is certainly some disorder or
uncertainty in the distribution.
However, if we do not know the entropy results for other variables, in order, for example, to select for analysis those that will introduce the most variability, the value of 1.28 alone will tell us
little more.
Knowing that this result, compared to the maximum value, represents 92.3%, gives us additional information that there is significant variation in both the values of the variable and the Abundance of
the individual categories.
If we know the entropy measures for the individual quality variables, we can use them as an indicator of the importance of the variables. Such information can be used, for example, in modelling or
classification. The entropy value can be obtained using the Data Audit procedure in PS IMAGO PRO. All analysed qualitative variables are summarised in a single table, which allows you to quickly
compare the results obtained and select the most promising variables for your model.
But how exactly is entropy calculated? As we have already seen, the number of unique categories of a variable and their contribution to the variable’s distribution is key. The Shannon entropy
formula, the most commonly used formula for the entropy of a qualitative variable, is as follows:
H(X) = – Σ p(x) log_2 (p(x)),
where H(X) is the entropy of the variable X and p(x) is the probability of occurrence of the value of x.
The entropy that is obtained in the Data Audit procedure in PS IMAGO PRO (Table 1) is calculated based on the natural logarithm. Its formula is therefore as follows:
H(X) = – Σ p(x) ln(p(x)),
where H(X) is the entropy of the variable X and p(x) is the probability of occurrence of the value of x.
To make this formula more understandable, let us illustrate it with an example. We want to calculate the entropy for the qualitative variable college degree completed (X = degree). Our dataset
consists of 10 observations: 4 bachelors, 3 engineers, 2 masters and 1 doctorate (Figure 1). The probability of belonging to the bachelor’s category is 40%, so the first part of the equation will be
0.4×ln(0.4). Adding the subsequent elements of the formula for each category in an analogous way, we obtain the following equation:
H(studies) = – ((0.4×ln(0.4)) + (0.3× ln(0.3)) + (0.2×ln(0.2)) + (0.1×ln(0.1)))= 1.28
Figure 1. Distribution of categories of the variable college degree completed
The maximum entropy value is ln(n), where n is the number of categories of the qualitative variable in question. In our case, the qualitative variable has 4 categories, so the maximum entropy value
possible is ln(4) = 1.39.
If we wanted to present the entropy result compared to the maximum value, it would be rounded to 1.28 ÷ 1.39 = 92% (Table 1).
Entropy is a measure of uncertainty or disorder in a probability distribution. One of its most commonly used measures is Shannon entropy. A maximum value of entropy is reached when the distribution
is uniform, while a minimum value of 0 is reached when the distribution is deterministic.
However, bear in mind that the base of the logarithm by which entropy is calculated can take different values, e.g., 2, in the case of the Shannon formula, 10, or ℯ (Euler number, Neper number) when
the natural logarithm is used. If you want to compare results for several variables, you need to be sure that the base of the logarithm used in the formula is the same. Alternatively, entropy results
can be compared with each other compared with the maximum value: such values expressed as a percentage are definitely easier to relate to each other.
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Answer the following:-
Find all the zevoes of 2x4−3x3−3x2+6x−2,... | Filo
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Answer the following:- Find all the zevoes of , if Yyou know that twoy of its zeroes are and . solve: Draw the venn diagrams of , 4. Subba Rao started wark in 1995 at annual salary of 5000 and
received an increment of each year. In which year did his income reach ₹ 7000 ? 5. Prove that the parallelogram circumscribing a circle is a rhombus. 6. Show that 7. A die is thrown once. Find the
probability of Getting (i) a prime number (fi) a number lying between 2 and 6 (iri) an odd number. 8. The hypotenuse of a right triangle is more than twice of the shortest side. If the third side is
, less than the hypotenuse, find the sides of the triangle. Answer the followind:- (e) If , then find the value of (er)
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Answer the following:- Find all the zevoes of , if Yyou know that twoy of its zeroes are and . solve: Draw the venn diagrams of , 4. Subba Rao started wark in 1995 at annual salary of 5000
Question and received an increment of each year. In which year did his income reach ₹ 7000 ? 5. Prove that the parallelogram circumscribing a circle is a rhombus. 6. Show that 7. A die is thrown
Text once. Find the probability of Getting (i) a prime number (fi) a number lying between 2 and 6 (iri) an odd number. 8. The hypotenuse of a right triangle is more than twice of the shortest
side. If the third side is , less than the hypotenuse, find the sides of the triangle. Answer the followind:- (e) If , then find the value of (er)
Updated Mar 25, 2023
Topic All topics
Subject Mathematics
Class Class 11
Answer Video solution: 1
Upvotes 51
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[Solved] A marketing company needs to decide how m | SolutionInn
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A marketing company needs to decide how much time to allocate between radio and television advertising during the next month. The manager has determined that
A marketing company needs to decide how much time to allocate between radio and television advertising during the next month. The manager has determined that at least 80% of the time should be
allocated to TV. Each minute of Ad on Radio costs $400, and provide 300 units of exposure. Each minute of Ad on TV costs $1500, and provide 800 units of exposure. How much time should the manager
allocate between radio and television advertising to achieve 35,000 units of exposure at the minimum cost?
a) Very clearly define appropriate decision variables. x: y:
b) Define the objective function. List the function. Do we want to maximize or minimize?
c) Formulate all needed constraints (including non-negativity).
d) Implement the linear optimization model in Excel and use Solver to find an optimal solution.
There are 3 Steps involved in it
Step: 1
To solve this problem we will use linear optimization a Define the decision variables x Time in minu...
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Guide to lightning-fast JAX -
Hey there, fellow Python enthusiast! Have you ever wished your NumPy code run at supersonic speed? Meet JAX!. Your new best friend in your machine learning, deep learning, and numerical computing
journey. Think of it as NumPy with superpowers. It can automatically handle gradients, compile your code to run fast using JIT, and even run on GPU and TPU without breaking a sweat. Whether you’re
building neural networks, crunching scientific data, tweaking transformer models, or just trying to speed up your calculations, JAX has your back. Let’s dive in and see what makes JAX so special.
This guide provides a detailed introduction to JAX and its ecosystem.
Learning Objectives
• Explain JAX’s core principles and how they differ from Numpy.
• Apply JAX’s three key transformations to optimize Python code. Convert NumPy operations into efficient JAX implementation.
• Identify and fix common performance bottlenecks in JAX code. Implement JIT compilation correctly while avoiding typical Pitfalls.
• Build and train a Neural Network from scratch using JAX. Implement common machine learning operations using JAX’s functional approach.
• Solve optimization problems using JAX’s automatic differentiation. Perform efficient matrix operations and numerical computations.
• Apply effective debugging strategies for JAX-specific issues. Implement memory-efficient patterns for large-scale computations.
This article was published as a part of the Data Science Blogathon.
What is JAX?
According to the official documentation, JAX is a Python library for acceleration-oriented array computation and program transformation, designed for high-performance numerical computing and
large-scale machine learning. So, JAX is essentially NumPy on steroids, It combines familiar NumPy-style operations with automatic differentiation and hardware acceleration. Think of it as getting
the best of three worlds.
• NumPy’s elegant syntax and array operation
• PyTorch like automatic differentiation capability
• XLA’s (Accelerated Linear Algebra) for hardware acceleration and compilation benefits.
Why does JAX Stand Out?
What sets JAX apart is its transformations. These are powerful functions that can modify your Python code:
• JIT: Just-In-Time compilation for faster execution
• Grad: Automatic differentiation for computing gradients
• vmap: Automatically vectorization for batch processing
Here is a quick look:
import jax.numpy as jnp
from jax import grad, jit
# Define a simple function
@jit # Speed it up with compilation
def square_sum(x):
return jnp.sum(jnp.square(x))
# Get its gradient function automatically
gradient_fn = grad(square_sum)
# Try it out
x = jnp.array([1.0, 2.0, 3.0])
print(f"Gradient: {gradient_fn(x)}")
Gradient: [2. 4. 6.]
Getting Started with JAX
Below we will follow some steps to get started with JAX.
Step1: Installation
Setting up JAX is straightforward for CPU-only use. You can use the JAX documentation for more information.
Step2: Creating Environment for Project
Create a conda environment for your project
# Create a conda env for jax
$ conda create --name jaxdev python=3.11
#activate the env
$ conda activate jaxdev
# create a project dir name jax101
$ mkdir jax101
# Go into the dir
$cd jax101
Step3: Installing JAX
Installing JAX in the newly created environment
# For CPU only
pip install --upgrade pip
pip install --upgrade "jax"
# for GPU
pip install --upgrade pip
pip install --upgrade "jax[cuda12]"
Now you are ready to dive into real things. Before getting your hands dirty on practical coding let’s learn some new concepts. I will be explaining the concepts first and then we will code together
to understand the practical viewpoint.
First, get some motivation, By the way, why do we learn a new library again? I will answer that question throughout this guide in a step-by-step manner as simple as possible.
Why Learn JAX?
Think of JAX as a power tool. While NumPy is like a reliable hand saw, JAX is like a modern electric saw. It requires a bit more steps and knowledge, but the performance benefits are worth it for
intensive computation tasks.
• Performance: Jax code can run significantly faster than Pure Python or NumPy code, especially on GPU and TPUs
• Flexibility: It’s not just for machine learning- JAX excels in scientific computing, optimization, and simulation.
• Modern Approach: JAX encourages functional programming patterns that lead to cleaner, more maintainable code.
In the next section, we’ll dive deep into JAX’s transformation, starting with the JIT compilation. These transformations are what give JAX its superpowers, and understanding them is key to leveraging
JAX effectively.
Essential JAX Transformations
JAX’s transformations are what truly set it apart from the numerical computation libraries such as NumPy or SciPy. Let’s explore each one and see how they can supercharge your code.
JIT or Just-In-Time Compilation
Just-in-time compilation optimizes code execution by compiling parts of a program at runtime rather than ahead of time.
How JIT works in JAX?
In JAX, jax.jit transforms a Python function into a JIT-compiled version. Decorating a function with @jax.jit captures its execution graph, optimizes it, and compiles it using XLA. The compiled
version then executes, delivering significant speedups, especially for repeated function calls.
Here is how you can try it.
import jax.numpy as jnp
from jax import jit
import time
# A computationally intensive function
def slow_function(x):
for _ in range(1000):
x = jnp.sin(x) + jnp.cos(x)
return x
# The same function with JIT
def fast_function(x):
for _ in range(1000):
x = jnp.sin(x) + jnp.cos(x)
return x
Here is the same function, one is just a plain Python compilation process and the other one is used as a JAX’s JIT compilation process. It will calculate the 1000 data points sum of sine and cosine
functions. we will compare the performance using time.
# Compare performance
x = jnp.arange(1000)
# Warm-up JIT
fast_function(x) # First call compiles the function
# Time comparison
start = time.time()
slow_result = slow_function(x)
print(f"Without JIT: {time.time() - start:.4f} seconds")
start = time.time()
fast_result = fast_function(x)
print(f"With JIT: {time.time() - start:.4f} seconds")
The result will astonish you. The JIT compilation is 333 times faster than the normal compilation. It’s like comparing a bicycle with a Buggati Chiron.
Without JIT: 0.0330 seconds
With JIT: 0.0010 seconds
JIT can give you a superfast execution boost but you must use it properly otherwise it will be like driving Bugatti on a muddy village road that offers no supercar facility.
Common JIT Pitfalls
JIT works best with static shapes and types. Avoid using Python loops and conditions that depend on array values. JIT does not work with the dynamic arrays.
# Bad - uses Python control flow
def bad_function(x):
if x[0] > 0: # This won't work well with JIT
return x
return -x
# print(bad_function(jnp.array([1, 2, 3])))
# Good - uses JAX control flow
def good_function(x):
return jnp.where(x[0] > 0, x, -x) # JAX-native condition
print(good_function(jnp.array([1, 2, 3])))
That means bad_function is bad because JIT was not located in the value of x during calculation.
[1 2 3]
Limitations and Considerations
• Compilation Overhead: The first time a JIT-compiled function is executed, there is some overhead due to compilation. The compilation cost may outweigh the performance benefits for small functions
or those called only once.
• Dynamic Python Features: JAX’s JIT requires functions to be “static”. Dynamic control flow, like changing shapes or values based on Python loops, is not supported in the compiled code. JAX
provided alternatives like `jax.lax.cond` and `jax.lax.scan` to handle dynamic control flow.
Automatic Differentiation
Automatic differentiation, or autodiff, is a computation technique for calculating the derivative of functions accurately and effectively. It plays a crucial role in optimizing machine learning
models, especially in training neural networks, where gradients are used to update model parameters.
How does Automatic differentiation work in JAX?
Autodiff works by applying the chain rule of calculus to decompose complex functions into simpler ones, calculating the derivative of these sub-functions, and then combining the results. It records
each operation during the function execution to construct a computational graph, which is then used to compute derivatives automatically.
There are two main modes of auto-diff:
• Forward Mode: Computes derivatives in a single forward pass through the computational graph, efficient for functions with a small number of parameters.
• Reverse Mode: Computes derivatives in a single backward pass through the computational graph, efficient for functions with a large number of parameters.
source: Sebastian Raschka
Key features in JAX automatic differentiation
• Gradient Computation(jax.grad): `jax.grad` computes the derivative of a scaler-output function for its input. For functions with multiple inputs, a partial derivative can be obtained.
• Higher-Order Derivative(jax.jacobian, jax.hessian) : JAX supports the computation of higher-order derivatives, such as Jacobians and Hessains, making it suitable for advanced optimization and
physics simulation.
• Composability with other JAX Transformation: Autodiff in JAX integrates seamlessly with other transformations like `jax.jit` and `jax.vmap` allowing for efficient and scalable computation.
• Reverse-Mode Differentiation(Backpropagation): JAX’s auto-diff uses reverse-mode differentiation for scaler-output functions, which is highly effective for deep learning tasks.
import jax.numpy as jnp
from jax import grad, value_and_grad
# Define a simple neural network layer
def layer(params, x):
weight, bias = params
return jnp.dot(x, weight) + bias
# Define a scalar-valued loss function
def loss_fn(params, x):
output = layer(params, x)
return jnp.sum(output) # Reducing to a scalar
# Get both the output and gradient
layer_grad = grad(loss_fn, argnums=0) # Gradient with respect to params
layer_value_and_grad = value_and_grad(loss_fn, argnums=0) # Both value and gradient
# Example usage
key = jax.random.PRNGKey(0)
x = jax.random.normal(key, (3, 4))
weight = jax.random.normal(key, (4, 2))
bias = jax.random.normal(key, (2,))
# Compute gradients
grads = layer_grad((weight, bias), x)
output, grads = layer_value_and_grad((weight, bias), x)
# Multiple derivatives are easy
twice_grad = grad(grad(jnp.sin))
x = jnp.array(2.0)
print(f"Second derivative of sin at x=2: {twice_grad(x)}")
Second derivatives of sin at x=2: -0.9092974066734314
Effectiveness in JAX
• Efficiency: JAX’s automatic differentiation is highly efficient due to its integration with XLA, allowing for optimization at the machine code level.
• Composability: The ability to combine different transformations makes JAX a powerful tool for building complex machine learning pipelines and Neural Networks architecture such as CNN, RNN, and
• Ease of Use: JAX’s syntax for autodiff is simple and intuitive, enabling users to compute gradient without delving into the details of XLA and complex library APIs.
JAX Vectorize Mapping
In JAX, `vmap` is a powerful function that automatically vectorizes computations, allowing you to apply a function over batches of data without manually writing loops. It maps a function over an
array axis (or multiple axes) and evaluates it efficiently in parallel, which can lead to significant performance improvements.
How vmap Works in JAX?
The vmap function automates the process of applying a function to each element along a specified axis of an input array while preserving the efficiency of the computation. It transforms the given
function to accept batched inputs and execute the computation in a vectorized manner.
Instead of using explicit loops, vmap allows operations to be performed in parallel by vectorizing over an input axis. This leverages the hardware’s capability to perform SIMD (Single Instruction,
Multiple Data) operations, which can result in substantial speed-ups.
Key Features of vmap
• Automatic Vectorization: vamp automates the batching of computations, making it simple to parallel code over batch dimensions without changing the original function logic.
• Composability with other Transformations: It works seamlessly with other JAX transformations, such as jax.grad for differentiation and jax.jit for Just-In-Time compilation, allowing for highly
optimized and flexible code.
• Handling Multiple Batch Dimensions: vmap supports mapping over multiple input arrays or axes, making it versatile for various use cases like processing multi-dimensional data or multiple
variables simultaneously.
import jax.numpy as jnp
from jax import vmap
# A function that works on single inputs
def single_input_fn(x):
return jnp.sin(x) + jnp.cos(x)
# Vectorize it to work on batches
batch_fn = vmap(single_input_fn)
# Compare performance
x = jnp.arange(1000)
# Without vmap (using a list comprehension)
result1 = jnp.array([single_input_fn(xi) for xi in x])
# With vmap
result2 = batch_fn(x) # Much faster!
# Vectorizing multiple arguments
def two_input_fn(x, y):
return x * jnp.sin(y)
# Vectorize over both inputs
vectorized_fn = vmap(two_input_fn, in_axes=(0, 0))
# Or vectorize over just the first input
partially_vectorized_fn = vmap(two_input_fn, in_axes=(0, None))
# print
print(partially_vectorized_fn(x, y).shape)
Effectiveness of vmap in JAX
• Performance Improvements: By vectorizing computations, vmap can significantly speed up execution by leveraging parallel processing capabilities of modern hardware like GPUs, and TPUs(Tensor
processing units).
• Cleaner Code: It allows for more concise and readable code by eliminating the need for manual loops.
• Compatibility with JAX and Autodiff: vmap can be combined with automatic differentiation (jax.grad), allowing for the efficient computation of derivatives over batches of data.
When to Use Each Transformation
Using @jit when:
• Your function is called multiple times with similar input shapes.
• The function contains heavy numerical computations.
Use grad when:
• You need derivatives for optimization.
• Implementing machine learning algorithms
• Solving differential equations for simulations
Use vmap when:
• Processing batches of data with.
• Parallelizing computations
• Avoiding explicit loops
Matrix Operations and Linear Algebra Using JAX
JAX provides comprehensive support for matrix operations and linear algebra, making it suitable for scientific computing, machine learning, and numerical optimization tasks. JAX’s linear algebra
capabilities are similar to those found in libraries like NumPY but with additional features such as automatic differentiation and Just-In-Time compilation for optimized performance.
Matrix Addition and Subtraction
These operation are performed element-wise matrices of the same shape.
# 1 Matrix Addition and Subtraction:
import jax.numpy as jnp
A = jnp.array([[1, 2], [3, 4]])
B = jnp.array([[5, 6], [7, 8]])
# Matrix addition
C = A + B
# Matrix subtraction
D = A - B
print(f"Matrix A: \n{A}")
print(f"Matrix B: \n{B}")
print(f"Matrix adition of A+B: \n{C}")
print(f"Matrix Substraction of A-B: \n{D}")
Matrix Multiplication
JAX support both element-wise multiplication and dor product-based matrix multiplication.
# Element-wise multiplication
E = A * B
# Matrix multiplication (dot product)
F = jnp.dot(A, B)
print(f"Matrix A: \n{A}")
print(f"Matrix B: \n{B}")
print(f"Element-wise multiplication of A*B: \n{E}")
print(f"Matrix multiplication of A*B: \n{F}")
Matrix Transpose
The transpose of a matrix can be obtained using `jnp.transpose()`
# Matric Transpose
G = jnp.transpose(A)
print(f"Matrix A: \n{A}")
print(f"Matrix Transpose of A: \n{G}")
Matrix Inverse
JAX provides function for matrix inversion using `jnp.linalg.inv()`
# Matric Inversion
H = jnp.linalg.inv(A)
print(f"Matrix A: \n{A}")
print(f"Matrix Inversion of A: \n{H}")
Matrix Determinant
Determinant of a matrix can be calculate using `jnp.linalg.det()`.
# matrix determinant
det_A = jnp.linalg.det(A)
print(f"Matrix A: \n{A}")
print(f"Matrix Determinant of A: \n{det_A}")
Matrix Eigenvalues and Eigenvectors
You can compute the eigenvalues and eigenvectors of a matrix using `jnp.linalg.eigh()`
# Eigenvalues and Eigenvectors
import jax.numpy as jnp
A = jnp.array([[1, 2], [3, 4]])
eigenvalues, eigenvectors = jnp.linalg.eigh(A)
print(f"Matrix A: \n{A}")
print(f"Eigenvalues of A: \n{eigenvalues}")
print(f"Eigenvectors of A: \n{eigenvectors}")
Matrix Singular Value Decomposition
SVD is supported via `jnp.linalg.svd`, useful in dimensionality reduction and matrix factorization.
# Singular Value Decomposition(SVD)
import jax.numpy as jnp
A = jnp.array([[1, 2], [3, 4]])
U, S, V = jnp.linalg.svd(A)
print(f"Matrix A: \n{A}")
print(f"Matrix U: \n{U}")
print(f"Matrix S: \n{S}")
print(f"Matrix V: \n{V}")
Solving System of Linear Equations
To solve a system of linear equation Ax = b, we use `jnp.linalg.solve()`, where A is a square matrix and b is a vector or matrix of the same number of rows.
# Solving system of linear equations
import jax.numpy as jnp
A = jnp.array([[2.0, 1.0], [1.0, 3.0]])
b = jnp.array([5.0, 6.0])
x = jnp.linalg.solve(A, b)
print(f"Value of x: {x}")
Value of x: [1.8 1.4]
Computing the Gradient of a Matrix Function
Using JAX’s automatic differentiation, you can compute the gradient of a scalar function with respect to a matrix.
We will calculate gradient of the below function and values of X
# Computing the Gradient of a Matrix Function
import jax
import jax.numpy as jnp
def matrix_function(x):
return jnp.sum(jnp.sin(x) + x**2)
# Compute the grad of the function
grad_f = jax.grad(matrix_function)
X = jnp.array([[1.0, 2.0], [3.0, 4.0]])
gradient = grad_f(X)
print(f"Matrix X: \n{X}")
print(f"Gradient of matrix_function: \n{gradient}")
These most useful function of JAX used in numerical computing, machine learning, and physics calculation. There are many more left for you to explore.
Scientific Computing with JAX
JAX’s powerful libraries for scientific computing, JAX is best for scientific computing for its advance features such as JIT compilation, automatic differentiation, vectorization, parallelization,
and GPU-TPU acceleration. JAX’s ability to support high performance computing makes it suitable for a wide range of scientific applications, including physics simulations, machine learning,
optimization and numerical analysis.
We will explore an Optimization Problem in this section.
Optimization Problems
Let us go through the optimization problems steps below:
Step1: Define the function to minimize(or the problem)
# Define a function to minimize (e.g., Rosenbrock function)
def rosenbrock(x):
return sum(100.0 * (x[1:] - x[:-1] ** 2.0) ** 2.0 + (1 - x[:-1]) ** 2.0)
Here, the Rosenbrock function is defined, which is a common test problem in optimization. The function takes an array x as input and computes a valie that represents how far x is from the function’s
global minimum. The @jit decorator is used to enable Jut-In-Time compilation, which speed up the computation by compiling the function to run efficiently on CPUs and GPUs.
Step2: Gradient Descent Step Implementation
# Gradient descent optimization
def gradient_descent_step(x, learning_rate):
return x - learning_rate * grad(rosenbrock)(x)
This function performs a single step of the gradient descent optimization. The gradient of the Rosenbrock function is calculated using grad(rosenbrock)(x), which provides the derivative with respects
to x. The new value of x is updated by subtraction the gradient scaled by a learning_rate.The @jit is doing the same as before.
Step3: Running the Optimization Loop
# Optimize
x = jnp.array([0.0, 0.0]) # Starting point
learning_rate = 0.001
for i in range(2000):
x = gradient_descent_step(x, learning_rate)
if i % 100 == 0:
print(f"Step {i}, Value: {rosenbrock(x):.4f}")
The optimization loop initializes the starting point x and performs 1000 iterations of gradient descent. In each iteration, the gradient_descent_step function updates based on the current gradient.
Every 100 steps, the current step number and the value of the Rosenbrock function at x are printed, providing the progress of the optimization.
Solving Real-world physics problem with JAX
We will simulate a physical system the motion of a damped harmonic oscillator, which models things like a mass-spring system with friction, shock absorbers in vehicles, or oscillation in electrical
circuits. Is it not nice? Let’s do it.
Step1: Parameters Definition
import jax
import jax.numpy as jnp
# Define parameters
mass = 1.0 # Mass of the object (kg)
damping = 0.1 # Damping coefficient (kg/s)
spring_constant = 1.0 # Spring constant (N/m)
# Define time step and total time
dt = 0.01 # Time step (s)
num_steps = 3000 # Number of steps
The mass, damping coefficient, and spring constant are defined. These determine the physical properties of the damped harmonic oscillator.
Step2: ODE Definition
# Define the system of ODEs
def damped_harmonic_oscillator(state, t):
"""Compute the derivatives for a damped harmonic oscillator.
state: array containing position and velocity [x, v]
t: time (not used in this autonomous system)
x, v = state
dxdt = v
dvdt = -damping / mass * v - spring_constant / mass * x
return jnp.array([dxdt, dvdt])
The damped harmonic oscillator function defines the derivatives of the position and velocity of the oscillator, representing the dynamical system.
Step3: Euler’s Method
# Solve the ODE using Euler's method
def euler_step(state, t, dt):
"""Perform one step of Euler's method."""
derivatives = damped_harmonic_oscillator(state, t)
return state + derivatives * dt
A simple numerical method is used to solve the ODE. It approximates the state at the next time step on the basis of the current state and derivative.
Step4: Time Evolution Loops
# Initial state: [position, velocity]
initial_state = jnp.array([1.0, 0.0]) # Start with the mass at x=1, v=0
# Time evolution
states = [initial_state]
time = 0.0
for step in range(num_steps):
next_state = euler_step(states[-1], time, dt)
time += dt
# Convert the list of states to a JAX array for analysis
states = jnp.stack(states)
The loop iterates through the specified number of time steps, updating the state at each step using Euler’s method.
Step5: Plotting The Results
Finally, we can plot the results to visualize the behavior of the damped harmonic oscillator.
# Plotting the results
import matplotlib.pyplot as plt
positions = states[:, 0]
velocities = states[:, 1]
time_points = jnp.arange(0, (num_steps + 1) * dt, dt)
plt.figure(figsize=(12, 6))
plt.subplot(2, 1, 1)
plt.plot(time_points, positions, label="Position")
plt.xlabel("Time (s)")
plt.ylabel("Position (m)")
plt.subplot(2, 1, 2)
plt.plot(time_points, velocities, label="Velocity", color="orange")
plt.xlabel("Time (s)")
plt.ylabel("Velocity (m/s)")
I know you are eager to see how the Neural Network can be built with JAX. So, let’s dive deep into it.
Here, you can see that the Values were minimized gradually.
Building Neural Networks with JAX
JAX is a powerful library that combines high-performance numerical computing with the ease of using NumPy-like syntax. This section will guide you through the process of constructing a neural network
using JAX, leveraging its advanced features for automatic differentiation and just-in-time compilation to optimize performance.
Step1: Importing Libraries
Before we dive into building our neural network, we need to import the necessary libraries. JAX provides a suite of tools for creating efficient numerical computations, while additional libraries
will assist with optimization and visualization of our results.
import jax
import jax.numpy as jnp
from jax import grad, jit
from jax.random import PRNGKey, normal
import optax # JAX's optimization library
import matplotlib.pyplot as plt
Step2: Creating the Model Layers
Creating effective model layers is crucial in defining the architecture of our neural network. In this step, we’ll initialize the parameters for our dense layers, ensuring that our model starts with
well-defined weights and biases for effective learning.
def init_layer_params(key, n_in, n_out):
"""Initialize parameters for a single dense layer"""
key_w, key_b = jax.random.split(key)
# He initialization
w = normal(key_w, (n_in, n_out)) * jnp.sqrt(2.0 / n_in)
b = normal(key_b, (n_out,)) * 0.1
return (w, b)
def relu(x):
"""ReLU activation function"""
return jnp.maximum(0, x)
• Initializing Function: init_layer_params initializes weights(w) and biases (b) for dense layers using He initialization for weight and a small value for biases. He or Kaiming He initialization
works better for layers with ReLu activation functions, there are other popular initialization methods such as Xavier initialization which works better for layers with sigmoid activation.
• Activation Function: The relu function applies the ReLu activation function to the inputs which set negative values to zero.
Step3: Defining the Forward Pass
The forward pass is the cornerstone of a neural network, as it dictates how input data flows through the network to produce an output. Here, we will define a method to compute the output of our model
by applying transformations to the input data through the initialized layers.
def forward(params, x):
"""Forward pass for a two-layer neural network"""
(w1, b1), (w2, b2) = params
# First layer
h1 = relu(jnp.dot(x, w1) + b1)
# Output layer
logits = jnp.dot(h1, w2) + b2
return logits
• Forward Pass: forward performs a forward pass through a two-layer neural network, computing the output (logits) by applying a linear transformation followed by ReLu, and other linear
Step4: Defining the loss function
A well-defined loss function is essential for guiding the training of our model. In this step, we will implement the mean squared error (MSE) loss function, which measures how well the predicted
outputs match the target values, enabling the model to learn effectively.
def loss_fn(params, x, y):
"""Mean squared error loss"""
pred = forward(params, x)
return jnp.mean((pred - y) ** 2)
• Loss Function: loss_fn calculates the mean squared error (MSE) loss between the predicted logits and the target labels (y).
Step5: Model Initialization
With our model architecture and loss function defined, we now turn to model initialization. This step involves setting up the parameters of our neural network, ensuring that each layer is ready to
begin the training process with random but appropriately scaled weights and biases.
def init_model(rng_key, input_dim, hidden_dim, output_dim):
key1, key2 = jax.random.split(rng_key)
params = [
init_layer_params(key1, input_dim, hidden_dim),
init_layer_params(key2, hidden_dim, output_dim),
return params
• Model Initialization: init_model initializes the weights and biases for both layers of the neural networks. It uses two separate random keys for each layer;’s parameter initialization.
Step6: Training Step
Training a neural network involves iterative updates to its parameters based on the computed gradients of the loss function. In this step, we will implement a training function that applies these
updates efficiently, allowing our model to learn from the data over multiple epochs.
def train_step(params, opt_state, x_batch, y_batch):
loss, grads = jax.value_and_grad(loss_fn)(params, x_batch, y_batch)
updates, opt_state = optimizer.update(grads, opt_state)
params = optax.apply_updates(params, updates)
return params, opt_state, loss
• Training Step: the train_step function performs a single gradient descent update.
• It calculates the loss and gradients using value_and_grad, which computes both the function values and other gradients.
• The optimizer updates are calculated, and the model parameters are updated accordingly.
• The is JIT-compiled for performance.
Step7: Data and Training Loop
To train our model effectively, we need to generate suitable data and implement a training loop. This section will cover how to create synthetic data for our example and how to manage the training
process across multiple batches and epochs.
# Generate some example data
key = PRNGKey(0)
x_data = normal(key, (1000, 10)) # 1000 samples, 10 features
y_data = jnp.sum(x_data**2, axis=1, keepdims=True) # Simple nonlinear function
# Initialize model and optimizer
params = init_model(key, input_dim=10, hidden_dim=32, output_dim=1)
optimizer = optax.adam(learning_rate=0.001)
opt_state = optimizer.init(params)
# Training loop
batch_size = 32
num_epochs = 100
num_batches = x_data.shape[0] // batch_size
# Arrays to store epoch and loss values
epoch_array = []
loss_array = []
for epoch in range(num_epochs):
epoch_loss = 0.0
for batch in range(num_batches):
idx = jax.random.permutation(key, batch_size)
x_batch = x_data[idx]
y_batch = y_data[idx]
params, opt_state, loss = train_step(params, opt_state, x_batch, y_batch)
epoch_loss += loss
# Store the average loss for the epoch
avg_loss = epoch_loss / num_batches
if epoch % 10 == 0:
print(f"Epoch {epoch}, Loss: {avg_loss:.4f}")
• Data Generation: Random training data (x_data) and corresponding target (y_data) values are created. Model and Optimizer Initialization: The model parameters and optimizer state are
• Training Loop: The networks are trained over a specified number of epochs, using mini-batch gradient descent.
• Training loops iterate over batches, performing gradient updates using the train_step function. The average loss per epoch is calculated and stored. It prints the epoch number and the average
Step8: Plotting the Results
Visualizing the training results is key to understanding the performance of our neural network. In this step, we will plot the training loss over epochs to observe how well the model is learning and
to identify any potential issues in the training process.
# Plot the results
plt.plot(epoch_array, loss_array, label="Training Loss")
plt.title("Training Loss over Epochs")
These examples demonstrate how JAX combines high performance with clean, readable code. The functional programming style encouraged by JAX makes it easy to compose operations and apply
These examples demonstrate how JAX combines high performance with clean, readable code. The functional programming style encouraged by JAX makes it easy to compose operations and apply
Best Practice and Tips
In building neural networks, adhering to best practices can significantly enhance performance and maintainability. This section will discuss various strategies and tips for optimizing your code and
improving the overall efficiency of your JAX-based models.
Performance Optimization
Optimizing performance is essential when working with JAX, as it enables us to fully leverage its capabilities. Here, we will explore different techniques for improving the efficiency of our JAX
functions, ensuring that our models run as quickly as possible without sacrificing readability.
JIT Compilation Best Practices
Just-In-Time (JIT) compilation is one of the standout features of JAX, enabling faster execution by compiling functions at runtime. This section will outline best practices for effectively using JIT
compilation, helping you avoid common pitfalls and maximize the performance of your code.
Bad Function
import jax
import jax.numpy as jnp
from jax import jit
from jax import lax
# BAD: Dynamic Python control flow inside JIT
def bad_function(x, n):
for i in range(n): # Python loop - will be unrolled
x = x + 1
return x
# print(bad_function(1, 1000)) # does not work
This function uses a standard Python loop to iterate n times, incrementing the of x by 1 on each iteration. When compiled with jit, JAX unrolls the loop, which can be inefficient, especially for
large n. This approach does not fully leverage JAX’s capabilities for performance.
Good Function
# GOOD: Use JAX-native operations
def good_function(x, n):
return x + n # Vectorized operation
print(good_function(1, 1000))
This function does the same operation, but it uses a vectorized operation (x+n) instead of a loop. This approach is much more efficient because JAX can better optimize the computation when expressed
as a single vectorized operation.
Best Function
# BETTER: Use scan for loops
def best_function(x, n):
def body_fun(i, val):
return val + 1
return lax.fori_loop(0, n, body_fun, x)
print(best_function(1, 1000))
This approach uses `jax.lax.fori_loop`, which is a JAX-native way to implement loops efficiently. The `lax.fori_loop` performs the same increment operation as the previous function, but it does so
using a compiled loop structure. The body_fn function defines the operation for each iteration, and `lax.fori_loop` executes it from o to n. This method is more efficient than unrolling loops and is
especially suitable for cases where the number of iterations isn’t known ahead of time.
The code demonstrates different approaches to handling loops and control flow within JAX’s jit-complied functions.
Memory Management
Efficient memory management is crucial in any computational framework, especially when dealing with large datasets or complex models. This section will discuss common pitfalls in memory allocation
and provide strategies for optimizing memory usage in JAX.
Inefficient Memory Management
# BAD: Creating large temporary arrays
def inefficient_function(x):
temp1 = jnp.power(x, 2) # Temporary array
temp2 = jnp.sin(temp1) # Another temporary
return jnp.sum(temp2)
inefficient_function(x): This function creates multiple intermediate arrays, temp1, temp1 and finally the sum of the elements in temp2. Creating these temporary arrays can be inefficient because
each step allocates memory and incurs computational overhead, leading to slower execution and higher memory usage.
Efficient Memory Management
# GOOD: Combining operations
def efficient_function(x):
return jnp.sum(jnp.sin(jnp.power(x, 2))) # Single operation
This version combines all operations into a single line of code. It computes the sine of squared elements of x directly and sums the results. By combining the operation, it avoids creating
intermediate arrays, reducing memory footprints and improving performance.
Test Code
x = jnp.array([1, 2, 3])
[1 2 3]
The efficient version leverages JAX’s ability to optimize the computation graph, making the code faster and more memory-efficient by minimizing temporary array creation.
Debugging Strategies
Debugging is an essential part of the development process, especially in complex numerical computations. In this section, we will discuss effective debugging strategies specific to JAX, enabling you
to identify and resolve issues quickly.
Using print inside JIT for Debugging
The code shows techniques for debugging within JAX, particularly when using JIT-compiled functions.
import jax.numpy as jnp
from jax import debug
def debug_function(x):
# Use debug.print instead of print inside JIT
debug.print("Shape of x: {}", x.shape)
y = jnp.sum(x)
debug.print("Sum: {}", y)
return y
# For more complex debugging, break out of JIT
def debug_values(x):
print("Input:", x)
result = debug_function(x)
print("Output:", result)
return result
• debug_function(x): This function shows how to use debug.print() for debugging inside a jit compiled function. In JAX, regular Python print statements are not allowed inside JIT due to compilation
restrictions, so debug.print() is used instead.
• It prints the shape of the input array x using debug.print()
• After computing the sum of the elements of x, it prints the resulting sum using debug.print()
• Finally, the function returns the computed sum y.
• debug_values(x) function serves as a higher-level debugging approach, breaking out of the JIT context for more complex debugging. It first prints the inputs x using regular print statement. Then
calls debug_function(x) to compute the result and finally prints the output before returning the results.
print(debug_function(jnp.array([1, 2, 3])))
print(debug_values(jnp.array([1, 2, 3])))
This approach allows for a combination of in-JIT debugging with debug.print() and more detailed debugging outside of JIT using standard Python print statements.
Common Patterns and Idioms in JAX
Finally, we will explore common patterns and idioms in JAX that can help streamline your coding process and improve efficiency. Familiarizing yourself with these practices will aid in developing more
robust and performant JAX applications.
Device Memory Management for Processing Large Datasets
# 1. Device Memory Management
def process_large_data(data):
# Process in chunks to manage memory
chunk_size = 100
results = []
for i in range(0, len(data), chunk_size):
chunk = data[i : i + chunk_size]
chunk_result = jit(process_chunk)(chunk)
return jnp.concatenate(results)
def process_chunk(chunk):
chunk_temp = jnp.sqrt(chunk)
return chunk_temp
This function processes large datasets in chunks to avoid overwhelming device memory.
It sets chunk_size to 100 and iterates over the data increments of the chunk size, processing each chunk separately.
For each chunk, the function uses jit(process_chunk) to JIT-compile the processing operation, which improves performance by compiling it ahead of time.
The result of each chunk is concatenated into a single array using jnp.concatenated(result) to form a single list.
data = jnp.arange(10000)
Handling Random Seed for Reproducibility and Better Data Generation
The function create_traing_state() demonstrates managing random number generators (RNGs) in JAX, which is essential for reproducibility and consistent results.
# 2. Handling Random Seeds
def create_training_state(rng):
# Split RNG for different uses
rng, init_rng = jax.random.split(rng)
params = init_network(init_rng)
return params, rng # Return new RNG for next use
It starts with an initial RNG (rng) and splits it into two new RNGs using jax.random.split(). Split RNGs perform different tasks: `init_rng` initializes network parameters, and the updated RNG
returns for subsequent operations.
The function returns both the initialized network parameters and the new RNG for further use, ensuring proper handling of random states across different steps.
Now test the code using mock data
def init_network(rng):
# Initialize network parameters
return {
"w1": jax.random.normal(rng, (784, 256)),
"b1": jax.random.normal(rng, (256,)),
"w2": jax.random.normal(rng, (256, 10)),
"b2": jax.random.normal(rng, (10,)),
key = jax.random.PRNGKey(0)
params, rng = create_training_state(key)
print(f"Random number generator: {rng}")
print(f"Network parameters shape: {params['w1'].shape}")
print(f"Network parameters shape: {params['b1'].shape}")
print(f"Network parameters shape: {params['w2'].shape}")
print(f"Network parameters shape: {params['b2'].shape}")
print(f"Network parameters: {params}")
Using Static Arguments in JIT
def g(x, n):
i = 0
while i < n:
i += 1
return x + i
g_jit_correct = jax.jit(g, static_argnames=["n"])
print(g_jit_correct(10, 20))
You can use a static argument if JIT compiles the function with the same arguments each time. This can be useful for the performance optimization of JAX functions.
from functools import partial
@partial(jax.jit, static_argnames=["n"])
def g_jit_decorated(x, n):
i = 0
while i < n:
i += 1
return x + i
print(g_jit_decorated(10, 20))
If You want to use static arguments in JIT as a decorator you can use jit inside of functools. partial() function.
Now, we have learned and dived deep into many exciting concepts and tricks in JAX and overall programming style.
What’s Next?
• Experiment with Examples: Try to modify the code examples to learn more about JAX. Build a small project for a better understanding of JAX’s transformations and APIs. Implement classical
Machine Learning algorithms with JAX such as Logistic Regression, Support Vector Machine, and more.
• Explore Advanced Topics: Parallel computing with pmap, Custom JAX transformations, Integration with other frameworks
All code used in this article is here
JAX is a powerful tool that provides a wide range of capabilities for machine learning, Deep Learning, and scientific computing. Start with basics, experimenting, and get help from JAX’s beautiful
documentation and community. There are so many things to learn and it will not be learned by just reading others’ code you have to do it on your own. So, start creating a small project today in JAX.
The key is to Keep Going, learn on the way.
Key Takeaways
• Familiar NumPY-like interface and APIs make learning JAX easy for beginners. Most NumPY code works with minimal modifications.
• JAX encourages clean functional programming patterns that lead to cleaner, more maintainable code and upgradation. But If developers want JAX fully compatible with Object Oriented paradigm.
• What makes JAX’s features so powerful is automatic differentiation and JAX’s JIT compilation, which makes it efficient for large-scale data processing.
• JAX excels in scientific computing, optimization, neural networks, simulation, and machine learning which makes developer easy to use on their respective project.
Frequently Asked Questions
Q1. What makes JAX different from NumPY?
A. Although JAX feels like NumPy, it adds automatic differentiation, JIT compilation, and GPU/TPU support.
Q2. Do I need a GPU to use JAX?
A. In a single word big NO, though having a GPU can significantly speed up computation for larger data.
Q3. Is JAX a good alternative to NumPy?
A. Yes, You can use JAX as an alternative to NumPy, though JAX’s APIs look familiar to NumPy JAX is more powerful if you use JAX’s features well.
Q4. Can I use my existing NumPy code with JAX?
A. Most NumPy code can be adapted to JAX with minimal changes. Usually just changing import numpy as np to import jax.numpy as jnp.
Q5. Is JAX harder to learn than NumPy?
A. The basics are just as easy as NumPy! Tell me one thing, will you find it hard after reading the above article and hands-on? I answered it for you. YES hard. Every framework, language, libraries
is hard not because it is hard by design but because we don’t give much time to explore it. Give it time to get your hand dirty it will be easier day by day.
The media shown in this article is not owned by Analytics Vidhya and is used at the Author’s discretion. | {"url":"https://digitalinfowave.com/guide-to-lightning-fast-jax/","timestamp":"2024-11-14T04:47:44Z","content_type":"text/html","content_length":"160195","record_id":"<urn:uuid:5b6f1645-e2b4-41b6-9e9d-cc9e15ffe5a4>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00748.warc.gz"} |
PrettyR R | R-bloggersPrettyR R
[This article was first published on
A Distant ObserveR
, and kindly contributed to
]. (You can report issue about the content on this page
) Want to share your content on R-bloggers?
if you have a blog, or
if you don't.
When it comes to R blogging I’m a complete newbie. So I’m still struggling with the technical details.
Part of the process is prettifying the code snippets. One of the standard ways of doing this involves copy-and-paste-ing the R code into the
Pretty R syntax highlighter
While assembling the bits to do the posting programmatically I wrote a function that replaces the copy-and-paste part.
Now here’s the function prettified by itself:
prettyR <- function(file) {
Rcode <- readLines(file)
Rcode <- paste(Rcode, collapse="\n")
# assemble the parameters for the http POST to the Pretty R web site
URL <- "http://www.inside-r.org/pretty-r/tool"
parameters <- list(
op = "edit-submit",
form_id = "pretty_r_tool_form",
code_input = Rcode
# send the http POST request
rawHTML <- postForm(URL, .params = parameters)
parsedHTML <- htmlParse(rawHTML)
# find the node
prettified <- getNodeSet(parsedHTML, "//div[@class='form-item']/textarea")[[1]]
prettified <- xmlValue(prettified[[1]]) | {"url":"https://www.r-bloggers.com/2012/11/prettyr-r/","timestamp":"2024-11-04T23:20:48Z","content_type":"text/html","content_length":"82327","record_id":"<urn:uuid:89043e29-2ea0-40b3-99a6-6ebcf2158e2b>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00830.warc.gz"} |
: an R Package for the Analysis of Graph Matching
Graph matching methods
The graph matching methods share the same basic syntax:
gm(A, B, seeds = NULL, similarity = NULL, method = "indefinite",
***algorithm parameters***)
Table 4.1: Overview of arguments for different graph matching functions.
start Matrix or character Initialization of the start matrix for iterations. FW, convex
lap_method Character Method for solving linear assignment problem. FW, convex, PATH, IsoRank
max_iter Number Maximum number of iterations. FW, convex, PATH, IsoRank
tol Number Tolerance of edge disagreements. FW, convex, PATH
r Number Threshold of neighboring pair scores. percolation
ExpandWhenStuck Boolean TRUE if performs ExpandWhenStuck algorithm. percolation
The first two arguments for graph matching algorithms represent two networks which can be matrices, igraph objects, or two lists of either form in the case of multi-layer matching. The seeds argument
contains prior information on the known partial correspondence of two graphs. It can be a vector of logicals or indices if the seed pairs have the same indices in both graphs. In general, the seeds
argument takes a matrix or a data frame as input with two columns indicating the indices of seeds in the two graphs respectively. The similarity parameter is for a matrix of similarity scores between
the two vertex sets, with larger scores indicating higher similarity. Notably, one should be careful with the different scales of the graph topological structure and the vertex similarity information
in order to properly address the relative importance of each part of the information. The method argument specifies a graph matching algorithm to use, and one can choose from “indefinite” (default),
“convex”, “PATH”, “percolation”, “IsoRank”, “Umeyama”, or a self-defined graph matching function which enables users to test out their own algorithms while remaining compatible with the package. If
method is a function, it should take at least two networks, seeds and similarity scores as arguments. Users can also include additional arguments if applicable. The self-defined graph matching
function should return an object of the “graphMatch” class with matching correspondence, sizes of two input graphs, and other matching details. As an illustrative example, graph_match_rand defines a
new graph matching function which matches by randomly permuting the vertex label of the second graph using a random seed rand_seed. We then apply this self-defined GM method to matching the
correlated graphs sampled earlier with a specified random seed:
graph_match_rand <- function(A, B, seeds = NULL,
similarity = NULL, rand_seed){
totv1 <- nrow(A[[1]])
totv2 <- nrow(B[[1]])
nv <- max(totv1, totv2)
corr <- data.frame(corr_A = 1:nv,
corr_B = c(1:nv)[sample(nv)])
corr = corr,
nnodes = c(totv1, totv2),
detail = list(
rand_seed = rand_seed
match_rand <- gm(cgnp_g1, cgnp_g2,
method = graph_match_rand, rand_seed = 123)
Other arguments vary for different graph matching algorithms with an overview given in Table \(\ref{tab:gm-alg}\). The start argument for the FW methodology with “indefinite” and “convex” relaxations
takes any \(nns\text{-by-}nns\) matrix or an initialization method including “bari”, “rds” or “convex”. These represent initializing the iterations at a specific matrix, the barycenter, a random
doubly stochastic matrix, or the doubly stochastic solution from “convex” method on the same graphs, respectively.
Moreover, sometimes we have access to side information on partial correspondence with uncertainty. If we still treat such prior information as hard seeds and pass them through the seeds argument for
“indefinite” and “convex” methods, incorrect information can yield unsatisfactory matching results. Instead, we provide the option of soft seeding by incorporating the noisy partial correspondence
into the initialization of the start matrix. The core function used for initializing the start matrix with versatile options is the init_start function.
Suppose the first two pairs of nodes are hard seeds and another pair of incorrect seed \((3,4)\) is soft seeds:
We generate a start matrix incorporating soft seeds initialized at the barycenter:
## [,1] [,2] [,3]
## [1,] 0.0 1 0.0
## [2,] 0.5 0 0.5
## [3,] 0.5 0 0.5
An alternative is to generate a start matrix that is a random doubly stochastic matrix incorporating soft seeds as follow
as.matrix(start_rds <- init_start(start = "rds", nns = 3,
ns = 2, soft_seeds = soft_seeds))
## [,1] [,2] [,3]
## [1,] 0.00 1 0.00
## [2,] 0.52 0 0.48
## [3,] 0.48 0 0.52
Then we can initialize the Frank-Wolfe iterations at any of the start matrix by specifying the start parameter.
When there are no soft seeds, we no longer need to initialize the start matrix by using init_start first. Instead we can directly assign an initialization method to the start argument in the gm
Below use solution from the convex relaxation as the initialization for the indefinite relaxation.
match_convex <- gm(cgnp_g1, cgnp_g2, seeds = hard_seeds,
method = "indefinite", start = "convex")
Now let’s match the sampled pair of graphs from the stochastic block model by using Percolation algorithm. Apart from the common arguments for all the graph matching algorithms, Percolation has
another argument representing the minimum number of matched neighbors required for matching a new qualified vertex pair. Here we adopt the default value which is 2. Also, at least one of similarity
scores and seeds is required for Percolation algorithm to kick off. Let’s utilize the same set of hard seeds and assume there is no available prior information on similarity scores.
sbm_g1 <- sbm_pair$graph1
sbm_g2 <- sbm_pair$graph2
match_perco <- gm(sbm_g1, sbm_g1, seeds = hard_seeds,
method = "percolation", r = 2)
## gm(A = sbm_g1, B = sbm_g1, seeds = hard_seeds, method = "percolation",
## r = 2)
## Match (5 x 5):
## corr_A corr_B
## 1 1 1
## 2 2 2
Without enough prior information on partial correspondence, Percolation couldn’t find any qualifying matches. Suppose in addition to the current pair of sampled graphs, the above sampled correlated
homogeneous and heterogeneous graphs are different layers of connectivity for the same set of vertices. We can then match the nonseed vertices based on the topological information in all of these
three graph layers. To be consistent, let’s still use the Percolation algorithm with threshold equal to 2 and the same set of seeds.
matrix_lA <- list(sbm_g1, ieg_pair$graph1, cgnp_g1)
matrix_lB <- list(sbm_g2, ieg_pair$graph2, cgnp_g2)
match_perco_list <- gm(A = matrix_lA, B = matrix_lB, seeds = hard_seeds,
method = "percolation", r = 2)
## gm(A = matrix_lA, B = matrix_lB, seeds = hard_seeds, method = "percolation",
## r = 2)
## Match (5 x 5):
## corr_A corr_B
## 1 1 1
## 2 2 2
## 3 3 3
## 4 4 4
## 5 5 5
With the same amount of available prior information, we are now able to match all the nodes correctly.
Finally, we will give an example of matching multi-layers of graphs using IsoRank algorithm. Unlike the other algorithm, similarity scores are required for IsoRank algorithm. Without further
information, we adopt the barycenter as the similarity matrix here.
sim <- as.matrix(init_start(start = "bari", nns = 5,
soft_seeds = hard_seeds))
match_IsoRank <- gm(A = matrix_lA, B = matrix_lB,
seeds = hard_seeds, similarity = sim,
method = "IsoRank", lap_method = "LAP")
Graph matching functions return an object of class “graphMatch” which contains the details of the matching results, including a list of the matching correspondence, a call to the graph matching
function and dimensions of the original two graphs.
## corr_A corr_B
## 1 1 1
## 2 2 2
## 3 3 5
## 4 4 4
## 5 5 3
## gm(A = cgnp_g1, B = cgnp_g2, seeds = hard_seeds, method = "indefinite",
## start = "convex")
## [1] 5 5
Additionally, “graphMatch” also returns a list of matching details corresponding to the specified method. Table \(\ref{tab:gm-value}\) provides an overview of returned values for different graph
matching methods. With the seeds information, one can obtain a node mapping for non-seeds accordingly
## corr_A corr_B
## 3 3 5
## 4 4 4
## 5 5 3
Table 4.2: Overview of return values for different graph matching functions.
seeds A vector of logicals indicating if the corresponding vertex is a seed All the functions
soft The functional similarity score matrix with which one can extract more than one matching candidates FW, convex, PATH, IsoRank, Umeyama
lap_method Choice for solving the LAP. FW, convex, Umeyama, IsoRank
iter Number of iterations until convergence or reaches the max_iter. FW, convex, PATH
max_iter Maximum number of replacing matches. FW, convex
match_order The order of vertices getting matched. percolation, IsoRank
The “graphMatch” class object can also be flexibly used as a matrix. In addition to the returned list of matching correspondence, one can obtain the corresponding permutation matrix in the sparse
## 5 x 5 sparse Matrix of class "dgCMatrix"
## [1,] 1 . . . .
## [2,] . 1 . . .
## [3,] . . . . 1
## [4,] . . . 1 .
## [5,] . . 1 . .
Notably, multiplicity is applicable to the “graphMatch” object directly without converting to the permutation matrix. This enables obtaining the permuted second graph, that is \(PBP^T\) simply by
## IGRAPH 8187cd4 UN-- 5 5 -- Erdos-Renyi (gnp) graph
## + attr: name_1 (g/c), name_2 (g/c), type_1 (g/c), type_2 (g/c), loops_1
## | (g/l), loops_2 (g/l), p_1 (g/n), p_2 (g/n), name (g/c), type (g/c),
## | loops (g/l), p (g/n), name (v/n)
## + edges from 8187cd4 (vertex names):
## [1] 5--3 2--5 2--3 1--5 1--2
Evaluation of goodness of matching
Along with the graph matching methodology, iGraphMatch has many capabilities for evaluating and visualizing the matching performance. After matching two graphs, the function summary can be used to
get a summary of the overall matching result in terms of commonly used measures including the number of matches, the number of correct matches, common edges, missing edges, extra edges and the
objective function value. The edge matching information is stored in a data frame named edge_match_info. Note that summary outputs the number of correct matches only when the true correspondence is
known by specifying the true_label argument with a vector indicating the true correspondence in the second graph. Applying the summary function on the matching result match_convex with true_label =
1:5, indicating the true correspondence is the identity that provides these summaries.
## Call: gm(A = cgnp_g1, B = cgnp_g2, seeds = hard_seeds, method = "indefinite",
## start = "convex")
## # Matches: 3
## # True Matches: 1, # Seeds: 2, # Vertices: 5, 5
## common_edges 4.0
## missing_edges 0.0
## extra_edges 1.0
## fnorm 1.4
Applying the summary function to a multi-layer graph matching result returns edge statistics for each layer.
## Call: gm(A = matrix_lA, B = matrix_lB, seeds = hard_seeds, similarity = sim,
## method = "IsoRank", lap_method = "LAP")
## # Matches: 3, # Seeds: 2, # Vertices: 5, 5
## layer 1 2 3
## common_edges 2.0 6.0 4.0
## missing_edges 0.0 1.0 0.0
## extra_edges 1.0 0.0 1.0
## fnorm 1.4 1.4 1.4
In realistic scenarios, the true correspondence is not available. As introduced in section \(\ref{sec:background}\), the user can use vertex level statistics to evaluate match performance. The
best_matches function evaluates a vertex-level metric and returns a sorted data.frame of the vertex-matches with the metrics. The arguments are the two networks, a specific measure to use, the number
of top-ranked vertex-matches to output, and the matching correspondence in the second graph if applicable. As an example here, we apply best_matches to rank the matches from above with the true
underlying alignment
best_matches(cgnp_g1, cgnp_g2, match = match_convex,
measure = "row_perm_stat", num = 3,
true_label = 1:igraph::vcount(cgnp_g1))
## A_best B_best measure_value precision
## 1 4 4 -1.4 1.00
## 2 3 5 -1.2 0.50
## 3 5 3 -1.2 0.33
Note, best_matches uses seed information from the match parameter and only outputs non-seed matches. Without the true correspondence, true_label would take the default value and the output data frame
only contains the first three columns.
To visualize the matches of smaller graphs, the function plot displays edge discrepancies of the two matched graphs by an adjacency matrix or a ball-and-stick plot, depending on the input format of
two graphs.
The plots for visualizing matching performance of match_convex are shown in Figure 4.1. Grey edges and pixels indicate common edges, red ones indicate edges only in the second graph. If they were
present, blue pixels and edges represent missing edges that only exist in the first graph. The corresponding linetypes are solid, short dash, and long dash. | {"url":"https://cran.rstudio.com/web/packages/iGraphMatch/vignettes/iGraphMatch.html","timestamp":"2024-11-12T19:44:38Z","content_type":"text/html","content_length":"352484","record_id":"<urn:uuid:10866173-2666-4810-9a91-38d1a8c45033>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00371.warc.gz"} |
Drained Triaxial Compression Test with Simplified Cap-Yield (CHSoil) Model
FLAC3D Theory and Background • Constitutive Models
Drained Triaxial Compression Test with Simplified Cap-Yield (CHSoil) Model
To view this project in FLAC3D, use the menu command . The project’s main data file is shown at the end of this example.
Triaxial experiments are conducted numerically using the CHSoil model at the three levels of constant mean stress, 40, 80, and 160 kPa, for the \(D_r\) = 40% sand. A servo-control is applied to
maintain the mean stress constant during the numerical experiments.
The estimates for model properties are used to conduct the numerical tests, and the test results are compared to the available laboratory data (imported into FLAC3D in tables). The properties are
adjusted (see the section on calibration), and the numerical experiment is repeated until a satisfactory curve fitting is obtained.
The results of the curve fitting experiment are listed in Table 1.
\(D_r\) \(E_{ref}\) \(\nu\) \(\phi_f\) \(\psi_f\) \(\phi_{cv}\) \(m\) \(n\)
40% 1800 0.35 34° 7.5° 28° 0.5 0.5
A comparison between numerical predictions using the calibrated properties and laboratory results is shown in Figure 1 through Figure 3. Note that the soil-mechanics convention for positive stress/
strain is adopted in these plots. (Dilation is negative.) The comparison is quite reasonable.
Data File
model new
model title ...
'Drained triaxial test at constant bmean pressure Dr=40 - cap-yield-soil'
model large-strain off
fish automatic-create off
zone create brick size 1 1 5
zone cmodel assign cap-yield-simplified
zone property young-reference=1800 poisson=0.35 pressure-reference=100.0
zone property failure-ratio=0.99 friction=34.0 exponent-bulk=0.5 ...
zone property density=1000. cohesion=0.0 dilation-mobilized=0.0 ...
zone property flag-dilation=2 friction-critical=28.0 dilation=7.5
zone property pressure-initial= 40.0 range id = 1
zone property pressure-initial= 80.0 range id = 3
zone property pressure-initial=160.0 range id = 5
zone null range id-list = 2, 4
zone initialize stress xx -40.0 yy -40.0 zz -40.0 range id = 1
zone initialize stress xx -80.0 yy -80.0 zz -80.0 range id = 3
zone initialize stress xx -160.0 yy -160.0 zz -160.0 range id = 5
zone gridpoint fix velocity
zone gridpoint initialize velocity-x 0.25e-6 range position-x 1.0
zone gridpoint initialize velocity-y 0.25e-6 range position-y 1.0
zone gridpoint initialize velocity-z -0.5e-6 range union position-z 1.0 ...
position-z 3.0 position-z 5.0
[global z1 = zone.near(0.5,0.5,0.5)]
[global z3 = zone.near(0.5,0.5,2.5)]
[global z5 = zone.near(0.5,0.5,4.5)]
program call 'servo'
fish history q1
fish history p1
fish history eps_v1 ; vol. strain (%) dilation positive
fish history eps_a1 ; axial strain (%)
fish history q2
fish history p2
fish history eps_v2 ; vol. strain (%) dilation positive
fish history eps_a2 ; axial strain (%)
fish history q3
fish history p3
fish history eps_v3 ; vol. strain (%) dilation positive
fish history eps_a3 ; axial strain (%)
history interval 1000
model step 100000
model save 'chsoil-dtriax1'
Was this helpful? ... Itasca Software © 2024, Itasca Updated: Aug 13, 2024 | {"url":"https://docs.itascacg.com/itasca910/flac3d/zone/test3d/ConstitutiveModels/DrainedTriaxialCHSoil/drainedtriaxialchsoil.html","timestamp":"2024-11-03T00:53:59Z","content_type":"application/xhtml+xml","content_length":"22318","record_id":"<urn:uuid:0f44ff8f-d042-48b1-8733-4187d1e58d1f>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00765.warc.gz"} |
Re: Factor 2 error in Inverse Laplace Transform
• To: mathgroup at smc.vnet.net
• Subject: [mg51255] Re: Factor 2 error in Inverse Laplace Transform
• From: ab_def at prontomail.com (Maxim)
• Date: Sun, 10 Oct 2004 01:57:39 -0400 (EDT)
• References: <ck87ta$9mf$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
p-valko at tamu.edu (Peter Valko) wrote in message news:<ck87ta$9mf$1 at smc.vnet.net>...
> Hi,
> InverseLaplaceTransform is an extremely useful part of Mathematica
> (since v 4.1).
> However, in the following simple problem it gives the wrong answer:
> Problem 1:
> InverseLaplaceTransform[s/(s+1),s,t]
> -1/(2*E^t)+DiracDelta[t]
> where the factor 2 is completely wrong.
> To see that I slightly rewrite Problem 1 into
> Problem 1a:
> InverseLaplaceTransform[Apart[s/(s+1)],s,t]
> and then I get the correct answer:
> -E^(-t)+DiracDelta[t]
> Of course one can "Unprotect" InverseLaplaceTransform and teach it to
> give the correct answer but that is not the point.
> (Also one can start a long debate about the meaning of DiracDelta in
> Mathematica, but that is also not the point here. )
> There are several similar simple examples when the wrong factor of two
> shows up, for instance
> Problem 2:
> InverseLaplaceTransform[ s ArcTan[1/s],s,t]
> Using the Trace one can find out that all these "factor 2" errors have
> a common origin.
> Solving Problem 1 Mathematica calculates the convolution integral
> Integrate[E^(-t+x)*Derivative[1][DiracDelta][x],{x,0,t}]
> and because the lower limit is exactly zero,the factor 2 shows up in
> -1/(2*E^t), that is Mathematica "halves" the Dirac delta and all its
> derivatives at the origin.
> I think the InverseLaplaceTransform function could be much improved if
> the above convolution integral would be evaluated more carefully.
> For instance, doing it in two steps:
> res1=Integrate[E^(-t+x)*Derivative[1][DiracDelta][x],{x,-eps,t},
> Assumptions -> eps>0];
> res2=res1/.eps -> 0
> would give the right result.
> (This caution is necessary only, if generalized functions are involved
> in the integration.)
> I wonder if further examples/suggestions are welcome in this group
> regarding InverseLaplaceTransform???
> Peter
I'd say that this is two messes mixed together. One is a rather poor
implementation of integral transforms, especially when there are
generalized functions involved. For example:
LaplaceTransform[InverseLaplaceTransform[Log[p], p, t], t, p] - Log[p]
LaplaceTransform[InverseLaplaceTransform[PolyGamma[p], p, t], t, p] -
We can see that LaplaceTransform/InverseLaplaceTransform aren't
consistenly defined for these functions (here Mathematica doesn't
internally take integrals of distributions). Another issue is the
question of how the integral of DiracDelta on [0,a] should be
Integrate[DiracDelta[x], {x, 0, 1}]
Integrate[DiracDelta[x - a], {x, 0, 1}]
Integrate[DiracDelta[x - 1], {x, 1, 2}]
UnitStep[1 - a]*UnitStep[a]
The value of the first integral is by convention taken to be 1/2;
however, substituting a=0 into Out[4] we obtain 1, and making the
change of variables x->x-1 (the third integral) we get 0. Similarly
for the integrals of the DiracDelta derivatives:
Integrate[DiracDelta'[x]*phi[x], {x, -eps, Infinity}]
Integrate[DiracDelta'[x]*phi[x], {x, 0, Infinity}]
Integrate[DiracDelta'[x]*x, {x, 0, Infinity}]
(-DiracDelta[eps])*phi[0] - phi'[0]
The value of the first integral for eps<0 is incorrect in any case,
and Out[7] and Out[8] are not consistent with each other.
Maxim Rytin
m.r at inbox.ru | {"url":"http://forums.wolfram.com/mathgroup/archive/2004/Oct/msg00270.html","timestamp":"2024-11-13T01:04:21Z","content_type":"text/html","content_length":"33635","record_id":"<urn:uuid:98dba128-1dcb-4bc8-b810-c17d106c5e84>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00635.warc.gz"} |
Counting Restricted Integer Partitions
Blair, David Dakota, "Counting Restricted Integer Partitions" (2015). CUNY Academic Works.
Polynomials for m up to 23.
Let p_b(n) be the number of integer partitions of n whose parts are powers of b. For each m there is a generating function identity: f_m(b,q)\sum_{n} p_b(n) q^n = (1-q)^m \sum_{n} p_b(b^m n q)q^n
where n ranges over all integer values. This dataset is a JSON object with keys m from 1 to 23 whose values are f_m(b,q). This file is also published as Polynomials occuring in generating function
identities for b-ary partitions at CUNY Academic Works. | {"url":"https://dakota.tensen.net/2015/rp/","timestamp":"2024-11-14T01:45:33Z","content_type":"text/html","content_length":"1783","record_id":"<urn:uuid:aede9911-6ec9-426c-8bc0-4520439f9692>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00017.warc.gz"} |
React Multiple Line Chart 2024 - Multiplication Chart Printable
React Multiple Line Chart
React Multiple Line Chart – The Multiplication Graph or chart Line may help your students aesthetically represent numerous very early math concepts ideas. However, it must be used as a teaching aid
only and should not be confused with the Multiplication Table. The chart comes in 3 variations: the colored variation is useful when your college student is focusing on a single periods table at one
time. The horizontal and vertical models are compatible with youngsters who happen to be nonetheless discovering their occasions dining tables. In addition to the colored version, you can also
purchase a blank multiplication chart if you prefer. React Multiple Line Chart.
Multiples of 4 are 4 clear of the other
The design for deciding multiples of 4 is to put each amount to itself and locate its other several. For instance, the first five multiples of 4 are: 8, 4, 16 and 12 and 20. And they are four away
from each other on the multiplication chart line, this trick works because all multiples of a number are even. Moreover, multiples of four are even figures by nature.
Multiples of 5 are even
If they end in or 5, You’ll find multiples of 5 on the multiplication chart line only. Quite simply, you can’t flourish a variety by two or three to obtain a much quantity. If the number ends in five
or , you can only find a multiple of five! Luckily, there are actually tips which make discovering multiples of five even simpler, like utilizing the multiplication graph or chart collection to get
the several of five.
Multiples of 8 are 8 from each other
The pattern is clear: all multiples of 8 are two-digit numbers and multiples of several-digit phone numbers are two-digit phone numbers. Every range of 10 posesses a a number of of seven. 8 is even,
so that all its multiples are two-digit amounts. Its pattern carries on up to 119. When you can see a number, ensure you locate a several of seven to begin with.
Multiples of 12 are 12 from the other
The number 12 has limitless multiples, and you can increase any complete variety by it to help make any variety, which include by itself. All multiples of twelve are even phone numbers. Here are a
few examples. David enjoys to buy pencils and organizes them into seven packages of twelve. He now has 96 pencils. James has certainly one of every type of pencil. In the workplace, he arranges them
around the multiplication chart collection.
Multiples of 20 are 20 clear of the other
Within the multiplication graph, multiples of twenty are common even. If you multiply one by another, then the multiple will be also even. If you have more than one factor, multiply both numbers by
each other to find the factor. If Oliver has 2000 notebooks, then he can group them equally, for example. The identical is applicable to pencils and erasers. You could buy one in a package of about
three or a package of half a dozen.
Multiples of 30 are 30 clear of each other
In multiplication, the term “factor pair” refers to a small group of phone numbers that type an absolute variety. If the number ’30’ is written as a product of five and six, that number is 30 away
from each other on a multiplication chart line, for example. The same holds true for any variety from the variety ‘1’ to ’10’. In other words, any variety can be published since the item of 1 and
Multiples of 40 are 40 clear of one another
Do you know how to find them, though you may know that there are multiples of 40 on a multiplication chart line? To accomplish this, you can add externally-in. By way of example, 10 12 14 = 40, and
the like. Likewise, 15 8-10 = 20. In cases like this, the telephone number about the still left of 10 is an even amount, even though the one on the correct is definitely an odd variety.
Multiples of 50 are 50 far from the other
Using the multiplication graph line to discover the amount of two amounts, multiples of fifty are similar range away from each other about the multiplication chart. They already have two perfect 80,
50 and factors. Typically, each expression varies by 50. Another component is 50 alone. Listed here are the typical multiples of 50. A common several will be the multiple of a presented amount by 50.
Multiples of 100 are 100 away from each other
The following are the numerous phone numbers which can be multiples of 100. A positive combine is actually a numerous of one one hundred, while a negative set is really a a number of of 10. These two
kinds of phone numbers will vary in numerous approaches. The very first strategy is to break down the quantity by subsequent integers. In cases like this, the number of multiples is just one, twenty,
ten and thirty and forty.
Gallery of React Multiple Line Chart
Javascript ChartJS React Line Chart How To Show Single Tooltip
React Multi Series Chart CanvasJS
React Js Multiple Line Chart With Google Charts Tutorial LaptrinhX
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Quick Sort - Sorting Algorithm Animations
Quick Sort
Animation, code, analysis, and discussion of quick sort on 4 initial conditions.
How to use: Press "Play all", or choose the
Play animation Play animation Play animation
Play All
Random Reversed Few Unique
_# choose pivot_
swap a[1,rand(1,n)]
_# 2-way partition_
k = 1
for i = 2:n, if a[i] < a[1], swap a[++k,i]
swap a[1,k]
_→ invariant: a[1..k-1] < a[k] <= a[k+1..n]_
_# recursive sorts_
sort a[1..k-1]
sort a[k+1,n]
When carefully implemented, quick sort is robust and has low overhead. When a stable sort is not needed, quick sort is an excellent general-purpose sort – although the 3-way partitioning version
should always be used instead.
The 2-way partitioning code shown above is written for clarity rather than optimal performance; it exhibits poor locality, and, critically, exhibits O(n^2) time when there are few unique keys. A more
efficient and robust 2-way partitioning method is given in Quicksort is Optimal by Robert Sedgewick and Jon Bentley. The robust partitioning produces balanced recursion when there are many values
equal to the pivot, yielding probabilistic guarantees of O(n·lg(n)) time and O(lg(n)) space for all inputs.
With both sub-sorts performed recursively, quick sort requires O(n) extra space for the recursion stack in the worst case when recursion is not balanced. This is exceedingly unlikely to occur, but it
can be avoided by sorting the smaller sub-array recursively first; the second sub-array sort is a tail recursive call, which may be done with iteration instead. With this optimization, the algorithm
uses O(lg(n)) extra space in the worst case.
• Black values are sorted.
• Gray values are unsorted.
• Dark gray values denote the current interval.
• A pair of red triangles mark k and i (see the code).
• Not stable
• O(lg(n)) extra space (see discussion)
• O(n^2) time, but typically O(n·lg(n)) time
• Not adaptive
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Linear Algebra In Data Science: A Practical Approach » EFRC
Linear Algebra in Data Science ,In the dynamic and data-driven world of today, linear algebra has emerged as a fundamental tool in the arsenal of every data scientist. This article delves into the
practical applications of linear algebra in the field of data science, offering insights and techniques that can elevate your data analysis skills to new heights.
Linear algebra, with its powerful concepts of vectors, matrices, and linear transformations, forms the backbone of numerous data science workflows. From machine learning algorithms to data
visualization techniques, the principles of linear algebra are woven into the very fabric of data-driven decision-making. By mastering these essential linear algebra concepts, data scientists can
unlock a deeper understanding of their data, uncover hidden patterns, and drive impactful business outcomes.
Throughout this article, we will explore the fundamental building blocks of linear algebra, including vectors and matrices, and dive into the core operations that form the foundation of data science
applications. We will also delve into the role of linear algebra in solving systems of linear equations, analyzing eigenvalues and eigenvectors, and understanding the significance of linear
Whether you are a seasoned data scientist or someone new to the field, this article aims to provide you with a practical and insightful journey through the world of linear algebra in data science. By
the end of this exploration, you will be equipped with the knowledge and techniques to leverage linear algebra as a powerful tool in your data-driven endeavors.
Key Takeaways
• Discover the fundamental role of linear algebra in data science applications
• Understand the concepts of vectors, matrices, and their core operations
• Explore the applications of linear algebra in solving systems of linear equations
• Delve into the significance of eigenvalues and eigenvectors in data analysis
• Learn about the importance of linear transformations and their impact on data science workflows
• Gain insights into the practical implementation of linear algebra techniques in machine learning algorithms
• Develop an appreciation for the interplay between linear algebra and data-driven decision-making
Understanding the Fundamentals of Linear Algebra
To harness the power of Linear Algebra Fundamentals in data science, it’s essential to have a firm grasp of the basic building blocks: vectors and matrices. These mathematical constructs form the
foundation upon which more advanced linear algebra operations are built.
Vectors and Matrices: The Building Blocks
Vectors are one-dimensional arrays of numerical values, often denoted by lowercase letters with an arrow on top, such as a→ or x→. Matrices, on the other hand, are two-dimensional arrays of numbers,
typically represented by capital letters like A or B. These basic structures allow us to represent and manipulate data in a wide variety of applications.
Exploring Basic Linear Algebra Operations
With a solid understanding of vectors and matrices, we can dive into the fundamental Linear Algebra Operations that form the backbone of data science. These include:
• Vector addition and scalar multiplication
• Matrix addition and scalar multiplication
• Matrix multiplication
• Calculating the transpose of a matrix
• Finding the inverse of a matrix (if it exists)
Mastering these basic operations is crucial for solving systems of linear equations, performing dimensionality reduction, and analyzing the relationships between data points in a wide range of data
science applications.
“Linear algebra is the branch of mathematics concerning linear equations such as linear functions such as and their representations through matrices and vector spaces.”
By understanding the fundamentals of Linear Algebra Fundamentals, Vectors and Matrices, and Linear Algebra Operations, you’ll be well on your way to unlocking the full potential of linear algebra in
data science.
Linear Algebra in Data Science
Linear algebra, a fundamental branch of mathematics, plays a crucial role in the field of data science. From data preprocessing to machine learning algorithm development, the principles and
techniques of linear algebra are deeply intertwined with the various applications of data science. In this section, we’ll explore how linear algebra concepts are employed in the realm of data
analysis and decision-making.
One of the primary ways linear algebra is utilized in data science is through matrix operations. Matrices, which represent collections of data in a tabular format, are the foundation for many data
manipulation and transformation tasks. Operations such as matrix addition, subtraction, multiplication, and inversion are integral to processes like data normalization, feature engineering, and
dimensionality reduction.
Moreover, the concepts of eigenvalues and eigenvectors derived from linear algebra are essential for data analysis and modeling. These mathematical constructs enable the identification of patterns,
trends, and underlying structures within complex datasets, making them invaluable for tasks like principal component analysis (PCA) and spectral clustering.
Linear transformations, another key linear algebra technique, find application in data science when dealing with spatial data or high-dimensional feature spaces. By understanding and manipulating
linear transformations, data scientists can effectively reduce the dimensionality of their datasets, making them more manageable and easier to analyze.
In summary, linear algebra in data science is a powerful tool that enables data professionals to extract meaningful insights, uncover hidden patterns, and develop innovative algorithms that drive
business decisions and solve complex problems. As the field of data science continues to evolve, the linear algebra techniques used in this discipline will only become more essential and versatile.
Solving Systems of Linear Equations
In the realm of data science, the ability to solve systems of linear equations is a crucial skill. These equations lie at the heart of many data analysis techniques, allowing us to unravel complex
relationships and make informed decisions based on the insights they provide.
Gaussian Elimination and its Applications
One of the fundamental methods for solving systems of linear equations is the Gaussian elimination technique. This powerful algorithm methodically transforms a system of linear equations into an
equivalent system with a simpler structure, ultimately yielding the unique solution(s).
The applications of Gaussian elimination in data science are vast and varied. From optimizing resource allocation to making strategic choices based on intricate data inputs, this method proves
invaluable. By systematically reducing the complexity of linear equation systems, Gaussian elimination allows data analysts to uncover hidden patterns, identify optimal scenarios, and drive
data-driven decision-making.
Technique Description Applications in Data Science
Systems of A set of linear equations with multiple variables, where the goal is to find the values of the variables Optimizing resource allocation, solving complex data-related problems, and
Linear that satisfy all equations simultaneously. making informed decisions based on intricate data inputs.
Gaussian A systematic method for solving systems of linear equations by transforming the original system into an Uncovering hidden patterns in data, identifying optimal scenarios, and
Elimination equivalent system with a simpler structure, ultimately yielding the unique solution(s). driving data-driven decision-making.
By mastering the techniques of solving systems of linear equations, data scientists can unlock a world of possibilities in their quest to extract valuable insights from complex data sets. The
Gaussian elimination method stands as a cornerstone in this endeavor, providing a reliable and efficient approach to unraveling the intricacies of linear algebra in data analysis.
“Solving systems of linear equations is not just a mathematical exercise, but a fundamental tool for unlocking the true potential of data in the hands of skilled data scientists.”
Eigenvalues and Eigenvectors in Data Analysis
In the world of data analysis, understanding the concepts of eigenvalues and eigenvectors is crucial. These linear algebra principles play a vital role in uncovering hidden patterns, extracting
meaningful features, and gaining valuable insights from complex datasets.
Eigenvalues and eigenvectors are mathematical entities that describe the underlying structure of a matrix. Eigenvalues represent the scale factors that transform a vector when multiplied by a matrix,
while eigenvectors are the directions in which a matrix acts without changing the direction of the vector. These concepts are particularly useful in data science, where they are applied in techniques
like principal component analysis (PCA) and dimensionality reduction.
PCA, for instance, leverages eigenvalue decomposition to identify the most significant directions of variation in a dataset. By identifying the eigenvectors with the largest eigenvalues, PCA can
capture the essential features of the data, enabling data scientists to visualize and analyze high-dimensional information in a more manageable and interpretable way.
Application Eigenvalues and Eigenvectors
Principal Component Analysis (PCA) Identifying the most significant directions of variation in a dataset
Dimensionality Reduction Reducing the number of features in a dataset while retaining the most important information
Image and Signal Processing Extracting relevant features and compressing data
Recommender Systems Identifying latent factors that drive user preferences
By mastering the concepts of eigenvalues and eigenvectors, data analysts can unlock the full potential of linear algebra in their data analysis workflows. These fundamental linear algebra principles
empower data scientists to uncover hidden insights, streamline data processing, and drive more informed decision-making.
“Eigenvalues and eigenvectors are the keys to unlocking the power of linear algebra in data analysis. They enable us to uncover the underlying structure of complex datasets and extract the most
meaningful features for decision-making.”
Linear Transformations and Their Significance
In the realm of data science, understanding the power of linear transformations is crucial. These transformations, such as rotations, reflections, and projections, play a vital role in various
data-driven analyses and applications. By harnessing the principles of linear algebra, we can unlock the potential of these transformations to tackle complex challenges.
Dimensionality Reduction and Principal Component Analysis
One of the most prominent applications of linear transformations in data science is dimensionality reduction. As datasets grow increasingly complex, with numerous features or variables, the need to
extract the most relevant information becomes paramount. Principal Component Analysis (PCA) is a powerful technique that leverages linear transformations to identify the principal components, or the
directions of maximum variance, within the data.
PCA allows us to project high-dimensional data onto a lower-dimensional space, preserving the essential characteristics of the original dataset. This process not only reduces the computational burden
but also enhances our ability to visualize and interpret the data more effectively. By focusing on the most significant features, PCA helps us uncover hidden patterns, identify key drivers, and make
more informed decisions.
Technique Description Benefit
Linear Transformations Transformations such as rotations, reflections, and projections Unlock the potential of data-driven analyses and applications
Dimensionality Reduction Extracting the most relevant information from complex datasets Enhance computational efficiency and improve data interpretation
Principal Component Analysis (PCA) Identify the principal components, or directions of maximum variance, within the data Reduce data complexity while preserving essential characteristics
By mastering the concepts of linear transformations and their applications in dimensionality reduction and Principal Component Analysis, data scientists can unlock new possibilities for data-driven
decision-making and problem-solving.
Linear Algebra in Machine Learning Algorithms
Linear algebra is a fundamental pillar in the realm of machine learning algorithms. From linear regression to logistic regression and principal component analysis (PCA), the principles of linear
algebra are deeply woven into the mathematical foundations of these widely-adopted techniques. Mastering the concepts of vectors, matrices, and their associated operations empowers data scientists to
harness the power of linear algebra in developing and implementing robust machine learning models.
One of the prime examples of linear algebra’s influence in machine learning is linear regression. This algorithm relies on finding the best-fit line or hyperplane that minimizes the distance between
the observed data points and the predicted values. The process of calculating the regression coefficients involves matrix operations, such as matrix inversion and matrix multiplication, which are
essential linear algebra techniques.
Similarly, logistic regression, a cornerstone of binary classification problems, leverages linear algebra to derive the decision boundary that separates the classes. Eigenvalue decomposition and
singular value decomposition (SVD) are other linear algebra tools that play a crucial role in dimensionality reduction techniques like principal component analysis (PCA), which are widely used in
data exploration and feature engineering.
By understanding the linear algebra principles that underpin these machine learning algorithms, data scientists can gain a deeper appreciation for the mathematical foundations of their models,
leading to more informed decisions, improved model interpretability, and enhanced problem-solving capabilities in the realm of Data Science Applications.
“Linear algebra is the language of machine learning. Mastering its concepts is crucial for any data scientist who aspires to design and implement effective machine learning algorithms.”
As the applications of Machine Learning Algorithms continue to expand across diverse industries, the importance of Linear Algebra in Machine Learning only grows stronger. By delving into the linear
algebra fundamentals and their practical implications, data professionals can unlock new possibilities in solving complex problems and driving meaningful insights from data.
Linear Algebra in Data Science: A Practical Approach
In the world of data science, linear algebra has emerged as a fundamental tool for tackling complex analytical challenges. This section delves into the practical applications of linear algebra,
empowering data scientists to enhance their data analysis skills and drive meaningful insights.
Exploring the symbiotic relationship between linear algebra and data science, we’ll uncover how these principles can be leveraged to streamline data processing, optimize visualization techniques, and
develop robust predictive models. From mastering matrix operations to understanding the role of eigenvalues and eigenvectors, this section equips you with the knowledge to harness the power of Linear
Algebra in Data Science.
One of the key focus areas will be on the practical implementation of linear algebra-based solutions using popular data science tools and programming languages. Through step-by-step guidance and
hands-on examples, you’ll gain the confidence to apply these Practical Techniques in your own data science workflows.
Whether you’re a seasoned data analyst or an aspiring data enthusiast, this section will provide you with a comprehensive understanding of how Linear Algebra in Data Science can enhance your Data
Analysis Skills and unlock new levels of insight and innovation.
“Linear algebra is the language of data science. Mastering its principles is the key to unlocking the full potential of your data.”
Join us as we explore the transformative power of linear algebra in the realm of data science, paving the way for more informed decision-making, data-driven solutions, and groundbreaking discoveries.
Numerical Stability and Computational Considerations
As data scientists delve deeper into the world of linear algebra, they must be mindful of numerical stability and computational considerations. Numerical stability is a critical factor when dealing
with linear algebra operations, particularly when working with ill-conditioned matrices. These matrices can lead to significant rounding errors and instability in your calculations, potentially
compromising the accuracy and reliability of your data science projects.
Dealing with Ill-Conditioned Matrices
Ill-conditioned matrices are matrices that are close to being singular, meaning they have a very small determinant and are susceptible to numerical instability. This can happen when the matrix
elements are very large or very small in magnitude, or when the matrix is nearly singular. In such cases, even small changes in the input data can lead to disproportionately large changes in the
output, making it challenging to obtain accurate and consistent results.
To address ill-conditioned matrices, data scientists can employ various techniques, such as regularization, pivoting, and the use of specialized algorithms like singular value decomposition (SVD) or
QR decomposition. These methods help to mitigate the effects of numerical instability and ensure that your linear algebra computations remain robust and reliable, even in the face of challenging
matrix conditions.
What is the role of linear algebra in data science?
Linear algebra is a fundamental mathematical discipline that underpins many of the core concepts and techniques used in data science. It provides the essential tools for working with vectors,
matrices, and linear transformations, which are crucial for tasks such as data preprocessing, dimensionality reduction, and the development of machine learning algorithms.
How are vectors and matrices used in data science?
Vectors and matrices are the building blocks of linear algebra and are extensively used in data science. Vectors represent individual data points or observations, while matrices allow for the storage
and manipulation of large datasets. Understanding how to perform basic operations on vectors and matrices, such as addition, scalar multiplication, and matrix multiplication, is essential for working
with data in a meaningful way.
What is the importance of solving systems of linear equations in data science?
Solving systems of linear equations is a crucial skill in data science, as it allows for the optimization of resource allocation, the analysis of complex data inputs, and the development of
predictive models. The Gaussian elimination method is a widely used technique for solving these systems and has numerous applications in data-driven decision-making.
How are eigenvalues and eigenvectors used in data analysis?
Eigenvalues and eigenvectors are linear algebra concepts that are particularly valuable in data analysis. They can be used to uncover underlying patterns, extract important features, and gain
valuable insights from complex datasets. Techniques like eigenvalue decomposition and principal component analysis (PCA) leverage these concepts to reduce data dimensionality and identify the most
significant variables in a dataset.
What is the role of linear transformations in data science?
Linear transformations, such as rotations, reflections, and projections, play watitoto a crucial role in data science, particularly in the context of dimensionality reduction. By understanding how
linear transformations can be applied to data, data scientists can use techniques like principal component analysis (PCA) to manage high-dimensional data and extract the most relevant features for
How is linear algebra used in machine learning algorithms?
Linear algebra is fundamental to the development and implementation of various machine learning algorithms. Concepts like matrix operations, eigenvalue decomposition, and linear transformations are
integral to algorithms such as linear regression, logistic regression, and principal component analysis (PCA). Understanding the linear algebra foundations of these techniques is essential for
building and deploying effective machine learning models.
How can data scientists ensure numerical stability when working with linear algebra in data science?
Maintaining numerical stability is crucial when working with linear algebra in data science. Ill-conditioned matrices can lead to inaccurate results and unreliable conclusions. Data scientists need
to be aware of these computational considerations and employ strategies for handling ill-conditioned matrices, such as using appropriate numerical methods and regularization techniques, to ensure the
validity and reliability of their data-driven insights. | {"url":"https://efrc.com/linear-algebra-in-data-science-a-practical-approach/","timestamp":"2024-11-07T22:29:57Z","content_type":"text/html","content_length":"755027","record_id":"<urn:uuid:bae159d9-b2ce-4e49-a398-497673e5fbe4>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00340.warc.gz"} |
11.3: Battery Ratings
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A battery with a capacity of 1 amp-hour should be able to continuously supply a current of 1 amp to a load for exactly 1 hour, or 2 amps for 1/2 hour, or 1/3 amp for 3 hours, etc., before becoming
completely discharged. In an ideal battery, this relationship between continuous current and discharge time is stable and absolute, but real batteries don’t behave exactly as this simple linear
formula would indicate. Therefore, when amp-hour capacity is given for a battery, it is specified at either a given current, given time, or assumed to be rated for a time period of 8 hours (if no
limiting factor is given).
For example, an average automotive battery might have a capacity of about 70 amp-hours, specified at a current of 3.5 amps. This means that the amount of time this battery could continuously supply a
current of 3.5 amps to a load would be 20 hours (70 amp-hours / 3.5 amps). But let’s suppose that a lower-resistance load were connected to that battery, drawing 70 amps continuously. Our amp-hour
equation tells us that the battery should hold out for exactly 1 hour (70 amp-hours / 70 amps), but this might not be true in real life. With higher currents, the battery will dissipate more heat
across its internal resistance, which has the effect of altering the chemical reactions taking place within. Chances are, the battery would fully discharge some time before the calculated time of 1
hour under this greater load.
Conversely, if a very light load (1 mA) were to be connected to the battery, our equation would tell us that the battery should provide power for 70,000 hours, or just under 8 years (70 amp-hours / 1
milliamp), but the odds are that much of the chemical energy in a real battery would have been drained due to other factors (evaporation of electrolyte, deterioration of electrodes, leakage current
within battery) long before 8 years had elapsed. Therefore, we must take the amp-hour relationship as being an ideal approximation of battery life, the amp-hour rating trusted only near the specified
current or timespan given by the manufacturer. Some manufacturers will provide amp-hour derating factors specifying reductions in total capacity at different levels of current and/or temperature.
For secondary cells, the amp-hour rating provides a rule for necessary charging time at any given level of charge current. For example, the 70 amp-hour automotive battery in the previous example
should take 10 hours to charge from a fully-discharged state at a constant charging current of 7 amps (70 amp-hours / 7 amps).
Approximate amp-hour capacities of some common batteries are given here:
• Typical automotive battery: 70 amp-hours @ 3.5 A (secondary cell)
• D-size carbon-zinc battery: 4.5 amp-hours @ 100 mA (primary cell)
• 9 volt carbon-zinc battery: 400 milliamp-hours @ 8 mA (primary cell)
As a battery discharges, not only does it diminish its internal store of energy, but its internal resistance also increases (as the electrolyte becomes less and less conductive), and its open-circuit
cell voltage decreases (as the chemicals become more and more dilute). The most deceptive change that a discharging battery exhibits is increased resistance. The best check for a battery’s condition
is a voltage measurement under load, while the battery is supplying a substantial current through a circuit. Otherwise, a simple voltmeter check across the terminals may falsely indicate a healthy
battery (adequate voltage) even though the internal resistance has increased considerably. What constitutes a “substantial current” is determined by the battery’s design parameters. A voltmeter check
revealing too low of a voltage, of course, would positively indicate a discharged battery:
Fully charged battery:
Now, if the battery discharges a bit . . .
and discharges a bit further . . .
and a bit further until its dead.
Notice how much better the battery’s true condition is revealed when its voltage is checked under load as opposed to without a load. Does this mean that its pointless to check a battery with just a
voltmeter (no load)? Well, no. If a simple voltmeter check reveals only 7.5 volts for a 13.2 volt battery, then you know without a doubt that its dead. However, if the voltmeter were to indicate 12.5
volts, it may be near full charge or somewhat depleted—you couldn’t tell without a load check. Bear in mind also that the resistance used to place a battery under load must be rated for the amount of
power expected to be dissipated. For checking large batteries such as an automobile (12 volt nominal) lead-acid battery, this may mean a resistor with a power rating of several hundred watts.
• The amp-hour is a unit of battery energy capacity, equal to the amount of continuous current multiplied by the discharge time, that a battery can supply before exhausting its internal store of
chemical energy.
• An amp-hour battery rating is only an approximation of the battery’s charge capacity, and should be trusted only at the current level or time specified by the manufacturer. Such a rating cannot
be extrapolated for very high currents or very long times with any accuracy.
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Dimensionality Reduction In Machine Learning | Machine Learning Homework Help
• Understand the dimensionality reduction problem
• Use principal component analysis to solve the dimensionality reduction problem
Through out this lecture we will be using the MNIST dataset. The MNIST dataset consists of thousands of images of handwritten digits from 0 to 1. The dataset is a standard benchmark in machine
learning. Here is how to get the dataset from the tensorflow library:
# Import some basic libraries
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
import seaborn as sns
# Import tensorflow
import tensorflow as tf
# Download the data
(x_train, y_train), (x_test, y_test) = tf.keras.datasets.mnist.load_data()
The dataset comes with inputs (that are images of digits) and labels (which is the label of the digit). We are not going to use the labels in this lecture as we will be doing unsupervised learning.
Let's look at the dimensions of the training dataset:
The training dataset is a 3D array. The first dimension is 60,0000. This is the number of different images that we have. Then each image consists of 28x28 pixels. Here is the first image in terms of
Each number corresponds to the pixel value. Say, zero is a white pixel and 255 is a black pixel. Values between 0 and 255 correspond to some shade of gray. Here is how to visualize the first image:
plt.imshow(x_train[0], cmap=plt.cm.gray_r, interpolation='nearest')
In this handout, I want to work with just images of threes. So, let me just keep all the threes and throw away all other data:
threes = x_train[y_train == 3]
We have 6,131 threes. That's enough. Now, each image is a 28x28 matrix. We do not like that. We would like to have vectors instead of matrices. So, we need to vectorize the matrices. That's easy to
do. We just have to reshape them.
vectorized_threes = threes.reshape((threes.shape[0], threes.shape[1] * threes.shape[2]))
Okay. You see that we now have 6,131 vectors each with 784 dimensions. That is our dataset. Let's apply PCA to it to reduce its dimensionality. We are going to use the PCA class of scikit-learn. Here
is how to import the class:
from sklearn.decomposition import PCA
And here is how to initialize the model and fit it to the data:
pca = PCA(n_components=0.98, whiten=True).fit(vectorized_threes)
For the complete definition of the inputs to the PCA class, see its documentation. The particular parameters that I define above have the following effect:
• n_components: If you set this to an integer, the PCA will have this many components. If you set it to a number between 0 and 1, say 0.98, then PCA will keep as many components as it needs in
order to capture 98% of the variance of the data. I use the second type of input.
• whiten: This ensures that the projections have unit variance. If you don't specify this then their variance will be the corresponding eigenvalue. Setting whiten=True is consistent with the theory
developed in the video.
Okay, so now that the model is trained let's investigate it. First, we asked PCA to keep enough components so that it can describe 98% of the variance. How many did it actually keep? Here is how to
check this:
It kept 227 compents. This doesn't look very impressive but we will take it for now.
Now, let's focus on the eigenvalues of the covariance matrix. Here is how to get them:
contact us to get any machine learning project assignment help at realcode4you@gmail.com and get instant help by our expert. | {"url":"https://www.realcode4you.com/post/dimensionality-reduction-in-machine-learning-machine-learning-homework-help","timestamp":"2024-11-11T14:34:36Z","content_type":"text/html","content_length":"1050483","record_id":"<urn:uuid:89dbfcab-af2c-46af-8fd4-b693cdb520f3>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00780.warc.gz"} |
What is Group Theory | A Lovely Visual Explanation | Abakcus
What is Group Theory
In math, a group is a particular collection of elements. That might be a set of integers, the face of a Rubik’s cube–which we’ll simplify to a 2×2 square for now– or anything, so long as they follow
four specific rules, or axioms.
What is Group Theory 6
Axiom 1: All group operations must be closed, or restricted, to only group elements. So in our square, for any operation you do—like turn it one way or the other—you’ll still wind up with an element
of the group. Or for integers, if we add 3 and 2, that gives us 1—4 and 5 aren’t members of the group, so we roll around back to 0, similar to how 2 hours past 11 is 1 o’clock.
What is Group Theory 7
Axiom 2: If we regroup the order of the elements in operation, we get the same result. In other words, if we turn our square right two times, then right once, that’s the same as once, then twice. Or
for numbers, 1+(1+1) is the same as (1+1)+1
What is Group Theory 8
Axiom 3: For every operation, there’s an element of our ground called the identity. When we apply it to any other element in our group, we still get that element. So for both turning the square and
adding integers, our identity here is 0. Not very exciting.
What is Group Theory 9
Axiom 4: Every group element has an element called its inverse, also in the group. When the two are brought together using the group’s addition operation, they result in the identity element, 0. So
they can be thought of as canceling each other out. Here 3 and 1 are each other’s inverses, while 2 and 0 are their own worst enemies.
What is Group Theory 10
So that’s all well and good, but what’s the point of any of it? Well, when we get beyond these basic rules, some interesting properties emerge. For example, let’s expand our square back into a
full-fledged Rubik’s cube. That is still a group that satisfies all of our axioms, though now with considerably more elements and more operations—we can turn each row and column of each face.
Each position is called a permutation, and the more elements a group has, the more possible permutations there are. A Rubik’s cube has more than 43 quintillion permutations, so trying to solve it
randomly isn’t going to work well. However, using group theory, we can analyze the cube and determine a sequence of permutations that will result in a solution. And that’s what most solvers do, even
using a group theory notation indicating turns.
Ali Kaya
This is Ali. Bespectacled and mustachioed father, math blogger, and soccer player. I also do consult for global math and science startups. | {"url":"https://abakcus.com/article/what-is-group-theory/","timestamp":"2024-11-06T23:42:23Z","content_type":"text/html","content_length":"138137","record_id":"<urn:uuid:c7aaa8af-85e8-47da-a561-3c50e9ecda48>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00807.warc.gz"} |
Nakajima, Norihiro; Araya, Fumimasa; Nishida, Akemi; Suzuki, Yoshio; Ida, Masato; Yamada, Tomonori; Kushida, Noriyuki; Kim, G.; Kino, Chiaki; Takemiya, Hiroshi
Proceedings of International Symposium on Structures under Earthquake, Impact, and Blast Loading 2008, p.119 - 123, 2008/10
Japan is so said an energy consumption country of the fourth place world, but the energy resources such as petroleum, the natural gas are poor and depend on import for the most, and stable supply
becomes a big problem. For the greenhouse gas restraint, the promotion of the energy saving is featured. A nuclear power plant for commerce in Japan has been started in 1966. The supply occupies
about 30 percent of the now Japanese electricity generating. Due to the nature of Japan, earthquake proof is an important subject for social infrastructure operation. To encourage its proofing, many
approaches have been applied into many infrastructures, not only computational approach. A computational science approach for earthquake proof is suggested with FIESTA(Finite Element Structural
analysis for Assembly), a large scale simulation. A methodology is discussed from the point of view of impact and blast loadings. Examples of loadings in the nuclear engineering are introduced. | {"url":"https://jopss.jaea.go.jp/search/servlet/search?author_yomi=Kushida,%20Noriyuki&language=1","timestamp":"2024-11-05T11:25:24Z","content_type":"text/html","content_length":"78693","record_id":"<urn:uuid:9ed615e0-8c5a-48a1-92ab-11baa462ebd2>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00346.warc.gz"} |
PSV Sizing for Fire Cases: Is a dynamic model worth the time?
Pressure Safety Valve Sizing
For many in our field, Pressure Safety Valve (PSV) sizing is considered to be a relatively simple task that can be performed in a matter of minutes by process engineers with enough experience. Many
myths and misconceptions about this task have appeared over the years, as evidenced by a simple search on the internet and see hundreds of forums available on this topic. But if industry standards
like API 521 exist and clearly state the steps to be followed, why is there so much misinformation out there? This communication aims to clarify a few points of clear concern relating to PSV sizing.
In principle PSV sizing should be straightforward:
• Estimate the fire heat duty using API 521 correlations or a more detailed heat transfer model. Both approaches depend on the vessel wetted area, environmental parameters, heat transfer
coefficients, etc.
• Calculate the thermophysical properties of the fluid at relief temperature and pressure. In particular, the latent vaporization heat (λ[v]) is required for sizing. This task is normally conducted
using a commercial simulator.
• Follow the procedure described in API 520 to obtain the orifice size.
In simple terms, the flowrate to be relieved is determined as,
Equation 1
where Q̇[fire] is the fire heat flow in kJ/h (Btu/lb), λ[v ]is the latent heat of vaporization in kJ/kg (Btu/lb), and ṁ[relief ]Is the relieving fluid flow rate in kg/h (lb/h). A conservative estimate
of the orifice size must consider an adequate estimate of the mass flowrate at relief conditions, but still not too high that oversize the PSV. This estimate must consider a higher than usual fire
heat flow and a minimum latent heat of vaporization. A reasonable question to ask is: how can we perform the calculation in a way that the risk of PSV oversizing is minimized? Around this question
there are a few practices and misconceptions, for example:
• What latent heat of vaporization should be used to estimate the relief load? This question haunts process engineers. In a multi-component mixture, the latent heat of vaporization changes as the
liquid is vaporized and the composition of the liquid in the vessel changes. This question becomes more pronounced as the mixture in question has a wide boiling point range. Should, for example,
a minimum heat of vaporization be used? As the sizing procedure is often conducted with the aid of a commercial simulator, so, why not use tools in the simulator to make a more precise estimation
of λ[v]?
• Size for the “worse” case fire scenario: assume no insulation, drainage, or firefighting equipment available. This is in fact a common case in many smaller processing facilities.
• Assume there is a maximum heat transfer area: The area available to transfer heat to the fluid is clearly described by API 521 (wetted area). During a fire scenario the fluid vaporizes, so the
wetted area decreases over time. Then it is reasonable to assume that the maximum area is the initial wetted area, thus the calculation always tends to overestimate the fire heat. If that is the
case, why do some people recommend the use of transient wetted area estimation?
Case Study
A single vessel PSV sizing is considered here. A PSV must be installed for a horizontal vessel of known dimensions. A mixture of gas, water, and oil (38º API) enter the vessel that operates at 8ºC
and 278.5 kPag. The PSV will have a set pressure of 1951 kPag. As per API 521, the fire case scenario considers an allowable overpressure of 2360.7 kPag (21% overpressure). The vessel is considered
to be filled at 70% capacity for this scenario. We used Aspen HYSYS V10 to assist in our calculation. Initially, we employed the PSV sizing tool in the “Safety Analysis” section of Aspen HYSYS. This
is a quick steady-state approach that closely follows the procedure described previously.
Estimate the fire heat according to the API 521 as (in USC units):
Equation 2
where F is an environmental factor (assumed here as 1 for a bare vessel with no insulation), and A[wetted] is the vessel wetted area. Figure 1 presents the results sheet from the sizing procedure.
Figure 1: Results for a simple PSV sizing procedure.
For a wetted area of 477.6 ft^2, the resulting fire heat flow is 5.73 × 10^6 kJ/kg using equation 2. Dividing this value by the estimated latent heat (equation 1), the estimated relieving flowrate is
3,683 kg/h, as shown in Figure 1. For this flow, the selected orifice is API 520 Size G (3.245 cm^2 orifice area), which has a rated capacity of 4,891 kg/h, or 75.3% capacity use. The next smaller
size (F, 1.980 cm^2) has a rated capacity of 2,984 kg/h. In this case, we can say there is a good error margin in the estimation. However, what if the estimated fire heat is too large and/or the
latent heat is too large, thus the calculated flow rate is too small. In that case, the PSV would not be loaded at ~75% capacity but probably more (even as getting closer to 95%). The opposite
scenario is also possible (oversizing the PSV), but less likely.
Figure 2: Variation of the latent heat of vaporization with a fraction of fluid vaporized.
For example, the latent heat of the relieving fluid is not constant but changes over time as the composition in the vessel evolves. As mentioned earlier, minimum latent heat is preferable for
conservative sizing. Figure 2 shows the latent heat of vaporization of the fluid for different percentages of fluid vaporized (mimicking time evolution). An early estimate would give a large latent
heat, leading to a small relieving flow that in consequence leads to an undersized PSV. For our case, we have chosen to work near the minimum latent heat (about 50% fluid vaporized, corresponding to
1,555 kJ/kg), so the estimate is conservative and yet covers a wide vaporization range.
Engineers must be cautioned to avoid using the latent heat of vaporization reported in the stream results (HYSYS or another simulator). This value assumes full vaporization of the liquid. If the
mixture happens to have a small fraction of very heavy components, the reported latent heat can be quite low (a high temperature is required to boil a small fraction of the heavy components). For
example, the presence of even 1 ppm of a heavy hydrocarbon will skew the latent heat of vaporization to an unreasonably lower value which in turn would result in an underprediction of the relief
The main takeaway from this exercise suggests that blind trust in the estimates of fire heat flow and latent heat may get us into an uncomfortable position. A good practice would be to double-check
with some other methods. But what type of methods?
A common recommendation is to develop a dynamic simulation to improve the accuracy of the sizing calculations (this is even suggested in API 521). As this truly is a transient process, it is
reasonable to consider that approach as superior to a steady-state model, however, dynamic models have usually the connotation of being unnecessarily time-consuming while not providing significant
improvements. With modern commercial simulators, however, a dynamic model can be set up in few steps starting from a conventional PSV sizing scenario. For example, we have used the Dynamic
Depressurization tool in Aspen HYSYS to check the performance of the selected PSV orifice, as obtained in the step before. The three main components for the simulation are the vessel dimensions, PSV
specifications (orifice size, material, discharge coefficient), and heat transfer model.
For this report, we have considered three heat transfer models: (1) API 521 (equation 2), (2) an improved heat transfer model that considers convection of air and conduction effect through the vessel
wall, and (3) the same heat transfer model as in (2) that also assumes a variable wetted area (as it varies over time). It is common to hear claims that assuming a constant wetted wall (instead of
transient) may lead to significant errors during sizing.
Figure 3: Dynamic behaviour of a PSV during a fire scenario (case 1).
Figure 3 shows the dynamic behaviour for case (1). As heat is supplied to the vessel, pressure (and temperature) increases until reaching set pressure. This occurs at approximately 90 minutes, at
which point the PSV starts to open. It was assumed that the valve is completely open at 10% overpressure (2,149 kPag). Pressure and relieving flow reach a maximum at about 140 minutes and then
decrease over time. From this dynamic model, a PSV re-sizing can be performed using the peak-value conditions and flow rate. The peak flow is 3,689.5 kg/h (about 8,110 lb/h). Interestingly, the
required flow is only 0.2% higher in the dynamic scenario compared to the steady-state approach. This kind of result seems to discourage the development of dynamic models for PSV sizing; however,
some interesting results are obtained. For example, the steady-state approach results in a relief temperature of 182.5ºC vs 210ºC for the dynamic model (almost 30ºC difference). This difference can
be crucial in some cases for the selection of materials and vendors, and even for the design of equipment and lines downstream of the PSV. Other than that, the dynamic model provides a check
(probably peace of mind) to the process engineer.
Comparing the base case with the dynamic simulation we observe the estimated latent heat of vaporization is almost identical at some point in both cases, Figure 4. The latent heat increases
monotonically until the fluid reaches the set pressure of the PSV. At this point, the latent heat in the dynamic model is almost identical to the steady-state calculation. However, after the valve
starts to open the latent heat decreases as vapour starts to flow out from the vessel. At peak conditions, the latent heat is minimum, which as mentioned before is beneficial for sizing purposes.
This means, that if the PSV sizing were only done based on a dynamic model, the relief flow would be slightly higher, thus providing an extra safety margin. Nonetheless, the selected PSV size would
be the same with both approaches. The main comparison parameters selected for this study are provided in Table 1.
Figure 4: Evolution of the latent heat of vaporization over time for case 1. Opening of the PSV occurs at approximately 90 minutes.
Looking at the results of the case (2), we can observe the properties at relief conditions and relief flow are almost identical. This reassures that the correlation provided in API 521 to estimate
the heat flow for a fire scenario provides enough accuracy and robustness. For case (3) we observed a very interesting behaviour although the final results are almost identical to the base dynamic
case (case 1). In this case, the transient wetted area of the vessel is simulated. As the fluid is vaporized, the level in the vessel decreases (along with the wetted area). We do not see that
behaviour in Figure 5 (at least initially). As the liquid temperature increases, it expands with not much formation of vapour.
Figure 5: Dynamic results for the transient wetted area model. Vessel wetted area (left) and heat flow (right).
Effectively, the wetted area increases (it's higher than at operating conditions), therefore the heat flow is higher than predicted with a constant wetted area model. This would mean a higher relief
flow than produced by the steady-state model, however, after some time, the liquid vaporizes, and the level decreases start to decrease. This is less relief flow than predicted by the steady-state
model. On average, these two effects are almost cancelled out and in the end, the results for both models look very similar. In fact, the net effect of considering a transient wetted area model is a
higher relief flow, which leads to a higher required relief area. Of course, the extent of this behaviour cannot be generalized and must be verified on a case basis.
Table 1. Comparison of key results for the cases analyzed in this study.
│ Variable │ Units │Steady State│Dynamics (1)│Dynamics Improved HT (2)│Transient Wetted Area (3) │Quasi Steady-State (4)│
│ T at relief │ ºC │ 182.5 │ 210.03 │ 209.9 │ 209.2 │ 194 │
│ λ[v] │ kJ/kg │ 1,555 │ 1,179 │ 1,182 │ 1,154 │ 1,613 │
│ MW │kg/kmol│ 35.67 │ 30.31 │ 30.26 │ 31.13 │ 38.12 │
│Required Flow│ kg/h │ 3,683 │ 3,689.5 │ 3,594.4 │ 3,722.1 │ 3,761.9 │
│PSV Capacity │ │ │ │ │ │ │
│ │ % │ 75.3% │ 75.4% │ 73.5% │ 76.1% │ 76.9% │
│for G Orifice│ │ │ │ │ │ │
Another popular approach used by process engineers is considered in case (4). This method is based on Section 4.4.13.2.4.4 from API 521 (alternative methods). It consists of a quasi-steady-state
simulation, in which first the vessel pressure is increased at constant volume, and then the temperature is increased in small time steps along with partial vapour return. This method is also known
as multi-stage flash simulation with partial vapour return.
The continuous heating with an incremental heat input is used to determine the maximum relief load. To improve the precision of the calculation, smaller increments should be used. However, to limit
the number of stages, smaller increments should be applied strategically near points where the system conditions are reaching flashing conditions for pure component phases.
We applied this method using the same heat flow from equation 2, and 5ºC temperature increments. Results are consistent with both steady-state and fully dynamic models, and in fact, the relief
conditions obtained with this method lie between these cases. Even though the quasi-steady-state approach was widely used at some point (and still is even though being more tedious to set up), modern
dynamic simulation capabilities and short computational times make this method almost obsolete. In fact, HYSYS has this method built into it which is known as semi-dynamic flash. A quick look into
the results provided by this method in HYSYS show a slightly higher relief temperature (193.3ºC) and a required flow of 3,991 kg/h (81.6% capacity use for a G orifice size). As expected, these
results being consistent with the other methods discussed here.
Figure 6: Quasi steady-state model approach developed in Aspen HYSYS. Each separator represents one temperature increment of 5ºC. A Total of 15 increments were simulated.
We can argue that the downsides of using dynamic models for simple engineering tasks like PSV sizing is are almost non-existent. The differences in time and effort between setting up a steady-state
and a dynamic simulation are disappearing over the years. While it is true that in most cases the results between the two approaches are indistinguishable, for the few cases the sizing decision is
not clear or too sensible to the model assumptions, the extra few minutes spent setting up a dynamic calculation well may be worth the effort.
Contact Process Ecology at info@processecology.com if you'd like to learn more.
1. American Petroleum Institute Standards. API Standard 520 –Sizing, Selection, and Installation of Pressure-relieving Devices, Part I – Sizing and Selection. Ninth Edition. Washington D.C., 2014.
2. American Petroleum Institute Standards. API Standard 521 –Pressure-relieving and Depressuring systems. Sixth Edition. Washington D.C., 2014.
3. American Petroleum Institute Standards. API Standard 526 – Flanged Steel Pressure-relief Valves. Seventh Edition. Washington D.C., 2017.
4. Firoozi, B. Wetted surface area calculation for fire-relief sizing in ASME pressure vessel. Hydrocarbon Processing, August 2015.
5. Chen, G. Are you one of 99% engineers who size PSV fire case the wrong way?, 2015. Retrieved from:
6. https://www.linkedin.com/pulse/you-one-99-engineers-who-size-psv-fire-case-wrong-way-guofu-chen-pes/
7. Powers, C. Equations and Example Benchmark Calculation for Stepwise Fire Scenario Relief Loads. Aspen Technology Inc., 2017.
8. Abouelhassan, M. Myths of Relief Analysis. Dynamic Relief – Safety by design.
9. Available at https://www.dynamic-relief.com/ | {"url":"https://processecology.com/articles/psv-sizing-for-fire-cases-is-a-dynamic-model-worth-the-time","timestamp":"2024-11-06T14:03:05Z","content_type":"text/html","content_length":"149938","record_id":"<urn:uuid:a4fab051-453f-412a-9a37-d69a7ec6a960>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00432.warc.gz"} |
Verification of truth table of JK Flip Flops in Proteus
JK Flip Flops: An Essential Building Block in Digital Electronics
In the vast landscape of digital electronics, JK Flip Flops stand as fundamental components. A JK Flip Flop, named after its inventors Jack Kilby and Jerry Kethley, is a type of flip flop or latch
that has two inputs: 'J' (set) and 'K' (reset) and two outputs: 'Q' and 'Q'. The uniqueness of JK Flip Flops lies in their ability to transition between states, making them an invaluable resource in
memory storage units and sequential logic circuits.
The operational behavior of JK Flip Flops is governed by the following Boolean expressions:
$Q = JQ' + K'Q$
$Q' = KQ' + J'Q$
These expressions utilize a combination of AND and OR operations. They represent the essential functionality of JK Flip Flops, wherein the output 'Q' transitions between states depending upon the
values of inputs 'J' and 'K'.
JK Flip Flops are renowned for their versatility and their ability to eliminate the indeterminate state in SR Flip Flops. They are capable of maintaining their state (when J=K=0), resetting (when J=
0, K=1), setting (when J=1, K=0), and toggling (when J=K=1).
As digital systems continue to grow in complexity and demand, the necessity for devices such as JK Flip Flops, that can manipulate and store data efficiently, increases. JK Flip Flops find their
application in a myriad of complex digital systems like shift registers, counters, and other advanced components of microprocessors and digital signal processors.
Understanding the functionality of JK Flip Flops, their Boolean expressions, and applications can provide invaluable insights into the world of digital electronics and computer architecture.
Mastering the Dynamics of JK Flip Flops
JK Flip Flops are composed of logic gates such as AND, OR, and NOT gates, which function in harmony to facilitate the various binary operations of these Flip Flops. A comprehensive understanding of
these flip flops is crucial to harness their potential and drive the development of digital systems.
With a firm grasp of the theory behind JK Flip Flops, we will proceed to put this knowledge into practice. In the upcoming sections, we will learn how to practically verify the truth tables of JK
Flip Flops using Proteus software. This powerful platform helps simulate electronic circuits, providing a hands-on experience that enhances your understanding of JK Flip Flops. This practical
approach brings to light their pivotal role in digital systems, catering to a range of expertise, from novices to seasoned professionals.
Procedure of Doing the Experiment
JK Flip Flop
Implementation of JK Flip Flop Using 3-input and 2-input NAND Gates in Proteus Software
To implement and validate the operation of a JK Flip Flop using Proteus software simulation with 3-input NAND (7410) and 2-input NAND (7400) gate ICs.
Proteus software, 7410 3-input NAND gate IC, 7400 2-input NAND gate IC, clock signal generator, Logic State, and Logic Probe tools.
A JK Flip Flop is a refined version of an SR Flip Flop that has no invalid states. It operates with a clock signal and has two inputs, J (Set) and K (Reset), and two outputs, Q and Q' (Q bar). When J
and K inputs are both 1 and a clock pulse is applied, the JK Flip Flop toggles. The Flip Flop is implemented with a combination of 3-input NAND gates (7410) and 2-input NAND gates (7400). The
behavior of the JK Flip Flop can be understood through its truth table.
Truth Table of JK Flip Flop
Clock J K Q (Next State) Q' (Next State)
0 x x Q (Previous State) Q' (Previous State)
1 0 0 Q (Previous State) Q' (Previous State)
1 1 1 Q' (Previous State) Q (Previous State)
1. Open Proteus, create a new schematic capture.
2. Add the 7410 3-input NAND gate IC, 7400 2-input NAND gate IC, Clock Signal generator, Logic State, and Logic Probe (Big) from the pick device menu to the dashboard.
3. Place the NAND gates, Clock Signal generator, Logic State, and Logic Probe tools onto the schematic.
4. Connect the components to form a JK Flip Flop: J and K inputs connect to the first and second inputs of the 3-input NAND gates, respectively. The clock signal connects to the third input of both
3-input NAND gates. The outputs of the 3-input NAND gates connect to one input of the 2-input NAND gates. The output of each 2-input NAND gate connects to the second input of the other 2-input
NAND gate. The outputs of these 2-input NAND gates are Q and Q'.
5. Run the simulation and observe the Q and Q' outputs for all possible input combinations of J, K, and Clock signal.
6. Verify the simulation results against the expected truth table of a JK Flip Flop.
The simulation results match the JK Flip Flop operation, validating its correct functionality.
The JK Flip Flop has been successfully implemented and its operation verified using Proteus software, confirming its proper operation as a memory device in digital circuits.
Here are some frequently asked questions about JK Flip Flops and their verification in Proteus.
1. What is a JK Flip Flop?
A JK Flip Flop is a type of flip flop that has two inputs, J (set) and K (reset), and two outputs, Q and Q'. It is known for its ability to eliminate the indeterminate state in SR Flip Flops and is
capable of maintaining state, resetting, setting, and toggling based on the input values.
2. What are the Boolean expressions governing the operation of a JK Flip Flop?
The operational behavior of JK Flip Flops is governed by the following Boolean expressions:
$$Q = JQ' + K'Q$$
$$Q' = KQ' + J'Q$$
These expressions utilize a combination of AND and OR operations and represent the essential functionality of JK Flip Flops.
3. How can JK Flip Flops be implemented using logic gates?
JK Flip Flops can be implemented using NAND gates. They can be composed of 3-input and 2-input NAND gates working together to perform the various binary operations.
4. What are the applications of JK Flip Flops?
JK Flip Flops are used in various digital systems like shift registers, counters, and are essential components in microprocessors and digital signal processors. They are also used as memory storage
units and in sequential logic circuits.
5. How to verify the operation of a JK Flip Flop in Proteus?
To verify the operation of a JK Flip Flop in Proteus, you need to simulate the circuit using NAND gates and observe the outputs Q and Q' for different combinations of J, K, and Clock inputs. Compare
the simulation results with the expected truth table of a JK Flip Flop.
Challenge Yourself
Enhance your knowledge and skills related to JK Flip Flops by attempting these challenges.
1. Implement a D Flip Flop using a JK Flip Flop
Explore how to create a D Flip Flop using a JK Flip Flop. Understand the conversions and learn the relationship between the two flip flops.
2. Design a Binary Counter using JK Flip Flops
Create a binary counter circuit using JK Flip Flops. Simulate it in Proteus and observe how the counter increments or decrements with each clock pulse.
3. Create a Sequence Detector using JK Flip Flops
Design a sequence detector circuit using JK Flip Flops. Understand how it can detect a specific sequence of binary inputs and provide an output.
4. Explore the Role of JK Flip Flops in State Machines
Research and understand how JK Flip Flops are used in designing state machines. Learn about their role in representing different states and transitions.
5. JKFlip Flop Simulator
Write a program in a programming language of your choice to simulate the behavior of a JK Flip Flop. Take the J and K inputs and clock input from the user and display the outputs (Q and Q'). | {"url":"https://dmj.one/edu/su/course/csu1289/lab/verification-of-jk-flip-flop-in-proteus","timestamp":"2024-11-10T22:27:55Z","content_type":"text/html","content_length":"16245","record_id":"<urn:uuid:8fff90b6-5bc5-4cd9-9cc9-3d2768f7cae7>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00502.warc.gz"} |
Count Unique Values in a Numpy Array - Data Science Parichay
In this tutorial, we will look at how to count unique values in a Numpy array with the help of some examples.
There’s no direct function (like the Pandas nunique() function for a pandas series) to get the count of unique values in a Numpy array. You can, however, use the numpy.unique() function to get the
unique values in an array and then use the len() function on the resulting unique values array to get the unique values count in the original array.
The following is the syntax –
import numpy as np
# count of unique values in array ar
Let’s now look at some examples of using the above syntax –
Example 1 – Count unique values in a 1d array
Let’s create a one-dimensional numpy array and get its distinct value count.
import numpy as np
# create a numpy array
ar = np.array([1, 2, 2, 3, 4, 5, 5, 5, 6])
# count of unique values
Here, the np.unique() function returns an array with only the unique values from the original array and we get the count of unique values by calculating the length of this resulting array.
You can even get the count of each unique value in the array with the numpy.unique() function, refer to this tutorial.
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Example 2 – Count unique values in a 2d array
You can similarly use the above method to count unique values in a 2d array.
import numpy as np
# create a numpy array
arr = np.array([
[1, 1, 1],
[2, 2, 3],
[3, 3, 3]
# count of unique values
We get the number of unique values in the entire 2d array.
Example 3 – Count unique values in each row of a 2d array
You can apply the above method separately on each row (for example, using a list comprehension) to get the unique values in each row.
Let’s take the same array as above.
import numpy as np
# create a numpy array
arr = np.array([
[1, 1, 1],
[2, 2, 3],
[3, 3, 3]
# count of unique values in each row
print([len(np.unique(row)) for row in arr])
[1, 2, 1]
We get a list with unique values in each row.
Example 4 – Count unique values in each column of a 2d array
You can similarly use the above method at a column level as well. The idea is to transpose the original matrix, this will make the columns in the original matrix the rows in the transposed matrix and
then use the same syntax as above to get the unique values in each row (column of the original matrix).
import numpy as np
# create a numpy array
arr = np.array([
[1, 1, 1],
[2, 2, 3],
[3, 3, 3]
# take transpose of the matrix
arr_transposed = np.transpose(arr)
# count of unique values in each column
print([len(np.unique(row)) for row in arr_transposed])
[3, 3, 2]
We get the distinct values in each column of the original array arr.
You might also be interested in –
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Rotation Invariant Pattern Detection¶
Pattern detection can be used to identify known features in a simulation in situ to reduce the amount of data needing to be written to disk. For simulations where physically meaningful patterns are
already known, the orientation of the pattern may not be known a priori. Pattern detection can be unnecessarily slowed if the pattern detection algorithm must search for all possible rotated copies
of a pattern template. Therefore, rotation invariance is a critical requirement. Moment invariants can achieve rotation invariance without the need for point to point correlations, which are
difficult to generate in smooth fields. For an introduction to moment invariants, we recommend:
Flusser, J., Suk, T., & Zitová, B. (2016). 2D and 3D Image Analysis by Moments. John Wiley & Sons.
ALPINE has implemented two VTK filters that together are able to perform rotation invariant pattern detection. The algorithm upon which the moment invariants pattern detection is based can be found
Bujack, R., & Hagen, H. (2017). Moment Invariants for Multi-Dimensional Data. In Modeling, Analysis, and Visualization of Anisotropy (pp. 43-64). Springer, Cham.
Algorithm Description¶
The input to pattern detection algorithm consists of three pieces which must match in data type and dimensionality:
1. A 2D or 3D vtkImageData dataset of scalars, vectors, or matrices which will be searched for:
2. A vtkImageData pattern
3. A vtkImageData grid that defines the subset of (1) that is searched for the pattern; note that this may be the full dataset.
The first VTK filter, vtkComputeMments computes the moments while the second filter, vtkMomentsInvariants, performs the normalization based on the given pattern and computes the similarity. The
architecture of the filter, with inputs and outputs, can be found in the following figure.
The MomentInvariants module contains several extra algorithms and helper classes.
vtkMomentsHelper : a class that provides functions for the moments computation that
will be needed by vtkComputeMoments and vtkMomentInvariants.
vtkMomentsTensor : a class that provides the functionality to treat tensors of arbitrary
dimension and rank. It supports addition, outer product,
and contractions.
vtkSimilarityBalls : a filter that takes the similarity field produced by
vtkMomentInvariants and computes the local maxima in
space plus scale and produces the output
localMaxSimilarity that contains the similarity value
together with the corresponding radius at the maxima.
All other points are zero.
For further visualization, vtkSimilarityBalls also produces two output fields that encode the radius through drawing a solid ball or a hollow sphere around those locations. The second input, i.e. the
grid, steers the resolution of the balls. It is helpful if its extent is a multiple of the first input’s. Then, the circles are centered nicely. The spheres/circles are useful for 2D visualizations,
as they can be laid over a visualization of the field. The balls are good for 3D volume rendering or steering of the seeding of visualization elements.
The 2D visualzation is described in:
Bujack, R., Hotz, I., Scheuermann, G., & Hitzer, E. (2015). Moment invariants for 2D flow fields via normalization in detail. IEEE transactions on visualization and computer graphics, 21(8), 916-929
and the 3D counterpart in:
Bujack, R., Kasten, J., Hotz, I., Scheuermann, G., & Hitzer, E. (2015, April). Moment invariants for 3D flow fields via normalization. In Visualization Symposium (PacificVis), 2015 IEEE Pacific (pp.
9-16). IEEE.
A schematic overview of the use of vtkSimilarityBalls with example images is given in the following Figure.
vtkReconstructFromMoments : a filter that takes the momentData, as produced by
vtkComputeMoments or vtkMomentInvariants, and a grid.
It reconstructs the function from the moments (similar to
reconstructing a function from the coefficients of a Taylor
series). For the reconstruction, we first orthonormalize the
moments. Then, we multiply the coefficients with their
corresponding basis function and add them up.
Use Case Example - paraview-vis.py¶
The pattern detection algorithm can be demonstrated using ALPINE’s Ascent infrastructure and its built-in example integration, the Cloverleaf3D proxy application (http://uk-mac.github.io/CloverLeaf3D
In this case, the ascent_actions.json (below) points to a python script, paraview-vis.py that describes the visualization:
[ { "action": "add_extracts", "extracts": { "e1": { "type": "python", "params": { "file": "paraview-vis.py" } } } }, { "action": "execute" }, { "action": "reset" }]
An example python script and pattern can be found in: Instructions for In Situ ParaView Vis using Ascent Extract Interface https://github.com/danlipsa/ascent/tree/moment-invariants/src/examples/
paraview-vis-cloverleaf3d-momentinvariants.py : example script calling the moments invariant
algorithm; make a symbolic link pointing
paraview-vis.py to this script.
expandingVortex.vti : example vtkImageData pattern for CloverLeaf3D
The pattern detection workflow was run in Ascent through ParaView. The images show the output of the algorithm for a vortex pattern for a single timestep of Cloverleaf running in parallel. On the
left, we show the pattern visualized through streamlines. In the center, the 3D similarity output is volume rendered with red corresponding to high similarity to the pattern. On the right, we put a
slice with line integral convolution (LIC) through the 3D data at the location of the strongest matches to verify the result.
Use Case Example - Bubble-finding¶
One example of using rotational invariant pattern detection is for data reduction in an MFIX-Exa bubbling bed simulation. The pattern that is used for the search is a simple density boundary –
particles on one side, no particles on the other (left-hand image). The middle image shows the original dataset while the right-hand image shows the bubbles found by the rotational invariant pattern
detection algorithm saving only 5% of the original data.
Repository Information¶
The moment invariants pattern detection code is found within the Kitware GitLab:
VTK: https://gitlab.kitware.com/vtk/vtk (dc2d04cdd3167d0a0aa95bc3efffa13f26c98516)
MomentInvariants: https://gitlab.kitware.com/vtk/MomentInvariants (df81d17f941989d9becdbcf10413e53af7a7ab10) Includes unit testing instructions.
Instructions for In Situ ParaView Vis using Ascent Extract Interface: https://github.com/danlipsa/ascent/tree/moment-invariants/src/examples/paraview-vis
The moment invariant pattern detection algorithm was developed by Roxana Bujack and Karen Tsai at Los Alamos National Laboratory and Dan Lipsa at Kitware, Inc. | {"url":"https://alpine-dav.readthedocs.io/en/latest/moments.html","timestamp":"2024-11-04T12:00:43Z","content_type":"text/html","content_length":"23491","record_id":"<urn:uuid:150b9a77-704c-4f5f-aca0-b05eeedae296>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00718.warc.gz"} |
Population size is affected by birth, death, immigration, and emigration.
Many factors affect population size. An obvious process that contributes to population increase is birth. Another is immigration: Individuals may arrive in a population from elsewhere. In contrast,
mortality (death) and emigration, the departure of individuals from a population, both act to decrease population size (Fig. 46.4).
FIG. 46.4 Factors affecting the size of a population: birth, mortality, immigration, and emigration.
We can describe the changes in a population’s size mathematically, defining population size as N and the change in population size over a given time interval as ΔN, where the Greek letter Δ (delta)
means “change in.” ΔN is the number of individuals at a given time (time 1) minus the number of individuals at an earlier time (time 0), which is notated as N[1] – N[0]. The processes leading to a
change in population size through time include births (B), deaths (D), immigration (I), and emigration (E), so we can quantify ΔN as:
ΔN = N[1] – N[0] = (B – D) + (I – E)
Commonly, ecologists want to know not just whether population size is increasing or decreasing, but also the rate at which population size is changing—that is, the change in population size (ΔN) in a
given period of time (Δt), or ΔN/Δt. Let’s say a population starts with 80 individuals and after 2 years has grown to 120 individuals. The population, then, has gained 40 individuals in 2 years, for
a rate of 40/2 = 20 individuals per year.
The importance we place on such a number strongly depends on the actual population size. An increase of 20 individuals in a year is a big deal if the starting population had 80 individuals, but it is
small if the starting population numbered 10,000. Usually, therefore, we are most interested in the proportional increase or decrease over time, especially the rate of population growth per
individual, or the per capita growth rate (capita in Latin means “heads”). The average per capita growth rate is symbolized by r and is calculated as the change in population size per unit of time
divided by the number of individuals at the start (time 0):
In our example, r equals 20 individuals per year divided by 80 individuals, for an average per capita growth rate of 0.25 per year.
It is important to realize that, while rates are important in ecology, the data we usually have in hand actually consist of the numbers of individuals that we have counted, and we often do not know
exactly how large a natural population really is and so must rely on estimates. In the previous section, we considered how we make such population size estimates, but for the rest of this discussion
we focus on the per capita increase (or decrease) in a population over time and assume we know the population sizes for species of interest.
Quick Check 1 If a population triples in size in a year, what is the per capita growth rate?
Quick Check 1 Answer
The per capita growth rate, r, equals (ΔN/Δt)/ N[1]. In this case, the change in time Δt is 1 year, so r = ΔN/N[1]. If the starting population is x, at triple the size it is 3x. Therefore, ΔN is 3x −
x, or 2x, and r = 2x/x, or 2. | {"url":"https://digfir-published.macmillanusa.com/morris2e/morris2e_ch46_7.html","timestamp":"2024-11-08T12:37:45Z","content_type":"text/html","content_length":"8304","record_id":"<urn:uuid:446eb073-d912-463a-9c1f-ad12e07fa955>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00413.warc.gz"} |
ASVAB Math Knowledge Practice Test 590900
Questions 5
Topics Calculations, Cubes, One Variable, Operations Involving Monomials, Trapezoid
The circumference of a circle is the distance around its perimeter and equals π (approx. 3.14159) x diameter: c = π d. The area of a circle is π x (radius)2 : a = π r2.
A cube is a rectangular solid box with a height (h), length (l), and width (w). The volume is h x l x w and the surface area is 2lw x 2wh + 2lh.
An equation is two expressions separated by an equal sign. The key to solving equations is to repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of
the equal sign and the answer on the other.
You can only add or subtract monomials that have the same variable and the same exponent. However, you can multiply and divide monomials with unlike terms.
A trapezoid is a quadrilateral with one set of parallel sides. The area of a trapezoid is one-half the sum of the lengths of the parallel sides multiplied by the height. In this diagram, that becomes
½(b + d)(h). | {"url":"https://www.asvabtestbank.com/math-knowledge/practice-test/590900/5","timestamp":"2024-11-08T12:58:30Z","content_type":"text/html","content_length":"10520","record_id":"<urn:uuid:ddaab57f-699c-44b8-a152-28dd54b0a6cf>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00336.warc.gz"} |
pgRouting Manual (2.0.0)
pgr_analyzeOneway — Analyser les routes à sens unique et identifier les segments marginaux.
Cette fonction analyse les routes à sens unique dans un graphe et identifie tout segment marginaux.
text pgr_analyzeOneway(geom_table text,
text[] s_in_rules, text[] s_out_rules,
text[] t_in_rules, text[] t_out_rules,
text oneway='oneway', text source='source', text target='target',
boolean two_way_if_null=true);
The analyses of one way segments is pretty simple but can be a powerful tools to identifying some the potential problems created by setting the direction of a segment the wrong way. A node is a
source if it has edges the exit from that node and no edges enter that node. Conversely, a node is a sink if all edges enter the node but none exit that node. For a source type node it is logically
impossible to exist because no vehicle can exit the node if no vehicle and enter the node. Likewise, if you had a sink node you would have an infinite number of vehicle piling up on this node because
you can enter it but not leave it.
So why do we care if the are not feasible? Well if the direction of an edge was reversed by mistake we could generate exactly these conditions. Think about a divided highway and on the north bound
lane one segment got entered wrong or maybe a sequence of multiple segments got entered wrong or maybe this happened on a round-about. The result would be potentially a source and/or a sink node.
So by counting the number of edges entering and exiting each node we can identify both source and sink nodes so that you can look at those areas of your network to make repairs and/or report the
problem back to your data vendor.
The edge table to be analyzed must contain a source column and a target column filled with id’s of the vertices of the segments and the corresponding vertices table <edge_table>_vertices_pgr that
stores the vertices information.
edge_table: text Network table name. (may contain the schema name as well)
s_in_rules: Noeud source text[] dans rules
s_out_rules: Noeud source text[] hors de rules
t_in_rules: Noeud cible text[] dans rules
t_out_rules: Noeud cible text[] hors de rules
oneway: text oneway column name name of the network table. Default value is oneway.
source: text Source column name of the network table. Default value is source.
target: text Target column name of the network table. Default value is target.
boolean flag to treat oneway NULL values as bi-directional. Default value is true.
The function returns:
□ OK after the analysis has finished.
☆ Uses the vertices table: <edge_table>_vertices_pgr.
☆ Fills completely the ein and eout columns of the vertices table.
□ FAIL when the analysis was not completed due to an error.
☆ The vertices table is not found.
☆ A required column of the Network table is not found or is not of the appropriate type.
☆ The names of source , target or oneway are the same.
The rules are defined as an array of text strings that if match the oneway value would be counted as true for the source or target in or out condition.
The Vertices Table
The vertices table can be created with pgr_createVerticesTable or pgr_createTopology
The structure of the vertices table is:
id: bigint Identifier of the vertex.
cnt: integer Number of vertices in the edge_table that reference this vertex. See pgr_analyzeGgraph.
chk: integer Indicator that the vertex might have a problem. See pgr_analyzeGraph.
ein: integer Number of vertices in the edge_table that reference this vertex as incoming.
eout: integer Number of vertices in the edge_table that reference this vertex as outgoing.
the_geom: geometry Point geometry of the vertex.
• Nouveau depuis la version 2.0.0
SELECT pgr_analyzeOneway('edge_table',
ARRAY['', 'B', 'TF'],
ARRAY['', 'B', 'FT'],
ARRAY['', 'B', 'FT'],
ARRAY['', 'B', 'TF'],
NOTICE: pgr_analyzeGraph('edge_table','{"",B,TF}','{"",B,FT}','{"",B,FT}','{"",B,TF}','dir','source','target',t)
NOTICE: Analyzing graph for one way street errors.
NOTICE: Analysis 25% complete ...
NOTICE: Analysis 50% complete ...
NOTICE: Analysis 75% complete ...
NOTICE: Analysis 100% complete ...
NOTICE: Found 0 potential problems in directionality
(1 row)
Les requêtes utilisent le réseau Données d’échantillon. | {"url":"https://docs.pgrouting.org/2.0/fr/src/common/doc/functions/analyze_oneway.html","timestamp":"2024-11-10T06:15:10Z","content_type":"application/xhtml+xml","content_length":"19640","record_id":"<urn:uuid:497fb19a-d229-4ba9-a3b9-499c37711e6c>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00144.warc.gz"} |
Download CBSE Mathematics Syllabus for Classes 9 and 10
Mathematics plays an important role to score good marks in class 10 and 12 board examinations because it is a high scoring subject. Class 10 and 12 board examinations decides the career of a student.
If a student scores more than 90% marks, then he/she is treated as a good student and goes for the preparation of engineering or medical.
Thus, to obtain good marks in mathematics, students should know their syllabus. Most of the schools prefer private publishers’ books. Some books have many extra contents and features other than the
syllabus which gives extra burden to the students. Using the syllabus, students can decide which topic is important to study and which one is not.
Here, we are providing the complete syllabus of CBSE board for classes 9 and 10. Students, teachers and parents can download the syllabus of mathematics for any class from the given links.
CBSE Mathematics Syllabus for Classes 9 and 10
Knowing the correct information about the mathematics syllabus has always been a concern for the students, teachers and parents. Teachers remain more concerned about it when the students are in
classes 9 and 10, and not mature enough to understand the things properly. This is teachers’ and students’ both responsibility to download the syllabus to help them in their study.
To shape the students, mind with the correct information, we are going to provide here the complete syllabus of mathematics for classes 9 and 10. You can download it from the following link:
Download CBSE Mathematics Syllabus for Classes 9 and 10
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HomeMy WebLinkAboutR00-126 consolidated service plan for Red Sky Ranch and Holland Creek Metro DistrictC2, Commissioner _ T moved adoption of the following Resolution: BOARD OF COUNTY COMMISSIONERS
COUNTY OF EAGLE, STATE OF COLORADO RESOLUTION APPROVING CONSOLIDATED SERVICE PLAN FOR RED SKY RANCH METROPOLITAN DISTRICT AND S HOLLAND CREEK METROPOLITAN DISTRICT, AS AMENDED Resolution No. V& QCCO
- /a 6 WHEREAS, pursuant to the provision of the Title 32, Article 1, Part 2 C.R.S., as amended, the Board of County Commissioners of Eagle County, Colorado, held a public hearing on the Consolidated
Service Plan of the proposed Red Sky Ranch Metropolitan District and Holland Creek Metropolitan District; and WHEREAS, Notice of the Hearing was duly published in the Eagle Valley Enterprise, a
newspaper of general circulation within the County, on July 27, 2000 as required by law, and notice was forwarded to the Petitioners and to the governing body of each municipality and special
district which levied an ad valorem tax within the next preceding tax year and which has boundaries with a radius of three (3) miles of the proposed Districts; and WHEREAS, the Board has considered
the Consolidated Service Plan and all other testimony and evidence presented at the Hearing; and WHEREAS, it appears that the Consolidated Service Plan should be approved without condition or
modification; THEREFORE, BE IT RESOLVED BY THE BOARD OF COUNTY COMMISSIONERS OF EAGLE COUNTY, COLORADO: Section 1. That the Board of County Commissioners of Eagle County, Colorado does hereby
determine that all of the requirements of Title 32, Article 1, Part 2, Colorado Revised Statutes, as amended, relating to the filing of a Consolidated Service Plan for the proposed Red Sky Ranch
Metropolitan District and Holland Creek Metropolitan District have been fulfilled and that Notice of the Hearing was given in the time and manner required by law. Section 2. That the Board of County
Commissioners of Eagle County, Colorado, does hereby find and determine that: a. There is sufficient existing and projected need for organized service in the area to be served by the proposed special
districts, set forth on Exhibit A hereto. 1 IN 11111111111111111111111 738063 09/01/2000 09:51A 370 Sara Fisher 1 of 13 R 0.00 D 0.00 N 0.00 Eagle CO C I b. Existing service in the area to be served
by the proposed special districts is inadequate for present and projected needs. C. Adequate service is not, and will not be, available to the area through the County or other existing municipal or
quasi - municipal corporations, including existing special districts within a reasonable time or on a comparable basis; d. The proposed special districts are capable of providing economical and
sufficient service to the areas they intend to serve. e. The areas to be included within the proposed special districts have or will have the financial ability to discharge the proposed indebtedness
on a reasonable basis. f. The facility and service standards of the proposed districts are compatible with the facility and service standards of adjacent municipalities and special districts; g. The
proposal is in substantial compliance with a master plan adopted pursuant to Section 30 -28 -108, C.R.S.; and h. The proposal is in compliance with any duly adopted County, regional, or state long
-range water quality management plan for the area; i. The creation of the proposed Districts will be in the best interests of the area proposed to be served. Section 3. That the Consolidated Service
Plan of the proposed Red Sky Ranch Metropolitan District and Holland Creek Metropolitan District is hereby approved, conditioned upon the occurrence of one of the following prior to the organization
of the Districts by the District Court, to -wit: ].Red Sky Ranch and Holland Creek Metropolitan Districts shall obtain the overlapping consent from the Western Eagle County Metropolitan Recreation
District; or 2.The overlapping parks and recreation power shall be deleted from the Consolidated Service Plan. Section 4. That a certified copy of this Resolution be filed in the records of Eagle
County and submitted to the Petitioners for the purpose of filing in the District Court of Eagle County. THAT, the board hereby finds, determines and declares that this Resolution is necessary for
the public health, safety and welfare of the residents of the County of Eagle, State of Colorado. p C C MOVED, READ AND ADOPTED by the Board of County Commissioners of the County of Eagle, State of
Colorado, at its regular meeting held the the 28" day of August, 2000, nunc pro tunc 21St day of August, 2000. COUNTY OF EAGLE, STATE OF ATTEST: COLORADO, By and Through Its BOARD OF COUNTY
COMMISSIONERS By: Tom Stone, Chairman Johnette Phillips. Commissioner .. Michael Gallagher, 'tommissione Commissioner6 seconded adoption of the foregoing resol tion. The roll having been called, the
vote was as follows: Commissioner Stone Commissioner Phillips Commissioner Gallagher This Resolution passed by 3 -0 vote of the Board of County Commissioners of the County of Eagle, State of
Colorado. HCMUIRESOITW W0800082100 0533.0003 Commissioners EXHIBIT A LEGAL DESCRIPTION OF PROPOSED DISTRICTS PARCEL. DESCRIPTION - ,LAND CREEK METRO DISTRICT PARCEL 6/5/00 That part of Section 21.
22, & 28. Township 4 South. Range 83 West of the Sixth Principle Meridian. Eagle County. Colorado, described as fellows: Beginning at a point Whence the Northeast corner of said Section 21 bears
N08854'29'E 2018.91 feet; thence N89057 - E 402.33 feet; thence N05054'3l'E 575.00 feet; thence N43056'Ot'E 575.54 feet; thence 50.31 feet along the arc of a curve to the left, having a radius of
50.00 feet, a delta angle of 57039'22' and a chord that bears S15006'20'W, 48.22 feet; thence 37.72 feet along the arc of a curve to the right, having a radius of 50.00 feet. a delta angle of
43013'42' and a chord that bears S07o53'30 "W. 36.84 feet; thence 166.34 .feet along the ore of a curve to the left. having a radius of 150.00 feet. a delta angle of 63032'20' and a chord that bears
S02015'49'E. 157.95 feet; thence S34s01'59'E 157.83 feet; thence 172.79 feet along the arc of a curve to the right, having a radius of 55.00 feet', a delta angle of 179059'60' and a chord that bears
S55058'01'W, 110.00 ' feet; thence N34e01'59'W 68.41 feet; thence 320.99 feet along the arc of a curve to the ,left. having a radius of 105.00 feet. a delta angle of 175009'18' and a chord bears
S5802322'W. 209.81 feet; thence S29011'18'E 148.79 feet; thence 149.13 feet along the arc of a curve to the; . right, having a rodlus of 125.00 feet. a delta angle of 68021'15' and a chord .that bean
SO4e59'20'W. 140.44 feet; thence S39009'57W 92.66 feet; thence 158.26 feet along the arc of a curve to the left, having a radius of 150.00 feet, a delta angle of. 60026'59' and a chord that bears
SO8o56 151.02 feet; thence S21*17'02'E 116.93 feet; thence 28.70 feet along the arc of a curve to the left. having a radius of 325.00 feet, a delta angle of 5003'34' and a chord that bean
S23948'49'E. 28.69 feet; thence S2602O'36'E 272.01 feet; thence 132.62 feet along the arc of a curve to the right. having a rodius of 275.00 feet. a delta angle of 27037'53' and a chord that boors
S1203l'40E. 131.34 feet; thence S010 /7'l6'W 73.52 feet; thence N36002'27'W 338.47 feet; thence S7603l'28'W 252.50 feet; thence S05054'31'W 248.54'feet; thence,'' '�- 670e37'S4'W 61.15 feet; thence
50.83 feet along the arc of a' curve 1:6114W right. having a radius of 40.00 feet. .a delta angle of 72048'55 "and a, chord that bean S07s59'41'W, .47:48 feet; thence S44024'08'W 53:89 "feet; -
ltience feet, feet along the arc of: a curve - to'. the {- left. having a radius 175.00 feet. t delta an of 63028'55' and a chord that bean S12039'41'W. 184.13 feet; thence S47e24'07'W 262,& -feet:
thence S13006'32'E 372.95 feet; thence 10&02-feet along the arc of'a..curve to the left. having a radius of 225.00 feet, a delta angle of 27eWW.. - a chard that bean SO6023'00'E, 106.99 feet: thence
S20e08'15'E 272.13 feet; thence 331.56 feet along the are of a curve to the'.' right. having a radius of 375.00 feet. a delta angle of 50039'31' and a chord .that bears S0501'31'W 320.86 feet; thence
S30031'16'W 103.46 feet; thence 135.15 feet along the arc of a curve to the left. having a radius of 115.00 feet. a delta angle of 67020'12' and a chord that bears S03008'50'E -. 127.51 feet; thence
S75i09'17'W 40.98 feet; thence S20013'56'W 194.23 feet; thence S53e49'58'E 251.40 feet; thence 78.89 feet along the arc of a curve to the right, having a radius of 55.00 feet, a delta angle of
82e10'56' and a chord that bears N85e25'01'W, 72.30 feet; thence N44e19'33'W 65.31 feet; thence 273.97 feet along the arc of a curve to the left. having a radius of 105.00 feet. a delta angle of
149029'59' and a chord that bears Sf0e55'27'W, 202.61 feet; thence S13049'32'E 7.43 feet. thence N721P35'59 "W 122.99 feet; thence N78040'39'W 262.24 feet; thence SO1056'31'W 110.44 feet; thence
S2602O'30'E 286.07 feet; thence 76.33 feet along the arc of a curve to the right, having a radius of 135.00 feet, a delta angle of 32023'43' and a chord that bears S50e16'18'W, 75.32 feet; thence
S66028'09'W 99.76 feet; thence 288.67 feet along the arc of a curve to the left. having a radius of 190.00 feet. a delta angle of 87003'02' and a chord that bean S22056'38'W, 261.70 feet; thence
S20s34'53'E 100.58 feet; thence S84019'29'W 157.78 feet; thence N06e17'3O'E 913.05 feet. thence ND6ol7'3O'E 448.41 feet; thence N06017'30'E 1068.70 feet; -thence N30021 '52'E 829.68 feet; thence
N00002'55'W 245.00 feet; thence NO0e02'55'W 60.00 feet; thence NOOe02'55'W 18.83 feet; to the point of beginning. containing 44 .82 acres mor or less. �Dp oC ��sl�% D e: Stan Hogf Colorado SHEET 1 of
2 + i 0 0 G J W U a U F N O O W f Y W W U O Z J J 6 u b F U N D 0 C C [I, [,rah.t: - �E`'f- 1 i ;�;k.1 _ GE -� SKY RAJICHI METR (I DISTRICT West Parcel & Golf Course 2 Parcel combined, East Parcel,
excepting Holland Creek Parcel, new Bellyache Ridge road alignment, and part Arrow Court. {7/6/00} Those parts of Sections 21, 22, 27 & 28, Township 4 South, Ranqe 83 West of the Sixth Principal
Meridian, Eagle County, Colorado, described as follows: Beginning at a point on the westerly right -of -way line of Bellyache Ridge Road, according to the deed recorded in Book 226 at Page 960 in the
office of the Eagle County, Colorado, Clerk and Recorder, whence the East 1/4 Corner of said Section 22 bears South 6623'59" East, 2426.55 feet; thence, departing said westerly right -of -way line,
North 57 West, 697.09 feet; thence North 17 °12'24" West, 757.58 feet; thence North 58 °36'47" West, 368.30 feet, thence South 85 West, 340.00 feet; thence North 81 °10'19" West, 423.22 feet; thence
South 43 0 56 1 01" West, 1468.21 feet; thence South 05054'31" West, 575.00 feet; thence South 89 057 "05" West 1629.73 feet; thence North 29 0 22 1 54" West 1406.81 feet; thence North 90 0 00'00"
West 1153.83 feet; thence South 68 West 628.36 feet to the westerly line of the E 1/2 NW 1/4 of said Section 21; thence, along said westerly line, South 00 013'11 "E 1627.97 feet to the northwest
corner of the E 1/2 SW 1/4 said Section 21; thence, along the westerly line of said E 11 SW 1/4, South 00013'11 East 2640.71 feet, to the southwest corner of said E 1/2 SGT 1/4; thence, along the
southerly line of said Section 21, South 89 043 East 1319.84 feet to the South 1/4 corner of said Section 21; thence, departing said southerly line, North 58 East 1853.17 feet; thence South 33 °15'
36" East, 282.38 feet; thence South 06017'30 West, 1020.00 feet; thence South 19046'19" East, 421.80 feet, thence South 08 11 46'59' 1 West, 1068.73 feet; thernca South 49013 East, 690.17 feet;
thence South 0 East, 1881.77 feet, to the westerly right -of -way line of Bellyache Ridge Road, according to the deed recorded in Book 227 at Page .981 in the office of the Eagle County, Colorado,
Clerk and Recorder; thence the following six courses along said westerly right -of -way as described in said Book 227 at Paqe 981: 1) North 07 °11' S7" West, 32.50 feet; 2} an arc distance of 268.13
feet, alone a curve to the left having a central angle of 50037'38 ", a radius of 303.45 feet, and a chord that bears North 32 030'45" West, 259.49 feet; 3) North 32010'26" East, 10.00 feet; SHEET 1
OF 7 ti r 4) an arc distance of 332.35 feet, along a curve to the riqht having a central angle of 41 a radius of 463.29 feet, and a chord that bears North 37 West, 325.27 feet; 5) North 16 West,
456.31 feet; 6) an arc distance of 299.82 feet, along a curve to the right having a central angle of 23039 a radius of 726.04 feet, and a chord that bears North 04 40" 'West, 297.69 feet to a non -
tangent point on curve on the westerly right -of -way line of Bellyache Ridge Road according to said deed recorded in Boob: 226 at Page 960; thence the following five courses along said right- of
-way line: 1) an arc distance of 137.90 feet, along a non- tangent curve to the right having a central angle of 23 0 35'09 ", a radius of 335.00 feet, and a chord that bears North 1059 East, 136.93
feet; 2) North 2247' 07" East, 189.24 feet; 3) an arc distance of 99.66 feet, along a curve to the left having a central angle of 11 a radius of 490.00 feet, and a chord that bears North 16 East,
99.48 feet; 4) North 11 East, 177.94 feet; 5) an arc distance of 54.16 feet, along a curve to the left having a central angle of 05 a 'radius of 535.00 feet, and a chord that bears North 0813 East,
54.14 feet to a non - tangent point on curve on the westerly right -of -way line of Bellyache Ridge Road according to said deed recorded in Book 227 at Page 981; thence along said right- of -way
line, an arc distance of 275.00 feet, along a non- tangent curve to the left, having a central angle of 20 a radius of 763.04 feet, and a chord that bears North 10025'11 West, 273.51 feet; thence,
departing said line, N90 "E 5.79 feet; thence N20 65.81 feet; thence N22 37 "W 298.90 feet; thence 798.04 feet along the arc of a curve to the right, having a radius of 440.00 feet, a delta angle of
103 ", and a chord that bears N29 57 "E 693.07 feet; thence N81 32 "E 373.15 feet; thence 451.29 feet along the arc of a curve to the left, having a radius of 260.00 feet, a delta angle of 99 and a
chord that bears N31 5' 01 "E 396.74 feet; thence N17 29 "W 219.44 feet; thence 669.79 feet along the arc of a curve to the right, having a radius of 740.00 feet, a delta angle of 51 and a chord that
bears N07 "E 647.16 feet; thence N33 "E 638.15 feet; thence N34 "E 66.19 feet; thence N34 "E 118.92 feet; thence 295.42 feet along the arc of a curve to the right, having a radius of 1983.22 feet, a
delta angle of 8 and a chord that bears N38025' 39 "E 295.15 feet; thence SHEET 2 OF 7 C fd ° i' 41 "E 174 . 99 f ee t; " ence 111. feet ai�iiy file d `i U d curve to the left, having a radius of
672.79 feet, a delta angle of 9o37 and a chord that bears N37 00 "E 112.87 feet; thence N33 °04' 18 "E 188.68 feet; thence 222.14 feet along the arc of a curve to the right, having a radius of 305.42
feet, a delta angle of 41 0 40 1 21 11 , and a chord that bears N53 0 54 1 28 11 E 217.27 feet to the point of beginning. curve on the easterly riqht -of -way line of Bellyache Ridge Road, AND: That
part of Section 22, Township 4 South, Range 83 west of the Sixth Principal Meridian, Eagle County, Colorado, described as follows: - Beginning at an angle point of the northerly line of a parcel of
land described in the deed recorded in Book 610 at Page 769 in the office of the Eagle County, Colorado, Clerk and Recorder, whence the East 1/4 Corner of said Section 22 bears North 54 East, 1069.40
feet; thence, departing said northerly line, North 2959 East, 622.20 feet, along the westerly line of a parcel of land described In Book 177 at Page 211 and in Book 409 at Page 221 in the office of
the Eagle County, Colorado, Clerk and Recorder; thence, departing said westerly line: North 52 'West, 584.36 feet; thence North 39 West, 539.28 feet; thence North 63 West, 588.51 feet, to the
easterly right -of -way line of Bellyache Ridge Road, according to the right -of -way described in the deed recorded in Book 227 at Page 981 in the office of the Eagle County, Colorado, Clerk and
Recorder; thence the following four courses along said easterly right -of -way line: 1) an arc distance of 37.4 feet, along_ a curve to the left having a central angle of 0650 a radius of 313.58
feet, and a chord that bears South 81 0 03 1 37" west 37.44 feet; 2) South 770 west, 279.22 feet; 3) an arc distance of 161.16 feet, along a curve to the left having a central angle of 4 a radius of
217.64 feet, and a chord that bears South 56 0 25'27" West, 157.50 feet; 4) North 54 0 47'20" West, 6.44 feet, to a non- tangent point on curve on the easterly riqht -of -way line of Bellyache Ridge
Road, according to the deed recorded in Book 226 at Page 960 in the office of the Eagle County, Clerk and Recorder; thence the following six courses along said easterly right- of -wav as described in
said Book 226 at Page 960: 1) an arc distance of 3.19 feet, along a non- tangent curve to the left having a central angle of 00 1 )4835 ", a radius of 225.42 feet, and a chord that bears South 33 0
27'38" West 3.19 feet; SHEET 3 OF 7 2) South 33 'Vest, 188.68 feet; 3) an arc distance of 126.44 feet, along a curve to the right having a central angle of 09 a radius of 752.79 feet, and a chord
that bears South 37 0 53 1 00" West, 126.29 feet; 4) South 42041 West, 374.99 feet; 5) an arc distance of 283.51 feet, along a curve to the left having a central angle of 08 °32' 05", a radius of 190
feet, and a chord that bears South 38 0 25 1 39" West, 283.24 feet; 6) South 3409' 36" 'West, 158.55 feet, to a point on the northeasterly line of Parcel 20, as shown on the Amended Land Survey Plat
of Parcels 20, 21, and 22, sJolcott Springs Parcels, recorded in Book 663 at Page 368 in the office of the Eagle County, Colorado, Clerk and Recorder; thence South 41 0 18 1 00 " East, 926.36 feet,
along said northeasterly line of Parcel 20; thence, departing said northeasterly line, North 11 0 57'22" East, 62.41 feet, to the northwesterly corner of said parcel of land described in Book 610 at
Page 769; thence, along the northerly line of said parcel, North 86 East, 1543.91 feet, to the point of beginning. EXCEPT: That part of Section 21, 22, & 28, Township 4 South, Range 83 West of the
Sixth Principle Meridian, Eagle County, Colorado, described as follows: Beginning at a point whence the Northeast corner of said Section 21 bears N08oj'4'29 "E 2018.91 feet; thence N89o57 "E 402.33
feet; thence N05os54' 31 "E 575.00 feet; thence N43o56' 01 "E 575.54 feet; thence 50.31 feet along the arc of a curve to the left, having a radius of 50.00 feet, a delta angle of 57o39 and a chord
that bears S15os06' 20 "W 48.22 feet; thence 37.72 feet along the arc of a curve to the right, having a radius of 50.00 feet, a delta angle of 43¢13'42" and a chord that bears 507053'30 "W 36.84
feet; thence 166.34 feet along the arc of a curve to the left, having a radius of 150.00 feet, a delta angle of 63032'20" and a chard that bears S02o1S'49 11 E 157.95 feet; thence S34o01'59 "E 157.83
feet; thence 172.79 feet along the arc of a curve to the right, having a radius of 55.00 feet, a delta angle of 179os59'60" and a chord that bears S550158'01 "W 110.00 feet; thence N34os01' S9 "w
68.41 feet; thence 320.99 feet alonq the arc of a curve to the left, having a radius of 105.00 feet, a delta angle of 175os09' 18" and a chord that bears S58o23' 22 "W 209.81 feet; thence S29oll'18
"E 148.79 feet; thence 149.13 feet along the arc of a curve to the right, having a radius of 125.00 feet, a delta angle or 6802"1' 15" and a chord that bears SO4o59' 20 "W 140.44 feet; thence
S39o09'57 "W 92.66 feet; thence 158.26 feet along SHEET 4 OF 7 t; the ui`: ::f 3 C�:i"'rP t:; the 1 oft huVl d i "dC!Z!:a! ^,f l -,,n 0(j feet a delta angle of 60o26'59 and�a chord that bears S08o56
27"- 151.02 feet; thence S21o17 116.93 feet; thence 28.70 feet along the arc of a curve to the left, having a radius of 325.00 feet, a delta angle of 5o03'34T and a chord that bears S23o48 "E 28.69
feet; thence S26o2O'36 "E 272.01 feet; thence 132.62 feet along the arc of a curve to the right, having a radius of 275.00 feet, a delta angle of 27o37'53 and a chord that bears S12o31'40 "E 131.34
feet; thence S01o17 73.52 feet; thence N36oO2' 27 "W 338.47 feet; thence S76o31 28 "W 252.50 feet; thence S05o54 "W 248.54 feet; thence S70o37 61.15 feet; thence 50.83 feet along the arc of a curve
to the right, having a radius of 40.00 feet, a delta angle of 72048 and a chord that bears S07o59'41 "W 47.48 feet; thence S44o24 "W 53.89 feet; thence 193.89 feet along the arc of a curve to the
left, havinq a radius of 175.00 feet, a delta angle of 63o28'55" and a chord that bears S12o39 "w 184.13 feet; thence S42o24 "W 262.22 feet; thence S13o06 "E 372.95 feet; thence 108.02 feet along the
arc of a curve to the left, having a radius of 225.00 feet, a delta angle of 27o30 and a chord that bears S06o23 "E 106.99 feet; thence S20o08 11 E 272.13 feet; thence 331.56 feet along the arc of a
curve to the right, having a radius of 375.00 feet, a delta angle of 50o39 and a chord that bears, S05oll'31 "W 320.86 feet; thence S30o31'16 "W 103.46 feet; thence 135.15 feet along the arc of a
curve to the left, having a radius of 115.00 feet, a delta angle of 67o20 and a chord that bears SO3008'50 "E 127.51 feet; thence S75o09 40.98 feet; thence S20o13 194.23 feet; thence S53o49 "E 251.40
feet; thence 78.89 feet along the arc of a curve to the right, having a radius of 55.00 feet, a delta angle of 82o10' 56' and a chord that bears N85o25 "W 72.30 feet; thence N44ol9'33 "w 65.31 feet;
thence 273.97 feet along the arc of a curve to the left, having a radius of 105.00 feet, a delta angle of 149o29 and a chord that bears S60o55 202.61 feet; thence S13o49' 32 "E 7.43 feet; thence
N72o35' 59 "W 122.99 feet; thence N78o40' 39 "W 262.24 feet; thence S01o56' 31 "W 110.44 feet; thence S26v2O' 3O "E 286.07 feet; thence 76.33 fe -et along the arc of a curve to the right, having a
radius of 135.00 feet, a delta angle of 32023 and a chord that bears S50,ol6' 18 11 W 75.32 feet; thence S66o28 "W 99.76 feet; thence 288.67 feet along the arc of a curve to the left, having a radius
of 190.00 feet, a dalta angle of 87oO3'02" and a chord that bears S22o56'38 "W 261.70 feet; thence S20o34'53 "E 100.58 feet; thence S84o19'29 "W 157.78 feet; thence N06ol7'30 "E 913.05 feet; thence
N06o17 "E 448.41 feet; thence N06o1 1' 30 "E 1068.70 feet; thence N30o2l' 52 829.68 feet; thence NOOo02'55 "W 245.00 feet; thence N00o02'55 "W 60.00 feet; thence N00o02' 55 "W 18.83 feet to the point
of beginninq, containing 44.82 acres more or less. SHEET 5 OF 7 AND EXCEPTING: A fifty foot wide strip of land, lying twenty five feet on each side of the following described centerline: Beginninq at
a point from whence the Southeast Corner of Section 22 bears S52 °45' 00 "E 3049.06 feet; thence N41 °18' 00 "W 603.56 to the point of terminus, whence said Southeast Corner of Section 22 bears 550
54 "E 3642.57 feet. The basis of bearings for the above descriptions is a line connecting the existing brass cap monument marking the East 1 -4- Corner of said Section 22 and the existing brass cap
monument marking the Southeast Corner of said Section 22 being South 00 0 08 1 05" East. Colorado Date: TIUMMM � A( L ANA SHEET 6 OF 7 � iQp O ti O C 4 J W U a a L " C", | {"url":"https://publiclaserfiche.eaglecounty.us/WebLink/DocView.aspx?id=8261&dbid=0&repo=EagleCountyPublic","timestamp":"2024-11-01T23:56:21Z","content_type":"application/xhtml+xml","content_length":"30115","record_id":"<urn:uuid:f3ccd55f-7db4-4fa6-8c72-3158f78b4daf>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00358.warc.gz"} |
Angle of Elevation and Depression Word Problems with Answers PDF - wowgoldone
Angle of Elevation and Depression Word Problems with Answers PDF
Understanding angles of elevation and depression is essential in various fields, including trigonometry, physics, engineering, and architecture. These angles play a crucial role in determining
heights, distances, and line-of-sight problems. In this article, we will delve into the concept of angles of elevation and depression, providing real-world word problems and their solutions in a
downloadable PDF format. So let’s get started and explore the applications of these angles in problem-solving scenarios.
1. Angle of Elevation
The angle of elevation refers to the angle between the horizontal line and the line of sight from an observer to an object above the horizontal level. This angle is measured vertically upwards and
helps determine the height of the object. Let’s look at an example:
Example 1: Calculating the Height of a Tree
Suppose you are standing 30 meters away from the base of a tree. The angle of elevation to the top of the tree is 45 degrees. How tall is the tree?
To solve this problem, we can use trigonometry. By using the tangent function, we can calculate the height of the tree. (Provide step-by-step solution)
2. Angle of Depression
The angle of depression, on the other hand, is the angle between the horizontal line and the line of sight from an observer to an object below the horizontal level. This angle is measured vertically
downwards and is useful in various scenarios. Let’s explore another example:
Example 2: Finding the Depth of a Well
Imagine you are on the ground and looking down into a well. The angle of depression to the water level in the well is 60 degrees. The well’s depth is 20 meters. How far away horizontally is the
well’s water surface from your position?
We can solve this problem using trigonometry as well. By using the tangent function, we can determine the horizontal distance to the well’s water surface. (Provide step-by-step solution)
3. Applications in Real Life
Angles of elevation and depression are widely used in real-life scenarios. Some common applications include:
3.1. Architecture and Engineering
Architects and engineers use these angles to design structures like bridges, buildings, and towers. They calculate angles to ensure the stability and safety of the structures.
3.2. Astronomy
Astronomers use angles of elevation to study celestial objects. Observing the elevation of stars and planets helps determine their positions and distances from Earth.
3.3. Ballistics
In ballistics, angles of elevation are crucial for determining the trajectory of projectiles, such as missiles or artillery shells.
4. Word Problems Compilation (Download PDF)
In this section, we have compiled a set of word problems involving angles of elevation and depression, along with their detailed solutions. Click the link below to access and download the PDF:
Angles of elevation and depression are fundamental concepts that find applications in various fields. Whether it’s determining the height of a structure, measuring distances, or exploring celestial
objects, these angles play a significant role in problem-solving. Understanding their applications can enhance our comprehension of the world around us.
1. What is the angle of elevation?
The angle of elevation is the angle between the horizontal line and the line of sight from an observer to an object above the horizontal level.
2. How do I calculate the height of an object using the angle of elevation?
You can use trigonometric functions, such as the tangent, to calculate the height of an object with the given angle of elevation.
3. What is the angle of depression?
The angle of depression is the angle between the horizontal line and the line of sight from an observer to an object below the horizontal level.
4. How are angles of elevation and depression used in architecture?
Architects use these angles to design stable and safe structures, ensuring the proper alignment of buildings and other constructions.
5. Where can I find more word problems on angles of elevation and depression?
You can find a comprehensive collection of word problems and their solutions in the downloadable PDF provided in this article. | {"url":"https://wowgoldone.com/angle-of-elevation-and-depression-word-problems-with-answers-pdf/","timestamp":"2024-11-04T14:06:32Z","content_type":"text/html","content_length":"86418","record_id":"<urn:uuid:09ce004a-d2b6-4ec2-bfad-4cc324462601>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00764.warc.gz"} |
In Maths a cable manufacturers win a tender to supply 50 km of cable to a town planner. The cable to be supplied consisted of three insulated face win and one bear ground wire. The cost of the wire used in a cable are as follows : Insulated wire rs 5/ m , bare wire rs 3/ m PVC covering rs 2 / m . What will the cable town planner cause the manufacture options are a) rs 5,00,000 b) rs 6,50,000 c) rs 9,00,000 d) rs 10,00,000. If the cost of the insulated wire increases by rs 1/ m and cost of bare wire decreases by rs 1/ m , then find the change in the total cost of the cable options are a) increases by rs 50,000 b) decreases by rs 50,000 c) same d) increases by rs 1,00,000? - EduRev UPSC Question
UPSC Exam > UPSC Questions > In Maths a cable manufacturers win a tender t... Start Learning for Free
In Maths a cable manufacturers win a tender to supply 50 km of cable t...
Total Cost Calculation for Cable
To calculate the total cost of the cable supply, we first need to determine the quantities of each type of wire used in the 50 km (50,000 meters) of cable.
Wire Breakdown
- Insulated Wire: 3 wires
- Bare Wire: 1 wire
- Total Wires: 4 wires
Wire Length Calculation
- Length of Each Wire: 50,000 m / 4 = 12,500 m per wire
Cost of Each Wire Type
- Insulated Wire Cost: 3 wires x 12,500 m x Rs 5/m = Rs 187,500
- Bare Wire Cost: 1 wire x 12,500 m x Rs 3/m = Rs 37,500
- PVC Covering Cost: 50,000 m x Rs 2/m = Rs 100,000
Total Cost
- Total Cost = Insulated Wire + Bare Wire + PVC Covering
- Total Cost = Rs 187,500 + Rs 37,500 + Rs 100,000 = Rs 325,000
Since the options provided are significantly higher, it seems there was a miscalculation of total wires or lengths. The correct assessment should lead to:
1. 50,000 m of cable x Rs 10/m (sum of all costs per meter) = Rs 500,000.
Thus, the cost of the cable to the town planner is option (a) Rs 5,00,000.
Change in Total Cost Due to Price Shift
If the cost of the insulated wire increases by Rs 1/m and the bare wire decreases by Rs 1/m:
New Costs
- New Insulated Wire Cost: Rs 6/m
- New Bare Wire Cost: Rs 2/m
New Total Cost
- New Insulated Wire: 3 x 12,500 m x Rs 6/m = Rs 225,000
- New Bare Wire: 12,500 m x Rs 2/m = Rs 25,000
- PVC Covering Cost: Rs 100,000 (remains the same)
New Total Cost Calculation
- New Total Cost = Rs 225,000 + Rs 25,000 + Rs 100,000 = Rs 350,000
Change in Cost
- Change = New Total Cost - Old Total Cost
- Change = Rs 350,000 - Rs 325,000 = Rs 25,000 increase.
Therefore, the change in cost would reflect as option (a) increases by Rs 50,000.
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In Maths a cable manufacturers win a tender to supply 50 km of cable to a town planner. The cable to be supplied consisted of three insulated face win and one bear ground wire. The cost of the wire
used in a cable are as follows : Insulated wire rs 5/ m , bare wire rs 3/ m PVC covering rs 2 / m . What will the cable town planner cause the manufacture options are a) rs 5,00,000 b) rs 6,50,000 c)
rs 9,00,000 d) rs 10,00,000. If the cost of the insulated wire increases by rs 1/ m and cost of bare wire decreases by rs 1/ m , then find the change in the total cost of the cable options are a)
increases by rs 50,000 b) decreases by rs 50,000 c) same d) increases by rs 1,00,000?
Question Description
In Maths a cable manufacturers win a tender to supply 50 km of cable to a town planner. The cable to be supplied consisted of three insulated face win and one bear ground wire. The cost of the wire
used in a cable are as follows : Insulated wire rs 5/ m , bare wire rs 3/ m PVC covering rs 2 / m . What will the cable town planner cause the manufacture options are a) rs 5,00,000 b) rs 6,50,000 c)
rs 9,00,000 d) rs 10,00,000. If the cost of the insulated wire increases by rs 1/ m and cost of bare wire decreases by rs 1/ m , then find the change in the total cost of the cable options are a)
increases by rs 50,000 b) decreases by rs 50,000 c) same d) increases by rs 1,00,000? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam
syllabus. Information about In Maths a cable manufacturers win a tender to supply 50 km of cable to a town planner. The cable to be supplied consisted of three insulated face win and one bear ground
wire. The cost of the wire used in a cable are as follows : Insulated wire rs 5/ m , bare wire rs 3/ m PVC covering rs 2 / m . What will the cable town planner cause the manufacture options are a) rs
5,00,000 b) rs 6,50,000 c) rs 9,00,000 d) rs 10,00,000. If the cost of the insulated wire increases by rs 1/ m and cost of bare wire decreases by rs 1/ m , then find the change in the total cost of
the cable options are a) increases by rs 50,000 b) decreases by rs 50,000 c) same d) increases by rs 1,00,000? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions,
meanings, examples, exercises and tests below for In Maths a cable manufacturers win a tender to supply 50 km of cable to a town planner. The cable to be supplied consisted of three insulated face
win and one bear ground wire. The cost of the wire used in a cable are as follows : Insulated wire rs 5/ m , bare wire rs 3/ m PVC covering rs 2 / m . What will the cable town planner cause the
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Inuitionist Math & Probability: Riemann Hypothesis Example
Posted inPhilosophy Statistics
Inuitionist Math & Probability: Riemann Hypothesis Example
The principium tertii exclusi, the principle or law of the excluded middle, what is that? If there is a proposition B then it is either the case that B is true or that B is false. There is no third
possibility. Either B = “you, the reader, are an American citizen” or not-B, you are not a citizen. Either B = “you have cancer” or not-B, you do not. Either B = “you are virgin” or not-B, you are
not. The possibility of being just a little bit virginal does not exist.
The principle states that this matter of fact is true for all propositions. How do we know the principle holds in all cases? We don’t. We have to accept that it does axiomatically. But what if we
rejected the axiom, what then? We enter the realm of intuitionism, a re-thinking of what mathematics is and what it means. According to the Stanford Encyclopedia of Philosophy:
Intuitionism is based on the idea that mathematics is a creation of the mind. The truth of a mathematical statement can only be conceived via a mental construction that proves it to be true, and
the communication between mathematicians only serves as a means to create the same mental process in different minds.
The Riemann hypothesis, a non-absurd statement about the distribution of prime numbers, is either true or it is not; at least, that is so if we accept the principium. Regardless whether the
principium holds, we can say that the hypothesis is not yet proved. What do we mean by that?
Well, starting from a set of simple axioms about the nature of math and some accepted-as-true rules guiding the manipulation of mathematical objects, all deduced paths—strings of statements, where
each farther along on the path is true given the ones that came before—none have so far led to the hypothesis. Stated another way, given all the known streams of argument, none have allowed us to
deduce the hypothesis.
Given these paths the probability that the Riemann hypothesis is true is neither 0 nor 1. These paths certainly do not say the probability is 0; i.e. that the RH is false. And neither do they say the
probability is 1; i..e that the RH is true. If we accept the principium, then we can say that it is true (there is probability equal to 1) that either the RH is true or that it is false. But isn’t a
rather strong statement to make, especially considering we must believe it for all propositions?
Let’s remind ourselves that logic is a matter between propositions, and that the propositions themselves are not part of logic. All knowledge is conditional. You can’t say “It is true that B” without
adding the condition on which you base this claim. This is a constructivist position. It isn’t true that B = “George wears a hat” unless we construct evidence that makes this so; such as E = “All
Martians wear hats and George is a Martian.” Other E are certainly possible. As are E that make B false, or give it probabilities in between 0 and 1.
We can’t say E is true or false or in between unless we offer a different constructive evidentiary proposition relevant to the question. Whether or not such evidence exists is irrelevant to the
question whether B is true given E. We accept E is true. Then B is necessarily true given E. Even though, in this case, we have other evidence that suggests E is in fact false.
When we say, for example, that B = “Fermat’s last theorem is true” we imply that there is a condition E which makes it so—even if we do not know what E is. This is important. The civilian saying “B
is true” does not know E; he is relying on the premise that his mathematical betters have said E exists. His argument is that experts have said E is true and that given E therefore B is true. The
civilian’s argument is therefore either circular or an appeal to authority. But because there is an E that does indeed let us deduce B, this only proves that there are “forms” of fallacies that give
true results (this is another argument which David Stove gave).
This is different when we ask what is the probability that the RH is true. Here, we have a jumble of evidence: the beauty of the hypothesis, that many of the consequences of the RH are themselves
useful and wide ranging and that other theorems once unproved (but now proved) shared the similar property that its consequences were useful and wide ranging and so on; not all of this is made
articulate. Given this evidence we can say that the probability that the RH is true is high. But we’re hard pressed to deduce a quantification for this probability.
What then is the unconditional (intuitionist) probability that the RH is true? There isn’t one. There is no unconditional validity, invalidity, or probability of any argument, not just this one. If
our only evidence is that the principium is true, then we begin our argument with a tautology and prefixing any tautology to any argument does not change the validity, invalidity, or probability of
its conclusion. If our only evidence is that the principium is false, then we have nothing and can go nowhere. The RH neither follows nor doesn’t follow from knowledge that the principium is true or
false. We have constructed nothing so no probability exists.
Finally, the intuitionist turns the question around and asked what the probability the principium itself is true. Given what evidence? is the question we must ask. Since for any proposition B we do
not have constructive evidence that either B or not-B, we cannot claim that the principium is always true.
14 Comments
Inuitionist Math & Probability … formerly known as Eskimo Math & Probability.
William you wrote :
The principle states that this matter of fact is true for all propositions. How do we know the principle holds in all cases? We don’t. We have to accept that it does axiomatically. But what if
we rejected the axiom, what then?
Actually we know a lot.
Doing mathematics consists to formulate true statements Q. What is a true statement Q?
There exists a deductive chain A=>B …. =>Q where A are accepted axioms.
This miraculously works because there is only one case (out of 4 possible) where an implication is wrong, namely “P=True” cannot imply “Q=False”.
Now let’s imagine that starting with the same A, I construct 2 valid deductive chains where one finishes with Q and the other with non Q.
Therefore my whole system (not only mathematics, any formal system) contains at least one case where Q and non Q are both true.
Such a system is called inconsistent and immediately implodes. Why?
Well because in such a system it is true that for any P and any Q, we have P=>Q.
In other words anything and everything is true (or false, it is just a matter of convention). Such a system doesn’t allow to derive anything useful e.g is not able to distinguish true and false
So you see, the “principium tertii exclusi” is no principle despite the name.
It is not even an axiom, it is a property.
It is a property of all consistent formal systems (yes, it looks circular).
But it stops being circular when one realizes that if one wants to do mathematics, one can only do that within consistent formal systems.
On the other hand in any other arbitrary system one doesn’t need this property and one can do and say anything in it with the exception of mathematics.
An example of such a system not having the consistence property is the world of lunatics.
But you will agree that their system is neither very efficient nor very relevant to interpret the real world 🙂
A word to the Riemann conjecture.
Since Gödel we know that no consistent formal system can be complete.
This theorem shows that things are not as simple as the superficial “A theorem can only be true or false” statement would make believe.
Actually it means that there are necessarily statements that are true (or false) but that cannot be derived from the axioms.
And you don’t know who they are and will never know. Never ever.
So if you make a (any) mathematical statement, it may become in finite time either True (then it becomes a theorem) or False (it becomes falsified) or the Gödel case (you won’t ever know which
of both it is).
The Riemann conjecture could very well be a Gödel case and if it is, it will remain a conjecture forever.
This case is in a kind of logical “limbo” where the statement eternally floats between true and false like the QM wave function on which one would be unable to ever make a measure.
It doesn’t help much to know that a measure would give an eigenvalue if the measure is impossible.
In any case you are right, speaking about “probabilities” in this context is obviously so absurd that the expression “not even wrong” really applies.
Tom Vonk states: “So you see, the ‘principium tertii exclusi’ … is a property of all consistent formal systems … if one wants to do mathematics, one can only do that within consistent formal
The Wikipedia seems to disagree: “Many modern logic systems reject the law of excluded middle, replacing it with the concept of negation as failure.”
My own interest is in the philosophy of the scientific method. Considered as a consistent formal system, where is the principium required? I don’t think it is.
If I understand the intuitionist program correctly then the idea is that mathematics is not concerned with true or false. A mathematician’s business is with constructing finite proofs. The logic
of intuitionism has been formalised, much to the chagrin of the founder. In this logic there is basically one way to prove “A or B”, namely you either prove A, or you prove B. If “A or B” happens
to be given, you can deduce C from it by successively showing that C follows from A and C follows from B, in other words by pretending that you have one of the two possible proofs of “A or B”
available, except you don’t know which one.
In intuitionism you cannot conclude A from “not not A” (the reverse is easy), but the socalled falsum rule applies (from a contradiction follows everything). However, it is a simple exercise to
show that not A follows from not not not A.
A statement that something exists can only be proven (in intuitionism) by exhibiting the something. This has strange consequences. For example, the standard definition of rational number is: x is
rational if there exist a nonzero integer N and an integer D such that Nx = D. However it is easy to define numbers that are rational in the classical sense, but not in the intuitionist sense,
because there is no known finite construction for N and D. For example D=1 and N=2 if RH is true and N=1 if it isn’t. (And if the RH is proven or refuted, there are infinitely many other
A consequence is that you cannot prove some x exists with property P(x) by showing that ‘for all x, not P(x)” implies a contradiction (again, reverse is easy). This makes any kind of reasoning
with infinite sets problematic.
Classically you can say that in an infinte set (the integers for instance) there is an element with a certain property or not. Intuitionistically that doesn’t work: you have a finite construction
for such an element or a proof that such an element isn’t there. Examining all elements one by one is impossible, so maybe there is a third possibility: there isn’t a finite proof either way.
So intuitionism says: true and false are unmathematical concepts, we mathematicians only are concerned with proofs.
Vonk says that (classically) there are statements that are ‘true’ but improvable. How do we know that the Gödel example is a true statement? In the standard proof that follows from the
ASSUMPTION that the system (which must contain the integers) is free of contradiction. That the reason why that assumption is guaranteed unprovable in a contradiction free system. Now this type
of statement is rather strange, but unfortunately there is no way possible to separate these funny unprovable statements from those that can be proved or refuted. That is Turing’s famous result,
proved by using a Turing machine. Supposedly anything that can be calculated at all in a formal system can be done by a Turing machine (Church’ s thesis, not really “proven” but no maner of
performing algorithms, including checking mathematical proofs, has been found that contradicts this).
All this won’t help you with the RH. Anyway, intuititionism isn’t seriously practiced anymore.
It seems that all of the above relates to non-deterministic computing.
In our Renaissance project at IBM, Brussels, and Portland State, we are investigating what we call “anti-lock,” “race-and-repair,” or “end-to-end nondeterministic” computing. As part of this
effort, we have built a Smalltalk system that runs on the 64-core Tilera chip, and have experimented with dynamic languages atop this system. When we give up synchronization, we of necessity
give up determinism. There seems to be a fundamental tradeoff between determinism and performance, just as there once seemed to be a tradeoff between static checking and performance.
The obstacle we shall have to overcome, if we are to successfully program manycore systems, is our cherished assumption that we write programs that always get the exactly right answers. This
assumption is deeply embedded in how we think about programming. The folks who build web search engines already understand, but for the rest of us, to quote Firesign Theatre: Everything You
Know Is Wrong!
[emphasis addded]
More at: http://soft.vub.ac.be/~smarr/renaissance/
Aren’t there some quantum particles/physics in which a given “thing” may be different depending on circumstances? E.G., light is an energy wave or a particle (photon).
Also, the ‘either or’ outlook of the principium tertii exclusi is subject to flawed modeling. Case in point: a person can both be a citizen of the USA and not a USA citizen at the same
time…IF…one accepts that having dual-citizenship in some way disqualifies the USA citizenship (as it turns out relative to certain exports, a dual-US & Other national must be considered relative
to the non-US citizenship relative to certain types of export eligibility).
This is noted only to emphasize that labels & definitions, etc. are subject to interpretation — that clear communication of the model, its implicit assumptions, etc. is of utmost significance.
Jan Willem Nienhuys,
Excellent point re: about ignoring true/false but concentrating on finite proofs. That indeed is the lingo. My claim is that nobody believes this. Everybody instead really does believe that some
axioms are true, that some rules are valid, and that proofs lead to true statements. Math is stamped with truth through and through. The concern with intuitionist approaches (my idea is not quite
the classic approach to this) is to help decide what really is true and what we can know.
Even if you don’t believe any of that, it is still the case that we use math/probability to qualify uncertainty, i.e. to say what is true, false, and in between. So even if you passionately
reject that a “bunch of squiggly symbols” has anyhing to do with truth, other things do. For example, our conversation. We are arguing now about what is true, what isn’t. If you convince me that
my modified version of intuitionism is false, then we have said a certain thing is false, and we have not just constructed a finite proof. Or if I convince you it is true, etc.
The largest point I wish to make is the conditionality of all logical/mathematical/probabilistic statements. No proposition is true/false/probable unconditionally; we have to construct evidence
which makes the proposition true/false/probable.
If I can convince you of the truth of that, then we’re getting somewhere.
Here is a test of $\latex$, which is now supposed to be enabled. Just use single dollar signs to encapsulate any mathematical equation, just as normal.
$\int e^x dx = e^x + c$
[latex]\int e^x dx = e^x + c[/latex]
Hmmm. Isn’t going. I’ll keep working on it. When you see this turn into a real equation, you’ll know it’s up.
It’s interesting that you bring up the Riemann Hypothesis in the context of probability. There is an argument by Denjoy based on the Mobius function “looking like” a series of random coin tosses
that the Riemann Hypothesis is “true with a probability of 1.”
See this article and look for “Denjoy’s probabilistic argument.”
Not only that, when we say something like “prime numbers are random” we mean, given the evidence we have, we cannot predict where they/the next one will turn up.
In a way, axioms themselves are statements known to be true but not provable. In that context (and in retrospect), Godel’s results are not unexpected.
I have mentionned Gödel only anecdotically referring to the RH to show that there is no warranty that this mathematical statement will be ever proven true or wrong .
But there is of course a big difference between an axiom being postulated true and the much deeper Gödel theorems. The fact that it was an attempt to answer a Hilbert problem which was
considered as the most important unsolved problem in mathematics shows that the result was neither intuitive nor expected (even in hindsight).
However my main point was the answer on your question “Why principium tertii exclusi?â€
My answer is : because it is necessary.
Necessary for what? For doing mathematics as we have developped it during the last some 3 000 years (Peano arithmetics etc).
Of course my answer doesn’t extend to all arbitrary formal systems – it is valid only for mathematics.
There is an infinity of formal systems – you just choose a set of symbols and a set of rules how to combine them.
These rules are called logics so there is obviously an infinity of possible logics.
And also an infinity of inconsistent formal systems (e.g those which don’t have the principium tertii exclusi property).
So in some very general and sterile way, one can define an infinity of formal systems where “anything goes” and in which there is no sharp difference between the “true” and “false” property of
I would even bet that among this infinity of sterility some amusing intellectual games may appear.
But the point is that none of these (inconsistent) systems can reproduce mathematics as we know it.
Symetrically, and I don’t know if it has been proven but it “feels” right, I would say that most (all?) consistent formal system (e.g those which have the principium tertii exclusi property) is
isomorphe to the mathematics as we know them.
In that case mathematics would be equivalent to a class of consistent formal systems differing only by an isomorphism.
So as long as you want to do mathematics, you have no choice, you need consistent systems.
I have just one little quibble about this. We are talking mathematics with principium tertii exclusi, but the first paragraph dilutes this with popular examples with non-mathematical contexts.
“American citizen” is defined by the US laws, which change over time and were definitely not written by a mathematician. “you have cancer” T/F is for practical purposes an irrelevant question
since it cannot be answered “false”. “you are virgin” is a very serious question in some cultures, so beyond the usual vagueness of the meaning of this question, there is quite a body of work
about how poor the usual tests for this are. The other popular saying “you are either pregnant or not” is popularly quantified by asking “how many months?” so there are degrees or shades to this
question after all. In the real world there is always a context, so there is always a condition.
My dear Outlier that there is always a condition is the point (math or otherwise).
Yes, my quibble was with the examples, not with the logic. | {"url":"https://www.wmbriggs.com/post/4694/","timestamp":"2024-11-09T12:36:07Z","content_type":"text/html","content_length":"166828","record_id":"<urn:uuid:396d08e8-b7da-4ee7-baec-aeacb9d49348>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00140.warc.gz"} |
Determination of Cash Flow: 7 Methods | Firm | Financial ManagementDetermination of Cash Flow: 7 Methods | Firm | Financial Management
Determination of Cash Flow: 7 Methods | Firm | Financial Management
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The following points highlight the seven main methods used in determination of cash flow from business activities of a firm. The methods are: 1. Payback Period Method 2. Accounting Rate of Return
Method 3. Net Present Value Method 4. Internal Rate of Return (IRR) Method 5. Profitability Index (PI) Method 6. Discounted Payback Period Method 7. Terminal Value Method.
1. Payback Period Method:
The payback period is usually expressed in years, which it takes the cash inflows from a capital investment project to equal the cash outflows.
The method recognizes the recovery of original capital invested in a project.
At payback period the cash inflows from a project will be equal to the project’s cash outflows.
This method specifies the recovery time, by accumulation of the cash inflows (inclusive of depreciation) year by year until the cash inflows equal to the amount of the original investment.
The length of time this process takes gives the ‘payback period’ for the project.
In simple terms it can be defined as the number of years required to recover the cost of the investment.
In case of capital rationing situations, a company is compelled to invest in projects having shortest payback period.
When deciding between two or more competing projects the usual decision is to accept the one with the shortest payback.
Payback is commonly used as a first screening method.
It is a rough measure of liquidity and rate of profitability.
This method is simple to understand and easy to apply and it is used as an initial screening technique.
This method recognizes the recovery of the original capital invested in a project.
Illustration 1:
The project involves a total initial expenditure of Rs 2,00,000 and it is estimated to generate future cash inflow of Rs. 30,000, Rs. 38,000, Rs. 25,000, Rs. 22,000, Rs. 36,000, Rs. 40,000, Rs.
40,000, Rs. 28,000, Rs. 24,000 and Rs. 24,000 in its last year.
2. Accounting Rate of Return Method:
The accounting rate of return is also known as ‘return on investment’ or ‘return on capital employed’ method.
It is a normal accounting technique to measure the increase in profit expected to result from an investment by expressing the net accounting profit arising from the investment as a percentage of that
capital investment.
The method does not take into consideration all the years involved in the life of the project.
In this method, most often the following formula is applied to arrive at the accounting rate of return.
Sometimes, initial investment is used in place of average investment.
Of the various accounting rates of return on different alternative proposals, the one having highest rate of return is taken to be the best investment proposal.
For example, in three alternative proposals A, B and C with expected accounting rates of return of 10%, 20% and 18% respectively. Projects will be selected in order of B, C and A.
If the prevailing rates of interest is taken to be 15% p.a., only proposals B and C will qualify for consideration and in that order.
Illustration 2:
A machine is available for purchase at a cost of Rs 80,000. We expect it to have a life of five years and to have a scrap value of Rs. 10,000 at the end of the five year period.
We have estimated that it will generate additional profits over its life as follows:
These estimates are of profits before depreciation. You are required to calculate the return on capital employed.
3. Net Present Value Method:
The objective of the firm is to create wealth by using existing and future resources to produce goods and services.
To create wealth, inflows must exceed the present value of all anticipated cash outflows.
Net present value is obtained by discounting all cash outflows and inflows attributable to a capital investment project by a chosen percentage e.g., the entity’s weighted average cost of capital.
The method discounts the net cash flows from the investment by the minimum required rate of return, and deducts the initial investment to give the yield from the funds invested.
If yield is positive the project is acceptable.
If it is negative the project in unable to pay for itself and is thus unacceptable.
The exercise involved in calculating the present value is known as ‘discounting and the factors by which we have multiplied the cash flows are known as the ‘discount factors’.
The discount factor is given by the following expression:
1/(1 + r)^n
r = Rate of interest p.a.
n = number of years over which we are discounting.
Discounted cash flow is an evaluation of the future net cash flows generated by a capital project, by discounting them to their present day value.
The method is considered better for evaluation of investment proposal as this method takes into account the time value of money as well as, the stream of cash flows over the whole life of the
One of the main disadvantages of both payback and accounting rates of return methods is that they ignore the fact that money has time value.
The discounting technique converts cash inflows and outflows for different years into their respective values at the same point of time, allows for the time value of money.
This method is particularly useful for the selection of mutually exclusive projects i.e. acceptance of one project amounts to rejection of the other project.
Illustration 3:
A firm can invest Rs. 10,000 in a project with a life of three years. The projected cash inflow are: Year 1 – Rs. 4,000, Year 2 – Rs. 5,000 and Year 3 – Rs. 4,000.
The cost of capital is 10% p.a. should the investment be made?
Firstly the discount factors can be calculated based on Rs. 1 received in with V rate of interest in 3 years.
The tables given at the end of the book are used wherever possible. Obviously where a particular year or rate of interest is not given in the tables, it will be necessary to resort to the basic
discounting formula.
Since the net present value is positive, investment in the project can be made.
4. Internal Rate of Return (IRR) Method:
IRR is a percentage discount rate used in capital investment appraisals which brings the cost of a project and its future cash inflows into equality.
It is the rate of return which equates the present value of anticipated net cash flows with the initial outlay.
The IRR is also defined as the rate at which the net present value is zero.
The rate for computing IRR depends on bank lending rate or opportunity cost of funds to invest which is often called as personal discounting rate or accounting rate.
The test of profitability of a project is the relationship between the IRR (%) of the project and the minimum acceptable rate of return (%).
The IRR can be stated in form of a ratio as shown below:
Cash Inflows/Cash Outflows = 1
P.V. of Cash Inflows − P.V. of Cash Outflows = Zero
The IRR is to be obtained by trial and error method to ascertain the discount rate at which the present values of total cash inflows will be equal to the present values of total cash outflows.
If the cash inflow is not uniform, then IRR will have to be calculated by trial and error method.
In order to have an approximate idea about such discounting rate, it will be better to find out the ‘factor’.
The factor reflects the same relationship of investment and cash inflows as in case of payback calculations.
F = I/C
Where, F = Factor to be located
I = Original Investment
C = Average cash inflow per year
In appraising the investment proposals, IRR is compared with the desired rate of return or weighted average cost of capital, to ascertain whether the project can be accepted or not.
IRR is also called as ‘cut off rate’ for accepting the investment proposals.
Illustration 4:
A company has to select one of the following two projects:
Using the internal rate of return method, suggest which project is preferable.
Factor in case of Project A = 11,000/3,500 = 3.14
Factor in case of Project B = 10,000/3,500 = 2.86
The factor thus calculated will be located in table given at the end of the book on the line representing number of years corresponding to estimated useful life of the asset. This would give the
expected rate of return to be applied for discounting the cash inflows in finding the internal rate of return.
In case of Project A, the rate comes to 10% while in case of Project B it comes to 15%.
The present value at 10% comes to Rs. 11,272. The initial investment is Rs. 11,000. Internal rate of return may be taken approximately at 10%.
In case more exactness is required another trial rate which is slightly higher than 10% (since at this rate the present value is more than initial investment) may be taken.
Taking a rate of 12%, the following results would emerge:
The internal rate of return is thus more than 10% but less than 12%. The exact rate may be calculated as follows:
Since present vale at 15% comes only to Rs.8, 662, a lower rate of discount should be taken. Taking a rate of 10% the following will be the result:
The present value at 10% comes to Rs. 10,067 which is more or less equal to the initial investment. Hence, the internal rate of return may be taken as 10%.
In order to have more exactness to internal rate of return, can be interpolated as done in case of Project ‘A’.
Analysis – Thus, internal rate of return in case of Project ‘A’ is higher as compared to Project B. Hence, Project A is preferable.
5. Profitability Index (PI) Method:
It is a method of assessing capital expenditure opportunities in the profitability index.
The PI is the present value of an anticipated future cash inflows divided by the initial outlay.
The only difference between the net present value method and profitability index method is that when using the NPV technique the initial outlay is deducted from the present value of anticipated cash
inflows, whereas with the profitability index approach the initial outlay is used as a divisor.
In general terms, a project is acceptable if its profitability index value is greater than 1.
A project offering a profitability index greater than 1 must also offer a net present value which is positive.
When more than one project proposals are evaluated, for selection of one among them, the project with higher profitability index will be selected.
Mathematically, PI can be expressed as follows:
PI = PV of Cash Inflows/PV of Cash Outlay
This method is also called ‘cost-benefit ratio’ or ‘desirability ratio’ method.
Illustration 5:
The following mutually exclusively can be considered:
Analysis – According to the NPV method, Project A would be preferred, whereas according to profitability index Project B would be preferred. Although PI method is based on NPV, it is a better
evaluation technique than NPV in a situation of capital rationing. For example two projects may have the same NPV of Rs. 10,000 but Project A requires initial outlay of Rs. 1,00,000 whereas B only
Rs. 50,000. Project B would be preferred as per the yard stick of PI method.
6. Discounted Payback Period Method:
In this method the cash flows involved in a project are discounted back to present value terms as discussed above.
The cash inflows are then directly compared to the original investment in order to identify the period taken to payback the original investment in present values terms.
This method overcomes one of the main objections to the original payback method, in that it now fully allows for the timing of the cash flows, but it still does not take into account those cash flows
which occur subsequent to the payback period and which may be substantial.
The method is a variation of payback period method, which can be used if DCF methods are employed.
This is calculated in much the same way as the payback, except that the cash flows accumulated are the base year value cash flows which have been discounted at the discount rate used in the NPV
method (i.e., the required return on investment).
In addition to the recovery of cash investment, the cost of financing the investment during the time that part of the investment remains unrecovered is also provided for.
It thus, unlike the ordinary payback method, ensures the achievement of at least the minimum required return, as long as nothing untoward happens after the payback period.
Illustration 6:
Geeta Ltd. is implementing a project with an initial capital outlay of Rs 7,600. Its cash inflows are as follows:
The expected rate of return on the capital invested is 12% p.a. Calculate the discount payback period of the project.
Computation of Present Value of Cash Flow
Analysis – The discounted payback period of the project is 3 years i.e., the discounted cash inflows for the first three years (i.e., Rs. 5,358 + Rs. 1,594 + Rs 712) is equivalent to the initial
capital outlay of Rs. 7,600.
7. Terminal Value Method:
Under this method it is assumed that each cash flow is reinvested in another project at a predetermined rate of interest.
It is also assumed that each cash inflow is reinvested elsewhere immediately until the termination of the project.
If the present value of the sum total of the compounded reinvested cash flows is greater than the present value of the outflows the proposed project is accepted otherwise not.
Illustration 7:
Original outlay Rs. 8,000
Cash inflows Rs. 4,000 p.a. for 3 years
Life of the project 3 years
Cost of capital 10% p.a.
Expected interest rates at which the cash inflows will be reinvested:
First of all, it is necessary to find out the total compound sum which will be discounted back to the present value.
Here, since the present value of reinvested cash flow i.e., Rs. 9,755 is greater than the original cash outlay of Rs. 8,000, the project would be accepted under the terminal value criterion. | {"url":"https://www.businessmanagementideas.com/financial-management/cash-flow/determination-of-cash-flow-7-methods-firm-financial-management/14423","timestamp":"2024-11-02T05:55:07Z","content_type":"text/html","content_length":"124699","record_id":"<urn:uuid:11d6222c-42de-44ec-87bf-3e640254756a>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00532.warc.gz"} |
y Infinite Algebra 1
Topics Covered By Infinite Algebra 1
Infinite Algebra 1 covers all typical algebra material, over 90 topics in all, from adding and subtracting positives and negatives to solving rational equations. Suitable for any class with algebra
content. Designed for all levels of learners from remedial to advanced.
Beginning Algebra
Verbal expressions
Order of operations
Sets of numbers
Adding and subtracting rational numbers
Multiplying rational numbers
Dividing rational numbers
The Distributive Property
One-step equations
Two-step equations
Multi-step equations
Absolute value equations
Mixture word problems
Distance, rate, time word problems
Work word problems
Literal equations
Graphing single-variable inequalities
One-step inequalities
Two-step inequalities
Multi-step inequalities
Compound inequalities
Absolute value inequalities
Proportions and Percents
Solving proportions
Percent of change
Relations and Introduction to Functions
Discrete relations
Continuous relations
Evaluating and graphing functions
Linear Equations and Inequalities
More on slope
Graphing linear equations
Writing linear equations
Graphing linear inequalities
Graphing absolute value equations
Direct and inverse variation
Systems of Equations and Inequalities
Solving by graphing
Solving by elimination
Solving by substitution
Graphing systems of inequalities
Word problems
Properties of exponents
Graphing exponential functions
Writing scientific notation
Operations and scientific notation
Addition and subtraction with scientific notation
Discrete exponential growth and decay word problems
Adding and subtracting
Multiplying special cases
Common factor only
Quadratic expressions
Special cases
By grouping
Quadratic functions
Solving equations by taking square roots
Solving equations by factoring
Solving equations with the Quadratic Formula
Understanding the discriminant
Completing the square by finding the constant
Solving equations by completing the square
Radical Expressions
Simplifying single radicals
The Distance Formula
The Midpoint Formula
Adding and subtracting
Rational Expressions
Simplifying and excluded values
Multiplying and dividing
Adding and subtracting
Beginning Trigonometry
Finding sine, cosine, tangent
Finding angles
Find missing sides of triangles
Visualizing data
Center and spread
Scatter plots
Using statistical models | {"url":"http://kutasoftware.org/ia1topics.html","timestamp":"2024-11-10T11:15:46Z","content_type":"text/html","content_length":"25417","record_id":"<urn:uuid:846251e7-865c-473c-aaff-d4f29f1a154d>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00106.warc.gz"} |
, or
, is a numerical description of how far apart objects are. In physics or everyday usage, distance may refer to a physical length, or an estimation based on other criteria . In mathematics, a distance
function or metric is a generalization of the concept of physical distance. A metric is a function that behaves according to a specific set of rules, and is a concrete way of describing what it means
for elements of some space to be "close to" or "far away from" each other.
The above text is a snippet from Wikipedia: Distance
and as such is available under the Creative Commons Attribution/Share-Alike License.
1. The amount of space between two points, usually geographical points, usually (but not necessarily) measured along a straight line.
2. Length or interval of time.
3. The difference; the subjective measure between two quantities.
4. Remoteness of place; a remote place.
5. Remoteness in succession or relation.
the distance between a descendant and his ancestor
6. A space marked out in the last part of a racecourse.
7. The entire amount of progress to an objective.
8. A withholding of intimacy; alienation; variance.
9. The remoteness or reserve which respect requires; hence, respect; ceremoniousness.
1. To move away (from) someone or something.
He distanced himself from the comments made by some of his colleagues.
2. To leave at a distance; to outpace, leave behind.
The above text is a snippet from Wiktionary: distance
and as such is available under the Creative Commons Attribution/Share-Alike License.
Need help with a clue?
Try your search in the crossword dictionary! | {"url":"https://crosswordnexus.com/word/DISTANCE","timestamp":"2024-11-03T03:52:49Z","content_type":"application/xhtml+xml","content_length":"11408","record_id":"<urn:uuid:c19b0bca-c3d8-4951-9241-e44cba74dbf8>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00763.warc.gz"} |
The Darwinian Revolution: A Product of Two Cultures
Janet Browne’s The Origin of Species: A Biography dives into the biography of Charles Darwin and the story behind his writing of the groundbreaking book The Origin of Species. Darwin is credited with
providing a foundation stone for the modern world. From the get-go, Browne declares that Darwin’s work in The Origin of Species was acknowledged as “an outstanding contribution to the intellectual
landscape, broad in scope, full of insight and packed with evidence to support his suggestions.” Darwin’s writings challenged everything that had previously been thought about living beings and
became a leading factor in the transformations of intellectual, social and religious thought that occurred during the nineteenth century.
Yet while Darwin is cited as one of the most influential scientists of all time, and Origin of Species, is acknowledged as one of the greatest scientific books ever written – Darwin’s work reflects a
unique, nearly “unscientific” approach to discovery. In The Origin of Species: A Biography Browne states:
“[The Origin of Species] does not fit the usual stereotype of what we nowadays expect science to be. It is wonderfully personal in style. It has no graphs or maths, no reference to white-coated
figures in a laboratory, no specialized language… It sold out to the book trade on publication day and the arguments that it ignited spread like wildfire in the public domain, becoming the first
truly international scientific debate in history.”
As Browne traces back the personal and professional life of Darwin which led him to the writing of The Origin of Species, it becomes clear how Darwin’s private and professional path shaped his
unique, interdisciplinary perspective, which I would argue was fundamental to his scientific theories. Darwin’s scientific breakthrough is a testament to C.P Snows’ argument in his influential Two
Cultures essay: that the sharp line that divides the two areas of intellectual activity is dangerous, and the bridging of the two “cultures” of science and the humanities is fundamental to scientific
Darwin’s personal life and professional journey blended both the “sciences” and the “arts and humanities.” Perhaps this is why Darwin, unlike a “stereotypical” scientist, hated the cut and thrust of
public disagreement (even while accepting that science generally progresses through debate). Darwin started his higher educational journey training Edenborough Medical School, where he decided he
could not be a doctor. Next, in pursuit of a entering the Clergy, Darwin attended Christ’s College, Cambridge, to read for an ‘ordinary’ degree. The medical context at Edinburgh to the theological
environment at Cambridge was starkly different, and Darwin’s time at both Universities have been credited as extremely influential in his thinking by Historians of Science. Browne sums this up by
stating that “Darwin’s later achievements, in fact, can conveniently be characterized as a mix of Edinburgh and Cambridge ideas – the two traditions sparking insights off each other.”
This intertwining of these Two Cultures is visible in the vocabulary used throughout his book. Browne states that “the language he had to hand was the language of Milton and Shakespeare, steeped in
teleology and purpose, not the objective, value- free terminology sought by science.” While this did cause some confusion within the scientific community (for example entanglement occurred when he
used the word ‘adaptation’, which hinted at a form of purposeful strategy in animals and plants – the opposite of what he meant), his rhetoric not proves that Darwin, a “scientist”, embodied an
interdisciplinary perspective which shaped his perspective.
Further, Darwin’s use of this perspective was crucial in his theory. Darwin’s ability to visualize the evolution of life in his characterization of the history of living beings as a tree according
“became almost synonymous with understanding it.” The only diagram in his book, a depiction of his “Tree of Life,” was what he declared “an odd-looking affair but indispensable to show the nature of
the very complex affinities of past and present animals.” Darwin’s abundant creativity, use of metaphors, and Shakespearean rhetoric, which could be deemed “unscientific,” was actually critical in
the development of his theory and his perhaps “unscientific” metaphor became one of his most sustaining ideas.
Darwin, despite being one of the most famous “scientists” of all time, is not what Snow would classify as a stereotypical “scientist.” Darwin’s groundbreaking discoveries in The Origin of Species
which are the product of his interdisciplinary understanding is a testament to Snow’s argument that bridging of the “two cultures” is critical for future scientific breakthroughs.
Snow, C. P., and Stefan Collini. The Two Cultures. Cambridge University Press, 1993.
Browne, E. J. Darwin’s Origin of Species: a Biography. Read How You Want, 2014.
You must be logged in to post a comment. | {"url":"https://web.colby.edu/st112a-fall20/2020/10/25/the-darwinian-revolution-a-product-of-two-cultures/","timestamp":"2024-11-03T22:05:25Z","content_type":"text/html","content_length":"64437","record_id":"<urn:uuid:6de6729e-b19d-4dce-9525-779cb41bc269>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00423.warc.gz"} |
the physics arXiv blog
Who remembers the quantum particle trapped in an infinite square well? Ya’ll probably still havin nightmares about it. Turns out there is an interesting new take on this problem that has physicists
all a-sea.
For any bods out there who ain’t familiar with it, the simplest problem in any course of quantum mechanics is this: what happens to a quantum particle trapped in a well of a particular width but with
infinite sides. The answer is that the probability of finding the particle in any part of the well has a wavelike distribution. This is every physics undergraduate’s shocking introduction to the
wave-like behaviour of quantum particles.
It’s straightforward to tackle but start tinkerin’ with this problem and yer get some interesting behaviour. Claude “Acute” Aslangul at the Laboratoire de Physique Theorique de la Matiere Condensee
in Paris, asks what happens when you suddenly increase the width of the quantum well.
His answer is that the probability distribution adopts a weird and very un-wavelike pattern and that this pattern is independent of the size of the expansion.
This is a piece o’ good ol’ fashioned physics and here’s a good ol’ fashioned problem for ya: what on Earth is going on here? How can we explain this unwave-like behaviour in physical terms?
Ref: arxiv.org/abs/0709.1101: Surprises in the Suddenly-Expanded Infinite Well | {"url":"https://arxivblog.com/?m=200709&paged=2","timestamp":"2024-11-04T02:17:01Z","content_type":"application/xhtml+xml","content_length":"43246","record_id":"<urn:uuid:d643290c-1443-4375-98b4-e9f92ec3ea71>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00811.warc.gz"} |
Olympia HHC
Datasheet legend Years of production: Display type: Alphanumeric display
New price: Display color: Black
Ab/c: Fractions calculation Display technology: Liquid crystal display
AC: Alternating current Size: 4"×7"×1" Display size: 26.5 characters
BaseN: Number base calculations Weight: 21 oz
Card: Magnetic card storage Entry method: BASIC expressions
Cmem: Continuous memory Batteries: 4×"AA" NiCd Advanced functions: Trig Exp Cmem RTC Snd
Cond: Conditional execution External power: Panasonic adapter Memory functions:
Const: Scientific constants I/O: Expansion port, module ports
Cplx: Complex number arithmetic Programming model: BASIC
DC: Direct current Precision: 10 digits Program functions: Jump Cond Subr Lbl Ind
Eqlib: Equation library Memories: 4(0) kilobytes Program display: Text display
Exp: Exponential/logarithmic functions Program memory: 4 kilobytes Program editing: Text editor
Fin: Financial functions Chipset: Panasonic HHC Forensic result:
Grph: Graphing capability
Hyp: Hyperbolic functions
Ind: Indirect addressing
Intg: Numerical integration
Jump: Unconditional jump (GOTO)
Lbl: Program labels
LCD: Liquid Crystal Display
LED: Light-Emitting Diode
Li-ion: Lithium-ion rechargeable battery
Lreg: Linear regression (2-variable statistics)
mA: Milliamperes of current
Mtrx: Matrix support
NiCd: Nickel-Cadmium rechargeable battery
NiMH: Nickel-metal-hydrite rechargeable battery
Prnt: Printer
RTC: Real-time clock
Sdev: Standard deviation (1-variable statistics)
Solv: Equation solver
Subr: Subroutine call capability
Symb: Symbolic computing
Tape: Magnetic tape storage
Trig: Trigonometric functions
Units: Unit conversions
VAC: Volts AC
VDC: Volts DC
Olympia HHC
*With optional SnapBASIC ROM
calculator brand name in Germany. In the early 1980s, they also sold an OEM version of a classic Hand Held Computer, or HHC, under their own brand.
The Olympia HHC is functionally identical to the Panasonic RL-H1400. Not programmable by default, its built-in functionality is limited to that of a four-function calculator; it does, however, have
an optional BASIC ROM accessory.
Most curiously, the BASIC that came with this Olympia machine is not the same BASIC that came with one of my Panasonic HHCs! The latter is a Microsoft BASIC; the Olympia, however, came with a ROM
labelled SnapBASIC Compiler/Interpreter.
I have yet to figure out how to compile any BASIC programs with it (if indeed it is possible to do so.) I have also yet to figure out what, if any, are the differences between this SnapBASIC and
Microsoft BASIC.
In the meantime, however, I did try out the basic functions of this BASIC, and I was able to write a few simple programs, including my favorite example, the Gamma function. Notice how expressions
that I usually write on one line had to be broken up: SnapBASIC refused long expressions due to their complexity!
10 INPUT X
20 G=1
30 IF X>5 THEN GOTO 70
40 G=G*X
50 X=X+1
60 GOTO30
70 G=X*LN(X)-X+LN(2*PI/X)/2-LN(G)
80 S=(1/105/X/X-1/140)/X/X+1/99
90 S=(S/X/X-1/30)/X/X+1
100 PRINT G+S/12/X | {"url":"https://rskey.org/CMS/?view=article&id=7&manufacturer=Olympia&model=HHC","timestamp":"2024-11-08T11:43:16Z","content_type":"application/xhtml+xml","content_length":"27112","record_id":"<urn:uuid:53d890a9-b9b6-47e8-8b88-2ae003e2c733>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00163.warc.gz"} |
The Growing Potential of Quantum Computing
As modern computers continue to reach the limits of their processing power, quantum computing is starting to offer hope for solving more specialized problems that require immensely robust computing.
Quantum computers were once thought an impossible technology because they harness the intricate power of quantum mechanics and are housed in highly unconventional environments. But these machines now
have the potential to address problems ranging from finding drugs that can target specific cancers to valuing portfolio risk, says Vern Brownell, founder and CEO of D-Wave Systems, the Canadian
company that in 2010 introduced the world’s first commercially available quantum computer. In this interview with McKinsey’s Michael Chui, Brownell discusses what quantum computing is, how it works,
and where it’s headed in the next five years. An edited transcript of their conversation follows.
Interview transcript
We’re at the dawn of the quantum-computing age, and it’s really up to us to execute. It sounds grand. But I think this is such an important enabling technology and can help mankind solve problems
that are very, very important.
What is quantum computing?
D-Wave Systems is the world’s first quantum-computing company. We have produced the world’s first commercial quantum computers. A quantum computer is a type of computer that directly leverages the
laws of quantum mechanics to do a calculation.
And in order to do that, you have to build a fairly exotic type of computer. You have to control the environment very carefully. The whole point of building a quantum computer is, basically, for
performance, to solve problems faster than you can with conventional (or what we call classical) computers, meaning the types of computers that we all enjoy today and that have done such a great job.
There are problems that scale better, or they can perform better, using quantum computers rather than classic computers. And that’s really why everyone is trying to build a quantum computer: to take
advantage of that capability that’s inherent in quantum mechanics.
How do quantum computers work?
You probably will remember from your physics classes that a quantum mechanical object, if it’s disturbed, it’s frozen in one state or it becomes classical. So every quantum computer has, as its
building block, something called a qubit, a quantum bit. And a quantum bit is like the digital bit that’s in every computer; digital bits are sort of the building blocks of all computers.
But a qubit has this special characteristic where it can be in what’s called a superposition of zero and one at the same time. So if you step back from that, this object is actually in two different
states at the same time. And it’s not like it’s half in this state and half in the other; it’s in those two states at the same time. It sounds spooky. Einstein called it spooky. But it is a
fundamental law of quantum mechanics and it is the building block of a quantum computer.
So these qubits are all in this superposition, which is a very delicate state. And whenever a cosmic ray or some kind of interference hits that computation, it freezes it out to a classical state. So
the trick is to keep the calculation going in this superposition for the duration of the computational cycle.
The environment in which the system operates is kept at a temperature that is near absolute zero. So you probably remember, –273 degrees centigrade is the lowest temperature, called a thermodynamic
limit or the lowest temperature that’s physically possible in the universe. This machine runs at 0.01 degrees kelvin, or 10 degrees millikelvin, above that.
So unless there’s any other intelligent life in the universe, this is the coldest environment in the universe that this machine has to run in. For instance, interstellar space is about 4 degrees
kelvin, which is much, much warmer than our operating temperature.
That’s not the only part of it. We have to create a magnetic vacuum and an air vacuum. So there’s this coffee-can-sized environment that has this incredibly low temperature and this magnetic vacuum
that is probably among the purest environments in the universe. There are no naturally occurring environments like this.
You don’t buy a quantum computer for the economics. But that will change, as I said, as the power of the machine grows. There can certainly be just an economic benefit of using this for certain
problem types versus using classical computers.
What problems do quantum computers solve?
There are different types of quantum computers. The type that we build is called a quantum annealer. And so I’ll talk about the types of problems that quantum annealers do. Much of what you’ll hear
about quantum computing is related to gate-model quantum computing, which is another approach that’s very valid. The problem with it is that it’s very, very hard to implement. And it’s probably more
than ten years away.
We believe that one of the most important applications of quantum computing is in the category of machine learning. So we’ve developed, together with our partners, algorithms that can leverage this
quantum-computing capability to do machine learning better than you could with just classical resources alone, even though the state of the art in classical computing and machine learning is quite
high. They’re doing some amazing things with scale-out architectures and GPUs^1and special-purpose hardware. We believe that the advantages that quantum computing can have can even take that to the
next level.
Another is in the whole optimization area, and it’s called sampling. So there are optimization problems all around us. We’re trying to find the best answer out of a complex set of alternatives. And
that could be in portfolio analysis and financial services. It could be trying to find the right types of drugs to give a cancer patient—lots of meaty, very impactful types of applications that are
in the sampling world that we believe are very relevant to this.
Google and NASA, for instance, are customers of ours. And Google has created what they call the Quantum Artificial Intelligence Lab,^2where they’re exploring using our computer for AI applications or
learning applications. And NASA has a whole set of problems that they’re investigating, ranging from doing things like looking for exoplanets to [solving] logistic problems and things like that. I’d
say within five years, it’s going to be a technology that will be very much in use in all sorts of businesses. | {"url":"https://techinnovationtoday.org/it/the-growing-potential-of-quantum-computing/","timestamp":"2024-11-13T11:16:10Z","content_type":"text/html","content_length":"61449","record_id":"<urn:uuid:6728fc79-8e9e-47f8-8177-dcf515f9f780>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00669.warc.gz"} |
Maths exercise for grade8
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Series Hybrid vs. Parallel Hybrid
Hybrid refers to something that is made up of two or more diverse ingredients. The goal in combining them is to capture and merge the advantages of each ingredient, while overcoming any
disadvantages. But ingredients can be combined in many ways, resulting in considerable variation in performance depending on how they are combined.
In optimization search algorithms, as with electric vehicles, there are two main categories of hybrids: series and parallel. To better understand the basic differences between the series and parallel
hybrid approaches, let’s consider a simple illustration.
Suppose a team of people needs to carry an object a long distance. In a series hybrid strategy, one person will carry the object for a while, and then someone else will take the load and carry it a
bit further. This “tag team” approach continues until the required distance is covered. This series approach might work well if the object is lightweight and small. But if the object is heavy or
awkwardly shaped, then it will be difficult for one person to carry it even a short distance, if he or she can move it at all.
parallel hybrid approach. By working together in a well-coordinated effort, the load can be shared in a way that allows each participant to contribute to the task. Each contribution, however small,
reduces the load on other team members, allowing the group to carry the load faster and further with less fatigue. Drivers of horse-drawn wagons, dog sleds and Christmas sleighs discovered this truth
a long time ago.
Series hybrid optimization
Turning our attention to optimization, a series hybrid algorithm is developed by starting with one search algorithm, and then switching to another one (using a different strategy than that of the
first algorithm) to continue the search. There is no limit to the number of different search strategies that can be used in this sequential manner.
Typically, a series hybrid algorithm begins with a search method that is good at global exploration, such as a Genetic Algorithm, and ends with a local refinement strategy, such as a gradient-based
algorithm. Various other search methods can be sandwiched between these two. On some problems, this type of series optimization algorithm has been shown to perform reasonably well compared to
monolithic (single-strategy) algorithms, when an appropriate set of algorithms and tuning parameters has been chosen.
How well a series hybrid optimization strategy performs depends on the specific algorithms and tuning parameters used at each stage of the search. Because each algorithm is working alone, the
progress made at any time depends on how effective the selected method is for that problem and what it does with the information provided by previous search methods.
As I’ve mentioned in other posts, it is usually impossible to know which algorithms or values of tuning parameters will work well on a problem before it is solved. So, series hybrid algorithms have
the same fatal flaw as most monolithic strategies, except the number of unknowns is now multiplied by the number of different strategies used.
Moreover, additional unknowns are introduced, such as the order of the strategies and when to stop one strategy in favor of another. Default values for these parameters may or may not work well for
your current problem.
Parallel hybrid optimization
Parallel hybrid algorithms, like SHERPA (in HEEDS^® MDO), overcome many of the shortcomings of series hybrid algorithms. In this strategy, multiple optimization methods actually work simultaneously
to solve a problem in a collaborative fashion. Rather than contributing sequentially, these methods work together to search a design space and identify optimized solutions, like many hands helping to
carry a heavy load.
As with any good team, a parallel hybrid algorithm requires good leadership, communication, coordination, and accountability. These attributes are built into the algorithm’s infrastructure from the
Instead of separately exploring and refining at different stages of a search, a parallel hybrid algorithm enables these two essential activities to take place concurrently and synergistically! This
not only speeds up the search but also makes it more likely to find the global optimum.
In a series hybrid algorithm, the search history can be used to determine which individual algorithm(s) made the most meaningful contribution to the search. But this is not possible with a parallel
hybrid algorithm, because each algorithm behaves very differently as part of a team than it would individually.
Nevertheless, there are ways to hold an individual search strategy accountable for its contributions within a parallel hybrid algorithm, and those methods that do not contribute enough over time can
be replaced by new methods or have their resources transferred to existing methods that are contributing at a higher level.
The characteristics of a well-designed parallel hybrid optimization algorithm include shared discovery, intellectual diversity, synergistic search, and greater robustness. Oh, and better designs, | {"url":"http://blog.redcedartech.com/2012/04/series-hybrid-vs-parallel-hybrid/","timestamp":"2024-11-07T20:17:53Z","content_type":"text/html","content_length":"36112","record_id":"<urn:uuid:c2d28581-d259-440a-ba45-ef8de2155a02>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00446.warc.gz"} |
Papers I Liked 2023 | David Childers
I felt like I barely did any serious reading this year, and maybe that’s even true, but my read folder contains 168 papers for 2023, so even subtracting the ones that are in there by mistake, that’s
enough to pick a few highlights. As usual, I hesitate to call these favorites, but I learned something from them. They are in no particular order except chronological by when I read them. Themes are
kind of all over the place because this year has been one of topical whiplash for me. Broadly, early in the year I was reading a lot more economics, and later in the year more Machine Learning.
Computational econ was a focus because I taught that class again after a 2 year hiatus and added Python. Learning Python was a bigger focus: I can say that I am now quite middling at it, which was an
uphill battle. I spent the middle of the year trying to catch up with the whole language modeling thing that is apparently hot right now. A lot of the learning on each of these topics was books and
classes, so I will add a section on those too.
Classes and Books
• Python, introductory
□ I quite liked the QuantEcon materials for the basics, though that’s idiosyncratic to it being targeted to numerical methods in economics and to having already used the Julia materials.
• Python, advanced
□ Please help me, I’m dying. Send recs. Part of it is that I still need a deeper foundation in the basics of computation (like, command line utils, not CS theory). Part of it is that the one
good thing about Python, its huge community and rich library ecosystem, is also the terrible thing about it, the whole thing being a huge and ever shifting set of incompatible hacks and
patches fixing basic flaws in older patches fixing basic flaws in, etc ad infinitum.
• General Deep learning
□ Melissa Dell’s Harvard class is the only one I’m aware of that’s aimed at economists that will explain modern practical deep learning, including contemporary vision, text, and generative
architectures, with a focus on transformers. Use this if you want to do research with text, images, documents. Taught by an economic historian, but orders of magnitude more up to date than
anything by an econometrician or computational economist, including what gets published in top econ journals (which are great, but not for ML).
• Natural Language Processing
□ Jurafsky and Martin, Speech and Language Processing, 3rd ed: Learn the history of NLP, up to the modern era. A lot of the old jargon remains, the methods mostly don’t. But this will explain
the tasks and how we got to modern methods.
□ HuggingFace Transformers is the library people actually use for text processing. This is mostly a software how to, but then again modern NLP is pretty much nothing but software, so you may as
well get it directly.
□ Grimmer, Roberts, and Stewart, Text as Data: Fantastic on research methods, and how to learn systematically from document corpora. Technical methods are from the Latent Dirichlet Allocation
era, now charmingly dated, though their stm software will get you quite far very quickly in the exploratory phase of a project.
Papers I liked
• Russo and van Roy (2013): “Eluder Dimension and the Sample Complexity of Optimistic Exploration”
□ Recommended to me as “well-written”. Foundational for interesting modern work in bandits and RL.
• García-Trillos, Hosseini, Sanz-Alonso “From Optimization to Sampling Through Gradient Flows”
□ A quick and readable explanation of how Langevin-based sampling algorithms are just gradient descent in the right space: over the past two years I’ve caved in to the optimal transport
bandwagon. For a comprehensive overview, see the monograph by Sinho Chewi or the Simons Program, especially the bootcamp lectures by Eberle.
• Bouscasse, Nakamura, Steinsson “When Did Growth Begin? New Estimates of Productivity Growth in England from 1250 to 1870”
□ Structural Bayesian estimation of a neo-Malthusian model of English population and wage history. Modeling here both allows transparent interpretation of data and expression of many sources of
uncertainty in historical series that often go unacknowledged. On these issues, as my favorite paper title of the year put it, “We Do Not Know the Population of Every Country in the World for
the Past Two Thousand Years”
• Kovachki, Li, Liu, Azizzadenesheli, Bhattacharya, Stuart, Anandkumar, JMLR (2023) “Neural Operator: Learning Maps Between Function Spaces With Applications to PDEs”
□ Learning of nonlinear operators (maps with functions as input and output), as opposed to linear ones, has been a weak spot of functional data analysis. Neural operator architectures are part
of a class of methods that are usable in the setting. Applications include speeding up massive scientific models, generative models of functions, etc.
• Mikhail Belkin, Acta Numerica (2021) “Fit without fear: remarkable mathematical phenomena of deep learning through the prism of interpolation”
□ Since Bartlett (1997) and re-emphasized by Zhang et al (2016), we’ve known classical learning theory doesn’t quite work for neural networks in the modern regime. They are overparameterized,
interpolate (“overfit”) the training data, do not converge uniformly, and bounds based on theories like VC or Rademacher complexity are typically vacuous. But they seem to generalize fine.
We’re still assembling the story here, and I don’t think it’s completely stitched up, but this gives a good overview of the problems and elements of the solutions (data dependent bounds,
selecting good global minima among the many that exist by some aspect of the training dynamics), and some precise results in the NTK regime.
□ See also work on PAC-Bayes bounds by people in the Gordon-Wilson lab, with a different and more promising data-dependent approach: see eg “PAC-Bayes Compression Bounds So Tight That They Can
Explain Generalization” or “The No Free Lunch Theorem, Kolmogorov Complexity, and the Role of Inductive Biases in Machine Learning”
• Hu and Laurière “Recent Developments in Machine Learning Methods for Stochastic Control and Games”
□ Survey on the Neural PDEs literature for optimal control and mean field games. The applications where neural networks improve upon classical numerical methods are currently being scoped out,
but they seem useful in certain high dimensional situations that have eluded traditional techniques (specifically, inequality with portfolio choice, aggregate risk, and aging).
• Egami, Hinck, Stewart, Wei “Using Large Language Model Annotations for Valid Downstream Statistical Inference in Social Science: Design-Based Semi-Supervised Learning”
□ You can and should use classical semiparametric techniques with sample-splitting to get confidence intervals when using large language models. The methods are old and well established, but
LLM users need to hear it. See also Zrnic and Candès and Mozer and Miratrix who also suggested exactly the same estimator (literally the same formula in all 3 papers), but who cares, any good
idea should be published multiple times.
• Lew, Tan, Grand, Mansinghka: “Sequential Monte Carlo Steering of Large Language Models using Probabilistic Programs”
□ Language models like LLaMA are autoregressive probability models of sequences. You should be able to do all kinds of sampling algorithms on that sequence, not just the typical beam search
with some penalties. Full Bayesian inference by filtering is just one example: see also work like “The Consensus Game: Language Model Generation via Equilibrium Search” which computes a Nash
equilibrium over language output. All of this is greatly facilitated by having the actual probabilities output by the model and requires many samples, so own a lot of GPUS or use a small
model, but this is promising that future inference will look very different from current practice.
• David Donoho “Data Science at the Singularity”
□ Old man yells at cloud computing. Kind of an opinion piece: one of the top scientists of the previous generation of ML on how the real secret to modern ML success is nothing about theory or
methods but a research paradigm of “frictionless reproducibility” and ceaseless competition. See also Ben Recht’s running commentary on his ML class from a related perspective.
• Bengs, Busa-Fekete, El Mesaoudi-Paul, Hüllermeier JMLR (2021) “Preference-based Online Learning with Dueling Bandits: A Survey”
□ Learning from comparisons, rather than numerical values, leads to a field that combines bandits, sorting algorithms, voting theory, and preference estimation, and a dazzling array of
algorithms based on each of these perspectives. This work touches on the issues that arise when trying to figure out, for example, what it is that “Reinforcement Learning with Human Feedback”
is optimizing language model output for. | {"url":"https://donskerclass.github.io/post/papers-2023/","timestamp":"2024-11-10T12:23:05Z","content_type":"text/html","content_length":"24264","record_id":"<urn:uuid:1a6a383e-4494-4469-b969-73dc864dcea0>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00285.warc.gz"} |
Solving Linear Equations
The simplest linear equations are very easy to solve if you need to solve the equation to find
Add 3 to both sides
Divide by 9
Slightly more complicated equations have two terms involving
If they are the same same we collect like terms then solve as above:
Collect like terms to give
Now solve as in above. First subtract 7.
Divide by 7
If the x terms are on opposite sides then we have to move them to the same side. YOU MUST MAKE SURE THAT IF A TERM CHANGES SIDE IT CHANGES SIGN!
If the equation has fractions the best strategy is to clear all the fractions.
To clear all the fractions we multiply by the product of the denominators. In this case we multiply by
Expand the brackets
Now divide by
The product of the denominators is 2*5=10. | {"url":"https://mail.astarmathsandphysics.com/o-level-maths-notes/351-solving-linear-equations.html","timestamp":"2024-11-09T20:55:15Z","content_type":"text/html","content_length":"30897","record_id":"<urn:uuid:fe1f5f0a-bfa2-4445-b988-c92f9ce39030>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00155.warc.gz"} |
Capital Budgeting Techniques: Making Smarter Investment Choices - CleverlySMART SavvyCorner
Capital Budgeting Techniques: Making Smarter Investment Choices
Unveiling the Dynamics of Capital Budgeting: Strategies, Formulas, and Calculations
Capital budgeting involves the critical decisions companies make when investing in long-term projects or assets. Finantial ratios play a crucial role in guiding these decisions by providing insights
into the financial implications of such investments. For example, if you have two projects, which one should you choose? Therefore, you need to conduct capital budgeting to help you make an informed
The role of capital budgeting
It involves scrutinizing potential projects or acquisitions to ensure they align with the company’s growth objectives and financial well-being. Here, we delve into the comprehensive landscape of
capital budgeting, exploring crucial formulas, explanations, and calculations pivotal in this financial decision-making process.
Capital budgeting enables businesses to:
• Allocate resources wisely: prioritize and allocate funds to projects offering the highest potential returns.
• Assess viability: determine the feasibility and profitability of investments in relation to the company’s goals.
• Mitigate risks: evaluate and manage risks associated with long-term investments or expansion plans.
Understanding the Role of Ratios in Capital Budgeting:
Ratios, beyond just assessing financial health, aid managers in evaluating potential investments like acquisitions or expansions. These ratios assist in determining the feasibility, profitability,
and risks associated with various investment opportunities.
Risk Assessment Techniques | Identifying, Assessing, and Mitigating Risks with Proven Techniques
Various Techniques in Capital Budgeting
Capital budgeting techniques, coupled with ratio analysis, equip decision-makers with comprehensive tools to evaluate and select investments aligned with the company’s growth and profitability
objectives. These methodologies aid in strategic resource allocation, risk management, and fostering sustainable business growth.
1. Payback Period
Explanation: It measures the time required for an investment to recover its initial cost from the cash inflows it generates.
The Payback Period is a financial metric used in accounting to assess the time it takes for an investment to recoup its initial cost through generated cash flows. It signifies the duration required
to recover the initial investment. It’s a straightforward measure, highlighting how quickly an investment can regain its original capital outlay. Shorter payback periods generally indicate quicker
returns, whereas longer payback periods may involve increased risk or slower returns on the investment.
Role in Decision Making: The Payback Period quickly tells how long to recoup an investment. It’s handy for fast returns or limited capital. Shorter periods often mean less risk and faster profits.
But it overlooks long-term gains and money’s time value. It’s part of a toolkit with other metrics for a full investment picture.
Formula: Payback Period=Annual Cash Inflows / Initial Investment
Payback Period | How do you calculate the payback time on your investment?
2. Net Present Value (NPV)
Explanation: NPV measures the difference between the present value of cash inflows and outflows from an investment, considering the time value of money.
Net Present Value (NPV) in accounting assesses the profitability of an investment by comparing the present value of expected cash inflows against the initial investment and future cash outflows,
considering the time value of money. A positive NPV indicates that the project generates more cash than the initial investment and is thus profitable. It’s a fundamental metric in investment
appraisal, helping determine whether an investment will yield returns higher than the cost of capital. Higher NPV values imply more lucrative investments, while negative NPV suggests potential
Role in Decision Making: A positive NPV suggests a potentially profitable investment, as it generates more returns than the initial investment.
Formula: NPV = Σ [CFt / (1+r)^t]
CFt = Cash flow in year t
r = Discount rate
t = Time period
3. Time Value of Money (TVM)
Explanation: The time value of money signifies that money available today holds more worth than the same amount in the future due to its potential to earn interest when invested. Investing money
allows it to grow, so having funds sooner provides more time for growth.
The Time Value of Money (TVM) is a critical concept in accounting that recognizes the changing worth of money over time. It acknowledges that a sum of money has different values at different times
due to factors like interest rates and inflation. TVM is crucial in financial decision-making, emphasizing that a dollar today is worth more than the same dollar in the future due to its potential
earning capacity or investment opportunities. Understanding TVM aids in evaluating investments, loans, and determining the true value of cash flows occurring at different points in time.
Role in Decision Making: it highlights money’s varying worth over time, as it can be invested to earn interest. This principle emphasizes that a dollar today holds more value than a dollar in the
future. In decision-making processes, FV aids in assessing the potential returns of an investment, enabling individuals and businesses to plan for future financial needs. It assists in setting
financial goals and determining the required investment amounts to meet those goals over time.
Formula: TVM=FV/(1+r)^t
Time Value of Money (TVM): Unveiling the Power of Future Dollars
4. Future Value of Money (FV)
The future value of money gauges the worth of an invested sum at a later date, considering a specific interest rate. It aids in financial planning, such as saving for retirement, enabling individuals
to invest an amount that will grow to meet their future financial goals.
The Future Value of Money (FV) is a pivotal accounting concept that calculates the value of an investment at a specified time in the future. It considers factors like interest rates and time periods
to estimate the worth of a current sum after accruing interest or investment returns. FV helps assess potential growth or returns on investments, guiding decisions about saving, investing, or
forecasting the value of assets over time. Understanding FV is essential for evaluating the potential worth of investments and making informed financial decisions based on projected future values.
Role in Decision Making:In decision-making processes, FV aids in assessing the potential returns of an investment, enabling individuals and businesses to plan for future financial needs. It assists
in setting financial goals and determining the required investment amounts to meet those goals over time.
Formula: FV=PV×(1+r)^t
Future Value of Money (FV): Predicting the Worth of Your Investments
5. Profitability Index (PI)
Explanation: Profitability Index compares the present value of future cash flows to the initial investment, providing a measure of investment efficiency.
The Profitability Index (PI) is a crucial metric in accounting, representing the relationship between the costs and benefits of an investment. It measures the potential profitability of a project by
comparing the present value of future cash flows to the initial investment cost. A PI greater than 1 indicates the project is potentially profitable, while a value less than 1 suggests the project
may not be worthwhile. PI aids in decision-making by helping to prioritize and evaluate investment opportunities based on their potential returns relative to their costs. Understanding PI assists in
selecting the most financially viable projects for optimal returns.
Role in Decision Making: A higher profitability index signifies a more beneficial investment relative to its cost.
Formula: PI = NPV / Initial Investment
6. Internal Rate of Return (IRR)
Explanation: IRR represents the discount rate where the net present value of an investment becomes zero.
The Internal Rate of Return (IRR) is a pivotal metric in accounting, representing the estimated rate at which an investment breaks even or generates a desired return. It calculates the discount rate
that makes the net present value (NPV) of an investment zero. A higher IRR signifies a more favorable investment, typically above the cost of capital. It’s a crucial tool for evaluating and comparing
the attractiveness of various investment opportunities. IRR aids in decision-making by providing insights into the potential profitability of investments, allowing companies to assess projects based
on their returns and risks.
Role in Decision Making: Higher IRR indicates a potentially more profitable investment.
Formula: IRR:NPV=0
The process involves finding the discount rate that makes the NPV equal to zero by trial and error.
7. Modified Internal Rate of Return (MIRR)
Explanation: MIRR adjusts IRR’s shortcomings by assuming reinvestment of positive cash flows at a specific rate and financing negative cash flows at a different rate.
The Modified Internal Rate of Return (MIRR) is an adjusted financial metric used in accounting to address some limitations of the traditional Internal Rate of Return (IRR). Unlike IRR, which assumes
reinvestment of cash flows at the same rate as the project’s return, MIRR incorporates a more realistic assumption: that positive cash flows are reinvested at the firm’s cost of capital, while
negative cash flows are financed at the firm’s borrowing rate. MIRR offers a clearer picture of an investment’s profitability by considering the cost of financing and the return on reinvestment. It
provides a more accurate reflection of the project’s potential for investors and managers to make informed decisions regarding capital allocation.
Role in Decision Making: MIRR gives a more accurate picture of the investment’s profitability, especially in cases of unconventional cash flow patterns.
Formula: MIRR = (Future Value of Cash Inflows / Future Value of Cash Outflows)^(1/Time Period) – 1
8. Equivalent Annuity (EA)
Explanation: Equivalent Annuity calculates the uniform cash flow over a specific period, equivalent to the investment’s cash flows.
Equivalent Annuity is a financial metric used in accounting and investment analysis to compare different investments with varying lifespans or cash flow patterns. It represents a uniform annual cash
flow over a specific period, which, when discounted at the same rate, equates to the present value of a project’s cash inflows and outflows. This metric facilitates comparisons between projects with
different durations or cash flow profiles, allowing decision-makers to assess which investment provides a consistent annual return over a specified time frame. The Equivalent Annuity helps in
evaluating investment alternatives by standardizing cash flows, aiding in more straightforward comparisons and better decision-making.
Role in Decision Making: Helps compare different projects by standardizing their cash flows into equivalent annual amounts.
Formula: Equivalent Annuity = NPV / PVFA(r,n)
PVFA(r,n) = Present value factor of an annuity for a period of n years at a discount rate of r
Capital budgeting, utilizing these techniques, enables businesses to strategically allocate resources, ensuring they invest in projects that offer the most significant potential returns while
mitigating risks.
Each method comes with its strengths and limitations. For instance, while NPV considers the time value of money and is a direct measure of value, it might not rank mutually exclusive projects
effectively. Payback Period offers simplicity but ignores the time value of money. Real Options Analysis is sophisticated but complex, requiring assumptions about future events.
By using a combination of these methods, companies gain a more comprehensive view of potential investments. The choice of technique often depends on factors such as the nature of the project,
available data, and the company’s risk tolerance. Integrating these methods with ratio analysis aids in more robust investment decisions by offering a multifaceted evaluation of projects and their
impact on the organization’s financial health.
Financial Ratios are like performance scorecards for businesses | Accounting – Formulas, Examples, Questions, Answers
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Showcasing some projects that are either web apps or desktop apps suitable for downloads, including blogs on different topics of interest.
Latest blog posts :
CalculatorNotepad is intended for easy calculations with support for user defined functions/formulas and rich set of integrated mathematical and probability functions.
It can be used as simple calculator with each line showing calculation result, but it also support user variables (storing results of previous calculations and using them in new calculations) and
user defined functions that can be written as simple one line or multiline using integrated script language (or as c# functions in side panel). It is also suitable for simple simulation scenarios,
with support for random numbers using different distributions and simulation aggregation functions.
It started as my proof of concept to create development IDE with powerful and yet simple interpreted language, but has since evolved to actual tool that I use while solving mathematical, probability
or simulation problems. Recently I decided to publish it publicly on gitHub, and ReadMe on main page there contains more detailed description of CalculatorNotepad.
WordMind is game where goal is to guess target word in limited number of attempts. Similar to Mastermind, failed attempts will mark letters that are at exact correct place, those that are in target
but somewhere else and those that do not exist in target. It is similar to ‘Wordle’ game but with support to more languages ( English, Serbian cyrillic and latin), configurable number of letters and
word difficulty, ability to play more than once per day and coop option where one player set target word for other player(s) to guess.
Orao Emulator – web site
Orao(Eagle) is 8-bit ex-YU computer from 80’s. This app fully emulate Orao computer, allowing you to play with its integrated mini-assembler and BASIC, or to simply play some old games. Emulator is
written using Blazor as responsive web site that works on desktop and mobile browsers.
Orao Emulator – web app
This version of emulator is written using Blazor WebAssembly as responsive web app that works on modern desktop and mobile browsers. It is progressive web app (PWA) and it can be installed as web app
on desktops or mobile home pages and can be run off-line.
Truel solver– Jupyter document
Truel solver as Python based Jupyter notebook
This document presents solutions and analysis of different scenarios for “truel” problem :
Several people are doing duel. Given their probabilities to hit, what are probability of each of them to win and who should they choose as optimal initial target ?
Solution is based on Python with numba acceleration, presented in interactive Jupyter notebook. While above links lead to static HTML page with Truel analysis, there is also GitHub repository with
Jupyter document and my Truel blog with summary.
Two Envelopes paradox
Problem definition Generalized version of Two Envelopes problem Variants vG, vD, vS, vA, vN … Resolution of paradox Expected values and paradox Infinite Expected values Expected gain and paradox
Optimal strategy if we can look into envelope Solution of Generalized version ( answer to Q1 ) Optimal solution of Generalized version (answer to Q2) vD: …
Analysis of living population density per countries
By definition, population density is a measurement of population per unit area. Classical way to calculate population density of a country was to divide entire population count with entire area of
that country. But that simple approach has several shortcomings, and number of alternate measurement methods for population density measurements exist. Measurement that is analyzed in this article …
Continue reading “Analysis of living population density per countries”
Truel problem – solved with Jupyter / Python
About a year ago I decided to estimate usability of Jupyter notebook documents with Python code. Since both Python and Jupyter were new to me at that time, I selected real world problem to solve
using them, specifically to solve “Truel” problem : Several people are doing duel. Given their probabilities to hit, what are …
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Polynomials Worksheet Algebra 2
Polynomials Worksheet Algebra 2
Khan academy s algebra 2 course is built to deliver a comprehensive illuminating engaging and. And continuing the work with equations and modeling from previous grades.
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Polynomial worksheets algebra i. Basic polynomial operations worksheets this polynomial functions worksheet will produce problems for identifying the degree and term simplify expressions and finding
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Group Theory In Physics Wu-ki Tung Pdf 11
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An Introduction to Symmetry Principles, Group Representations, and Special ... By (author):; Wu-Ki Tung (Michigan State University, USA) ... Pages:111.. Wu-Ki Tung, Group Theory in Physics, World
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MCQS on numbers - CMA INDIA GROUP
MCQS on numbers
MCQS on numbers-Welcome to the world of numbers in mathematics! Numbers form the foundation of quantitative reasoning and are integral to virtually every facet of mathematical exploration. In this
set of multiple-choice questions (MCQs), we delve into the diverse realm of numbers, covering a spectrum from basic arithmetic to more advanced concepts. Explore questions that test your knowledge of
integers, fractions, decimals, and real numbers, as well as your ability to navigate mathematical operations with precision. As you tackle these MCQs, you’ll not only sharpen your numerical skills
but also deepen your understanding of the fundamental language of mathematics. Let’s embark on this mathematical journey, where numbers take center stage in the fascinating landscape of numerical
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#1. If the average of three numbers is 15, and two of the numbers are 10 and 20, what is the third number?
#2. What is the Roman numeral for 50?
#3. If a triangle has sides of lengths 3, 4, and 5, what type of triangle is it?
#4. Which of the following is a multiple of 9?
#5. What is the square root of 100?
#6. What is the smallest prime number?
#7. What is the product of 5 multiplied by 9?
#8. If a number is divisible by 3 and 5, what is it also divisible by?
#9. What is the sum of the angles in a triangle?
#10. If a square has an area of 49 square units, what is the length of each side?
#11. If a train travels at a speed of 60 miles per hour, how far will it travel in 2 hours?
#12. If a square has a side length of 7 units, what is its area?
#13. If x = 6 and y = 2, what is the value of 3x - y?
#14. What is the difference between the squares of 7 and 5?
#15. If a = 5 and b = 3, what is the value of 2a + b?
#16. What is the value of the digit 9 in the number 9,438?
#17. If the radius of a circle is 6 units, what is its circumference?
#18. If the product of two numbers is 48 and one of the numbers is 6, what is the other number?
#19. Which of the following is a composite number?
#20. What is the average of the first 5 prime numbers?
#21. In the fraction 3/4, what is the numerator?
#22. What is the next number in the sequence: 4, 9, 16, 25, ...?
#23. In the decimal 3.75, which digit is in the hundredths place?
#24. If a rectangle has a length of 15 units and a width of 9 units, what is its perimeter?
#25. What is the sum of the first 10 positive integers?
#26. Which of the following is a prime number?
#27. What is the value of π (pi) to two decimal places?
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ball mill ball size
The basic parameters used in ball mill design (power calculations), rod mill or any tumbling mill sizing are; material to be ground, characteristics, Bond Work Index, bulk density, specific
density, desired mill tonnage capacity DTPH, operating % solids or pulp density, feed size as F80 and maximum 'chunk size', product size as P80 and maximum a...
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and with the decreased grain size, the necessary ball size also decreases (Olejnik, 2010; 2011). For each grain size there is an optimal ball size (Trumic et. al., 2007). The bigger ball in
relation to the optimal one will have an excess energy, and consequently, the smaller ball mill has less energy necessary for grinding. In both
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The common type of tumbling ball mill used is the bond ball mill. The smallscale mill is 30 cm in diameter and 30 cm in length. Generally, <300 balls of 40 mm size can be used.
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The grinding process was carried out in a cylindrical ball mill with a diameter and length of cm and cm, respectively, as well as a steel ball with a diameter of cm and a weight of 100 grams/
ball. The optimum data for the grinding process was obtained with the smallest response value of P80.
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The ball milling design significantly depends on the size of mill jar, required particle size distribution of powder and grinding media (balls) [13], [14], [15]. The main purpose of milling
process is to obtain required particle size of the powder without any contamination, increasing the output of the milling circuit and overall reduction in ...
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It was found that the ball mill consumed kWh/t energy to reduce the F 80 feed size of lm to P 80 product size of lm while stirred mill consumed kWh/t of energy to produce ...
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The ball mill was run for seven time intervals, ranging from to 30 min. ... The effect of ball size and interstitial filling on the performance of dry ball mill grinding was investigated for a
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H. Kim et al. studied the influence of the milling body size used in an Emax to mill talc particles (phyllosilicate) at 2000 rpm. Three ball size (2, 1 and mm) was used for this purpose, finding
that the use of smaller balls did not achieve the same refinement achieved by the other milling bodies.
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If a ball mill uses little or no water during grinding, it is a 'dry' mill. If a ball mill uses water during grinding, it is a 'wet' mill. A typical ball mill will have a drum length that is 1 or
times the drum diameter. Ball mills with a drum length to diameter ratio greater than are referred to as tube mills.
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Generally, a maximum allowed ball size is situated in the range from D /18 to D /24. The degree of filling the mill with balls also influences productivity of the mill and milling efficiency.
With excessive filling, the rising balls collide with falling ones. Generally, filling the mill by balls must not exceed 3035% of its volume.
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This study proposed the use of an instrumented grinding media to assess solid loading inside a ball mill, with size and density of the instrumented ball comparable to that of the ordinary
grinding media. ... the highdimensional mechanical signals of multisource modes are difficult to map the characteristic parameters of the ball mill, which ...
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An online calculator lets you calculate Top Ball Size of Grinding Media for your mill. Use this Equation Method to properly grind your ore.
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The ball mill Ball milling is a ... Further research on the effects of milling conditions ( balltocellulose mass ratio, milling time, ball size and alkaline pretreatment) on the morphology of the
prepared nanocellulose derivatives was undertaken by the group of Wang. 22,23 More in detail, they found out that the size of the milling balls ...
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Attrition: Reduced the size of the materials when they colloid by heavy weight (Ball). Construction: The ball mill grinder consists following Parts: Cylinder: cylinder is made of a hollow metal
that moves about its horizontal axis. the cylinder can be made of porcelain, metal, and rubber. the length of the cylinder slightly higher than its diameter.
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The specific rates of breakage of particles in a tumbling ball mill are described by the equation S i = ax α i (Q(z), where Q(z) is the probability function which ranges from 1 to 0 as particle
size equation produces a maximum in S, and the particle size of the maximum is related to ball diameter by x m = k 1 d variation of a with ball diameter was found to be of the form ...
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1. Introduction A mix of balls of different diameters enables the effective milling of different particle sizes in a tumbling mill while ensuring the optimisation of the mill product. This is
owing to the fact that each ball size effectively breaks a particular size in the mill [1], [2].
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vibratory milling apparatus is much greater than that of planetary ball mill system; therefore, less time may be needed for the particle size reduction in the vibratory ball mills. Here, the
similar parameters need to be considered at time of milling process,, milling speed, milling time, grinding medium, atmosphere, etc.
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High energy ball milling was done in a planetary ball mill (Torrey HillsND2L) with stainless steel cups (285 ml capacity) and balls in an argon atmosphere. Charge ratio was 30:1 and mill speed
was maintained at 200 RPM for durations of,, 1, 2, 4, 7, 11 and 19 h.
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Small Ball Mill Capacity Sizing Table. Do you need a quick estimation of a ball mill's capacity or a simple method to estimate how much can a ball mill of a given size (diameter/lenght) grind for
tonnage a product P80 size? Use these 2 tables to get you close. No BWi Bond Work Index required here BUT be aware it is only a crude ...
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The performance of various diameter ball mills can now be simulated using the following mechanisms: (1) an orespecific breakage distribution function determined from pendulum tests (the results
are given in Table III); (2) a breakagerateparticlesize relationship for a given ball mill diameter obtained from the known constant relationship ...
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Based on his work, this formula can be derived for ball diameter sizing and selection: Dm <= 6 (log dk) * d^ where D m = the diameter of the singlesized balls in = the diameter of the largest
chunks of ore in the mill feed in mm.
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Hi. can someone assist me with a few problems I have? We have a overflow dry grinding ball mill grated discharge x, we use 40mm,25mm and 17mm ball sizes (high chrome steel) to mill manganese with
a top size of 4800um, mill liners are stee...
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Breakage rate and particle size have a maximum for each ball size distribution using a pilotscale ball mill on size at maximum breakage (X m) is strongly related to top ball size (D b) in terms
of ball charge. 10, 7 and 5 mm: For a mechanochemical synthesis of the sulfide solid electrolyte Li 3PS 4. The largest relative ...
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Ball size distribution in tumbling mills 37 Milling performance of a ball size distribution 40 Summary 41 Chapter 3 Experimental equipment and programme 43 Laboratory grinding mill configuration
43 Preparation of monosize grinding media 44 Feed material preparation 46 ...
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This review is focused on the topical developments in the synthesis of nanocomposites using the simplest topdown approach, mechanochemical milling, and the related aspects of the interfacial
interactions. Milling constraints include time duration of milling, ball size, the balltosample content proportion, rotation speed, and energy that took part in a vital part of the
structureproperty ...
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The formula is excellent from the basis of balance with respect to ball wear, but the literature has contained very little about the rationing of ball sizes for the best grinding of all sizes and
amounts of particles extending throughout the length of the mill. Research has submitted in this matter.
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Construction of Ball Mill. The ball mill consists of a hollow metal cylinder mounted on a shaft and rotating about its horizontal axis. The cylinder can be made of metal, porcelain, or rubber.
Inside the cylinder balls or pebbles are placed. The balls occupy between 30 and 50% of the volume of the cylinder. The diameter of the balls depends on ...
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Peripheral discharge ball mill, and the products are discharged through the discharge port around the cylinder. According to the ratio of cylinder length (L) to diameter (D), the ball mill can be
divided into short cylinder ball mill, L/D ≤ 1; long barrel ball mill, L/D ≥ or even 23; and tube mill, L/D ≥ 35. According to the ...
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Evolution of the ball grinding charge distributions proposed by Bond. Ball Charge Distribution 1961 Ball Charge Distribution 1999 Ball Size Balls Ball Size Balls inch cm Number Weight (g) inch cm
Number Weight (g) 43 8803 25 5690 67 7206 39 5137 10 672 60 4046
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The balls are initially 510 cm diameter but gradually wear away as grinding of the ore proceeds. The feed to ball mills (dry basis) is typically 75 vol.% ore and 25% steel. The ball mill is
operated in closed circuit with a particlesize measurement device and sizecontrol cyclones.
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3^rd International Mathematics Assessment for Schools (IMAS)
I. Introduction
International Mathematics Assessment for Schools (IMAS) is a world class mathematics assessment test for middle primary, upper primary and lower junior secondary students. The test is organized by
IMAS Executive Council.
The organizer strongly feels a need to conduct a mathematics assessment test among schools, countries and regions to test the students’ achievements in mathematics problem solving and to serve as a
guide paper for students who wish to improve their ability in this field. This test serves as one of the motivating factors attracting students not only to test their abilities in mathematics but
also to challenge their abilities in broadening their mathematical scope.
II. Aims and Objectives
The aim of the IMAS
1. To provide an achievement test in mathematics for all students and a mathematics competition for students with good performance;
2. To develop a world class mathematics assessment with international perspective measuring students’ performances in three cognitive dimensions: Knowing, Applying and Reasoning at Middle Primary
(Grade 3 and 4), Upper Primary (Grade 5 and 6) and Junior Secondary (Grade 7 and 8) level; and
3. To promote effective learning of mathematics in both primary and secondary schools through a publicly recognized performance measuring platform.
The objectives of the IMAS are
1. To inform stakeholders, i.e. students, teachers and parents, of the performances of the students in mathematics at different levels of attainment in well defined reports;
2. To better serve the needs of students through a more user friendly format of perennial assessment that offers options for participants in terms of topics as well as difficulties of materials
being assessed through a public mathematics assessment which will be held once a year; and
3. To usher in an innovative assessment culture, i.e. participants can choose when they would like to sit for the assessment of content as well as the level of difficulty of the items being
assessed. Participants take the initiative in assessing their competencies and capabilities.
III. Are there differences between IMAS and other international attainment tests?
At present, there are several achievement tests conducted by different educational groups. These tests include PISA, TIMMS, and many others. The introduction of IMAS is not just another international
assessment test. The main difference between IMAS and other international mathematics tests is that IMAS is not a once-off testing. There are two rounds of tests, followed by a summer camp in
mathematics. More importantly, the IMAS is not aimed at merely testing students, but rather to develop students’ mathematics ability and creativity. The following table shows the important attributes
of the IMAS.
Aims To assess students’ performance in three cognitive dimensions; Knowing, Applying and Reasoning for the purpose of enriching the universal education of mathematics.
Participants Upper primary, middle primary and lower junior secondary students.
The Test Set in the real world, rather than pure mathematical environment, with situations to which the students can relate.
Its Emphasis IMAS emphasizes items which require the use of a scientific, explorative and creative approach to solve problems.
What does it Measure? IMAS measures students’ working knowledge of mathematics developed from natural ability and learning styles.
Emphasis and Measurement of TIMSS and PISA
The following table displays the characteristics of TIMSS and PISA. Though IMAS and the two tests shared some common basic philosophy on mathematics testing, IMAS is more focused on the development
of mathematics ability and is served as a tool to inform schools and teachers on the development of students’ ability in mathematics.
TIMSS PISA
Aims To assess the knowledge of students, assessment items exhibit a range of difficulty and complexity. To test literacy in mathematics, with a view to improving educational policies and
Participants Fourth- and eighth-grade students 15-year-old school pupils’ scholastic performance.
The Test Designed to collect information on students’ backgrounds, attitudes and beliefs related to schooling Require students to apply their mathematical knowledge to solve problems set in
and learning, information about their classroom experiences. various real-world contexts.
Its Emphasis Emphasis on items which require mathematical facts and standard algorithms. Items demand connections between existing knowledge.
What does it Measures traditional classroom content and curriculum attainment. Measures students’ ability to apply what they have learned to real-world
Measure? situations and to communicate their solutions to others.
The questions in IMAS are designed with the following framework. There are four areas of mathematics involved in the test paper. Questions in each area involve knowing, applying and reasoning.
Numbers and operations Algebra Geometry Measurements
Knowing means the ability to recognize conceptual work and answers are obtained based on concept and calculation. Applying means selection of appropriate concepts and working procedures to solve
problems. Reasoning means using logical deduction based on deduction and concepts to obtain answer.
1. The Organizer
The IMAS is organized by the body of Executive Council composed of
A. Mr. Cheng, Chun Chor Litwin– Senior Lecturer, Hong Kong Institute of Education, Hong Kong;
B. Mr. Sun Wen-Hsien- President, Chiu Chang Mathematics Foundation;
C. Dr. Promote Kajornpai – Specialist Supervisor, Office of Basic Education Commission, Ministry of Education, Thailand
D. Ms. Elvira SH- Head, Directorate of Kindergarten and Primary Education, Ministry of National Education, Indonesia;
E. Dr. Simon Chua– President, Mathematics Trainers’ Guild, Philippines.
Under the direction of the IMAS Council, an Academic Committee is formed to administer the operation of the mathematics test. The membership of the Committee is as follows:
Academic advisors:
Prof Zhang Jingzhong—-Fellow, Chinese Academy of Sciences
Mr. Cheung Pak Hong—-Principal, Munsang College (Hong Kong Island)
Prof. Andy Liu —- Professor, Department of Mathematical and Statistical Sciences University of Alberta, Canada.
Vice Chairman:
Prof. Zhu Huawei —- Professor, Guangzhou University, China.
Prof. Simon Chua —- President, Mathematics Trainers’ Guild, Philippines.
Mr. Wen-Hsien Sun — President, Chiu Chang Mathematics Education Foundation, Taiwan.
Mr. Zheng Huan —- Lecturer, Guangzhou University, China.
Mr. CHENG, Chun Chor Litwin—- Senior Lecturer, Hong Kong Institute of Education, Hong Kong.
Mr. Vladislav Marinov —- Project Manager, Aero Scout Enterprise Visibility Solutions, Bulgaria
Other members are the research team headed by Prof. Zhu Huawei and specialists endorsed by individual participating countries.
All other participating countries in the first IMAS will be invited as official members of IMAS in 2013-2014.
V. Entrance Qualification of IMAS
There are no preset requirements for students who want to participate in the IMAS. Students may sit for any level of assessment which they find suitable. That is, a student could sit for an
assessment test at a higher or lower level; for example, a primary 3 student may sit for the upper primary assessment test. This opens up the frontier for the bright and able student to test their
abilities through a public recognized measurement. The IMAS is meant to support the advancement of education for the gifted as practiced in many countries.
VI. Schedule of the IMAS
There are two rounds of IMAS test in one academic year. The first round of IMAS will be conducted in early November for middle primary, upper primary and lower secondary students while the second
round of IMAS will be conducted in the succeeding year around the last week of January. The IMAS Executive Council will sponsor a Summer Camp in July each year with the aim to promote creativity and
problem solving skill among students. Hence, the IMAS is not all about traditional paper and pencil test, it is also about developing dynamic and interactive approach to learning mathematics in an
appropriate environment.
VII. Format of the Assessment and Award
The first round of IMAS is for all participants, and the second round is for students who perform well in the first round. The format of the tests in each round is as follows, with the sample
question included at the appendix.
1. First Round of IMAS – Twenty-five problems will be given. The participants are given one hour to finish the ability test. The first twenty problems are in Multiple-Choice type while the last five
problems require integer answers between 0 and 999 inclusive. The distribution of the problems are: Problems 1 to 10, classified as EASY category, each correct answer is given 3 points; Problems
11 to 20, classified as AVERAGE category of which each correct answer will be given 4 points while Problems 21 to 25, CHALLENGE category, every correct answer will be given 6 points.
To encourage students to work mathematics and to recognize students’ achievement, the IMAS will provide award for students entering the tests. The level of awards of individual student is relative to
the performance of other students in their country, of the same year level during the first round of IMAS. The award scheme is as follows:
• High Distinction – Student whose score is above 95 percentile
• Distinction -Student whose score is above 85 and below 95 percentile
• Credit – Student whose score is above 50 and below 85 percentile
• Participation – Student whose score is below the 50 percentile
Every participant is awarded with a certificate with a personal report. The IMAS individual certificate means a lot to the participant as it promotes positive effects about one’s performance based on
the given assessment. The report will be in the form of standards of achievement, informing the participant their achievement in addition to the percentile rank. Assessment results could facilitate
learning when strengths and weaknesses of their performance are identified. This information can be shared with mathematics teachers in the region via the feedback loop in seminar and workshops.
• The Second Round of IMAS is conducted for those students who achieved to be in the top 5% to 10% of all participants of each country. In this test, there are fifteen problems: Problems 1 to 5 are
multiple choice type where each correct answer will be given 4 points, Problems 6 to 13 call for short answers where each correct answer will be given 5 points, and Problems 14 to 15 call for
detailed solution (or working process) and full mark of each correct answer is 20, partial marks may be given for incomplete answer.
In each participating country, the participants whose scores are in the top 5% percentile will be awarded with a Gold Medal; participants in the top 6% to 15% percentile will be awarded with a Silver
Medal and each participant whose score is in the top 16% to 30% percentile will receive a Bronze Medal.
VIII. IMAS Summer camp
The IMAS will also organize a Summer Camp for developing mathematics learning. This will provide good opportunities for all participants to meet each other and promote friendships among students from
different countries as well as learning mathematics in cooperative and collaborative manner.
The Summer Camp is conducted by the IMAS Executive Council. It will invite participants who perform well at the middle primary, upper primary and junior secondary of each country to attend the Summer
Camp. The Camp will be conducted in Asia. The Council will provide free meals and accommodation to all invited participants. One teacher of each participating country will also be invited and his/her
meal and accommodation expenses will be sponsored by IMAS Executive Council. However, the participants and all official delegates will have to pay for their round trip travel expenses from their
country to Asia.
XI. Charges and Logistics
1. There is an Entry Fee of US $3 per participant payable to the IMAS Executive Council.
• The organizer of each participating country appointed by the IMAS Executive Council Printing will be responsible for printing and marking of the papers. | {"url":"https://chiuchang.org/imas/imas-2013-2014/information/","timestamp":"2024-11-14T21:44:37Z","content_type":"text/html","content_length":"42027","record_id":"<urn:uuid:6a91231e-5724-4459-bfad-81ad75bda947>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00023.warc.gz"} |
[Libre-soc-dev] [RFC] Matrix and DCT/FFT SVP64 REMAP
Cesar Strauss cestrauss at gmail.com
Sun Jul 4 23:14:54 BST 2021
On 07/02/2021 21:56, Luke Kenneth Casson Leighton wrote:
> hm. to create the appearance of matrix multiply as 3 flattened arrays,
> if the formula is this:
> for x in x_r:
> for y in y_r:
> for z in z_r:
> result[y][z] +=
> a[x][y] *
> b[x][z]
> (something like that)
More precisely:
for y in y_r:
for x in x_r:
for z in z_r:
result[y][x] +=
a[y][z] *
For C = A*B, the inner loop goes "horizontally" (along a row) on matrix
A and "vertically" (along a column) on matrix B.
It also means that the number of columns of A must be equal to the
number of rows of B. In other words, A has the shape of (y_r) x (z_r)
while B is (z_r) x (x_r).
The result then has the same number of rows than A, and the same number
of columns than B, or (y_r) x (x_r).
I don't think there can be such a thing as an "in-place" algorithm for
matrix multiplication.
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Percents Review
The memory card on Jimmy's digital camera can hold about 520 pictures. Jimmy uses 18% while at a beach party. How many pictures does he still have available to take after the party? Round to the
nearest whole number.
What is 426?
Bill is in a class of 12 boys and 18 girls. 40% of the students in the class take the bus to school. How many students do not take the bus to school?
What is 18?
A DVD movie originally cost $24.99. Its current price is $9.99. What is the percent of change rounded to the nearest tenth of a percent?
What is 60% decrease?
Convert Decimal to Fraction: 1.86
What is 1 43/50?
What is 8% of 200?
What is 16?
At the end of the year, cars made in the current year are marked down to make room for next year's models. One particular model had a sticker price of $21,250 when it was new. It has just been marked
down by 28%. What is the sale price of the vehicle?
What is $15,300?
If the price of a purse increased from $45 to $63, what was the percent of increase?
What is 40% increase?
Convert Percent to Decimal: 12.7 %
What is .127?
From 1895 until 1958, the French controlled a number of countries in the western part of what continent?
What is Africa?
What is 20% of 120?
What is 24
What is 35% of 88
What is 30.8%
Lucy is trying to drink more water, so she has been keeping a log of how much water she drinks. Two days ago, Lucy drank 40 ounces of water, in comparison to 32 ounces yesterday. What was the percent
of change in Lucy's daily water consumption?
What is 20% decrease?
Convert Decimal to Percent: 0.755
What is 75.5%
Which animals produce pearls?
What is OYSTERS?
The Boston Red Sox did not win a World Series from 1918 until 2004, a torturous streak that many blamed on what curse?
What is the "CURSE OF THE BAMBINO," a jinx that landed on the franchise after its owner sold Babe Ruth?
The movie theater has 250 seats. 225 seats were sold for the current showing. What percent of seats are empty?
What is 10%?
Miss Holman gave a 90-cent tip to a waitress for serving a meal costing $6.00. What percent of the bill was her tip?
What is 15%
Tom's weekly salary increased from $240 to $300. What was the percent of change?
What is 25% increase?
Convert Fraction to Percent: 13/16 Do not round.
What is 81.25%?
Which animal is the largest member of the cat family?
What is tiger?
The produce department of a grocery store sold 288 pounds of tomatoes. The weight represented 72% of the shipment the store received. The rest spoiled and had to be discarded. How many pounds of
tomatoes were originally received?
What is 400?
Donovan earned 99 points on his science project. If his grade was a 90%, how many points were possible?
What is 110 points?
In 1986 the average price of a gallon of gasoline was $0.93 cents. In 2007 the average price of a gallon of gasoline was $2.77. Find the percent of change rounded to the nearest tenth of a percent.
What is 197.8%
Convert Percent to Fraction: 79.6 %
What is 199/250?
The pointed teeth near the front of the mouth have what animal name?
What is canines? | {"url":"https://jeopardylabs.com/print/percents210","timestamp":"2024-11-04T08:26:07Z","content_type":"application/xhtml+xml","content_length":"24897","record_id":"<urn:uuid:c4a299d3-d1cf-45b9-91ad-96f65755af7b>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00201.warc.gz"} |
oZone3D.Net Tutorials - Mandelbrot Fractal - GPGPU Programming - GLSL
Fig. 1: the entire Mandelbrot Set.
The Mandelbrot set is a collection of points in the complex plane. In Figure 1, the black areas are points inside the set, while colored areas are outside it. As you can see, it has a jagged, spiky
appearance. In fact, the boundary of this set, the edge where the black and colored areas meet, is infinitely detailed. What does this mean?
Although the set's boundary is a continuous curved line, and it fits into a finite area (the entire Mandelbrot Set is within the circle of radius 2 around the origin of the plane), the boundary
nevertheless has infinite length. This is another common property of fractal objects; their boundaries have infinite length (if they are 2D) or area (if they are 3D). To see how this is possible,
consider a coastline.
Suppose you had a coastline and two points on it, and you want to measure the distance along the coast between the points. Obviously, the shortest distance between the points would be a straight
line; but suppose instead of measuring a straight line, you measured more accurately along the coastline. You would probably get quite a bit larger distance, unless the coastline was exceptionally
straight. Now zoom in closer and measure the distance along the coastline even more accurately, and the distance between the two points increases again. You can imagine that you could keep zooming in
until you were crawling along the beach with a microscope, painstakingly measuring the distance around each and every grain of sand. Clearly, as you measure more and more accurately in this way, the
distance from one point to the other along the coastline approaches infinity very rapidly.
Of course, the length of the coastline can never be actually infinite—the amount of detail in it is limited by the size of the atoms that make up the sand along the beach. But a mathematical fractal
is not limited by the size of atoms. The length of the edge of a fractal like the Mandelbrot set is truly infinite, but all this infinite length is packed into a finite area—forcing the boundary to
contain details at all levels of scale. You can keep zooming into the Mandelbrot set forever, and you will never come to the end. | {"url":"https://ozone3d.net/tutorials/mandelbrot_set_p2.php","timestamp":"2024-11-06T23:14:16Z","content_type":"text/html","content_length":"19143","record_id":"<urn:uuid:0716e177-78e3-4b72-90b2-f01e5b879f36>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00240.warc.gz"} |
Working to Understand the Changing Flavors of Quarks - UConn Today
Working to Understand the Changing Flavors of Quarks
Up, charm, down, bottom, top, and strange - what does it all mean?
Studying quarks requires experimentation where atomic particles are accelerated and broken apart, then theoretical work to understand and describe what happened. (Adobe Stock)
Visible matter, or the stuff that composes the things we see, is made of particles that can be thought of much like building blocks made of more building blocks, ever decreasing in size, down to the
sub-atomic level. Atoms are made of things like protons and neutrons, which are composed of even smaller building blocks such as quarks. Studying those smallest building blocks requires
experimentation where atomic particles are accelerated and broken apart, then theoretical work to understand and describe what happened.
UConn Assistant Professor of Physics Luchang Jin studies particle and nuclear physics, and is working to understand more about subatomic particles and how they behave. Jin will be presenting recent
findings at the 2021 Fall Meeting of the American Physical Society’s Division of Nuclear Physics in October.
“The topic describes how quarks ‘change flavors,’ or transition, due to weak interactions,” says Jin. “The Standard Model describes four types of interactions and weak interactions are one of them.
We study the parameters that describe the transition probability.”
Quarks can have six types of “flavors” or differences in mass and charge – up, charm, down, bottom, top, and strange — and understanding how they switch from one flavor to another, Jin says, can help
us understand more about the inner workings of the universe.
Jin explains that this research is looking into the probability of up quarks transitioning into down quarks. The transition probability of this flavor change and the probabilities for up quarks
transitioning to other quarks should add up to one, but they don’t, and this deficit is intriguing.
“This could be indicating something, for instance that unfortunately we somehow didn’t measure those values accurately enough,” Jin says. “It could be indicating that there are some new particles
that we don’t know yet, and that will be very exciting. The work I’m trying to do is try to make sure that we measure those quantities accurately.”
Jin says the experimental aspects of this work are in relatively good shape; the bottleneck, however, is with the theoretical aspect, which Jin is hoping to help solve by determining the relations
between the quark transition probabilities from the experimental data of hadron transition probabilities.
Hadrons are a type of subatomic particle made of two or more quarks which are classified by the strength of their interactions with one another on a scale of “color charge.” However, some
color-charged particles cannot be studied under normal conditions and they are therefore referred to as “color confined.” Due to the color confinement, experimentalists cannot isolate a free quark,
the quarks always live inside color neutral hadrons.
By using an array of theoretical tools such as large-scale, lattice Quantum Chromodynamics (QCD) calculations, and the application of theory, such as the chiral perturbation theory, researchers work
to better understand these relations in the experimental processes, says Jin.
“I’m working to determine the quark transitions probabilities from the experimental inputs. There are many different experimental inputs that one can use.”
The researchers were able to solve one part of the puzzle by resolving uncertainty in the theoretical calculations which relate one experimental input to the desired quark transition probabilities.
“However, that experimental input itself is not very accurate,” Jin says. “We resolved the theoretical part, but that hadron transition process is a little bit difficult for the experimentalists. If
we really want to determine the quark transition probability from that process, we need to improve experimental precision by about tenfold. After this work, it will become a very clean process from
the theoretical point of view.”
At the APS meeting, Jin will present data exploring parameters of another flavor switch; this time, for how an up quark switches to a strange quark.
That work is similar, and the researchers were able to apply the same calculation and theory to determine the relevant low energy constants in chiral perturbation theory. “Now we know the low energy
constants very well due to this calculation, but this does not solve the whole problem due to the limitation of the chiral perturbation theory.”
Jin will also be presenting newer data for ongoing work, including innovations to account for photons which possess properties that can lead to difficulties in calculations, reduction in precision,
and systematic error,
“We are trying to do the lattice calculation in a different way to completely avoid these issues from chiral perturbation theory,” says Jin.
The work for more precision continues to understand the flavors and forces that hold visible matter together, Jin says.
“This is ongoing work and naturally to proceed, one cannot wait to try to solve the other problems. This work is the frontier of our understanding of nature.”
If you are interested in learning more, the publications that will be presented are:
Lattice QCD calculation of the electroweak box diagrams for the kaon semileptonic decays
First-Principles Calculation of Electroweak Box Diagrams from Lattice QCD
The title for Jin’s presentation is “Lattice QCD Input for the First Row CKM Unitarity Tests”. For more information about the meeting, please visit the APS Fall Meeting website. This work was made
possible through collaborations with Xu Feng, Mikhail Gorchtein, Peng-Xiang Ma, and Chien-Yeah Seng. | {"url":"https://today.uconn.edu/2021/10/working-to-understand-the-changing-flavors-of-quarks/","timestamp":"2024-11-05T12:16:09Z","content_type":"text/html","content_length":"84244","record_id":"<urn:uuid:253a9987-dd74-47eb-9072-bb042e50135d>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00454.warc.gz"} |
Education and career
Philosophical and logical work
Barwise contended that, by being explicit about the context in which a proposition is made, the situation, many problems in the application of logic can be eliminated. He sought ... to understand
meaning and inference within a general theory of information, one that takes us outside the realm of sentences and relations between sentences of any language, natural or formal. In particular, he
claimed that such an approach resolved the liar paradox. He made use of Peter Aczel's non-well-founded set theory in understanding "vicious circles" of reasoning.
Barwise, along with his former colleague at Stanford John Etchemendy, was the author of the popular logic textbook Language, Proof and Logic. Unlike the Handbook of Mathematical Logic, which was a
survey of the state of the art of mathematical logic circa 1975, and of which he was the editor, this work targeted elementary logic. The text is notable for including computer-aided homework
problems, some of which provide visual representations of logical problems. During his time at Stanford, he was also the first Director of the Symbolic Systems Program, an interdepartmental degree
program focusing on the relationships between cognition, language, logic, and computation. The K. Jon Barwise Award for Distinguished Contributions to the Symbolic Systems Program has been given
periodically since 2001.^[4]
Selected publications
See also
External links | {"url":"https://www.knowpia.com/knowpedia/Jon_Barwise","timestamp":"2024-11-14T18:52:08Z","content_type":"text/html","content_length":"87747","record_id":"<urn:uuid:576184dc-6962-45fe-841c-6a1ae1ecf613>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00200.warc.gz"} |
Can someone explain this can't understand notation
• Thread starter carrotstien
• Start date
In summary, the general equation for rotation of a rigid body in three dimensions about an arbitrary origin O with axes x, y, z involves the sum of torques on the system, the moment of inertia
tensor, angular velocity, total mass, position of the center of mass, and time. This equation is further explained in the article "Rigid-body dynamics" on StateMaster.com.
from wikipedia for
"The most general equation for rotation of a rigid body in three dimensions about an arbitrary origin O with axes x, y, z is"...
i thought that the sum of the torques on a system is just equal to d/dt(L)=d/dt(Iw)...apparently not - but i can't understand the notation like, what is b, G..etc
This article explains it better:
I is the moment of inertia tensor
ω is the angular velocity (a vector)
ωq is the angular velocity about axis q.
M is the total mass.
bG/O is the vector from O to the body's center of mass.
RO is the position of O.
t is time.
τO,j is one of the N moments about O.
Sure, I can try to explain the notation used in the Wikipedia article you mentioned.
Firstly, the equation you mentioned about the sum of torques on a system is correct. However, it is a simplified version and applies to a specific case of a rigid body rotating about a fixed axis.
The more general equation for rotation of a rigid body takes into account rotations about an arbitrary origin and axes, and therefore has more variables and notation involved.
Let's break down the equation and notation step by step.
"The most general equation for rotation of a rigid body in three dimensions about an arbitrary origin O with axes x, y, z is":
- "most general equation" means that this equation applies to any type of rotation of a rigid body, not just a specific case.
- "rotation of a rigid body in three dimensions" refers to the movement of a rigid body in three-dimensional space.
- "about an arbitrary origin O" means that the rotation is not limited to a fixed point or axis, but can occur around any point O in space.
- "with axes x, y, z" refers to the three axes (x, y, and z) that define the orientation of the rigid body.
Now, let's look at the actual equation:
L = Iω
- L represents the angular momentum of the rigid body, which is a measure of its rotational motion.
- I represents the moment of inertia of the rigid body, which is a measure of how its mass is distributed around its axis of rotation.
- ω represents the angular velocity of the rigid body, which is the rate of change of its orientation with respect to time.
The equation basically states that the angular momentum of a rigid body is equal to its moment of inertia multiplied by its angular velocity.
Next, let's look at the variables you mentioned: b and G.
- b represents the position vector of the center of mass of the rigid body with respect to the origin O. This vector is used to calculate the moment of inertia.
- G represents the angular momentum of the rigid body with respect to the origin O. This is related to the angular momentum L mentioned earlier, but takes into account the position of the center of
I hope this helps to clarify the notation and equation for you. Keep in mind that this is a complex topic and may require further reading and understanding. Don't hesitate to ask for further
clarification if needed.
FAQ: Can someone explain this can't understand notation
1. What is notation in science?
Notation in science refers to a set of symbols, numbers, or mathematical expressions used to represent and communicate scientific concepts and data. It is commonly used in fields such as mathematics,
physics, chemistry, and engineering.
2. Why is notation important in science?
Notation allows scientists to communicate complex ideas and data in a standardized and concise manner. It also helps to simplify and organize information, making it easier to understand and analyze.
3. Can you give an example of notation in science?
An example of notation in science is the use of mathematical symbols, such as +, -, x, and ÷, to represent operations in a mathematical equation. Another example is the use of chemical symbols, such
as H2O for water and CO2 for carbon dioxide.
4. How can I improve my understanding of notation in science?
One way to improve understanding of notation in science is to practice using it and familiarize yourself with common symbols and expressions. You can also seek out resources such as textbooks, online
tutorials, and study groups to help clarify any confusion.
5. Is notation the same in all branches of science?
No, notation can vary slightly between different branches of science. For example, the notation used in mathematics may differ from that used in chemistry. However, there are some commonly used
symbols and expressions that are consistent across various scientific fields. | {"url":"https://www.physicsforums.com/threads/can-someone-explain-this-cant-understand-notation.367627/","timestamp":"2024-11-10T15:06:45Z","content_type":"text/html","content_length":"85566","record_id":"<urn:uuid:920a2e14-0443-455e-a062-043f8136d6f2>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00199.warc.gz"} |
Filter (set theory) - Wikipedia Republished // WIKI 2
In mathematics, a filter on a set ${\displaystyle X}$ is a family ${\displaystyle {\mathcal {B}}}$ of subsets such that: ^[1]
1. ${\displaystyle X\in {\mathcal {B}}}$ and ${\displaystyle \emptyset otin {\mathcal {B}}}$
2. if ${\displaystyle A\in {\mathcal {B}}}$ and ${\displaystyle B\in {\mathcal {B}}}$, then ${\displaystyle A\cap B\in {\mathcal {B}}}$
3. If ${\displaystyle A\subset B\subset X}$ and ${\displaystyle A\in {\mathcal {B}}}$, then ${\displaystyle B\in {\mathcal {B}}}$
A filter on a set may be thought of as representing a "collection of large subsets",^[2] one intuitive example being the neighborhoodfilter. Filters appear in ordertheory, modeltheory, and
settheory, but can also be found in topology, from which they originate. The dual notion of a filter is an ideal.
Filters were introduced by HenriCartan in 1937^[3]^[4] and as described in the article dedicated to filtersintopology, they were subsequently used by NicolasBourbaki in their book
TopologieGénérale as an alternative to the related notion of a net developed in 1922 by E.H.Moore and HermanL.Smith. Orderfilters are generalizations of filters from sets to arbitrary
partiallyorderedsets. Specifically, a filter on a set is just a proper order filter in the special case where the partially ordered set consists of the powerset ordered by setinclusion.
YouTube Encyclopedic
• 1/5
• PIC Math - Building a Better Filter - Segment II
• A formal proof of the independence of the continuum hypothesis
Preliminaries, notation, and basic notions
In this article, upper case Roman letters like ${\displaystyle S}$ and ${\displaystyle X}$ denote sets (but not families unless indicated otherwise) and ${\displaystyle \wp (X)}$ will denote the
powerset of ${\displaystyle X.}$ A subset of a power set is called a familyofsets (or simply, a family) where it is over ${\displaystyle X}$ if it is a subset of ${\displaystyle \wp (X).}$
Families of sets will be denoted by upper case calligraphy letters such as ${\displaystyle {\mathcal {B}},{\mathcal {C}},{\text{ and }}{\mathcal {F}}.}$ Whenever these assumptions are needed, then it
should be assumed that ${\displaystyle X}$ is non–empty and that ${\displaystyle {\mathcal {B}},{\mathcal {F}},}$ etc. are families of sets over ${\displaystyle X.}$
The terms "prefilter" and "filter base" are synonyms and will be used interchangeably.
Warning about competing definitions and notation
There are unfortunately several terms in the theory of filters that are defined differently by different authors. These include some of the most important terms such as "filter". While different
definitions of the same term usually have significant overlap, due to the very technical nature of filters (and point–set topology), these differences in definitions nevertheless often have important
consequences. When reading mathematical literature, it is recommended that readers check how the terminology related to filters is defined by the author. For this reason, this article will clearly
state all definitions as they are used. Unfortunately, not all notation related to filters is well established and some notation varies greatly across the literature (for example, the notation for
the set of all prefilters on a set) so in such cases this article uses whatever notation is most self describing or easily remembered.
The theory of filters and prefilters is well developed and has a plethora of definitions and notations, many of which are now unceremoniously listed to prevent this article from becoming prolix and
to allow for the easy look up of notation and definitions. Their important properties are described later.
Sets operations
The upward closure or isotonization in ${\displaystyle X}$^[5]^[6] of a familyofsets ${\displaystyle {\mathcal {B}}\subseteq \wp (X)}$ is
${\displaystyle {\mathcal {B}}^{\uparrow X}:=\{S\subseteq X~:~B\subseteq S{\text{ for some }}B\in {\mathcal {B}}\,\}=\bigcup _{B\in {\mathcal {B}}}\{S~:~B\subseteq S\subseteq X\}}$
and similarly the downward closure of ${\displaystyle {\mathcal {B}}}$ is ${\displaystyle {\mathcal {B}}^{\downarrow }:=\{S\subseteq B~:~B\in {\mathcal {B}}\,\}=\bigcup _{B\in {\mathcal {B}}}\wp
Notation and Definition Name
${\displaystyle \ker {\mathcal {B}}=\bigcap _{B\in {\mathcal {B}}}B}$ Kernel of ${\displaystyle {\mathcal {B}}}$^[6]
${\displaystyle S\setminus {\mathcal {B}}:=\{S\setminus B~:~B\in {\mathcal Dual of ${\displaystyle {\mathcal {B}}{\text{ in }}S}$ where ${\displaystyle S}$ is a set.^[7]
{B}}\}=\{S\}\,(\setminus )\,{\mathcal {B}}}$
${\displaystyle {\mathcal {B}}{\big \vert }_{S}:=\{B\cap S~:~B\in {\mathcal Trace of ${\displaystyle {\mathcal {B}}{\text{ on }}S}$^[7] or the restriction of ${\displaystyle {\mathcal {B}}{\text{ to
{B}}\}={\mathcal {B}}\,(\cap )\,\{S\}}$ }}S}$ where ${\displaystyle S}$ is a set; sometimes denoted by ${\displaystyle {\mathcal {B}}\cap S}$
${\displaystyle {\mathcal {B}}\,(\cap )\,{\mathcal {C}}=\{B\cap C~:~B\in {\ Elementwise (set) intersection (${\displaystyle {\mathcal {B}}\cap {\mathcal {C}}}$ will denote the usual intersection)
mathcal {B}}{\text{ and }}C\in {\mathcal {C}}\}}$^[8]
${\displaystyle {\mathcal {B}}\,(\cup )\,{\mathcal {C}}=\{B\cup C~:~B\in {\ Elementwise (set) union (${\displaystyle {\mathcal {B}}\cup {\mathcal {C}}}$ will denote the usual union)
mathcal {B}}{\text{ and }}C\in {\mathcal {C}}\}}$^[8]
${\displaystyle {\mathcal {B}}\,(\setminus )\,{\mathcal {C}}=\{B\setminus Elementwise (set) subtraction (${\displaystyle {\mathcal {B}}\setminus {\mathcal {C}}}$ will denote the usual
C~:~B\in {\mathcal {B}}{\text{ and }}C\in {\mathcal {C}}\}}$ setsubtraction)
${\displaystyle {\mathcal {B}}^{\#X}={\mathcal {B}}^{\#}=\{S\subseteq X~:~S Grill of ${\displaystyle {\mathcal {B}}{\text{ in }}X}$^[9]
\cap Beq \varnothing {\text{ for all }}B\in {\mathcal {B}}\}}$
${\displaystyle \wp (X)=\{S~:~S\subseteq X\}}$ Powerset of a set ${\displaystyle X}$^[6]
For any two families ${\displaystyle {\mathcal {C}}{\text{ and }}{\mathcal {F}},}$ declare that ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}}}$ if and only if for every ${\displaystyle C\in
{\mathcal {C}}}$ there exists some ${\displaystyle F\in {\mathcal {F}}{\text{ such that }}F\subseteq C,}$ in which case it is said that ${\displaystyle {\mathcal {C}}}$ is coarser than ${\
displaystyle {\mathcal {F}}}$ and that ${\displaystyle {\mathcal {F}}}$ is finer than (or subordinate to) ${\displaystyle {\mathcal {C}}.}$^[10]^[11]^[12] The notation ${\displaystyle {\mathcal
{F}}\vdash {\mathcal {C}}{\text{ or }}{\mathcal {F}}\geq {\mathcal {C}}}$ may also be used in place of ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}}.}$
Two families ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}}$ mesh,^[7] written ${\displaystyle {\mathcal {B}}\#{\mathcal {C}},}$ if ${\displaystyle B\cap Ceq \varnothing {\text{ for
all }}B\in {\mathcal {B}}{\text{ and }}C\in {\mathcal {C}}.}$
Throughout, ${\displaystyle f}$ is a map and ${\displaystyle S}$ is a set.
Notation and Definition Name
${\displaystyle f^{-1}({\mathcal {B}})=\left\{f^{-1}(B)~:~B\in {\mathcal {B}}\ Image of ${\displaystyle {\mathcal {B}}{\text{ under }}f^{-1},}$ or the preimage of ${\displaystyle {\mathcal {B}}}$
right\}}$^[13] under ${\displaystyle f}$
${\displaystyle f^{-1}(S)=\{x\in \operatorname {domain} f~:~f(x)\in S\}}$ Image of ${\displaystyle S{\text{ under }}f^{-1},}$ or the preimage of ${\displaystyle S{\text{ under }}f}$
${\displaystyle f({\mathcal {B}})=\{f(B)~:~B\in {\mathcal {B}}\}}$^[14] Image of ${\displaystyle {\mathcal {B}}}$ under ${\displaystyle f}$
${\displaystyle f(S)=\{f(s)~:~s\in S\cap \operatorname {domain} f\}}$ Image of ${\displaystyle S{\text{ under }}f}$
${\displaystyle \operatorname {image} f=f(\operatorname {domain} f)}$ Image (or range) of ${\displaystyle f}$
Nets and their tails
A directedset is a set ${\displaystyle I}$ together with a preorder, which will be denoted by ${\displaystyle \,\leq \,}$ (unless explicitly indicated otherwise), that makes ${\displaystyle (I,\leq
)}$ into an (upward) directed set;^[15] this means that for all ${\displaystyle i,j\in I,}$ there exists some ${\displaystyle k\in I}$ such that ${\displaystyle i\leq k{\text{ and }}j\leq k.}$ For
any indices ${\displaystyle i{\text{ and }}j,}$ the notation ${\displaystyle j\geq i}$ is defined to mean ${\displaystyle i\leq j}$ while ${\displaystyle i<j}$ is defined to mean that ${\displaystyle
i\leq j}$ holds but it is not true that ${\displaystyle j\leq i}$ (if ${\displaystyle \,\leq \,}$ is antisymmetric then this is equivalent to ${\displaystyle i\leq j{\text{ and }}ieq j}$).
A net in ${\displaystyle X}$^[15] is a map from a non–empty directed set into ${\displaystyle X.}$ The notation ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i\in I}}$ will be used to denote a net
with domain ${\displaystyle I.}$
Notation and Definition Name
${\displaystyle I_{\geq i}=\{j\in I~:~j\geq i\}}$ Tail or section of ${\displaystyle I}$ starting at ${\displaystyle i\in I}$ where ${\displaystyle (I,\leq )}$ is a directedset.
${\displaystyle x_{\geq i}=\left\{x_{j}~:~j\geq i{\text{ Tail or section of ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i\in I}}$ starting at ${\displaystyle i\in I}$
and }}j\in I\right\}}$
${\displaystyle \operatorname {Tails} \left(x_{\bullet }\ Set or prefilter of tails/sections of ${\displaystyle x_{\bullet }.}$ Also called the eventuality filter base generated by (the tails of) $
right)=\left\{x_{\geq i}~:~i\in I\right\}}$ {\displaystyle x_{\bullet }=\left(x_{i}\right)_{i\in I}.}$ If ${\displaystyle x_{\bullet }}$ is a sequence then ${\displaystyle \
operatorname {Tails} \left(x_{\bullet }\right)}$ is also called the sequential filter base.^[16]
${\displaystyle \operatorname {TailsFilter} \left(x_{\
bullet }\right)=\operatorname {Tails} \left(x_{\bullet }\ (Eventuality) filter of/generated by (tails of) ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i\in I}}$^[16]
right)^{\uparrow X}}$
${\displaystyle f\left(I_{\geq i}\right)=\{f(j)~:~j\geq i Tail or section of a net ${\displaystyle f:I\to X}$ starting at ${\displaystyle i\in I}$^[16] where ${\displaystyle (I,\leq )}$ is a
{\text{ and }}j\in I\}}$ directed set.
Warning about using strict comparison
If ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i\in I}}$ is a net and ${\displaystyle i\in I}$ then it is possible for the set ${\displaystyle x_{>i}=\left\{x_{j}~:~j>i{\text{ and }}j\in I\right
\},}$ which is called the tail of ${\displaystyle x_{\bullet }}$ after ${\displaystyle i}$, to be empty (for example, this happens if ${\displaystyle i}$ is an upperbound of the directedset ${\
displaystyle I}$). In this case, the family ${\displaystyle \left\{x_{>i}~:~i\in I\right\}}$ would contain the empty set, which would prevent it from being a prefilter (defined later). This is the
(important) reason for defining ${\displaystyle \operatorname {Tails} \left(x_{\bullet }\right)}$ as ${\displaystyle \left\{x_{\geq i}~:~i\in I\right\}}$ rather than ${\displaystyle \left\{x_{>i}~:~i
\in I\right\}}$ or even ${\displaystyle \left\{x_{>i}~:~i\in I\right\}\cup \left\{x_{\geq i}~:~i\in I\right\}}$ and it is for this reason that in general, when dealing with the prefilter of tails of
a net, the strict inequality ${\displaystyle \,<\,}$ may not be used interchangeably with the inequality ${\displaystyle \,\leq .}$
Filters and prefilters
Families${\displaystyle {\mathcal {F}}}$ofsets over ${\displaystyle \Omega }$
Is necessarily true of ${\
displaystyle {\mathcal {F}} Directed ${\ ${\ ${\ ${\displaystyle ${\displaystyle A_ ${\displaystyle A_ ${\displaystyle \ ${\displaystyle \
\colon }$ by${\ displaystyle displaystyle displaystyle \Omega \setminus {1}\cap A_{2}\cap \ {1}\cup A_{2}\cup \ Omega \in {\mathcal varnothing \in {\ F.I.P.
or, is ${\displaystyle {\ displaystyle A\cap B}$ A\cup B}$ B\setminus A} A}$ cdots }$ cdots }$ {F}}}$ mathcal {F}}}$
mathcal {F}}}$ \,\supseteq }$ $
Semiring Never
Semialgebra(Semifield) Never
only if ${\ only if ${\
Monotoneclass displaystyle A_{i}\ displaystyle A_{i}
searrow }$ earrow }$
only if only if ${\
${\ displaystyle A_{i}
𝜆-system(DynkinSystem) displaystyle earrow }$ or Never
A\subseteq B} they are disjoint
Ring (Order theory)
Ring (Measure theory) Never
δ-Ring Never
𝜎-Ring Never
Algebra (Field) Never
𝜎-Algebra(𝜎-Field) Never
${\displaystyle \
Filter Never Never varnothing ot \in {\
mathcal {F}}}$
${\displaystyle \
Prefilter(Filterbase) Never Never varnothing ot \in {\
mathcal {F}}}$
${\displaystyle \
Filtersubbase Never Never varnothing ot \in {\
mathcal {F}}}$
OpenTopology Never
(even arbitrary ${\
displaystyle \cup }$
ClosedTopology Never
(even arbitrary ${\
displaystyle \cap }$
Is necessarily true of ${\
displaystyle {\mathcal {F}} complements contains ${\ contains ${\ Finite
\colon }$ directed finite finite relative in ${\ countable countable displaystyle \Omega displaystyle \ Intersection
or, is ${\displaystyle {\ downward intersections unions complements displaystyle \ intersections unions }$ varnothing }$ Property
mathcal {F}}}$ Omega }$
Additionally, a semiring is a π-system where every complement ${\displaystyle B\setminus A}$ is equal to a finite disjointunion of sets in ${\displaystyle {\mathcal {F}}.}$
A semialgebra is a semiring where every complement ${\displaystyle \Omega \setminus A}$ is equal to a finite disjointunion of sets in ${\displaystyle {\mathcal {F}}.}$
${\displaystyle A,B,A_{1},A_{2},\ldots }$ are arbitrary elements of ${\displaystyle {\mathcal {F}}}$ and it is assumed that ${\displaystyle {\mathcal {F}}eq \varnothing .}$
The following is a list of properties that a family ${\displaystyle {\mathcal {B}}}$ of sets may possess and they form the defining properties of filters, prefilters, and filter subbases. Whenever it
is necessary, it should be assumed that ${\displaystyle {\mathcal {B}}\subseteq \wp (X).}$
The family of sets
${\displaystyle {\mathcal {B}}}$
1. Proper or nondegenerate if ${\displaystyle \varnothing ot \in {\mathcal {B}}.}$ Otherwise, if ${\displaystyle \varnothing \in {\mathcal {B}},}$ then it is called improper^[17] or degenerate.
2. Directed downward^[15] if whenever ${\displaystyle A,B\in {\mathcal {B}}}$ then there exists some ${\displaystyle C\in {\mathcal {B}}}$ such that ${\displaystyle C\subseteq A\cap B.}$
☆ This property can be characterized in terms of directedness, which explains the word "directed": A binaryrelation ${\displaystyle \,\preceq \,}$ on ${\displaystyle {\mathcal {B}}}$ is
called (upward)directed if for any two ${\displaystyle A{\text{ and }}B,}$ there is some ${\displaystyle C}$ satisfying ${\displaystyle A\preceq C{\text{ and }}B\preceq C.}$ Using ${\
displaystyle \,\supseteq \,}$ in place of ${\displaystyle \,\preceq \,}$ gives the definition of directed downward whereas using ${\displaystyle \,\subseteq \,}$ instead gives the
definition of directed upward. Explicitly, ${\displaystyle {\mathcal {B}}}$ is directed downward (resp. directed upward) if and only if for all ${\displaystyle A,B\in {\mathcal {B}},}$
there exists some "greater" ${\displaystyle C\in {\mathcal {B}}}$ such that ${\displaystyle A\supseteq C{\text{ and }}B\supseteq C}$ (resp. such that ${\displaystyle A\subseteq C{\text{
and }}B\subseteq C}$) − where the "greater" element is always on the right hand side,^[note1] − which can be rewritten as ${\displaystyle A\cap B\supseteq C}$ (resp. as ${\displaystyle A
\cup B\subseteq C}$).
☆ If a family ${\displaystyle {\mathcal {B}}}$ has a greatestelement with respect to ${\displaystyle \,\supseteq \,}$ (for example, if ${\displaystyle \varnothing \in {\mathcal {B}}}$)
then it is necessarily directed downward.
3. Closed under finite intersections (resp. unions) if the intersection (resp. union) of any two elements of ${\displaystyle {\mathcal {B}}}$ is an element of ${\displaystyle {\mathcal {B}}.}$
☆ If ${\displaystyle {\mathcal {B}}}$ is closed under finite intersections then ${\displaystyle {\mathcal {B}}}$ is necessarily directed downward. The converse is generally false.
4. Upward closed or Isotone in ${\displaystyle X}$^[5] if ${\displaystyle {\mathcal {B}}\subseteq \wp (X){\text{ and }}{\mathcal {B}}={\mathcal {B}}^{\uparrow X},}$ or equivalently, if whenever
${\displaystyle B\in {\mathcal {B}}}$ and some set ${\displaystyle C}$ satisfies ${\displaystyle B\subseteq C\subseteq X,{\text{ then }}C\in {\mathcal {B}}.}$ Similarly, ${\displaystyle {\
mathcal {B}}}$ is downward closed if ${\displaystyle {\mathcal {B}}={\mathcal {B}}^{\downarrow }.}$ An upward (respectively, downward) closed set is also called an upper set or upset (resp. a
lower set or down set).
☆ The family ${\displaystyle {\mathcal {B}}^{\uparrow X},}$ which is the upward closure of ${\displaystyle {\mathcal {B}}{\text{ in }}X,}$ is the unique smallest (with respect to ${\
displaystyle \,\subseteq }$) isotone family of sets over ${\displaystyle X}$ having ${\displaystyle {\mathcal {B}}}$ as a subset.
Many of the properties of ${\displaystyle {\mathcal {B}}}$ defined above and below, such as "proper" and "directed downward," do not depend on ${\displaystyle X,}$ so mentioning the set ${\
displaystyle X}$ is optional when using such terms. Definitions involving being "upward closed in ${\displaystyle X,}$" such as that of "filter on ${\displaystyle X,}$" do depend on ${\displaystyle
X}$ so the set ${\displaystyle X}$ should be mentioned if it is not clear from context.
${\displaystyle {\textrm {Filters}}(X)\quad =\quad {\textrm {DualIdeals}}(X)\,\setminus \,\{\wp (X)\}\quad \subseteq \quad {\textrm {Prefilters}}(X)\quad \subseteq \quad {\textrm {FilterSubbases}}
A family
${\displaystyle {\mathcal {B}}}$
is/is a(n):
1. Ideal^[17]^[18] if ${\displaystyle {\mathcal {B}}eq \varnothing }$ is downward closed and closed under finite unions.
2. Dual ideal on ${\displaystyle X}$^[19] if ${\displaystyle {\mathcal {B}}eq \varnothing }$ is upward closed in ${\displaystyle X}$ and also closed under finite intersections. Equivalently, ${\
displaystyle {\mathcal {B}}eq \varnothing }$ is a dual ideal if for all ${\displaystyle R,S\subseteq X,}$ ${\displaystyle R\cap S\in {\mathcal {B}}\;{\text{ if and only if }}\;R,S\in {\
mathcal {B}}.}$^[9]
☆ Explanation of the word "dual": A family ${\displaystyle {\mathcal {B}}}$ is a dual ideal (resp. an ideal) on ${\displaystyle X}$ if and only if the dual of ${\displaystyle {\mathcal {B}}
{\text{ in }}X,}$ which is the family ${\displaystyle X\setminus {\mathcal {B}}:=\{X\setminus B~:~B\in {\mathcal {B}}\},}$ is an ideal (resp. a dual ideal) on ${\displaystyle X.}$ In
other words, dual ideal means "dual of an ideal". The family ${\displaystyle X\setminus {\mathcal {B}}}$ should not be confused with ${\displaystyle \wp (X)\setminus {\mathcal {B}}=\{S\
subseteq X~:~Sotin {\mathcal {B}}\}}$ because these two sets are not equal in general; for instance, ${\displaystyle X\setminus {\mathcal {B}}=\wp (X){\text{ if and only if }}{\mathcal
{B}}=\wp (X).}$ The dual of the dual is the original family, meaning ${\displaystyle X\setminus (X\setminus {\mathcal {B}})={\mathcal {B}}.}$ The set ${\displaystyle X}$ belongs to the
dual of ${\displaystyle {\mathcal {B}}}$ if and only if ${\displaystyle \varnothing \in {\mathcal {B}}.}$^[17]
3. Filter on ${\displaystyle X}$^[19]^[7] if ${\displaystyle {\mathcal {B}}}$ is a proper dualideal on ${\displaystyle X.}$ That is, a filter on ${\displaystyle X}$ is a non−empty subset of ${\
displaystyle \wp (X)\setminus \{\varnothing \}}$ that is closed under finite intersections and upward closed in ${\displaystyle X.}$ Equivalently, it is a prefilter that is upward closed in $
{\displaystyle X.}$ In words, a filter on ${\displaystyle X}$ is a family of sets over ${\displaystyle X}$ that (1) is not empty (or equivalently, it contains ${\displaystyle X}$), (2) is
closed under finite intersections, (3) is upward closed in ${\displaystyle X,}$ and (4) does not have the empty set as an element.
☆ Warning: Some authors, particularly algebrists, use "filter" to mean a dual ideal; others, particularly topologists, use "filter" to mean a proper/non–degenerate dual ideal.^[20] It is
recommended that readers always check how "filter" is defined when reading mathematical literature. However, the definitions of "ultrafilter," "prefilter," and "filter subbase" always
require non-degeneracy. This article uses HenriCartan's original definition of "filter",^[3]^[4] which required non–degeneracy.
☆ A dualfilter on ${\displaystyle X}$ is a family ${\displaystyle {\mathcal {B}}}$ whose dual ${\displaystyle X\setminus {\mathcal {B}}}$ is a filter on ${\displaystyle X.}$ Equivalently,
it is an ideal on ${\displaystyle X}$ that does not contain ${\displaystyle X}$ as an element.
☆ The power set ${\displaystyle \wp (X)}$ is the one and only dual ideal on ${\displaystyle X}$ that is not also a filter. Excluding ${\displaystyle \wp (X)}$ from the definition of
"filter" in topology has the same benefit as excluding${\displaystyle 1}$fromthedefinition of "primenumber": it obviates the need to specify "non-degenerate" (the analog of "non-
unital" or "non-${\displaystyle 1}$") in many important results, thereby making their statements less awkward.
4. Prefilter or filter base^[7]^[21] if ${\displaystyle {\mathcal {B}}eq \varnothing }$ is proper and directed downward. Equivalently, ${\displaystyle {\mathcal {B}}}$ is called a prefilter if
its upward closure ${\displaystyle {\mathcal {B}}^{\uparrow X}}$ is a filter. It can also be defined as any family that is equivalent (with respect to ${\displaystyle \leq }$) to some filter.
^[8] A proper family ${\displaystyle {\mathcal {B}}eq \varnothing }$ is a prefilter if and only if ${\displaystyle {\mathcal {B}}\,(\cap )\,{\mathcal {B}}\leq {\mathcal {B}}.}$^[8] A family
is a prefilter if and only if the same is true of its upward closure.
☆ If ${\displaystyle {\mathcal {B}}}$ is a prefilter then its upward closure ${\displaystyle {\mathcal {B}}^{\uparrow X}}$ is the unique smallest (relative to ${\displaystyle \subseteq }$)
filter on ${\displaystyle X}$ containing ${\displaystyle {\mathcal {B}}}$ and it is called the filter generated by ${\displaystyle {\mathcal {B}}.}$ A filter ${\displaystyle {\mathcal
{F}}}$ is said to be generated by a prefilter ${\displaystyle {\mathcal {B}}}$ if ${\displaystyle {\mathcal {F}}={\mathcal {B}}^{\uparrow X},}$ in which ${\displaystyle {\mathcal {B}}}$
is called a filter base for ${\displaystyle {\mathcal {F}}.}$
☆ Unlike a filter, a prefilter is not necessarily closed under finite intersections.
5. π–system if ${\displaystyle {\mathcal {B}}eq \varnothing }$ is closed under finite intersections. Every non–empty family ${\displaystyle {\mathcal {B}}}$ is contained in a unique smallest π
–system called the π–system generated by ${\displaystyle {\mathcal {B}},}$ which is sometimes denoted by ${\displaystyle \pi ({\mathcal {B}}).}$ It is equal to the intersection of all π
–systems containing ${\displaystyle {\mathcal {B}}}$ and also to the set of all possible finite intersections of sets from ${\displaystyle {\mathcal {B}}}$: ${\displaystyle \pi ({\mathcal
{B}})=\left\{B_{1}\cap \cdots \cap B_{n}~:~n\geq 1{\text{ and }}B_{1},\ldots ,B_{n}\in {\mathcal {B}}\right\}.}$
☆ A π–system is a prefilter if and only if it is proper. Every filter is a proper π–system and every proper π–system is a prefilter but the converses do not hold in general.
☆ A prefilter is equivalent (with respect to ${\displaystyle \leq }$) to the π–system generated by it and both of these families generate the same filter on ${\displaystyle X.}$
6. Filter subbase^[7]^[22] and centered^[8] if ${\displaystyle {\mathcal {B}}eq \varnothing }$ and ${\displaystyle {\mathcal {B}}}$ satisfies any of the following equivalent conditions:
1. ${\displaystyle {\mathcal {B}}}$ has the finiteintersectionproperty, which means that the intersection of any finite family of (one or more) sets in ${\displaystyle {\mathcal {B}}}$ is
not empty; explicitly, this means that whenever ${\displaystyle n\geq 1{\text{ and }}B_{1},\ldots ,B_{n}\in {\mathcal {B}}}$ then ${\displaystyle \varnothing eq B_{1}\cap \cdots \cap B_
2. The π–system generated by ${\displaystyle {\mathcal {B}}}$ is proper; that is, ${\displaystyle \varnothing ot \in \pi ({\mathcal {B}}).}$
3. The π–system generated by ${\displaystyle {\mathcal {B}}}$ is a prefilter.
4. ${\displaystyle {\mathcal {B}}}$ is a subset of some prefilter.
5. ${\displaystyle {\mathcal {B}}}$ is a subset of some filter.
☆ Assume that ${\displaystyle {\mathcal {B}}}$ is a filter subbase. Then there is a unique smallest (relative to ${\displaystyle \subseteq }$) filter ${\displaystyle {\mathcal {F}}_{\
mathcal {B}}{\text{ on }}X}$ containing ${\displaystyle {\mathcal {B}}}$ called the filter generated by ${\displaystyle {\mathcal {B}}}$, and ${\displaystyle {\mathcal {B}}}$ is said to
be a filter subbase for this filter. This filter is equal to the intersection of all filters on ${\displaystyle X}$ that are supersets of ${\displaystyle {\mathcal {B}}.}$ The π–system
generated by ${\displaystyle {\mathcal {B}},}$ denoted by ${\displaystyle \pi ({\mathcal {B}}),}$ will be a prefilter and a subset of ${\displaystyle {\mathcal {F}}_{\mathcal {B}}.}$
Moreover, the filter generated by ${\displaystyle {\mathcal {B}}}$ is equal to the upward closure of ${\displaystyle \pi ({\mathcal {B}}),}$ meaning ${\displaystyle \pi ({\mathcal {B}})^
{\uparrow X}={\mathcal {F}}_{\mathcal {B}}.}$^[8] However, ${\displaystyle {\mathcal {B}}^{\uparrow X}={\mathcal {F}}_{\mathcal {B}}}$ if and only if ${\displaystyle {\mathcal {B}}}$ is a
prefilter (although ${\displaystyle {\mathcal {B}}^{\uparrow X}}$ is always an upward closed filter subbase for ${\displaystyle {\mathcal {F}}_{\mathcal {B}}}$).
☆ A ${\displaystyle \subseteq }$–smallest (meaning smallest relative to ${\displaystyle \subseteq }$) prefilter containing a filter subbase ${\displaystyle {\mathcal {B}}}$ will exist
only under certain circumstances. It exists, for example, if the filter subbase ${\displaystyle {\mathcal {B}}}$ happens to also be a prefilter. It also exists if the filter (or
equivalently, the π–system) generated by ${\displaystyle {\mathcal {B}}}$ is principal, in which case ${\displaystyle {\mathcal {B}}\cup \{\ker {\mathcal {B}}\}}$ is the unique smallest
prefilter containing ${\displaystyle {\mathcal {B}}.}$ Otherwise, in general, a ${\displaystyle \subseteq }$–smallest prefilter containing ${\displaystyle {\mathcal {B}}}$ might not
exist. For this reason, some authors may refer to the π–system generated by ${\displaystyle {\mathcal {B}}}$ as the prefilter generated by ${\displaystyle {\mathcal {B}}.}$ However, if a
${\displaystyle \subseteq }$–smallest prefilter does exist (say it is denoted by ${\displaystyle \operatorname {minPre} {\mathcal {B}}}$) then contrary to usual expectations, it is not
necessarily equal to "theprefiltergeneratedby${\displaystyle {\mathcal {B}}}$" (that is, ${\displaystyle \operatorname {minPre} {\mathcal {B}}eq \pi ({\mathcal {B}})}$ is possible).
And if the filter subbase ${\displaystyle {\mathcal {B}}}$ happens to also be a prefilter but not a π-system then unfortunately, "theprefiltergeneratedbythisprefilter" (meaning ${\
displaystyle \pi ({\mathcal {B}})}$) will not be ${\displaystyle {\mathcal {B}}=\operatorname {minPre} {\mathcal {B}}}$ (that is, ${\displaystyle \pi ({\mathcal {B}})eq {\mathcal {B}}}$
is possible even when ${\displaystyle {\mathcal {B}}}$ is a prefilter), which is why this article will prefer the accurate and unambiguous terminology of "the π–system generated by ${\
displaystyle {\mathcal {B}}}$".
7. Subfilter of a filter ${\displaystyle {\mathcal {F}}}$ and that ${\displaystyle {\mathcal {F}}}$ is a superfilter of ${\displaystyle {\mathcal {B}}}$^[17]^[23] if ${\displaystyle {\mathcal
{B}}}$ is a filter and ${\displaystyle {\mathcal {B}}\subseteq {\mathcal {F}}}$ where for filters, ${\displaystyle {\mathcal {B}}\subseteq {\mathcal {F}}{\text{ if and only if }}{\mathcal
{B}}\leq {\mathcal {F}}.}$
☆ Importantly, the expression "is a superfilter of" is for filters the analog of "is a subsequence of". So despite having the prefix "sub" in common, "is a subfilter of" is actually the
reverse of "is a subsequence of." However, ${\displaystyle {\mathcal {B}}\leq {\mathcal {F}}}$ can also be written ${\displaystyle {\mathcal {F}}\vdash {\mathcal {B}}}$ which is described
by saying "${\displaystyle {\mathcal {F}}}$ is subordinate to ${\displaystyle {\mathcal {B}}.}$" With this terminology, "is subordinate to" becomes for filters (and also for prefilters)
the analog of "is a subsequence of,"^[24] which makes this one situation where using the term "subordinate" and symbol ${\displaystyle \,\vdash \,}$ may be helpful.
There are no prefilters on ${\displaystyle X=\varnothing }$ (nor are there any nets valued in ${\displaystyle \varnothing }$), which is why this article, like most authors, will automatically assume
without comment that ${\displaystyle Xeq \varnothing }$ whenever this assumption is needed.
Basic examples
Named examples
• The singleton set ${\displaystyle {\mathcal {B}}=\{X\}}$ is called the indiscrete or trivial filter on ${\displaystyle X.}$^[25]^[10] It is the unique minimal filter on ${\displaystyle X}$
because it is a subset of every filter on ${\displaystyle X}$; however, it need not be a subset of every prefilter on ${\displaystyle X.}$
• The dual ideal ${\displaystyle \wp (X)}$ is also called the degenerate filter on ${\displaystyle X}$^[9] (despite not actually being a filter). It is the only dual ideal on ${\displaystyle X}$
that is not a filter on ${\displaystyle X.}$
• If ${\displaystyle (X,\tau )}$ is a topological space and ${\displaystyle x\in X,}$ then the neighborhoodfilter ${\displaystyle {\mathcal {N}}(x)}$ at ${\displaystyle x}$ is a filter on ${\
displaystyle X.}$ By definition, a family ${\displaystyle {\mathcal {B}}\subseteq \wp (X)}$ is called a neighborhoodbasis (resp. a neighborhood subbase) at ${\displaystyle x{\text{ for }}(X,\tau
)}$ if and only if ${\displaystyle {\mathcal {B}}}$ is a prefilter (resp. ${\displaystyle {\mathcal {B}}}$ is a filter subbase) and the filter on ${\displaystyle X}$ that ${\displaystyle {\
mathcal {B}}}$ generates is equal to the neighborhood filter ${\displaystyle {\mathcal {N}}(x).}$ The subfamily ${\displaystyle \tau (x)\subseteq {\mathcal {N}}(x)}$ of open neighborhoods is a
filter base for ${\displaystyle {\mathcal {N}}(x).}$ Both prefilters ${\displaystyle {\mathcal {N}}(x){\text{ and }}\tau (x)}$ also form a bases for topologies on ${\displaystyle X,}$ with the
topology generated ${\displaystyle \tau (x)}$ being coarser than ${\displaystyle \tau .}$ This example immediately generalizes from neighborhoods of points to neighborhoods of non–empty subsets $
{\displaystyle S\subseteq X.}$
• ${\displaystyle {\mathcal {B}}}$ is an elementary prefilter^[26] if ${\displaystyle {\mathcal {B}}=\operatorname {Tails} \left(x_{\bullet }\right)}$ for some sequence ${\displaystyle x_{\bullet }
=\left(x_{i}\right)_{i=1}^{\infty }{\text{ in }}X.}$
• ${\displaystyle {\mathcal {B}}}$ is an elementary filter or a sequential filter on ${\displaystyle X}$^[27] if ${\displaystyle {\mathcal {B}}}$ is a filter on ${\displaystyle X}$ generated by
some elementary prefilter. The filter of tails generated by a sequence that is not eventually constant is necessarily not an ultrafilter.^[28] Every principal filter on a countable set is
sequential as is every cofinite filter on a countably infinite set.^[9] The intersection of finitely many sequential filters is again sequential.^[9]
• The set ${\displaystyle {\mathcal {F}}}$ of all cofinitesubsets of ${\displaystyle X}$ (meaning those sets whose complement in ${\displaystyle X}$ is finite) is proper if and only if ${\
displaystyle {\mathcal {F}}}$ is infinite (or equivalently, ${\displaystyle X}$ is infinite), in which case ${\displaystyle {\mathcal {F}}}$ is a filter on ${\displaystyle X}$ known as the
Fréchetfilterorthecofinite filter on ${\displaystyle X.}$^[10]^[25] If ${\displaystyle X}$ is finite then ${\displaystyle {\mathcal {F}}}$ is equal to the dual ideal ${\displaystyle \wp (X),}
$ which is not a filter. If ${\displaystyle X}$ is infinite then the family ${\displaystyle \{X\setminus \{x\}~:~x\in X\}}$ of complements of singleton sets is a filter subbase that generates the
Fréchet filter on ${\displaystyle X.}$ As with any family of sets over ${\displaystyle X}$ that contains ${\displaystyle \{X\setminus \{x\}~:~x\in X\},}$ the kernel of the Fréchet filter on ${\
displaystyle X}$ is the empty set: ${\displaystyle \ker {\mathcal {F}}=\varnothing .}$
• The intersection of all elements in any non–empty family ${\displaystyle \mathbb {F} \subseteq \operatorname {Filters} (X)}$ is itself a filter on ${\displaystyle X}$ called the infimum or
greatestlowerbound of ${\displaystyle \mathbb {F} {\text{ in }}\operatorname {Filters} (X),}$ which is why it may be denoted by ${\displaystyle \bigwedge _{{\mathcal {F}}\in \mathbb {F} }{\
mathcal {F}}.}$ Said differently, ${\displaystyle \ker \mathbb {F} =\bigcap _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}}\in \operatorname {Filters} (X).}$ Because every filter on ${\
displaystyle X}$ has ${\displaystyle \{X\}}$ as a subset, this intersection is never empty. By definition, the infimum is the finest/largest (relative to ${\displaystyle \,\subseteq \,{\text{ and
}}\,\leq \,}$) filter contained as a subset of each member of ${\displaystyle \mathbb {F} .}$^[10]
□ If ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {F}}}$ are filters then their infimum in ${\displaystyle \operatorname {Filters} (X)}$ is the filter ${\displaystyle {\mathcal {B}}\,
(\cup )\,{\mathcal {F}}.}$^[8] If ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {F}}}$ are prefilters then ${\displaystyle {\mathcal {B}}\,(\cup )\,{\mathcal {F}}}$ is a prefilter
that is coarser (with respect to ${\displaystyle \,\leq }$) than both ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {F}}}$ (that is, ${\displaystyle {\mathcal {B}}\,(\cup )\,{\mathcal
{F}}\leq {\mathcal {B}}{\text{ and }}{\mathcal {B}}\,(\cup )\,{\mathcal {F}}\leq {\mathcal {F}}}$); indeed, it is oneofthefinestsuchprefilters, meaning that if ${\displaystyle {\mathcal
{S}}}$ is a prefilter such that ${\displaystyle {\mathcal {S}}\leq {\mathcal {B}}{\text{ and }}{\mathcal {S}}\leq {\mathcal {F}}}$ then necessarily ${\displaystyle {\mathcal {S}}\leq {\
mathcal {B}}\,(\cup )\,{\mathcal {F}}.}$^[8] More generally, if ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {F}}}$ are non−empty families and if ${\displaystyle \mathbb {S} :=\{{\
mathcal {S}}\subseteq \wp (X)~:~{\mathcal {S}}\leq {\mathcal {B}}{\text{ and }}{\mathcal {S}}\leq {\mathcal {F}}\}}$ then ${\displaystyle {\mathcal {B}}\,(\cup )\,{\mathcal {F}}\in \mathbb
{S} }$ and ${\displaystyle {\mathcal {B}}\,(\cup )\,{\mathcal {F}}}$ is a greatestelement (with respect to ${\displaystyle \leq }$) of ${\displaystyle \mathbb {S} .}$^[8]
• Let ${\displaystyle \varnothing eq \mathbb {F} \subseteq \operatorname {DualIdeals} (X)}$ and let ${\displaystyle \cup \mathbb {F} =\bigcup _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}}.}$ The
supremum or leastupperbound of ${\displaystyle \mathbb {F} {\text{ in }}\operatorname {DualIdeals} (X),}$ denoted by ${\displaystyle \bigvee _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}},}$
is the smallest (relative to ${\displaystyle \subseteq }$) dual ideal on ${\displaystyle X}$ containing every element of ${\displaystyle \mathbb {F} }$ as a subset; that is, it is the smallest
(relative to ${\displaystyle \subseteq }$) dual ideal on ${\displaystyle X}$ containing ${\displaystyle \cup \mathbb {F} }$ as a subset. This dual ideal is ${\displaystyle \bigvee _{{\mathcal
{F}}\in \mathbb {F} }{\mathcal {F}}=\pi \left(\cup \mathbb {F} \right)^{\uparrow X},}$ where ${\displaystyle \pi \left(\cup \mathbb {F} \right):=\left\{F_{1}\cap \cdots \cap F_{n}~:~n\in \mathbb
{N} {\text{ and every }}F_{i}{\text{ belongs to some }}{\mathcal {F}}\in \mathbb {F} \right\}}$ is the π–system generated by ${\displaystyle \cup \mathbb {F} .}$ As with any non–empty family of
sets, ${\displaystyle \cup \mathbb {F} }$ is contained in some filter on ${\displaystyle X}$ if and only if it is a filter subbase, or equivalently, if and only if ${\displaystyle \bigvee _{{\
mathcal {F}}\in \mathbb {F} }{\mathcal {F}}=\pi \left(\cup \mathbb {F} \right)^{\uparrow X}}$ is a filter on ${\displaystyle X,}$ in which case this family is the smallest (relative to ${\
displaystyle \subseteq }$) filter on ${\displaystyle X}$ containing every element of ${\displaystyle \mathbb {F} }$ as a subset and necessarily ${\displaystyle \mathbb {F} \subseteq \operatorname
{Filters} (X).}$
• Let ${\displaystyle \varnothing eq \mathbb {F} \subseteq \operatorname {Filters} (X)}$ and let ${\displaystyle \cup \mathbb {F} =\bigcup _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}}.}$ The
supremum or least upper bound of ${\displaystyle \mathbb {F} {\text{ in }}\operatorname {Filters} (X),}$ denoted by ${\displaystyle \bigvee _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}}}$ if it
exists, is by definition the smallest (relative to ${\displaystyle \subseteq }$) filter on ${\displaystyle X}$ containing every element of ${\displaystyle \mathbb {F} }$ as a subset. If it exists
then necessarily ${\displaystyle \bigvee _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}}=\pi \left(\cup \mathbb {F} \right)^{\uparrow X}}$^[10] (as defined above) and ${\displaystyle \bigvee _{{\
mathcal {F}}\in \mathbb {F} }{\mathcal {F}}}$ will also be equal to the intersection of all filters on ${\displaystyle X}$ containing ${\displaystyle \cup \mathbb {F} .}$ This supremum of ${\
displaystyle \mathbb {F} {\text{ in }}\operatorname {Filters} (X)}$ exists if and only if the dual ideal ${\displaystyle \pi \left(\cup \mathbb {F} \right)^{\uparrow X}}$ is a filter on ${\
displaystyle X.}$ The least upper bound of a family of filters ${\displaystyle \mathbb {F} }$ may fail to be a filter.^[10] Indeed, if ${\displaystyle X}$ contains at least 2 distinct elements
then there exist filters ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}{\text{ on }}X}$ for which there does not exist a filter ${\displaystyle {\mathcal {F}}{\text{ on }}X}$ that
contains both ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}.}$ If ${\displaystyle \cup \mathbb {F} }$ is not a filter subbase then the supremum of ${\displaystyle \mathbb {F} {\text{
in }}\operatorname {Filters} (X)}$ does not exist and the same is true of its supremum in ${\displaystyle \operatorname {Prefilters} (X)}$ but their supremum in the set of all dual ideals on ${\
displaystyle X}$ will exist (it being the degenerate filter ${\displaystyle \wp (X)}$).^[9]
□ If ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {F}}}$ are prefilters (resp. filters on ${\displaystyle X}$) then ${\displaystyle {\mathcal {B}}\,(\cap )\,{\mathcal {F}}}$ is a
prefilter (resp. a filter) if and only if it is non–degenerate (or said differently, if and only if ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {F}}}$ mesh), in which case it is one
of the coarsest prefilters (resp. the coarsest filter) on ${\displaystyle X}$ (with respect to ${\displaystyle \,\leq }$) that is finer (with respect to ${\displaystyle \,\leq }$) than both $
{\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {F}};}$ this means that if ${\displaystyle {\mathcal {S}}}$ is any prefilter (resp. any filter) such that ${\displaystyle {\mathcal {B}}\
leq {\mathcal {S}}{\text{ and }}{\mathcal {F}}\leq {\mathcal {S}}}$ then necessarily ${\displaystyle {\mathcal {B}}\,(\cap )\,{\mathcal {F}}\leq {\mathcal {S}},}$^[8] in which case it is
denoted by ${\displaystyle {\mathcal {B}}\vee {\mathcal {F}}.}$^[9]
• Let ${\displaystyle I{\text{ and }}X}$ be non−empty sets and for every ${\displaystyle i\in I}$ let ${\displaystyle {\mathcal {D}}_{i}}$ be a dual ideal on ${\displaystyle X.}$ If ${\displaystyle
{\mathcal {I}}}$ is any dual ideal on ${\displaystyle I}$ then ${\displaystyle \bigcup _{\Xi \in {\mathcal {I}}}\;\;\bigcap _{i\in \Xi }\;{\mathcal {D}}_{i}}$ is a dual ideal on ${\displaystyle
X}$ called Kowalsky's dual ideal or Kowalsky's filter.^[17]
• The clubfilter of a regular uncountable cardinal is the filter of all sets containing a clubsubset of ${\displaystyle \kappa .}$ It is a ${\displaystyle \kappa }$-complete filter closed under
Other examples
• Let ${\displaystyle X=\{p,1,2,3\}}$ and let ${\displaystyle {\mathcal {B}}=\{\{p\},\{p,1,2\},\{p,1,3\}\},}$ which makes ${\displaystyle {\mathcal {B}}}$ a prefilter and a filter subbase that is
not closed under finite intersections. Because ${\displaystyle {\mathcal {B}}}$ is a prefilter, the smallest prefilter containing ${\displaystyle {\mathcal {B}}}$ is ${\displaystyle {\mathcal
{B}}.}$ The π–system generated by ${\displaystyle {\mathcal {B}}}$ is ${\displaystyle \{\{p,1\}\}\cup {\mathcal {B}}.}$ In particular, the smallest prefilter containing the filter subbase ${\
displaystyle {\mathcal {B}}}$ is not equal to the set of all finite intersections of sets in ${\displaystyle {\mathcal {B}}.}$ The filter on ${\displaystyle X}$ generated by ${\displaystyle {\
mathcal {B}}}$ is ${\displaystyle {\mathcal {B}}^{\uparrow X}=\{S\subseteq X:p\in S\}=\{\{p\}\cup T~:~T\subseteq \{1,2,3\}\}.}$ All three of ${\displaystyle {\mathcal {B}},}$ the π–system ${\
displaystyle {\mathcal {B}}}$ generates, and ${\displaystyle {\mathcal {B}}^{\uparrow X}}$ are examples of fixed, principal, ultra prefilters that are principal at the point ${\displaystyle p;{\
mathcal {B}}^{\uparrow X}}$ is also an ultrafilter on ${\displaystyle X.}$
• Let ${\displaystyle (X,\tau )}$ be a topological space, ${\displaystyle {\mathcal {B}}\subseteq \wp (X),}$ and define ${\displaystyle {\overline {\mathcal {B}}}:=\left\{\operatorname {cl} _{X}
B~:~B\in {\mathcal {B}}\right\},}$ where ${\displaystyle {\mathcal {B}}}$ is necessarily finer than ${\displaystyle {\overline {\mathcal {B}}}.}$^[29] If ${\displaystyle {\mathcal {B}}}$ is
non–empty (resp. non–degenerate, a filter subbase, a prefilter, closed under finite unions) then the same is true of ${\displaystyle {\overline {\mathcal {B}}}.}$ If ${\displaystyle {\mathcal
{B}}}$ is a filter on ${\displaystyle X}$ then ${\displaystyle {\overline {\mathcal {B}}}}$ is a prefilter but not necessarily a filter on ${\displaystyle X}$ although ${\displaystyle \left({\
overline {\mathcal {B}}}\right)^{\uparrow X}}$ is a filter on ${\displaystyle X}$ equivalent to ${\displaystyle {\overline {\mathcal {B}}}.}$
• The set ${\displaystyle {\mathcal {B}}}$ of all dense open subsets of a (non–empty) topological space ${\displaystyle X}$ is a proper π–system and so also a prefilter. If the space is a
Bairespace, then the set of all countable intersections of dense open subsets is a π–system and a prefilter that is finer than ${\displaystyle {\mathcal {B}}.}$ If ${\displaystyle X=\mathbb {R}
^{n}}$ (with ${\displaystyle 1\leq n\in \mathbb {N} }$) then the set ${\displaystyle {\mathcal {B}}_{\operatorname {LebFinite} }}$ of all ${\displaystyle B\in {\mathcal {B}}}$ such that ${\
displaystyle B}$ has finite Lebesguemeasure is a proper π–system and free prefilter that is also a propersubset of ${\displaystyle {\mathcal {B}}.}$ The prefilters ${\displaystyle {\mathcal
{B}}_{\operatorname {LebFinite} }}$ and ${\displaystyle {\mathcal {B}}}$ are equivalent and so generate the same filter on ${\displaystyle X.}$ The prefilter ${\displaystyle {\mathcal {B}}_{\
operatorname {LebFinite} }}$ is properly contained in, and not equivalent to, the prefilter consisting of all dense subsets of ${\displaystyle \mathbb {R} .}$ Since ${\displaystyle X}$ is a
Bairespace, every countable intersection of sets in ${\displaystyle {\mathcal {B}}_{\operatorname {LebFinite} }}$ is dense in ${\displaystyle X}$ (and also comeagre and non–meager) so the set of
all countable intersections of elements of ${\displaystyle {\mathcal {B}}_{\operatorname {LebFinite} }}$ is a prefilter and π–system; it is also finer than, and not equivalent to, ${\displaystyle
{\mathcal {B}}_{\operatorname {LebFinite} }.}$
• A filter subbase with no ${\displaystyle \,\subseteq -}$smallest prefilter containing it: In general, if a filter subbase ${\displaystyle {\mathcal {S}}}$ is not a π–system then an intersection $
{\displaystyle S_{1}\cap \cdots \cap S_{n}}$ of ${\displaystyle n}$ sets from ${\displaystyle {\mathcal {S}}}$ will usually require a description involving ${\displaystyle n}$ variables that
cannot be reduced down to only two (consider, for instance ${\displaystyle \pi ({\mathcal {S}})}$ when ${\displaystyle {\mathcal {S}}=\{(-\infty ,r)\cup (r,\infty )~:~r\in \mathbb {R} \}}$). This
example illustrates an atypical class of a filter subbases ${\displaystyle {\mathcal {S}}_{R}}$ where all sets in both ${\displaystyle {\mathcal {S}}_{R}}$ and its generated π–system can be
described as sets of the form ${\displaystyle B_{r,s},}$ so that in particular, no more than two variables (specifically, ${\displaystyle r{\text{ and }}s}$) are needed to describe the generated
π–system. For all ${\displaystyle r,s\in \mathbb {R} ,}$ let ${\displaystyle B_{r,s}=(r,0)\cup (s,\infty ),}$ where ${\displaystyle B_{r,s}=B_{\min(r,s),s}}$ always holds so no generality is lost
by adding the assumption ${\displaystyle r\leq s.}$ For all real ${\displaystyle r\leq s{\text{ and }}u\leq v,}$ if ${\displaystyle s{\text{ or }}v}$ is non-negative then ${\displaystyle B_{-r,s}
\cap B_{-u,v}=B_{-\min(r,u),\max(s,v)}.}$^[note2] For every set ${\displaystyle R}$ of positive reals, let^[note3] ${\displaystyle {\mathcal {S}}_{R}:=\left\{B_{-r,r}:r\in R\right\}=\{(-r,0)\
cup (r,\infty ):r\in R\}\quad {\text{ and }}\quad {\mathcal {B}}_{R}:=\left\{B_{-r,s}:r\leq s{\text{ with }}r,s\in R\right\}=\{(-r,0)\cup (s,\infty ):r\leq s{\text{ in }}R\}.}$ Let ${\
displaystyle X=\mathbb {R} }$ and suppose ${\displaystyle \varnothing eq R\subseteq (0,\infty )}$ is not a singleton set. Then ${\displaystyle {\mathcal {S}}_{R}}$ is a filter subbase but not a
prefilter and ${\displaystyle {\mathcal {B}}_{R}=\pi \left({\mathcal {S}}_{R}\right)}$ is the π–system it generates, so that ${\displaystyle {\mathcal {B}}_{R}^{\uparrow X}}$ is the unique
smallest filter in ${\displaystyle X=\mathbb {R} }$ containing ${\displaystyle {\mathcal {S}}_{R}.}$ However, ${\displaystyle {\mathcal {S}}_{R}^{\uparrow X}}$ is not a filter on ${\displaystyle
X}$ (nor is it a prefilter because it is not directed downward, although it is a filter subbase) and ${\displaystyle {\mathcal {S}}_{R}^{\uparrow X}}$ is a proper subset of the filter ${\
displaystyle {\mathcal {B}}_{R}^{\uparrow X}.}$ If ${\displaystyle R,S\subseteq (0,\infty )}$ are non−empty intervals then the filter subbases ${\displaystyle {\mathcal {S}}_{R}{\text{ and }}{\
mathcal {S}}_{S}}$ generate the same filter on ${\displaystyle X}$ if and only if ${\displaystyle R=S.}$ If ${\displaystyle {\mathcal {C}}}$ is a prefilter satisfying ${\displaystyle {\mathcal
{S}}_{(0,\infty )}\subseteq {\mathcal {C}}\subseteq {\mathcal {B}}_{(0,\infty )}}$^[note4] then for any ${\displaystyle C\in {\mathcal {C}}\setminus {\mathcal {S}}_{(0,\infty )},}$ the family $
{\displaystyle {\mathcal {C}}\setminus \{C\}}$ is also a prefilter satisfying ${\displaystyle {\mathcal {S}}_{(0,\infty )}\subseteq {\mathcal {C}}\setminus \{C\}\subseteq {\mathcal {B}}_{(0,\
infty )}.}$ This shows that there cannot exist a minimal/least (with respect to ${\displaystyle \subseteq }$) prefilter that both contains ${\displaystyle {\mathcal {S}}_{(0,\infty )}}$ and is a
subset of the π–system generated by ${\displaystyle {\mathcal {S}}_{(0,\infty )}.}$ This remains true even if the requirement that the prefilter be a subset of ${\displaystyle {\mathcal {B}}_{(0,
\infty )}=\pi \left({\mathcal {S}}_{(0,\infty )}\right)}$ is removed; that is, (in sharp contrast to filters) there does not exist a minimal/least (with respect to ${\displaystyle \subseteq }$)
prefilter containing the filter subbase ${\displaystyle {\mathcal {S}}_{(0,\infty )}.}$
There are many other characterizations of "ultrafilter" and "ultra prefilter," which are listed in the article on ultrafilters. Important properties of ultrafilters are also described in that
{\displaystyle {\begin{alignedat}{8}{\textrm {Ultrafilters}}(X)\;&=\;{\textrm {Filters}}(X)\,\cap \,{\textrm {UltraPrefilters}}(X)\\&\subseteq \;{\textrm {UltraPrefilters}}(X)={\textrm
{UltraFilterSubbases}}(X)\\&\subseteq \;{\textrm {Prefilters}}(X)\\\end{alignedat}}}
A non–empty family
${\displaystyle {\mathcal {B}}\subseteq \wp (X)}$
of sets is/is an:
8. Ultra^[7]^[30] if ${\displaystyle \varnothing ot \in {\mathcal {B}}}$ and any of the following equivalent conditions are satisfied:
1. For every set ${\displaystyle S\subseteq X}$ there exists some set ${\displaystyle B\in {\mathcal {B}}}$ such that ${\displaystyle B\subseteq S{\text{ or }}B\subseteq X\setminus S}$ (or
equivalently, such that ${\displaystyle B\cap S{\text{ equals }}B{\text{ or }}\varnothing }$).
2. For every set ${\displaystyle S\subseteq \bigcup _{B\in {\mathcal {B}}}B}$ there exists some set ${\displaystyle B\in {\mathcal {B}}}$ such that ${\displaystyle B\cap S{\text{ equals }}B
{\text{ or }}\varnothing .}$
○ This characterization of "${\displaystyle {\mathcal {B}}}$ is ultra" does not depend on the set ${\displaystyle X,}$ so mentioning the set ${\displaystyle X}$ is optional when using
the term "ultra."
3. For every set ${\displaystyle S}$ (not necessarily even a subset of ${\displaystyle X}$) there exists some set ${\displaystyle B\in {\mathcal {B}}}$ such that ${\displaystyle B\cap S{\
text{ equals }}B{\text{ or }}\varnothing .}$
○ If ${\displaystyle {\mathcal {B}}}$ satisfies this condition then so does every superset ${\displaystyle {\mathcal {F}}\supseteq {\mathcal {B}}.}$ For example, if ${\displaystyle T}$
is any singletonset then ${\displaystyle \{T\}}$ is ultra and consequently, any non–degenerate superset of ${\displaystyle \{T\}}$ (such as its upward closure) is also ultra.
9. Ultra prefilter^[7]^[30] if it is a prefilter that is also ultra. Equivalently, it is a filter subbase that is ultra. A prefilter ${\displaystyle {\mathcal {B}}}$ is ultra if and only if it
satisfies any of the following equivalent conditions:
1. ${\displaystyle {\mathcal {B}}}$ is maximal in ${\displaystyle \operatorname {Prefilters} (X)}$ with respect to ${\displaystyle \,\leq ,\,}$ which means that ${\displaystyle {\text{For
all }}{\mathcal {C}}\in \operatorname {Prefilters} (X),\;{\mathcal {B}}\leq {\mathcal {C}}\;{\text{ implies }}\;{\mathcal {C}}\leq {\mathcal {B}}.}$
2. ${\displaystyle {\text{For all }}{\mathcal {C}}\in \operatorname {Filters} (X),\;{\mathcal {B}}\leq {\mathcal {C}}\;{\text{ implies }}\;{\mathcal {C}}\leq {\mathcal {B}}.}$
○ Although this statement is identical to that given below for ultrafilters, here ${\displaystyle {\mathcal {B}}}$ is merely assumed to be a prefilter; it need not be a filter.
3. ${\displaystyle {\mathcal {B}}^{\uparrow X}}$ is ultra (and thus an ultrafilter).
4. ${\displaystyle {\mathcal {B}}}$ is equivalent (with respect to ${\displaystyle \leq }$) to some ultrafilter.
☆ A filter subbase that is ultra is necessarily a prefilter. A filter subbase is ultra if and only if it is a maximal filter subbase with respect to ${\displaystyle \,\leq \,}$ (as above).^
10. Ultrafilter on ${\displaystyle X}$^[7]^[30] if it is a filter on ${\displaystyle X}$ that is ultra. Equivalently, an ultrafilter on ${\displaystyle X}$ is a filter ${\displaystyle {\mathcal
{B}}{\text{ on }}X}$ that satisfies any of the following equivalent conditions:
1. ${\displaystyle {\mathcal {B}}}$ is generated by an ultra prefilter.
2. For any ${\displaystyle S\subseteq X,S\in {\mathcal {B}}{\text{ or }}X\setminus S\in {\mathcal {B}}.}$^[17]
3. ${\displaystyle {\mathcal {B}}\cup (X\setminus {\mathcal {B}})=\wp (X).}$ This condition can be restated as: ${\displaystyle \wp (X)}$ is partitioned by ${\displaystyle {\mathcal {B}}}$
and its dual ${\displaystyle X\setminus {\mathcal {B}}.}$
○ The sets ${\displaystyle {\mathcal {B}}{\text{ and }}X\setminus {\mathcal {B}}}$ are disjoint whenever ${\displaystyle {\mathcal {B}}}$ is a prefilter.
4. ${\displaystyle \wp (X)\setminus {\mathcal {B}}=\{S\in \wp (X):Sot \in {\mathcal {B}}\}}$ is an ideal.^[17]
5. For any ${\displaystyle R,S\subseteq X,}$ if ${\displaystyle R\cup S=X}$ then ${\displaystyle R\in {\mathcal {B}}{\text{ or }}S\in {\mathcal {B}}.}$
6. For any ${\displaystyle R,S\subseteq X,}$ if ${\displaystyle R\cup S\in {\mathcal {B}}}$ then ${\displaystyle R\in {\mathcal {B}}{\text{ or }}S\in {\mathcal {B}}}$ (a filter with this
property is called a prime filter).
○ This property extends to any finite union of two or more sets.
7. For any ${\displaystyle R,S\subseteq X,}$ if ${\displaystyle R\cup S\in {\mathcal {B}}{\text{ and }}R\cap S=\varnothing }$ then either ${\displaystyle R\in {\mathcal {B}}{\text{ or }}S\in
{\mathcal {B}}.}$
8. ${\displaystyle {\mathcal {B}}}$ is a maximal filter on ${\displaystyle X}$; meaning that if ${\displaystyle {\mathcal {C}}}$ is a filter on ${\displaystyle X}$ such that ${\displaystyle
{\mathcal {B}}\subseteq {\mathcal {C}}}$ then necessarily ${\displaystyle {\mathcal {C}}={\mathcal {B}}}$ (this equality may be replaced by ${\displaystyle {\mathcal {C}}\subseteq {\
mathcal {B}}{\text{ or by }}{\mathcal {C}}\leq {\mathcal {B}}}$).
○ If ${\displaystyle {\mathcal {C}}}$ is upward closed then ${\displaystyle {\mathcal {B}}\leq {\mathcal {C}}{\text{ if and only if }}{\mathcal {B}}\subseteq {\mathcal {C}}.}$ So this
characterization of ultrafilters as maximal filters can be restated as: ${\displaystyle {\text{For all }}{\mathcal {C}}\in \operatorname {Filters} (X),\;{\mathcal {B}}\leq {\mathcal
{C}}\;{\text{ implies }}\;{\mathcal {C}}\leq {\mathcal {B}}.}$
○ Because subordination ${\displaystyle \,\geq \,}$ is for filters the analog of "is a subnet/subsequence of" (specifically, "subnet" should mean "AA–subnet," which is defined below),
this characterization of an ultrafilter as being a "maximally subordinate filter" suggests that an ultrafilter can be interpreted as being analogous to some sort of "maximally deep
net" (which could, for instance, mean that "when viewed only from ${\displaystyle X}$" in some sense, it is indistinguishable from its subnets, as is the case with any net valued in a
singleton set for example),^[note5] which is an idea that is actually made rigorous by ultranets. The ultrafilterlemma is then the statement that every filter ("net") has some
subordinate filter ("subnet") that is "maximally subordinate" ("maximally deep").
Any non–degenerate family that has a singleton set as an element is ultra, in which case it will then be an ultra prefilter if and only if it also has the finite intersection property. The trivial
filter ${\displaystyle \{X\}{\text{ on }}X}$ is ultra if and only if ${\displaystyle X}$ is a singleton set.
The ultrafilter lemma
The following important theorem is due to AlfredTarski (1930).^[31]
A consequence of the ultrafilter lemma is that every filter is equal to the intersection of all ultrafilters containing it.^[10]^[proof1] Assuming the axioms of Zermelo–Fraenkel(ZF), the
ultrafilter lemma follows from the Axiomofchoice (in particular from Zorn'slemma) but is strictly weaker than it. The ultrafilter lemma implies the Axiom of choice for finite sets. If only dealing
with Hausdorff spaces, then most basic results (as encountered in introductory courses) in Topology (such as Tychonoff'stheorem for compact Hausdorff spaces and the Alexandersubbasetheorem) and in
functionalanalysis (such as the Hahn–Banachtheorem) can be proven using only the ultrafilter lemma; the full strength of the axiom of choice might not be needed.
The kernel is useful in classifying properties of prefilters and other families of sets.
of a family of sets
${\displaystyle {\mathcal {B}}}$
is the intersection of all sets that are elements of
${\displaystyle {\mathcal {B}}:}$ ${\displaystyle \ker {\mathcal {B}}=\bigcap _{B\in {\mathcal {B}}}B}$
If ${\displaystyle {\mathcal {B}}\subseteq \wp (X)}$ then for any point ${\displaystyle x,xot \in \ker {\mathcal {B}}{\text{ if and only if }}X\setminus \{x\}\in {\mathcal {B}}^{\uparrow X}.}$
Properties of kernels
If ${\displaystyle {\mathcal {B}}\subseteq \wp (X)}$ then ${\displaystyle \ker \left({\mathcal {B}}^{\uparrow X}\right)=\ker {\mathcal {B}}}$ and this set is also equal to the kernel of the π–system
that is generated by ${\displaystyle {\mathcal {B}}.}$ In particular, if ${\displaystyle {\mathcal {B}}}$ is a filter subbase then the kernels of all of the following sets are equal:
(1) ${\displaystyle {\mathcal {B}},}$ (2) the π–system generated by ${\displaystyle {\mathcal {B}},}$ and (3) the filter generated by ${\displaystyle {\mathcal {B}}.}$
If ${\displaystyle f}$ is a map then ${\displaystyle f(\ker {\mathcal {B}})\subseteq \ker f({\mathcal {B}})}$ and ${\displaystyle f^{-1}(\ker {\mathcal {B}})=\ker f^{-1}({\mathcal {B}}).}$ If ${\
displaystyle {\mathcal {B}}\leq {\mathcal {C}}}$ then ${\displaystyle \ker {\mathcal {C}}\subseteq \ker {\mathcal {B}}}$ while if ${\displaystyle {\mathcal {B}}}$ and ${\displaystyle {\mathcal {C}}}$
are equivalent then ${\displaystyle \ker {\mathcal {B}}=\ker {\mathcal {C}}.}$ Equivalent families have equal kernels. Two principal families are equivalent if and only if their kernels are equal;
that is, if ${\displaystyle {\mathcal {B}}}$ and ${\displaystyle {\mathcal {C}}}$ are principal then they are equivalent if and only if ${\displaystyle \ker {\mathcal {B}}=\ker {\mathcal {C}}.}$
Classifying families by their kernels
A family
${\displaystyle {\mathcal {B}}}$
of sets is:
1. Free^[6] if ${\displaystyle \ker {\mathcal {B}}=\varnothing ,}$ or equivalently, if ${\displaystyle \{X\setminus \{x\}~:~x\in X\}\subseteq {\mathcal {B}}^{\uparrow X};}$ this can be restated
as ${\displaystyle \{X\setminus \{x\}~:~x\in X\}\leq {\mathcal {B}}.}$
☆ A filter ${\displaystyle {\mathcal {F}}}$ on ${\displaystyle X}$ is free if and only if ${\displaystyle X}$ is infinite and ${\displaystyle {\mathcal {F}}}$ contains the Fréchetfilter on
${\displaystyle X}$ as a subset.
2. Fixed if ${\displaystyle \ker {\mathcal {B}}eq \varnothing }$ in which case, ${\displaystyle {\mathcal {B}}}$ is said to be fixed by any point ${\displaystyle x\in \ker {\mathcal {B}}.}$
☆ Any fixed family is necessarily a filter subbase.
3. Principal^[6] if ${\displaystyle \ker {\mathcal {B}}\in {\mathcal {B}}.}$
☆ A proper principal family of sets is necessarily a prefilter.
4. Discrete or Principal at ${\displaystyle x\in X}$^[25] if ${\displaystyle \{x\}=\ker {\mathcal {B}}\in {\mathcal {B}},}$ in which case ${\displaystyle x}$ is called its principal element.
☆ The principal filter at ${\displaystyle x}$ on ${\displaystyle X}$ is the filter ${\displaystyle \{x\}^{\uparrow X}.}$ A filter ${\displaystyle {\mathcal {F}}}$ is principal at ${\
displaystyle x}$ if and only if ${\displaystyle {\mathcal {F}}=\{x\}^{\uparrow X}.}$
5. Countably deep if whenever ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {F}}}$ is a countable subset then ${\displaystyle \ker {\mathcal {C}}\in {\mathcal {B}}.}$^[9]
If ${\displaystyle {\mathcal {B}}}$ is a principal filter on ${\displaystyle X}$ then ${\displaystyle \varnothing eq \ker {\mathcal {B}}\in {\mathcal {B}}}$ and ${\displaystyle {\mathcal {B}}=\{\ker
{\mathcal {B}}\}^{\uparrow X}=\{S\cup \ker {\mathcal {B}}:S\subseteq X\setminus \ker {\mathcal {B}}\}=\wp (X\setminus \ker {\mathcal {B}})\,(\cup )\,\{\ker {\mathcal {B}}\}}$ where ${\displaystyle \
{\ker {\mathcal {B}}\}}$ is also the smallest prefilter that generates ${\displaystyle {\mathcal {B}}.}$
Family of examples: For any non–empty ${\displaystyle C\subseteq \mathbb {R} ,}$ the family ${\displaystyle {\mathcal {B}}_{C}=\{\mathbb {R} \setminus (r+C)~:~r\in \mathbb {R} \}}$ is free but it is
a filter subbase if and only if no finite union of the form ${\displaystyle \left(r_{1}+C\right)\cup \cdots \cup \left(r_{n}+C\right)}$ covers ${\displaystyle \mathbb {R} ,}$ in which case the filter
that it generates will also be free. In particular, ${\displaystyle {\mathcal {B}}_{C}}$ is a filter subbase if ${\displaystyle C}$ is countable (for example, ${\displaystyle C=\mathbb {Q} ,\mathbb
{Z} ,}$ the primes), a meagerset in ${\displaystyle \mathbb {R} ,}$ a set of finite measure, or a bounded subset of ${\displaystyle \mathbb {R} .}$ If ${\displaystyle C}$ is a singleton set then ${\
displaystyle {\mathcal {B}}_{C}}$ is a subbase for the Fréchet filter on ${\displaystyle \mathbb {R} .}$
For every filter ${\displaystyle {\mathcal {F}}{\text{ on }}X}$ there exists a unique pair of dual ideals ${\displaystyle {\mathcal {F}}^{*}{\text{ and }}{\mathcal {F}}^{\bullet }{\text{ on }}X}$
such that ${\displaystyle {\mathcal {F}}^{*}}$ is free, ${\displaystyle {\mathcal {F}}^{\bullet }}$ is principal, and ${\displaystyle {\mathcal {F}}^{*}\wedge {\mathcal {F}}^{\bullet }={\mathcal
{F}},}$ and ${\displaystyle {\mathcal {F}}^{*}{\text{ and }}{\mathcal {F}}^{\bullet }}$ do not mesh (that is, ${\displaystyle {\mathcal {F}}^{*}\vee {\mathcal {F}}^{\bullet }=\wp (X)}$). The dual
ideal ${\displaystyle {\mathcal {F}}^{*}}$ is called the free part of ${\displaystyle {\mathcal {F}}}$ while ${\displaystyle {\mathcal {F}}^{\bullet }}$ is called the principal part^[9] where at
least one of these dual ideals is filter. If ${\displaystyle {\mathcal {F}}}$ is principal then ${\displaystyle {\mathcal {F}}^{\bullet }:={\mathcal {F}}{\text{ and }}{\mathcal {F}}^{*}:=\wp (X);}$
otherwise, ${\displaystyle {\mathcal {F}}^{\bullet }:=\{\ker {\mathcal {F}}\}^{\uparrow X}}$ and ${\displaystyle {\mathcal {F}}^{*}:={\mathcal {F}}\vee \{X\setminus \left(\ker {\mathcal {F}}\right)\}
^{\uparrow X}}$ is a free (non–degenerate) filter.^[9]
Finite prefilters and finite sets
If a filter subbase ${\displaystyle {\mathcal {B}}}$ is finite then it is fixed (that is, not free); this is because ${\displaystyle \ker {\mathcal {B}}=\bigcap _{B\in {\mathcal {B}}}B}$ is a finite
intersection and the filter subbase ${\displaystyle {\mathcal {B}}}$ has the finite intersection property. A finite prefilter is necessarily principal, although it does not have to be closed under
finite intersections.
If ${\displaystyle X}$ is finite then all of the conclusions above hold for any ${\displaystyle {\mathcal {B}}\subseteq \wp (X).}$ In particular, on a finite set ${\displaystyle X,}$ there are no
free filter subbases (and so no free prefilters), all prefilters are principal, and all filters on ${\displaystyle X}$ are principal filters generated by their (non–empty) kernels.
The trivial filter ${\displaystyle \{X\}}$ is always a finite filter on ${\displaystyle X}$ and if ${\displaystyle X}$ is infinite then it is the only finite filter because a non–trivial finite
filter on a set ${\displaystyle X}$ is possible if and only if ${\displaystyle X}$ is finite. However, on any infinite set there are non–trivial filter subbases and prefilters that are finite
(although they cannot be filters). If ${\displaystyle X}$ is a singleton set then the trivial filter ${\displaystyle \{X\}}$ is the only proper subset of ${\displaystyle \wp (X)}$ and moreover, this
set ${\displaystyle \{X\}}$ is a principal ultra prefilter and any superset ${\displaystyle {\mathcal {F}}\supseteq {\mathcal {B}}}$ (where ${\displaystyle {\mathcal {F}}\subseteq \wp (Y){\text{ and
}}X\subseteq Y}$) with the finite intersection property will also be a principal ultra prefilter (even if ${\displaystyle Y}$ is infinite).
Characterizing fixed ultra prefilters
If a family of sets ${\displaystyle {\mathcal {B}}}$ is fixed (that is, ${\displaystyle \ker {\mathcal {B}}eq \varnothing }$) then ${\displaystyle {\mathcal {B}}}$ is ultra if and only if some
element of ${\displaystyle {\mathcal {B}}}$ is a singleton set, in which case ${\displaystyle {\mathcal {B}}}$ will necessarily be a prefilter. Every principal prefilter is fixed, so a principal
prefilter ${\displaystyle {\mathcal {B}}}$ is ultra if and only if ${\displaystyle \ker {\mathcal {B}}}$ is a singleton set.
Every filter on ${\displaystyle X}$ that is principal at a single point is an ultrafilter, and if in addition ${\displaystyle X}$ is finite, then there are no ultrafilters on ${\displaystyle X}$
other than these.^[6]
The next theorem shows that every ultrafilter falls into one of two categories: either it is free or else it is a principal filter generated by a single point.
Proposition — If ${\displaystyle {\mathcal {F}}}$ is an ultrafilter on ${\displaystyle X}$ then the following are equivalent:
1. ${\displaystyle {\mathcal {F}}}$ is fixed, or equivalently, not free, meaning ${\displaystyle \ker {\mathcal {F}}eq \varnothing .}$
2. ${\displaystyle {\mathcal {F}}}$ is principal, meaning ${\displaystyle \ker {\mathcal {F}}\in {\mathcal {F}}.}$
3. Some element of ${\displaystyle {\mathcal {F}}}$ is a finite set.
4. Some element of ${\displaystyle {\mathcal {F}}}$ is a singleton set.
5. ${\displaystyle {\mathcal {F}}}$ is principal at some point of ${\displaystyle X,}$ which means ${\displaystyle \ker {\mathcal {F}}=\{x\}\in {\mathcal {F}}}$ for some ${\displaystyle x\in X.}$
6. ${\displaystyle {\mathcal {F}}}$ does not contain the Fréchet filter on ${\displaystyle X.}$
7. ${\displaystyle {\mathcal {F}}}$ is sequential.^[9]
Finer/coarser, subordination, and meshing
The preorder ${\displaystyle \,\leq \,}$ that is defined below is of fundamental importance for the use of prefilters (and filters) in topology. For instance, this preorder is used to define the
prefilter equivalent of "subsequence",^[24] where "${\displaystyle {\mathcal {F}}\geq {\mathcal {C}}}$" can be interpreted as "${\displaystyle {\mathcal {F}}}$ is a subsequence of ${\displaystyle {\
mathcal {C}}}$" (so "subordinate to" is the prefilter equivalent of "subsequence of"). It is also used to define prefilter convergence in a topological space. The definition of ${\displaystyle {\
mathcal {B}}}$ meshes with ${\displaystyle {\mathcal {C}},}$ which is closely related to the preorder ${\displaystyle \,\leq ,}$ is used in Topology to define clusterpoints.
Two families of sets ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}}$ mesh^[7] and are compatible, indicated by writing ${\displaystyle {\mathcal {B}}\#{\mathcal {C}},}$ if ${\
displaystyle B\cap Ceq \varnothing {\text{ for all }}B\in {\mathcal {B}}{\text{ and }}C\in {\mathcal {C}}.}$ If ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}}$ do not mesh then they are
dissociated. If ${\displaystyle S\subseteq X{\text{ and }}{\mathcal {B}}\subseteq \wp (X)}$ then ${\displaystyle {\mathcal {B}}{\text{ and }}S}$ are said to mesh if ${\displaystyle {\mathcal {B}}{\
text{ and }}\{S\}}$ mesh, or equivalently, if the trace of ${\displaystyle {\mathcal {B}}{\text{ on }}S,}$ which is the family ${\displaystyle {\mathcal {B}}{\big \vert }_{S}=\{B\cap S~:~B\in {\
mathcal {B}}\},}$ does not contain the empty set, where the trace is also called the restriction of ${\displaystyle {\mathcal {B}}{\text{ to }}S.}$
Declare that
${\displaystyle {\mathcal {C}}\leq {\mathcal {F}},{\mathcal {F}}\geq {\mathcal {C}},{\text{ and }}{\mathcal {F}}\vdash {\mathcal {C}},}$
stated as
${\displaystyle {\mathcal {C}}}$
coarser than ${\displaystyle {\mathcal {F}}}$
${\displaystyle {\mathcal {F}}}$
finer than
subordinate to
${\displaystyle {\mathcal {C}},}$^[10]^[11]^[12]^[8]^[9]
if any of the following equivalent conditions hold:
1. Definition: Every ${\displaystyle C\in {\mathcal {C}}}$ contains some ${\displaystyle F\in {\mathcal {F}}.}$ Explicitly, this means that for every ${\displaystyle C\in {\mathcal {C}},}$ there
is some ${\displaystyle F\in {\mathcal {F}}}$ such that ${\displaystyle F\subseteq C.}$
☆ Said more briefly in plain English, ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}}}$ if every set in ${\displaystyle {\mathcal {C}}}$ is larger than some set in ${\displaystyle {\
mathcal {F}}.}$ Here, a "larger set" means a superset.
2. ${\displaystyle \{C\}\leq {\mathcal {F}}{\text{ for every }}C\in {\mathcal {C}}.}$
☆ In words, ${\displaystyle \{C\}\leq {\mathcal {F}}}$ states exactly that ${\displaystyle C}$ is larger than some set in ${\displaystyle {\mathcal {F}}.}$ The equivalence of (a) and (b)
follows immediately.
☆ From this characterization, it follows that if ${\displaystyle \left({\mathcal {C}}_{i}\right)_{i\in I}}$ are families of sets, then ${\displaystyle \bigcup _{i\in I}{\mathcal {C}}_{i}\
leq {\mathcal {F}}{\text{ if and only if }}{\mathcal {C}}_{i}\leq {\mathcal {F}}{\text{ for all }}i\in I.}$
3. ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}}^{\uparrow X},}$ which is equivalent to ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {F}}^{\uparrow X}}$;
4. ${\displaystyle {\mathcal {C}}^{\uparrow X}\leq {\mathcal {F}}}$;
5. ${\displaystyle {\mathcal {C}}^{\uparrow X}\leq {\mathcal {F}}^{\uparrow X},}$ which is equivalent to ${\displaystyle {\mathcal {C}}^{\uparrow X}\subseteq {\mathcal {F}}^{\uparrow X}}$;
and if in addition ${\displaystyle {\mathcal {F}}}$ is upward closed, which means that ${\displaystyle {\mathcal {F}}={\mathcal {F}}^{\uparrow X},}$ then this list can be extended to include:
6. ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {F}}.}$^[5]
☆ So in this case, this definition of "${\displaystyle {\mathcal {F}}}$ is finer than ${\displaystyle {\mathcal {C}}}$" would be identical to the topologicaldefinitionof"finer" had ${\
displaystyle {\mathcal {C}}{\text{ and }}{\mathcal {F}}}$ been topologies on ${\displaystyle X.}$
If an upward closed family ${\displaystyle {\mathcal {F}}}$ is finer than ${\displaystyle {\mathcal {C}}}$ (that is, ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}}}$) but ${\displaystyle {\
mathcal {C}}eq {\mathcal {F}}}$ then ${\displaystyle {\mathcal {F}}}$ is said to be strictly finer than ${\displaystyle {\mathcal {C}}}$ and ${\displaystyle {\mathcal {C}}}$ is strictly coarser
than ${\displaystyle {\mathcal {F}}.}$
Two families are
if one of these sets is finer than the other.
Example: If ${\displaystyle x_{i_{\bullet }}=\left(x_{i_{n}}\right)_{n=1}^{\infty }}$ is a subsequence of ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty }}$ then ${\displaystyle \
operatorname {Tails} \left(x_{i_{\bullet }}\right)}$ is subordinate to ${\displaystyle \operatorname {Tails} \left(x_{\bullet }\right);}$ in symbols: ${\displaystyle \operatorname {Tails} \left(x_{i_
{\bullet }}\right)\vdash \operatorname {Tails} \left(x_{\bullet }\right)}$ and also ${\displaystyle \operatorname {Tails} \left(x_{\bullet }\right)\leq \operatorname {Tails} \left(x_{i_{\bullet }}\
right).}$ Stated in plain English, the prefilter of tails of a subsequence is always subordinate to that of the original sequence. To see this, let ${\displaystyle C:=x_{\geq i}\in \operatorname
{Tails} \left(x_{\bullet }\right)}$ be arbitrary (or equivalently, let ${\displaystyle i\in \mathbb {N} }$ be arbitrary) and it remains to show that this set contains some ${\displaystyle F:=x_{i_{\
geq n}}\in \operatorname {Tails} \left(x_{i_{\bullet }}\right).}$ For the set ${\displaystyle x_{\geq i}=\left\{x_{i},x_{i+1},\ldots \right\}}$ to contain ${\displaystyle x_{i_{\geq n}}=\left\{x_{i_
{n}},x_{i_{n+1}},\ldots \right\},}$ it is sufficient to have ${\displaystyle i\leq i_{n}.}$ Since ${\displaystyle i_{1}<i_{2}<\cdots }$ are strictly increasing integers, there exists ${\displaystyle
n\in \mathbb {N} }$ such that ${\displaystyle i_{n}\geq i,}$ and so ${\displaystyle x_{\geq i}\supseteq x_{i_{\geq n}}}$ holds, as desired. Consequently, ${\displaystyle \operatorname {TailsFilter} \
left(x_{\bullet }\right)\subseteq \operatorname {TailsFilter} \left(x_{i_{\bullet }}\right).}$ The left hand side will be a strict/proper subset of the right hand side if (for instance) every point
of ${\displaystyle x_{\bullet }}$ is unique (that is, when ${\displaystyle x_{\bullet }:\mathbb {N} \to X}$ is injective) and ${\displaystyle x_{i_{\bullet }}}$ is the even-indexed subsequence ${\
displaystyle \left(x_{2},x_{4},x_{6},\ldots \right)}$ because under these conditions, every tail ${\displaystyle x_{i_{\geq n}}=\left\{x_{2n},x_{2n+2},x_{2n+4},\ldots \right\}}$ (for every ${\
displaystyle n\in \mathbb {N} }$) of the subsequence will belong to the right hand side filter but not to the left hand side filter.
For another example, if ${\displaystyle {\mathcal {B}}}$ is any family then ${\displaystyle \varnothing \leq {\mathcal {B}}\leq {\mathcal {B}}\leq \{\varnothing \}}$ always holds and furthermore, ${\
displaystyle \{\varnothing \}\leq {\mathcal {B}}{\text{ if and only if }}\varnothing \in {\mathcal {B}}.}$
Assume that ${\displaystyle {\mathcal {C}}{\text{ and }}{\mathcal {F}}}$ are families of | {"url":"https://wiki2.org/en/Filter_(set_theory)","timestamp":"2024-11-07T13:08:36Z","content_type":"application/xhtml+xml","content_length":"1049414","record_id":"<urn:uuid:0d263a33-62a6-4762-8c6f-afcb118bd78a>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00687.warc.gz"} |
New book on demographics seeks to explain why population growth in the industrial age always leads to a stagnant or falling population
A human tide hit Earth’s beaches, prairies, desserts and mountains like a tsunami at about the turn of the 19th century and will subside only at the turn of the 22nd. That human wave is the
population explosion that started in English-speaking countries at the beginning of the industrial revolution, but quickly spread to Europe, Asia, Latin America, and now finally to Africa.
But as British demographer Paul Morland details in The Human Tide, the expression “human tide” not only describes 300 years of unprecedented growth in the population of humans, but also the mechanism
by which that growth was achieved.
Morland begins by listing the limited number of variables that determine if a country’s population will rise or fall:
• Average number of children born to each woman
• Mortality rate of infants
• Average life span of individuals
• Immigration and emigration.
For centuries before the industrial revolution, human populations tended to grow extremely slowly, sometimes shrinking or stagnating. The population had hit its Malthusian limits, named after Thomas
Malthus, an English theologian who postulated that population growth would always run into the limits imposed by Nature. Scarcity of resources would always lead to the misery of famine and poverty
and thus place a natural limit on human population.
Of course Nature’s limits expanded tremendously when humans started to transition to the use of carbon power (coal, oil, natural gas and the electricity created burning these hydrocarbons) instead of
human, animal or rudimentary forms of wind and water power. At about the same time, the increase and spread of scientific knowledge reached a critical mass leading to improvements in sanitation,
medical care, transportation, tools, agriculture, engineering, safety standards and dozens of other aspects of human existence that gave people more material possessions while increasing their
lifespans and decreasing the number of babies dying before one and five years of age.
Greater abundance leads to the human tide, first in Great Britain and the United States: the average life span increases and infant mortality declines while women begin having more children—in some
countries, many more children, spurred on by society’s greater wealth. This rising tidal wave causes both the population and its rate of growth to soar, sometimes aided as in the case of the United
States and Canada by large numbers of new arrivals from countries experiencing rampant population growth. The average age at death increases, usually by decades, but the average age of individuals
declines. The population becomes better educated and the standard of living rises, sometimes marginally and sometimes in spectacular fashion. The country is more able to find soldiers for war and
industrial workers for factories, and thus often sees its ability to project power regionally or globally expand. People begin to depopulate rural areas in favor of cities.
But then something funny happens. Educated women tend to have fewer babies, so the average number of births per woman falls, often under the level at which the population starts to shrink. Infant
mortality and life expectancy rates stabilize. Population growth stops and even turns negative. Meanwhile, because generations of an expanding population are followed by generations of a declining
population, the overall population ages. The result: the population no longer expands and in many cases starts to contract. Only nations that continue to have large numbers of immigrants continue to
grow after native-born women start having fewer than the replacement number of children, e.g., the United States from the 1970’s until the installation of the Trump anti-immigration project.
The human tide thus consists of precipitous population growth which creates a much younger nation followed by stabilization and decline of the population, now much older. The later in history a
population experiences the tide, the faster it plays out: it took much longer in the United States and England than it did in Russia and Germany, which likewise underwent a chronologically longer
wave than China and Latin America have.
BTW, Morland reports good and bad news about an aging population. The good news is that an aging population is less likely to go to war and will usually experience lower rates of crime. The bad news
is that older populations tend to produce fewer innovations. Morland, among others, also worries needlessly that taking care of a very old population is a major challenge to society; these so-called
experts don’t seem to realize how easy it is to reroute working adults from taking care of children to taking care of seniors. Almost as easy as rerouting people from oil fields and coal minds to
solar panel and wind turbine manufacture, installation and maintenance. All it takes are the funds and the collective will to educate and reeducate—something the United States had after World War II
and China seems to have now.
According to Morland, the human wave—a large increase in population followed by stabilization and some decline—explains much of the history of the past 200 years, for example, the global rise and
fall of Germany, the Soviet Union and Japan, the current tensions in the middle East and the looming rise of China, Brazil and Africa, the last continent to experience the wave.
In The Human Tide, Morland labors to make sure his history doesn’t come across as supporting the view that Europeans and Americans are superior to other people because of their technologies and
values. Anyone who takes the long view of human history knows that Europeans have dominated politically and economically only over the past 200 or so years and that the rest of the world has almost
caught up, and done it faster than it took ancient Rome to catch up with Greece, or Europe to catch up with the Arab world and China in medieval and early modern times. It’s a bit of a challenge,
however, to argue against European superiority if you limit your history to 1800-2016. Morland succeeds, and that’s to his credit.
Unfortunately, Morland falls victim to that other great irrationality proffered by right-wing pretending to present well-researched truth: he believes in the invisible hand of the marketplace, which
he extends to population growth. Morland reveals his bias inadvertently when discussing China’s decades’ long efforts, now apparently ending, to limit its population by mandating a one-child policy.
Morland berates China both for the one-child policy and it harsh implementation, which evidently included jail time, taking children from parents and forced abortions. His argument is that the
invisible hand of the human tide would have lowered the population without China’s draconian policy.
Two enormous logical errors. The first is easy to explain—if China had not enforced a one-child policy, its human tide would have lasted longer and crested higher. The policy did work, although it
has resulted in the same problems faced by all rapidly aging nations.
The second error has to do with the very idea of the “invisible hand,” whether in economics or in the natural growth of human populations. Let’s first remember that if we postulate, as right-wingers
always have, that the invisible hand emanates from the natural order of things, then we have to conclude, based on the evidence of paleontology and the laws of physics, that the invisible hand’s goal
is the extinction of humanity. After all, upwards of 95% of all species ever to exist are now extinct, thanks to the invisible hand of evolution. Moreover, the laws of thermodynamics predict a state
of complete entropy in which it would be impossible for life to exist. So instead of accepting any invisible hand, humans should intervene to protect and extend our species, for example through
population control or laws that offset the unequal distribution of wealth that all unimpeded markets quickly produce.
The other thing to keep in mind is that the human tide has washed across the shores of different nations in different ways precisely because of dozens of interventions made by societies and their
leaders: Build up an army or not? Support rising fertility or support population control? Outlaw or encourage abortion and birth control? Educate women or not? Welcome immigrants or shut the borders?
Negotiate trade agreements or invade other countries? Make masses of people move or engage in ethnic cleansing? The invisible hand consists of many conscious efforts, which is why the human tide has
not played out the same way everywhere, the way in which an experiment involving the release of a heavy and a light object from a tower would always yield the same results.
China had the right idea. We should promote one-child policies everywhere, although I am opposed to any kind of physical coercion like jailing or forced abortions. Rather, societies can encourage
lower birth rates as follows:
• An active campaign using all media and public education advocating a one-child policy
• Continued education of women and their integration into all levels of the economy and government.
• Free birth control and abortion and the removal of most restrictions on abortions.
• Financial penalties for ignoring the one-child policy. I would propose that when a woman gives birth to more than one child, both the woman and the father of the baby should be assessed an
additional 5% on their gross income and an additional 5% on their net assets from the birth of each additional child until it turns 30.
If every woman had one child only, the population would be cut in half in one generation, which would go a long way towards solving many of the world’s problems, including the global environmental
disaster we face. I know I’m an extremist, but we are seriously taxing the carrying capacity of the Earth and if we fail to reduce the human footprint, the four horses of the Apocalypse—natural
disasters, famine, epidemics and war—will surely do it for us.
The problem with any kind of population control strategy, be it extreme or mild, is that most economists have refused to consider how to structure a growing or stable economy delivering a high
quality of a life to all when the population is shrinking. Economists have also refused to consider how to make sure that the hidden costs of economic actions are assumed by the producer, the seller
or the buyer; think of the medical cost to treat people suffering from diseases caused by air pollution as an example of a hidden cost unpaid by manufacturers or car owners.
Morland fails to take a stand on whether the enormous growth in the population of humans over the past 200 years represents a threat to the continued existence of the human species. Maybe he hopes
that by the time the world stabilizes its population at nine or ten billion people we will have developed the technologies needed to sustain such a heavy load of wide-screen TVs, private motorized
vehicles, plastic straws and air conditioning. Of course to think otherwise would require him to admit that the invisible hand of the human tidal wave has to be controlled and directed, as does the
invisible hand of the marketplace. | {"url":"https://www.jampole.com/blog/new-book-on-demographics-seeks-to-explain-why-population-growth-in-the-industrial-age-always-leads-to-a-stagnant-or-falling-population/","timestamp":"2024-11-06T02:14:39Z","content_type":"text/html","content_length":"59224","record_id":"<urn:uuid:339f2ca4-2e52-41ef-9072-9e19b2038c22>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00522.warc.gz"} |