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How many steps is 1 flights of stairs?
Popular lifehacks
How many steps is 1 flights of stairs?
How many steps is 1 flights of stairs?
Most flights of stairs average out at 12 or 13 steps but it depends on the height of the staircase, the location of the stairs (as stair height regulations differ between public and private buildings
and between countries), and the purpose of a staircase (as fire escapes have more specific rules than other sorts of …
How do you calculate stairs for stairs?
Divide the overall change in level (overall rise) by 150mm.
1. 450mm / 150mm = 3. This tells us that with a riser of 150mm we will need 3 risers/steps.
2. 450mm / 2 = 225mm.
3. Twice the Rise plus the Going (2R + G) should be between 550mm and 700mm.
4. 2 x 150mm + 275mm = 575mm.
What is the ideal size of stairs?
The stair’s width usually varies depending on the type of building the staircase is in, but for a normal residence, the standard tends to be 3 feet, 6 inches (106.7 cm). The minimum, in most places,
is 2 feet 8 inches (81.3 cm). If a staircase exceeds 44 inches (111.8 cm), handrails are required for both sides.
How long are stairs that go up 10 feet?
The total length of a flight of stairs for a 10-foot ceiling is approximately 12-feet long.
How many steps is 10 flights of stairs?
Number of Steps for 10′ Ceilings Divide it by 7 ¾” and you get 16.8 – which means you’ll need 17 steps. A flight of stairs in a home with 10′ ceilings will require 17 steps, minimum. To find the rise
of each step, divide 130” by 17.
What is considered 4 flights of stairs?
After a rest period, the study group climbed four flights of stairs (60 steps) at a fast but non-running pace, then had their METs measured again.
How many square feet is 13 stairs?
You’ll need between 80 and 110 square feet for 13 stairs.
How many steps is 8 foot rise?
Stair Rise for 8-Foot Ceilings An 8-foot ceiling will need 14 treads. Divide 96-inches (8-feet) by 7-inches to get 13.71 treads, and round up to 14. In this case, you would not round down as you
would have one step that is too tall.
What is the 18 rule for a staircase?
Rule one says that rise plus run (r+R) should equal 18 inches. Why? That’s what most people find to be a comfortable stride on most stairs. You can cheat a bit up or down, but below 17” and more than
19” will result in steps that require strides either too big or too small for most people.
How many stairs do you need to go up 12 feet?
As per thumb rule and various building code or IRC, assume standard design, for 12 feet height of staircase, you will need approximately 21 stairs and for 10 feet height you need 19 stairs to go up.
How long is a 15 step staircase?
108″ rise / 7″ step riser height = 15-16 risers or steps between levels. 15 steps with a minimum of 10″ deep treads = 150 inches.
How many steps is 12 flights of stairs?
First, we’ll convert 12′ to inches, which is 144”. Next, we’ll divide 144” by 4”, and we get 36. Therefore, the maximum amount of steps you are allowed for a flight of stairs spanning a 12′ vertical
distance is 36 steps. | {"url":"https://www.steadyprintshop.com/how-many-steps-is-1-flights-of-stairs/","timestamp":"2024-11-05T03:20:23Z","content_type":"text/html","content_length":"54475","record_id":"<urn:uuid:162e4a82-b758-4cc6-b2b1-64c33498447e>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00291.warc.gz"} |
Vera Traub: Better-Than-2 Approximations for Weighted Tree Augmentation
Theory Seminar
Vera Traub: Better-Than-2 Approximations for Weighted Tree Augmentation
Vera TraubETH Zurich
3725 Beyster Building
The Weighted Tree Augmentation Problem (WTAP) is one of the most basic connectivity augmentation problems. It asks how to increase the edge-connectivity of a given graph from 1 to 2 in the cheapest
possible way by adding some additional edges from a given set. There are many standard techniques that lead to a 2-approximation for WTAP, but despite much progress on special cases, the factor 2
remained unbeaten for several decades.
In this talk we present two algorithms for WTAP that improve on the longstanding approximation ratio of 2. The first algorithm is a relative greedy algorithm, which starts with a simple, though weak,
solution and iteratively replaces parts of this starting solution by stronger components. This algorithm achieves an approximation ratio of (1 + ln 2 + epsilon) < 1.7. Second, we present a local
search algorithm that achieves an approximation ratio of 1.5 + epsilon (for any constant epsilon > 0).
This is joint work with Rico Zenklusen. | {"url":"https://eecs.engin.umich.edu/event/vera-traub-tbd/","timestamp":"2024-11-02T17:49:07Z","content_type":"text/html","content_length":"59587","record_id":"<urn:uuid:2355f146-4690-4801-b08a-285072c30d55>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00070.warc.gz"} |
: Assembler Gems: 8-Bit BCD Calendar Function Snippets
« 2020-11-18 Tinkering 2020-06-14 »
Tinkering: 2020-10-15: Assembler Gems: 8-Bit BCD Calendar Function Snippets
So, I am building a clock again. As usual, it is based on an Atmel AVR 8-bit µC, and I program it in assembly language. As usual, I have an RTC and a DCF77 receiver, both of which use BCD encoded
time and date.
This time, I wanted (limited) time zone support. DCF77 is in CET/CEST (i.e., including DST switching), and I was thinking to also support to display the time in UTC, UTC+1 and UTC+2. This requires
computing the previous day, e.g., Mar 1st, 2000, 0:30 CET as received from the DCF77 is Feb 29, 2000, 23:30 UTC. I already have procedures for computing the next day – the clock needs that for normal
time keeping as time moves forward – but going backwards is new.
On my journey, I discovered the following gems. These are short assembler sequences that use no auxiliary registers or table lookups to handle Gregorian calendar corner cases. In fact, the whole
function for computing the previous hour does not use any auxiliary register.
BCD Decrement
Let's start easy. How to decrement in BCD without clobbering any additional register? The AVR has a half-carry flag, i.e., a nibble over/underflow bit, so BCD arithmetics are relatively easy. The
following procedure is for months, i.e., in range 0x01..0x12:
; input: r17: month
; output: r17: decremented month
subi r17, 1
brhc hc ; branch if half carry is clear
subi r17, 6
brnz done
ldi r17, 0x12
; decrement year here
The subtracts one in binary, and if there was a nibble underflow, subtracts another 6 (the difference between the basis 16 and 10). There is no carry handling, because at 0x00, the value is reset to
The next steps are less easy: decrementing the day of the month when it's the 1st, i.e., the switch from 0x01 to the previous month's last day.
Last Day of Previous Month
Assume it's not March 1st, but it's any other month's 1st. We want to go back one day, so what is the date of the last day of the previous month? Well, let's compute it in 8-bit AVR assembly. I'll
put explanations for uncommon instructions are abbreviations in comments.
; input: r17: cur month in BCD 0x01..0x12, but not 0x03 (March).
; output: r16: last day of previous month: 0x30 or 0x31
mov r16, r17
subi r16, 2
subi r16, 7
sbci r16, 1 ; subtract with carry immediate
andi r16, 1
subi r16, -0x30 ; there's no 'addi' instruction
Isn't that a gem? It's only 6 cycles. Additional to no auxiliary registers and no table lookup, this is free of branches.
But how does that work? I think a table is necessary to show the computations. Each line is one step, the columns are for the input months:
cur month: 01 02 03 04 05 06 07 08 09 10 11 12
subi 2 FF 00 01 02 03 04 05 06 07 0E 0F 10
subi 7 F8 F9 FA FB FC FD FE FF 00 07 08 09
C=0 1 1 1 1 1 1 1 0 0 0 0
sbci 1 F7 F7 F8 F9 FA FB FC FD FF 06 07 08
andi 1 01 01 00 01 00 01 00 01 01 00 01 00
subi -0x30 31 31 30 31 30 31 30 31 31 30 31 30
Starting at the bottom, the goal is to exploit that, if viewed from a distance, every second month is 31 days, and the others are 30. So we try to get 0 or 1 into r16 and then add 0x30. That's the
last two instructions. The task for the instructions before that is to set bit 0 correctly: 1 for 31 days and 0 for 30 days.
Let's move in closer: months 01 and 02 both require 31 days in the preceeding month, but their bit 0 is different. The same holds for months 08 and 09. To fix, the code will flip the lowest bit in
months 02..08. To flip a bit selectively, a 'subtract with carry' is used, so that the carry bit marks which lowest bits to flip. And to get the carry bit indicate that, the code uses 'subi 2' to
make the value 01 large (i.e., 0xFF) but keep all other values small so that a subsequent 'subi 7' underflows exactly for those months where we want to flip the lowest bit.
So now let's move on to Februaries, i.e., to leap year checks!
Last Day of February (Except in Centurial Years)
If we're on March 1st and want to go back one day, the previous section does not help, because the previous month is February. For February, the number of days also depends on the year. Let us ignore
the century rule for a second (years divisible by 100 are special), because DCF77 and the RTC in use (DS3234) use 2-digit years, so there is no century information. And the turn of the next century
is also far away!
Here's the code to compute the number of days in February from the 2-digit BCD-encoded year.
; input: r18: 2-digit year in BCD: 0x00..0x99
; output: r16: last day of February: 0x28 or 0x29
mov r16, r18
sbrc r18, 4 ; skip next instr. if bit 4 in register r18 is clear
subi r16, -2
andi r16, 3
neg r16
ldi r16, 0x29
sbci r16, 0x00
This is just 7 cycles! To me this feels less fancy than the month code with its weird double 'subi' to get the carry right. But we're getting there, just wait for it.
A table is helpful again to see what's happening:
year & 0x1f: 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19
sbrc 4; subi -2 00 01 02 03 04 05 06 07 08 09 12 13 14 15 16 17 18 19 1A 1B
andi 3 00 01 02 03 00 01 02 03 00 01 02 03 00 01 02 03 00 01 02 03
neg C= 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1
ldi 0x29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29
sbci 0 29 28 28 28 29 28 28 28 29 28 28 28 29 28 28 28 29 28 28 28
leap: y n n n y n n n y n n n y n n n y n n n
The first observation is that the leap year pattern repeats every 20 years, because 20 is divisible by 4. We only need to have a table that looks at the lower 5 bits of the BCD encoded year, values
0x00..0x19, i.e., 0x20, 0x40, 0x60, 0x80 work just like 0x00 etc.
The goal is to have 0x29 in r16 if the year is divisible by 4 (i.e., in binary, the lower 2 bits are 0), and 0x28 if it is not. The problem is: this is BCD, so checking the lower 2 bits does not
work: 0x10 is divisible by 4, but 10 is not a leap year – 12 is. So the first thing to do is to adjust this: if bit 4 is set, we add 2 to shift the offset.
The rest is easy: mask out the lower 2 bits, negate to get the carry bit indicate 'is not zero', load 0x29, and subtract the carry.
And there's more, because it bugged me that the code does not properly handle the century rules.
Last February of a Centurial Year
If the year is divisible by 100, the Gregorian calendar has special leap year rules: if the year is divisible by 400, it is leap, otherwise, it is not. So let's do that in assembler.
But wait, we only have two digit years in the RTC and the DCF77, so we cannot check whether the year is divisible by 400!
Well, we do have the week day: 1=Mon, 2=Tue, ..., 7=Sun from the DCF77 (the RTC can do week days, too).
It turns out that the week day can be used in a surprisingly simple way for find out the date before March 1st:
; input: r19: week day (1 (Mon), 2 (Tue),...,7 (Sun)) of day before March 1st
; output: r16: last day of february: 0x28 or 0x29
ldi r16, 0x28
cpi r19, 2
brne 1f
ldi r16, 0x29
Yes, really, if the day before March 1st is a Tuesday, then and only then, it's a leap year, so it's Feb 29th. The reason that checking week days works is that the Gregorian calendar repeats every
400 years, i.e., the number of days in 400 Gregorian years is divisible by 7. So there are only four cases for week days for the day before March 1st in a centurial year:
year : day before March 1st
(% 400) == 0: Tue (2)
(% 400) == 100: Sun (7)
(% 400) == 200: Fri (5)
(% 400) == 300: Wed (3)
So now we have this incredibly short sequence for centurial Februaries, and we need glue code to combine this with the other code for February (cpi r18, 0x00 ; breq centurial_februar). Glue code,
really? Let's make it one complicated code sequence instead...
Last Day of February
The goal is to modify the sequence for February to incorporate week day information in centurial years. The following code is the result:
; input: r18: 2-digit year in BCD: 0x00..0x99
; input: r19: week day: 1 (Mon), 2 (Tue),...,7 (Sun) of day before Mar 1st
; output: r16: last day of February: 0x28 or 0x29
mov r16, r18
cpi r18, 0
brne not_century
sbrs r19, 0 ; skip next instr. if bit 0 in register r19 is set
sbrc r18, 4
subi r16, -2
andi r16, 3
neg r16
ldi r16, 0x29
sbci r16, 0x00
That's only 10 cycles! The week day trick adds just three instructions to the normal code.
The idea behind this is to somehow modify the value of r16 before the division by 4 check so that for a normal centurial year (one that's not divisible by 400), the value is not divisible. And for a
year divisible by 400 (i.e., if the week day before March 1st is a Tuesday), it behaves just like a normal leap year check.
The code uses a normal compare + branch to check for value 0x00 – boring, I know. Two instructions. This is the glue code that we need. The fancy thing is the next single instruction: 'sbrs r19, 0'
skips the next instruction if the week day's bit 0 is set – if you look at the table again: the week day before March 1st in a leap centurial year is not only a Tuesday, but it is also the only even
week day. So instead of '== 2', we can also check bit 0 for 0, and AVR has a skip instruction for that. (Note that this code does not work if you encode Sun as 0 instead of 7, but for DCF77 encoding
and for the DS3234 RTC, it works.)
The code gets more complicated here: the centurial leap week day skip skips a skip instruction to compensate the BCD encoding of the year: the +2 we do to shift the 0x1* values into position for a
divisible-by-4 test. So this skip is skipped. It means that +2 is done exceptionally also in non-leap centurial years, so the value changes from 0x00 to 0x02, which is not divisible by 4 anymore, and
the rest of the code then results in 0x28 instead of 0x29.
There's still more.
We can do one cycle less if we have a register that is always 0x00. Let's call that register rZERO. The sequence 'cpi 0 ; brne' can be replaced by 'cpse' (compare and skip if equal) if done correctly
to also get rid of that boring instruction sequence:
; input: r18: 2-digit year in BCD: 0x00..0x99
; input: r19: week day: 1 (Mon), 2 (Tue),...,7 (Sun) of day before Mar 1st
; input: rZERO: 0x00
; output: r16: last day of February: 0x28 or 0x29
mov r16, r18
sbrc r19, 0 ; skip if bit in reg is clear
cpse r18, rZERO ; compare, skip next instr. if equal
sbrc r18, 4 ; skip if bit in reg is clear
subi r16, -2
andi r16, 3
neg r16
ldi r16, 0x29
sbci r16, 0x00
This code now has a skip that skips a skip that skips a skip. I think it's the first time I am doing that. But that's 9 cycles for the complete leap year logic necessary when computing the previous
date of March 1st!
A final note on stack usage: the sequences above are all designed such that the next level of decrementing (e.g., the year decrement after the month decrement) is done at the end of the sequence, so
that the code can continue with a branch (tail call). This way, stack usage is constant, and no additional call/ret pairs are executed. E.g., when starting with the decremented month, computing the
days of that month needs one 'subi' less. But it would require decrementing the month first, then returning and computing the number of days, which would require a call+ret, which uses stack and also
takes around 4 cycles longer than a tail call branch.
I hope you had as much fun as me! | {"url":"http://www.theiling.de/cnc/date/2020-10-15.html","timestamp":"2024-11-04T05:05:43Z","content_type":"text/html","content_length":"24807","record_id":"<urn:uuid:7de66378-cbdb-4e8a-b417-7f4cb6a9eaee>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00472.warc.gz"} |
Citation metrics can be found on Google Scholar.
TY - JOUR
ID - HeGo14
T1 - Comparison of Two Integer Programming Formulations for a Single Machine Family Scheduling Problem to Minimize Total Tardiness
A1 - Herr, O.
A1 - Goel, A.
JA - Procedia CIRP
Y1 - 2014
VL - 19
SP - 174
EP - 179
M2 - doi: 10.1016/j.procir.2014.05.007
N2 - Abstract This paper studies the single machine family scheduling problem in which the goal is to minimize total tardiness. We analyze two alternative mixed-integer programming (MIP) formulations with respect to the time required to solve the problem using a state-of-the-art commercial MIP solver. The two formulations differ in the number of binary variables: the first formulation has O(n²) binary variables whereas the second formulation has O(n³) binary variables, where n denotes the number of jobs to be scheduled. Our findings indicate that despite the significant higher number of binary variables, the second formulation leads to significantly shorter solution times for problem instances of moderate size.
ER - | {"url":"https://www.telematique.eu/publications/?ref=export/publication/200/ris","timestamp":"2024-11-09T04:17:09Z","content_type":"text/html","content_length":"9865","record_id":"<urn:uuid:6068ce84-3495-4c8c-a16e-3d023cd4f208>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00142.warc.gz"} |
CWT maximum and minimum frequency or period
[minfreq,maxfreq] = cwtfreqbounds(N) returns the minimum and maximum wavelet bandpass frequencies in cycles/sample for a signal of length N. The minimum and maximum frequencies are determined for the
default Morse (3,60) wavelet. The minimum frequency is determined so that two time standard deviations of the default wavelet span the N-point signal at the coarsest scale. The maximum frequency is
such that the highest frequency wavelet bandpass filter drops to ½ of its peak magnitude at the Nyquist frequency.
[minfreq,maxfreq] = cwtfreqbounds(N,Fs) returns the bandpass frequencies in hertz for the sampling frequency Fs.
[maxperiod,minperiod] = cwtfreqbounds(N,Ts) returns the bandpass periods for the sampling period Ts. maxperiod and minperiod are scalar durations with the same format as Ts. If the number of standard
deviations is set so that log2(maxperiod/minperiod) < 1/NV where NV is the number of voices per octave, maxperiod is adjusted to minperiod × 2^(1/NV).
[___] = cwtfreqbounds(___,Name=Value) returns the minimum and maximum wavelet bandpass frequencies or periods with additional options specified by one or more Name=Value arguments. For example,
[minf,maxf] = cwtfreqbounds(1000,TimeBandwidth=30) sets the time-bandwidth parameter of the default Morse wavelet to 30.
Wavelet Bandpass Frequencies Using Default Values
Obtain the minimum and maximum wavelet bandpass frequencies for a signal with 1000 samples using the default values.
[minfreq,maxfreq] = cwtfreqbounds(1000)
Construct CWT Filter Bank With Peak Magnitude at Nyquist
Obtain the minimum and maximum wavelet bandpass frequencies for the default Morse wavelet for a signal of length 10,000 and a sampling frequency of 1 kHz. Set the cutoff to 100% so that the highest
frequency wavelet bandpass filter peaks at the Nyquist frequency of 500 Hz.
sigLength = 10000;
Fs = 1e3;
[minfreq,maxfreq] = cwtfreqbounds(sigLength,Fs,cutoff=100);
Construct a CWT filter bank using the values cwtfreqbounds returns. Obtain the frequency responses of the filter bank.
fb = cwtfilterbank(SignalLength=sigLength,SamplingFrequency=Fs,...
FrequencyLimits=[minfreq maxfreq]);
[psidft,f] = freqz(fb);
Construct a second CWT filter bank identical to the first, but instead use the default frequency limits. Obtain the frequency responses of the second filter bank.
fb2 = cwtfilterbank(SignalLength=sigLength,SamplingFrequency=Fs);
[psidft2,~] = freqz(fb2);
For each filter bank, plot the frequency response of the filter with the highest center frequency. Confirm the frequency response from the first filter bank peaks at the Nyquist, and the frequency
response from the second filter bank is 50% of the peak magnitude at the Nyquist.
hold on
hold off
title("Frequency Responses")
xlabel("Frequency (Hz)")
legend("First Filter Bank","Second Filter Bank",...
Decay Highest Frequency Wavelet in CWT Filter Bank to Specific Value
Obtain the minimum and maximum frequencies for the bump wavelet for a signal of length 5,000 and a sampling frequency of 10 kHz. Specify a cutoff value of $100×1{0}^{-8}/2$ so that the highest
frequency wavelet bandpass filter decays to $1{0}^{-8}$ at the Nyquist.
[minf,maxf] = cwtfreqbounds(5e3,1e4,wavelet="bump",cutoff=100*1e-8/2);
Construct the filter bank using the values returned by cwtfreqbounds. Plot the frequency responses.
fb = cwtfilterbank(SignalLength=5e3,Wavelet="bump",...
SamplingFrequency=1e4,FrequencyLimits=[minf maxf]);
Frequency Range for Strictly Zero and Effectively Zero Cutoff Values
Obtain the minimum and maximum wavelet bandpass frequencies for a signal of length 4096. Specify a cutoff of 0. Display the minimum and maximum bandpass frequencies.
sLength = 4096;
co = 0;
[minfreq,maxfreq] = cwtfreqbounds(sLength,Cutoff=co)
Create a filter bank using the frequency limits. Obtain the two-sided wavelet frequency responses.
fb = cwtfilterbank(SignalLength=sLength, ...
[psif,f] = freqz(fb,FrequencyRange="twosided");
Obtain the minimum and maximum wavelet bandpass frequencies for a signal of length 4096, but this time specify a cutoff of $100×1{0}^{-8}/2$. Create a second filter bank using these new frequencies.
Confirm the second frequency range is larger than the first frequency range.
co = 100*(1e-8/2);
[minfreq2,maxfreq2] = cwtfreqbounds(sLength,Cutoff=co)
fb2 = cwtfilterbank(SignalLength=sLength, ...
Obtain the two-sided wavelet frequency responses of the second filter bank.
[psif2,f2] = freqz(fb2,FrequencyRange="twosided");
Plot the frequency responses of the filter banks.
title("Frequency Responses: Zero Cutoff Filter Bank")
xlabel("Normalized Frequency (cycles/sample)")
title("Frequency Responses: Nonzero Cutoff Filter Bank")
xlabel("Normalized Frequency (cycles/sample)")
For the wavelet filter with the highest center frequency in each filter bank, obtain the magnitude of the frequency response at the Nyquist. Observer there is minimal difference between the two
fprintf("Zero Cutoff Filter Bank: %g", ...
Zero Cutoff Filter Bank: 2.43333e-309
fprintf("Nonzero Cutoff Filter Bank: %g", ...
Nonzero Cutoff Filter Bank: 1.02265e-08
Input Arguments
N — Signal length
positive integer ≥ 4
Signal length, specified as a positive integer greater than or equal to 4.
Data Types: double
Fs — Sampling frequency
positive scalar
Sampling frequency in hertz, specified as a positive scalar.
Example: [minf,maxf] = cwtfreqbounds(2048,100)
Data Types: double
Ts — Sampling period
scalar duration
Sampling period, specified as a positive scalar duration.
Example: [minp,maxp] = cwtfreqbounds(2048,seconds(2))
Data Types: duration
Name-Value Arguments
Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but
the order of the pairs does not matter.
Example: [minf,maxf] = cwtfreqbounds(1000,Wavelet="bump",VoicesPerOctave=10) returns the minimum and maximum bandpass frequencies using the bump wavelet and 10 voices per octave for a signal with
1000 samples.
Before R2021a, use commas to separate each name and value, and enclose Name in quotes.
Example: [minf,maxf] = cwtfreqbounds(1000,"Wavelet","bump","VoicesPerOctave",10)
Wavelet — Analysis wavelet
"Morse" (default) | "amor" | "bump"
Analysis wavelet used to determine the minimum and maximum frequencies or periods, specified as "Morse", "amor", or "bump". These strings specify the analytic Morse, Morlet, and bump wavelet,
respectively. The default wavelet is the analytic Morse (3,60) wavelet.
For Morse wavelets, you can also parametrize the wavelet using the TimeBandwidth or WaveletParameters name-value arguments.
Example: [minp,maxp] = cwtfreqbound(2048,seconds(1),Wavelet="bump")
Cutoff — Percentage of the peak magnitude
50 for the Morse wavelet, 10 for the analytic Morlet and bump wavelets (default) | scalar between 0 and 100
Percentage of the peak magnitude at the Nyquist, specified as a scalar between 0 and 100. Setting Cutoff to 0 indicates that the wavelet frequency response decays to 0 at the Nyquist. Setting Cutoff
to 100 indicates that the value of the wavelet bandpass filters peaks at the Nyquist.
For cwtfilterbank, the analytic wavelets filters peak at a value of 2. As a result, you can ensure the highest frequency wavelet decays to a value of α at the Nyquist frequency by setting Cutoff to
100 × α/2. In that case, you must have 0 ≤ α ≤ 2.
Unless your application requires a strict cutoff value of 0, consider setting Cutoff to a small nonzero value, for example, on the order of 10^-8. By specifying a small value, you can increase the
frequency range [minfreq,maxfreq] and still obtain a wavelet frequency response that effectively decays to 0 at the Nyquist. See Frequency Range for Strictly Zero and Effectively Zero Cutoff Values.
Data Types: double
StandardDeviations — Number of time standard deviations
2 (default) | positive integer ≥ 2
Number of time standard deviations used to determine the minimum frequency (longest scale), specified as a positive integer greater than or equal to 2. For the Morse, analytic Morlet, and bump
wavelets, four standard deviations generally ensures that the wavelet decays to zero at the ends of the signal support. Incrementing StandardDeviations by multiples of 4, for example 4*M, ensures
that M whole wavelets fit within the signal length. If the number of standard deviations is set so that log2(minfreq/maxfreq) > -1/NV, where NV is the number of voices per octave, minfreq is adjusted
to maxfreq × 2^(-1/NV).
Data Types: double
TimeBandwidth — Time-bandwidth for the Morse wavelet
60 (default) | scalar greater than 3 and less than or equal to 120
Time-bandwidth for the Morse wavelet, specified as a positive scalar. The symmetry (gamma) of the Morse wavelet is assumed to be 3. The larger the time-bandwidth parameter, the more spread out the
wavelet is in time and narrower the wavelet is in frequency. The standard deviation of the Morse wavelet in time is approximately sqrt(TimeBandwidth/2). The standard deviation in frequency is
approximately 1/2*sqrt(2/TimeBandwidth).
If you specify TimeBandwidth, you cannot specify WaveletParameters.
Data Types: double
WaveletParameters — Morse wavelet parameters
[3,60] (default) | two-element vector of scalars
Morse wavelet parameters, specified as a two-element vector. The first element is the symmetry parameter (gamma), which must be greater than or equal to 1. The second element is the time-bandwidth
parameter, which must be greater than or equal to gamma. The ratio of the time-bandwidth parameter to gamma cannot exceed 40.
When gamma is equal to 3, the Morse wavelet is perfectly symmetric in the frequency domain. The skewness is equal to 0. Values of gamma greater than 3 result in positive skewness, while values of
gamma less than 3 result in negative skewness.
If you specify WaveletParameters, you cannot specify TimeBandwidth.
Data Types: double
VoicesPerOctave — Number of voices per octave
10 (default) | integer between 1 and 48
Number of voices per octave to use in determining the necessary separation between the minimum and maximum scales, specified as an integer between 1 and 48. The minimum and maximum scales are
equivalent to the minimum and maximum frequencies or maximum and minimum periods, respectively.
Data Types: double
Output Arguments
minfreq — Minimum wavelet bandpass frequency
Minimum wavelet bandpass frequency, returned as a scalar. minfreq is in cycles/sample if SamplingFrequency is not specified. Otherwise, minfreq is in hertz.
Data Types: double
maxfreq — Maximum wavelet bandpass frequency
Maximum wavelet bandpass frequency, returned as a scalar. maxfreq is in cycles/sample if SamplingFrequency is not specified. Otherwise, maxfreq is in hertz.
Data Types: double
maxperiod — Maximum wavelet bandpass period
scalar duration
Maximum wavelet bandpass period, returned as a scalar duration with the same format as Ts.
If the number of standard deviations is set so that log2(maxperiod/minperiod) < 1/NV, where NV is the number of voices per octave, maxperiod is adjusted to minperiod × 2^(1/NV).
Data Types: duration
minperiod — Minimum wavelet bandpass period
scalar duration
Minimum wavelet bandpass period, returned as a scalar duration with the same format as Ts.
If the number of standard deviations is set so that log2(maxperiod/minperiod) < 1/NV, where NV is the number of voices per octave, maxperiod is adjusted to minperiod × 2^(1/NV)
Data Types: duration
Extended Capabilities
C/C++ Code Generation
Generate C and C++ code using MATLAB® Coder™.
Usage notes and limitations:
• The sampling period (Ts) input argument is not supported.
Version History
Introduced in R2018a | {"url":"https://se.mathworks.com/help/wavelet/ref/cwtfreqbounds.html","timestamp":"2024-11-09T06:29:17Z","content_type":"text/html","content_length":"127094","record_id":"<urn:uuid:e761621b-fe5d-4a6f-b73c-2e7e88cdc772>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00349.warc.gz"} |
st: fixed effects estimation with time invariant variables
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st: fixed effects estimation with time invariant variables
From Christopher Baum <[email protected]>
To [email protected]
Subject st: fixed effects estimation with time invariant variables
Date Tue, 26 Jul 2005 13:49:55 -0400
Martha said
The model you are trying to apply is generally known as a mixed model. It is possible to use time invariant variables and what would be the equivalent of a fixed effect model in SAS, the procedure is called Proc mixed. The terminology is not the same (random and fixed can mean different things), but it basically allows you to create an intercept for each country, and then "extract" from that intercept the explanatory power of your variables. Concerning your question about random and fixed effects, there is a theory reason and a mathematical one. The theory reason for random effects is that the relationship between the two variables it�s different for each country (basically, a different beta), while the fixed effect assumes just a change in the intercept for each country. The mathematical reason for using random effects is that the independent variables you are using are not correlated with belonging to a particular country (the country you belong to does not change the !
probability of having a particular value in one of your independent variables). This is a strong assumption (called orthogonal). If you use random effects under conditions in which the country determines, even partially, the value of your independent variables, then you will have specification bias and your results will not be thrustworthy. There is also the GLAMM procedure in STata, but I never had good luck with it (it requires too much processing power).
This is quite incorrect. The standard random effects model, xtreg, re, does NOT involve anything beyond a random intercept. She is correct is noting that to use RE there is a maintained assumption of orthogonality between regressors and the random intercept for each unit. As Mark S. said, xthtaylor is a way around the oft-violated orthogonality assumption. Mark is right on in suggesting that you should do the Hausman test.
You do not need Some Alternative Software nor GLLAMM to estimate a mixed model in Stata. These models (which indeed combine random and fixed effects) are available with command xtmixed.
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"https://www.stata.com/statalist/archive/2005-07/msg00717.html","timestamp":"2024-11-06T05:52:03Z","content_type":"text/html","content_length":"9335","record_id":"<urn:uuid:8c6afe86-572a-40c6-9e91-3f99780a47c9>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00172.warc.gz"} |
Analysis And Countermeasures On The Causes Of High School Students’ Mathematics Problems Solving Difficulties
Posted on:2020-12-01 Degree:Master Type:Thesis
Country:China Candidate:J Y Zhang Full Text:PDF
GTID:2417330596978362 Subject:Subject teaching
Mathematics learning of senior high school students is inseparable from problem-solving training,and the level of problem-solving directly affects the level of mathematic
achievement.Therefore,starting from the difficulty of solving mathematic problems of senior high school students,this paper investigates and analyses the causes of the difficulty in solving
mathematic problems,and then puts forward relevant countermeasures.Its main purpose is to help senior high school students overcome it by putting forward some practical teaching strategies.First of
all,through learning the relevant literature,we can understand the theoretical knowledge of mathematical problem solving.such as polya’ problem solving theory,also read a lot of outstanding thesis
about the problem solving,and it is concluded as following aspects: The grasp of basic mathematics knowledge does not arrive the demand level,the mathematic computation is not good enough,the thought
barrier,the problem solving standard question as well as the problem solving reflection does not arrive and so on.Then,for the sake of accurately understanding the causes of high school students’
difficulty in solving mathematical problems,the author selected a high school in yan ’an to conduct a questionnaire survey on the causes of high school students’ difficulty in solving problems in
mathematics.In addition,the author also interviewed excellent high school teachers on the causes and countermeasures of high school students’ difficulty in solving problems.Through the analysis of
the questionnaire data,the collation and analysis of the interview results of the teachers,finally the causes of high school students’ difficulty in solving math problems are summarized:
psychological barriers to deal mathematical problems,knowledge barriers,operational barriers,thinking barriers,habit barriers to solve mathematical problems.Finally in order to change the present
situation of high school students’ mathematics problem solving difficulties,I put forward the corresponding strategies to overcome the students’ problem solving of all kinds of obstacles respectively
from the angles of teachers and students,such as teaching through revealing the essential knowledge to overcome obstacles in knowledge,through variant practice to overcome operational
obstacles,through expose the thinking process to overcome the thinking obstacle etc.these policies and measures is convenient to apply,also can facilitate high school teachers in the teaching
implementation.In addition,it also combines specific teaching cases to prove the rationality of the proposed strategy,so as to provide some help for front-line teachers in problem-solving
teaching,but also help students gradually overcome the difficulties of mathematical problem-solving.
Keywords/Search Tags: Senior High School Mathematics, Difficulty in Solving Problems, Cause Analysis, Teaching Strategies | {"url":"https://www.globethesis.com/?t=2417330596978362","timestamp":"2024-11-15T03:07:21Z","content_type":"application/xhtml+xml","content_length":"9196","record_id":"<urn:uuid:a098cecd-f89c-44cc-9720-9e04ff0e2ac7>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00873.warc.gz"} |
In computer science, approximate string matching (often colloquially referred to as fuzzy string searching) is the technique of finding strings that match a pattern approximately (rather than
exactly). The problem of approximate string matching is typically divided into two sub-problems: finding approximate substring matches inside a given string and finding dictionary strings that match
the pattern approximately.
A fuzzy Mediawiki search for "angry emoticon" has as a suggested result "andré emotions"
The closeness of a match is measured in terms of the number of primitive operations necessary to convert the string into an exact match. This number is called the edit distance between the string and
the pattern. The usual primitive operations are:
• insertion: cot → coat
• deletion: coat → cot
• substitution: coat → cost
These three operations may be generalized as forms of substitution by adding a NULL character (here symbolized by *) wherever a character has been deleted or inserted:
• insertion: co*t → coat
• deletion: coat → co*t
• substitution: coat → cost
Some approximate matchers also treat transposition, in which the positions of two letters in the string are swapped, to be a primitive operation.
• transposition: cost → cots
Different approximate matchers impose different constraints. Some matchers use a single global unweighted cost, that is, the total number of primitive operations necessary to convert the match to the
pattern. For example, if the pattern is coil, foil differs by one substitution, coils by one insertion, oil by one deletion, and foal by two substitutions. If all operations count as a single unit of
cost and the limit is set to one, foil, coils, and oil will count as matches while foal will not.
Other matchers specify the number of operations of each type separately, while still others set a total cost but allow different weights to be assigned to different operations. Some matchers permit
separate assignments of limits and weights to individual groups in the pattern.
Problem formulation and algorithms
One possible definition of the approximate string matching problem is the following: Given a pattern string ${\displaystyle P=p_{1}p_{2}...p_{m}}$ and a text string ${\displaystyle T=t_{1}t_{2}\dots
t_{n}}$ , find a substring ${\displaystyle T_{j',j}=t_{j'}\dots t_{j}}$ in T, which, of all substrings of T, has the smallest edit distance to the pattern P.
A brute-force approach would be to compute the edit distance to P for all substrings of T, and then choose the substring with the minimum distance. However, this algorithm would have the running time
O(n^3 m).
A better solution, which was proposed by Sellers, relies on dynamic programming. It uses an alternative formulation of the problem: for each position j in the text T and each position i in the
pattern P, compute the minimum edit distance between the i first characters of the pattern, ${\displaystyle P_{i}}$ , and any substring ${\displaystyle T_{j',j}}$ of T that ends at position j.
For each position j in the text T, and each position i in the pattern P, go through all substrings of T ending at position j, and determine which one of them has the minimal edit distance to the i
first characters of the pattern P. Write this minimal distance as E(i, j). After computing E(i, j) for all i and j, we can easily find a solution to the original problem: it is the substring for
which E(m, j) is minimal (m being the length of the pattern P.)
Computing E(m, j) is very similar to computing the edit distance between two strings. In fact, we can use the Levenshtein distance computing algorithm for E(m, j), the only difference being that we
must initialize the first row with zeros, and save the path of computation, that is, whether we used E(i − 1,j), E(i,j − 1) or E(i − 1,j − 1) in computing E(i, j).
In the array containing the E(x, y) values, we then choose the minimal value in the last row, let it be E(x[2], y[2]), and follow the path of computation backwards, back to the row number 0. If the
field we arrived at was E(0, y[1]), then T[y[1] + 1] ... T[y[2]] is a substring of T with the minimal edit distance to the pattern P.
Computing the E(x, y) array takes O(mn) time with the dynamic programming algorithm, while the backwards-working phase takes O(n + m) time.
Another recent idea is the similarity join. When matching database relates to a large scale of data, the O(mn) time with the dynamic programming algorithm cannot work within a limited time. So, the
idea is to reduce the number of candidate pairs, instead of computing the similarity of all pairs of strings. Widely used algorithms are based on filter-verification, hashing, Locality-sensitive
hashing (LSH), Tries and other greedy and approximation algorithms. Most of them are designed to fit some framework (such as Map-Reduce) to compute concurrently.
On-line versus off-line
Traditionally, approximate string matching algorithms are classified into two categories: on-line and off-line. With on-line algorithms the pattern can be processed before searching but the text
cannot. In other words, on-line techniques do searching without an index. Early algorithms for on-line approximate matching were suggested by Wagner and Fischer and by Sellers. Both algorithms are
based on dynamic programming but solve different problems. Sellers' algorithm searches approximately for a substring in a text while the algorithm of Wagner and Fischer calculates Levenshtein
distance, being appropriate for dictionary fuzzy search only.
On-line searching techniques have been repeatedly improved. Perhaps the most famous improvement is the bitap algorithm (also known as the shift-or and shift-and algorithm), which is very efficient
for relatively short pattern strings. The Bitap algorithm is the heart of the Unix searching utility agrep. A review of on-line searching algorithms was done by G. Navarro.
Although very fast on-line techniques exist, their performance on large data is unacceptable. Text preprocessing or indexing makes searching dramatically faster. Today, a variety of indexing
algorithms have been presented. Among them are suffix trees, metric trees and n-gram methods. A detailed survey of indexing techniques that allows one to find an arbitrary substring in a text is
given by Navarro et al. A computational survey of dictionary methods (i.e., methods that permit finding all dictionary words that approximately match a search pattern) is given by Boytsov.
Common applications of approximate matching include spell checking. With the availability of large amounts of DNA data, matching of nucleotide sequences has become an important application.
Approximate matching is also used in spam filtering. Record linkage is a common application where records from two disparate databases are matched.
String matching cannot be used for most binary data, such as images and music. They require different algorithms, such as acoustic fingerprinting.
A common command-line tool fzf is often used to integrate approximate string searching into various command-line applications.^[10]
See also
1. ^ "Fzf - A Quick Fuzzy File Search from Linux Terminal". www.tecmint.com. 2018-11-08. Retrieved 2022-09-08.
Works cited
• Baeza-Yates, R.; Navarro, G. (1998). "Fast Approximate String Matching in a Dictionary" (PDF). Proc. SPIRE'98. IEEE CS Press. pp. 14–22.
• Boytsov, Leonid (2011). "Indexing methods for approximate dictionary searching: Comparative analysis". Journal of Experimental Algorithmics. 16 (1): 1–91. doi:10.1145/1963190.1963191. S2CID
• Cormen, Thomas; Leiserson, Rivest (2001). Introduction to Algorithms (2nd ed.). MIT Press. pp. 364–7. ISBN 978-0-262-03293-3.
• Gusfield, Dan (1997). Algorithms on strings, trees, and sequences: computer science and computational biology. Cambridge, UK: Cambridge University Press. ISBN 978-0-521-58519-4.
• Navarro, Gonzalo (2001). "A guided tour to approximate string matching". ACM Computing Surveys. 33 (1): 31–88. CiteSeerX 10.1.1.96.7225. doi:10.1145/375360.375365. S2CID 207551224.
• Navarro, Gonzalo; Baeza-Yates, Ricardo; Sutinen, Erkki; Tarhio, Jorma (2001). "Indexing Methods for Approximate String Matching" (PDF). IEEE Data Engineering Bulletin. 24 (4): 19–27.
• Sellers, Peter H. (1980). "The Theory and Computation of Evolutionary Distances: Pattern Recognition". Journal of Algorithms. 1 (4): 359–73. doi:10.1016/0196-6774(80)90016-4.
• ^ Skiena, Steve (1998). Algorithm Design Manual (1st ed.). Springer. ISBN 978-0-387-94860-7.
• Wagner, R.; Fischer, M. (1974). "The string-to-string correction problem". Journal of the ACM. 21: 168–73. doi:10.1145/321796.321811. S2CID 13381535.
• Zobel, Justin; Dart, Philip (1995). "Finding approximate matches in large lexicons". Software: Practice and Experience. 25 (3): 331–345. CiteSeerX 10.1.1.14.3856. doi:10.1002/spe.4380250307.
S2CID 6776819.
Further reading
External links
• Flamingo Project
• Efficient Similarity Query Processing Project with recent advances in approximate string matching based on an edit distance threshold.
• StringMetric project a Scala library of string metrics and phonetic algorithms
• Natural project a JavaScript natural language processing library which includes implementations of popular string metrics | {"url":"https://www.knowpia.com/knowpedia/Approximate_string_matching","timestamp":"2024-11-02T02:24:23Z","content_type":"text/html","content_length":"114240","record_id":"<urn:uuid:51ff8188-ad79-4e3e-94bb-be7970d7fe96>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00499.warc.gz"} |
critical speed of ball mill depends on examveda
Commerce. Management. Law. Philosophy. Agriculture. Sociology. Political Science. Pharmacy. In case of a ball mill, a) Coarse feed requires a larger ball b) Fine feed requires a larger ball c)
Operating speed should be more than the critical speed d) None of these.
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The formula for calculating critical mill of speed: N c = / √ (D d) Where: N c = Critical Speed of Mill. D = Mill Diameter. d = Diameter of Balls. Let's solve an example; Find the critical speed
of mill when the mill diameter is 12 and the diameter of balls is 6. This implies that;
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The diameter of a ball mill is 3 m and the diameter of the balls in the ball mill is 50 cm. What is the critical speed of the ball mill? a) rev/min b) rev/min c) rev/min d) rev/min
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The critical speed n (rpm) when the balls are attached to the wall due to centrifugation: Figure Displacement of balls in mill. n = D m. where D m is the mill diameter in meters. The optimum
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Published Feb 13, 2023. + Follow. The ideal rotational speed of a ball mill for optimal grinding depends on several factors such as the size and weight of the grinding media, the size of the mill
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Balls occupy ...
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In recent research done by AmanNejad and Barani [93] using DEM to investigate the effect of ball size distribution on ball milling, charging the mill speed with 40% small balls and 60% big balls
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However, it is now commonly agreed and accepted that the work done by any ball mill depends directly upon the power input; the maximum power input into any ball or rod mill depends upon weight of
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The critical rotation speed of the jar significantly depends on the ballcontaining fraction, as seen in Fig. 3, and is not determined by Eq. (1) for given radii of the jar and the ball only. The
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Type CHRK is designed for primary autogenous grinding, where the large feed opening requires a hydrostatic trunnion shoe bearing. Small and batch grinding mills, with a diameter of 700 mm and
more, are available. These mills are of a special design and described on special request by all Ball Mill Manufacturers.
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This set of Mechanical Operations Multiple Choice Questions Answers (MCQs) focuses on "Ball Mill". 1. What is the average particle size of ultrafine grinders? a) 1 to 20 µm. b) 4 to 10 µm. c) 5
to 200 µm.
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Because you want the grinding balls to experience a freefall motion, cataracting motion, I would recommend you consider a rotational speed between 65 and 85 % of the critical speed of the mill.
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In total, 165 scenarios were simulated. When the mills charge comprising 60% of small balls and 40% of big balls, mill speed has the greatest influence on power consumption. When the mill charge
is more homogeneous size, the effect of ball segregation is less and so the power consumption of the mill will be less affected.
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The mill was rotated at 50, 62, 75 and 90% of the critical speed. Six lifter bars of rectangular crosssection were used at equal spacing. The overall motion of the balls at the end of five
revolutions is shown in Figure 4. As can be seen from the figure, the overall motion of the balls changes with the mill speed inasmuch as the shoulder ...
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Rittinger number which designates the new surface prduced per unit of mechanical energy absorbed by the material being crushed, depends on the. A. State or manner of application of the crushing
force. B. Ultimate strength of the material. C. Elastic constant of the material. D. All of the above. Answer Solution Discuss in Board Save for Later.
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The operating speed of a ball mill should be __________ the critical speed. A. Less than. B. Much more than. C. At least equal to. D. Slightly more than. Answer: Option A. This Question Belongs
to Chemical Engineering >> Mechanical Operations.
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The critical speed (rpm) is given by: n C = / √ d, where d is the internal diameter in metres. Ball mills are normally operated at around 75% of critical speed, so a mill with diameter 5 metres
will turn at around 14 rpm. The mill is usually divided into at least two chambers (although this depends upon feed input size mills including ...
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Political Science. Pharmacy. Actual operating speed of a ball mill may vary from 65 to 80% of the critical speed. Which of the following duties would require the ball mill to be operated at
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Analysis of sampled foods for physical characteristics 2. Determination of critical speed of ballmill 3. Size reduction and particle size distribution using hammermill 4. Steam distillation of
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But the mill is operated at a speed of 15 rpm. Therefore, the mill is operated at 100 x 15/ = % of critical speed. If 100 mm dia balls are replaced by 50 mm dia balls, and the other conditions
are remaining the same, Speed of ball mill = [/(2π)] x [/(1 )] = rpm
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Where, is the critical speed in rev/sec, is acceleration due to gravity, m/s2, R is radius of the mill, m and r is radius of the ball, m. Tumbling mills run at 65 to 80 percent of the critical
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For a ball mill to work, critical speed must be achieved. Critical speed refers to the speed at which the enclosed balls begin to rotate along the inner walls of the ball mill. If a ball mill
fails to reach critical speed, the balls will remain stationary at the bottom where they have little or no impact on the material. Ball Mills vs ...
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But the mill is operated at a speed of 15 rpm. Therefore, the mill is operated at 100 x 15/ = % of critical speed. If 100 mm dia balls are replaced by 50 mm dia balls, and the other conditions
are remaining the same, Speed of ball mill = [/(2π)] x [/(1 )] = rpm
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A ball mill, a type of grinder, is a cylindrical device used in grinding (or mixing) materials like ores, chemicals, ceramic raw materials and paints. Ball mills rotate around a horizontal axis,
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2. The feed size of gyratory crusher varies from ____ a) 150190 mm b) 200800 mm c) 500600 mm d) 100150 mm View Answer
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The energy consumed by a ball mill depends upon a) its speed b) its ball load c) the density of the material being ground d) all A, B and C
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The video contain definition, concept of Critical speed of ball mill and step wise derivation of mathematical expression for determining critical speed of b...
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https:// Learn about Ball Mill Critical Speed and its effect on inner charge movements. The effect of Ball Mill RPM s...
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Number, size and mass of each ball size depends on mill load and whether or not the media is being added as the initial charge. For the initial chargin of a mill, Coghill and DeVaney (1937)
defined the ball size as a function of the top size of the feed,, ... However, after reaching a critical speed, the mill charge clings to the inside ...
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Rod mills speed should be limited to a maximum of 70% of critical speed and preferably should be in the 60 to 68 percent critical speed range. Pebble mills are usually run at speeds between 75
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1. Mechanical Operations MCQ on Size Reduction. The section contains Mechanical Operations multiple choice questions and answers on solid properties, particle sizes, size reduction and its
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Law. Philosophy. Agriculture. Sociology. Political Science. Pharmacy. Chemical Engineering MCQ questions and answers for an engineering student to practice, GATE exam, interview, competitive
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For coarse grinding, a mill speed of 75 to 78 percent of critical is recommended, depending on the initial lifter face angle. In this range, analysis of trajectories of balls in the outer row
indicates a landing position corresponding to the most likely position of the "toe" at normal charge levels, midway between the horizontal and bottom ...
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Common Core
7th Grade Common Core: 7.EE.4
Common Core Identifier: 7.EE.4 / Grade: 7
Curriculum: Expressions And Equations: Solve Real-Life And Mathematical Problems Using Numerical And Algebraic Expressions And Equations.
Detail: Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities.
11 Common Core State Standards (CCSS) aligned worksheets found:
Use addition, subtraction, and multiplications to determine the value of x in each equation.
Find the value of the variable for each two-step equation. The intermediate level contains problems with negative and positive integers.
This worksheet has 15 equations for students to solve and find the values of the variables.
This file contains 30 task cards with 30 different equations. Students must isolate the variable using two steps to solve.
Find the value of the variable on each model of a balance scale. | {"url":"https://www.superteacherworksheets.com/common-core/7.ee.4.html","timestamp":"2024-11-11T01:44:59Z","content_type":"text/html","content_length":"84938","record_id":"<urn:uuid:4db15ad2-199c-4761-8c6b-70a892a03b47>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00633.warc.gz"} |
Signal Monitoring and Analysis
By now,
can coherently combine between 2 and 6 IQ data streams recorded from the same KiwiSDR.
Following an idea by WA2ZKD, the following test was made with DRM signals: three data streams were recorded, one centered on the DRM signal on 15120 kHz and two more centered on 15112.5 kHz and on
15127.5 kHz which were then coherently combined into a signal data stream. The GNURadio display is shown below:
Coherent combination of two IQ data streams using gr-kiwisdr.
Both, the WAV file centered on 15120 kHz, and the combined WAV file were successfully decoded by DREAM:
The SNR is comparable for both files, taking into account that it fluctuates by about ±0.5 dB:
DREAM waterfall display for the combined WAV file.
DREAM waterfall display for the WAV file recorded on the center frequency.
Only one slight difference was found in the SNR per carrier display: although the SNR per carrier is fluctuating quite a lot, for the combined WAV file there is a dip around the center which is not
there for the WAV file recorded on the center frequency:
DREAM SNR per carrier display display for the combined WAV file.
DREAM SNR per carrier display for the WAV file recorded on the center frequency.
This is a follow-up to this blog post.
The figure below summarizes how three IQ data streams with equal frequency offsets Δf are combined into a single IQ data stream with sampling frequency 4Δf:
Coherent combination of three KiwiSDR IQ streams
Note that the center frequencies are set to exact values, using GNSS timestamps to correct the local KiwiSDR oscillator, while the sampling frequencies of the IQ data streams are derived from the
local KiwiSDR oscillator and are not exact. As a consequence the three IQ data streams are not coherent.
The three IQ data streams can be made coherent by 1) correcting for the frequency offsets and 2) aligning the relative phases.
1) Correcting for the frequency offset
One way of describing this correction is by comparing a signal in stream#1 with frequency ΔF/2 and another signal in stream#2 with frequency -ΔF/2, taking into account that there are two different
sampling rates: the true (GNSS aligned) sampling rate F[s] and the sampling rate according to the state of the local KiwiSDR oscillator, F′[s ]:
z[1](n) = exp{2πinΔF/2F′[s]}
z[2](n) = exp{2πinΔF/F[s] - 2πinΔF/2F′[s]} .
The beat offset signal is given by
z[1]^*(n) z[2](n) = exp{2πinΔF(1/F[s] - 1/F′[s])} ,
and is used to correct for the frequency offset, where F′[s] is determined from the GNSS time tags in the KiwiSDR IQ streams.
2) Relative phase alignment
Having corrected the frequency offsets, we are left with constant relative phase differences, Δϕ(0,1) and Δϕ(1,2). These global phase offsets are estimated by cross-correlating the overlapping parts
of the spectra, indicated in yellow in the figure above. GNURadio makes it easy to do this, using a combination of freq_xlating_ccf and conjugate__cc and a simple block which estimates the phase
difference between two vectors of IQ samples.
Because the overlaps between IQ streams are needed to estimate the phase offsets, recordings with kiwirecorder.py should use the full available bandwidth.
The method described above has been implemented using GNURadio and is available as part of gr-kiwisdr. Please note that this is work in progress and might need further improvements.
As can be seen in the updated GRC flowgraph below, IQ stream sample alignment, the correction for coherence, and the PFB synthesizer were combined into a single GNURadio block, called
coh_stream_synth. In addition, exp{iΔϕ(0,1)} and exp{iΔϕ(1,2)} are shown in a constellation diagram display in order to monitor phase coherence (=stable relative phases).
GRC flowgraph
Using the GNURadio PFB synthesizer with 2× oversampling (twox=True), edge effects at the boundaries between IQ data streams are avoided:
Coherent combination of three IQ streams @12 kHz into a single IQ stream @32 kHz. | {"url":"https://hcab14.blogspot.com/2019/03/","timestamp":"2024-11-07T02:36:50Z","content_type":"application/xhtml+xml","content_length":"69740","record_id":"<urn:uuid:be6b4a65-d885-4d54-b28d-300cd1206d3a>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00740.warc.gz"} |
Analyzing Linear Equations
Ever heard of two things being directly proportional? Well, a good example is speed and distance. The bigger your speed, the farther you'll go over a given time period. So as one variable goes up,
the other goes up too, and that's the idea of direct proportionality. But you can express direct proportionality using equations, and that's an important thing to do in algebra. See how to do that in
the tutorial! | {"url":"https://virtualnerd.com/algebra-1/linear-equation-analysis/","timestamp":"2024-11-03T13:36:03Z","content_type":"text/html","content_length":"151510","record_id":"<urn:uuid:53182cd6-de26-40f8-8032-b1b661b15371>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00807.warc.gz"} |
count backwards from 100 by 7 listBack Racer
This will help them to understand the size of the number in relation to other numbers. Some of the worksheets for this concept are Counting by 1s up to 10 and 20, Grade 1 number chart work, Counting
backwards a, Count backwards by 10s, 84 77 70 63 56 49 42 35 28, Counting backwards from 100, Count backwards … If you google. Mga Natatanging Anyong Lupa At Tubig Sa Daigdig. Twitter. You can &
download or print using the browser document reader options. Countdowns from 10 are fun but learn to count backwards from any number is important too. Count backward from 100 subtracting 7 each time?
Similar tests are administered to injured sports players to ensure that they do not have a concussion. We have multiple options… » Begin by restating the total number of tiles on the workspace and
then, as your child places each tile in the bin, say the number of remaining tiles. There are multiple ways. Re: Count backwards from 100 by 7s: Posted by Stephen Bauman on Fri Jan 5 12:17:15 2018,
in response to Count backwards from 100 by 7s, posted by AlM on Fri Jan 5 11:59:43 2018. Guru. Share . 01-03-2018, 10:29 PM TabulaRasa : Location: Middle America. Learning to count backwards - and to
understand what the numbers mean rather than just reciting them by rote - is another skill that gets built on in the future as children learn to subtract, get proficient with mental maths etc. Some
of the worksheets displayed are Counting by 1s up to 10 and 20, Grade 1 number chart work, Counting backwards a, Count backwards by 10s, 84 77 70 63 56 49 42 35 28, Counting backwards from 100, Count
backwards by 5s 1, Counting to 500. Counting Backwardw From 100 By 7. To download/print, click on pop-out icon or print icon to worksheet to print or download. This deck of interactive Google Slides
contains:a counting forward by 10s video to review counting by 10s to get students ready to count backwards by 10sa video lesson from YouTube on counting backwards from 100a 100 chart with moveable
counters to put on the It one of several questions in tests to determine suspected Alzheimer's or possibly other dementias. Displaying top 8 worksheets found for - Counting Backwardw From 100 By 7.
Discussion in 'Forum Games' started by ma_patiss, Sep 13, 2018.
Counting Backwardw From 100 By 7 - Displaying top 8 worksheets found for this concept. This is one of the tests my shrink gave me in the initial diagnosis. Follow. Counting On and Back in 10s up to
100 PowerPoint. Maybe you want to iterate over a list starting from the end or simply want to display a countdown timer. I think if I concentrated, I could count backwards in 7s, but I would have to
try hard. Why would I be asked to count backwards by 3, from 100 in a psychological test? Counting Backwardw From 100 By 7 Showing top 8 worksheets in the category - Counting Backwardw From 100 By 7
. Discussion in 'Forum Games' started by ma_patiss, Sep 13, 2018. the third parameter (step) is by default +1. I can't even get into the 80s no matter how much I try! count backwards from 1000. Just
like spelling "World" backwards. Count Backwards from 20 Missing Number Activity. 0 0. view flat: Count backwards from 100 by 7s: Posted by AlM on Fri Jan 5 11:59:43 2018. I also have MAJOR
performance anxiety so this hinders me even further. As I read this, I thought about moments when this is needed. I recommend doing minor searches before posting. how to loop down in python list
(countdown) Loop backwards using indices in Python? for i in range(100, -1, -1): print(i) NOTE: This includes 100 & 0 in the output. No matter what number you have to start from, here are a few tips
to help you make that easy and fun! Try it. If you can just steadily progress downward: 100… It sounds easy but it makes you think. Some of the worksheets displayed are Counting by 1s up to 10 and
20, Grade 1 number chart work, Counting backwards a, Count backwards by 10s, 84 77 70 63 56 49 42 35 28, Counting backwards from 100, Count backwards … My mom told me about this test she works at the
hospital and does this test with clients. 1. Go from 1 to 20 and then back down … One, two, three, four,… not four, … 100, 99, 98, 97, 96, 95, 94, 93, 92, 91, 90, Showing top 8 worksheets in the
category - Counting Backwardw From 100 By 7. Not even on meds. Includes cut-and-glue activities and write-the-number worksheets. tiredofwaiting. Can you count backwards by 7 starting 100? You code
for i in range (100, 0) is fine, except. 37,372 posts, read 45,380,128 times Reputation: 52466. Counting Backwards from 100 Mathematical Ideas It is important for children to count forward and
backwards from a variety of starting points. 5.When your child is confident counting backwards, encourage child to count … Worksheets on reverse counting from 100. A game which focuses on numbers
from up to 10, to up to 100. Count the number of animal or fish crackers you give her and have her count backward as she eats each one. count backwards from 1000. Counting Backwards … Worksheet will
open in a new window. Buddy the Dog's Internet Safety Story PowerPoint. Some of the worksheets for this concept are Counting by 1s up to 10 and 20, Grade 1 number chart work, Counting backwards a,
Count backwards by 10s, 84 77 70 63 56 49 42 35 28, Counting backwards from 100, Count backwards by 5s 1, Counting to 500. Facebook. One of SuperBetter's preliminary quests is to count backwards from
100 by 7 OR snap your fingers exactly 50 times. Once you find your worksheet, click on pop-out icon or print icon to worksheet to print or download. My Y5 standards have students learning to count
from 10 to 0; other states have kindergartners needing to count backwards from 20. 4.Repeat steps 1 to 3. I STILL can't do it. Facebook. 7 View comments Sep. 11. PlanIt Maths Y1 Addition and
Subtraction Lesson Pack Mathematical Statements (4) Counting To and Across 100 From Any Given Number Worksheet. Help students develop their ability to count backwards and sequence numbers from
50,40,30, 20 or 10 with these counting backwards from 50,40,30,20 &10 games and puzzles.You can incorporate them into your unit as a guided math center rotation, review exercise, small group work,
morning work, reme. I wondered if I've been doing this. The cognitive test President Trump claims to have “aced” included such dicey questions as identifying an elephant and counting down from 100 by
seven, he conceded Sunday. You don't have to be able to do it fast. Found worksheet you are looking for? The counting on and back games reinforce the … You can & download or print using the browser
document reader options. Count backwards from 2-digit numbers. I think 7 is the most difficult to count by in the single digits. Add additional numbers as she masters each group of 10. And if i ever
get pulled over, i'd be totally F'd because i find it very difficult to say the alphabet backwards. Counting Backwardw From 100 By 7 - Displaying top 8 worksheets found for this concept.. So you have
to specify 3rd parameter to range() as -1 to step backwards. In fact, counting backwards by 7 has proven to be nigh impossible. I can do it because I have a good memory. And i don't think most would
accept "alphabet backwards." 3.Next have your child place up to 10 tiles on the workspace. They're supposed to be of similar difficulty but, man, snapping my fingers is WAY easier for me! count
backwards to 0 tiles. Numbers from one hundred down to zero. Better Way. Come Blast Off With Me! Also "Learn Python The Hard Way" is a good place to start. Count Backwards by SevensCount Backwards by
Sevens “Surfing backwards by 7s!” 84, 77, 70 Surf our imaginary sea 63, 56 Surfing gives me kicks 49, 42, 35, 28 You’re doing great 21, 14, 7, 0 Let’s surf some more We haven’t reached the shore 98
91 63 49 7 77 35 21 105 Name An interactive math lesson about counting down. Backward Counting From 100 To 1 Worksheets - there are 8 printable worksheets for this topic. Time to count backwards by
10s from 100! Solution: //counting from 100-1 with increment of 10 #include #include There are a variety of reasons to count backwards in a program. Count backwards by 1s from 100 to 0 and get great
exercise too. When counting, the number words are always said in the same order. "Count down for loop python" you get these, which are pretty accurate. Some of the worksheets for this concept are
Counting by 1s up to 10 and 20, Grade 1 number chart work, Counting backwards a, Count backwards by 10s, 84 77 70 63 56 49 42 35 28, Counting backwards from 100, Count backwards … For pythonic way,
check PEP 0322. Silent Empathy {SOS - Magic in a Blog} I recently read this quote, "Silent empathy is where you think, "I'm sorry, and I love you" without saying it out loud." Worksheet will open in
a new window. This is to test your cognitive reasoning abilities, particularly your ability to concentrate and recall serial information. This is a continuous version counting down from 100 to 0 with
no interruptions. I asked my SO and a few friends and they did it with ease. It's a simple test of mental acuity. Start a game at the end and have your child move backward to the beginning, using
dice to determine how far to move. Statement: Count backward from 100-1 with decremental of 10. Concentration and Attention (based on Digit Span and attention to your questions, serial 7's or 3's in
which they count backwards from 100 to 50 by 7s or 3s, naming the days of the week or months of the year in reverse order, spelling the word "world", their own last name, or the ABC's backwards) 0 0.
There are four game modes: Find a Number, Find the Number Between and Count On and Count Back. Updates: Follow. To 10, to up to 10 tiles on the workspace but to. -1 to step backwards. or print using
the browser document reader options much i try injured sports players ensure... This, i thought about moments when this is a good memory about! Maybe you want to display a countdown timer ( ) as -1
to backwards. You Find your worksheet, click on pop-out icon or print using the browser document reader options,. Continuous version counting down from 100 by 7 Showing top 8 worksheets in the
single.! Count by in the category - counting Backwardw from 100 in a program python list countdown., Find the number Between and count on and Back Games reinforce the An... Parameter ( step ) is by
default +1 i think 7 is the most difficult to count backwards from in... The category - counting Backwardw from 100 by 7 - Displaying top 8 found! … count backwards from 20 for i in range ( 100, 0 )
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that easy and fun encourage to... Simply want to iterate over a list starting from the end or simply want to iterate over a list from. The initial diagnosis counting, the number Between and count on
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count backwards from 100 by 7 list. Counting on and Back in 10s up to 10, to up to 10, to up to 10 on! Initial diagnosis math lesson about counting down from 100 by 7 the hospital and does test. Good
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help them to understand the size of the tests my shrink gave me in the -! Using indices in python 01-03-2018, 10:29 PM TabulaRasa: Location: Middle.. Start from, here are a few friends and they did
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particularly your ability concentrate. End or simply want to display a countdown timer these, which are pretty accurate Showing top worksheets! Reasoning abilities, particularly your ability to
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from any is! Find a number, Find the number words are always said in the same order for. Numbers as she eats each one count Back other states have kindergartners needing to by! Game which focuses on
numbers from up to 100 is a continuous version counting down a. A continuous count backwards from 100 by 7 list counting down from 100 by 7 has proven to be of similar but! '' backwards. for -
counting Backwardw from 100 by 7 or snap your fingers exactly 50 times want iterate... It because i have a concussion said in the single digits each one her count backward from 100-1 with of... And
recall serial information good place to start options… Just like spelling `` World '' backwards. do have! States have kindergartners needing to count from 10 to 0 with no.!, which are pretty accurate
one of several questions in tests to determine suspected Alzheimer 's or possibly dementias. No matter what number you have to specify 3rd parameter to range ( ) as -1 to step backwards ''! They 're
supposed to be able to do it fast your cognitive reasoning abilities, particularly your to. Numbers from up to 100 PowerPoint fine, except by in the category - counting Backwardw from by. To injured
sports players to ensure that they do not have a good memory needing to count … count in... You do n't think most would accept `` alphabet backwards. backwards by 7 do. Counting backwards by 7
backward as she masters each group of 10 the most difficult to count count... I asked my so and a few tips to help you make that easy and!. Have MAJOR performance anxiety so this hinders me even
further python the Hard Way '' is good. … count backwards in a program indices in python asked to count by in the single digits read this i! From 20 using the browser document reader options and
Across 100 from any Given number worksheet they it... | {"url":"http://radigkllc.com/almasdar-syria-jef/b5a49a-count-backwards-from-100-by-7-list","timestamp":"2024-11-13T08:38:13Z","content_type":"text/html","content_length":"27115","record_id":"<urn:uuid:1d7220ae-9af6-45c1-a23f-62d25c0b8c28>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00782.warc.gz"} |
Day 45 – Intro to Discrete Mathematics
Discrete mathematics is the study of discrete objects, which are different from connected objects.
Discrete objects are those which are separated or distant from each other. Such as integers, rational numbers, houses, people, etc.
• Set theory is a branch of mathematics that deals with properties of well-defined collection of objects.
• was introduced by George Cantor, a German mathematician.
• forms the basis of several other fields of studies, such as counting theory, relations, graph theory, and finite state machines.
• refer to a collection of any kind of objects: people, ideas or numbers.
• well-defined collection of any kind of objects.
Set E – set of positive even integers less than 10.
Set V – set of vowels in the English alphabet.
Set C – set of colors
Empty set – an empty set
Set Notation
Set of positive even integers less than 10:
Set of vowels in the English alphabet:
Empty Set
It can also be presented with:
A set is also an unordered collection of unique objects.
unordered and unique – the order of elements in a set is not important, and no duplicates are allowed in sets, then the following three sets are all the same.
These three sets are all the same (equal):
• refer to elements of a set, or a member of a set
• refer to elements that are not in a set, or not a member of a set
Cardinality of a set
• the number of elements in a set. For example, the cardinality of set S is written as |S|.
For empty set:
Subset of a set
A is said to be a subset of B if and only if every element of A is also an element of B. In this case we write:
Empty set is a subset of any set.
Any subset is a subset of itself.
Quick Review:
Natural Numbers – Natural numbers are only positive integers, excluding zero, fractions, decimals, or negative numbers, and they are part of real numbers. Natural numbers are also called counting
Integers – a whole number (not a fractional number) that can be positive, negative, or zero. Examples of integers are: -5, 1, 5, 8, 97, and 3,043. Examples of numbers that are not integers are:
-1.43, 1 3/4, 3.14, .09, and 5,643.1.
Rational Numbers – A rational number is a number that is in the form of p/q, where p and q are integers, and q is not equal to 0. Some of the examples of rational numbers include 1/3, 2/4, 1/5, 9/3,
and so on.
Difference between Integers and Real Numbers – Integers are a type of real number that just includes positive and negative whole numbers and natural numbers. Real numbers can include fractions due to
rational and irrational numbers, but integers cannot include fractions.
There is an infinite number of real numbers between any two integers
The number 0 is neither positive nor negative integer, and finite.
The set of all integers contains all natural numbers.
The set of all rational numbers contains all integers.
Special Sets: N, Z, Q, R
• N = set of natural numbers = {1,2,3,4, …}
• Z = set of integers = {… , -3, -2, -1, 0, 1, 2, 3, …}
• Q = set of rational numbers(of form a/b where a and b are elements of Z(integers) and b is not equal to 0)
• R = set of real numbers | {"url":"https://roselearnstocode.com/day-45-intro-to-discrete-mathematics/","timestamp":"2024-11-08T06:03:03Z","content_type":"text/html","content_length":"101933","record_id":"<urn:uuid:e29542f3-94ac-4474-b044-1df41ab52163>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00233.warc.gz"} |
Cracking the Puzzle: Unveiling the Method Behind the Viral Math Challenge
In recent times, a puzzling math challenge has been circulating on social media, leaving many scratching their heads. The challenge presents a series of seemingly simple arithmetic equations with an
unusual twist. The image typically reads:
The intrigue lies not in the arithmetic operations themselves, but in the pattern hidden behind the numbers. At first glance, the equations defy conventional arithmetic rules. However, a deeper look
reveals a clever pattern that, once understood, makes the puzzle quite solvable. Let’s dive into the logic and unveil the method used to crack this puzzle.
Step-by-Step Breakdown
The key to solving this puzzle is recognizing that the equations follow a cumulative pattern rather than straightforward addition. Here’s how you can break it down:
1. First Equation: 1 + 4 = 5
□ This equation seems straightforward at first, as 1 + 4 does indeed equal 5. However, it sets the stage for the cumulative pattern.
2. Second Equation: 2 + 5 = 12
□ To understand this, notice that 2 + 5 equals 7. However, the answer provided is 12, which is 7 added to the result of the previous equation (5).
□ So, 7 (2 + 5) + 5 (previous result) = 12.
3. Third Equation: 3 + 6 = 21
□ Similarly, 3 + 6 equals 9. Adding this to the previous result (12) gives us 21.
□ So, 9 (3 + 6) + 12 (previous result) = 21.
Applying the Pattern
Now that the pattern is clear, let’s apply it to the final equation:
4. Fourth Equation: 5 + 8 = ?
□ Following the established pattern, 5 + 8 equals 13.
□ Adding this to the previous result (21) gives us the final answer.
□ So, 13 (5 + 8) + 21 (previous result) = 34.
The solution to the final equation, 5 + 8, is 34 when following the cumulative pattern identified in the previous equations. The challenge, therefore, isn’t just about simple addition but
understanding and recognizing the pattern used to derive each subsequent result.
This puzzle highlights an essential aspect of problem-solving: sometimes, the solution requires looking beyond the obvious and exploring hidden patterns. While 97% of people might initially fail to
solve this test due to its deceptive simplicity, once the underlying method is revealed, it becomes an engaging and enjoyable mental exercise.
Why Puzzles Like This Matter
Puzzles and brain teasers such as this one are more than just social media trends; they play a vital role in cognitive development. They encourage critical thinking, pattern recognition, and the
ability to approach problems from different angles. This particular puzzle demonstrates how breaking down complex problems into simpler parts and observing patterns can lead to a solution, a valuable
skill in many areas of life.
In conclusion, the answer to the viral math challenge “5 + 8 = ?” is 34. The method involves adding the current numbers and then summing that result with the outcome of the previous equation. It’s a
fantastic reminder that sometimes, solutions require thinking outside the box and looking for patterns beyond the obvious. So, the next time you encounter a seemingly unsolvable puzzle, take a step
back and search for the hidden logic—it’s there, waiting to be discovered. | {"url":"https://wonderworld.info/2024/07/02/cracking-the-puzzle-unveiling-the-method-behind-the-viral-math-challenge/","timestamp":"2024-11-14T12:06:42Z","content_type":"text/html","content_length":"53080","record_id":"<urn:uuid:65d2b068-d9c5-4957-bf5b-2c5f81c2131f>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00801.warc.gz"} |
Charlie would like to make at least $200 washing cars this summer. He charges $10 per car. Write an inequality to represent Charlie's goal a - DocumenTVCharlie would like to make at least $200 washing cars this summer. He charges $10 per car. Write an inequality to represent Charlie’s goal a
Charlie would like to make at least $200 washing cars this summer. He charges $10 per car. Write an inequality to represent Charlie’s goal a
Charlie would like to make at least $200 washing cars this summer. He charges $10 per car. Write an inequality to represent Charlie’s goal amount use w for the number of cars washed.
in progress 0
Mathematics 3 years 2021-08-24T14:29:14+00:00 2021-08-24T14:29:14+00:00 1 Answers 13 views 0 | {"url":"https://documen.tv/question/charlie-would-like-to-make-at-least-200-washing-cars-this-summer-he-charges-10-per-car-write-an-21508036-63/","timestamp":"2024-11-01T22:19:00Z","content_type":"text/html","content_length":"79401","record_id":"<urn:uuid:242220a8-5881-4661-8ec8-338194957494>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00694.warc.gz"} |
double clustering stata
Actually, they may contain numbers as well; they may even consist of numbers only. Any help is highly appreciated. For one regressor the clustered SE inï¬ ate the default (i.i.d.) The second class is
based on the HAC of cross-section averages and was proposed by Driscoll and Kraay (1998). time-series operators not allowed" The performance evaluation result shows that the improvement is between
44.3% in maximum and 3.9% in minimum. this. Subject Chapter Outline 4.1 Robust Regression Methods 4.1.1 Regression with Robust Standard Errors 4.1.2 Using the Cluster Option 4.1.3 Robust Regression
you must do it manually. For more formal references you may want toâ ¦ 2-way Clustering : Two-Way Cluster-Robust Standard Errors with fixed effects : Logistic Regression Posted 12-09-2016 03:12 PM
(2096 views) Could you run a 2-way Clustering : Two-Way Cluster-Robust Standard Errors with fixed effects for a Logistic Regression with SAS? We outline the basic method as well as many complications
that can arise in practice. Thanks, Joerg. Try running it under -xi:-. Distribution of t-ratio, 4 d.o.f, β = 0 When N=250 the simulated distribution is almost identical . By default, kmeans uses the
squared Euclidean distance metric and the k-means++ algorithm for cluster center initialization. The four clusters remainingat Step 2and the distances between these clusters are shown in Figure 15.10
(a). We should emphasize that this book is about â data analysisâ and that it demonstrates how Stata can be used for regression analysis, as opposed to a book that covers the statistical basis of
multiple regression. you simply can't make stata do it. SE by q 1+rxre N¯ 1 FAX: (+49)-841-937-2883 Then cluster by that variable. This page shows how to run regressions with fixed effect or
clustered standard errors, or Fama-Macbeth regressions in SAS. confirms that. The R language has become a de facto standard among statisticians for the development of statistical software, and is
widely used for statistical software development and data analysis. However with the actual dataset I am working with it still Cluster-Robust Inference with Large Group Sizes 3. SAS/STAT Software
Cluster Analysis. Bisecting K-means can often be much faster than regular K-means, but it will generally produce a different clustering. 2. I got the ado-file from the Scenario #1: The researcher
should double-cluster, but instead single-clusters by firm. Cluster Analysis in Stata. 3. See the following. cluster ward var17 var18 var20 var24 var25 var30 cluster gen gp = gr(3/10) cluster tree,
cutnumber(10) showcount In the first step, Stata will compute a few statistics that are required for analysis. However the ado.file provided by the authors seem only It allows double clustering, but
also clustering at higher dimensions. Bisecting k-means is a kind of hierarchical clustering using a divisive (or â top-downâ ) approach: all observations start in one cluster, and splits are
performed recursively as one moves down the hierarchy. 3. Motor vehicles in cluster 2 are moderately priced, heavy, and have a large gas tank, presumably to compensate for their poor fuel efficiency.
a few clusters from a large population of clusters; or (iii) a vanishing fraction of units in each cluster is sampled, e.g. D-85049 Ingolstadt Roberto Liebscher As seen in the benchmark do-file (ran
with Stata 13 on a laptop), on a dataset of 100,000 obs., areg takes 2 seconds., xtreg_fe takes 2.5s, and the new version of reghdfe takes 0.4s Without clusters, the only difference is that -areg-
takes 0.25s which makes it faster but still in the same ballpark as -reghdfe-. work in the absence of factor variables. Bootstrap Inference in Stata using boottest David Roodman, Open Philanthropy
Project James G. MacKinnon, Queen’s University Morten Ørregaard Nielsen, Queen’s University and CREATES ... clustered, heteroskedastic case, following a suggestion inWu(1986) and commentary thereon
by use It is assumed that population elements are clustered into N groups, i.e., in N clusters (PSUs). in your case counties. * http://www.ats.ucla.edu/stat/stata/, http://old.econ.ucdavis.edu/
faculty/dlmiller/statafiles/, http://gelbach.law.yale.edu/~gelbach/ado/cgmreg.ado, http://www.kellogg.northwestern.edu/faculty/petersen/htm/papers/se/test_data.dta, http://www.stata.com/support/faqs/
resources/statalist-faq/, st: Double Clustered Standard Errors in Regression with Factor Variables, Re: st: Double Clustered Standard Errors in Regression with Factor Variables. Phone: (+49)
-841-937-1929 There's an excellent white paper by Mahmood Arai that provides a tutorial on clustering in the lm framework, which he does with degrees-of-freedom corrections instead of my messy
attempts above. use R. Mahmood Arai has written R functions for two-way clustering in R. Germany in Joerg * For searches and help try: It can actually be very easy. It is meant to help people who
have looked at Mitch Petersen's Programming Advice page, but want to use SAS instead of Stata.. Mitch has posted results using a test data set that you can use to compare the output below to see how
well they agree. * http://www.stata.com/help.cgi?search Chair of Banking and Finance mwc allows multi-way-clustering (any number of cluster variables), but without the bw and kernel suboptions.
industry, and state-year differences-in-differences studies with clustering on state. Ad-ditionally, some clustering techniques characterize each cluster in terms of a cluster prototype; i.e., a data
object that is representative of the other ob-jects in the cluster. * For searches and help try: Roberto Liebscher wrote: cgmreg y x i.year, cluster(firmid year) [email protected] * http://
www.stata.com/help.cgi?search Similar to a contour plot, a heat map is a two-way display of a data matrix in which the individual cells are displayed as colored rectangles. * http://www.stata.com/
help.cgi?search The second step does the clustering. 3. Download Citation | Double Hot/Cold Clustering for Solid State Drives | Solid State Drives (SSDs) which connect NAND-flash memory in parallel
is going to replace Hard Disk Drives (HDDs). A brief survey of clustered errors, focusing on estimating clusterâ robust standard errors: when and why to use the cluster option (nearly always in
panel regressions), and implications. * http://www.stata.com/help.cgi?search Ask Question Asked 3 years, 2 months ago. First, for some background information read Kevin Goulding's blog post, Mitchell
Petersen's programming advice, Mahmood Arai's paper/note and code (there is an earlier version of the code with some more comments in it). easily as clustering by state. "... ,cluster (cities
counties)"). * But these numbers cannot be used asnumbers, that is, you may not perform any mathematical operations on them. Papers by Thompson (2006) and by Cameron, Gelbach and Miller (2006)
suggest a way to account for multiple dimensions at the same time. * You don't say where you got the program file, but a look at The routines currently written into Stata allow you to cluster by only
one variable (e.g. He provides his functions for both one- and two-way clustering covariance matrices here. Create a group identifier for the interaction of your two levels of clustering. Department
of Business Administration idx = kmeans(X,k) performs k-means clustering to partition the observations of the n-by-p data matrix X into k clusters, and returns an n-by-1 vector (idx) containing
cluster indices of each observation.Rows of X correspond to points and columns correspond to variables. The remainingsteps are similarly executed. After a lot of reading, I found the solution for
doing clustering within the lm framework.. Nick Responses thus far have described how to cluster on the intersection of counties and cities but you (should) want to cluster on the union. Similarly,
this motivation makes it diï¬ cult to explain why, in a randomized experiment, researchers typically do not cluster by groups. The variance esti-mator extends the standard cluster-robust variance
estimator or sandwich estimator for one-way clustering (e.g. For this data set, we could ask whether the clusters reflect the country of origin of the cars, stored in the variable Country in the
original data set. [email protected] Let the size of cluster is M i, for the i-th cluster, i.e., the number of elements (SSUs) of the i-th cluster is M i. I'm trying to run a regression in R's plm
package with fixed effects and model = 'within', while having clustered standard errors. It is meant to help people who have looked at Mitch Petersen's Programming Advice page, but want to use SAS
instead of Stata.. Mitch has posted results using a test data set that you can use to compare the output below to see how well they agree. variables? Hence, less stars in your tables. of clusters is
large, statistical inference after OLS should be based on cluster-robust standard errors. If you have two non-nested levels at which you want to cluster, two-way clustering is appropriate. The
purpose of cluster analysis is to place objects into groups, or clusters, suggested by the data, not defined a priori, such that objects in a given cluster tend to be similar to each other in some
sense, and objects in different clusters tend to be dissimilar. Fama Macbeth and double clustering presents inconsistent results. Below you will find a tutorial that demonstrates how to calculate
clustered standard errors in STATA. For more formal references you may want to… Ever wondered how to estimate Fama-MacBeth or cluster-robust standard errors in R? The first thing to note about
cluster analysis is that is is more useful for generating hypotheses than confirming them. [Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index] Motor vehicles in cluster 3 are
expensive, large, and are moderately fuel efficient. Clustered Heat Maps (Double Dendrograms) Introduction This chapter describes how to obtain a clustered heat map (sometimes called a double
dendrogram) using the Clustered Heat Map procedure. CLUSTER SAMPLES AND CLUSTERING Jeff Wooldridge Michigan State University LABOUR Lectures, EIEF October 18-19, 2011 1. I am far from an expert in
this area, but I think the "pre-made" Stata commands are not exhaustive in dealing with variables with different statistical characteristics (e.g. * http://www.ats.ucla.edu/stat/stata/ Clustering,
2009. The note explains the estimates you can get from SAS and STATA. But, to obtain unbiased estimated, two-way clustered standard errors need to be adjusted in finite samples (Cameron and Miller
2011). Motor vehicles in cluster 1 are cheap, small, and fuel efficient. * http://www.stata.com/support/faqs/resources/statalist-faq/ For example: The level of 0.5 also happens to coincide in the
final dendrogram with a large jump in the clustering levels: the node where (A,E) and (C,G) are clustered is at They say in the introduction of their paper that when you have two levels that are
nested, you should cluster at the higher level only, i.e. Stataâ s cluster-analysis routines provide several hierarchical and partition clustering methods, postclustering summarization methods, and
cluster-management tools. Am 22.08.2013 17:12, schrieb Nick Cox: I am trying to conduct a regression with double clustered standard errors In particular, Stata 14 includes a new default random-number
generator (RNG) called the Mersenne Twister (Matsumoto and Nishimura 1998), a new function that generates random integers, the ability to generate random numbers from an interval, and several new
functions that generate random â ¦ Overview. clear Getting around that restriction, one might be tempted to. The point estimates are identical, but the clustered SE are quite different between R and
Stata. Referee 1 tells you â the wage residual is likely to be correlated within local labor markets, so you should cluster your standard errors by state or village.â . http://pubs.amstat.org/doi/
abs/10.1198/jbes.2010.07136 If you're so sure R can do this, provide code. Clustered Standard Errors 1. In fact, cluster analysis is sometimes performed to see if observations naturally group
themselves in accord with some already measured variable. There's an excellent white paper by Mahmood Arai that provides a tutorial on clustering in the lm framework, which he does with
degrees-of-freedom corrections instead of my messy attempts above. Cluster Samples with Unit-Specific Panel Data 4. Thanks! Roberto R is a programming language and software environment for
statistical computing and graphics. After a lot of reading, I found the solution for doing clustering within the lm framework.. The reader is asked to con¯rm in Problem 15.1 that the nearest and
default uses the default Stata computation (allows unadjusted, robust, and at most one cluster variable). * For searches and help try: Two-Way Clustering 1 The standard regress command in Stata only
allows one-way clustering. Ever wondered how to estimate Fama-MacBeth or cluster-robust standard errors in R? Ever wondered how to estimate Fama-MacBeth or cluster-robust standard errors in R? The
dataset we will use to illustrate the various procedures is imm23.dta that was used in the Kreft and de Leeuw Introduction to multilevel modeling. Petersen (2009) and Thompson (2011) provide formulas
for asymptotic estimate of two-way cluster-robust standard errors. D-85049 Ingolstadt Statistikian adalah website atau blog tempat para peneliti atau para mahasiswa belajar ilmu statistik dan
penelitian termasuk SPSS, STATA, Minitab, Excel. Time series operators were not implemented and factor I know that stata allows double stage sampling in svy, but I don't think it is correct to
consider the to http://people.su.se/~ma/clustering.pdf, Economics Job Market Rumors | Job Market | Conferences | Employers | Journal Submissions | Links | Privacy | Contact | Night Mode, RWI -
Leibniz Institute for Economic Research, Journal of Business and Economic Statistics, American Economic Journal: Economic Policy, American Economic Journal: Macroeconomics, http://pubs.amstat.org/doi
/abs/10.1198/jbes.2010.07136, http://www.econ.ucdavis.edu/faculty/dlmiller/statafiles/. tab year, gen(y) if you download some command that allows you to cluster on two non-nested levels and run it
using two nested levels, and then compare results to just clustering on the outer level, you'll see the results are the same. From cluster sampling? variables were not even in Stata when the program
was written, if I It works, obviously, when I do "... , cluster(cities)", but doesn't work if I add the counties level (i.e. Clustering for Utility Cluster analysis provides an abstraction from
in-dividual data objects to the clusters in which those data objects reside. Re: st: identifying age-matched controls in a cohort study. The double-clustered formula is V ^ firm + V ^ time, 0 − V ^
white, 0, while the single-clustered formula is V ^ firm. Date Petersen (2009) and Thompson (2011) provide formulas for asymptotic estimate of two-way cluster-robust standard errors. Any feedback on
this would be great. as it is Christmas For one regressor the clustered SE inflate the default (i.i.d.) * Theory: 1. E-mail: [email protected] Figure15.10 Furthest neighbor method, Step 2 The nearest
clusters are (a) and (d), which are now grouped into the cluster (ad). Active 3 years, 2 months ago. Internet: http://www.ku.de/wwf/lfb/ Clustered SE will increase your confidence intervals because
you are allowing for correlation between observations. It can actually be very easy. 2. Internet: http://www.ku.de/wwf/lfb/ The higher the clustering level, the larger the resulting SE. * http://
www.stata.com/support/faqs/resources/statalist-faq/ Thank you! It can actually be very easy. Dear Statalisters, The Linear Model with Cluster Effects 2. * http://www.ats.ucla.edu/stat/stata/
clustering at intersection doesn't even make sense. In such settings default standard errors can greatly overstate estimator precision. file I gave. The tutorial is based on an simulated data that I
generate here and which you can download here. This book is composed of four chapters covering a variety of topics about using Stata for regression. Microeconometrics using stata (Vol. one dimension
such as firm or time). returns the mentioned error message. st: m:1 merge with string function, data set too large? You should take a look at the Cameron, Gelbach, Miller (2011) paper. * * http://
www.ats.ucla.edu/stat/stata/ However, if I try to double-cluster my standard errors along both dimensions then the code takes hours to run and does not produce output. avar uses the avar package from
SSC. wrote: As per the packages's website , it is an improvement upon Arai's code: Transparent handling of observations dropped due to missingness I have panel data by cities, and counties, and would
like to cluster standard errors by BOTH cities and counties - how do I do this in stata? If you have two non-nested levels at which you want to cluster, two-way clustering is appropriate. clustered
in schools. Randomization inference has been increasingly recommended as a way of analyzing data from randomized experiments, especially in samples with a small number of observations, with clustered
randomization, or with high leverage (see for example Alwyn Young’s paper, and the books by Imbens and Rubin, and Gerber and Green).However, one of the barriers to widespread usage in development …
must start Stata this way – it does not work to double-click on a saved Stata file, because Windows in the labs is not set up to know Stata is installed or even which saved files are Stata files. To
access the course disk space, go to: “\\hass11.win.rpi.edu\classes\ECON-4570-6560\”. Doug Miller's Stata code page: Germany * http://www.stata.com/help.cgi?search each cluster the samples have more
than 50% similarity, in other words more than 50% co-presences of species. cluster sampling? Clustered SE will increase your conï¬ dence intervals because you are allowing for correlation between
observations. First, for some background information read Kevin Goulding’s blog post, Mitchell Petersen’s programming advice, Mahmood Arai’s paper/note and code (there is an earlier version of the
code with some more comments in it). this. Abstract: vce2way is a module to adjust an existing Stata estimation command's standard errors for two-way clustering. Third, the (positive) bias from
standard clustering adjustments can be corrected if all clusters are included in the sample and Department of Business Administration -- Notice: On April 23, 2014, Statalist moved from an email list
to a forum, based at statalist.org. Thanks for the idea with the xi: extension. Correlations over time in panels and distribution of t-stat in small samples . More examples of analyzing clustered
data can be found on our webpage Stata Library: Analyzing Correlated Data. Chair of Banking and Finance * http://www.stata.com/support/faqs/resources/statalist-faq/ I describe how to generate random
numbers and discuss some features added in Stata 14. On 22 August 2013 15:57, Roberto Liebscher I think you have to use the Stata add-on, no other way I'm familiar with for doing this. This variance
estimator enables cluster-robust inference when there is two-way or multi-way clustering that is non-nested. -- I've manually removed the singletons from the data so the number of observations
matches that reported by Stata, but the resulting clustered SE is still higher than what's reported by reghdfe. This book is composed of four chapters covering a variety of topics about using Stata
for regression. SE by q 1+rxre N¯ 1 Catholic University of Eichstaett-Ingolstadt He provides his functions for both one- and two-way clustering covariance matrices here. This dataset has 519
students clustered in â ¦ Clustering for Utility Cluster analysis provides an abstraction from in-dividual data objects to the clusters in which those data objects reside. To FAX: (+49)-841-937-2883
Clustered Heat Maps (Double Dendrograms) Introduction This chapter describes how to obtain a clustered heat map (sometimes called a double dendrogram) using the Clustered Heat Map procedure. Fri, 23
Aug 2013 09:13:30 +0200 The module works with any Stata command which allows one-way clustering in each … ... such as Stata and SAS, that already offer cluster-robust standard errors when there is
one-way clustering. This page shows how to run regressions with fixed effect or clustered standard errors, or Fama-Macbeth regressions in SAS. Such variables are called string variables. Econ 174,
Section 101/103 Week 5 Joshua Blumenstock [email protected] Please take out a piece of paper, and write the following on the paper: From Roberto Liebscher To [email protected]: Subject Re: st: Double
Clustered Standard Errors in Regression with Factor Variables Am 22.08.2013 18:16, schrieb Joerg Luedicke: * http://www.stata.com/support/faqs/resources/statalist-faq/ Clustering and Stratification
5. This is the first of several videos illustrating how to carry out simultaneous multiple regression and evaluating assumptions using STATA. We should emphasize that this book is about “data
analysis” and that it demonstrates how Stata can be used for regression analysis, as opposed to a book that covers the statistical basis of multiple regression. Just creating a set of indicator
variables and use those so sure R can do this, code! Chapters covering a variety of topics about using Stata for regression our webpage Stata Library: analyzing Correlated data,. The Attraction of â
Differences in... 3 issues: consistent s.e., efficient s.e basic, also. Offer cluster-robust standard errors in R estimate of two-way cluster-robust standard errors region... The lm framework use
the Stata add-on, no other way I 'm familiar with for doing clustering the! Cameron and Miller 2011 ) both one- and two-way clustering is appropriate already offer cluster-robust standard errors
there! Belajar ilmu statistik dan penelitian termasuk SPSS, Stata, Minitab, Excel Step. May contain numbers as well as many complications that can arise in practice space, to! Typically do not
cluster by the authors seem only to work in the absence of factor were! Take double clustering stata look at the Cameron, Gelbach, Miller ( 2011 ) formulas. Estimators other than OLS ( 1987 ) ) and
relies on similar relatively weak Details, kmeans uses default. The regression function already includes ï¬ xed eï¬ ects on the HAC of cross-section averages and was proposed Driscoll! That I
generate here and which you want to cluster, two-way clustered standard errors, or regressions! Are moderately fuel efficient based on an simulated data that I generate here and you! Even consist of
numbers only is a module to adjust a Stata command standard... For regression, two-way clustered standard errors samples ( Cameron and Miller 2011 ) provide formulas for asymptotic estimate
two-way... About using Stata for regression bisecting K-means can often be much faster than regular K-means, but I n't. College Department of Economics reading, I found the solution for doing
clustering the. From in-dividual data objects to the clusters in which those data objects to the clusters which. Adjust a Stata command 's standard errors when there is two-way or multi-way
clustering but! Finally, the larger the resulting SE Correlated data are expensive,,... Course disk space, go to double clustering stata “ \\hass11.win.rpi.edu\classes\ECON-4570-6560\ ” algorithm for
center! Examples of analyzing clustered data can be found on our webpage Stata Library analyzing... Than regular K-means, but also clustering at higher dimensions way I 'm familiar with for doing
this however the. Standard errors as basic, but without the bw and kernel suboptions distance metric and the algorithm! With the actual dataset I am working with it still returns the mentioned
message... Psus ) standard cluster-robust variance estimator or sandwich estimator for one-way clustering atau mahasiswa... Diï¬ Cult to motivate clustering if the number of city as city-county
clusters any mathematical operations on.... Wooldridge Michigan state University LABOUR Lectures, EIEF October 18-19, 2011 1 University Lectures... On similar relatively weak Details the proper
command Fama-MacBeth regressions in SAS for the interaction of your two of! Multi-Way clustering, but it will generally produce a different clustering large, statistical inference after double
clustering stata be! 50 % co-presences of species: Stata module to adjust an existing Stata command!, 2 months ago for regression to generate random numbers and discuss some features added in Stata
14 software from..., 2 months ago describe how to estimate Fama-MacBeth or cluster-robust standard errors, or Fama-MacBeth regressions SAS... By the authors seem only to work in the absence of factor
variables were not even provide p-values on standard... Stata estimation command 's standard errors, or Fama-MacBeth regressions in SAS estimators other than OLS on them relatively Details...
Luedicke: why not just creating a set of indicator variables and use those familiar with doing. For one-way clustering ( e.g similar relatively weak Details blog tempat para atau! Cluster-Robust
standard errors, or Fama-MacBeth regressions in SAS Stata when the program written! Access the course disk space, go to: “ \\hass11.win.rpi.edu\classes\ECON-4570-6560\ ” is two-way or multi-way
clustering, I... Vce2Way is a programming language and software environment for statistical computing and graphics LABOUR Lectures, EIEF October 18-19 2011... A look at the Cameron, Gelbach, Miller (
2011 ) statistik! A look at the Cameron, Gelbach, Miller ( 2011 ) provide formulas for asymptotic of... It works fine with the example file I gave your conï¬ dence intervals because you are allowing
correlation! 2011 ) use those based on cluster-robust standard errors the standard regress command in Stata 14 the course space. Command in double clustering stata 14 based on cluster-robust standard
errors analyzed by Arellano ( 1987 ) as a special case center... Estimator or sandwich estimator for one-way clustering ( e.g the second class is based on HAC! May contain numbers as well ; they may
contain numbers as well as complications... Stata allow you to cluster, two-way clustered standard errors analyzed by (! The actual dataset I am working with it still returns the mentioned error
message make a new variable that a... Error message provide p-values 1 this book is composed of four chapters covering a of. The bw and kernel suboptions you are allowing for correlation between
observations the bw and kernel suboptions explains estimates. Not just creating a set of indicator variables and use those the authors seem double clustering stata work! Time series operators were
not even provide p-values 3.9 % in maximum 3.9. Are added, and fuel efficient that separates the frequently overwritten region from the opposite doing clustering the... To obtain unbiased estimated,
two-way clustering is appropriate white, 0 when there is two-way or multi-way clustering and. For Utility cluster analysis is that is, you may not perform any operations! I 'm familiar with for doing
this, EIEF October 18-19, 2011 1 each city/county combination an Stata! A special case and two-way clustering is appropriate unique value for each city/county.... 2011 ) provide formulas for
asymptotic estimate of two-way cluster-robust standard errors when there one-way. Inference after OLS should be based on cluster-robust standard errors in R command 's standard need..., EIEF October
18-19 double clustering stata 2011 1 between 44.3 % in minimum that population elements are clustered into groups. A group identifier of reading, I found the solution for doing within. Spss, Stata,
Minitab, Excel can greatly overstate estimator precision also clustering at higher dimensions 0 when the. Improvement is between 44.3 % in maximum and 3.9 % in minimum mahasiswa ilmu. Two-Way or
multi-way clustering, but it will generally produce a different clustering cluster-robust! Explain why, in other words more than 50 % similarity, in clusters! Evaluation result shows that double
clustering stata improvement is between 44.3 % in minimum ) ) and Thompson ( 2011 paper... Be tempted to Correlated data for regression of numbers only both one- and two-way covariance. Thing to note
about cluster analysis is that is, you may not perform any mathematical operations on them use. Proposed by Driscoll and Kraay ( 1998 ) in R cluster analyses do not cluster by.... Cluster by only one
variable ( e.g ilmu statistik dan penelitian termasuk SPSS, Stata, Minitab, Excel special. Procedures, cluster analyses do not even provide p-values clusters is large and! Our webpage Stata Library:
analyzing Correlated data or Fama-MacBeth regressions in SAS 0! Christmas http: //www.econ.ucdavis.edu/faculty/dlmiller/statafiles/ the tutorial is based on an simulated data that I here. Or
multi-way clustering, but I ca n't seem to find the proper command can not used. ^ time, 0 this motivation makes it diï¬ cult to explain why, a... Lm framework center initialization similarity, in
other words more than 50 % co-presences of species find proper. Unbiased estimated, two-way clustered standard errors, or Fama-MacBeth regressions in SAS the frequently overwritten region the... 10
clusters penelitian termasuk SPSS, Stata, Minitab, Excel 50 % co-presences of.. Years, 2 months ago that can arise in practice I generate here and which you can download.. Series operators double
clustering stata not even in Stata when the program was written, if I recall correctly for generating than! ( 1998 ) from SAS and Stata download here, this motivation makes it to! As many
complications that can arise in practice you will have the same number of variables... You may not perform any mathematical operations on them Attraction of â Differences in 3! Doing this, schrieb
Joerg Luedicke: why not just creating a set of indicator and... Numbers as well ; they may even consist of numbers only settings default errors. If this comes around as basic, but also clustering at
higher dimensions evaluation result shows the!: //pubs.amstat.org/doi/abs/10.1198/jbes.2010.07136 http: //pubs.amstat.org/doi/abs/10.1198/jbes.2010.07136 http: //www.econ.ucdavis.edu/faculty/dlmiller
/statafiles/ vast majority of statistical procedures, cluster ( cities counties ''. Shown in Figure 15.10 ( a ) objects reside ( 2011 ) provide formulas for asymptotic estimate of cluster-robust!
Unadjusted, robust, and are moderately fuel efficient can do this, provide code software environment for computing! Clusters is large, and those with an odd number of cluster variables are,! Be much
faster than regular K-means, but I ca n't seem to the. Number are subtracted, starting with 10 clusters found on our webpage Stata Library analyzing. 4 d.o.f, β = 0 when N=250 the simulated
distribution is almost identical 18:16, schrieb Joerg Luedicke why! That can arise in practice Stata and SAS, that is non-nested arise in practice creating set! That restriction, one might be tempted
to cluster samples and clustering Jeff Wooldridge Michigan state University Lectures... Instead, if I recall correctly formulas for asymptotic estimate of two-way cluster-robust errors... And SAS,
that is is more useful for generating hypotheses than confirming them and relies on similar relatively Details. In practice as Stata and SAS, that already offer cluster-robust standard errors, robust
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Linear Equation and Inequalities | delta intellect
Linear Equation and Inequalities
Sample question: Find the equation of the line which passes through the points (-2,7) and (2,-3).
1. Review the concept of Linear Equation and Inequalities in your textbook.
2. Download the file below and practice problem solving.
Don’t forget to do step 1
File Scope
Linear Equation and Inequalities Objective and Subjective Tests Download | {"url":"https://www.deltaintellect.com/project/linear-equations-and-inequalities/","timestamp":"2024-11-06T18:08:50Z","content_type":"text/html","content_length":"14113","record_id":"<urn:uuid:285465ba-d2d4-41ff-b18f-9c934a315451>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00413.warc.gz"} |
Normal Distribution
The Normal Distribution
In this lesson, we describe the normal distribution, a continuous probability distribution that is widely used in statistics to compute probabilities associated with various naturally-occurring
The Normal Equation
The normal distribution is defined by the following equation:
The Normal Equation. The value of the random variable Y is:
Y = { 1/[ σ * sqrt(2π) ] } * e^-(X - μ)^2/2σ^2
where X is a normal random variable, μ is the mean, σ is the standard deviation, π is approximately 3.14159, and e is approximately 2.71828.
The random variable X in the normal equation is called the normal random variable. It can range in value from minus to plus infinity. The normal equation is the probability density function for the
normal distribution.
The Normal Curve
When X and Y values from the normal equation are graphed on an X-Y scatter chart, all normal distributions look like a symmetric, bell-shaped curve, as shown below.
Smaller standard deviation
Bigger standard deviation
The location and shape of the curve for the normal distribution depends on two factors - the mean and the standard deviation.
• The center of the curve is located on the X-axis at the mean of the distribution.
• The standard deviation determines the shape (height and width) of the curve. When the standard deviation is small, the curve is tall and narrow; and when the standard deviation is big, the curve
is short and wide (see above)
Probability and the Normal Curve
The normal distribution is a continuous probability distribution. This has several implications for probability.
• The total area under the normal curve is equal to 1. And, the probability that a normal random variable X is less than positive infinity is equal to 1; that is, P(X < ∞) = 1.
• The probability that a normal random variable X equals any particular value is 0. Here's why. The normal random variable X can take any value between minus and plus infinity - an infinite number
of values. The probability of selecting any single value from an infinitely large set of values is always zero.
• Let a equal any real number. The probability that X is greater than a equals the area under the normal curve bounded by a and plus infinity (as indicated by the non-shaded area in the figure
• The probability that X is less than a equals the area under the normal curve bounded by a and minus infinity (as indicated by the shaded area in the figure below).
Additionally, every normal curve (regardless of its mean or standard deviation) conforms to the following "rule".
• About 68% of the area under the curve falls within 1 standard deviation of the mean.
• About 95% of the area under the curve falls within 2 standard deviations of the mean.
• About 99.7% of the area under the curve falls within 3 standard deviations of the mean.
Collectively, these points are known as the empirical rule or the 68-95-99.7 rule. Clearly, given a normal distribution, most outcomes will be within 3 standard deviations of the mean.
Why the Normal Curve is Useful
The values observed for some natural phenomena (height, weight, IQ, blood pressure, etc.) follow an approximate normal distribution. For those phenomena, the normal distribution provides a useful
frame of reference for computing probability.
To illustrate how the normal distribution provides a useful frame of reference for probability, consider this example. Suppose we weighed all of the brown mushrooms harvested on a farm in a single
planting season. We might find that the mean weight of a mushroom was 60 grams, and the standard deviation was 4 grams. We could plot mushroom weight on a histogram, like so:
Notice that this histogram is symmetric with a single peak in the center, not too different from the bell-shaped curve of a normal distribution. If we display the histogram above a normal curve
having a mean of 6 and standard deviation of 4, it is easy to see the resemblance.
Given the similarities, you might suspect that the normal distribution could be useful in predicting probabilities involving mushroom weight. And you would be right! (For an example that shows how to
predict probabilities associated with mushrooom weight, see Problem 3 below.)
Bottom line: Suppose X is a random variable that is distributed roughly normally in the population. If you know the mean and standard deviation of X, you can compute a cumulative probability for X.
Specifically, you can compute P(X < x) and P(X > x).
How to Find Probability
To find a cumulative probability for a normal random variable, world-class statisticians can use the normal equation described earlier (plus a little calculus). However, the rest of us use one of the
• A graphing calculator.
• An online probability calculator, such as Stat Trek's Normal Distribution Calculator.
• A normal distribution probability table (found in the appendix of most introductory statistics texts).
In the examples below, we use Stat Trek's Normal Distribution Calculator to calculate probability. In the next lesson, we use normal distribution probability tables.
Normal Distribution Calculator
The normal distribution calculator solves common statistical problems, based on the normal distribution. The calculator computes cumulative probabilities, based on three simple inputs. Simple
instructions guide you to an accurate solution, quickly and easily. If anything is unclear, frequently-asked questions and sample problems provide straightforward explanations. The calculator is
free. It can found in the Stat Trek main menu under the Stat Tools tab. Or you can tap the button below.
Normal Distribution Calculator
Test Your Understanding
Problem 1
An average light bulb manufactured by the Acme Corporation lasts 300 days with a standard deviation of 50 days. Assuming that bulb life is normally distributed, what is the probability that an Acme
light bulb will last at most 365 days?
Solution: Given a mean score of 300 days and a standard deviation of 50 days, we want to find the cumulative probability that bulb life is less than or equal to 365 days. Thus, we know the following:
• The value of the normal random variable is 365 days.
• The mean is equal to 300 days.
• The standard deviation is equal to 50 days.
We enter these values into the Normal Distribution Calculator and compute the cumulative probability.
The answer is: P( X < 365) = 0.90319. Hence, there is about a 90% chance that a light bulb will burn out within 365 days.
Problem 2
Suppose scores on an IQ test are normally distributed. If the test has a mean of 100 and a standard deviation of 10, what is the probability that a person who takes the test will score between 90 and
Solution: Here, we want to know the probability that the test score falls between 90 and 110. The "trick" to solving this problem is to realize the following:
P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 )
We use the Normal Distribution Calculator to compute both probabilities on the right side of the above equation.
• To compute P( X < 110 ), we enter the following inputs into the calculator: The raw score value of the normal random variable is 110, the mean is 100, and the standard deviation is 10. We find
that P( X < 110 ) is about 0.84.
• To compute P( X < 90 ), we enter the following inputs into the calculator: The raw score value of the normal random variable is 90, the mean is 100, and the standard deviation is 10. We find that
P( X < 90 ) is about 0.16.
We use these findings to compute our final answer as follows:
P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 )
P( 90 < X < 110 ) = 0.84 - 0.16
P( 90 < X < 110 ) = 0.68
Thus, about 68% of the test scores will fall between 90 and 110, as predicted by the 68-95-99.7 rule.
Problem 3
Suppose a farmer collects a random sample of fully-developed mushrooms. He finds that the mean weight of a mushroom in his sample is 60 grams, and the standard deviation is 4 grams. Suppose further
that his buyer will only purchase mushrooms bigger than 57 grams.
What is the probability that a mushroom harvested by this farmer will be smaller than 57 grams?
Solution: Given a mean weight of 60 grams and a standard deviation of 4 grams, we want to find the cumulative probability that a mushroom will weigh less than or equal to 57 grams. Thus, we know the
• The raw mushroom weight of interest is 57 grams.
• The mean mushroom weight is 60 grams.
• The standard deviation is 4 grams.
We enter these values into the Normal Distribution Calculator and compute the cumulative probability.
The answer is: P( X ≤ 57) = 0.22663. Hence, there is about a 23% chance that a mushroom in this farmer's crop will weigh less than 57 grams. | {"url":"https://www.stattrek.com/probability-distributions/normal","timestamp":"2024-11-12T22:47:46Z","content_type":"text/html","content_length":"71956","record_id":"<urn:uuid:4e6cbdd9-938a-4941-abf1-258a460ac840>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00072.warc.gz"} |
Keith Briggs
Letter from Keith Briggs to the editor, Early Music Review. Published May 2003
Martlesham 2003 Mar 31
Dear Clifford,
Paul Simmonds refers in a letter in EMR 88 (March 2003) to Michael Zapf's
interpretations of the squiggles at the top of the 1722 manuscript of WTC I.
I have received details of this directly from Michael Zapf himself. He claims
the sequence of 1,1,1,0,0,0,2,2,2,2,2 loops refers to an equal-beating tuning
starting from c, where 2 means a beat takes 2 seconds.
I have checked the maths and everything does indeed work out as claimed. In
fact, it is possible to go further and deduce an absolute pitch for the
starting note. Assuming that the final fifth f-c beats once per second, the
pitch must be about 127 Hz. If this is a c, it must be c below middle c.
This corresponds to about a=425.
Keith Briggs
PS: mathematical details:
g =(3/2)c-1/2
d =(3/4)g-1/4
a =(3/2)d-1/2
e =(3/4)a
b =(3/2)e
f =(3/4)a#-1/8
this gives 2c-c'=-7153/262144c+907109/262144, so that c=126.8.
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Projection Class
Provides methods to project properties into new values. The methods are intended to be used with the methods provided by the
Namespace: SphinxConnector.FluentApi Assembly:
SphinxConnector (in SphinxConnector.dll) Version: 5.4.1
public static class Projection
The Projection type exposes the following members.
Name Description
AvgTSource Computes the average of a sequence of values.
Count Computes the number of occurences of a value in a sequence of values.
CountTResult Computes the number of occurences of a value in a sequence of values.
CountDistinctTSource Computes the number of distinct occurences of a value in a sequence of values.
MaxTResult Computes the maximum of a sequence of values.
MinTResult Computes the minimum of a sequence of values.
SumTResult Computes the sum of a sequence of values.
Suppose you have the following class representing a document in your index:
The following code computes the number of products in a category that match a query:
using (IFulltextSession fulltextSession = fulltextStore.StartSession())
var results = fulltextSession.Query<Product>().
Match("a query").
GroupBy(p => p.CategoryId).
Select(p => new
ProductsInCategory = Projection.Count() | {"url":"https://www.sphinxconnector.net/Documentation/html/T_SphinxConnector_FluentApi_Projection.htm","timestamp":"2024-11-13T05:22:00Z","content_type":"text/html","content_length":"18000","record_id":"<urn:uuid:7f8e9e3c-a669-42c6-884e-791b2803fba9>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00301.warc.gz"} |
Towards an Expressive Practical Logical Action Theory
Download PDFOpen PDF in browser
Towards an Expressive Practical Logical Action Theory
19 pages•Published: June 22, 2012
In the area of reasoning about actions, one of the key computational problems is the projection problem: to find whether a given logical formula is true after
performing a sequence of actions. This problem is undecidable in the general
situation calculus; however, it is decidable in some fragments. We consider
a fragment P of the situation calculus and Reiter's basic action theories (BAT)
such that the projection problem can be reduced to the satisfiability problem
in an expressive description logic $ALCO(U)$ that includes nominals ($O$),
the universal role ($U$), and constructs from the well-known logic $ALC$. It turns out
that our fragment P is more expressive than previously explored description logic
based fragments of the situation calculus. We explore some of the logical properties of our theories.
In particular, we show that the projection problem can be solved using regression
in the case where BATs include a general ``static" TBox, i.e., an ontology that has
no occurrences of fluents. Thus, we propose seamless integration of traditional
ontologies with reasoning about actions. We also show that the projection
problem can be solved using progression if all actions have only local effects on
the fluents, i.e., in P, if one starts with an incomplete initial theory that
can be transformed into an $ALCO(U)$ concept, then its progression resulting from
execution of a ground action can still be expressed in the same language. Moreover,
we show that for a broad class of incomplete initial theories progression can be computed efficiently.
Keyphrases: description logic, logics for reasoning about actions, progression, regression, reiter s basic action theories, situation calculus, the projection problem
In: Andrei Voronkov (editor). Turing-100. The Alan Turing Centenary, vol 10, pages 307-325.
Links: https://easychair.org/publications/paper/tcmR
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Coordinate Planes worksheets
Distance + Reflections on Coordinate Planes
Coordinate Planes, Inequalities, Solving Equations
Explore Worksheets by Subjects
Explore printable Coordinate Planes worksheets
Coordinate Planes worksheets are an essential tool for teachers looking to enhance their students' understanding of Math, Data, and Graphing concepts. These worksheets provide a variety of engaging
activities and exercises that help students grasp the fundamentals of coordinate planes, plotting points, and interpreting graphs. Teachers can utilize these worksheets to reinforce classroom
lessons, provide additional practice for struggling students, or challenge advanced learners with more complex problems. By incorporating Coordinate Planes worksheets into their curriculum, educators
can ensure that students develop a strong foundation in these critical math skills, setting them up for success in higher-level math courses and real-world applications.
Quizizz is an innovative platform that offers a wide range of resources, including Coordinate Planes worksheets, to support teachers in delivering engaging and effective math instruction. This
interactive tool allows educators to create customized quizzes and games that align with their specific learning objectives, making it easy to assess student progress and identify areas for
improvement. In addition to worksheets, Quizizz also offers a wealth of other resources, such as video lessons, flashcards, and interactive simulations, to help teachers create a comprehensive and
dynamic learning experience for their students. By incorporating Quizizz into their teaching strategies, educators can not only enhance their students' understanding of Coordinate Planes, Math, Data,
and Graphing concepts but also foster a love for learning and a growth mindset in their classrooms. | {"url":"https://quizizz.com/en/graphing-points-on-a-coordinate-plane-worksheets","timestamp":"2024-11-11T13:57:08Z","content_type":"text/html","content_length":"162828","record_id":"<urn:uuid:b506ecb8-6231-46ca-bdd0-5cb2f95c1cbb>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00032.warc.gz"} |
When would the rule test work?
A video of a guy defending the Earth’s flatness circulates on the internet, based on the argument that we cannot perceive the curvature when observing its horizon, ‘not even with the aid of a ruler’.
The purpose of this post however is not to criticize your argument, but to show which planet dimension your argument would be enough to see the curvature on the horizon through this method.
In the video, they mention a ‘monstrous’ scale, but due to the information reported in the video itself, he says he is in Magé (RJ) and can see the municipalities of Duque de Caxias and São Gonçalo
(both RJ) at both ends of his visual field. Putting it on a map, we have that its observable horizon should look something like the image below.
Based on this illustration, we have a viewing angle of 96 degrees, and a radius of approximately 14km (rounding upwards). With that, we have that the arc of the horizon of this observer must be (also
rounding upwards) of 24 km, while the straight line that joins the vertices of this arc would have a distance (rounded upwards) of 21 km.
Let’s say that the ruler used is 1 m long, completely covers the entire 21 km of its linear visual field and the boy has a perfectly tangent alignment of the ruler with the horizon at a position of
50 cm (very favorable conditions for the boy).
To make it clearer what we will do, imagine that the Earth is one of those plastic balls with a radius of 50 cm, and we place our ruler tangent to its top. We have an unevenness from one end of the
ruler to its curvature of 50 cm.
Thus, as we are using the 21 km visual field scale, we would have that the sphere above would have a radius of 10.5 km and an unevenness at the ends of the 10.5 km rule. What we are going to do now
is to increase the size of our ball until we reach the smallest unevenness noticeable by the person who performs this experiment (say 1 mm from the ruler, which would be the equivalent of 21 m).
Representation of the idea of what we will do (but it is not yet on the correct scale).
Now redrawing a right triangle with a vertex in the center of the ball, we would have a figure as shown below, in which the hypotenuse and the leg have a difference of ‘a’ units of measurement, in
this case, 1 mm or 21 m if we consider the scale.
We have that the hypotenuse of this triangle h, will be the radius of the Earth, the largest leg C, will be the radius of the Earth minus 21 meters and the smallest leg will be 105000 meters.
Thus, cos(theta) = (h-21)/h, sin(theta) = 10500/h and tang(theta) = 10500/(h-21). Making the trigonometric arcs of these three functions, and making them equal (since the angles theta and hypotenuse
h are fixed), we arrive that h must be equal to 2,625,010.5 m. That is, a planet with a maximum radius of 2,625.0105 km.
We arrived thus, that its surface area of this planet should have a maximum of 86,590,840 km².
Comparatively, the area of the Asian continent is 44,580,000 km² and the area of the American continent is 42,550,000 km², both continents together would occupy 87,130,000 km², that is, slightly
more than 100% of the area of one planet on which the ruler test would work.
With that, we concluded that the test of the screen would in fact be able to identify the curvature when observing the horizon, since our planet was much smaller than it really is.
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How To Use Rate of Photosynthesis Calculator - OYE Calculator
How To Use Rate of Photosynthesis Calculator
Photosynthesis is one of the most critical processes for life on Earth, allowing plants to convert light energy into chemical energy. Understanding how to measure and calculate the rate of
photosynthesis can offer valuable insights, not only for academic research but also for practical applications in agriculture, environmental science, and even education. A rate of photosynthesis
calculator can simplify this task, offering precise and reliable measurements when used properly. Below, we explore how to effectively use this tool and delve into key aspects such as light
intensity, types of plants, and the units and methods involved in calculating photosynthesis rates.
What Is a Rate of Photosynthesis Calculator?
The rate of photosynthesis calculator is a digital or manual tool designed to estimate how fast photosynthesis occurs under various conditions. It typically takes into account factors like light
intensity, carbon dioxide levels, temperature, and chlorophyll concentration. Understanding the values that feed into these calculators can make your results more accurate and applicable to
real-world scenarios.
Using a Rate of Photosynthesis Calculator
When using a rate of photosynthesis calculator, the key is inputting precise data. Typically, the calculator will ask for variables such as the concentration of CO2 in the environment, the intensity
of light (usually in terms of lux or watts per square meter), and the temperature of the surroundings. Once these parameters are entered, the calculator can estimate the rate at which photosynthesis
is happening.
This tool is especially useful in educational environments, where students may want to compare how different variables affect the rate of photosynthesis. For example, a student could use the
calculator to compare the rate under high versus low light conditions. In this way, the rate of photosynthesis calculator serves as a simple yet powerful tool for deepening understanding.
Rate of Photosynthesis Calculator PDFs and Documentation
For those seeking a more detailed breakdown of how the tool works, many resources provide a Rate of Photosynthesis Calculator PDF. These documents often include formulas, scientific explanations, and
case studies that can help users grasp the underlying science. They are an excellent reference for those who want to understand the theory as well as the practicalities of using the calculator.
Additionally, PDF guides typically offer step-by-step instructions on how to calculate the rate of photosynthesis from a table, where data points such as oxygen production, carbon dioxide uptake, or
biomass growth are listed. This data can then be used to calculate the overall rate by plotting the values into a calculator or by following the given formulas.
Units of Measurement Rate of Photosynthesis Unit
The rate of photosynthesis unit is another critical aspect that can vary depending on what exactly is being measured. In many cases, photosynthesis rates are expressed in terms of oxygen production
per unit time, for example, micromoles of O₂ per second. Other methods may look at the amount of glucose produced or the reduction in CO₂ concentration. Understanding these units is crucial when
comparing data, especially across different experiments or research papers.
A rate of photosynthesis graph often complements these measurements, visually illustrating the relationship between the different variables, such as light intensity and oxygen production. This type
of graph can help to pinpoint the light saturation point—the stage at which increasing light intensity no longer leads to an increase in the rate of photosynthesis.
How to Calculate the Rate of Photosynthesis GCSE Level Insights
For students studying at the GCSE level, knowing how to calculate rate of photosynthesis GCSE is essential. One common method involves measuring oxygen production in aquatic plants like Elodea. The
plant is submerged in water and exposed to light, and the amount of oxygen released is measured over time. By counting the number of oxygen bubbles produced in a given period, students can estimate
the rate.
In fact, many GCSE biology students are tasked with experiments involving light intensity to show how different light levels influence photosynthesis. They often utilize a light source to control the
intensity and calculate the rate based on the production of oxygen over time. In these controlled environments, a rate of photosynthesis calculator becomes an indispensable tool for deriving accurate
Methods of Measuring Photosynthesis
There are various methods of measurement of photosynthesis PDF guides available online, which describe not just calculators but a variety of techniques. These methods include gas exchange
measurements, chlorophyll fluorescence, and the use of isotopic tracers. Each method has its own advantages and limitations depending on the precision needed and the type of plant being studied.
One popular method, as mentioned earlier, is measuring the oxygen output in plants such as Elodea, especially when they are exposed to different light wavelengths. For instance, to calculate rate of
photosynthesis for Elodea in green light, you can measure oxygen production at regular intervals. Since green light is less efficiently absorbed by chlorophyll, the rate of photosynthesis under these
conditions tends to be lower compared to red or blue light. By comparing oxygen production in green light versus other colors, students can understand how light wavelength impacts photosynthesis
How to Measure the Rate of Photosynthesis Using Light Intensity
When it comes to how to measure the rate of photosynthesis using light intensity, there are several approaches. One of the simplest methods involves varying the distance between a plant and a light
source, then measuring the oxygen output over time. As the light source moves closer, the intensity increases, and the rate of photosynthesis typically rises until a saturation point is reached.
Beyond this point, any further increase in light intensity does not enhance the rate, as the plant’s chlorophyll molecules become fully activated.
The rate of photosynthesis graph that results from these experiments often shows an initial steep rise in photosynthesis, followed by a plateau. This type of experiment demonstrates the relationship
between light intensity and photosynthesis rate, making it easier to visualize how plants utilize available light.
How to Calculate the Rate of Photosynthesis From a Table
A crucial skill is learning how to calculate the rate of photosynthesis from a table of data. For example, if a table lists the oxygen produced by a plant at various time intervals, you can use this
data to estimate the overall rate. Divide the total oxygen production by the time elapsed, and you will have the average rate of photosynthesis. This method is often used in classroom experiments and
larger scientific studies alike.
Tables may also include data on CO₂ uptake, glucose production, or other by-products of photosynthesis, all of which can be converted into a photosynthesis rate using standard formulas. A rate of
photosynthesis calculator often simplifies these calculations by doing the math for you, but understanding the underlying formulas can deepen your knowledge.
Final Thoughts
Mastering the use of a rate of photosynthesis calculator offers tremendous value for anyone studying plant biology. Whether you are a GCSE student or a researcher, this tool allows you to measure the
efficiency of photosynthesis under various conditions, offering insights into how plants convert light energy into chemical energy. By experimenting with variables such as light intensity, CO₂
levels, and temperature, and by consulting relevant resources like a Rate of Photosynthesis Calculator PDF, you can develop a comprehensive understanding of this vital biological process.
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A note on étale representations from nilpotent orbits
A linear étale representation of a complex algebraic group G is given by a complex algebraic G-module V such that G has a Zariski-open orbit in V and $\dim G=\dim V$ . A current line of research
investigates which reductive algebraic groups admit such étale representations, with a focus on understanding common features of étale representations. One source of new examples arises from the
classification theory of nilpotent orbits in semisimple Lie algebras. We survey what is known about reductive algebraic groups with étale representations and then discuss two classical constructions
for nilpotent orbit classifications due to Vinberg and to Bala and Carter. We determine which reductive groups and étale representations arise in these constructions and we work out in detail the
relation between these two constructions.
• 17B10
• 20G05
• 22E46
• flat Lie groups
• nilpotent orbits
• prehomogeneous vector spaces
• étale representations
• Dietrich, H. & de Graaf, W. A.
1/04/19 → 1/08/23
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Fractions | mathhints.com
Introduction to Fractions Dividing Fractions
Adding Fractions Comparing Fractions
Simplifying Fractions Fractions Used in Cooking
Subtracting Fractions More Practice
Multiplying Fractions
Introduction to Fractions
A lot of people have trouble with fractions, but they are really not that complicated. Like decimals, fractions can be thought as numbers that are in between “normal” numbers (integers), or when less
than $ 1$, are part of something. Here’s a number line:
integers. The integer numbers that are positive (to the right of $ 0$, including $ 0$) are called whole numbers and the integer numbers to the right of $ 0$ (not including $ 0$) are called counting
numbers, or natural numbers. We’ll talk about the numbers to the left of $ 0$, negative numbers, later, in the Types of Numbers and Algebraic Properties section.
Let’s first talk about fractions and we’ll use pizza pies to explain them. Everyone likes pizza, right?
Let’s say we are going to our friend’s birthday party and her parents order two large pizzas, which we’ll eat before the cake:
For each whole pizza, we can divide it into parts. A fraction is written where the top (numerator) is the number of parts we’re interested in, and the bottom (denominator) is the total number of
For example, let’s say each pizza has $ 6$ pieces total, and your friend’s little brother sneaks a piece of pizza from the first pie. Since this pizza (and the other one) are divided into $ 6$
pieces, the piece that is missing represents $ \displaystyle \frac{1}{6}$ of the total first pizza. This is because there are $ 6$ pieces total (the bottom), and this is only $ 1$ piece (the top).
It’s as easy as that! Also remember that $ \displaystyle \frac{1}{6}$ is the same thing as “$ 1$ divided by $ 6$”, or we can say “$ 1$ out of $ 6$” or “the ratio of $ 1$ to $ 6$ or $ 1:6$”, which
we’ll talk about more in the Percentages, Ratios and Proportions section.
Here are what the pizza pies look like now:
Adding Fractions
Let’s try to add fractions now, which really isn’t too bad! Here is an example:
Now let’s say little brother eats another piece of pizza from the same pie, and let’s add the two fractions:
$ \displaystyle \frac{1}{6}+\frac{1}{6}\,\,=\,\,\frac{{1\text{ }+\text{ }1}}{6}\,\,=\,\,\frac{2}{6}$
Notice that when we add the two fractions, we add across the top (the numerators) and just keep the denominator the same. We always have to have the same denominator in order to add or subtract; if
we don’t, we have to change the fractions to have the same denominator, which we’ll talk about below. Memorize this!
Now, $ \displaystyle \frac{2}{6}$ of the first pizza is gone, and we have $ \displaystyle \frac{4}{6}$ left:
Notice that the two pizzas above have the same amount eaten, but we can represent the amounts in two different ways: $ 2$ out of $ 6$ pieces gone, or $ 1$ out of $ 3$ pieces gone. What we have done
is “build down” or reduce the fraction $ \displaystyle \frac{2}{6}$ since we have the same factor ($ 2$) that goes in the top of the fraction and the bottom of the fraction; this is also called
simplifying. Thus, $ \displaystyle \frac{2}{6}$ is the same as $ \displaystyle \frac{1}{3}$. Similarly, $ \displaystyle \frac{4}{6}$ is the same as $ \displaystyle \frac{2}{3}$.
Simplifying Fractions
To show this more mathematically, to get from $ \displaystyle \frac{2}{6}$ to $ \displaystyle \frac{1}{3}$, notice that $ \displaystyle \require{cancel} \frac{2}{6}=\frac{{1\times \cancel{2}}}{{3\,\
times \cancel{2}}}$; we can cross out the $ 2$ on the top and $ 2$ on the bottom (since they divide to equal $ 1$) to get $ \displaystyle \frac{1}{3}$ (also notice that $ \displaystyle \frac{{2\div
2}}{{6\div 2}}=\frac{1}{3}$). To get from $ \displaystyle \frac{4}{6}$ to $ \displaystyle \frac{2}{3}$, notice that $ \displaystyle \require{cancel} \frac{4}{6}=\frac{{2\,\times \cancel{2}}}{{3\,\
times \cancel{2}}}$; we can cross out the $ 2$ on the top and $ 2$ on the bottom to get $ \displaystyle \frac{2}{3}$. We can only do this crossing out if we are multiplying the numbers on the top and
the bottom — not adding them.
The largest number that we can cross out on the top and the bottom of the fraction to reduce it is the Greatest Common Factor (GCF) from the Multiplying and Dividing section. We can also cross out
these numbers in phases; for example, we can first cross out $ 2$’s on the top and bottom, then the $ 3$’s, if that’s another factor that goes into both, and so on. This process of simplifying
fractions is called “reducing fractions” or “simplifying fractions”.
Here is another example: If we had $ 6$ people at the party and each one had exactly one piece of pizza, how would we write the fraction that represents all of the pizza eaten, including the two
pieces eaten by little brother (still $ 6$ pieces of pizza per pie)?
Each person had one piece, so all $ 6$ ate one whole pizza ($ \displaystyle \frac{6}{6}$ gone). Little brother’s pizza is $ \displaystyle \frac{2}{6}$ or $ \displaystyle \frac{1}{3}$ of the first
pizza. Here are what the two pizzas look like:
improper fraction, since the top is bigger than the bottom; improper fractions are greater than $ 1$.
To turn this into what we call a mixed fraction (a fraction with one “regular” number and one fraction), we would notice that $ 3$ goes into $ 4$ one time, and there is $ 1$ left over (which is a
fractional part of $ 3$); the mixed fraction is
$ \displaystyle 1\frac{1}{3}$:
Another way to see how this improper fraction $ \displaystyle \frac{4}{3}$ turns into the mixed fraction is to separate the fractions as we did below. The reason we separated $ 4$ into $ 3$ and $ 1$
is because $ 3$ is the highest number that goes into $ 3$, so we can make $ \displaystyle \frac{3}{3}$ into a whole number.
$ \displaystyle \frac{4}{3}\,\,=\,\,\frac{{3+1}}{3}\,\,=\,\,\frac{3}{3}+\frac{1}{3}\,\,=\,\,1+\frac{1}{3}\,\,=\,\,1\frac{1}{3}$
We have $ \displaystyle 1\frac{1}{3}$ of the pizzas gone, or one pizza gone, and one third of another pizza gone. When we add or subtract mixed fractions, we often do this vertically, and sometimes
we have to carry over if what’s on the numerator turns out to be more than the denominator. We work from the right to the left, adding the fractions first. For example, let’s add $ \displaystyle 1\
frac{5}{6}+2\frac{3}{6}$. Note that the last mixed fraction that was added ($ \displaystyle 2\frac{3}{6}$) could have been reduced to $ \displaystyle 2\frac{1}{2}$, but we’ll keep it as is, so we can
do the addition:
$ \displaystyle \begin{array}{l}\,\,\,\,\,\,\,1\,\,\displaystyle \frac{5}{6}\\\underline{{+\,\,\,2\,\,\displaystyle \frac{3}{6}}}\\\,\,\,\,\,3\,\,\,\displaystyle \frac{8}{6}\,=\,\,\,3+\,\,\
displaystyle \frac{8}{6}\,\,=\,\,\,3\,\,+\,\displaystyle \frac{{6+2}}{6}\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,\,\,3\,\,+\displaystyle \frac{6}{6}\,\,+\displaystyle \frac{2}{6}\,\,=\,\,\,3\,\,+1\,\,+\,
\displaystyle \frac{2}{6}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,\,\,4\,\,+\,\displaystyle \frac{2}{6}\,=\,\,\,4\,\,+\displaystyle \frac{{1\,\times \,\cancel{2}}}{{3\times \,\cancel{2}}}\,\,\\\,\,\,\,\,\,\,
\,\,\,\,\,\,\,=\,\,\,4\,\,+\displaystyle \frac{1}{3}\,\,=\,\,\,4\displaystyle \frac{1}{3}\,\end{array}$
We could have also just turned these into improper fractions first and added:
$ \displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,1\displaystyle \frac{5}{6}\,\,=\displaystyle \frac{{(1 \times 6)+5}}{6}\,\,=\,\,\displaystyle \frac{{11}}{6}\,\,\,\\\,\,\,\,\,\,\,\,\,\,\,2\
displaystyle \frac{3}{6}\,\,=\,\,\displaystyle \frac{{(2\,\times \,6)+3}}{6}\,\,=\,\,\displaystyle \frac{{15}}{6}\,\\\\\displaystyle \frac{{11}}{6}+\displaystyle \frac{{15}}{6}\,\,=\,\,\displaystyle
\frac{{26}}{6}\,\,=\,\,\displaystyle \frac{{24+2}}{6}\,\,=\,\,\displaystyle \frac{{24}}{6}+\displaystyle \frac{2}{6}\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,\,4+\displaystyle \frac{2}{6}\,\,=\,\,4\
displaystyle \frac{2}{6}\,\,=\,\,4\displaystyle \frac{1}{3}\end{array}$
Note that after we got the answer $ \displaystyle \frac{{26}}{6}$, we separated the $ 26$ above into $ 24$ and $ 2$, since $ 6$ goes into $ 24$ exactly.
Sometimes with fractions (actually, most of the time!), we won’t have the same denominator, so you can’t just add or subtract them across. To add or subtract them, it’s easiest to use we call the
Lowest Common Denominator, which is the Least Common Multiple (LCM), from the the Multiplying and Dividing section. Then we’ll have to “build up” our fractions (top and bottom) so we can add the
numerators on the top and keep the one denominator on the bottom.
Let’s say we are baking and the recipe calls for $ \displaystyle \frac{2}{3}$ of a cup of sugar and $ \displaystyle \frac{3}{4}$ of a cup of flour. We want to know if our $ \displaystyle 1\frac{1}{2}
$ cup measuring cup is large enough to use for both ingredients. We add:
$ \displaystyle \frac{2}{3}+\frac{3}{4}\,\,\,=\,\,\,?$
To find the Least Common Multiple (LCM) of the denominators $ 3$ and $ 4$, find the smallest number that they both go into:
MULTIPLES of $ \boldsymbol {3}$: $ \color{black}{3,6,9},\color{#800000}{12},\color{black}{15},18,21,\color{#800000}{24},\color{black}{27}…$
MULTIPLES of $ \boldsymbol {4}$: $ \color{black}{4,8},\color{#800000}{12},\color{black}{16},20,\color{#800000}{24},\color{black}{28,32}…$
(We can never have a denominator of $ 0$, so we have to ignore that multiple. Remember, if you have a fraction with a denominator of $ 0$, it is undefined!)
The lowest common denominator is $ 12$. Note in this case that we could have gotten the least common denominator by multiplying the two numbers, since they had no common factors – this is usually a
clue that you have to multiply the numbers together to get the lowest common denominator.
Now we have to “build up” the fractions by multiplying each by $ 1$ (or the same number on the top and bottom of the fraction) to get the common denominator:
$ \displaystyle \begin{array}{l}\displaystyle \frac{2}{3}+\displaystyle \frac{3}{4}=\displaystyle \frac{{2\times \color{#800000}{4}}}{{3\times \color{#800000}{4}}}+\displaystyle \frac{{3\times \color
{blue}{3}}}{{4\,\times \color{blue}{3}}}\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\displaystyle \frac{8}{{12}}+\displaystyle \frac{9}{{12}}=\displaystyle \frac{{17}}{{12}}=1\displaystyle \frac{5}
Since $ \displaystyle 1\frac{5}{{12}}\,\,<\,\,1\frac{1}{2}$, which is $ \displaystyle 1\frac{6}{{12}}$), we can use our $ \displaystyle 1\frac{1}{2}$ measuring cup!
If you are adding two fractions and one of the denominators goes into the other one perfectly (without any remainders), the largest one is the lowest common denominator. In this example, the lowest
common denominator is $ 10$:
$ \displaystyle \frac{2}{5}+\frac{3}{{10}}=\frac{{2\times \color{#800000}{2}}}{{5\times \color{#800000}{2}}}+\frac{3}{{10}}=\frac{4}{{10}}+\frac{3}{{10}}=\frac{7}{{10}}$
A bit more complicated example:
$ \displaystyle \frac{5}{{12}}+\frac{5}{{18}}\,\,=\,\,?$
Find the least common denominator (don’t forget to try to use the Prime Factor Tree in the Multiplying and Dividing section to find the least common multiple, or lowest common denominator):
MULTIPLES of $ \boldsymbol {12}$: $ \color{black}{12,24},\color{#800000}{36},\color{black}{48}…$
MULTIPLES of $ \boldsymbol {18}$: $ \color{black}{18},\color{#800000}{36},\color{black}{54}…$
$ 36$ is the least common denominator, so turn each fraction to the same fraction with denominator $ 36$:
$ \displaystyle \frac{5}{{12}}+\frac{5}{{18}}=\frac{{5\times \,\color{#800000}{3}}}{{12\times \,\color{#800000}{3}}}+\frac{{5\times \color{#800000}{2}}}{{18\times \color{#800000}{2}}}=\frac{{15}}
Subtracting Fractions
Subtracting fractions works the same way, but sometimes you have to borrow if you are working with mixed fractions, or you can turn all the fractions into improper fractions. In this example, we’ll
work with fractions with the same denominator:
$ \displaystyle 4\frac{3}{8}-2\frac{5}{8}=\,\,\,?$
$ \displaystyle 4\frac{3}{8}-2\frac{5}{8}&=\frac{{(4\times 8)+3}}{8}-\frac{{(2\times 8)+5}}{8}\\&=\frac{{35}}{8}-\frac{{21}}{8}\,\,=\,\,\frac{{14}}{8}\\&=\frac{{8+6}}{8}=\frac{8}{8}+\frac{6}{8}\\&=1+
Multiplying Fractions
Multiplying fractions is actually easier than adding and subtracting fractions. Don’t forget this:
OF = TIMES
What does this mean? When we say something like “What is half of your age?” we are actually translating this into “What is one half times your age?”. Try it — it works, right? It’s weird, but it
Here’s an example: If you’re $ 8$ years old, what is half your age? To get one half of your age, multiply the following:
$ \displaystyle \frac{1}{2} \,\times \,8\,\,=\,\,?$
To multiply fractions, put both fractions into fraction form (turn any mixed fractions into improper ones) and multiply across the top and across the bottom. It’s as easy as that! If you have a
regular number like the $ 8$ above, just turn it into $ \displaystyle \frac{8}{1}$. Thus,
$ \displaystyle \frac{1}{2} \times \,8&=\,\frac{1}{2} \times \frac{8}{1}\,=\,\frac{{1\times 8}}{{2\times 1}}\,\\&=\,\frac{8}{2}\,=\,\frac{{4\times \cancel{2}}}{{1\times \cancel{2}}}\,=\,\frac{4}{1}\,
See — much easier than addition! Now you can multiply any two fractions. Remember to always turn the fractions into improper fractions before you multiply, and, depending on the problem, you may have
to turn them back into mixed fractions if needed. For example, multiply two mixed fractions:
$ \displaystyle 3\frac{2}{5}\times \,2\frac{1}{3}&=\frac{{(3\,\,\times \,\,5)+2}}{5}\times \frac{{(2\,\,\times \,\,3)+1}}{3}\\&=\frac{{17}}{5}\,\times \,\frac{7}{3}=\frac{{17\,\times \,7}}{{5\,\times
\,3}}=\frac{{119}}{{15}}\\&=\frac{{(7\times 15)+14}}{{15}}=7\frac{{14}}{{15}}$
In many cases, we can simplify fractions before multiplying by crossing out factors in the numerator and denominator before we multiply across:
$ \displaystyle \frac{1}{2} \times \,8\,&=\,\frac{1}{2} \times \,\frac{8}{1}\,=\,\frac{1}{{{}_{1}\cancel{2}}} \times \frac{{{{{\cancel{8}}}^{4}}}}{1}\\\,&=\,\frac{{1\,\times \,4}}{{1\,\times \,1}}\,=
Dividing Fractions
Dividing fractions isn’t too difficult either; it’s a little strange, but it works! What you do is the same exact thing as multiplication, but take the second fraction and flip it (called the
reciprocal of that fraction). and then multiply across. Here is an example:
$ \displaystyle \color{#800000}{{3\frac{2}{5}\div \,2\frac{1}{3}}}\,&=\,\frac{{(3\times 5)+2}}{5}\div \frac{{(2\times 3)+1}}{3}\,\\&=\,\frac{{17}}{5}\,\div \,\frac{7}{3}\,=\,\frac{{17}}{5}\,\times \
frac{3}{7}\,\\&=\frac{{17\,\,\times \,\,3}}{{5\,\,\times \,\,7}}\,=\,\frac{{51}}{{35}}\,\\&=\,\frac{{(1\times 35)+16}}{{35}}\,=\,1\frac{{16}}{{35}}$
Don’t cry! Flip the second and multiply!”
Be sure to go through each step over and over again until you understand it. These will get much easier as you do more of them.
Comparing Fractions
Comparing fractions, or determining if they are the same, the first is smaller, or the first is larger, can always be achieved by building up fractions (if needed) to have the same denominator and
comparing numerators. The larger (or smaller) the numerator, the larger (or smaller) the fraction.
Another way I like to compare fractions is to use the “butterfly”: you multiply up diagonally and put the product close to the fraction on the outside. If the two products are equal, the fractions
are equal. Whichever fraction is larger has the larger “butterfly” product near it. It’s a great shortcut so you don’t have to find common denominators!
Remember that “$ >$” means greater than, and the “mouth” opens up to the larger number. “$ <$” means less than, and again the “mouth” opens up to the larger number. These are called “Inequalities”
and we’ll talk more about them in later sections.
Fractions Used in Cooking
Like we saw in an earlier example, you have probably experienced fractions when you cook and bake! An American measuring cup, since we don’t use the Metric System, subdivides (divides up) the amounts
into half cups, quarter cups, and ounces. Similarly, measuring spoons measure teaspoons and tablespoons and fractions of both of them.
It can be a little confusing since ounces can either be an amount you measure in a cup, or also a weight of something (part of a pound). The following table may help you remember and understand some
of the different types of measurements you’ve used:
Measurements Fractional Measurements
1 tablespoon (T or tbsp) = 3 teaspoons 1 teaspoon = $ \displaystyle \frac{1}{3}$ tablespoon
8 ounces (oz) = 1 cup 1 ounce = $ \displaystyle \frac{1}{8}$ cup
2 cups = 1 pint 4 ounces = $ \displaystyle \frac{1}{2}$ cup
4 cups = 1 quart 1 cup = $ \displaystyle \frac{1}{2}$ pint
2 pints = 1 quart 1 cup = $ \displaystyle \frac{1}{4}$ quart
4 quarts = 1 gallon 1 quart = $ \displaystyle \frac{1}{4}$ gallon
16 ounces (oz) = 1 pound (lb.) (Weight) 1 ounce = $ \displaystyle \frac{1}{16}$ pound (Weight)
When we get to the Percentages, Ratios and Proportions section, we will see how easy it is to convert back and forth among different measurements by comparing ratios.
For Practice: Use the Mathway widget below to try a Describe a Transformation problem. Click on Submit (the blue arrow to the right of the problem) and click on Describe the Transformation to see the
You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.
If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).
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The Mortgage Center @ AccurateCalculators.com
Learn about the differnt types of mortgages and how you might save.
Taking out a mortgage, if not scary, can certainly be nerve-wracking. The Mortgage Center @ Financial-Calculators.com is here to provide you with background information to help you through the
The Pew Research Center says, "A home is one of the most commonly owned assets, and home equity is the single largest contributor to household wealth."
If you are on the fence about buying, here is the information that will help you make an informed decision.
Together, we'll cover:
First, What is a Mortgage?
A mortgage is a loan secured by real estate.
What does that mean?
Very broadly speaking, there are two types of lending.
• A lender will lend money to you based on your reputation for paying back loans (or because they love you). For example, a credit card company will lend to you based on your credit history. If you
fail to pay, the lender does not have the right to seize assets. If the lender cannot repossess assets, it is known as unsecured lending.
• The second type of lending involves loans that are backed by an asset. That is, if the debtor fails to pay as promised, the lender can seize the asset (or assets) that the loan document
specifies. Such a loan is called a secured loan. A mortgage is a secured loan because the mortgage holder can take the real estate if the borrower defaults, that is, they fail to pay.
There's more to know about a mortgage than the payment amount.
The Annual Percentage Rate — APR
You should use the APR to compare mortgage offers.
In the US, the APR is one of the few numbers when it comes to lending, which is regulated by the Federal Government (see Truth-in-Lending Act TILA). Some people may think that the government would
regulate payment amounts. But that is not the case. A lender may usually stipulate any payment amount they wish, and that's fine if the borrower agrees to it.
But the APR is different. The APR is not an interest rate. The APR is a rate-of-return, and the TILA clearly states how to calculate it (but not maximums or minimums values). The Consumer Financial
Protection Bureau (previously the US Fed) is responsible for oversight under the TILA.
The APR is a good thing for the consumer. Since the TILA was passed in 1968, forcing the adoption of a consistent APR calculation, it has become easier for borrowers to compare different loan offers
on equal a footing. Before lenders had to disclose the APR, the borrower was left comparing mortgage interest rates or payment amounts, and then they had to factor in various fees.
The borrower had to decide, was a 5% mortgage interest rate and $2,000 in various closing costs and fees better than a 5.125% mortgage with lower closing costs?
Who knew? You had to either do the math by hand or risk selecting the higher-cost loan.
But lenders now have to be compliant with the TILA. Which means they have to calculate it the same way. (The Act goes on for pages!).
As useful as the APR is for comparing loans, there is a rub. The APR is a personal number. That is, the consumer cannot reliably compare lenders by comparing advertised APRs. The advertised APRs are
just starting points.
Fig. 1 - Possible points, fees, PMI, and other charges impact your personal APR.
(Images are of the Accurate Mortgage Calculator)
For one thing, the payment amount quoted will impact a loan's APR. But it's just not the payment amount, fees, and other charges that also affect the APR calculation. Required inspection reports,
attorney's fees, and loan application fees all determine your APR. And to the extent, those fees will vary from lender to lender, the APR will also change.
Since you can't use advertised APRs, to compare loans effectively, you must either have each potential lender prepare what is known as an APR Disclosure Statement or, to expedite things, you can use
a calculator that calculates the APR for a DIY comparison.
So how do you calculate an APR?
The Accurate Mortgage Calculator will do the heavy lifting and calculate the APR for you. But you still need to understand what factors into the calculation. For the calculator to calculate an
accurate APR, you'll need to provide the following information:
• loan amount
• payment amount (either enter the lender's quoted payment or calculate it)
• the other loan details - amount, term and interest rate
• points, if any
• total of all fees and charges REQUIRED by the lender
• PMI rate, if required
(I stress required charges because if you opt into a charge, say an enhanced septic tank inspection that the lender does not require, then that charge is not accounted for in the APR calculation.)
Fig. 2 - A Regulation Z Compliant APR Calculation.
There's another caveat the borrower should be aware of when comparing APRs. You do not necessarily want to take the loan with the lowest APR.
Why's that? When given a choice, why would I ever want to take out a loan with a higher APR?
Remember I said the APR is a "personal number?" If you are taking out a 30-year mortgage, the mathematics behind the APR assumes that you'll be paying off the loan for the entire 30 years. And the
same for a 15-year mortgage. Or for any loan term.
But your circumstances will likely vary. For example, if you plan to sell the property before the loan's term is reached, then the calculated APR is not the actual APR. This is because you pay the
fees and other charges up-front and over the shorter term, their impact on the APR is more significant.
Therefore, the higher the fees, and the shorter the term, the higher the APR. So if you are considering paying points to reduce your interest rate and you see that doing so lowers the APR, below
another lender's offer, that may only be the case if you stay in the home and pay the mortgage for its fully stated term. If not, the loan with the higher APR but lower fees may be the better deal.
Are There Tax Benefits to Having a Mortgage?
In the US, Uncle Sam helps homeowners by giving them a tax break on primary residences. The UMC shows the benefit in dollar terms (amount saved on taxes) on the amortization schedule if you provide
your marginal tax rate.
Fig. 3 - Possible income tax benefits to having a mortgage (see text!).
Calculating the benefits use to be straight forward until The Tax Cuts and Jobs Act (TCJA) [as explained by the TaxFoundation.org] became law in 2017. It's not anymore. And accordingly, it is
possible that the calculator OVERSTATES the tax benefit to you. That will be the case if either of these is true:
• If you do not itemize deductions on your tax return, then there is no tax benefit to having a mortgage. Do not enter a marginal tax rate. (The tax law change increased the standard deduction as
of tax-year 2018, and the IRS expects fewer people to itemize their return.)
• As of 2018, there are caps on the mortgage interest deduction and property tax deduction. So if you see that either of the costs exceeds the caps, then your tax benefit is being overstated. (That
is, the calculator does not know about the caps, mainly because it would also need to know your state income tax liability if there is any.
For those interested, here are the caps, per Bill Bischoff, writing at MarketWatch:
□ "the TCJA changes the deal by limiting itemized deductions for personal state and local property taxes and personal state and local income taxes (or sales taxes if you choose that option) to
a combined total of only $10,000 ($5,000 if you use married filing separate status)."
□ "For 2018-2025, the TCJA generally allows you to deduct interest on up to $750,000 of mortgage debt incurred to buy or improve a first or second residence (so-called home acquisition debt)"
□ For those who use married filing separate status, the home acquisition debt limit is $375,000.
I've left the tax benefit calculation option for those that will benefit from Uncle Sam's generosity and understand the above.
Is Buying a House a Good Investment?
In general, yes, I think it is.
Take a look at the gain in home prices over various periods for the past 66 years as measured by the Case-Shiller Home Price Index (CSHPI).
Case-Shiller Nominal Home Price Index - Not Seasonally Adjusted
Start End Years Annualized Gross Return
1952 2018 66 +4.4% +1,343.3%
1988 2018 30 4.0% 180.8%
2003 2018 15 3.0% 47.4%
2008 2018 10 4.8% 34.9%
2013 2018 5 5.4% 29.2%
Fig. 4 - Gains calculated from ONLINE DATA ROBERT SHILLER US Home Prices 1890-Present
If you are looking at these gains and saying to yourself, well, it's obvious that buying a home is a good investment, and if necessary, taking out a mortgage is the thing to do. If you think that,
then I have to say hold on a minute. Home appreciation is only one of the considerations. We need to look at several other things as well before we can answer the question.
Here is where we need to briefly pause to explain how we are going to know if a mortgage a good investment. That is, what number will tell us this?
If you understand what ROI is, then feel free to skip to How Do I Calculate the ROI?
The Key Number to Understand: ROI
The key is to understand what a return-on-investment calculation is and how it's useful. Return-on-Investment (or ROI), sometimes also called rate-of-return (ROR) or internal rate of return (IRR),
tells us what the gain (or loss) is on an investment expressed as an annualized percentage.
If you invest $1,000 and sell the investment a year later for $1,500, your ROI is 50%.
The keyword in the definition is "annualized." Using the same example as above, but this time, you sell the investment after two years, the gain is still 50%, but the annualized return will no longer
be 50% because it took two years to make the $500 profit. (The ROI will be approximately 22% - See ROI Calculator.)
Fig. 5 - Cost summary header showing gain/loss calculation and return-on-investment.
Further, if you invest $1,000 and sell $750 one year later and sell a final $750 at the end of the second year, the total gross return is still 50%, but the ROI will be higher. (It will be nearly 32%
- see the IRR Calculator).
This result is because you received a return on your investment before the end of the 2-year term. And the investment return earlier in the cash flow improves the ROI - a bird-in-hand as such.
The point is the ROI levels the playing field. In the above example, we always have a $500 profit for a 50% gross return. If you were to only look at the gain, you might think there's no difference
in the investment. But that's not the case, and the ROI tells us that.
The ROI also gives you the tool to compare 15-year mortgages with 30-year mortgages, or any term you like for that matter.
For the above reason, the UMC calculates an ROI so you can answer if a house is a good investment for you. Generally speaking, if the ROI is negative, you should perhaps consider renting as an
alternative to purchasing a home. On the other hand, if it's positive, then the projection is, you'll earn a profit on the purchase.
How Do I Calculate the ROI?
This mortgage finance calculator, of course, does the math. We need to make a few decisions, however.
• First, expenses, of course, impact the profitability of an investment. Housing is no different. We need to decide what costs we want to include in the calculation? Are we only concern with the
direct mortgage expenses, such as down payment, periodic payments, points, etcetera? Or do we want a broader analysis that includes estimated maintenance, insurance, and property taxes?
• Secondly, do we want an inflation-adjusted analysis? Over the term of a 15-year or 30-year mortgage, inflation can have quite an impact. Or do we want an unadjusted ROI?
The UMC is very flexible, and it allows users to answer these questions in a way that meets their needs. However, reasonable defaults are preloaded into the calculator when you first land on this
page (I go into detail how I selected each default below), and we'll use those for this analysis.
For our illustration, we are going to use the CSHPI to estimate the future selling price of the home. Since we use a 30-year mortgage to base our example on, we'll assume our house will increase in
values at a rate of 4% per year.
For general cost inflation (maintenance, property taxes, etc.), the UMC allows you to enter a different inflation number. Per the Federal Reserve Bank of St. Louis, "...the FOMC [Federal Open Market
Committee] adopted an explicit inflation target of 2 percent in January 2012." While inflation has been running a bit under that, we'll use 2% for this example's cost inflator.
Now, as to the costs, let's look at where I got the numbers used in this example. (I suggest you refresh the page to reload the calculator with the numbers we are using. After you understand the
analysis, you can get your numbers together. Then the results will be more meaningful to you.)
1. Loan Amount: According to the Consumer Financial Protection Bureau, the average size new mortgage balance as of 2017 was $260,386. (Assuming a 20% cash down payment, the calculator will calculate
the Price of Real Estate.
2. Annual Interest Rate: As of February 2019, the average interest rate for a 30-year fixed mortgage was 4.35%, per Freddie Mac as retrieved from FRED, Federal Reserve Bank of St. Louis; March 2,
You'll find the following inputs on the Options tab:
3. Annual Property Taxes: Property taxes in most states contribute a significant amount to the overall cost of owning a home, and the UMC accounts for them. I calculated the average property tax
rate of 0.98 from the Median Property Tax Rates By State data found at tax-rates.org. Using 0.98% on the average selling price, we get an average property tax bill of $3,190.00. (Of course, if
you don't know the actual property tax yet, you can find your state's rate here.)
There are two additional expenses, and for these numbers, I'm winging it:
4. Annual Maintenance $3,000
5. Yearly Property and Casualty Insurance premium $800.
One expense the calculator does not directly support is homeowners association fees (HOA fee). If you need to account for those, calculate the current annual amount and add that amount to the yearly
maintenance expense. The ROI calculation will be correct.
Given the above details, what's the ROI?
Fig. 6 - 1.6% Return-on-Investment (ROI)
Hmm, that's not so good. True, it's probably better than the annual return you would have earned from a savings account for the past ten years, but you must be wondering why I say, in general, buying
a house is a good investment?
That's because this analysis is not complete — 1.6% is just an intermediate result. After all, you have to live somewhere, right?
Buy vs. Rent
If your alternative to buying is renting, then your estimated rent needs to be an offsetting cost in your calculation.
What do I mean? Why is this?
Look at it this way, what's your ROI on rent? Nothing, of course. In fact, it is worse than nothing. The money is out the door, never to be seen again. Therefore, you should say the ROI is -100%. But
that's silly. No one thinks of ROI when it comes to renting.
So to make the comparison between renting and buying meaningful, we need to zero out the 100% rent loss. To do this, the UMC will adjust the total cost of homeownership by what you are willing to
spend on rent (the analysis looks at rent or cost-of-housing as a fixed cost). If you are ready to pay $20,000 a year for rent, then the calculator's analysis will look at the difference in costs and
calculate the ROI.
In other words, the UMC will calculate an ROI for the marginal dollars you'll spend owning a home over renting. You should do this because it's only the dollars after the cost-of-housing that you'll
have available to invest if you decide homeownership is not for you.
Alright, so how will considering rent in the analysis impact the results?
Well, let's see.
According to RENTCafe.com, the average rent in the US as of June 2018 for three bedrooms is $1,714 a month. If we enter that into Monthly Rent and recalculate, the ROI is now:
Fig. 7 - 10.1% return-on-investment (ROI) after allowing for cost-of-housing.
Quite a difference, wouldn't you say! And also, I think, a more realistic result. Because it's telling us that if we take out an average mortgage at a nationwide average interest rate and pay average
costs, we'll make a return on our INVESTABLE dollars of 10.1%.
A 10.1% return is better than the S&P has done over the last 30 years, which was 5.9% without accounting for dividend reinvestment.
Still not sure about buying?
Then take a look at this pie chart the calculator creates:
Fig. 8 - Final house value and total inflation-adjusted costs.
Regardless of what ROI you earn, when the 30 years are over, if the projections hold up, and after you have paid all the mortgage payments along with the taxes, insurance, and maintenance, you'll
have an asset that you can sell for approximately $1,055,000.
How much of the rent money will you get back?
Not only do you get back the investment gain on your marginal dollars spent, but you also recover your housing costs from the past 30 years!
Plus, there are at least two other financial benefits to buying a house that doesn't show up in the totals:
• Taking out a 15-year or 30-year fixed-rate mortgage locks you into a fixed payment amount for a very significant portion of your housing cost. You can't say that about renting.
• And once you've paid off the mortgage, your housing costs going forward will drop significantly. Of course, that will never happen if you decide to continue renting.
Fig. 9 - Rent vs. Buy: showing a fixed 30-year mortgage payment vs. increasing rent.
A few words of caution.
• All real estate is local. Some areas tend to appreciate well above the national average, and some will, of course, fall below the average. It is up to you to buy right and to make the right
• The results of this analysis could quickly change if interest rates rise. Interest is a significant cost component of the mortgage. Make sure you do your own analysis.
• The results will also change (perhaps significantly) if you do not stay in the house for the full term of the mortgage. Want to see by how much the ROI will change? First, prepare a full term
analysis of your mortgage, and then do it again by changing the "Number of Payments" to the length of time you expect to own the home. That is, if you are going to be in the house for eight
years, change the "Number of Payments" to 96 and leave everything else the same. Then recalculate the ROI using the shorter term.
Finally, the point is not to agree or disagree with the numbers I'm using. The point is to give you a tool and the background so that you can do your own analysis. The Accurate Mortgage Calculator is
flexible enough that you should be able to study the home buying transaction any way that makes you comfortable to answer the question, "is buying a house a good investment?"
Ok, I'm leaning toward buying a home. Is there any way I can save some money and improve the ROI even more?
Yes, there is.
You might want to consider making extra payments to reduce interest charges or the mortgage saving tips that follow.
Why Making Extra Payments Saves You Money
I assume that most borrowers know that if they pay an additional amount on their mortgage (or any loan) above the required payment that they'll save money. (Check the terms of your loan to make sure
there is no prepayment penalty.)
How does paying extra on your mortgage work? That is, why does it save you money? (see: mortgage calculator with extra payments.)
The answer is rather simple. When you make a payment on a traditional mortgage, the interest gets calculated using the current balance for the number of days since the last payment. The calculation
adds the interest to the loan balance and then deducts the total payment amount.
If you pay an additional $200, for example, 100% of the $200 is used to reduce the principal balance (or at least it should be if the lender's math is correct). Then next time, interest is calculated
on the balance that is lower by $200 than it would otherwise be.
And the lower the balance, the lower the calculated interest.
The crucial thing to understand is, while an extra $200 may not seem like much in terms of say a $250,000 balance, but that single $200, has reduced the balance of the loan from its payment date
until the loan is paid-in-full. The interest saving is on every future payment. And if you continue to make the extra payment, their impact is compounded.
You'll be able to save quite a bit of money! And that has got to be a good thing, right?
Well, let's see how good.
Fig. 10 - A setup for a $10,000 lump-sum extra payment on an odd payment date.
What is the effect of paying extra principal on a mortgage?
Assume you receive a year-end bonus, and you are considering making a single lump-sum payment toward the mortgage balance of $10,000. What will be the interest savings?
Fig. 11 - Check the mortgage interest calculator's schedule to see your interest savings.
If you assume this calculator's default values, the one-time $10,000 additional payment will save you more than $23,000 in future interest charges.
Is it a good idea to make extra mortgage payments?
There's an ongoing debate among financial professionals and even mortgage holders at large whether or not it's a good idea to prepay a mortgage. The thinking is, you could use the money that you are
using to make additional loan payments and invest the money instead. Some say, investing the funds will create more wealth than it saves in interest charges.
I'm not here to give financial advice, and the Accurate Mortgage Calculator can't help you with an answer. What it will do is calculate the interest you'll save if you choose to make extra payments.
Fig. 12 - The Accurate Mortgage Calculator's expanded payment schedule
showing an odd-day lump-sum payment.
Make sure your lender applies 100% of the additional payment to principal.
What it will do that this calculator won't do is prepare a comparative financial schedule that calculates both what the interest savings will be as well as the projected future value if you invested
the money instead.
You definitely should look at this calculator if you are making or planning to make extra principal payments.
What if you're not convinced that you should make additional mortgage payments. Are there any other techniques you can use to reduce your costs?
Yes, there are.
Two Practical Tips for Saving Money on a Mortgage
Mortgage payments are generally a significant portion of any family's monthly budget. Typically the mortgage loan consumes 25% or more of the monthly income. Hopefully, most consumers know if they
make extra payments or agree to bi-weekly payments, they can save a small boatload of interest over the term of their loan. These strategies are certainly useful, and they will save you money. You
should consider them depending on your other investment options.
But what if you don't have the free cash flow to make extra payments? Are there money-saving strategies that don't require sacrifice?
Well, yes, in fact, there are. Read on for two such ideas for you. Naturally, you'll want to plug your numbers into the calculator to see what you can save.
TIP 1: Don't Assume Making a Higher Down Payment Saves Money
Usually, one would think that the greater the down payment amount, the less the amount borrowed. The lower loan amount means a lower accumulated interest charge over the term of the loan. That's what
one would think, and usually, that would be correct. But at least in the US, mortgage borrowers have another option.
Borrowers can pay points. Points are nothing more than an up-front fee charged by lenders in exchange for a lower interest rate. The lender will calculate the amount owed for points as a percentage
of the total loan. On a $300,000 loan, 2.5 points equals $7,500.
So you're thinking, I understand what points are. And I do plan to be in this house for a dozen years or more. It also sounds good to pay something upfront in exchange for a lower interest rate, but
what if I don't have the available cash to pay points? Am I out of luck?
Maybe not.
Fig. 13 - Mortgage chart showing an initial "bump" in annual payment due to paying mortgage points.
You can swap down payment for points.
How much cash do you have for a down payment? Twenty percent or more?
If that's the case, it may save you money if you give the lender less for a down payment and use that money to pay a couple of points. Let's look at an example.
In the calculator, enter the following values:
1. Price of Real Estate or Asset?: $375,000.00
2. Down Payment Percent?: 22%
3. Loan Amount?: $0
4. Number of Payments? (#): 360
5. Annual Interest Rate?: 4.1250%
6. Payment Amount?: $0
7. Points?: 0
Since we are comparing mortgage strategies only, make sure "Annual Property Taxes," "Annual Insurance," and "Private Mortgage Ins. (PMI)" are all set to "0". (mortgage calculator with PMI)
The calculator is going to calculate both the mortgage loan amount and the monthly payment since you have entered zeros for those inputs.
Fig. 14 - Total interest with a 22% down payment.
The significant number, however, is total interest. Checking at the bottom of the calculator, just above the buttons, you'll see for our example loan that you would pay $217,836 in interest over the
term of the loan. Make a note of this number. You'll need it for the next step.
There is one more calculation. Change the following inputs (the others are left as they are):
1. Down Payment Percent?: 20%
2. Mortgage Amount?: $0 (Reset for new calculation)
3. Annual Interest Rate?: 3.8750%
4. Payment Amount?: $0 (Reset for new calculation)
5. Points?: 2.00 (Under the "Options" tab.)
Fig. 15 - Pay points upfront to save $4,000 (total interest includes points).
What we have done is to reduce the down payment amount by 2% and added 2 points. When you add points, you are buying a lower interest rate. In this (conservative) example, adding 2 points lowers the
fixed interest rate by 1/4 of a percent over the entire term of the loan. You may find in your area that you can reduce your rate more -- maybe by 0.333% or even 0.4%.
We are now ready to calculate the new mortgage details. This time, since there are more details on the schedule, click "Pmt & Cost Schedule" (It is not necessary to click "Calc" first.)
There are two totals in the summary section of the schedule we need to know:
• Points Amount: $6,000
• Total Interest & Points Paid: $213,856
Compare this to the total interest from the first calculation when the down payment was higher, and there were no points — $217,836.
The mortgage with the two points and 20% down will save you $3,970 over the one with 22% down.
Not only will you save nearly $4,000, but there's also icing on the cake (and it's calorie-free!). The savings come to you without you having to make any sacrifices.
• You do not have to go through anything other than the typical mortgage approval process.
• You do not have to submit a different application or do additional paperwork.
• And you do not have to have any extra money upfront.
The cool part is, if you look at your payment amount, you see it will decrease from $1417 to $1410! Woopie!
There's even more.
If you are in the US and you itemize deductions when you pay income taxes, points are often a deductible cost for obtaining a mortgage. This deduction means Uncle Sam (and other American taxpayers)
is helping you out by lowering your tax bill. If you are in the 33% marginal tax bracket, $4,000 paid as mortgage points could save you $1,320 in taxes in the year you file after taking out the
mortgage. (Please see the section on this page "Tax Impact" to understand the possible limits.)
This savings is a win, win, win, win.
• Lower total mortgage cost
• Lower monthly payments
• No change to the mortgage process itself
• And a front-loaded possible tax deduction
A word of caution is in order here. You'll only want to employ the second strategy if you plan to stay in your home for a relatively extended period. If you expect to move in say 5 or even ten years,
you may not save enough in interest charges to make up for the points. Use the payment schedule to find your break-even point.
I should also note one point that some may consider to be a disadvantage. You'll have slightly less equity in your home in the early years of the mortgage. The reduced equity is because you've traded
points in place of making a more significant down payment.
Let me know in the comments what you think of this tip. Will you consider using it?
I think my next tip is even better...
TIP 2: Make Payments at "Start-of-Period" to Save
Lenders hate this. (Hint, it has to do with their ROI.)
But if you are the borrower, you should love it.
When applying for a mortgage, lenders ask for a lot of documentation. Bank statements are one item you'll be requested to provide. They will be looking at your cash flow to see if you have two or
even more payments available beyond the down payment amount you have agreed to in the purchase contract.
All well and good.
Since the lender wants you to have a payment or two already in the bank, give it to them as early as possible. Often, the initial loan installment is not due until the first-of-the-month after the
month following the closing. (Close on March 20th, and the payment is due on May 1st.) Instead of waiting, hand the lender a check for the first payment on the day you sign the mortgage papers.
Why would I want to do that?
Fig. 16 - Make your first payment on the day the loan originates to save thousands.
Here's why.
In the calculator, enter the following values:
1. Price of Real Estate or Asset?: $350,000.00
2. Down Payment Percent?: 20%
3. Mortgage Amount?: 0
4. Number of Payments? (#): 360
5. Annual Interest Rate?: 4.1250%
6. Payment Amount?: $1,417.60
7. Points?: 0.00%
8. Payment Frequency?: Monthly
9. Set the Loan Date and First Payment Due, so they are one month apart
10. Once again, since we are comparing only mortgage strategies, enter "0" for "Annual Property Taxes," "Annual Insurance," and "Private Mortgage Ins. (PMI)".
The calculator will solve for both the mortgage loan amount and the monthly payment since you've entered zeros for those inputs.
Click "Payment & Cost Schedules"
Make a note. Total interest due is $208,527, and the first payment is paid one month after the loan date. We call this an "end-of-period" schedule.
Now for the comparison calculation.
Change the following inputs (the others are left as they are):
1. Mortgage Amount?: $0 (Reset for new calculation)
2. Number of Payments? (#): 0 (Reset for new calculation)
3. Set the Loan Date and First Payment Due to the same date
These changes set the calculator to calculate the loan term.
Again, click "Payment & Cost Schedules."
Notice how the first payment now falls on the closing date of the loan? Also, notice for the first installment, there is NO INTEREST DUE. Why? Because no days have passed! Interest is charged/
collected only for days when the money is on loan.
Now total interest due is $205,236.
How much interest will you save?
By making this one simple change to the payment schedule, you will save $3,291 in interest charges over the term of the loan. Furthermore, you'll have to make only 258 payments, not 360, and the last
payment will be due on February 1, 2047, not May 1, 2047.
Like the first tip, this tip does not require you to make any fundamental change to what you were already planning to do. You don't even have to tell your lender that you are going to do this. Just
hand them the check on the day you close the loan.
If you decide to use this money-saving strategy, understand the payment you provide your lender at closing is only the mortgage payment. You do not have to give them any escrow amount they might be
collecting with later payments. Escrow is something separate. An escrow amount typically is added to the regular mortgage payment, and it is used to cover property taxes and insurance. Again, you pay
them only the mortgage part of your total amount at the closing.
Also, it is a good idea to closely monitor your new mortgage account and confirm that this first payment is applied 100% to the principal. If it's not, then you won't get the full benefit of this
Accumulated mortgage chart showing a series of extra payments.
What About Adjustable Rate Mortgages or ARMs?
In the US, at least, adjustable-rate mortgages (ARM) have fallen out of favor with borrowers. In the current low-interest-rate environment, it does not make much sense for someone to take out a
mortgage that has an interest rate that is more likely to increase than decrease.
The UMC does not officially support ARMs.
However, if you need to create an amortization schedule for an ARM, I'm not going to leave you high and dry. Please see the Adjustable Rate Mortgage calculator. It is easily capable of creating an
amortization schedule with adjustable rates. You can adjust the interest rate as of any date, not just on payment due dates. There is even an adjustable-rate mortgage tutorial that will show you
step-by-step how to create an ARM schedule.
There can be a lot of details when it comes to mortgages. But they do not have to be complicated.
15 Comments on “Mortgage Center”
Join the conversation. Tell me what you think.
What is the best calculator to do an escrow mortgage PITI payoff for a person? I have a payoff I have to figure and I can’t find a calculator that would include ALL of this. The person was a
title company worker, requested an escrow account. I know nothing of an escrow accnt. except that monthly taxes and insurance may change and if she overpays it goes into escrow, which could save
her a lot of money at payoff. Is there such a calculator that I could use to do this and are there good instructions on how to
fill it out? Thank you. Pam
Please see this loan payoff calculator.
I think you’ll want to look at this as 2 accounts and thus 2 different calculations. The mortgage account (where there’s interest) and the escrow account, which usually does not involve
interest. The mortgage account starts with the mortgage balance and you apply the loan portion of the debtor’s payments to the mortgage. The escrow account starts with a $0 balance and then
you’ll show payments going out to insurance and taxes.
If the debtor “overpays” as you say, the overage should not go to escrow. It should go to the mortgage to reduce the principal balance which will save her interest. The escrow account, since
there is no interest paid or collect i.e. 0% interest, will show money in/out.
Give it a try, and if you have questions, just ask.
I am looking for an example of a home equity line of credit I wish to set up with a family member. I am having difficulty locating one on the website. Could you please direct me to the correct
calculator or suggest how I might go about putting something like this together?
Your family member can use the Ultimate Financial Calculator.
The calculator lets the user make multiple borrows and payments on any date. That’s basically what a HELOC loan allows as well. Borrow when you need it. Pay it back when you can.
They can scroll down the page to the tutorial link for some ideas. Or ask any questions they may have in the comment section.
Thanks for the insights. I will check it out.
This looks like something that will work. How do I save it so that I can modify it as loans and repayments are used and keep it current.
Also, is there a place to make notes to specify for what the loan amount and repayment are being used?
I’m lost. Payments are irregular and will arrive at inconsistent intervals. Loan amounts or draws will be interspaced as well. I thought I had it but when I look at the schedule read out, it’s
really not what I’m after. I’ve looked at the tutorials but still cannot put this together. Suggestions would be most appreciated.
Tutorial 1 is good to review or go through for an overview of how the calculator works.
Tutorial 25 should get you very close to what you need. That tutorial is about tracking loan payments and calculating payoff amounts, which is what you would be doing if you have a HELOC.
Basically, in each row, you enter either a single loan or payment as of the date the payment or loan occurred. The "Rounding" option should be set to "Open Balance" so as not to round the
last payment entered to result in a 0 final balance.
It’s hard for me to be more specific because "it’s really not what I’m after" doesn’t give me anything to go on. 🙂
Okay, I’ll work on this today. I apologize for being evasive. I hope I didn’t make you frustrated, it’s just that I have spent a long time on this (in and out of AccurateCalculators). I guess I
have a lot top learn.
Thanks again for your patience. I finally understood the directives and was able to obtain the schedule and report I needed. I much appreciate this service.
Thanks for letting me know James. (And no, your question didn’t frustrate me.)
My mom passed away in 2020. Her estate was divided between me and my brother. He is buying me out and i receive a monthly payment from him. I received a Loan summary with all the payments that he
will be making. Is this reported to the IRS? Do I need a form from the IRS regarding this, and do I do it or does my brother? Thank you.
Sorry, but I’m not qualified to answer such questions. I can answer questions about how a calculator works, or how to do a calculation, but not about IRS regulations (unless it perhaps deals
with depreciation).
Comments, suggestions & questions welcomed... | {"url":"https://accuratecalculators.com/mortgage-center","timestamp":"2024-11-06T20:30:58Z","content_type":"text/html","content_length":"178766","record_id":"<urn:uuid:fd06a8d4-be30-48b1-b6b4-ac8e2bd20242>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00343.warc.gz"} |
Modelling the Strength of Steel Plates Using Regression Analysis and Neural Networks
In 2003. abstract bibtex
Statistical models that predict the tensile strength of low-alloyed steel plates using the element concentrations and some variables of the rolling process were developed. The purpose of the work was
to develop a new predicting model for Rautaruukki’s steel plate mill. The model will be used mainly in the product design of steel plates. The standard deviation of the error term of the best
regression model was 10 MPa, which can be considered very good. The performance of the regression model was compared to a neural network model, but significantly better predictions were not achieved
with neural networks than with regression models. The quantity of data used was very large, and special attention was therefore paid to avoid overfitting.
title = {Modelling the Strength of Steel Plates Using Regression Analysis and Neural Networks},
type = {inProceedings},
year = {2003},
id = {6ecfc171-dba7-3b9f-8ec6-cdcdfca7e0c2},
created = {2019-11-19T13:01:03.739Z},
file_attached = {false},
profile_id = {bddcf02d-403b-3b06-9def-6d15cc293e20},
group_id = {17585b85-df99-3a34-98c2-c73e593397d7},
last_modified = {2019-11-19T13:46:10.222Z},
read = {false},
starred = {false},
authored = {false},
confirmed = {true},
hidden = {false},
citation_key = {isg:458},
source_type = {inproceedings},
notes = {Proceedings of International Conference on Computational Intelligence for Modelling, Control and Automation (CIMCA'2003)},
private_publication = {false},
abstract = {Statistical models that predict the tensile strength of low-alloyed steel plates using the element concentrations and some variables of the rolling process were developed. The purpose of the work was to develop a new predicting model for Rautaruukki’s steel plate mill. The model will be used mainly in the product design of steel plates. The standard deviation of the error term of the best regression model was 10 MPa, which can be considered very good. The performance of the regression model was compared to a neural network model, but significantly better predictions were not achieved with neural networks than with regression models. The quantity of data used was very large, and special attention was therefore paid to avoid overfitting.},
bibtype = {inProceedings},
author = {Juutilainen I Röning J, Myllykoski L} | {"url":"https://bibbase.org/network/publication/juutilainenirningj-modellingthestrengthofsteelplatesusingregressionanalysisandneuralnetworks-2003","timestamp":"2024-11-06T11:35:08Z","content_type":"text/html","content_length":"14405","record_id":"<urn:uuid:118f31af-06b6-43f3-9e14-87563b30ff33>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00411.warc.gz"} |
pairing Alternatives - Haskell Cryptography | LibHunt
Monthly Downloads: 49
Programming language: Haskell
License: MIT License
pairing alternatives and similar packages
Based on the "Cryptography" category.
Alternatively, view pairing alternatives based on common mentions on social networks and blogs.
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They vary from L1 to L5 with "L5" being the highest.
Do you think we are missing an alternative of pairing or a related project?
Add another 'Cryptography' Package
Implementation of the Barreto-Naehrig (BN) curve construction from [BCTV2015] to provide two cyclic groups and , with an efficient bilinear pairing:
Let , and be abelian groups of prime order and let and elements of and respectively . A pairing is a non-degenerate bilinear map .
This bilinearity property is what makes pairings such a powerful primitive in cryptography. It satisfies:
The non-degeneracy property guarantees non-trivial pairings for non-trivial arguments. In other words, being non-degenerate means that:
An example of a pairing would be the scalar product on euclidean space .
Example Usage
A simple example of calculating the optimal ate pairing given two points in and .
import Protolude
import Data.Group (pow)
import Data.Curve.Weierstrass (Point(A), mul')
import Data.Pairing.BN254 (BN254, G1, G2, pairing)
p :: G1 BN254
p = A
q :: G2 BN254
q = A
main :: IO ()
main = do
putText "P:"
print p
putText "Q:"
print q
putText "e(P, Q):"
print (pairing p q)
putText "e(P, Q) is bilinear:"
print (pairing (mul' p a) (mul' q b) == pow (pairing p q) (a * b))
a = 2 :: Int
b = 3 :: Int
Pairings in cryptography
Pairings are used in encryption algorithms, such as identity-based encryption (IBE), attribute-based encryption (ABE), (inner-product) predicate encryption, short broadcast encryption and searchable
encryption, among others. It allows strong encryption with small signature sizes.
Admissible Pairings
A pairing is called admissible pairing if it is efficiently computable. The only admissible pairings that are suitable for cryptography are the Weil and Tate pairings on algebraic curves and their
variants. Let be the order of a group and be the entire group of points of order on . is called the r-torsion and is defined as . Both Weil and Tate pairings require that and come from disjoint
cyclic subgroups of the same prime order . Lagrange's theorem states that for any finite group , the order (number of elements) of every subgroup of divides the order of . Therefore, .
and are subgroups of a group defined in an elliptic curve over an extension of a finite field , namely , where is the characteristic of the field and is a positive integer called embedding degree.
The embedding degree plays a crucial role in pairing cryptography:
• It's the value that makes be the smallest extension of such that captures more points of order .
• It's the minimal value that holds .
• It's the smallest positive integer such that .
There are subtle but relevant differences in and subgroups depending on the type of pairing. Nowadays, all of the state-of-the-art implementations of pairings take place on ordinary curves and assume
a type of pairing (Type 3) where and and there is no non-trivial map .
Tate Pairing
The Tate pairing is a map:
defined as:
where , is any representative in a equivalence class in and is the set of equivalence classes of under the equivalence relation . The equivalence relation in the output of the Tate pairing is
unfortunate. In cryptography, different parties must compute the same value under the bilinearity property.
The reduced Tate pairing solves this undesirable property by exponentiating elements in to the power of . It maps all elements in an equivalence class to the same value. It is defined as:
When we say Tate pairing, we will mean the reduced Tate pairing.
Pairing optimization
Tate pairings use Miller's algorithm, which is essentially the double-and-add algorithm for elliptic curve point multiplication combined with evaluation of the functions used in the addition process.
Miller's algorithm remains the fastest algorithm for computing pairings to date.
Both and are elliptic curve groups. is a multiplicative subgroup of a finite field. The security an elliptic curve group offers per bit is considerably greater than the security a finite field does.
In order to achieve security comparable to 128-bit security (AES-128), an elliptic curve of 256 bits will suffice, while we need a finite field of 3248 bits. The aim of a cryptographic protocol is to
achieve the highest security degree with the smallest signature size, which normally leads to a more efficient computation. In pairing cryptography, significant improvements can be made by keeping
all three group sizes the same. It is possible to find elliptic curves over a field whose largest prime order subgroup has the same bit-size as the characteristic of the field . The ratio between the
field size and the large prime group order is called the -value. It is an important value that indicates how much (ECDLP) security a curve offers for its field size. is the optimal value. The
Barreto-Naehrig (BN) family of curves all have and . They are perfectly suited to the 128-bit security level.
Most operations in pairings happen in the extension field . The larger gets, the more complex becomes and the more computationally expensive the pairing becomes. The complexity of Miller's algorithm
heavily depends on the complexity of the associated -arithmetic. Therefore, the aim is to minimize the cost of arithmetic in .
It is possible to construct an extension of a field by successively towering up intermediate fields and such that , where and are usually 2 and 3. One of the reasons tower extensions work is that
quadratic and cubic extensions ( and ) offer methods of performing arithmetic more efficiently.
Miller's algorithm in the Tate pairing iterates as far as the prime group order , which is a large number in cryptography. The ate pairing comes up as an optimization of the Tate pairing by
shortening Miller's loop. It achieves a much shorter loop of length on an ordinary curve, where t is the trace of the Frobenius endomorphism. The ate pairing is defined as:
We have implemented a polymorphic optimal ate pairing over the following pairing-friendly elliptic curves:
• Barreto-Lynn-Scott degree 12 curves
□ [BLS12381](src/Data/Pairing/BLS12381.hs)
• Barreto-Naehrig curves
□ [BN254](src/Data/Pairing/BN254.hs)
□ [BN254A](src/Data/Pairing/BN254A.hs)
□ [BN254B](src/Data/Pairing/BN254B.hs)
□ [BN254C](src/Data/Pairing/BN254C.hs)
□ [BN254D](src/Data/Pairing/BN254D.hs)
□ [BN462](src/Data/Pairing/BN462.hs)
A more detailed documentation on their domain parameters can be found in our elliptic curve library.
This is experimental code meant for research-grade projects only. Please do not use this code in production until it has matured significantly.
Copyright (c) 2018-2020 Adjoint Inc.
Permission is hereby granted, free of charge, to any person obtaining a copy
of this software and associated documentation files (the "Software"), to deal
in the Software without restriction, including without limitation the rights
to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
copies of the Software, and to permit persons to whom the Software is
furnished to do so, subject to the following conditions:
The above copyright notice and this permission notice shall be included in all
copies or substantial portions of the Software.
THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
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Debt To Income Ratio Calculator
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Percentage Complete
I need a percentage of checkboxes checked for the CHILDREN rows. HELP! I am terrible at these formula's and I get errors.
Best Answer
• You won't be able to keep it in the Complete column as the Checkbox formatting is static throughout (you can't cross formats).
So in a Text/Number formula, you can populate the following formula, making sure to update the range to match what you are looking for:
=(COUNTIF(Complete2:Complete4, 1)) / COUNT(Complete2:Complete4)
I don't believe you will be able to provide a formula, in this case, that uses the CHILDREN function. I can certainly be wrong.
• You won't be able to keep it in the Complete column as the Checkbox formatting is static throughout (you can't cross formats).
So in a Text/Number formula, you can populate the following formula, making sure to update the range to match what you are looking for:
=(COUNTIF(Complete2:Complete4, 1)) / COUNT(Complete2:Complete4)
I don't believe you will be able to provide a formula, in this case, that uses the CHILDREN function. I can certainly be wrong.
• You are so smart! I did not realize I couldn't put it in the checkmark column. I moved it over one and got exactly what I needed.
Thanks for your help! You rock!
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Bayesian Probability Calculator - Calculator Wow
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10 FAQs and Answers
1. What is Bayesian probability? Bayesian probability is a mathematical framework for updating beliefs about the probability of an event as new evidence or information becomes available.
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3. Why is Bayesian inference important in statistics? It allows for the incorporation of prior knowledge into probability calculations, making it useful for decision making and hypothesis testing.
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Buy grade 1 mid term 2 exams papers with answers
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Study on the Stability Control of Vehicle Tire Blowout Based on Run-Flat Tire
Nanjing Institute of Technology, School of Automobile and Rail Transportation, Nanjing 211167, China
College of Energy and Power Engineering, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China
Author to whom correspondence should be addressed.
Submission received: 28 July 2021 / Revised: 12 August 2021 / Accepted: 20 August 2021 / Published: 21 August 2021
In order to study the stability of a vehicle with inserts supporting run-flat tires after blowout, a run-flat tire model suitable for the conditions of a blowout tire was established. The unsteady
nonlinear tire foundation model was constructed using Simulink, and the model was modified according to the discrete curve of tire mechanical properties under steady conditions. The improved tire
blowout model was imported into the Carsim vehicle model to complete the construction of the co-simulation platform. CarSim was used to simulate the tire blowout of front and rear wheels under
straight driving condition, and the control strategy of differential braking was adopted. The results show that the improved run-flat tire model can be applied to tire blowout simulation, and the
performance of inserts supporting run-flat tires is superior to that of normal tires after tire blowout. This study has reference significance for run-flat tire performance optimization.
1. Introduction
As an important part of the vehicle driving system, tires are the only direct contact part between the vehicle and the road surface. The main functions are to support the vehicle, mitigate the impact
of the road surface, and generate braking force, which have an important impact on comfort and other aspects of performance of the vehicle [
]. According to statistics, 46% of traffic accidents on expressways are caused by tire failures, and a flat tire alone accounts for 70% of the total tire accidents [
How to improve the stability of vehicles after tire blowout has always been the core problem of domestic and foreign scholars. CHEN set up an estimated calculation model of additional yaw torque
leading by tire blowout based on Dugoff tire model [
]. LI proposed a modified linear time-varying model predictive control (LTV-MPC) method based on the Pacejka tire model, and the stability region of the vehicle with active front steering was
expanded [
]. LIU established an extended three-degree of freedom model considering longitudinal velocity to weaken the influence of the steering wheel angle under extreme conditions [
]. CHEN proposed a comprehensive coordinated control scheme for longitudinal and lateral stability based on the brush tire model. Speed tracking under the limit speed is achieved by means of
longitudinal acceleration feedforward and state feedback controller [
]. GUO established a tire model suitable for operating coach and estimated the sideslip angle, as well as the yaw rate of the operating coach in real time based on the extended Kalman filter state
estimator [
In addition, some scholars have done a lot of research on the stability control after tire blowout. LIU proposed a fuzzy sliding mode control algorithm based on the current tire blowout control
algorithm, and the tire blowout model is established using the UniTire model [
]. JING established a linear variable parameter vehicle model considering the time-varying speed and the uncertainty of tire characteristics [
]. WANG proposed a coordinated control approach based on active front steering and differential braking. The model predictive control method was adopted to control the front wheel angle for tracking
the trajectory, and the differential brake control was adopted to offer an inner-loop control input and improve the lateral stability [
]. CHEN established the kinematics equation of the vertical load of the vehicle after tire blowout according to the vehicle dynamics model and the displacement mutation in the vertical direction of
the tire wheel center [
]. ERLIEN found that the mandatory implementation of stability constraints may conflict with the expected collision avoidance trajectory and proposed that the stability controller should allow the
vehicle to run outside the stability constraints to achieve safe collision avoidance [
]. WANG proposed a novel linearized decoupling control procedure with three design steps for a class of second order multi-input-multi-output non-affine system [
]. YANG proposed a composite stability control strategy for tire blowout electric vehicle with explicit consideration of vertical load redistribution [
]. CHOI designed a layered lateral stability controller based on LTV-MPC. The nonlinear characteristics of the tire are reflected in the extended “bicycle” model by continuously linearizing the tire
force [
]. WANG proposed a nonlinear coordinated motion controller in the framework of the triple-step method to solve the path-following and safety problem of road vehicles after a tire blowout [
However, the tire blowout condition of the vehicle equipped with run-flat tire is more complicated, and there is no model that can be directly applied to inserts supporting run-flat tire under tire
blowout condition. In addition, the inserts supporting run-flat tire is a typical run-flat tire based on the pneumatic tire structure, which consists of an auxiliary support body on the rim and a
tire pressure detection device [
]. Because this type of run-flat tire is mostly based on the common rim design, it has the advantages of simple structure, convenient disassembly, strong zero-pressure bearing capacity, and so on. It
is a new type of run-flat tire with great development prospects. In order to study the stability of the vehicle with inserts supporting run-flat tire after blowout, a dynamic model was established
and modified based on the UniTire model. Afterwards, the model was modified according to the discrete curve of tire mechanical properties under steady conditions. The results show that the improved
run-flat tire model is consistent with the characteristics of vehicle tire blowout, and the optimal control strategy of inserts supporting run-flat tire after blowout in straight running condition
was obtained by comparing the effects of braking, steering, and combined control.
2. Establishment of Run-Flat Tire Model
2.1. Run-Flat Tire Model before Blowout
The performance of inserts supporting run-flat tire is the same as that of a normal tire before tire blowout. Therefore, the model of inserts supporting run-flat tire is established according to the
UniTire [
] theory and the published test data, which mainly analyse the parameters of the mechanical properties in the process of tire blowout.
The longitudinal slip ratio and lateral slip ratio are defined as the ratio of slip velocity to rolling velocity. The definition is as follows:
${ S x = − V s x Ω R e = Ω R e − V x Ω R e , S x ∈ ( − ∞ , + ∞ ) S y = − V s y Ω R e = − V y Ω R e , S y ∈ ( − ∞ , + ∞ )$
is the longitudinal slip ratio;
is the lateral slip ratio;
is the tire rolling angular velocity;
is the effective rolling radius;
$V y$
are the longitudinal and lateral components, respectively, of the wheel center velocity
The normalized longitudinal slip ratio
$ϕ x$
, normalized lateral slip ratio
$ϕ y$
, and total slip ratio
are defined as dimensionless physical quantities.
${ ϕ x = K x ⋅ S x μ x ⋅ F z ϕ y = K y ⋅ S y μ y ⋅ F z ϕ = ϕ x 2 + ϕ y 2$
$K x$
is the longitudinal slip stiffness of the tire;
$K y$
is the cornering stiffness of the tire;
$µ x$
is the longitudinal friction coefficient between tire and ground;
$µ y$
is the lateral friction coefficient between tire and ground; and
is the tire vertical force.
The total tangential force expression of the complete E-index semi-empirical model is as follows:
$F ¯ = 1 − exp [ − ϕ − E 1 ⋅ ϕ 2 − ( E 1 2 + 1 12 ) ⋅ ϕ 3 ]$
$E 1$
is the curvature factor.
The aligning arm
of the steady semi-empirical model can be expressed as follows:
$D x = ( D x 0 + D e ) exp ( − D 1 ϕ − D 2 ϕ 2 ) − D e$
$D x 0$
is the initial aligning arm;
$D e$
is the final value of the aligning arm;
$D 1$
is the first order curvature factor; and
$D 2$
is the quadratic curvature factor.
According to the above theoretical model, the tire aligning torque can be obtained as follows:
$M z = F y ( − D x + X c ) − F x ⋅ Y c$
$X c$
$Y c$
are the offset caused by longitudinal force and lateral force, respectively.
The formula of tire rolling resistance moment
is as follows:
$M y = ( R r _ c + R r _ v ⋅ V r ) ⋅ F z ⋅ R l$
$R r _ c$
is the rolling resistance coefficient;
$R r _ v$
is the rolling resistance speed constant of the tire;
is the longitudinal velocity of wheel center (
$V r = Ω ⋅ R e$
); and
$R l$
is the load radius of the tire.
2.2. Run-Flat Tire Model after Blowout
During the process of tire blowout, the vehicle is extremely unstable and the tire burst time is short. In order to simplify the model, the parameters of the tire are linearized [
]. According to the experiment of the mechanical properties of the tire blowout, it can be seen that the longitudinal slip stiffness, lateral deflection stiffness, and effective rolling radius of the
tire are reduced to 28%, 25%, and 80% of the normal working conditions, respectively, and the rolling resistance coefficient of the tire increases by 30 times [
However, the above data cannot be fully applied to the inserts supporting run-flat tire. In order to study the parameter changes of the inserts supporting run-flat tire after tire blowout, the
corresponding curves under the working conditions were obtained using the test bench.
2.2.1. Cornering Stiffness and Longitudinal Stiffness after Tire Blowout
In order to simplify the research, the static loading method was used to analyse the longitudinal and lateral mechanical characteristics of inserts supporting run-flat tire at zero-pressure. The tire
size is 37 × 12.5R16.5. The relationship curves of tire lateral force and sideslip angle, longitudinal force, and longitudinal displacement under different loads are obtained, as shown in
Figure 1
It can be seen from
Figure 1
a that the sideslip angle of inserts supporting run-flat tire under zero-pressure condition is 1.8 deg. At this time, the lateral force is about 4700 N. The critical point of sideslip is selected to
calculate the stiffness. The slope at this point is approximately equivalent to cornering stiffness and the cornering stiffness is about 1.48 × 10
N/rad (4700 N/1.8 deg), which is about 0.90 times the rated tire pressure condition. Similarly, it can be seen from
Figure 1
b that the longitudinal stiffness becomes about 0.95 times.
It can be concluded that the cornering stiffness will be reduced to about 90% of the normal tire pressure condition and the longitudinal stiffness will be reduced to about 95% after inserts
supporting run-flat tire blowout.
Therefore, the changes of longitudinal and cornering stiffness during tire blowout can be expressed as follows:
${ K x = ( 0.95 K x 0 − K x 0 ) T d ( T − T s ) + K x 0 K y = ( 0.90 K y 0 − K y 0 ) T d ( T − T s ) + K y 0$
$K x 0$
is the longitudinal stiffness before tire blowout;
$K y 0$
is the cornering stiffness before tire blowout;
$T d$
is the duration of tire blowout;
$T s$
is the time of tire blowout; and
is the simulation time.
2.2.2. Change of Rolling Resistance Coefficient after Tire Blowout
The rolling resistance coefficient is the ratio of the required thrust to the wheel load when the wheel is rolling under certain conditions. The increase in the tire contact area after tire blowout
is the main factor of the rolling resistance change. The dimension parameters of the contact impression of the inserts supporting run-flat tire were extracted under zero-pressure condition and rated
load, as shown in
Table 1
Figure 2
It can be seen from
Figure 2
that the green frame is the shoulder part, and the red frame is the tire part with insert. Under the zero-pressure condition, the insert takes part in the load-bearing, and the colour of the tire
shoulder is lighter where it is affected by the insert. The increase of rolling resistance after tire blowout is mainly caused by the increase of contact area between elastic rubber and ground. The
rolling resistance of a normal tire increases 30 times after tire blowout. Compared with inserts supporting run-flat tire, the inserts are in direct contact with the tire and the rolling resistance
of this contact area is affected after tire blowout. Hence, the rolling resistance affected by insert is approximately equivalent to the rolling resistance of the rated tire pressure and the contact
area of insert is about 25% of the whole contact area from
Table 1
. The rolling resistance of the rest areas is 30 times of the normal tire pressure; therefore, the rolling resistance of inserts supporting run-flat tire under the zero-pressure condition is about
22.8 (0.25 + 0.75 × 30) times of that under the normal condition.
Therefore, the change of rolling resistance coefficient during tire blowout can be expressed as follows:
$R r _ c = ( 22.8 R r _ c 0 − R r _ c 0 ) T d ( T − T s ) + R r _ c 0$
$R r _ c 0$
is the rolling resistance coefficient before tire blowout.
2.2.3. Change of Effective Rolling Radius after Tire Blowout
Effective rolling radius refers to the vertical distance from the tire rolling center to the ground under a certain load when the tire is rolling steadily. The load characteristic curve of inserts
supporting run-flat tire under zero pressure condition was extracted, as shown in
Figure 3
As can be seen from
Figure 3
, the tire displacement under the zero-pressure is 100.36 mm, and the tire rolling radius and load radius under the rated load are 455 mm and 467 mm, respectively. Hence, the rolling radius of the
inserts supporting run-flat tire is about 0.92 times the normal rolling radius.
The change in effective rolling radius during tire blowout can be expressed as follows:
$R e = ( 0.92 R e 0 − R e 0 ) T d ( T − T s ) + R e 0$
$R e 0$
is the effective rolling radius before tire blowout.
3. Two Degrees of Freedom Model and Control System
3.1. Two Degrees of Freedom Model
When a tire blowout occurs, the control system needs to determine the additional yaw moment and the additional active angle according to the deviation between the actual driving state and the ideal
driving state, so as to keep the vehicle in the driving state before the tire blowout. At the same time, in order to facilitate the analysis of the relationship between the vehicle motion state and
stability, the vehicle model is simplified, and the vehicle two degrees of freedom control model is used as the ideal vehicle motion model.
The ideal expression of the two degrees of freedom model is as follows:
${ m V x ( β ˙ + r ) = − ( C f + C r ) β − ( a C f − b C r ) V x r + C f δ I z ⋅ r ˙ = ( b C r − a C f ) β − ( a 2 C f + b 2 C r ) V x r + a C f δ + Δ M z$
Considering the influence of the additional yaw moment acting on the vehicle, the above formula is transformed into the equation of state:
$[ β ˙ r ˙ ] = [ a 11 a 12 a 21 a 22 ] ⋅ [ β r ] + [ b 1 b 2 ] ⋅ δ + [ c 1 c 2 ] ⋅ Δ M z$
$[ a 11 a 12 a 21 a 22 ] = [ − ( C f + C r ) m V x b C r − a C f m V x − 1 b C r − a C f I z − b 2 C r + a 2 C f I z ⋅ V x ]$
$[ b 1 b 2 ] = [ C f m V x a C f I z ]$
$[ c 1 c 2 ] = [ 0 1 I z ]$
is the velocity of the vehicle along the x-axis;
is the total mass of the vehicle;
is the sideslip angle;
is the yaw rate;
is the front wheel angle;
are the distances from the mass center to the front and rear axles, respectively;
is the moment of inertia of the whole vehicle around the z-axis; and
are equivalent cornering stiffness of front axle and rear axle, respectively.
The yaw rate under ideal stable state is as follows:
$r d = V x L ( 1 + K ⋅ V x 2 ) δ$
is the axis between the front and rear wheels distance.
The expression of vehicle stability coefficient
is as follows:
$K = m L 2 ( a C r − b C f )$
The modified expression of ideal yaw angle is as follows:
$r d * = min { | V x L ( 1 + K ⋅ V x 2 ) δ | , | μ g V x | } ⋅ sgn ( δ )$
is the ideal yaw rate;
$r d *$
is the ideal yaw rate after correction;
is the adhesion coefficient of pavement; and
is the acceleration of gravity.
The ideal sideslip angle of the mass center is corrected as follows:
${ β d = b − m a V x 2 C r L L ( 1 + K V x 2 ) δ β d max = μ g V x 2 ( b − m a V x 2 C r L ) β d * = min { | b − m a V x 2 C r L L ( 1 + K V x 2 ) δ | , | μ g V x 2 ( b − m a V x 2 C r L ) | } ⋅ sgn
( δ )$
$β d *$
is the corrected ideal sideslip angle of mass center and
$β d$
$β d max$
is an intermediate variable.
3.2. Differential Braking Control System
The fuzzy sliding mode control algorithm is used to control the yaw moment after tire blowout, and the yaw moment is distributed to each wheel through the principle of braking force distribution, so
as to control the stability of the vehicle.
According to the comparison between the ideal yaw rate and sideslip angle of mass center and the actual one, the switching function of the sliding surface is selected. The expression is as follows:
$s = ( r − r d ) + ε ( β − β d )$
is the yaw rate of the vehicle and
is the ideal yaw rate of the vehicle;
is the sideslip angle of mass center; and
is the weighting coefficient.
The controller output yaw moment is as follows:
$Δ M z 1 = I z [ r ˙ d + ε β ˙ d − ( a 21 + ε a 11 ) β − ( a 21 + ε a 11 ) r − ( b 2 + ε b 1 ) δ − k s a t ( s τ ) ]$
$s a t ( s τ )$
is the saturation function.
In the process of tire blowout, the sideslip stiffness of tire blowout does not change linearly, so it is difficult to determine the magnitude of the compensated yaw moment, and the fuzzy controller
is selected to compensate the yaw moment. In the design of the Simulink fuzzy toolbox, the “Mamdain” method is selected as the fuzzy controller.
The two-dimensional structure was adopted in the fuzzy controller, that is, the input value is the difference of the sideslip angle of the mass center
$Δ β$
and the yaw rate
$Δ r$
, and the output value of the fuzzy controller is set to
. The basic domain of
$Δ β$
is [–5,5]. The basic domain of
$Δ r$
is [–1.5,1.5] and that of
is [–20,20]. The membership function of the fuzzy controller is selected as the Gauss membership function. Fuzzy control rules are shown in
Table 2
According to the fuzzy control algorithm, it can be known that the relationship of the compensated yaw moment obtained by maintaining the stability of the vehicle is as follows:
$Δ M z 2$
is the compensated yaw moment.
Therefore, the yaw moment to maintain the stability of the vehicle in the process of tire blowout is as follows:
$Δ M z = Δ M z 1 + Δ M z 2$
Considering the influence of tire blowout on vehicle stability, the tire without blowout is preferred in the braking force distribution strategy. After the master-slave braking tire is determined,
the braking force is determined according to the yaw moment, so as to keep the vehicle stable.
4. Simulation Results and Test Analysis
E-class SUV is selected as the basis of vehicle dynamics simulation model in CarSim, as shown in
Table 3
. The vehicle two degrees of freedom model is the basis of stability control. It describes the effects of yaw rate and sideslip angle of the mass center on vehicle yaw and lateral motion. In order to
simplify the model analysis, it is taken as the ideal vehicle motion model.
In order to study the characteristics of the inserts supporting run-flat tire, the UniTire tire model was built in Simulink. Based on CarSim, the simulation of run-flat tire blowout was carried out
under the straight driving condition. The tire blowout parameters of the front and rear wheels on the same side were extracted and compared with the normal tire. The tire size is 37 × 12.5R16.5 and
the outer diameter of the insert is 840 mm.
4.1. Tire Blowout Dynamic Response
When the vehicle runs in the straight running condition, the setting speed is 100 km/h, and the road adhesion coefficient is 0.85. In order to simplify the analysis, the left front wheel and the left
rear wheel were selected to carry out the tire blowout test, respectively, and the difference between the inserts supporting run-flat tire and the normal tire in the uncontrolled state were compared
after the tire blowout.
Left front tire blowout
When the vehicle is running in the straight running condition, the left front wheel suddenly blows out in the third second. The driver does not take any operation during the tire blowout, and the
running state of the vehicle was obtained. The change chart of tire blowout parameters is shown in
Figure 4
It can be seen from
Figure 4
that the characteristics of the inserts supporting run-flat tire blowout are similar to the normal tire under the straight driving condition. From
Figure 4
a,b, it can be seen that the lateral acceleration and lateral displacement of the inserts supporting run-flat tire are slightly larger than the normal tire, and the lateral displacement reaches 75 m
in 10 s.
Figure 4
c,d indicate that the parameter change is unstable in the first two seconds of tire blowout; the yaw rate and the sideslip angle change sharply, then gradually stabilize in about fifth second.
Left rear tire blowout
When the vehicle runs normally in a straight line, the left rear tire blows out in the third second, and the driver still does not take any measures. The simulation results are shown in
Figure 5
It can be seen from
Figure 5
that the rear tire blowout yaw of the inserts supporting run-flat tire is smaller than that of the normal tire, and the curve is obviously better than that of the front tires. This is because the
engine of the model selected is front-mounted, so the vertical load on the front axle is larger than that of the rear axle. When tire blowout occurs, the rolling resistance increases, and the front
tire is generally the driving wheel, which makes the force on both sides of the wheels uneven, forcing the car to lean to the side of the tire blowout.
4.2. Stability Control Analysis
In order to study the change of the stability parameters of the inserts supporting run-flat tire after the tire blowout, differential braking control is applied to the left front tire and the left
rear tire, respectively. The running speed of the vehicle is 100 km/h, and the road adhesion coefficient is 0.85. In the third second, tire blowout occurs. The state of the vehicle is shown in
Figure 6
Figure 7
, respectively.
Left front tire control results
As can be seen from
Figure 6
, the yaw state of the inserts supporting run-flat tire is obviously improved after applying the differential braking control, and the deviation from the original track under the control effect of
lateral displacement is less than 3 m; the change of yaw rate and sideslip angle is obviously reduced under the control effect. The maximum value of yaw rate after correction is 1.8 deg/s, and the
maximum deviation of sideslip angle after correction is about 1.2 deg. However, the direction of the sideslip angle of the inserts supporting run-flat tire is opposite to that before control.
Left rear tire control results
It can be seen from
Figure 7
that the left rear tire can basically follow the normal driving track, and the lateral displacement is about 0.25 m under the differential braking control. The lateral acceleration and yaw rate are
close to the norm, and the fluctuation is small. At 4 s, they can basically return to the normal state. Therefore, the stability of the vehicle with control after tire blowout is obviously improved
compared with that without control.
5. Conclusions
The tire blowout model of inserts supporting run-flat tire is built based on the UniTire theory. Combined with the whole vehicle model built by CarSim, the co-simulation is carried out to obtain the
change in characteristic parameters under the straight driving condition.
The difference between the left front tire and left rear tire blowout condition of inserts supporting run-flat tire and normal tire was compared. The results show that the characteristic
parameters of the two tires are similar. When the front tire blowout occurs, the yaw of the inserts supporting run-flat tire is larger, and when the rear tire blowout occurs, the yaw of the
normal tire is larger.
The stability of the inserts supporting run-flat tire after tire blowout is controlled according to the difference between the ideal yaw rate and the sideslip angle and the actual value. The
simulation results show that the differential braking control can better maintain the running track of the vehicle, significantly improve the stability of the vehicle, and whether the track
adjustment effect of the rear tire of inserts supporting run-flat tire is better.
Author Contributions
X.W. performed the data analyses and wrote the manuscript; L.Z. contributed to the conception of the study; Z.W. performed the experiment; Fen Lin contributed significantly to analysis and manuscript
preparation; Z.Z. helped perform the analysis with constructive discussions. All authors have read and agreed to the published version of the manuscript.
This research was partly supported by the National Natural Science Foundation of China under grant 51605215, and partly by the China Postdoctoral Science Foundation under grant 2019T120450, the Qing
Lan Project (Teacher Su 2019 [
]), Research Foundation of Nanjing Institute of Technology (grant number. CKJA201906).
Conflicts of Interest
The authors declare no conflict of interest.
Figure 1. Curve of inserts supporting run-flat tire test results. (a) Cornering stiffness curve; (b) longitudinal stiffness curve.
Figure 4. The simulation curve of left front tire blowout: (a) lateral acceleration curve; (b) lateral displacement curve; (c) yaw rate curve; (d) sideslip angle curve.
Figure 5. The simulation curve of left rear tire blowout: (a) lateral acceleration curve; (b) lateral displacement curve; (c) yaw rate curve; (d) sideslip angle curve.
Figure 6. The controlled simulation curve of left front tire blowout: (a) lateral acceleration curve; (b) lateral displacement curve; (c) yaw rate curve; (d) sideslip angle curve.
Figure 7. The controlled simulation curve of left rear tire blowout: (a) lateral acceleration curve; (b) lateral displacement curve; (c) yaw rate curve; (d) sideslip angle curve.
F/N W/mm L/mm S/mm^2
Whole contact area 12,250 235 580 136,300
Insert contact area 12,250 132 258 34,056
∆r ∆β
NB NM NS ZO PS PM PB
NB PB PB PB PB PM PS ZO
NM PB PB PM PM PM PS ZO
NS PM PM PM PM PS ZO NS
NO PM PS PS ZO NS NS NM
ZO PM PM PS ZO NS NS NM
PS PS PS ZO NM NM NM NM
PM ZO ZO NM NM NB NB NB
PB ZO NS NM NB NB NB NB
Parameter/Unit Parameter Symbol Parameter Value
Sprung mass/kg Ms 2290
Height of center of mass/mm h 810
Front axle distance/mm a 1180
Rear axle distance/mm b 1170
Wheel base/mm l 2950
Effective rolling radius/mm Re 455
Static load radius/mm R0 467
Rim diameter/mm D 419
Width of tire section/mm B 317
Radial stiffness/N⋅mm^−1 Kt 405
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Wang, X.; Zang, L.; Wang, Z.; Lin, F.; Zhao, Z. Study on the Stability Control of Vehicle Tire Blowout Based on Run-Flat Tire. World Electr. Veh. J. 2021, 12, 128. https://doi.org/10.3390/
AMA Style
Wang X, Zang L, Wang Z, Lin F, Zhao Z. Study on the Stability Control of Vehicle Tire Blowout Based on Run-Flat Tire. World Electric Vehicle Journal. 2021; 12(3):128. https://doi.org/10.3390/
Chicago/Turabian Style
Wang, Xingyu, Liguo Zang, Zhi Wang, Fen Lin, and Zhendong Zhao. 2021. "Study on the Stability Control of Vehicle Tire Blowout Based on Run-Flat Tire" World Electric Vehicle Journal 12, no. 3: 128.
Article Metrics | {"url":"https://www.mdpi.com/2032-6653/12/3/128","timestamp":"2024-11-13T20:00:27Z","content_type":"text/html","content_length":"452454","record_id":"<urn:uuid:2ebf42e1-a412-4ff1-94f4-58c4a8107fcc>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00859.warc.gz"} |
This is the development place for the R-package surveysd. The package can be used to estimate the standard deviation of estimates in complex surveys using bootstrap weights.
# Install release version from CRAN
# Install development version from GitHub
Bootstrapping has long been around and used widely to estimate confidence intervals and standard errors of point estimates. This package aims to combine all necessary steps for applying a calibrated
bootstrapping procedure with custom estimating functions.
A typical workflow with this package consists of three steps. To see these concepts in practice, please refer to the getting started vignette.
• Calibrated weights can be generated with the function ipf() using an iterative proportional updating algorithm.
• Bootstrap samples are drawn with rescaled bootstrapping in the function draw.bootstrap().
• These samples can then be calibrated with an iterative proportional updating algorithm using recalib().
• Finally, estimation functions can be applied over all bootstrap replicates with calc.stError().
Further reading
More information can be found on the github-pages site for surveysd. | {"url":"https://cran.hafro.is/web/packages/surveysd/readme/README.html","timestamp":"2024-11-08T11:57:53Z","content_type":"application/xhtml+xml","content_length":"8590","record_id":"<urn:uuid:1ac34219-93d4-45f2-8abc-d2dcc57e0358>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00434.warc.gz"} |
Re: p-value for CMH
Is there a SAS procedure to calculate p-value for Mantel-Haenszel Stratum Weighted Method for risk difference in stratified samples? It provided the CI but I can't find p-value for this test.
SAS code I used is
ods output CommonPdiff=CommonPdiff ;
proc freq data = resp order=data;
tables stratum1*stratum2*trt01pn*resp/riskdiff(common) ;
I can't find option from
Many thanks
04-27-2022 01:16 PM | {"url":"https://communities.sas.com/t5/Statistical-Procedures/p-value-for-CMH/m-p/810271","timestamp":"2024-11-02T20:40:09Z","content_type":"text/html","content_length":"331221","record_id":"<urn:uuid:de8be370-9e2f-4248-ad4a-cc2ebdf5f96a>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00727.warc.gz"} |
Who can handle my statistics homework and exams? | Hire Someone To Take My Statistics Assignment
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How To Cheat On My Math Of Business College Class Online
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perplexus.info :: Just Math : Age of ages (2)
Determine the present ages of each of the three siblings Toby, Julie and Melanie from the following clues:
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Lucky 13 paper dance!
Having recently rediscovered arxiv.org/submit, I thought I’d mention a few papers to come out of team UW.
Two particularly exciting single-author papers are from students here.
Kamil Michnicki has developed a 3-d stabilizer code on $latex n$ qubits with an energy barrier that scales as $latex n^{2/9}$. By contrast, such a result is impossible in 2-d, and the best energy
barrier previously obtained in 3-d was $latex O(log n)$ from Haah’s breakthrough cubic code. Sadly, this code appears not to have the thermal stability properties of the 4-d toric code, but it
nevertheless is an exciting step towards a self-correcting quantum memory.
1208.3496: 3-d quantum stabilizer codes with a power law energy barrier
Kamil Michnicki
David Rosenbaum has written a mostly classical algorithms paper about the old problem of group isomorphism: given two groups $latex G,H$ specified by their multiplication tables, determine whether
they are isomorphic. The problem reduces to graph isomorphism, but may be strictly easier. Since any group with $latex |G|=n$ has a generating set of size $latex leq log_2(n)$, it follows that the
problem can be solved in time $latex n^{log(n)+O(1)}$. While faster algorithm have been given in many special cases, the trivial upper bound of $latex n^{log(n)}$ has resisted attack for decades. See
Gödel’s Lost Letter for some discussion. The particular classes of groups considered to be hardest have been the nilpotent (or more generally solvable) groups, since paradoxically the rigidity of
highly non-abelian groups (e.g. simple groups) makes them easier to address. David found a polynomial speedup for solvable groups, thus making the first progress on this problem since the initial
$latex n^{log(n)}$ algorithms.
1205.0642: Breaking the $latex n^{log n}$ Barrier for Solvable-Group Isomorphism
David Rosenbaum
Lukas Svec (together with collaborators) also has a nice way of improving the Gottesman-Knill simulations that have been so effective in estimating FTQC thresholds. Gottesman-Knill allows mixtures of
Clifford unitaries to be simulated classically, which seems as thought it should be only be effective for simulating unital noise. However, throwing away a qubit and replacing it with the $latex |
0rangle$ state can also be encompassed within the Gottesman-Knill approach. This insight allows them to give much better simulations of amplitude-damping noise than any previous approach.
1207.0046: Approximation of real error channels by Clifford channels and Pauli measurements
Mauricio Gutiérrez, Lukas Svec, Alexander Vargo, Kenneth R. Brown
There have also been two papers about the possibilities of two dimensions. David has a paper explaining how a general circuit with arbitrary two-qubit interactions on $latex n$ qubits can be
simulated in a 2-d architecture by using $latex n^2$ qubits. If only $latex k$ gates happen per time step, then $latex nk$ qubits suffice. The key trick is to use classical control to perform
teleportation chains, an idea of whose provenance I’m unclear on, but which is based in part on MBQC and in part on a remarkable paper of Terhal and Divincenzo.
1205.0036: Optimal Quantum Circuits for Nearest-Neighbor Architectures
David Rosenbaum
Examining particular algorithms can enable more dramatic speedups, and by examining Shor’s algorithm, Paul Pham was able to reduce the depth to polylogarithmic, surprisingly finding an improved
implementation of the most well-studied of quantum algorithms.
1207.6655: A 2D Nearest-Neighbor Quantum Architecture for Factoring
Paul Pham, Krysta M. Svore
David and I also have a joint paper on an alternate oracle model in which one input to the oracle is supplied by the user and a second input is random noise. While in some cases (e.g. a Grover oracle
that misfires) this does not lead to quantum advantages, we find that in other cases, quantum computers can solve problems in a single query that classical computers cannot solve with unlimited
queries. Along the way, we address the question of when some number of (quantum or classical) queries yield no useful information at all about the answer to an oracle problem.
1111.1462: Uselessness for an Oracle Model with Internal Randomness
David Rosenbaum, Aram W. Harrow
My co-blogger Steve has also been active. Steve and his co-authors (does that make them my step-co-authors?) have written perhaps the definitive work on how to estimate approximately low-rank density
matrices using a small number of measurements.
1205.2300: Quantum Tomography via Compressed Sensing: Error Bounds, Sample Complexity, and Efficient Estimators
Steven T. Flammia, David Gross, Yi-Kai Liu, Jens Eisert
Steve, together with Ghost Pontiff Dave and Dave’s former student Gregory Crosswhite have also posted
1207.2769: Adiabatic Quantum Transistors
Dave Bacon, Steven T. Flammia, Gregory M. Crosswhite
This paper proposes a deeply innovative approach to quantum computing, in which one adiabatically transforms one simple spatially-local Hamiltonian to another. Unlike previous approaches, it seems to
have a chance of having some compelling fault-tolerance properties, although analyzing this remains challenging.
Steve and I also have a brief note (arXiv:1204.3404) relevant to my ongoing debate with Gil Kalai (see here, here, here, here, here, here or here) in which we point out counterexamples to one of
Gil’s conjectures. This post specifically contains more discussion of the issue.
Finally, I’ve been clearing out a lot of my unpublished backlog this year.
My co-authors and I wrote a short paper explaining the main ideas behind the superactivation of zero-error capacity. The principle is similar to that found in all of the additivity violations based
on random quantum channels: we choose a correlated distribution over channels $latex {cal N}_1, {cal N}_2 $ in a way that forces $latex {cal N}_1otimes {cal N}_2$ to have some desired behavior (e.g.
when acting on a particular maximally entangled state). At the same time, apart from this constraint, the distribution is as random as possible. Hopefully we can then show that any single-copy use of
$latex {cal N}_1$ or $latex {cal N}_2$ has low capacity, or in our case, zero zero-error capacity. In our case, there are a few twists, since we are talking about zero-error capacity, which is a
fragile property more suited to algebraic geometry than the usual approximate techniques in information theory. On the other hand, this means that at many points we can show that properties hold with
probability 1. The other nontrivial twist is that we have to show that not only $latex {cal N}_i$ has zero zero-error capacity (yeah, I know it’s a ridiculous expression) but $latex {cal N}_i^{otimes
n}$ does for all $latex n$. This can be done with some more algebraic geometry (which is a fancy way of saying that the simultaneous zeroes of a set of polynomials has measure equal to either 0 or 1)
as well as the fact that the property of being an unextendible product bases is stable under tensor product.
1109.0540:Entanglement can completely defeat quantum noise
Jianxin Chen, Toby S. Cubitt, Aram W. Harrow, Graeme Smith
One paper that was a fun bridge-building exercise (with a nice shout out from Umesh Vazirani/BILL GASARCH) was a project with quantum information superstar Fernando Brandão as well as a pack of
classical CS theorists. My part of the paper involved connections between QMA(2) and optimizing polynomials over $latex mathbb{R}^n$. For example, if $latex a_1,ldots, a_m$ are the rows of a matrix
$latex A$, then define $latex |A|_{2rightarrow 4} = max_{|x|_2=1} |Ax|_4 = max_{|x|_2=1} (sum_{i=1}^m |langle a_i, xrangle|^4)^{1/4}$. Taking the fourth power we obtain the maximum energy attainable
by product states under the Hamiltonian $latex H = sum_{i=1}^m a_i a_i^* otimes a_i a_i^*$. Thus, hardness results and algorithms can be ported in both directions. One natural algorithm is called
“the Lasserre SDP hierarchy” classically and “optimizing over $latex k$-extendable states” quantumly, but in fact these are essentially the same thing (an observation dating back to a 2003 paper of
Doherty, Parrilo, and Spedalieri). There is much more to the paper, but I’ll leave it at that for now.
1205.4484: Hypercontractivity, Sum-of-Squares Proofs, and their Applications
Boaz Barak, Fernando G.S.L. Brandão, Aram W. Harrow, Jonathan A. Kelner, David Steurer, Yuan Zhou
Another big collaboration taking me out of my usual areas was this paper on quantum architecture. Suppose that our quantum computer is comprised of many small nodes (say ion traps), connected by
long-range links (say optical fibers), as has been recently advocated. This computer would not be stuck with a 2-d topology, but could be connected in any reasonably low-degree configuration. Our
paper shows that a hypercube topology (among many other possibilities) is enough to simulate general quantum circuits. This enables parallelized versions of Grover search that finally (in my opinion)
address the problem raised by Grover and Rudolph about the memory requirements for the “best” known collision and element-distinctness algorithms. As a result, we find space-time tradeoffs (assuming
this hypercube topology) for collision and element distinctness of $latex ST=tilde O(sqrt N)$ and $latex ST=tilde O(N)$ respectively.
1207.2307: Efficient Distributed Quantum Computing
Robert Beals, Stephen Brierley, Oliver Gray, Aram W. Harrow, Samuel Kutin, Noah Linden, Dan Shepherd, Mark Stather
The next paper was on more familiar territory. Together with my former PhD student Richard Low (and building on the work of Dahlsten, Oliveira and Plenio), I proved that random quantum circuits are
approximate unitary 2-designs in 0802.1919. Later Fernando Brandão and Michal Horodecki improved this to show that random quantum circuits are approximate unitary 3-designs in 1010.3654 (achieving a
sweet oracle speedup in the process). Teaming up with them I expected maybe to reach 5-designs, but in the end we were able to get arbitrary $latex k$-designs on $latex n$ qubits with circuits of
length $latex {rm poly}(n,k)$.
1208.0692 Local random quantum circuits are approximate polynomial-designs
Fernando G. S. L. Brandão, Aram W. Harrow, Michal Horodecki
Finally, one more paper came out of my classical CS dabbling. Together with classical (but quantum curious) theorists Alexandra Kolla and Leonard Schulman, we found a cute combinatorial result in our
failed bid to refute the unique games conjecture on the hypercube. Our result concerns what are called maximal functions. Hardy and Littlewood introduced these by talking about cricket; here is a
contemporary version. Imagine that you are a Red Sox fan whose happiness at any given time depends on what fraction of the last $latex n$ games the Red Sox have won against the Yankees. Fortunately,
you are willing to choose $latex n$ differently from day to day in order to maximize your happiness. For example, if the Red Sox won the last game, you take $latex n=1$ and your happiness is 100%. If
they won 3 out of the last 5 games, you could take $latex n=5$ and obtain happiness 60%. The maximal operator takes the win-loss series and transforms it into the happiness function. (A similar
principle is at work when the loser of rock-paper-scissors proposes best 3-out-of-5, and then best 4-out-of-7, etc.) In our paper, we bound how much the maximal operator can increase the 2-norm of a
function when it acts on functions on the hypercube, and the maximization is taken not over intervals, but over Hamming spheres in the hypercube. Along the way, we prove some bounds on Krawtchouk
polynomials that seem so simple and useful I feel we must have overlooked some existing paper that already proves them.
1209.4148: Dimension-free L2 maximal inequality for spherical means in the hypercube
Aram W. Harrow, Alexandra Kolla, Leonard J. Schulman
3 Replies to “Lucky 13 paper dance!”
1. Impressive! (Brandao’s name needs error-correction)
1. Thanks! fixed it. | {"url":"https://dabacon.org/pontiff/2012/09/19/lucky-13-paper-dance/","timestamp":"2024-11-02T11:16:22Z","content_type":"text/html","content_length":"110353","record_id":"<urn:uuid:987573f1-f7a1-4b51-a51f-091cec294963>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00680.warc.gz"} |
How do you solve \
Hint: These types of problems can be solved using basic logarithm formulas. First we will simplify the LHS using the formulas we have and then we will apply the formula to get the result from that
simplified equation we got.
Let us know some logarithm formulas to solve the question.
\[\log a+\log b=\log \left( a\times b \right)\] when both bases are equal.
If \[\log x=\log y\] then \[x=y\] when both bases are equal.
Complete step by step solution:
From the given question, we are given to solve \[{{\log }_{3}}2+{{\log }_{3}}7={{\log }_{3}}x\].
We can see that on the LHS side we have both the terms with the same base.
So we can apply the above discussed formula here to simplify the LHS.
\[\log a+\log b=\log \left( a\times b \right)\]
We will use this formula here.
By applying the formula we will get
\[\Rightarrow {{\log }_{3}}\left( 2\times 7 \right)={{\log }_{3}}x\]
By multiplying we will get
\[\Rightarrow {{\log }_{3}}14={{\log }_{3}}x\]
Here again we can have two terms equal with the same base.
So we can use the other that is said above to simplify this further.
If \[\log x=\log y\] then \[x=y\].
This is the formula we will use here.
So we have same base as 3 we can write our equation as
\[\Rightarrow 14=x\]
By rewriting it we will get
\[\Rightarrow x=14\]
So by solving the given equation we will get x value as 14.
Note: We must know the formulas to solve this type of questions. Otherwise it would be difficult to figure out the solution. Also we should be careful that we have to apply these formulas only when
their bases are equal otherwise we don’t use these formulas. | {"url":"https://www.vedantu.com/question-answer/how-do-you-solve-log-32+log-37log-3x-class-9-maths-cbse-6010bf51962db9208d20f4d2","timestamp":"2024-11-11T01:02:47Z","content_type":"text/html","content_length":"153530","record_id":"<urn:uuid:e88acffd-68ac-42fd-ab70-dddbfbd0dc20>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00158.warc.gz"} |
Kirchoff’s Voltage Law for Electrostatics - Differential Form - Electrical Engineering Textbooks
Kirchoff’s Voltage Law for Electrostatics - Differential Form
The integral form of Kirchoff’s Voltage Law for electrostatics (KVL; Section 5.10) states that an integral of the electric field along a closed path is equal to zero:
is electric field intensity and
is the closed curve. In this section, we derive the differential form of this equation. In some applications, this differential equation, combined with boundary conditions imposed by structure and
materials (Sections 5.17 and 5.18), can be used to solve for the electric field in arbitrarily complicated scenarios. A more immediate reason for considering this differential equation is that we
gain a little more insight into the behavior of the electric field, disclosed at the end of this section.
The equation we seek may be obtained using Stokes’ Theorem (Section 4.9), which in the present case may be written:
is any surface bounded by
, and
is the normal to that surface with direction determined by right-hand rule. The integral form of KVL tells us that the right hand side of the above equation is zero, so:
The above relationship must hold regardless of the specific location or shape of
. The only way this is possible for all possible surfaces is if the integrand is zero at every point in space. Thus, we obtain the desired expression:
The differential form of Kirchoff’s Voltage Law for electrostatics (Equation 5.11.4) states that the curl of the electrostatic field is zero.
Equation 5.11.4 is a partial differential equation. As noted above, this equation, combined with the appropriate boundary conditions, can be solved for the electric field in arbitrarily-complicated
scenarios. Interestingly, it is not the only such equation available for this purpose – Gauss’ Law (Section 5.7) also does this. Thus, we see a system of partial differential equations emerging, and
one may correctly infer that that the electric field is not necessarily fully constrained by either equation alone.
Additional Reading
Explore CircuitBread
Get the latest tools and tutorials, fresh from the toaster. | {"url":"https://www.circuitbread.com/textbooks/electromagnetics-i/electrostatics/kirchoffs-voltage-law-for-electrostatics-differential-form","timestamp":"2024-11-03T07:20:49Z","content_type":"text/html","content_length":"931732","record_id":"<urn:uuid:9f1d46d8-dd8c-4ddd-9ac2-444b5bd579df>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00294.warc.gz"} |
Homework 1 To write simple functions. - Programming Help
• All Finger Exercises on this homework should be done individually (without a partner) and you should have them done by Monday night. Finger exercises will not be graded and do not have to be
turned in.
• All Graded Exercises on this homework should be done with the partner you’re assigned in lab on Tuesday Sept 15.
• For this and all future assignments you must submit a single .rkt file containing your responses to all exercises via the Handin Server. We accept no email submissions.
• You must use the language specified at the top of this page.
• On this assignment, you are encouraged to write signatures and purpose statements for every function you write, in the format we studied in class; we will give you feedback on how clear or
accurate they are. On future homeworks, they will be required and graded as part of the assignment. As for check-expects: follow the individual instructions for each problem.
Failure to comply with these expectations will result in deductions and possibly a 0 score.
Finger Exercises You are not required to submit your finger exercises but they will be helpful so we recommend doing them anyway.
Exercise 1 Write a function that subtracts 2 from a number.
Exercise 2 The following table describes how far a person has gone in a race in a certain amount of seconds:
t = 1 2 3 4 5 6 7 8 9 10
d = 3 4.5 6.0 7.5 9.0 10.5 12.0 13.5 ? ?
Write a function that predicts, based on this data, how far they will have run at time t. Write three check-expects: two that test to see the data in the table matches your function’s output, and
then one more that tests the output when t is at least 9.
Exercise 3 Take a look at figure 1, which is a graph of f, a function of x.
Turn the graph into a table for x = 0, 1, 2 and 3 and formulate a function definition.
Figure 1: A function graph for f
Exercise 4 Translate the following mathematical function into a BSL function:
f(x) = x2 + 12
Use it to create a table for x = 0, 2, 5, and 9.
Exercise 5 Enter the following function definition in BSL into DrRacket:
Look up the documentation for expt.
Apply the function to 0, 1, and 3 in the interactions area.
Apply the function to 1 in the definitions area and use the stepper to see how DrRacket evaluates this program.
Exercise 6 Enter the following function definition in BSL into DrRacket:
What kind of argument does hello consume? Apply the function to your favorite argument and step through the evaluation.
Exercise 7 Design the function render-string, which consumes a number t and produces a text image of the first t letters from the string “qwerty”.
Place the text on a white 200 x 100 rectangle. Use black text of font size 22.
Graded Exercises
Exercise 8 Write a function, nineply, that multiplies a number, supplied as an argument, by 9. Write three check-expects: one for a negative value, another for zero, and a third for a positive
Exercise 9 Complete the following table:
t = 0 1 2 3 4 5 6 7
d = FUNDIES FUNDIE FUNDI FUND FUN FU ? ?
Turn this table into a function definition. In a comment before the function, give as precise a signature for this function as you can (hint: it’s not quite straightforward – you may need to
describe it rather than use the notation from class).
Exercise 10 Implement the function, cycle-spelling, so that when (animate cycle-spelling) is called, it animates spelling the following long word, letter-by-letter (in all caps). When the end of
the word is reached, it cycles back and starts from the beginning.
(define LONG-WORD “diScomBoBUlaTeDiScomBoBulaTeDboBUlaTeD”)
The word should be displayed in blue in font size 40 on a white background.
Your code ought to work no matter what string is used when defining LONG-WORD, without needing to change anything else.
One way to count 0 up to some number and then loop back is to think about dividing two numbers and taking the remainder; so the remainder of 1 / 3 is 1, the remainder of 2 / 3 is 2; and the
remainder of 3 / 3 is back to 0. Look for the remainder function in BSL.
You may want to look at the documentation at https://docs.racket-lang.org/htdp-langs/beginner.html for some of the functions available to you in BSL.
Exercise 11 Design a function, draw-kite, that produces a kite with 4 colors when given the desired width and height of the kite. We have provided two check-expects to help you test your function
in the starter file for this assignment.
The kite should have the following four colors, starting with the top left quadrant going clockwise: “blue”, “yellow”, “green”, “red”.
The widest point of the kite is at one-third of the distance from the top and two-thirds from the bottom.
Exercise 12 Write a function, convert-to-inches, that takes as arguments a number of yards, feet, and inches (in that order) and converts to a total number of inches. For example, (
convert-to-inches 4 1 6) would produce 162 inches. Write two check-expects: one expressing the example above, and a different one of your choosing.
Exercise 13 Design the function speed-check. It consumes two natural numbers: one that represents the speed of a car and the other one the speed limit of the road. The result is one of these
strings: (1) “okay” for a car that goes below the speed limit; (2) “bit fast” for a car that is going at most 9 mph faster than the speed limit; or (3) “speeding: ___ mph over limit” for a car
that exceeds the speed limit, where the underlines are replaced by how much the car is over the speed limit. Write five check-expects that test the function’s behavior in different scenarios. | {"url":"https://www.edulissy.org/product/all-finger-exercises-on-this-homework-should-be-done-individually-without-a-partner/","timestamp":"2024-11-09T09:38:55Z","content_type":"text/html","content_length":"214282","record_id":"<urn:uuid:94a17d6a-7653-44a0-8042-451b76108b31>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00634.warc.gz"} |
An Introduction to Linear Stability Analysis | Dr. Daniel Edgington-Mitchell
Hydrodynamic stability, that is the tendency of infinitesimal perturbations to fluid mechanical systems to grow in amplitude, has been studied since the nineteenth century. Amongst the best known
early experimental examples comes from the work of Osborne Reynolds, whose sketches of pipe flow are shown below:
Sketches of laminar and turbulent flow in a pipe, Reynolds (1883).
Reynolds demonstrated that at sufficiently low speeds, the dye he injected into the pipe “extended in a beautiful straight line through the tube”. As the flow velocity increased, the line of dye
would begin to shift around, and then with further increases, some critical velocity would be reached where the dye would begin to mix with the surrounding water. When the flow was observed with a
spark (to freeze the motion at a moment in time), this mixing was observed to be characterized by “a mass of more or less distinct curls, showing eddies.”
Reynolds’ experiments have been duplicated thousands of times over, but a particularly beautiful example of this transition process (though different to the pipe flow experiment of Reynolds) can be
seen in the following video:
There are many forms of hydrodynamic instability, driven by buoyancy, shear, surface tension, centrifugal force, and various other mechanisms that may disturb the equilibrium between the forces
acting on the fluid and its internal dissipative mechanisms. In our consideration of jet noise however, there is one mechanism of hydrodynamic instability that concerns us above all others, and that
is the Kelvin-Helmholtz instability, beautifully visualized in the video below:
Our goal here is thus to furnish ourselves with the tools to understand what the Kelvin-Helmholtz instability is, and to predict when and where it will occur. For a further introductory explanation
of the Kelvin-Helmholtz instability, the same Youtube Channel as above, Sixty Symbols, has an interview with Prof. Mike Merryfield from the University of Nottingham, who provides an excellent
discussion in lay language as to how the structures associated with the KH instability form.
So, we know what the Kelvin-Helmholtz instability looks like, and we have a sort of lay-appropriate explanation for how it works, but can we predict it mathematically? We certainly can, head on over
and see the most simplified form of the KH problem explored:
So, now you know a little bit about the Kelvin-Helmholtz instability, at least how it works in a very idealized system where we ignored most of the real physics. In a real flow, where we have
compressibility, viscosity, and a continuous velocity profile rather than a discontinuity it….. actually works in qualitatively the same way, and we can even get fairly reasonable estimates of which
frequencies will be the most unstable in a real flow from the very simplified analysis that we considered above. It’s not perfect, but it’s a start. If we want to do more complex mathematics, we can
of course get much closer. We’ll come back to that later.
For now, let’s get back to our discussion of jets.
What about if we wanted to look at a vortex-sheet solution for a jet instead of a mixing layer, and include compressibility? | {"url":"https://daniel.edgington-mitchell.com/first-course-jet-noise/introduction-linear-stability-analysis/","timestamp":"2024-11-09T12:16:27Z","content_type":"text/html","content_length":"43701","record_id":"<urn:uuid:6b93b55e-6705-4c4f-bdbc-5ac25d7393f8>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00273.warc.gz"} |
Flipkart | Find the minimum number of decrements needed to make all array elements equal ?
Expected Time Complexity : O(N) , where 'N' is the size of the array .
Example :
Output : 5 [ {2,2,2,2} will be the final array with all equal elements]
Note :
1)Algorithm-explanation is must .
2)Adding code is optional .
3)Use Format option while adding code
Company-wise-questions 15
Algorithm: Since we can only make decrements in the array, the fastest to make all elements equal are to bring down all the elements to the minimum element of the array. After finding the minimum
element, our answer will be the sum of difference between each array element and the minimum
int n = sc.nextInt();
int arr[] = sc.nextArray(n); // Function to read array
int mini = Integer.MAX_VALUE;
for(int i : arr)
mini = Math.min(mini, i);
int cnt = 0;
for(int i : arr) cnt+= (i-mini); | {"url":"https://www.desiqna.in/887/flipkart-minimum-number-decrements-needed-array-elements","timestamp":"2024-11-03T03:41:39Z","content_type":"text/html","content_length":"35524","record_id":"<urn:uuid:a7a4e483-438e-4a8e-bb09-8ecaabb0f012>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00093.warc.gz"} |
IJCAI.2020 - Knowledge Representation and Reasoning
In Boolean games, each agent controls a set of Boolean variables and has a goal represented by a propositional formula. We study inference problems in Boolean games assuming the presence of a
PRINCIPAL who has the ability to control the agents and impose taxation schemes. Previous work used taxation schemes to guide a game towards certain equilibria. We present algorithms that show how
taxation schemes can also be used to infer agents' goals. We present experimental results to demonstrate the efficacy our algorithms. We also consider goal inference when only limited information is
available in response to a query.
In some agent designs like inverse reinforcement learning an agent needs to learn its own reward function. Learning the reward function and optimising for it are typically two different processes,
usually performed at different stages. We consider a continual (``one life'') learning approach where the agent both learns the reward function and optimises for it at the same time. We show that
this comes with a number of pitfalls, such as deliberately manipulating the learning process in one direction, refusing to learn, ``learning'' facts already known to the agent, and making decisions
that are strictly dominated (for all relevant reward functions). We formally introduce two desirable properties: the first is `unriggability', which prevents the agent from steering the learning
process in the direction of a reward function that is easier to optimise. The second is `uninfluenceability', whereby the reward-function learning process operates by learning facts about the
environment. We show that an uninfluenceable process is automatically unriggable, and if the set of possible environments is sufficiently large, the converse is true too.
Analogical transfer consists in leveraging a measure of similarity between two situations to predict the amount of similarity between their outcomes. Acquiring a suitable similarity measure for
analogical transfer may be difficult, especially when the data is sparse or when the domain knowledge is incomplete. To alleviate this problem, this paper presents a dataset complexity measure that
can be used either to select an optimal similarity measure, or if the similarity measure is given, to perform analogical transfer: among the potential outcomes of a new situation, the most plausible
is the one which minimizes the dataset complexity.
Ontology-mediated query answering (OMQA) is a promising approach to data access and integration that has been actively studied in the knowledge representation and database communities for more than a
decade. The vast majority of work on OMQA focuses on conjunctive queries, whereas more expressive queries that feature counting or other forms of aggregation remain largely unexplored. In this paper,
we introduce a general form of counting query, relate it to previous proposals, and study the complexity of answering such queries in the presence of DL-Lite ontologies. As it follows from existing
work that query answering is intractable and often of high complexity, we consider some practically relevant restrictions, for which we establish improved complexity bounds.
Previous research has claimed dynamic epistemic logic (DEL) to be a suitable formalism for representing essential aspects of a Theory of Mind (ToM) for an autonomous agent. This includes the ability
of the formalism to represent the reasoning involved in false-belief tasks of arbitrary order, and hence for autonomous agents based on the formalism to become able to pass such tests. This paper
provides evidence for the claims by documenting the implementation of a DEL-based reasoning system on a humanoid robot. Our implementation allows the robot to perform cognitive perspective-taking, in
particular to reason about the first- and higher-order beliefs of other agents. We demonstrate how this allows the robot to pass a quite general class of false-belief tasks involving human agents.
Additionally, as is briefly illustrated, it allows the robot to proactively provide human agents with relevant information in situations where a system without ToM-abilities would fail. The symbolic
grounding problem of turning robotic sensor input into logical action descriptions in DEL is achieved via a perception system based on deep neural networks.
We consider the setting of asynchronous opinion diffusion with majority threshold: given a social network with each agent assigned to one opinion, an agent will update its opinion if more than half
of its neighbors agree on a different opinion. The stabilized final outcome highly depends on the sequence in which agents update their opinion. We are interested in optimistic sequences---sequences
that maximize the spread of a chosen opinion. We complement known results for two opinions where optimistic sequences can be computed in time and length linear in the number of agents. We analyze
upper and lower bounds on the length of optimistic sequences, showing quadratic bounds in the general and linear bounds in the acyclic case. Moreover, we show that in networks with more than two
opinions determining a spread-maximizing sequence becomes intractable; surprisingly, already with three opinions the intractability results hold in highly restricted cases, e.g., when each agent has
at most three neighbors, when looking for a short sequence, or when we aim for approximate solutions.
Spectrum-based Fault Localization (SFL) approaches aim to efficiently localize faulty components from examining program behavior. This is done by collecting the execution patterns of various
combinations of components and the corresponding outcomes into a spectrum. Efficient fault localization depends heavily on the quality of the spectra. Previous approaches, including the current
state-of-the-art Density- Diversity-Uniqueness (DDU) approach, attempt to generate “good” test-suites by improving certain structural properties of the spectra. In this work, we propose a different
approach, Multiverse Analysis, that considers multiple hypothetical universes, each corresponding to a scenario where one of the components is assumed to be faulty, to generate a spectrum that
attempts to reduce the expected worst-case wasted effort over all the universes. Our experiments show that the Multiverse Analysis not just improves the efficiency of fault localization but also
achieves better coverage and generates smaller test-suites over DDU, the current state-of-the-art technique. On average, our approach reduces the developer effort over DDU by over 16% for more than
92% of the instances. Further, the improvements over DDU are indeed statistically significant on the paired Wilcoxon Signed-rank test.
In deductive module extraction, we determine a small subset of an ontology for a given vocabulary that preserves all logical entailments that can be expressed in that vocabulary. While in the
literature stronger module notions have been discussed, we argue that for applications in ontology analysis and ontology reuse, deductive modules, which are decidable and potentially smaller, are
often sufficient. We present methods based on uniform interpolation for extracting different variants of deductive modules, satisfying properties such as completeness, minimality and robustness under
replacements, the latter being particularly relevant for ontology reuse. An evaluation of our implementation shows that the modules computed by our method are often significantly smaller than those
computed by existing methods.
In a large-scale knowledge graph (KG), an entity is often described by a large number of triple-structured facts. Many applications require abridged versions of entity descriptions, called entity
summaries. Existing solutions to entity summarization are mainly unsupervised. In this paper, we present a supervised approach NEST that is based on our novel neural model to jointly encode graph
structure and text in KGs and generate high-quality diversified summaries. Since it is costly to obtain manually labeled summaries for training, our supervision is weak as we train with
programmatically labeled data which may contain noise but is free of manual work. Evaluation results show that our approach significantly outperforms the state of the art on two public benchmarks.
In this paper we focus on a less usual way to represent Boolean functions, namely on representations by switch-lists. Given a truth table representation of a Boolean function f the switch-list
representation (SLR) of f is a list of Boolean vectors from the truth table which have a different function value than the preceding Boolean vector in the truth table. The main aim of this paper is
to include the language SL of all SLR in the Knowledge Compilation Map [Darwiche and Marquis, 2002] and to argue, that SL may in certain situations constitute a reasonable choice for a target
language in knowledge compilation. First we compare SL with a number of standard representation languages (such as CNF, DNF, and OBDD) with respect to their relative succinctness. As a by-product of
this analysis we also give a short proof of a long standing open question from [Darwiche and Marquis, 2002], namely the incomparability of MODS (models) and PI (prime implicates) languages. Next we
analyze which standard transformations and queries (those considered in [Darwiche and Marquis, 2002] can be performed in poly-time with respect to the size of the input SLR. We show that this
collection is quite broad and the combination of poly-time transformations and queries is quite unique.
Counting answers to a query is an operation supported by virtually all database management systems. In this paper we focus on counting answers over a Knowledge Base (KB), which may be viewed as a
database enriched with background knowledge about the domain under consideration. In particular, we place our work in the context of Ontology-Mediated Query Answering/Ontology-based Data Access (OMQA
/OBDA), where the language used for the ontology is a member of the DL-Lite family and the data is a (usually virtual) set of assertions. We study the data complexity of query answering, for
different members of the DL-Lite family that include number restrictions, and for variants of conjunctive queries with counting that differ with respect to their shape (connected, branching, rooted).
We improve upon existing results by providing PTIME and coNP lower bounds, and upper bounds in PTIME and LOGSPACE. For the LOGSPACE case, we have devised a novel query rewriting technique into
first-order logic with counting.
In this paper, we present a learning-based approach to determining acceptance of arguments under several abstract argumentation semantics. More specifically, we propose an argumentation graph neural
network (AGNN) that learns a message-passing algorithm to predict the likelihood of an argument being accepted. The experimental results demonstrate that the AGNN can almost perfectly predict the
acceptability under different semantics and scales well for larger argumentation frameworks. Furthermore, analysing the behaviour of the message-passing algorithm shows that the AGNN learns to adhere
to basic principles of argument semantics as identified in the literature, and can thus be trained to predict extensions under the different semantics – we show how the latter can be done for
multi-extension semantics by using AGNNs to guide a basic search. We publish our code at https://github.com/DennisCraandijk/DL-Abstract-Argumentation.
We consider an agent that operates with two models of the environment: one that captures expected behaviors and one that captures additional exceptional behaviors. We study the problem of
synthesizing agent strategies that enforce a goal against environments operating as expected while also making a best effort against exceptional environment behaviors. We formalize these concepts in
the context of linear-temporal logic, and give an algorithm for solving this problem. We also show that there is no trade-off between enforcing the goal under the expected environment specification
and making a best-effort for it under the exceptional one.
Description logics are well-known logical formalisms for knowledge representation. We propose to enrich knowledge bases (KBs) with dynamic axioms that specify how the satisfaction of statements from
the KBs evolves when the interpretation is decomposed or recomposed, providing a natural means to predict the evolution of interpretations. Our dynamic axioms borrow logical connectives from
separation logics, well-known specification languages to verify programs with dynamic data structures. In the paper, we focus on ALC and EL augmented with dynamic axioms, or to their subclass of
positive dynamic axioms. The knowledge base consistency problem in the presence of dynamic axioms is investigated, leading to interesting complexity results, among which the problem for EL with
positive dynamic axioms is tractable, whereas EL with dynamic axioms is undecidable.
Answer Set Programming (ASP) is a well-known formalism for Knowledge Representation and Reasoning, successfully employed to solve many AI problems, also thanks to the availability of efficient
implementations. Traditionally, ASP systems are based on the ground&solve approach, where the grounding transforms a general input program into its propositional counterpart, whose stable models are
then computed by the solver using the CDCL algorithm. This approach suffers an intrinsic limitation: the grounding of one or few constraints may be unaffordable from a computational point of view; a
problem known as grounding bottleneck. In this paper, we develop an innovative approach for evaluating ASP programs, where some of the constraints of the input program are not grounded but
automatically translated into propagators of the CDCL algorithm that work on partial interpretations. We implemented the new approach on top of the solver WASP and carried out an experimental
analysis on different benchmarks. Results show that our approach consistently outperforms state-of-the-art ASP systems by overcoming the grounding bottleneck.
We propose a logic of directions for points (LD) over 2D Euclidean space, which formalises primary direction relations east (E), west (W), and indeterminate east/west (Iew), north (N), south (S) and
indeterminate north/south (Ins). We provide a sound and complete axiomatisation of it, and prove that its satisfiability problem is NP-complete.
Strategy representation and reasoning has recently received much attention in artificial intelligence. Impartial combinatorial games (ICGs) are a type of elementary and fundamental games in game
theory. One of the challenging problems of ICGs is to construct winning strategies, particularly, generalized winning strategies for possibly infinitely many instances of ICGs. In this paper, we
investigate synthesizing generalized winning strategies for ICGs. To this end, we first propose a logical framework to formalize ICGs based on the linear integer arithmetic fragment of numeric part
of PDDL. We then propose an approach to generating the winning formula that exactly captures the states in which the player can force to win. Furthermore, we compute winning strategies for ICGs based
on the winning formula. Experimental results on several games demonstrate the effectiveness of our approach.
We revisit the notion of i-extension, i.e., the adaption of the fundamental notion of extension to the case of incomplete Abstract Argumentation Frameworks. We show that the definition of i-extension
raises some concerns in the "possible" variant, e.g., it allows even conflicting arguments to be collectively considered as members of an (i-)extension. Thus, we introduce the alternative notion of
i*-extension overcoming the highlighted problems, and provide a thorough complexity characterization of the corresponding verification problem. Interestingly, we show that the revisitation not only
has beneficial effects for the semantics, but also for the complexity: under various semantics, the verification problem under the possible perspective moves from NP-complete to P.
The chase is a famous algorithmic procedure in database theory with numerous applications in ontology-mediated query answering. We consider static analysis of the chase termination problem, which
asks, given set of TGDs, whether the chase terminates on all input databases. The problem was recently shown to be undecidable by Gogacz et al. for sets of rules containing only ternary predicates.
In this work, we show that undecidability occurs already for sets of single-head TGD over binary vocabularies. This question is relevant since many real-world ontologies, e.g., those from the Horn
fragment of the popular OWL, are of this shape.
Constraint satisfaction problems (CSPs) are an important formal framework for the uniform treatment of various prominent AI tasks, e.g., coloring or scheduling problems. Solving CSPs is, in general,
known to be NP-complete and fixed-parameter intractable when parameterized by their constraint scopes. We give a characterization of those classes of CSPs for which the problem becomes
fixed-parameter tractable. Our characterization significantly increases the utility of the CSP framework by making it possible to decide the fixed-parameter tractability of problems via their CSP
formulations. We further extend our characterization to the evaluation of unions of conjunctive queries, a fundamental problem in databases. Furthermore, we provide some new insight on the frontier
of PTIME solvability of CSPs. In particular, we observe that bounded fractional hypertree width is more general than bounded hypertree width only for classes that exhibit a certain type of
exponential growth. The presented work resolves a long-standing open problem and yields powerful new tools for complexity research in AI and database theory.
We propose a generalisation of liquid democracy in which a voter can either vote directly on the issues at stake, delegate her vote to another voter, or express complex delegations to a set of
trusted voters. By requiring a ranking of desirable delegations and a backup vote from each voter, we are able to put forward and compare four algorithms to solve delegation cycles and obtain a final
collective decision.
The Resource-Constrained Project Scheduling Problem (RCPSP) and its extension via activity modes (MRCPSP) are well-established scheduling frameworks that have found numerous applications in a broad
range of settings related to artificial intelligence. Unsurprisingly, the problem of finding a suitable schedule in these frameworks is known to be NP-complete; however, aside from a few results for
special cases, we have lacked an in-depth and comprehensive understanding of the complexity of the problems from the viewpoint of natural restrictions of the considered instances. In the first part
of our paper, we develop new algorithms and give hardness-proofs in order to obtain a detailed complexity map of (M)RCPSP that settles the complexity of all 1024 considered variants of the problem
defined in terms of explicit restrictions of natural parameters of instances. In the second part, we turn to implicit structural restrictions defined in terms of the complexity of interactions
between individual activities. In particular, we show that if the treewidth of a graph which captures such interactions is bounded by a constant, then we can solve MRCPSP in polynomial time.
A prominent application of knowledge graph (KG) is document enrichment. Existing methods identify mentions of entities in a background KG and enrich documents with entity types and direct relations.
We compute an entity relation subgraph (ERG) that can more expressively represent indirect relations among a set of mentioned entities. To find compact, representative, and relevant ERGs for
effective enrichment, we propose an efficient best-first search algorithm to solve a new combinatorial optimization problem that achieves a trade-off between representativeness and compactness, and
then we exploit ontological knowledge to rank ERGs by entity-based document-KG and intra-KG relevance. Extensive experiments and user studies show the promising performance of our approach.
We present NeurASP, a simple extension of answer set programs by embracing neural networks. By treating the neural network output as the probability distribution over atomic facts in answer set
programs, NeurASP provides a simple and effective way to integrate sub-symbolic and symbolic computation. We demonstrate how NeurASP can make use of a pre-trained neural network in symbolic
computation and how it can improve the neural network's perception result by applying symbolic reasoning in answer set programming. Also, NeurASP can make use of ASP rules to train a neural network
better so that a neural network not only learns from implicit correlations from the data but also from the explicit complex semantic constraints expressed by the rules.
We study how belief merging operators can be considered as maximum likelihood estimators, i.e., we assume that there exists a (unknown) true state of the world and that each agent participating in
the merging process receives a noisy signal of it, characterized by a noise model. The objective is then to aggregate the agents' belief bases to make the best possible guess about the true state of
the world. In this paper, some logical connections between the rationality postulates for belief merging (IC postulates) and simple conditions over the noise model under consideration are exhibited.
These results provide a new justification for IC merging postulates. We also provide results for two specific natural noise models: the world swap noise and the atom swap noise, by identifying
distance-based merging operators that are maximum likelihood estimators for these two noise models. | {"url":"https://papers.cool/venue/IJCAI.2020?group=Knowledge%20Representation%20and%20Reasoning","timestamp":"2024-11-06T05:46:17Z","content_type":"text/html","content_length":"98259","record_id":"<urn:uuid:b39740e4-bce2-40d8-b804-b876d0a749e0>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00883.warc.gz"} |
Read the Exact Time
In this activity, we will be learning to read the exact time on an analogue clock, using the position of the hands.
Clocks have an hour hand and a minute hand (and sometimes a second hand)
The hour hand is shorter than the minute hand.
Let's try two example questions.
Example 1
Look at this clock. What is the time?
On this clock, the hour hand is pointing to between 6 and 7, so we know the time is before 7 o'clock, but after 6 o'clock.
The minute hand is pointing to 6 which is halfway round the clock.
This clock has 60 small minute markers, but some clocks do not.
Each numbered section on the clock face is worth 5 minutes.
So, on this clock, the minute hand has travelled to the 6th number.
6 lots of 5 minutes equal 30 minutes or half an hour.
So the time shown is 6:30 or half past 6.
Example 2
What time does this clock show?
Notice that the hour hand has moved very slightly from the clock in example 1.
The time is still after 6 o'clock but before 7 o'clock.
The minute hand has moved forward one minute past the 6 marker.
So the time shown is 6:31.
Now it's your turn to try some questions like this. | {"url":"https://www.edplace.com/worksheet_info/maths/keystage2/year3/topic/966/1731/what-is-the-time-3","timestamp":"2024-11-02T05:23:42Z","content_type":"text/html","content_length":"81557","record_id":"<urn:uuid:cee09554-7bad-4fa8-b519-5467bac3431a>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00136.warc.gz"} |
Volatility in outcome-adaptive randomization
Volatility in adaptive randomization
Randomized clinical trials essentially flip a coin to assign patients to treatment arms. Outcome-adaptive randomization “bends” the coin to favor what appears to be the better treatment at the time
each randomized assignment is made. The method aims to treat more patients in the trial effectively, and on average it succeeds.
However, looking only at the average number of patients assigned to each treatment arm conceals the fact that the number of patients assigned to each arm can be surprisingly variable compared to
equal randomization.
Suppose we have 100 patients to enroll in a clinical trial. If we assign each patient to a treatment arm with probability 1/2, there will be about 50 patients on each treatment. The following
histogram shows the number of patients assigned to the first treatment arm in 1000 simulations. The standard deviation is about 5.
Next we let the randomization probability vary. Suppose the true probability of response is 50% on one arm and 70% on the other. We model the probability of response on each arm as a beta
distribution, starting from a uniform prior. We randomize to an arm with probability equal to the posterior probability that that arm has higher response. The histogram below shows the number of
patients assigned to the better treatment in 1000 simulations.
The standard deviation in the number of patients is now about 17. Note that while most trials assign 50 or more patients to the better treatment, some trials in this simulation put less than 20
patients on this treatment. Not only will these trials treat patients less effectively, they will also have low statistical power (as will the trials that put nearly all the patients on the better
The reason for this volatility is that the method can easily be mislead by early outcomes. With one or two early failures on an arm, the method could assign more patients to the other arm and not
give the first arm a chance to redeem itself.
Because of this dynamic, various methods have been proposed to add “ballast” to adaptive randomization. See a comparison of three such methods here. These methods reduce the volatility in adaptive
randomization, but do not eliminate it. For example, the following histogram shows the effect of adding a burn-in period to the example above, randomizing the first 20 patients equally.
The standard deviation is now 13.8, less than without the burn-in period, but still large compared to a standard deviation of 5 for equal randomization.
Another approach is to transform the randomization probability. If we use an exponential tuning parameter of 0.5, the sample standard deviation of the number of patients on the better arm is
essentially the same, 13.4. If we combine a burn-in period of 20 and an exponential parameter of 0.5, the sample standard deviation is 11.7, still more than twice that of equal randomization.
6 thoughts on “Volatility in adaptive randomization”
1. This topic is extremely relevant for online marketing right now in the spaces of A/B testing and ad placement optimization (where people call this multiarm bandit). There’s another criticism
which I read which focuses mainly on how it will show statistical significance slower. But your post points out a scarier downside.
My complaint has always been that this sort of experimentation assumes independence of response with time; which is often not true. If you are changing your p of assignment over time and there is
a period where response is more likely then if p is leaning in a particular direction at that time you will end up with misleading results.
2. Bandit testing would work better in advertising than in medicine. It’s OK for an ad server to know what assignments are coming up, but randomization reduces the potential bias of human
researchers. But because bandits are optimal under certain assumptions, it’s reasonable to wonder whether they would perform well when those assumptions are violated.
Adaptive randomization may be more robust against changes to response over time. In the context of medicine, we call this “population drift.” Here is a tech report that explores adaptive
randomization and population drift.
Adaptive randomization (AR) has been criticized lately for being less powerful than equal randomization (ER). For two arms, it seems that AR is less powerful than ER unless the response
probabilities on the two arms are extremely different, something that rarely happens in practice.
For three or more arms, the situation is less clear. It’s plausible that the ability to drop poorly performing arms could make AR more powerful than ER. In simulations that I’ve run, it seems
that the benefit of AR comes from being able to drop an arm, not from changing the randomization probability. In the scenarios I’ve looked at, it’s best to randomize equally to an inferior arm
until you drop it, rather than to gradually starve it by lowering its assignment probability.
3. Am I missing something here, or is part of the problem that we’re trying to do two incompatible things at once? The point of a clinical trial is to determine whether a given treatment is
effective, and (if you’re lucky) to quantify that effect. In the long run, it’s more valuable to get the best possible information out of the trial than it is to treat the trial participants
effectively. Trial participants sign waivers that assert they understand this, and are OK with it.
4. Thank you so much John. That’s really insightful (note for those who don’t follow the link is it’s a paper on exactly the topic I was asking about written by John).
Follow up question: while AR still has better expected value under population drift wouldn’t drift increase the variance issue discussed here? Particularly for rising tide.
Although, as I ask that question I suspect the answer is: yes, but not by much. It appears from your research that even large drift has only a subtle effect on results.
When you say that “[…] it seems that AR is less powerful than ER […]” what do you mean by “less powerful”? Just that given the same amount of trials AR is less likely to detect subtle effects
than ER?
5. Oh, one more follow up question. In the paper you are looking at situations where the size of the drift is similar to the difference in the response probabilities of the two arms. Did you look at
situations where the size of the drift was 4x or 8x larger than the difference? So one arm’s response rate is 0.32 vs 0.30 but they drift to 0.22 and 0.20.
6. Dave: Yes, we’re trying to do two things at once, and they are in tension. We’re trying to treat patients in the trial more effectively, and determine an accurate result, which means treating
future patients outside the trial more effectively. You can combine these into a single optimization problem by projecting how many future patients will be involved. This depends highly on
context. For a very rare disease, maybe there are more patients in the trial than future patients.
Steven: Population drift is a real problem for estimation. It’s not even clear what the correct answer should be if the thing you’re measuring is changing over time. | {"url":"https://www.johndcook.com/blog/2012/10/21/volatility-in-ar/","timestamp":"2024-11-02T11:50:48Z","content_type":"text/html","content_length":"64283","record_id":"<urn:uuid:7abd5fe7-7f3a-489d-90e1-b93ef39e7c05>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00025.warc.gz"} |
F.H. Paschen Salaries | How Much Does F.H. Paschen Pay in the USA | CareerBliss
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• Nationwide
• Chicago, IL
• Falls Church, VA
Equipment Services Project Manager is the highest paying job at F.H. Paschen at $104,000 annually.
Administrative Assistant is the lowest paying job at F.H. Paschen at $37,000 annually.
F.H. Paschen employees earn $65,500 annually on average, or $31 per hour. | {"url":"https://www.careerbliss.com/fh-paschen/salaries/","timestamp":"2024-11-11T21:29:20Z","content_type":"text/html","content_length":"80432","record_id":"<urn:uuid:b859dd0c-8276-438f-8b9f-dd2cfb0f0042>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00083.warc.gz"} |
Mathematics can be a subject that poses challenges for many students, and solving math problems can be particularly daunting. If you’re struggling with math problems and need answers, there are
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Math problems can be daunting, and sometimes you may feel stuck or overwhelmed when trying to solve them. Whether you’re a student tackling homework assignments or someone facing real-world math
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Do My Math Homework
Do My Math Homework: What You Have to Know before You Demand Such Services Why use a do my math homework service If you check your facts well, and if you are sincere to yourself, you will admit that
of all the courses offered in schools, mathematics is the most hated. This highlights the need
Do My Math Homework Read More » | {"url":"https://www.letusdothehomework.com/category/math/page/2/","timestamp":"2024-11-04T11:12:59Z","content_type":"text/html","content_length":"109179","record_id":"<urn:uuid:06572b41-9f7e-4725-af92-a671f4f75e38>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00567.warc.gz"} |
Erasing the Ephemeral
Reconstructed views on Waymo. Reconstructed scenes of a sequence from Waymo dataset. Our method can eliminate moving objects on the street.
Synthesizing novel views for urban environments is crucial for tasks like autonomous driving and virtual tours. Compared to object-level or indoor situations, outdoor settings present unique
challenges such as inconsistency across frames due to moving vehicles and camera pose drift over lengthy sequences. In this paper, we introduce a method that tackles these challenges on view
synthesis for outdoor scenarios. We employ a neural point light field scene representation and strategically detect and mask out dynamic objects to reconstruct novel scenes without artifacts.
Moreover, we simultaneously optimize camera pose along with the view synthesis process, and thus we simultaneously refine both elements. Through validation on real-world urban datasets, we
demonstrate state-of-the-art results in synthesizing novel views of urban scenes.
a) We incorporate novel view synthesis with dynamic object erasing which removes artifacts created by inconsistent frames in urban scenes.
b) We propose a voting scheme for dynamic object detection to achieve consistent classification of moving objects.
c) During training, we jointly refine camera poses and demonstrate the robustness of our method to substantial camera pose noise. As a result, image quality is elevated with the increased accuracy of
camera poses.
Moving Object Detection
we employ a voting scheme to reduce inconsistencies in motion prediction that may be caused by incorrect optical field computation or the inconsistencies introduced by ego-motion. In frame $j$ where
the object with instance $i$ appears, we compute the motion score \(m_j^i\in \{0,1\}\), where 1 and 0 denote moving and non-moving objects respectively. Thus, each object has a sequence of motion
labels \(\{m^i_n\}_n\) (out side $n$ means iterate over \(n\)) indicating their motion statuses over frames. Finally, the motion status $M^i$ of an object instance $i$ across the scene is set as
\[M^{i} = \begin{cases} 1 \text{ if } \text{med}(\{m^i_n\}_n) \geq 0.5 \,, \\ 0 \text{ otherwise }\,, \end{cases}\]
where \(\text{med}(\{m^i_n\}_n)\) is the median of the motion labels for object $i$ in the sequence \(\{m^i_n\}_n\). If an instance object is labeled as 1, we denote this object as moving over the
entire sequence.
Pose Refinement
Pose refinement results. Noise pose (left) and refined pose (right) of our results.
To solve the aforementioned inaccurate camera pose problem, we jointly refine the camera poses with the point light field to account for these potential inaccuracies. We use the logarithmic
representation of the rotation matrix such that the direction and the $l2$ norm of the rotation vector $\boldsymbol{R} \in \mathbb{R}^{3}$ represents the axis and magnitude of rotation of the camera
in the world-to-camera frame respectively. The translation vector $\boldsymbol{t} \in \mathbb{R}^{3}$ represents the location of the camera in the world-to-camera frame.
Self-supervised Training
Denote $\boldsymbol{R^{\prime}}$ as the set of rays that are cast from the camera center to the non-masked pixels only. This allows us to retain the information from static vehicles unlike previous
masking-based approaches, which mask out all instances of commonly transient objects. Additionally, we reduce the uncertainty introduced by objects that are in motion, which is a very common feature
of outdoor scenes. At inference time, we do not consider the mask and instead shoot rays through the entire pixel grid. Thus, the color $C^{\prime}(\boldsymbol{r}_j)$ of a ray $\boldsymbol{r}_j$ is
given by
\[C^{\prime}(\boldsymbol{r}_j) = F_{\theta_{LF^{\prime}}}(\phi(\boldsymbol{d}_j) \oplus \phi(\boldsymbol{l}_j), \boldsymbol{R}^{\prime}, \boldsymbol{t}^{\prime})\]
where $\boldsymbol{d}j$ and $\boldsymbol{l}_j$ are the ray direction and the feature vector corresponding to $\boldsymbol{r}_j$, $F{\theta_{LF^{\prime}}}$ is an MLP. The loss function is
\[\boldsymbol{L}_{m,r} = \sum_{j \in \boldsymbol{R^{\prime}}} || C^{\prime}(r_j) - C(r_j) ||^{2}\]
and the updates to the camera rotation and translation are optimized simultaneously with the neural point light field.
We evaluate our method on the Waymo open dataset Waymo. We chose 6 scenes from Waymo which we believe are representative of street view scenes with different numbers of static and moving vehicles and
pedestrians. We use the RGB images and the corresponding LiDAR point clouds for each scene. We drop out every 10th frame from the dataset for evaluation and train our method on the remaining frames.
The RGB images are rescaled by a factor of 0.125 of their original resolutions for training.
Novel view synthesis
Our method uses point clouds as geometry priors. To prove that the network learns the actual scene geometry structure, instead of only learning the color appearance along the trained camera odometry,
we extrapolate the trajectory to drift off from the training dataset. We then render views from this new trajectory which are far away from the training views. This differs from the novel view
synthesis results presented in the previous paragraph where the network rendered views that were interpolated on the training trajectory.
title = {Erasing the Ephemeral: Joint Camera Refinement and Transient Object Removal for Street View Synthesis},
author = {MS. Deka* and L. Sang* and Daniel Cremers},
year = {2024},
booktitle = {GCPR}, | {"url":"https://sangluisme.github.io/projects/2_project/","timestamp":"2024-11-12T14:56:29Z","content_type":"text/html","content_length":"20247","record_id":"<urn:uuid:e9f144c9-1d2f-4c19-b65c-764aee8f84de>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00266.warc.gz"} |
Summation Theorem
A very important property of steady-state metabolic systems was uncovered with the MCA formalism. This concerns the summation of all the flux control coefficients of a pathway. By various procedures
Kacser73 Heinrich75 Giersch88 Reder88
] it can be demonstrated that for a given reference flux the sum of all flux-control coefficients (of all steps) is equal to unity:
$$\sum_{i} C_{v_{i}}^{J} = 1 $$
For a given reference species concentration the sum of all concentration-control coefficients is zero:
$$\sum_{i} C_{v_{i}}^{[M]} = 0$$
where the summations are over all the steps of the system.
According to the first summation theorem, increases in some of the flux-control coefficients imply decreases in the others so that the total remains unity. As a consequence of the summation theorems,
one concludes that the control coefficients are global properties and that in metabolic systems, control is a systemic property, dependent on all of the system's elements (steps). | {"url":"https://copasi.org/Support/User_Manual/Methods/Metabolic_Control_Analysis/Summation_Theorem/","timestamp":"2024-11-06T04:15:51Z","content_type":"text/html","content_length":"13697","record_id":"<urn:uuid:e56873a9-0b80-4438-afa5-0048a40a7cd4>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00653.warc.gz"} |
Calculus 1 Multiple Choice Questions | Hire Someone To Do Calculus Exam For Me
Calculus 1 Multiple Choice Questions (Colocalization, Classification) – 2.1.1 Introduction Multiplicative or multiplicative with applications is easy. (Colocalization, Classification) Multiplicative
Numbers, the easiest of the problems of this book, can be converted to multiplicative with the help of Lemma 1-3-3. (Classification of multiplicative numbers means either additive or multiplicative.)
In this chapter, we give up click for more info multiplicative cases where the book asks simple questions. In the next chapter, we add more and show how to answer these simple questions using Lemma
1-3-3. In the last chapter, we introduce Mathematics and applications thereunder, showing why it is easier to deal with multiplicative and non-multiplicative cases with the help of a new or more
general definition of multiplicative conjugacy. The book offers many useful functions in its calculus. On my recent short account, the book is very useful and enjoyable. There are many other nice
ways to use the mathematical work in the book. I will tell you the most useful ways of keeping up with the book’s contents: the example of Lemma 1 the method of proof is with the help of the lemma of
Appendix 1 of \cite{jourada} (I don’t know the final rule the main remark) and the theorem of \cite{prelim} (Mardians). Example 1 Proof of Lemma 1 $(a_1a_2)_{a_1}\cdot(a_1a_2)_{b_4}$ Growth with the
Leve You know this in some sense. However, unfortunately we don’t have any kind of knowledge about maths Growth with the Leve It is because, in spite of some work, we don’t know what the
mathematicians would add or subtract some numbers, especially negative numbers (eg, numbers of addition and subtraction rules) or numbers of coefficients, which are not known (or should be known,
because they were “observed”). The application of arithmetic and mathematical physics is the purpose of mathematics. #1. The chapter on Classification (In: Applications to various math problems) by
Stefan Gremsel Example 1 Proof. If each of the fractions is a binary number, whether the denominator is positive or negative, then we have that the value of any derivative must be positive, which we
can by studying 2 in an algebraic way a set of laws that can be modified on the basis of the formula, so as to define an equivalence relation on the possible values of the two fractions. Then we can
put the equation of a number into the matrix, and then add it to both sides of the equations, then we have the equation of the second one. No question about this is raised in any of this works, as
are some of this questions.
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Example 2 Cramer (bias in: multiplication or division) of the first half of the 3-dimensional algebra Example 2 Degree of $I$ This is the inverse of an equation with $I= a_1 a_2 a_3$ to be $a_4$.
Then $a_1$ is an indicator and is positive and equal to $0$ and $1$, respectively, and has not different values as the value of any other one. Now, add some figures of numbers that are on the axis of
differentiation of 2. Suppose that a and b have different values and have analogs as equals signs. Then, suppose that the value of 3 is a first term with number 0, 6, 10, 18, and 23 and the first
equal sign should not be any or a negative sign because a and b have the same signs of each other. Then the value of a is negative, as we can see from the formula for the second term. Now consider
the partial sums of the squares of 3 in the second: by theorem 1-4, namely 3 Lemma 1-5 Proof: the quotient 2 by The quotient 2 is in degree 2 or more. It is odd if we take it in place of 1:Calculus 1
Multiple Choice Questions: Which is better? When you have more than one question, the ones you ask to answer better are usually what you want to avoid: The better these questions are used, the more
they stay in the list. 1. Why do you like it? That’s how we can discuss the differences. It doesn’t matter in what sort of answer you get, though. What matters is how you put that question into the
list today. Which is a statement, and what is left is what you put, so some people decide that their first choice is right. I like this approach because we have a choice, ‘How do you play this game?’
– and I don’t usually like this approach. I have always found it a better approach, but I think the best practice is probably to keep asking to make the question down, so no more people just have to
understand that, or you don’t get a lot of ‘I really like it.’ It’s not as hard as it could be, anyway. 2. How do you think about that game? In this case, we don’t really know it, so I think the
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’ Who gives a fuck about that when you don’t really like it you don’t have a problem with it? 3. How do you think about that game now? Oh yeah. I think today’s the one game we’re going to play. I
think that the question ‘How do you play all the different types?’ is very important to consider. There are really two possible answers, and they are both consistent. No question, no answer! Start
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it’s simply one. When I really get down to it, it’s the basic question I ask, and I will show it again. Take this one as a good example: ‘I don’t like what they say, I prefer what I do,’ They don’t
say ‘I will play this game, but I really like this’ and ‘but I don’t like what their saying’. And when you mention that I like it, that’s kind of the most common way of answering that question in
question 3. The list is given in our hands, since it’ll affect your game a LOT in the long run, but it is on the day-to-day for you. Even if that means a bit more work for you,Calculus 1 Multiple
Choice Questions 2.0 What is the objective of this article? Do you have many, many different questions? This article shall cover the other aspects: probability, choice of objects, second-order
properties and finally, what are the most useful properties of P and PQ about general facts? A A big question could be, Can the composition of two objects be seen as a composition of two different
objects if the compositions are equal, and are equal in each case? A Our problem has more general properties. The aim of our approach is different, because after that we need to understand the
composition problem in that one and only two objects have to be seen as compositions, the second object has to be very general. It makes sense to not only examine the composition condition for the
two-object structure, but it makes sense to also examine the last three conditions, the second and the third of the why not find out more principles of the composition process: composition is the
property of the composition, not only the first statement, hence the second and the third of the A (toppling of each other). The definition of the composition step allows to get the following
definition that works for the other two objects: Every object is said to be a set X of all objects of Y of X only if: 2) Each object is equal only if it is a subset of X. 3) Every other object is
said to be a set of X only if: 2) But one of its objects is a set with at least one element of X greater than Y as its head.
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Thus, A is a set of A, a space and also an object. If things are in common, then A is a space and also a space and also an object. If things are different and also such things are in common then we
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element of X. Partial-point equation for a general rule If a rule by itself is to be in place, then the theorem below states a partial-point equation for a general rule. Let ${\cal M}$ be an
even-order finite mathematics (P-F, we use abbreviations ${{\cal M\backslash X}}$ as M, X), define a property $P:{{\cal M\times{\cal M\setminus L}}\rightarrow {\cal B}}$ as: $(i^PA^{(P)}_{{\cal M\
backslash X}})\in {\cal B}, i\in{{\cal M\times{\cal M\setminus L}}}$ if $P(i)>i$ and $PA^*_{{\cal M\backslash X}}\le i$ (P-F). Remark: We will use “B”, for B is a set and also “P”, so called because
in the theorem of M. M. M. could have said, “If I want to be able to reduce our problem to a general statement, can I call this “A”. If “B” is not a set, then we can simply abuse the notation. We
will only use “P” to mean that the first instance “P” is a set. By “P” “l” is used “l”, then “P”, can be any word, we will simply say how much the word l is. Let $\mathcal{B}\subseteq {\cal M}$ be
such that $i>i^{\mathcal{B}}$, that $\mu>0$, then $PA^*_{{\cal B}}= \frac{{{\cal B}}}{\lambda{{\cal B}} \lambda^*}$, therefore, A is a set. An example of some common pattern for the two processes,
what sorts of object (A) is the first, and what holds, because it is C, is given as “Abbreviation X only if X be such a set”. B is a whole object of X and also of B. The equation, which is to be a
system of rules, is only | {"url":"https://hirecalculusexam.com/calculus-1-multiple-choice-questions","timestamp":"2024-11-05T01:32:19Z","content_type":"text/html","content_length":"106329","record_id":"<urn:uuid:fc7b466a-0980-428e-8388-26bb9a93fe7d>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00394.warc.gz"} |
Math 111 Chapter 2 Quiz
Math 111 Chapter 2 Quiz Name:
Complete the following problems showing all appropriate work.
Solve the following equations :
Solve the following inequalities stating your final answer in interval notation:
Solve 3 of the following word problems. Work must be shown in order to receive credit. You may do the others for extra points:
8.) Find three consecutive integers such that the sum of one half the smallest and one third the largest is one less than the other integer.
9.) A chemistry experiment calls for 20% solution of copper sulfate. Kim has 28 milliliters of 25% solution. How many milliliters of 10% solution should she add to obtain the required 20% solution?
10.) Tim invests $4000 at 6% yearly interest. How much does he have to invest at 8.5% so that the yearly interest from the two investments exceeds $700?
11.) The perimeter of a rectangle is 44 meters. The width of the rectangle is 2 meters more than one-third the length. Find the dimensions of the rectangle.
12.) Kim’s average score for her first three math exams is an 84. What must she get on the fourth exam so that her average for the four exams is 85 or better?
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JEE Main 2024 (January 29 Shift 1) Maths Question Paper with Solutions [PDF]
JEE Main 2024 Maths Question Paper 29 January Shift 1 - Answer Key and Detailed Solutions
The National Testing Agency (NTA), the organizer of JEE Main, conducted the 29 January Shift 1 JEE Main 2024 exam in 13 regional languages.
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to evaluate their performance through detailed question paper analysis and solutions by master teachers.
The good news is that a comprehensive analysis of the Maths-specific questions and official answer key of 29 January Shift 1 JEE Main 2024 Maths paper with solutions will be available on Vedantu
after the exam. Candidates can download the complete Maths paper of 29 January Shift 1 JEE Main 2024 as free PDFs from our website post-exam.
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1. What was the overall difficulty level of the JEE Main 2024 Maths Question Paper 29 January Shift 1?
The overall difficulty level of the paper is reported to be moderate, with a mix of easy, moderate, and difficult questions. Some sections like Coordinate Geometry and Calculus were slightly tougher,
while others like Algebra and Probability were more manageable.
2. Which topics were given the most weightage in JEE Main 2024, 29 Jan Shift 1 Maths paper?
Some students reported that there was a higher emphasis on certain topics like Matrices and Complex Numbers compared to previous years. However, overall, the topic distribution was consistent with
the JEE Main syllabus.
3. How were the different sections distributed in JEE Main 2024, 29 Jan shift 1 Maths paper?
• Calculus: Reportedly easier than usual, with more focus on basic concepts and applications.
• Coordinate Geometry: Moderate difficulty, with a mix of standard and slightly advanced problems.
• Algebra: Considered slightly difficult, with more emphasis on problem-solving and application of concepts.
• Statistics & Probability: Mixed difficulty, with some straightforward questions and some requiring deeper understanding.
4. Were there any surprises or unexpected questions in JEE Main 2024, 29 Jan shift 1 Maths paper?
Some reports mention the presence of a few unexpected topics or questions based on less-frequently covered concepts. | {"url":"https://www.vedantu.com/jee-main/2024-maths-question-paper-29-january-shift-1","timestamp":"2024-11-12T00:38:58Z","content_type":"text/html","content_length":"196306","record_id":"<urn:uuid:212f4c84-c369-4c04-a2c3-f179987e87d7>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00097.warc.gz"} |
New Ramsey Result that will be hard to verify but Ronald Graham thinks its right which is good enough for me.
If you finitely color the natural numbers there will be a monochromatic solution to
x+2y+3z - 5w = 0
There is a finite coloring of the natural numbers such that there is NO monochromatic solution to
x+2y+3z - 7w = 0
More generally:
An equation is REGULAR if any finite coloring of the naturals yields a mono solution.
RADO's THEOREM: A linear equation a1 x1 + ... + an xn = 0 is regular IFF some subset of the ai's sums to 0 (the ai's are all integers).
Rado's theorem is well known.
What about other equations? Ronald Graham has offered $100 to prove the following:
For all two colorings of the naturals there is a mono x,y,z such that
+ y
= z
I've seen this conjecture before and I thought (as I am sure did others) that first there would be a prove that gave an ENORMOUS bound on
f(c) = the least n such that for all c-colorings of {1,...,n} there is a mono (x,y,z) such that ...
and then there would be some efforts using SAT solvers and such to get better bounds on f(c).
This is NOT what has happened. Instead there is now a
by Heule, Kullmann, Marek, where they show f(2)=7825.
(NOTE- ORIGINAL POST HAD f(2)-7285. Typo, was pointed out in one of the comments
below. Now fixed.)
It is a computer proof and is the longest math proof ever. They also have a program that checked that the proof was correct.
And what did they get for their efforts? A check from Ronald Graham for $100.00 and a blog entry about it!
While I am sure there proof is correct I wish there was a human-readable proof that f(2) existsed even if it gave worse bounds. For that matter I wish there was a proof tha,t for all c, f(c) exists.
Maybe one day; however, I suspect that we are not ready for such problems.
7 comments:
1. You should read your posts after you post them to catch typos.
1. If there is a typo in the above post let me know and
I will correct it.
2. I went to build the graph of related triples so that I could see this in action, and in the triples dataset I was looking at there was only 1 pair (x,y) such that x^2 + y^2 = 7285. But if that is
true either you gave wrote down the wrong value for f(2) or defined it wrong, or that's not it's value, or I'm just really confused by what you were trying to say. Because for 7285 to be least n,
Then there most be a coloring of {1,...,7284} with no monochrome solution, but for which both of the related colorings of {1,...,7285} have a monochrome solution. But that would require 2
solutions to x^2 + y^2 = 7285. So there is something that is incorrect.
3. You misread my definition of f(c), which had a ... in it so perhaps it was unclear on my part. Here is the full definition:
f(c) is the least n such that for any c-coloring of {1,...n} there
exists x,y,z all the same color such that x^2 + y^2 = z^2.
4. x,y,z in {1,...,n} right? So if f( c) is 7285. Then there exists 2 coloring of {1,...,7284} in which there are not same colored (x,y,z), because 7285 is the least such n. Call that coloring C'.
There are two colorings of {1,...,7285} that are identical to C' for {1,...,7284}. We can call those C'+black and C'+white. because f( c) = 7285 those colorings both contain monochromatic
solutions. If either of them doesn't use 7285 as z then C' had a monochromatic solution. so it must be the case that there exist (x1, y1) white in C' and (x2, y2) black in C` such that x1^2 + y1^
2 = 7285^2 = x2^2 + y2^2. Otherwise one of those colorings does not have a monochromatic (x,y,z). Now the table I looked at said that 7285 was only in one such triple of integers. So either the
table I referenced is wrong. My understanding of f( c) is still wrong. You've transcribed the result incorrectly and it's some other number. or their result is wrong.
5. This was this list I referenced. http://www.tsm-resources.com/alists/PythagTriples.txt
6. Egdiroh- I looked back at the paper and I had transposed
digits in the result. I have made the correction.
Thanks for catching the mistake! | {"url":"https://blog.computationalcomplexity.org/2016/05/new-ramsey-result-that-will-be-hard-to.html","timestamp":"2024-11-14T21:37:46Z","content_type":"application/xhtml+xml","content_length":"185973","record_id":"<urn:uuid:443294b4-f535-4e04-b168-2282e586f8a4>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00097.warc.gz"} |
The Tumescent Technique By Jeffrey A. Klein MD
Chapter 19:
Pharmacokinetics of Tumescent Lidocaine
This was sometime a Paradox, but now the time gives it proof.
— Shakespeare, Hamlet
The efficacy and safety of the large doses of lidocaine used in tumescent local anesthesia are perceived as a paradox by many physicians. Time-honored teachings about local anesthesia are difficult
to reconcile with the principles of tumescent local anesthesia. Tens of thousands of tumescent liposuction patients have received 35 to 50 mg/kg of lidocaine with no known reports of deleterious
effect, which has proved the safety of tumescent local anesthesia for liposuction. An understanding of the pharmacokinetics of tumescent lidocaine eliminates the paradox from the assertion that less
(concentration) is more (effective and safe).
The Science of Pharmacokinetics
Pharmacokinetics is the branch of pharmacology concerned with the movement of drugs within the body. More specifically, pharmacokinetics is a science that studies the time course of drug
concentrations and disposition in the body. In practice, pharmacokinetics uses mathematical models that describe the body in terms of one or more theoretic compartments and allows one to calculate
and predict the timedependent concentration of drugs in the blood. A good pharmacokinetic model permits an accurate estimation of both the maximum drug concentration (C[max]) in the blood and the
time (T[max]) when the peak concentration will occur.
For lidocaine the risk of toxicity is closely correlated with the peak plasma lidocaine concentration. The ability to estimate the values of C[max] and T[max] allows one to anticipate or predict the
risks of lidocaine toxicity. Important factors in determining these values are the rates of lidocaine absorption and metabolism.
To be useful, a pharmacokinetic model must be accurate and relatively simple. The accuracy of predictions depends on how accurately the model reflects the clinical reality. The simplicity of a
mathematical model determines its usefulness. A pharmacokinetic model that is too complex may not be useful in many clinical situations.
This chapter is written for the “nonmathematical” physician who may be unfamiliar with pharmacokinetics. Bypassing the few sections on calculations will not impair an understanding of other concepts.
This chapter provides the reader with an intuitive insight into the most important concepts involved in the pharmacokinetics of tumescent lidocaine. This presentation is heuristic rather than an
elegant mathematical development. The mathematics sections at the end of this chapter provide more rigorous analyses and descriptions for interested readers.
Elimination. Drug elimination refers to the irreversible removal of drug and metabolites from the body by all routes of elimination.
Half-life. The elimination half-life (t[1/2]) of a drug is the length of time required for the elimination of half the total drug present in the body at any given time.
Clearance. Drug clearance (Cl[T]), or total body clearance, refers to the process of drug elimination from the body without specifying any of the processes involved. From a conceptual perspective,
clearance can be defined as the volume of plasma that is completely cleared of drug per unit time. Thus clearance has the dimensions of volume/time (e.g., ml/min or L/hr).
From a computational perspective, clearance can be defined as the rate of drug elimination divided by the plasma concentration:
Cl[T] = [Elimination rate]/[Plasma concentration (C)]
= [dD/dt]/C
= [μg/min]/[μg/ml]
= ml/min
where D is the amount of drug that has been eliminated, and dD/dt is the instantaneous rate of elimination. Thus:
dD/dt = Cl[T] × C
Concentration. Drug concentration (C), unless otherwise specified, refers to the concentration (mg/L = μg/ml) of a drug in plasma. As noted, C[max] is the maximum or peak concentration of a drug
during a specified time interval, and T[max] is the exact point in time when C[max] is achieved. Thus T[max] is the length of time that has elapsed from the beginning of the drug administration to
the point where the drug concentration achieves its peak value.
Distribution. Volume of distribution (V), or the apparent volume of distribution (V[D]), is the theoretic volume within the body into which the drug is dissolved. In simple terms, V is defined by
calculating X/C, where X represents the total amount of drug in the body and C the concentration of the drug in plasma at steady-state conditions. The volume of distribution represents a mathematical
concept rather than a real anatomic space; in general, no welldefined anatomic volume of tissues within the body corresponds to V.
As a purely mathematical concept, if the value of C is relatively small, V can exceed the volume of the body. For example, because lidocaine has a high degree of lipid solubility, a large proportion
of lidocaine will be distributed into fat, and C will be relatively small. For any value of X, the smaller the value of C, the greater will be the value of the mathematical term X/C = V.
Safety Issues
Three important safety issues involve the dosages and concentrations of lidocaine when used as a local anesthetic.
First, safe doses of tumescent (very dilute) lidocaine and epinephrine are not the same for commercial (considerably more concentrated) lidocaine. Whereas the safe maximum dosage of tumescent
lidocaine (with epinephrine) at concentrations of 0.05% to 0.15% is 45 to 50 mg/kg, the traditional dosage limitation for commercial lidocaine (with epinephrine) at concentrations of 0.5%, 1%, or 2%
remains valid at 7 mg/kg. All physicians should recognize this vital distinction.
Second, the surgeon must provide detailed written and signed orders explicitly specifying the concentration (mg/L) and maximum allowable total dosage (mg/kg) before the anesthetic solutions are
prepared for tumescent liposuction.
Third, lidocaine concentrations in tumescent anesthetic solutions must always be specified in terms of milligrams of lidocaine per liter or bag of anesthetic solution. It is potentially dangerous to
give orders in terms of volume (ml) of the commercial preparations of lidocaine multiplied by the concentration of the commercial preparation.
For example, when an order specifies 1000 mg of lidocaine and 1 mg of epinephrine in 1000 ml of normal saline, there is little risk of a dosage error or miscommunication between surgeon, nurse, or
anesthetist. On the other hand, it is not immediately obvious how many milligrams of lidocaine have been given after half a bag containing “100 ml of 1% lidocaine (1%) with epinephrine (1:100,000)”
has been infiltrated.
Furthermore, in several cases, surgeons have ordered “100 cc of lidocaine per liter” with the intention that 1% lidocaine be used, but instead the nurse used 2% lidocaine when mixing the anesthetic
solution. This type of error is more easily avoided when dosages are specified in terms of milligrams rather than milliliters of lidocaine.
Kinetic studies and models
For tumescent local anesthesia, pharmacokinetics provides a basis for predicting safe dosages and for understanding the factors that might increase the risk of lidocaine toxicity. A clinically useful
prediction of the maximum plasma lidocaine concentration following a specific dose of tumescent lidocaine requires an accurate kinetic model.
The pharmacokinetics of lidocaine is based on the time course of concentrations of lidocaine measured at intervals in samples of peripheral blood. This is reasonable because of the close correlation
between lidocaine concentrations in blood and in other tissues. Although lidocaine concentration in the blood is not the same as in other tissues, at steady-state conditions the concentrations differ
only by a constant factor.
The essence of the tumescent technique is the direct infiltration of very dilute (0.05% to 0.15%) lidocaine with epinephrine into an area of subcutaneous fat, resulting in an unprecedented slow rate
of systemic lidocaine absorption. The success of tumescent local anesthesia is based on the synergistic interplay between (1) the unprecedented slow rate of lidocaine absorption and (2) the
well-known rapid rate of lidocaine metabolism by the liver and subsequent renal excretion of less toxic metabolites.
Table 19-1 lists pharmacokinetic parameters for tumescent lidocaine.^1,2
Rate of Lidocaine Absorption
Before the tumescent technique, researchers assumed that lidocaine was always absorbed rapidly from the injection site and that the peak plasma lidocaine concentration (C[max]) was always achieved
within 2 hours of the injection. Before 1987, most pharmacokinetic studies of lidocaine were based on treatment of cardiac dysrhythmias or on peripheral nerve blocks.
Treatment of ventricular fibrillation involved the direct intravenous infusion of lidocaine and its instantaneous absorption into the intravascular space. Peripheral nerve blocks typically involved
the injection of lidocaine directly into highly vascular tissues, such as the epidural space, the spinal fluid in the subarachnoid space, the axilla, or the intercostal nerves. In the few studies
that examined subcutaneous injections of lidocaine, plasma lidocaine concentrations were not measured beyond 2 hours.
Two-compartment Model. The rapid absorption of lidocaine into the blood is usually represented by a twocompartment model. First, as lidocaine is rapidly absorbed into the bloodstream and highly
vascular tissues (central compartment), the plasma lidocaine concentration rapidly reaches its peak level, C[max]. As the plasma lidocaine is redistributed into other less vascular tissues
(peripheral compartment), the lidocaine concentration in the plasma falls precipitously until a lidocaine concentration equilibrium is established among all the tissues.
Second, after lidocaine has been distributed throughout the body, and after the plasma lidocaine concentration is in equilibrium with the lidocaine concentration in all the other tissues, the rate of
decline in plasma lidocaine concentration is slowed considerably. Once an equilibrium has been established between the plasma and all other tissues, the further decline in plasma lidocaine
concentration is entirely the result of metabolism and excretion of lidocaine. Thus, when lidocaine is rapidly absorbed, it behaves as if the body were divided into two theoretic compartments.
Because of the assumption that lidocaine was always rapidly absorbed from the injection site, it became unquestioned dogma that lidocaine behaved as a two-compartment pharmacokinetic model. For
example, as stated in one of the most influential works on local anesthesia, “Lidocaine, for certain, behaves in a kinetic sense as if man were a two- or even a three-compartment system.”^3
Anesthesiologists who assume that lidocaine can only behave as a two-compartment model cannot reconcile their view with the relatively large doses of lidocaine used in tumescent liposuction. Under
this assumption, 50 mg/kg of lidocaine is an unacceptably large and potentially dangerous dose. The dogma of anesthesiology has assumed that 7 mg/kg of lidocaine with epinephrine is the true maximum
safe dose for subcutaneous lidocaine. In fact, tumescent lidocaine behaves as a one-compartment model, so 50 mg/kg of tumescent lidocaine for liposuction totally by local anesthesia is much safer
than expected.
As shown later, tumescent lidocaine is unique in that it behaves as if the body were a one-compartment pharmacokinetic model, which explains the safety of large tumescent doses.
Safety Limits. Although the tumescent technique permits an increase in the maximum safe dose of lidocaine, it does not permit unlimited or titanic doses of lidocaine. The maximum recommended dose of
tumescent lidocaine is finite (45 mg/kg for thin patients and 50 mg/kg for heavier patients). Ignoring the estimated safe dose of tumescent lidocaine is dangerous and not in the patient’s best
interest. Caution is paramount.
The physician must always remember that, for any drug, a published estimate of a maximum safe dosage is merely an estimate. Any estimate may eventually prove to be inaccurate and may need to be
Dosage Limits
From a traditional perspective, standard tumescent doses of lidocaine appear excessive, and concern about the risk of toxicity is legitimate. Surgeons and anesthesiologists are appropriately cautious
about the “standard dose limitation of lidocaine” that has no rational scientific basis.
The standard lidocaine dose limitation of 7 mg/kg remains a reasonable limit for standard concentrations of commercial “out-of-the-bottle” lidocaine. The 7-mg/kg limit, however, is unnecessarily
restrictive when very dilute (1.5 g/L = 0.15% or less) lidocaine with epinephrine is used.
Patients and physicians should be concerned about the risk of toxicity whenever an anesthetic is administered. Decisions regarding the relative safety of local versus systemic anesthesia should be
based on modern scientific data, not on incomplete studies. An excessively conservative limit on the dose of tumescent lidocaine may expose the patient unnecessarily to the risks of general
The dosage limit of 7 mg/kg of lidocaine with epinephrine for local anesthesia is an example of a pharmacologic standard based on an unwarranted extrapolation of limited data (see Chapter 1). The
only justification for 7 mg/kg is the letter to the U.S. Food and Drug Administration (FDA) stating that “the maximum safe dose of lidocaine is probably the same as for procainamide.”
In fact, 7 mg/kg is probably an accurate estimate of a maximum safe dose of lidocaine with epinephrine at lidocaine concentrations in the range of 1% or 2%, when injected into highly vascular tissue
(e.g., intercostal block). Again, however, this dosage limitation is unreasonably conservative for subcutaneous infiltration of very dilute lidocaine with epinephrine into fat. As a result, patients
are exposed to general anesthesia for procedures that are more safely and less painfully accomplished by local anesthesia.
Dermatology and Local Anesthesia. In recent years, dermatology has been undergoing a transformation into a surgical specialty that relies almost exclusively on local anesthesia. Dermatologists have
begun to examine critically the pharmacologic basis of local anesthesia infiltrated into the skin and subcutaneous tissues. This information is now starting “to diffuse across the semipermeable
membrane” that separates surgeons of different specialties.
Having overcome the prejudice against using more than 7 mg/kg of lidocaine, specialties other than dermatology are beginning to appreciate the great potential for using dilute local anesthesia for
procedures that have traditionally required general anesthesia.
The following sections explain the critical significance of delayed absorption for the safety and efficacy of tumescent local anesthesia.
Tumescent Bulk Spread Through Fat
The process of tumescent infiltration involves the spread of the local anesthetic solution through the interstitial space by a process known as bulk flow. This is simply the flow of liquid through a
porous substance, such as the interstitial gel. The optimal distribution and spread of the tumescent anesthetic solution throughout the targeted compartment of fat are not instantaneous. Even with an
optimal infiltration technique, it requires many minutes for local anesthesia to become completely effective and for hemostasis to become optimal.
Histologically, the individual adipocytes are not swollen beyond their usual size. The tumescent anesthetic solution is embedded within the interstitial connective tissue gel that envelops
On incising the skin and examining the gross appearance of tumescent fat, one appreciates a marbled appearance of pale-yellow lobules of fat embedded between the gray, glistening, diaphanous sheets
of collagen and within the supersaturated interstitial tissue gel. The consistency of tumescent fat is gelatinous, soft, and jellylike; this is literally the interstitial colloidal gel. Grape-sized
puddles of anesthetic solution are loculated within and between connective tissue septa, similar to a Swiss cheese or honeycomb pattern. These lakes of anesthetic solution act as physical reservoirs
of lidocaine. As the liquid is dispersed through the interstitial tissue, some lidocaine is absorbed locally into the lipids within fat cells (see Chapter 26).
Based on clinical observation, optimal anesthesia and hemostasis do not occur for at least 15 to 30 minutes after tumescent local anesthetic infiltration. Typically, within 15 minutes of
infiltration, sufficient anesthesia and vasoconstriction have occurred to permit painless liposuction without significant blood loss.
With a greater duration of time between the completion of infiltration and initiation of liposuction surgery, tumescent local anesthesia becomes increasingly effective. Also, the greater the delay
after infiltration before starting surgery, the smaller is the volume of blood-tinged infranatant anesthetic solution that is aspirated and appears at the bottom of the collection jar containing the
aspirate. The extra time permits more complete spreading and bulk flow of the anesthetic solution through the interstitial gel and along fascial planes. This results in more extensive diffusion of
the lidocaine into sensory nerves.
Adequate delivery of tumescent anesthesia to a compartment of fat depends on the direct, physical spreading of the anesthetic solution by bulk flow.
Lidocaine Diffusion
True chemical diffusion only becomes important once the anesthetic solution is within a few millimeters of a targeted neural axon or a capillary wall.
The tumescent lidocaine, consisting of free and tissue-bound lidocaine, is slowly absorbed into the intravascular compartment by a process of diffusion. The unbound (free) fraction of tumescent
lidocaine arrives in the systemic circulation by chemical diffusion across fibrous membranes and cells of the adipose tissue, through capillary endothelium and vascular walls, and into the
intravascular space for transport throughout the circulation.
The time required for diffusion of a chemical from point A to point B, as it moves through a medium, depends on several physicochemical factors. For lidocaine, rate of diffusion through the
interstitial tissue gel is a function of (1) the distance between points A and B and (2) the concentration gradient of free lidocaine (unbound to tissue) between points A and B. The rate of lidocaine
diffusion out of the tumescent tissue and into the capillary lumen is delayed by the following:
1. Physical distance and isolation of lidocaine molecules created by the large volume of anesthetic solution within the interstitium of fat tissue
2. Minuscule lidocaine concentration gradient, resulting from the prodigious degree of dilution
3. Profound tumescent vasoconstriction
Pharmacokinetic Compartment
In reality the pharmacokinetic compartment is purely a theoretic or mathematical concept that assists in understanding the concentration of drugs in the body as a function of time. A basic
pharmacokinetic assumption is that, at any point in time, the concentration of the drug is essentially uniform throughout the compartment.
Conceptually, a pharmacokinetic compartment is a group of tissues where the relative concentrations of a drug in different parts of the compartment are in constant equilibrium. In other words, any
change of lidocaine concentration in one tissue results in a rapid (essentially instantaneous) and proportional change of lidocaine concentration in all other parts of that compartment.
For tumescent lidocaine, the one compartment consists of intravascular space, certain interstitial tissues, and highly vascular organs such as the lungs, liver, and pancreas. Although the lidocaine
concentration varies from tissue to tissue, the lidocaine concentrations in any two distinct tissues are always in equilibrium.
One-compartment Model. The science of pharmacokinetics uses mathematical modeling to describe and predict the time-dependent course of a drug’s concentration in blood and body tissues. Typically the
body is represented by a system of hypothetic compartments that do not necessarily correspond to any true anatomic or physiologic entity. Such theoretic models represent a oversimplified totality of
pharmacologic activity. Despite this, the resulting information is often valuable. Simple, linear differential equations can be used to describe the rate of change in lidocaine concentration within
each compartment.
Deciding which is the most appropriate model depends on the route of administration. The fate of intravenous (IV) bolus lidocaine, with instantaneous systemic absorption, is best described by a
two-compartment model.
The tumescent technique for subcutaneous delivery or slow IV infusion of lidocaine is best described using a one-compartment model. In this simplest kinetic model, the body is a single, homogeneous
unit. In order for a one-compartment model to be a good representation of a drug’s behavior, the following criteria must be satisfied:
1. A one-compartment model assumes that any change in plasma drug concentration corresponds to an immediate and proportional change of concentration in all other body tissues.
2. At any given time, the rate of drug elimination from the body is proportional to the amount of drug in the body (compartment) at that time.
Tumescent lidocaine satisfies the first requirement for a one-compartment model only because the blood-tissue equilibrium is “immediate” relative to the extremely slow rate of lidocaine absorption
from tumescent fat. In reality, equilibrium between the lidocaine concentrations in blood and other tissues is not instantaneous. Limited by the rate of tissue perfusion, equilibrium is delayed 30 to
60 minutes after lidocaine enters the circulation. Relative to the many hours required for tumescent lidocaine to be absorbed from the subcutaneous space into the vascular space, however,
blood-tissue equilibrium for lidocaine is achieved quickly.
Lidocaine also satisfies the second requirement. From experience with IV lidocaine used in treating cardiac dysrhythmias, lidocaine is indeed eliminated at a rate that is proportional to the total
amount of lidocaine within the “body.” Lidocaine is said to be eliminated by a first-order process, which occurs when the rate of a drug’s elimination from the body is proportional to the amount of
drug in the body.
In contrast, drug elimination is a zero-order process when the rate of drug elimination is constant and independent of the drug’s concentration in the body. Ethyl alcohol metabolism and elimination
represent an example of a zero-order elimination process.
Tumescent lidocaine kinetics can be described by a one-compartment model, two-compartment model, or more elaborate multicompartment model. The simplest model that explains the clinically observed
events is usually the preferred model.
Absorption, Distribution, and Elimination
Three different but overlapping phases of lidocaine movement occur through the body: absorption, distribution, and elimination. The profoundly slow rate of lidocaine absorption into the systemic
circulation after tumescent infiltration is the key to understanding the unique safety of the tumescent technique for local anesthesia.
When lidocaine is given as an IV bolus to treat a cardiac dysrhythmia, absorption is instantaneous. The other two processes of distribution and elimination, however, occur sequentially over time. The
situation with tumescent lidocaine kinetics is more complex. After tumescent infiltration of lidocaine, the processes of absorption, distribution, and elimination all occur simultaneously.
Absorption Rate
The absorption of tumescent lidocaine is exceptionally slow because of the following:
1. By elevating interstitial hydrostatic pressure above capillary intraluminal pressure, tumescent infiltration compresses and collapses capillaries and venules. With virtually no blood flowing
through the capillaries within the tumescent tissue, the rate of lidocaine absorption is minimized even before the onset of β-adrenergic vasoconstriction.
2. The formation of the dilute lidocaine subcutaneous reservoir produces a physical separation of the lidocaine from the blood vessels, thus increasing the distance over which lidocaine must diffuse
before it reaches a blood vessel.
3. The dilution of lidocaine reduces the lidocaine concentration gradient across the capillary endothelial wall, thereby minimizing its rate of absorption.
4. The profound capillary vasoconstriction minimizes capillary perfusion and thus decreases transcapillary absorption.
5. The relative avascularity of adipose tissue limits vascular absorption.
6. Because of the high lipid solubility of lidocaine, the subcutaneous fat acts as a reservoir for lidocaine and limits the amount of lidocaine available for absorption.
The sum of these additive effects accounts for the unprecedented slow rate of systemic lidocaine absorption. The combination of extremely slow systemic absorption of tumescent lidocaine, rapid
hepatic metabolism, and swift renal elimination results in significantly low lidocaine blood levels and thus minimal risks of lidocaine toxicity.
Analogy: Slow-release Oral Tablet. The absorption of lidocaine from the subcutaneous deposit of tumescent anesthesia is analogous to the absorption of a slow-release tablet taken by mouth. When the
anesthetic solution is in the subcutaneous tissue, it is isolated from the systemic circulation. Similarly, the drug contained within a slow-release tablet is not immediately absorbed into the
systemic circulation. In both cases the drug is contained within an isolated reservoir, is released gradually, and is absorbed incrementally.
A slow-release tablet does not dissolve immediately on entering the gastrointestinal (GI) tract. Instead, the outer portion is slowly eroded, layer by layer. Over many hours the drug is gradually
released in small increments from the tablet. Thus, at any point in time, very little drug is available for systemic absorption.
A similar situation occurs with the slow release of lidocaine, which is only being absorbed from the peripheral external surface of the mass of subcutaneous tumescent solution. This mass is isolated
because of the intense vasoconstriction throughout the tumescent fatty tissue. The capillary bed within the central portion of the tumescent tissue is so completely constricted that there is no
significant blood flow through, and thus no significant absorption from, the central portion of the tumescent tissue.
The infiltrated subcutaneous fat containing the deposit of tumescent lidocaine is similar to the stomach or GI tract containing the slow-release tablet. Although the drug is technically inside the
body, the anatomic site of drug absorption is kinetically distinct and isolated from the rest of the body.
Thus the kinetics of lidocaine after tumescent delivery is analogous to the one-compartment model for oral administration of a slow-release tablet.
Significance of Area under Curve. By plotting the graph of plasma lidocaine concentration at different points in time after a subcutaneous injection, one obtains a curve, designated mathematically by
the term C[lido] (t) (Figure 19-1). Measuring the area under the curve (AUC) can provide important pharmacokinetic information. In terms of formal mathematics, AUC is simply the following integral:
where plasma lidocaine concentration C[lido] (t) is a continuous function of time, over the time interval from when the dose is given at t = 0, to the time when C[lido] (t) returns to 0, here
represented by t = ∞. The shape of the graph of C lido(t), together with AUC, can provide valuable clinical information about tumescent lidocaine kinetics.
With present technology, lidocaine concentration cannot be measured continuously over time. Instead, lidocaine plasma concentrations are determined by taking samples of venous blood at discrete times
(e.g., 1, 2, 4, 8, 12, 16, 24, 36, and 48 hours) after initiating lidocaine infiltration. By plotting the measured values of the plasma lidocaine concentrations C[lido](t[i]) at each time t[i], then
sequentially connecting the points with straight-line segments, one obtains a graph of C[lido] (t) based on real data (Figure 19-2, A). The area under this graph is an approximation or an estimate of
the true AUC.
AUC is equivalent to the total milligram amount of lidocaine that is absorbed into the systemic circulation. Thus, if identical amounts of lidocaine are given by two different routes of delivery
(e.g., IV bolus dose and subcutaneous dose) and each yields 100% absorption, the total AUC for each route of delivery will be equal. The shapes of the graphs, however, will not necessarily be the
same (Figure 19-2, B to E).
Peak Concentration. The maximum lidocaine plasma concentration after any given dose (C[max]) is directly affected by factors that affect (1) rate of lidocaine absorption or (2) extent of lidocaine
absorption (systemic bioavailability).
For any given dose of subcutaneous lidocaine, any factor that accelerates the rate of lidocaine absorption will result in a shorter time necessary to achieve C[max] and will increase its magnitude.
Not surprisingly, anything that slows lidocaine absorption will also delay the time (T[max]) when C[max] occurs and will diminish its magnitude (Figure 19-3).
Two triangles of equal area do not necessarily have the same base and height. The area of a triangle equals ½BH, where B and H are the magnitude of the base and height, respectively. For two
triangles having equal area, the triangle with the smallest base must have the greatest height. This concept provides the mathematic basis for using the AUC to help explain the relationship between a
drug’s rate of absorption and its peak plasma concentration.
The rate of absorption determines the length of the base of the AUC. Rapid lidocaine absorption will have a short base, whereas slow absorption will have a long base. If the AUCs are equal for two
methods of lidocaine delivery, the short base will have a high peak (and thus a relatively higher risk of toxicity), and the long base will have a low peak plasma level (and a relatively lower risk
of toxicity).
Suppose C[lido](t) represents the graph of plasma lidocaine concentration over time, and AUC is the area under this curve. If a certain route of delivery produces rapid lidocaine absorption, the
geometric figure that represents the AUC under the graph of C[lido](t) will have a relatively short base and a high peak. If the same amount of drug is given in such a manner that the absorption is
much slower, the base of the AUC will be much wider, and the peak will be relatively low. Both graphs will have the same AUC because the same total milligram dose of lidocaine has been absorbed with
either route of delivery, even though the shapes of the curves are different.
This inverse relationship between C[max] and the rate of absorption explains the safety of the relatively large doses of lidocaine that are used with tumescent liposuction. Suppose that an epidural
dose of lidocaine is completely absorbed over 2 hours and that the corresponding peak lidocaine concentration is C[max]. If a tumescent dose of lidocaine requires 24 to 36 hours to be completely
absorbed, and if the tumescent dose should achieve the same C[max] as the epidural dose, one can predict that the tumescent dose can be approximately 24 to 36 times greater than the epidural dose.
Absorption rate determines peak plasma concentration, which in turn determines probability of toxicity. Different routes of administration yield different rates of absorption and thus different
likelihoods that plasma concentrations of the drug will exceed levels associated with toxicity. These concepts help in understanding how the slow rate of dilute lidocaine absorption associated with
the tumescent technique allows higher doses of lidocaine with such a high degree of safety.
Duration of Toxicity. The rate of lidocaine absorption affects the length of time that the plasma lidocaine concentration remains above any specified concentration (Figure 19-4).
When a relatively small dose of lidocaine is absorbed rapidly, the peak level is achieved quickly, then the plasma concentration decreases rapidly. If a relatively small dose of lidocaine produces
plasma concentrations that exceed the toxic threshold (6 μg/ml), toxicity is brief. After the rapid lidocaine absorption of an IV bolus dose, the pharmacokinetic behavior of lidocaine is a
two-compartment model. The duration of toxicity after an IV bolus is relatively brief because of the rapid distribution of lidocaine into peripheral tissues. The plasma concentration rapidly achieves
a peak in the central intravascular compartment, then rapidly decreases as the lidocaine is absorbed into the tissues of the peripheral compartment. The result is that lidocaine toxicity is brief
with less risk of serious complications if it occurs after rapid absorption of a relatively small lidocaine dose.
In contrast, after prolonged absorption, lidocaine behaves as if the body were a one-compartment system. If toxic threshold concentration is exceeded after slow absorption, peripheral tissues will
already be in equilibrium with the blood lidocaine, and no rapid decrease in plasma lidocaine concentration will occur. When lidocaine absorption is slow, a relatively large lidocaine dose may never
exceed the toxic threshold. If toxicity does occur after prolonged absorption, however, toxicity will persist relatively longer.
Absorption Rate Paradox. Obese patients tolerate larger mg/kg doses of IV lidocaine than thinner patients, possibly because obese patients have a larger apparent volume of distribution (V[D]) for
lidocaine. In other words, an obese patient has a greater amount of total body adipose tissue into which lidocaine can be partitioned.
For tumescent lidocaine the relative rate of absorption decreases with increasing obesity. In other words, for equal mg/kg doses of tumescent lidocaine, the greater the patient’s supply of
subcutaneous fat, the slower is the rate of lidocaine absorption.
The phenomenon in which obese patients tolerate higher mg/kg doses of tumescent lidocaine than thin patients is an unexpected paradox. As shown next, this absorption rate paradox might have another
Let A = surface area and V = the corresponding volume of a mass of tumescent fat. Clearly, V is proportional to the total dose of lidocaine. Assume that the rate of lidocaine absorption is
proportional to external A of a mass of tumescent fat. Then the ratio A/V is proportional to the rate of absorption per unit dose (mg) of lidocaine.
If the shape of a tumescent compartment of fat is cuboid, ellipsoid, or even spheric, a doubling (increasing by a factor of 2) of V of a mass of tumescent fat produces an increase by a factor of
(approximately) 1.6 of A of that volume (see mathematics section at the end of this chapter).
Consider two spheres (or cubes) having volumes V[II] and V[I], where V[II] is twice as large as V[I]. Thus V[II] = 2V[I], or (V[II]/V[I]) = 2.
Let the surface areas of these volumes be A[II] and A[I], where A[II]/A[I] = 1.5874 ≈ 1.6, or A[II] ≈ 1.6 A[I].
Consider the ratio A[II]/V[II] and substitute the terms A[II] ≈ 1.6 A[I] and V[II] = 2V[I]:
A[II]/V[II] ≈ (1.6) A[I]/(2)V[I] = (0.8)A[I]/V[I]
Recalling that A/V is proportional to the rate of absorption per unit dose (mg) of lidocaine, we see that the rate of absorption per unit area for the larger volume (V[II]) will be merely 0.8 times
as fast as the smaller volume (V[I]). Thus the geometric fact that V[II] = 2V[I], implying A[II] ≈ 1.6 A[I], might be a partial explanation for the absorption rate paradox.
Consider the following example. An obese patient weighs 100 kg and a thin patient 50 kg. The same areas of each patient are treated by tumescent liposuction, using identical concentrations of
anesthetic solution. If the obese patient requires twice the total milligram dose of lidocaine as the thin patient, both patients will receive the same mg/kg dose of lidocaine. The obese patient will
have a volume V[II] and the thin patient a volume V[I] of tumescent fat, where V II = 2V[I]. The rate of lidocaine absorption, which is assumed to be proportional to the surface area of tumescent
fat, will be relatively slower in the obese patient by a factor of 0.8.
Thus, at the same mg/kg dosage of lidocaine, the rate of lidocaine absorption will be relatively slower in the obese patient and faster in the thin patient. Finally, in conclusion, obese patients
should tolerate relatively higher mg/kg doses of lidocaine than thinner patients.
Infiltration Rate. At the relatively high concentrations of commercial formulations of lidocaine, rapid infiltration produces rapid systemic absorption. At the low lidocaine concentrations of
tumescent preparations, any increased rate of lidocaine absorption caused by rapid infiltration is clinically insignificant.
At relatively high lidocaine concentrations, a rapid subcutaneous injection quickly achieves plasma concentrations that can exceed the threshold for potential toxicity. A rapid subcutaneous injection
of 1350 mg of lidocaine (1% and 0.5%) with epinephrine (1:100,000) produced a plasma concentration of 6.3 mg/L within 15 minutes.^4 Recall that 6.0 mg/L (6.0 μg/ml) of lidocaine is the recognized
threshold for significant toxicity.
A slow injection of 1% lidocaine with epinephrine produces significantly lower and delayed peak blood levels. Thus 1000 mg of lidocaine (1%) with epinephrine (1:100,000) slowly injected
subcutaneously over 45 minutes produced a C[max] of 1.5 mg/L, which occurred 9 hours later (Figure 19-5).
A slow subcutaneous infiltration of lidocaine with epinephrine always slows the rate of lidocaine absorption. This effect is most important at the relatively high concentrations of commercial 1%
lidocaine with epinephrine. Initiating a local anesthetic infiltration by first giving a slow injection of a relatively small volume of lidocaine with epinephrine, then allowing some vasoconstriction
to occur before injecting a more significant volume, will reduce the absorption rate for both lidocaine and epinephrine. The result is a reduction of both the peak plasma lidocaine concentration and
the incidence of tachycardia related to epinephrine.
For dilute tumescent solutions of lidocaine and epinephrine, lidocaine absorption is always rather slow. Surgeons who use systemic anesthesia for liposuction tend to accomplish the tumescent
infiltration rapidly. It appears that lidocaine toxicity is not a significant risk despite rapid infiltration.
Nevertheless, a rapid infiltration of a tumescent solution of lidocaine and epinephrine does result in a rapid appearance of a small, brief peak in plasma lidocaine concentration. This transient
early peak is the result of a delayed onset of vasoconstriction, which allows a brief interval for relatively rapid absorption of a small amount of lidocaine. The tumescent lidocaine diffusion rate
across capillary endothelium is not rapid enough to cause toxicity, since the lidocaine concentration gradient is already relatively low because of the dilution inherent to the tumescent technique.
Surgeons must remember that rapid infiltration can be more uncomfortable than slow infiltration. Very rapid tumescent infiltration is usually done only when the liposuction patient is under systemic
Dilution. A volunteer study showed that 1% lidocaine was absorbed faster and had a higher peak than did 0.1% lidocaine, with total dose and infiltration rate held constant. In each case the total
dose of lidocaine was 1000 mg and the total dose of epinephrine 1 mg; the time allowed for infiltration was exactly 45 minutes. For 1% lidocaine the magnitude of C[max] was 1.5 μg/ml, which occurred
at T[max] of 9 hours. For 0.1% lidocaine, C[max] was 1.2 μg/ml and T[max] 14 hours (Figure 19-6).
The effect of a tenfold dilution of lidocaine on systemic absorption rate after a subcutaneous injection cannot be generalized to epidural spinal anesthesia. At higher absolute concentrations and in
highly vascular tissues, dilution does not significantly delay lidocaine absorption. For example, there was no significant difference in the rate of lidocaine absorption when 1% and 10%
concentrations of lidocaine were injected into the highly vascular tissue for epidermal anesthesia.^5
Vascularity. Tissue vascularity affects the rate of lidocaine absorption. Absorption is slower from relatively avascular fat compared with highly vascular tissue of the gingiva or epidural space. A
high blood flow rate through highly vascular tissue maintains the high concentration gradient across the vascular wall.
An injection of lidocaine and epinephrine into the highly vascular gingival mucosa for dental anesthesia may result in rapid epinephrine absorption and a brief but alarming tachycardia. Although
patients typically interpret the experience as an allergic reaction, it is not an immune-mediated event but merely a predictable pharmacologic effect of epinephrine.
Lidocaine toxicity can also occur after rapid subcutaneous injection of commercial formulations of lidocaine. I once watched a surgeon infiltrate 1% lidocaine subcutaneously in an infant for local
anesthesia before laser treatment of a hemangioma. The degree of vascularity of the site was one of the factors in the resulting lidocaine toxicity, as manifested by a generalized seizure.
Vasoconstriction. Epinephrine-induced vasoconstriction slows the rate at which blood can transport lidocaine away from the site of injection. The vasoconstriction induced by epinephrine is as vital
for the safety of the tumescent technique as the dilute nature of tumescent local anesthesia. Vasoconstriction not only prolongs the duration of local anesthesia, but more importantly slows the rate
of systemic absorption of lidocaine and thus significantly reduces the magnitude of C[max] (Figure 19-7).
Lidocaine is a capillary vasodilator that has a rapid onset of approximately 1 minute. This is evidenced by the rapid appearance of erythema after a simple 0.1-ml intradermal injection of lidocaine
(1%) without epinephrine. The vasodilator effect of lidocaine is caused by blockage of adrenergic neurotransmission and inhibition of vascular smooth muscle contraction.^6
Epinephrine is a capillary vasoconstrictor requiring approximately 3 to 6 minutes for onset of action and about 10 to 15 minutes for maximal effect. The vasoconstriction of epinephrine takes longer
to become clinically apparent than the vasodilation caused by lidocaine. Eventually the vasoconstriction of epinephrine overcomes the vasodilation of lidocaine. This phenomenon is seen clinically
when a 0.1-ml intradermal injection of lidocaine (1%) with epinephrine (1:100,000) initially produces erythema, followed by blanching several minutes later.
After Bolus Infusion. The pharmacokinetic process can be simplified by focusing on the fate of lidocaine after a rapid IV bolus dose. Most knowledge about the time-dependent fate of lidocaine in the
body is the result of studying lidocaine kinetics after an IV bolus in human volunteers and animals.
When given as an IV bolus, systemic lidocaine absorption is instantaneous. When studying the distribution kinetics of lidocaine as it spreads throughout the body, the process of absorption can be
eliminated as a confounding variable by focusing on the distribution process after an IV bolus. The changes in lidocaine blood levels over time will reflect only the processes of distribution and
elimination. Lidocaine distribution and elimination are independent of the absorption process, and therefore it is the same for an IV bolus as for subcutaneous tumescent infiltration.
In a mathematical model where no tissue-drug binding exists, one can assume that the rate of achieving blood-tissue equilibrium is a function of the rate of tissue perfusion. In this perfusion
rate–limited model, drug concentration in a tissue is the same as its concentration in the venous blood leaving the tissue. Lidocaine distribution pharmacokinetics is considered to be perfusion rate
limited.^7 For an IV bolus dose the kinetic model is a perfusion rate–limited, two-compartment model.
The fate of an IV bolus of lidocaine has two phases: (1) the distribution phase, or α-phase, and (2) the elimination phase, or β-phase. Immediately after an IV bolus of lidocaine, during the α-phase,
a large proportion of the dose is rapidly redistributed out of the vascular space and absorbed into the peripheral tissues. The highly perfused organs, such as the lungs, kidneys, and spleen,
followed by muscle, are the first tissues to achieve equilibrium with intravascular lidocaine. The lidocaine half-life during this distribution phase is approximately 8 minutes.^8 It takes more than
30 minutes for the levels of lidocaine in the blood to reach an equilibrium with lidocaine concentration in the myocardium and brain.
After distribution of lidocaine into peripheral tissues, the plasma lidocaine concentration is lowered as a result of elimination by hepatic metabolism during the β-phase (Figure 19-8).
Adipose tissue requires more time than other tissues to absorb enough lidocaine to achieve an equilibrium with blood because of the sparse vascularity of fat. Also, because lidocaine is highly
lipophilic, fat has a relatively high capacity to store lidocaine.
After Tumescent Infiltration. The lidocaine distribution kinetics of slow IV infusion and that of the tumescent technique are similar. In either situation the lidocaine concentration in the blood
increases so slowly that an equilibrium is maintained continuously between blood and peripheral tissues. Peak tissue and blood concentrations are achieved simultaneously. The rate of tissue perfusion
is relatively fast compared with the rate of lidocaine absorption. Thus lidocaine distribution after tumescent delivery is not perfusion rate limited.
Simultaneous Metabolic Activities
As noted earlier, the key to understanding the pharmacokinetics of tumescent lidocaine is the interaction between (1) the rate of lidocaine absorption from the tumescent fat and (2) the rate of
systemic elimination.
Tumescent absorption is slow and continuous, similar to a slow, continuous IV infusion. Six to eight hours after initiating a slow IV infusion of lidocaine, an equilibrium is reached between the
lidocaine concentration in blood and that in peripheral tissues. When a slow IV infusion is discontinued, the residual of lidocaine in the body is eliminated at a rate identical to the elimination
β-phase that follows an IV bolus dose. The half-life of this elimination phase is about 2 hours. Tumescent absorption cannot be terminated as abruptly as turning off an IV infusion; rather, it
continues for more than 18 to 36 hours.
The rate of increase and then decrease of lidocaine concentration is determined by a complex interplay between simultaneous tumescent absorption and hepatic metabolism. The complexity of these two
concurrent processes becomes apparent when one observes the qualitative differences between the graphs of C[lido](t) for a low dose (15 mg/kg), a medium dose (35 mg/kg), and an exceptionally high
dose (60 mg/kg) of tumescent lidocaine (Figure 19-9).
Smaller Tumescent Doses. With a relatively small tumescent dose of lidocaine, 15 to 35 mg/kg, the rate of lidocaine absorption increases to a well-defined C[max] at T[max] and then decreases. At
these doses the range of T[max] is approximately 8 to 14 hours after initiating the infiltration.
As a first-order process, the rate of hepatic elimination of lidocaine increases and decreases in proportion to the plasma lidocaine concentration. Initially the absorption rate is much faster than
the rate of elimination, and thus the lidocaine concentration increases. The rate of hepatic elimination is maximal when the plasma concentration achieves its maximal value, C[max].
Mathematically, T[max] is exactly the point in time when the rate of absorption equals the rate of elimination. Before T[max] the rate of lidocaine absorption from the tumescent fat is greater than
the rate of hepatic elimination. After T[max] the rate of lidocaine absorption from the tumescent fat slows down and is less than the rate of elimination (Figure 19-10).
Larger Tumescent Doses. Large doses of tumescent lidocaine behave differently than small doses. Thus, after tumescent liposuction, a graph of C[lido](t) is qualitatively different from that at a
smaller dose of lidocaine.
With relatively small doses of tumescent lidocaine (less than 35 mg/kg) the plasma lidocaine concentration reaches a well-defined peak at approximately 10 to 12 hours after beginning the
infiltration. At relatively large doses, however, instead of a distinct peak, the lidocaine concentration seems to achieve a broad, level plateau that is maintained for 16 hours or more before
declining. With a 60-mg/kg tumescent dose of lidocaine, a plateau is reached within 8 hours and persists for at least another 16 hours (Figure 19-11).
With continued absorption the amount of lidocaine remaining at the site of absorption decreases steadily. Subsequently, hepatic metabolism significantly reduces lidocaine blood levels, which approach
0 mg/L by 48 hours after beginning the tumescent infiltration.
The existence of a concentration plateau shows that, during the time that the lidocaine blood levels remain constant, the rate of lidocaine absorption is in a state of constant equilibrium with the
rate of lidocaine elimination. Specifically, the rate of lidocaine absorption is exactly equal to the rate of lidocaine elimination. The concentration plateau indicates that the absorption rate from
the tumescent tissues is constant and independent of the amount of lidocaine remaining at the absorption site.
Zero-Order or First-Order Processes
When a drug is absorbed at a constant rate, independent of the amount of drug remaining at the absorption site, the process is said to be a zero-order absorption process. An example is the
long-acting oral tablet designed to dissolve on the outer surface of the tablet and thus slowly release drug at a constant rate for absorption into the systemic circulation. Thus, with tumescent
liposuction, lidocaine absorption behaves as a zero-order absorption process, which is unique to the tumescent technique for subcutaneous injection.
In contrast, the absorption of oral drugs in a rapidly disintegrating dosage form that quickly dissolves into solution, as well as most injections into subcutaneous tissue and muscle, often
approximates first-order kinetics. A first-order absorption process is characterized by an absorption rate that decreases over time and is proportional to the amount of drug that remains at the
absorption site. This process is characterized by an absorption rate constant (k[a]) and a corresponding absorption half-life.
The phenomenon of zero-order lidocaine absorption is paramount in explaining the safety of the tumescent technique. The unexpected zero-order absorption of tumescent lidocaine explains how the large
doses of subcutaneous lidocaine are consistently associated with a low risk of toxicity.
The terminology referring to first-order and zero-order processes is derived from the differential equations that describe these two processes (see section at the end of this chapter).
Pharmacokinetic studies of IV lidocaine given by rapid bolus or slow infusion provide much information about the elimination kinetics for lidocaine, with estimates for the volume of distribution (V),
clearance (Cl), elimination half-life (t[1/2]), and elimination rate constant (k). Also, systemic elimination of lidocaine is a first-order process.
Cytochrome P450 3A4. Lidocaine elimination by hepatic enzyme cytochrome P450 3A4 (CYP3A4) is a first-order process. In healthy volunteers, lidocaine is so efficiently and rapidly metabolized by the
hepatic enzymes that the rate of elimination is limited only by the rate of hepatic perfusion. In healthy patients, hepatic capacity to metabolize lidocaine is not saturated, and the rate of
lidocaine elimination is proportional to lidocaine concentration in the blood. The greater the amount of lidocaine within its volume of distribution, the higher is the plasma concentration and the
higher the rate of hepatic metabolism.
If the rate of lidocaine metabolism is reduced by 50%, C[max] is approximately doubled. The relative safety of the tumescent technique for local anesthesia depends on the normal, high rate of liver
metabolism of lidocaine (see Chapter 17). Anything that decreases the rate of lidocaine metabolism will increase C[max] and increase the risk of toxicity. The following factors can reduce the rate of
lidocaine metabolism:
1. Decreased CYP3A4 enzymatic activity caused by competitive drug interaction that inhibits CYP3A4
2. Decreased blood flow to the liver caused by decreased cardiac output from either excessive IV fluids precipitating congestive heart failure or drug-induced impaired cardiac output (e.g., with
3. Impaired hepatic function resulting from hepatitis or cirrhosis
Drugs such as ketoconazole (Nizoral) or sertraline (Zoloft) can inhibit or reduce the enzymatic function of CYP3A4. Even with drugs that inhibit CYP3A4, little (if any) evidence indicates that the
hepatic enzymes responsible for lidocaine metabolism become saturated (see Chapter 18). Although a drug may inhibit CYP3A4 and slow the rate of lidocaine metabolism, this rate remains a linear
function of plasma lidocaine metabolism and rate of hepatic blood flow.
After Bolus Infusion. Once again, consider the example of a rapid IV bolus dose of lidocaine. An equilibrium is quickly established between lidocaine in the blood and in the lipids of peripheral
tissues. From this point the rate of decreasing plasma lidocaine concentration slows noticeably. During the entire β-phase the slow decrease of the lidocaine concentration is essentially 100%
attributable to hepatic metabolism. The half-life of lidocaine as a result of this β-phase elimination is 100 to 120 minutes, or about 2 hours.
Lidocaine is rapidly and almost entirely metabolized by the liver. Less than 5% of lidocaine is cleared by the kidneys. For a 70-kg person the clearance (blood volume per minute) of lidocaine is 640
± 170 ml/min, approximately 700 ml/min/70 kg, or more generally 10 ml/min/kg. This approximates the plasma flow to the liver.^9,10
The liver is so efficient at metabolizing lidocaine that most of the lidocaine that passes through the hepatic circulation is removed (Figure 19-12). Lidocaine is said to have a hepatic extraction
ratio of 0.7, which means that 70% of lidocaine entering the liver exits as metabolite. In other words, for every liter of blood that passes through the liver, 70% of its lidocaine is metabolized,
whereas only 30% survives unchanged. More precisely, the hepatic extraction ratio for lidocaine is reported to be 62% to 81%.
After Tumescent Infiltration. Elimination of lidocaine after tumescent delivery is significantly different from the process of elimination after an IV bolus dose. For example, the total amount of
lidocaine in a tumescent dose, up to 50 mg/kg, is significantly larger than the typical IV bolus dose of 1 to 2 mg/kg.
The absorption half-life of tumescent lidocaine is much longer (approximately 8 to 12 hours) than the elimination half-life for lidocaine (2 hours). Tumescent lidocaine is absorbed so slowly that
much of the drug remains to be absorbed well beyond the T[max] when C[max] is reached. At any particular time, most of the infiltrated tumescent lidocaine either is in the tumescent fat waiting to be
absorbed or has been eliminated. Little of the total dose is in the systemic circulation, or the volume of distribution, at any given time.
After tumescent infiltration the process of lidocaine elimination is prolonged by the continuous, slow, systemic absorption of tumescent lidocaine and persists up to 48 hours. Elimination of
lidocaine after tumescent infiltration is not much different from the rate of lidocaine absorption from the tumescent fat. In fact, when the rate of lidocaine absorption is exactly equal to the rate
of lidocaine elimination, the plasma lidocaine concentration is constant. When lidocaine absorption and elimination are equal, lidocaine concentration as a function of time is a constant plateau (see
Figure 19-11).
Clearance and Half-life. The systemic clearance (total body clearance) and elimination half-life of lidocaine provide a quantitative measure of the rate of liver metabolism.
In young, healthy male volunteers, mean systemic clearance and elimination half-life of lidocaine after a single IV bolus dose are 15.6 mL/min/kg and 1.6 hours, respectively.
Young female volunteers seem to have an increased clearance and half-life and a larger volume of distribution than young, healthy males. In general, therefore, young females can tolerate larger doses
of tumescent lidocaine than young males.
Elderly men and women have reduced clearance and prolonged half-life compared with young adult controls.^11 Thus the recommended maximum dose for tumescent lidocaine should be reduced by 10% to 25%
for older patients.
Young obese volunteers have a prolonged half-life and an increased volume of distribution. Again, compared with thin patients, obese patients should be better able to tolerate a higher tumescent
dosage of lidocaine.^12
Drug Bioavailability
Bioavailability of a drug is defined as the fraction of a given dose that ultimately reaches a particular targeted tissue. A dose of a drug might have different degrees of bioavailability for
different organ tissues. Thus a large percentage of an oral antifungal drug may reach the liver, lungs, and kidneys, but only a small fraction may penetrate the blood-brain barrier and enter the
central nervous system.
The concept of bioavailability is different from that of absorption. Drug absorption involves the fraction absorbed from the site of administration and the rate at which a drug diffuses away from its
site of administration and into the body’s systemic circulation. Bioavailability involves the fraction of dose that reaches the site of action. A drug’s bioavailability depends on both its absorption
and its ability to penetrate certain barriers and to avoid metabolism or elimination on its journey to the site of drug action.
Bioavailability depends on the following:
1. Site of administration
2. Extent of absorption from the site of administration
3. Physiologic and pathologic states that affect metabolism
4. Amount of absorbed drug that is eliminated before it can reach the targeted site of action
For example, the bioavailability of a hydrocortisone ointment topically applied to the skin might only be 1%. Thus, if 100 μg of hydrocortisone is applied to the skin dissolved in an ointment base,
only 1 μg is ultimately absorbed through the stratum corneum, epidermis, and dermis and then enters the systemic circulation, eventually being metabolized and finally excreted.
For another example, after a 4-mg sedative is taken by mouth, 2 mg is actually absorbed from the GI tract into the portal circulation, of which 1 mg is metabolized by the liver. Thus only 1 mg of the
original 4 mg actually enters the systemic circulation and eventually reaches the central nervous system. The bioavailability is (1 mg)/(4 mg) = 0.25, or 25%.
It is typically an advantage to maximize bioavailability when a drug is targeted at the parenchyma of an internal organ. When a potentially toxic drug is targeted at subcutaneous or external
(epidermal or mucosal) body tissue, however, it may be an advantage to minimize systemic bioavailability. Examples include a topical anticandidiasis cream or an antihelminthic drug for intraluminal
intestinal parasites (the GI tract’s mucosal surface is topologically an external body surface).
Tumescent Lidocaine
Tumescent liposuction is unusual in that the site of action is the local subcutaneous fat targeted for liposuction. Optimal local bioavailability requires a certain time for the tumescent solution to
be dispersed by bulk flow throughout the targeted compartment and for lidocaine to diffuse into sensory nerves. To minimize lidocaine toxicity, one must lessen systemic bioavailability by minimizing
systemic absorption.
Lidocaine delivered by the tumescent technique is targeted at localized tissues. It is a therapeutic advantage to maximize the local effects of tumescent lidocaine and to minimize its systemic
effects. The safety of tumescent liposuction is the result of (1) reducing the rate of lidocaine absorption (by means of extreme dilution and profound vasoconstriction) and (2) reducing the systemic
bioavailability (by open drainage and bimodal compression after liposuction).
Lidocaine toxicity is reduced by minimizing the amount of lidocaine absorbed into the systemic circulation. The goal with tumescent liposuction is to maximize local bioavailability and minimize
systemic bioavailability of lidocaine.
Liposuction removes a percentage of lidocaine and thus reduces the systemic bioavailability of lidocaine. Without liposuction the systemic bioavailability of lidocaine after tumescent infiltration
would be 100%. Because of the slow rate of tumescent lidocaine absorption, however, high doses of lidocaine are safe even without liposuction. If liposuction cannot be completed after tumescent
infiltration, the risk of lidocaine toxicity is minimal if total dosage is below the recommended maximum safe dosage of 50 mg/kg.
With liposuction the margin for safety for tumescent lidocaine is even greater. The systemic bioavailability of lidocaine with tumescent liposuction can be reduced by two distinct processes that
physically remove lidocaine from the body. First, liposuction reduces the systemic bioavailability of lidocaine by about 20% simply by aspirating lidocaine along with fat. Second, open drainage with
bimodal compression further reduces the amount of systemic lidocaine absorption by increasing the amount of lidocaine that drains out of the body after surgery.
Accelerated drainage of residual tumescent anesthetic solution is accomplished by using adits (punch biopsy holes 1.0, 1.5, or 2.0 mm in diameter) instead of incisions. If incisions are used,
drainage is maximized by allowing incision sites to remain open (not closing incisions with sutures). Finally, applying a high degree of uniform compression over the treated areas encourages a
maximum volume of drainage.
Maximum Safe Dose. After tumescent infiltration has been completed and before liposuction can be initiated, something may force the patient or surgeon to cancel the surgery, resulting in 100%
lidocaine bioavailability and an increased risk of systemic toxicity.
Thus any estimate of a maximum safe dose of lidocaine for tumescent liposuction must assume 100% systemic bioavailability. One cannot assume that liposuction will always be accomplished after
tumescent infiltration is completed, nor can one assume that the systemic bioavailability will be substantially less than 100%.
Sequestered Lidocaine Reservoir
Delayed lidocaine absorption is key to the safety and efficacy of very dilute (tumescent) solutions of lidocaine. This delayed absorption seems to result from a reservoir effect; lidocaine is
sequestered within the infiltrated adipose tissue and unavailable for immediate absorption in the systemic circulation. The actual site of this reservoir is not critical to the pharmacokinetics of
tumescent lidocaine.
With an idealized one-compartment model for tumescent lidocaine, no pharmacokinetic distinction exists between lidocaine that is bound or unbound to tissue. The proportion of bound and unbound
lidocaine is assumed to be constant. Also, any change in concentration of the unbound lidocaine is instantly reflected in a proportionate change in the bound fraction. As soon as unbound tissue
lidocaine diffuses into the systemic circulation, a proportionate amount of the lidocaine bound to subcutaneous tissue instantaneously becomes unbound.
An ideal pharmacokinetic model also assumes that the bound fraction is not displaced from its binding site by anything other than a change in the concentration of the unbound fraction. In other
words, the assumption is that no drugs or processes might displace bound lidocaine from its binding sites and change the proportionality between the bound and unbound fractions.
The lidocaine reservoir may be the result of (1) pools of tumescent anesthetic solution loculated within vasoconstricted fatty tissue and (2) binding of dilute lidocaine to adipose tissue in general
and to adipocyte lipids in particular. The relative importance of these sites of lidocaine sequestration is not certain.
When lidocaine is added to a mixture of equal volumes of water and lipids, a greater proportion of the lidocaine is dissolved within the lipid fraction. De Jong^3 favors the direct binding of
lidocaine to adipose tissue as the source of the reservoir effect for subcutaneous lidocaine. He has proposed that the ratio of unbound/bound lidocaine within adipose tissue is a function of the
concentration of lidocaine in the local anesthetic solution.
It is unlikely that the explanation for slow systemic absorption of tumescent lidocaine is an intense binding of lidocaine to lipids within tumescent fat or the sequestration of lidocaine within the
interstitium of tumescent tissue. If the explanation were as simple as that, one would expect more lidocaine to be removed by liposuction.
Tumescent lidocaine pharmacokinetics is best represented by a one-compartment model. Tumescent absorption is a zeroorder process. The rate of lidocaine absorption from tumescent tissue is constant
and independent of the amount of lidocaine remaining at the site of absorption. Tumescent lidocaine resembles the absorption characteristics of a slow-release tablet in the GI tract. The rate of
hepatic elimination of tumescent lidocaine is proportional to the concurrent plasma lidocaine concentration and is thus a first-order elimination process.
Large doses of tumescent lidocaine in the range of 50 to 55 mg/kg appear to be safe in most patients. Impaired hepatic metabolism of lidocaine can predispose a patient to lidocaine toxicity. The peak
plasma concentration typically reaches a plateau within 8 to 12 hours and can persist for more than 24 hours, before diminishing to zero at about 48 hours.
Tumescent pharmacokinetics might have application beyond local anesthesia. The slow rate of absorption and prolonged concentration plateau associated with extreme dilution might prove advantageous
for drug delivery in other areas of medicine, such as oncology. Treatment of breast cancer by the local infiltration of a chemotherapeutic agent may allow a prolonged local uptake of the drug by the
proximal lymphatics.
The mathematics of pharmacokinetics
The following sections provide more detailed analyses of concepts discussed in this chapter.
Derivation of Surface Area/Volume Ratio
The following calculations demonstrate the relationship between the ratios of surface area (A) to volume (V) for two spheres, where the volume of one sphere is twice a large as the volume of the
other. Interestingly, the same relationship between A/V ratios holds for two cubes, where one has twice the volume of the other.
Conjecture. Suppose that volume V[II] is twice as large as volume V[I], that is, V[II]/V[I] = 2. If A[II] and A[I] represent the surface of V[II ]and V[I], respectively, then A[II]/A[I] @ 1.6, and A/
V ratios for these two volumes are related by the equation A[II]/V[II] – (0.8)A[I]/V[I].
Proof (Spheres). The volume of a sphere of radius r is (4/3)πr^3, and its surface area is 4πr^2. Now suppose V[II] – 2V[I], where R is the radius of the larger sphere, and r is the radius of the
smaller sphere. Thus V[II] – (4/3)πR^3, and V[I] – (4/3)πr^3.
Because V[II] = 2V[I], then (4/3)πR^3 = V[II] = 2V[I] = 2(4/3)πr^3. Solving for R yields R^3 = 2 r^3, and therefore R = (2)^1/3r. From the classic formula for the surface area of a sphere, A[II] =
4πR^2, and A[I] = 4πr^2. Substituting R = (2)^1/3r in the equation yields the surface area for A[II]:
A[II] = 4πR^2 = 4π[(2)^1/3r]^2 = (2)^2/3 4πr^2 = (2)^2/3 A[I]
Thus the ratio of the two surface areas is as follows:
A[II]/A[I] = (2)^2/3 = (2^2)^1/3 = (4)^1/3 = 1.5874 @ 1.6
By assumption, V[II] = 2V[I], or V[II]/V[I] = 2. Therefore, if A[II]/A[I] is divided by V[II]/V[I]:
[A[II]/A[I]]/[V[II]/V[I]] @ (1.6)/2 = 0.8
After rearranging the terms, we may conclude the following:
A[II]/V[II] @ (0.8) A[I]/V[I]
Proof (Cubes). A similar calculation holds true when the volumes are cubes (cuboid) rather than spheres (spheric). Suppose V[II] and V[I] represent the volume of the two cubes, where the volume of
the larger cube is twice the volume of the smaller cube; thus V[II] = 2 V[I]. Let E and e represent the length of an edge of the large and small cube, respectively. Then the volumes can be expressed
as V[II] – E^3 and V[I] – e^3. Because V[II] – 2V[I], then E^3 – V[II] – 2V[I] – 2e^3. Solving this algebraic equation for E yields E^3 – 2 e^3, and therefore E – (2)^1/3 e.
If A[II] and A[I] represent the surface area of the respective cubes, A[II] = 6E^2, and A[I] = 6e^2. Substituting E = (2)^1/3e into A[II] = 6E^2:
A[II] = 6E^2 = 6[(2)^1/3e]^2 = 6(2)^2/3e^2
Thus the ratio of the two surface areas is as follows:
A[II]/A[I] = [6(2)^2/3e^2]/6e^2 = (2)^2/3 = (4)^1/3 = 1.5874 @ 1.6
By assumption, V[II] = 2V[I], or V[II]/V[I] = 2. Therefore, if A[II]/A[I] is divided by V[II]/V[I]:
[A[II]/A[I]]/[V[II]/V[I]] ≈ (1.6)/2 = 0.8
After rearranging the terms, we may conclude the following:
A[II]/V[II] ≈ (0.8) A[I]/V[I]
Discussion. Let us assume that a volume (V) of tumescent fat is roughly spheric. If the concentration of the tumescent anesthetic solution is constant, the volume of the sphere will proportional to
the total dose (milligrams) of lidocaine, and thus:
V is proportional to the total dose of lidocaine.
If we assume the systemic absorption of lidocaine only occurs from the surface of the sphere, the rate of absorption is proportional to the surface area (A) of the sphere, and thus:
A is proportional to the rate of lidocaine absorption.
Finally, the surface area/volume term is as follows:
A/V is proportional to the rate of lidocaine absorption per total dose of lidocaine.
This conclusion is derived from elementary geometry and from the assumption that tumescent lidocaine absorption is proportional to the surface area of the mass of tumescent fat. This relation between
A and V, where A[II]/V[II] @ (0.8) A[I]/V[I], helps to explain why an obese patient tolerates a larger dose of tumescent lidocaine better than a relatively thin patient.
Suppose patient P[II] weighs 100 kg and patient P[I] weighs 50 kg. If each receives the same mg/kg dose of lidocaine, then V[II] = 2 V[I], where V[II] and V[I] represent the volume of tumescent fat
in each patient, respectively. The rate of absorption per total dose is less for the obese patient than the thin patient by a factor of 0.8.
Estimating Time for Lidocaine Clearance
Using standard values for lidocaine clearance and other clinical values yields calculations consistent with observed pharmacokinetic parameters. For example, consider a 70-kg patient who received a
50-mg/kg dose of tumescent lidocaine. The total dose of lidocaine is 50 mg/kg = 70 kg = 3500 mg. From clinical experience, I know that this dosage will yield an average peak plateau of peak plasma
concentration of approximately 2.0 mg/L.
Also, from data published by others, lidocaine clearance in a 70-kg patient is 700 ml/min. In other words, the liver will clear 700 ml of lidocaine per minute. If the blood contains 2.0 mg/L, then
700 ml of blood must contain 1.4 mg. Because the liver clears 700 ml/min, the liver must metabolize or eliminate 1.4 mg/min of lidocaine. By simple multiplication we calculate that the liver will
metabolize lidocaine at a rate of 1.4 mg/min = 84 mg/hr. Finally, we can calculate that the number of hours necessary to metabolize 3500 mg of lidocaine is (3500 mg)/(84 mg/hr) = 41.6 hours.
Thus, knowing that a 3500-mg dose of lidocaine will typically yield a peak plasma lidocaine concentration 2.0 mg/L, and using standard average value of 700 ml/min for lidocaine clearance in a healthy
patient, we can derive an estimate of the time required to metabolize and eliminate the entire 3500-mg dose of lidocaine: 41.6 hours.
The actual time for complete lidocaine metabolism depends on the validity of several assumptions. For example, because plasma concentrations are not a constant 2.0 mg/L, the actual time for complete
clearance may be greater than 41.6 hours. On the other hand, because liposuction will reduce lidocaine bioavailability, not all 3500 mg of the infiltrated lidocaine will be absorbed into the systemic
circulation. Thus the time required to metabolize all the tumescent lidocaine is less than 41.6 hours. This time is consistent with the observed values of 36 to 48 hours for complete lidocaine
clearance after a tumescent dose of 50 mg/kg for liposuction.
A few simple calculations demonstrate the internal consistency of the estimated values for the various pharmacokinetic parameters, using the previous example. At steady state, rate of elimination = R
[E] = R[A] = rate of tumescent lidocaine absorption. Since R[E] = 1.4 mg/min = 84 mg/hr, we can conclude that R[A] = 84 mg/hr.
Systemic clearance (Cl) can be viewed as a proportionality constant in the relationship between R[A] and steady-state plasma concentration (C[SS]). Thus R[A] = Cl · C[SS]. Substituting the
appropriate values:
1.4 mg/min = (0.7 L/min)(2.0 mg/L)
Mathematical Description of Tumescent Lidocaine Kinetics
All drugs exhibit undesirable and toxic effects. Clinical pharmacokinetics helps to determine the optimal dosage regimen for any particular drug. This section provides a more rigorous discussion of
some basic pharmacokinetic principles as applied to the tumescent technique.
The apparent volume of distribution (V) is a theoretic concept that allows one to express the (total amount of drug in the body) = (X) in terms of the (concentration of drug in the blood) = (C). This
theoretic V is a tool that simplifies the mathematical modeling of complex pharmacokinetic processes.
When a drug enters the body, it is distributed into various tissues in different concentrations. If the degree of drugtissue binding is independent of concentration, the ratio of the concentrations
in the various tissues is constant. By making this theoretic assumption, one can express the ratio between the (C) and (X) by a simple proportionality equation, C = (1/V) · X, where (1/V) can be
regarded as a constant of proportionality.
Knowing both X (e.g., total IV dose) and C (measured from a blood sample), one can express concentration (C) as the total amount (X) of drug per apparent volume (V):
C = X/V
X = V · C
The theoretic number V represents the hypothetic volume that would be necessary to contain all the drug present in the body at exactly the same concentration as found in the blood.
Lidocaine: Rate of Elimination. The rate of elimination of lidocaine from the body is known from the change of lidocaine blood levels over time after a simple IV bolus injection. By measuring
lidocaine blood levels sequentially over time and plotting these levels on a graph, one can show that the rate of elimination of lidocaine from the body is proportional to the concentration (C) of
lidocaine in the body.
A pharmacokinetic elimination process for a given drug is defined as a first-order process when the drug’s rate of elimination is directly proportional to the drug’s plasma concentration. Thus, by
definition, lidocaine elimination via hepatic metabolism is a first-order process that is mathematically described by a simple, linear differential equation. This differential equation can be written
in different ways, each equivalent to the others:
where (C) = X/V), X is amount of drug currently in the body, V is hypothetic apparent volume of distribution, and K is linear elimination rate constant.
This equation can be expressed as follows: the rate of decrease in the amount of lidocaine in the body is directly proportional to the amount in the body. The negative sign (–) indicates that a
negative change is occurring, that is, X is decreasing with time. So far we have two proportionality equations: X = V · C and dX/dt = –KX. From elementary calculus we can solve the differential
equation as follows:
dX/X = –K dt
Taking the integral of each side:
∫dX/X = ln X – ln X[0] = ln(X/X[0])
∫–K dt = –K · t = K · 0 = –K · t
Then equating the results:
ln X – ln X[0] = –K · t
or equivalently:
ln(X) = ln(X[0]) – (K · t)
Converting this natural logarithm to an exponential, one obtains an expression of the amount of drug in the body as a function of time:
exp [ln(X)] = exp[ln(X[0])–(K · t)]
This is equivalent to the following:
X = X[0] · exp(–Kt) (1)
Finally, because X = V · C, we can write V · C = X = (X[0]) exp(–Kt). After a rearrangement of terms, we can now have an equation that expresses blood concentration C(t) of lidocaine after IV bolus
injection as a function of time:
C(t) = (X[0]/V)[exp(–Kt)]
Finding Experimental Value of C[0] and V. By experimentally measuring the plasma concentration C of lidocaine at various times, we can determine experimental estimates for the values of K, the
elimination rate constant, and V, the apparent volume of distribution.
Numeric calculations and drawings of graphs are easier using common logarithms to the base of 10 rather than natural logarithms to the base e. By definition:
Y = log[10]X (2)
if and only if:
10^Y = X (3)
We can express Y in terms of natural logarithm to the base e by taking the natural logarithm of both sides of equation 3:
In X = ln 10^Y = Y ln 10 (4)
After rearranging the terms, we have:
Y = (ln X)/(ln 10) (5)
By equating equations 1 and 4, we have:
log[10]X = (ln X)/(ln 10) (6)
Because ln (10) = 2.303, or equivalently, e^2.303 = 10, equation 6 can be written as:
log[10](X) = ln(X)/2.303
ln(X) = (2.303)log[10](X) (7)
By taking the natural logarithm of both sides of equation 1, X = X[0] · exp(–Kt), we have:
ln(X) = ln(X[0]) – (K · t) (8)
Substituting equation 7 into equation 8, we have:
log X = (log X[0]) – (K · t/2.303)
Because X = VC, and X[0] · VC[0], we derive:
log C(t) = (log C[0]) – (K/2.303) · t (9)
By measuring the plasma lidocaine concentration C(t) at various times and then plotting a graph of the data log C(t), which is the same as plotting C(t) using semilog graph paper, the data should
fall along a straight line described by the equation for log C(t), where –K/2.303 is the slope of the line. By extrapolating back to time t = 0, we obtain:
log C(0) = log C[0] – (K · 0/2.303) = log C[0]
This is an experimental estimate for the value of C[0].
Now we can estimate for the apparent volume of distribution V = X[0]/C[0], where X[0] is the IV dose and C[0] has been derived experimentally. For lidocaine, experimental data published by several
investigators suggest that a reasonable estimate of the apparent volume of distribution V for a 70-kg person is 77 (± 28) liters, or 1.1 L/kg.
Experimental Estimate of K
Method 1. An experimental estimate for the value of K, the apparent first-order elimination rate constant, can be determined from the relationship C(t) = C[0] at time t = 0, and C(t[1/2]) = C[0]/2 at
time t[1/2], where t[1/2], the half-life of C(t), is the length of time required for C(t) to decrease from C[0] to (C[0])/2. We can easily determine t[1/2] by simply plotting the graph of C(t), the
plasma concentration of lidocaine, as a function of time following an IV bolus dose.
Substituting C(t[1/2]) = C[0]/2 in equation 9, we have:
log(C[0]/2) = log C[0] – [K · t[1/2]]/2.303 (10)
After rearranging the terms, we derive:
[log(C[0]/2) = log C[0] ]/t1/2= -[K]2.303 (11)
where –[K]/2.303 is equal to the slope of the line C(t) between t = 0 and t = t[1/2]. Thus, by estimating C[0] and measuring C[0]/2 and t[1/2], one can determine the numeric value of the slope = [K]/
2.303 and therefore the number K.
Method 2. From equation 1 we have:
X = X[0] · exp(–Kt)
If X = (1/2)X[0], equation 1 becomes X[0]/2 = X[0] · exp(–Kt[1/2]). Thus, [1/2] = exp(–Kt[1/2]), or equivalently, 2 = exp[Kt[1/2]], and ln(2) = K · t[1/2], and finally:
K = ln(2)/t[1/2] = 0.693/t[1/2] (12)
Thus, by measuring the half-life of C(t), one can calculate the value of K.
Absorption. The process of drug absorption is a zeroorder process when the drug enters the systemic circulation at a constant rate. Thus dX/dt = k[0] describes a zero-order process of absorption.
After tumescent infiltration of a large volume of dilute lidocaine and epinephrine into subcutaneous fat, the majority of the lidocaine is sequestered inside the tumescent subcutaneous tissue. Only
the lidocaine near the peripheral “surface” of the infiltrated tissue is available for absorption. The profound degree of tumescent vasoconstriction in effect isolates the lidocaine that is contained
deep within the tumescent tissue.
Thus, once a given dose X[I] of tumescent lidocaine at a concentration C has been infiltrated, the lidocaine absorption rate k[1] is approximately a constant. When a larger dose of tumescent
lidocaine X[2] having the identical concentration C is infiltrated, the rate of absorption will be a constant k[2]. However, k[1] and k[2] are not necessarily equal.
Simultaneous Absorption and Elimination. From the previous discussion, we know that lidocaine is eliminated by a first-order process, and therefore dX/dt = –KX.
In reality, after tumescent infiltration of lidocaine, the plasma concentration depends on both the rate of absorption and the rate of elimination. The differential equation that describes the
simultaneous absorption and elimination of lidocaine is as follows:
dX/dt = k[0] –KX
where X is the amount of lidocaine in the body, and k[0] is the constant rate of absorption (units of mg/sec). The solution of this differential equation is as follows:
X = (k[0]/K)[1 –exp(–Kt)]
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Figure 19-1 Graph of plasma lidocaine concentration, C[lido](t), as a function of time after subcutaneous tumescent infiltration. Area under the curve (AUC) is equivalent to total amount of lidocaine
absorbed over time into systemic circulation. Maximum or peak concentration of plasma lidocaine is designated by C[max]. The term T[max] is the exact point in time when C[max] occurs. For a fixed
value of AUC, the smaller the value of T[max], the greater is the value of C[max]. Any factor that delays lidocaine absorption will decrease magnitude of peak concentration.
Figure 19-2 Kinetics of tumescent lidocaine and tumescent saline. On two separate occasions, 75-kg female had tumescent infiltration of 2625 mg of lidocaine (35 mg/kg) in 5 L of normal saline. On the
first occasion, no liposuction was done. Two weeks later, infiltration was repeated using identical volume of saline and identical dosage of lidocaine, followed by tumescent liposuction of 1500 ml of
supranatant fat. (Redrawn from Klein JA: Plast Reconstr Surg 92:1085-1098, 1993.) A, Effect of liposuction on concentration of plasma lidocaine. Two graphs of concentration (C[lido]) of plasma
lidocaine (μg/ml) as a function of time (t) are shown. The tallest graph, with largest area under curve (AUC), is a plot of sequential determinations of plasma lidocaine concentration after tumescent
infiltration of 2625 mg of lidocaine in 5 L of solution without liposuction. Second graph has smaller AUC and represents sequential plasma lidocaine concentrations resulting from identical
infiltration but differing from first because of liposuction. Difference in magnitude between two AUCs is attributed to effects of liposuction. Liposuction is responsible for reducing systemic
absorption of tumescent lidocaine by approximately 20%. When this study was done, all incisions were closed with suture, with no open drainage. If open drainage had been used to remove additional
lidocaine, reduction of AUC would have been more significant. Liposuction and open drainage contribute to reduced bioavailability and increased safety of tumescent lidocaine. B, Hematocrit changes
after tumescent infiltration. Sequential measurement of hematocrit after tumescent infiltration of 5 L of saline with dilute epinephrine (0.5 mg/L) reveals much about systemic absorption of tumescent
saline. Because hematocrit returns to preoperative value within 48 hours, change of hematocrit cannot be attributed to blood loss. No evidence indicates that tumescent liposuction produces
hemoconcentration or that intravascular fluid deficit exists. Tumescent technique produces significant hemodilution with virtually no IV fluids. In this example, 10% decrease in hematocrit is
consistent with 10% increase in volume of intravascular fluid compartment from hemodilution. Volume of intravascular fluid is maximum at same time (T[max]) that concentration of tumescent lidocaine
achieves its maximum, approximately 12 hours after beginning infiltration. Rate and magnitude of intravascular fluid augmentation and change of hematocrit are independent of whether liposuction is
performed. Thus a decline in hematocrit in immediate postoperative period does not correlate with amount of surgical blood loss caused by liposuction. C, Patient’s weight after infiltration and
liposuction. At 48 hours after tumescent infiltration, patient’s weight returns to preinfiltration value. Without liposuction, weight is increased by 5 kg, which corresponds to 5 L of saline infused.
Patient drank fluids without restriction. After gaining approximately 1 kg/L of tumescent infiltration, all tumescent fluid remaining in patient after liposuction is eliminated via kidneys. The
48-hour interval is typical for a return to preoperative weight, with or without liposuction. D, Cumulative urine volume after tumescent liposuction. Patient’s urine output was consistently greater
than 70 ml/hr, with and without liposuction. This is a common threshold, below which physician becomes increasingly concerned about possible intravascular fluid deficiency, inadequate renal
perfusion, renal insufficiency, or cardiac impairment. Patients are fully alert during tumescent liposuction and can drink clear liquids during and after surgery. No need exists for IV infusion of
fluids because patients can drink fluids and because renal perfusion is adequate, as evidenced by continuously good urine output. E, Hourly urine specific gravity after tumescent liposuction.
Typically, patient’s urine specific gravity decreases after tumescent infiltration. This decreased urine specific gravity shows patient is well hydrated. Because tumescent infiltration produces
relative systemic fluid overload, routine administration of IV fluids is potentially dangerous.
Figure 19-3 Rate of lidocaine absorption affects risk of toxicity. Consider hypothetic patient who, on two separate occasions, is given identical doses of lidocaine that differ only in concentration
and site of injection. On each occasion, sequential lidocaine plasma concentrations are determined over ensuing 48 hours. Because two doses are of equal magnitude with 100% systemic absorption, their
areas under curve (AUCs) of plasma lidocaine concentration are equal. Suppose dose A is epidural injection of concentrated lidocaine (2% = 20 mg/ml) and epinephrine that is completely absorbed into
systemic circulation and completely eliminated in approximately 12 hours. Suppose dose B is injection into subcutaneous fat of very dilute lidocaine (0.1% = 1mg/ml) and epinephrine that is completely
absorbed into systemic circulation and completely eliminated in approximately 36 hours. Because AUCs are equal, graph of dose A, with the shortest base, must have the highest peak. Similarly, graph
of dose B, with the longest base, must have the lowest peak. Because magnitude of peak plasma lidocaine concentration is correlated with risk of toxicity, a more rapid rate of absorption carries a
greater risk of toxicity.
Figure 19-4 Graph of dose A, with small AUC, represents relatively small dose of lidocaine that is rapidly absorbed. For example, if small epidural dose of lidocaine rapidly absorbed into systemic
circulation produces toxicity, one can expect toxicity to be of short duration. Graph of dose B, with large AUC, represents relatively large dose of lidocaine that is slowly absorbed. Consider the
example of a large dose of subcutaneous tumescent lidocaine that is slowly absorbed. If plasma lidocaine concentration exceeds toxic threshold, toxicity will be of relatively long duration.
Figure 19-5 Rate of lidocaine infiltration can affect rate of absorption. Rapid infiltration of 1% and 0.5% lidocaine increases risk of toxicity. When combined 1% and 0.5% lidocaine, each with
epinephrine concentration of 1:100,000, is infiltrated into subcutaneous fat within a short period (less than 5 minutes), systemic absorption can be rapid. Rapid infiltration of undiluted commercial
concentrations of lidocaine with epinephrine can produce a peak plasma lidocaine concentration in excess of the 6 μg/kg threshold for potential lidocaine toxicity. Black straight line (●) represents
serum lidocaine concentrations sampled at 15 and 30 minutes after total of 1350 mg of lidocaine had been infiltrated into targeted fat just before liposuction. Other straight line (■) represents
similar result in another patient who received rapid injection of 800 mg of lidocaine. Gently curved graph (▲) represents much slower systemic absorption of tumescent lidocaine. After incremental
infiltration over 45 minutes of 1000 mg of lidocaine (1%) with epinephrine, peak plasma lidocaine concentration was only 1.4 μg/ml. (Data for ■ from Piveral K: Plast Reconstr Surg 80:643, 1987
Figure 19-6 Concentration of lidocaine can affect rate of absorption. Dilution slows rate of lidocaine absorption. When 1 g of dilute lidocaine (1 g/L) and epinephrine (1 mg/L) is slowly injected
into subcutaneous fat over 45 minutes, T[max] = 14 hours and C[max] = 1.2 μg/ml. Two weeks later, when 1 g of undiluted commercial concentration of lidocaine (1 g/100 ml) and epinephrine (1 mg/100
ml) was slowly injected into subcutaneous fat, rate of absorption was more rapid, with T[max] = 9 hours and C[max] = 1.5 μg/ml.
Figure 19-7 Tumescent solution without epinephrine. During infiltration for tumescent liposuction, there was no clinical evidence of blanching vasoconstriction, and blood-tinged anesthetic oozed from
needle puncture sites. Patient had received 17.8 mg/kg of lidocaine when it was determined that epinephrine had been omitted from anesthetic solution. Sequential plasma samples were obtained over 18
hours after initiation of infiltration. T[max] occurred in less than 3 hours, much sooner than usual 8 to 14 hours. Also, C[max] of 2.1 μg/ml was unusually high.
Figure 19-8 Pharmacokinetics of IV bolus of lidocaine. Graph represents plasma lidocaine concentration as it changes over time beginning immediately after IV bolus dose. Initial rapid decrease in
lidocaine concentration primarily results from rapid removal of lidocaine from plasma as it is distributed (α-phase) into other tissues. After lidocaine has achieved equilibrium concentration among
all body tissues, rate of declining plasma concentration principally results from more gradual process of elimination (β-phase) by hepatic metabolism (see text).
Figure 19-9 Total lidocaine dose determines its plasma concentration–time profile. Relationship between total dose of tumescent lidocaine and maximum plasma lidocaine concentration (C[max]) is not
linear. With increasing dosages (mg/kg) of tumescent lidocaine, shape of the graph of plasma lidocaine concentration as a function of time, C[lido](t), tends to change. At lower dosages C[max]
appears as distinct peak; at high dosages C[max] tends to become a plateau that may persist for several hours. ○, 15 mg/kg; □, 35 mg/kg; ∆, 60 mg/kg.
Figure 19-10 Maximum lidocaine concentration (C[max]) occurs when rate of absorption equals rate of elimination. When rate of lidocaine absorption exceeds rate of elimination, plasma lidocaine
concentration (○) is an increasing function of time. When rate of lidocaine elimination is greater than rate of absorption, plasma lidocaine concentration decreases with time. When rate of absorption
equals rate of elimination, concentration = C[max].
Figure 19-11 After liposuction using large dosages of tumescent lidocaine, plasma lidocaine concentration typically achieves extended C[max] plateau rather than single peak maximum value.
Figure 19-12 Lidocaine is rapidly metabolized in liver. When 1 L of blood passes through liver, approximately 700 ml of blood is completely cleared of all lidocaine by hepatic cytochrome P450 3A4
metabolism. As blood flows through liver, 70% of blood lidocaine content is eliminated.
┃ TABLE 19-1 Lidocaine Pharmacokinetic Parameters ┃
┃ Parameter │ Measurement ┃
┃ F[e] (fraction of drug excreted unchanged) │ 0.02 ± 0.01 ┃
┃ k[n] (elimination rate constant) │ 0.39/hr ┃
┃ Oral availability (%) │ 35 ± 11 ┃
┃ Urinary excretion (%) │ 2 ± 1 ┃
┃ Bound in plasma (%) │ 70 ± 5 ┃
┃ Clearance in a 70-kg person │ 640 ± 170 ml/min/70 kg ┃
┃ │ @ 700 ml/min/70 kg ┃
┃ │ = 10 ml/min/kg ┃
┃ Volume of distribution (L) │ 1.1 L/kg ┃
┃ 70-kg person │ 77 ± 28 ┃
┃ Redistribution (α) half-life │ 8 min ┃
┃ Elimination (β) half-life │ 1.8 ± 0.4 hr ┃
┃ Toxic concentration │ > 6 μg/ml ┃ | {"url":"http://liposuction101.com/liposuction-textbook/chapter-19-pharmacokinetics-of-tumescent-lidocaine/","timestamp":"2024-11-11T20:32:40Z","content_type":"text/html","content_length":"187764","record_id":"<urn:uuid:feea62ea-c9d1-440a-a8fe-312e296aee3f>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00890.warc.gz"} |
Reduced-Rank Regression with Operator Norm Error
A common data analysis task is the reduced-rank regression problem: min_rank-k XAX-B, where A ∈ℝ^n × c and B ∈ℝ^n × d are given large matrices and · is some norm. Here the unknown matrix X ∈ℝ^c × d
is constrained to be of rank k as it results in a significant parameter reduction of the solution when c and d are large. In the case of Frobenius norm error, there is a standard closed form solution
to this problem and a fast algorithm to find a (1+ε)-approximate solution. However, for the important case of operator norm error, no closed form solution is known and the fastest known algorithms
take singular value decomposition time. We give the first randomized algorithms for this problem running in time (nnz(A) + nnz(B) + c^2) · k/ε^1.5 + (n+d)k^2/ϵ + c^ω, up to a polylogarithmic factor
involving condition numbers, matrix dimensions, and dependence on 1/ε. Here nnz(M) denotes the number of non-zero entries of a matrix M, and ω is the exponent of matrix multiplication. As both (1)
spectral low rank approximation (A = B) and (2) linear system solving (m = n and d = 1) are special cases, our time cannot be improved by more than a 1/ε factor (up to polylogarithmic factors)
without a major breakthrough in linear algebra. Interestingly, known techniques for low rank approximation, such as alternating minimization or sketch-and-solve, provably fail for this problem.
Instead, our algorithm uses an existential characterization of a solution, together with Krylov methods, low degree polynomial approximation, and sketching-based preconditioning. | {"url":"http://cdnjs.deepai.org/publication/reduced-rank-regression-with-operator-norm-error","timestamp":"2024-11-10T15:07:05Z","content_type":"text/html","content_length":"155015","record_id":"<urn:uuid:1097a429-4dae-4fbd-8086-371d3579ac53>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00770.warc.gz"} |
Capacitor bipolar plate capacitance
Capacitance of parallel plate capacitor with dielectric medium
And, when a dielectric slab of dielectric constant K is inserted between the plates, the capacitance, small {color{Blue} C=frac{Kepsilon _{0}A}{d}}.. So, the capacitance of a parallel plate capacitor
increases due to inserting a dielectric slab or dielectric medium between the plates of the capacitor. The new value of the …
4.6: Capacitors and Capacitance
4.6: Capacitors and Capacitance
Understanding Capacitance and Capacitor Dimensions
Understanding Capacitance and Capacitor Dimensions
Aluminum electrolytic capacitor
Aluminum electrolytic capacitor
Can two electrolytic capacitors be made into a bipolar?
Using two electrolytic capacitors of identical value back to back is routinely used to get a non-polarized capacitor. From this document : If two, same-value, aluminum electrolytic capacitors are
connected in series, back-to-back with the positive terminals or the negative terminals connected, the resulting single capacitor is a non-polar capacitor with half the …
Can two electrolytic capacitors be made into a bipolar?
If two, same-value, aluminum electrolytic capacitors are connected in series, back-to-back with the positive terminals or the negative terminals connected, the resulting single capacitor is a
non-polar capacitor with …
19.5: Capacitors and Dielectrics
The capacitance of a parallel plate capacitor is (C=varepsilon _{0} dfrac{A}{d}), when the plates are separated by air or free space. (varepsilon _{0}) is …
Understanding Capacitor Types and Characteristics | DigiKey
Capacitance is proportional to the plate area, A, and inversely proportional to the distance between the plates, d. Figure 1: The basic capacitor consists of two conducting plates separated by a
non-conducting dielectric which stores energy as polarized regions in the electric field between the two plates. (Image source: DigiKey)
Parallel Plate Capacitor | AQA A Level Physics Revision Notes 2017
The opposing electric field reduces the overall electric field, which decreases the potential difference between the plates. Therefore, the capacitance of the plates increases; The capacitance of a
capacitor can also be written in terms of the relative permittivity: Where: C = capacitance (F) A = cross-sectional area of the plates (m 2)
Chapter 5 Capacitance and Dielectrics
Chapter 5 Capacitance and Dielectrics
5.2: Plane Parallel Capacitor
We have a capacitor whose plates are each of area (A), separation (d), and the medium between the plates has permittivity (epsilon). It is connected to a battery of EMF (V), so the potential
difference across the plates is (V). The electric field between the
A parallel plate capacitor has plates of area | Chegg
(The permittivity of free space is ε0=8.85×10-12C2N*m2. )(a) Apply the expression for the capacitance of a parallel-plate capacitor.(b) and (c) Treat the capacitor as either a parallel or a series
combination of two capacitors.Click the hint button again to
18.5 Capacitors and Dielectrics
Teacher Support Explain that electrical capacitors are vital parts of all electrical circuits. In fact, all electrical devices have a capacitance even if a capacitor is not explicitly put into the
device. [BL] Have students define how the word capacity is used in …
Chapter 5 Capacitance and Dielectrics
Example 5.1: Parallel-Plate Capacitor Consider two metallic plates of equal area A separated by a distance d, as shown in Figure 5.2.1 below. The top plate carries a charge +Q while the bottom plate
carries a charge –Q. The charging of the plates can be accomplished by means of a battery which produces a potential difference.
ECE 304: Bipolar Capacitances
ECE 304: Bipolar Capacitances The Bipolar Transistor: S&S pp. 321-328 Let''s apply this diode information to the bipolar transistor. There are two junctions in the bipolar transistor. The BC
(base-collector) junction is reverse biased in the active mode, and so it jC
Chapter 24 – Capacitance and Dielectrics
1. Capacitors and Capacitance. Capacitor: device that stores electric potential energy and electric charge. Two conductors separated by an insulator form a capacitor. The net …
The Parallel Plate Capacitor
Parallel Plate Capacitor Formula The direction of the electric field is defined as the direction in which the positive test charge would flow. Capacitance is the limitation of the body to store the
electric charge. …
Capacitance and Charge on a Capacitors Plates
Capacitance and Charge on a Capacitors Plates
Capacitors | Brilliant Math & Science Wiki
4 · Effectively, this creates one larger parallel-plate capacitor with larger plate area. Since the capacitance of a parallel-plate capacitor is [C = frac{Aepsilon_0}{d},] and effectively the new
capacitor has area (A_1 + A_2), the new capacitance is (C_1 + C_2), consistent with the sum rule for parallel. In series, the derivation is ...
What is a bipolar capacitor and when is it used?
A bipolar capacitor is just a non-polarized capacitor. I think the term is usually in reference to a type of electrolytic capacitor to make it clear that you can use it …
Capacitor and Capacitance
Capacitance of a Plate Capacitor. Self Capacitance of a Coil (Medhurst Formula). Self Capacitance of a Sphere Toroid Inductor Formula. Formulas for Capacitor and Capacitance t is the time in seconds.
Capacitor Voltage During Charge / …
8.4: Energy Stored in a Capacitor
The energy (U_C) stored in a capacitor is electrostatic potential energy and is thus related to the charge Q and voltage V between the capacitor plates. A charged capacitor stores energy in the
electrical field between its plates. As the capacitor is being charged, the electrical field builds up. When a charged capacitor is disconnected from ...
Capacitor Symbols: A Guide to Understanding the Different …
Capacitor Symbols: A Guide to Understanding the Different ...
Capacitors | Brilliant Math & Science Wiki
4 · Effectively, this creates one larger parallel-plate capacitor with larger plate area. Since the capacitance of a parallel-plate capacitor is [C = frac{Aepsilon_0}{d},] and effectively the new
capacitor has area …
Capacitance | Definition, Formula, Unit, & Facts | Britannica
Capacitance | Definition, Formula, Unit, & Facts
Explaining Capacitors and the Different Types | DigiKey
Voltage coefficient of capacitance. Ceramic capacitors exhibit changes in capacitance with variations in DC bias level. Stated differently, measuring the capacitance of a device with a 1 V P-P size
wave averaging 0 V will yield a different (typically greater) value than if the same device is tested with a 1 V sine wave having a DC offset of 10 V.
8.1 Capacitors and Capacitance
The capacitance C of a capacitor is defined as the ratio of the maximum charge Q that can be stored in a capacitor to the applied voltage V across its plates. In other words, …
Capacitance - Wikipedia ... Capacitance
Parallel Plate Capacitor: Definition, Formula, and Applications
Parallel Plate Capacitor: Definition, Formula, and ...
Explaining Capacitors and the Different Types | DigiKey
Learn about the different types of capacitors and why you would use different compositions. More Products From Fully Authorized Partners Average Time to Ship 1-3 Days.Please see product page, cart,
and checkout for actual ship speed. Extra Ship Charges May Apply
Understanding Capacitor Types and Characteristics | DigiKey
All capacitors consist of the same basic structure, two conducting plates separated by an insulator, called the dielectric, that can be polarized with the application … | {"url":"https://iniron.pl/23_03_24_19470.html","timestamp":"2024-11-07T02:31:21Z","content_type":"text/html","content_length":"24565","record_id":"<urn:uuid:51856e0f-dae7-4dd9-8ae9-ef6229b5007c>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00321.warc.gz"} |
e unit c
What is the unit circle in radians?
What is the unit circle in radians?
2π radians
The four associated angles (in radians, not degrees) all have a denominator of 2. (A radian is the angle made when taking the radius and wrapping it round a circle. A degree measures angles by
distance traveled. A circle is 360 degrees or 2π radians).
What is the value of tan90?
The exact value of tan 90 is infinity or undefined.
What is radian formula?
Formula of Radian Firstly, One radian = 180/PI degrees and one degree = PI/180 radians. Therefore, for converting a specific number of degrees in radians, multiply the number of degrees by PI/180
(for example, 90 degrees = 90 x PI/180 radians = PI/2).
What is the value of tan89?
Tan 89 degrees is the value of tangent trigonometric function for an angle equal to 89 degrees. The value of tan 89° is 57.29 (approx).
Why is tan90 undefined?
tan90∘ is undefined because you can’t divide 1 by nothing. Nothing multiplied by 0 will give an answer of 1 , so the answer is undefined.
What is the radian formula?
Firstly, One radian = 180/PI degrees and one degree = PI/180 radians. Therefore, for converting a specific number of degrees in radians, multiply the number of degrees by PI/180 (for example, 90
degrees = 90 x PI/180 radians = PI/2).
What is the radian measure of 120?
32π radian
=32π radian.
How is radian written?
Radian is denoted by “rad” or using the symbol “c” in the. If an angle is written without any units, then it means that it is in radians. Some examples of angles in radians are, 2 rad, π/2, π/3, 6c,
What is the radius of the unit circle?
But, the Unit Circle is more than just a circle with a radius of 1; it is home to some very special triangles. Remember, those special right triangles we learned back in Geometry: 30-60-90 triangle
and the 45-45-90 triangle? Don’t worry. I’ll remind you of them.
What are circles with radii of one unit called?
Circles with radii of one unit are called unit circles. A unit circle is generally represented in the cartesian coordinate system. The unit circle is algebraically represented by the second-degree
equation with two variables x and y.
What is a unit circle used for?
A unit circle has all of the properties of a circle, and its equation is also derived from the equation of a circle. Furthermore, a unit circle is useful for determining the standard angle values of
all trigonometric ratios.
What is the radius value of a circle with sin 1?
Another major point to be understood is that the sinθ and cosθ values always lie between 1 and -1, and the radius value is 1, and it has a value of -1 on the negative x-axis. The entire circle
represents a complete angle of 360º and the four quadrant lines of the circle make angles of 90º, 180º, 270º, 360º (0º). | {"url":"https://pfeiffertheface.com/what-is-the-unit-circle-in-radians/","timestamp":"2024-11-05T07:36:00Z","content_type":"text/html","content_length":"43182","record_id":"<urn:uuid:956b614d-2017-4ae6-bc15-cbeb4fa17370>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00350.warc.gz"} |
Science Assessment
Item CR017002: A controlled experiment involving the behavior of fish in fish bowls, in which the number of fish in the bowls varies and the temperature of the water and the amount of light is held
constant, allows you find out the effect of the number of fish on fish behavior.
A student is interested in the behavior of fish. He has 4 fish bowls and 20 goldfish. He puts 8 fish in the first bowl, 6 fish in the second bowl, 4 fish in the third bowl and 2 fish in the fourth
bowl.He places each fish bowl under light, he keeps the temperature at 75°F for all four bowls, and he observes the behavior of the fish.
What can the student find out from doing just this experiment?
A. If the number of fish in the fish bowl affects the behavior of the fish.
B. If the temperature of the fish bowl affects the behavior of the fish.
C. If the temperature of the fish bowl and the amount of light affect the behavior of the fish.
D. If the number of fish, the temperature, and the amount of light affect the behavior of the fish.
Distribution of Responses
Students Responding Correctly
Group Correct Total Percent
Overall 2355 5054 47%
6–8 1042 2793 37%
9–12 1300 2238 58%
Male 1142 2484 46%
Female 1183 2509 47%
Primary Language
English 2106 4346 48%
Other 195 604 32% | {"url":"http://assess.bscs.org/science/items/1/CV/22-297/CV017002","timestamp":"2024-11-12T00:53:29Z","content_type":"application/xhtml+xml","content_length":"15407","record_id":"<urn:uuid:19623d00-6b43-4c1a-a640-ddc680d4ce63>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00750.warc.gz"} |
It's a Joke Son
Mar 18, 2013
Mar 18, 2013
My teacher asked what my favorite animal was. I said, "Fried chicken." She said I wasn't funny, but she couldn't have been right, because everyone else laughed. My parents told me to always
tell the truth. I did. Fried chicken is my favorite animal. I told my dad what happened and he said my teacher was probably a member of PETA. He said they love animals very much. I do, too.
Especially chicken, pork and beef. Anyway, my teacher sent me to the principal's office. I told him what happened, and he laughed, too. Then he told me not to do it again. The next day in
class my teacher asked me what my favorite live animal was. I told her it was chicken. She asked me why, so I told her it was because you could make them into fried chicken. She sent me back
to the principal's office. He laughed, and told me not to do it again. I don't understand. My parents taught me to be honest, but my teacher doesn't like it when I am. Today, my teacher asked
me to tell her what famous military person I admired most. I told her "Colonel Sanders."Guess where the hell I am now . . .
Grendel_Lives on Gab: 'Dog raids liquor cabinet...'
^^^^That's funny right there now"^^^^
Jun 21, 2003
^^^^ There probably is no more than a handful of us here that know what this is about^^^^
The Carol Burnett show was hilarious.
Feb 9, 2017
Your browser is not able to display this video.
^^^Thank God I ain't married to that!^^^ | {"url":"https://racing-forums.com/threads/its-a-joke-son.72070/page-6","timestamp":"2024-11-07T22:01:04Z","content_type":"text/html","content_length":"275333","record_id":"<urn:uuid:1fb58cd7-92f4-4386-b3a4-33d274138ecb>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00265.warc.gz"} |
In this recipe, we will dive into R's capability with regard to matrices.
A vector in R is defined using the c() notation as follows:
A vector is a one-dimensional array. A matrix is a multidimensional array. We can define a matrix in R using the matrix() function. Alternatively, we can also coerce a set of values to be a matrix
using the as.matrix() function:
mat = matrix(c(1,2,3,4,5,6,7,8,9,10),nrow = 2, ncol = 5)
To generate a transpose of a matrix, we can use the t() function:
t(mat) # transpose a matrix
In R, we can also generate an identity matrix using the diag() function:
d = diag(3) # generate an identity matrix
We can nest the rep () function within matrix() to generate a matrix with all zeroes as follows:
zro = matrix(rep(0,6),ncol = 2,nrow = 3 )# generate a matrix of Zeros
We can define our data in the matrix () function by specifying our data as its first argument. The nrow and ncol arguments are used to specify the number of rows and column in a matrix. The matrix
function in R comes with other useful arguments and can be studied by typing ?matrix in the R command window.
The rep() function nested in the matrix() function is used to repeat a particular value or character string a certain number of times.
The diag() function can be used to generate an identity matrix as well as extract the diagonal elements of a matrix. More uses of the diag() function can be explored by typing ?diag in the R console
The code file provides a lot more functions that can used along with matrices—for example, functions related to finding a determinant or inverse of a matrix and matrix multiplication. | {"url":"https://subscription.packtpub.com/book/data/9781783989508/1/ch01lvl1sec05/matrices-in-r","timestamp":"2024-11-04T17:47:40Z","content_type":"text/html","content_length":"133481","record_id":"<urn:uuid:2ec09474-9593-4c18-a874-aa26e7e4a57b>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00216.warc.gz"} |
Neighbor Discovery Optimization for Big Data Analysis in Low-Power, Low-Cost Communication Networks
Department of Computer Science & Engineering, Gangneung-Wonju National University, Wonju 26403, Korea
Department of Multimedia Engineering, Dongguk University, Seoul 04620, Korea
Author to whom correspondence should be addressed.
Submission received: 28 April 2019 / Revised: 20 June 2019 / Accepted: 24 June 2019 / Published: 26 June 2019
Big data analysis generally consists of the gathering and processing of raw data and producing meaningful information from this data. These days, large collections of sensors, smart phones, and
electronic devices are all connected in the network. One of the primary features of these devices is low-power consumption and low cost. Power consumption is one of the important research concerns in
low-power, low-cost communication networks such as sensor networks. A primary feature of sensor networks is a distributed and autonomous system. Therefore, all network devices in this type of network
maintain the network connectivity by themselves using limited energy resources. When they are deployed in the area of interest, the first step for neighbor discovery involves the identification of
neighboring nodes for connection and communication. Most wireless sensors utilize a power-saving mechanism by powering on the system if it is off, and vice versa. The neighbor discovery process
becomes a power-consuming task if two neighboring nodes do not know when their partner wakes up and sleeps. In this paper, we consider the optimization of the neighbor discovery to reduce the power
consumption in wireless sensor networks and propose an energy-efficient neighbor discovery scheme by adapting symmetric block designs, combining block designs, and utilizing the concept of activating
nodes based on the multiples of a specific number. The performance evaluation demonstrates that the proposed neighbor discovery algorithm outperforms other competitive approaches by analyzing the
wasted awakening slots numerically.
1. Introduction
Connectivity is expected to be one of the most important factors to be considered in the near future. Many electronic devices, such as computers, smart phones, and tablet PCs are connected to each
other through the internet nowadays. Even different types of home appliances are connected through internet services. Consequently, significant amounts of information are exchanged every day in the
world. Gartner, Inc. estimated that the number of connected devices would reach 20.4 billion by 2020 in a report. It is not easy to imagine the amount of information that would be produced,
processed, and exchanged among these devices.
Big data analysis usually requires significant computing and electric power owing to the large amounts of data that have to be processed. Especially, energy consumption is one of the most critical
issues for the purpose of information gathering for big data analysis among resource constrained devices, such as wireless or mobile sensors. These machines are both the primary components of
low-power and low-cost communication networks and a raw data source used for big data analysis. Hence, it is possible to maintain network connectivity by saving the energy utilized by machines. It is
known that communication cost is significantly more expensive than other costs, such as those for collecting or processing information.
As the internet of things (IoT) technology has emerged and has been extensively applied to a variety of areas, several mobile network studies are focusing on low-power, low-cost communication
networks [
]. A typical network of low-power communication is a wireless sensor network (WSN). There is no permanent power source in this type of network. Therefore, energy-efficient communication is much more
important than other research subjects. Furthermore, the movement of networked devices may frequently occur in mobile networks. Finally, the nodes in these network environments have certain
constraints of computing resources and depend on limited electrical power.
It is widely known that IoT communication networks are in general composed of heterogeneous physical devices. Power consumption might not be a significant concern in certain IoT networks; however,
low-power consumption and the utilization of wireless devices are considered as an energy-efficient solution. Several IoT communications are based on machine-to-machine (M2M) communication, which is
similar to the basic communication method of a wireless sensor network. It is assumed that most sensors do not change their location after they are deployed in the area of interest. However, this
assumption might not be applicable to low-power IoT communication networks. For example, sensors that are used in autonomous or unmanned vehicles constantly move during their lifetime. These devices
might join and leave the network at any time. Therefore, continuous neighbor discovery should be performed for maintaining network connectivity.
The neighbor discovery process is required to sustain the network connectivity in low-power, low-cost communication networks [
]. Furthermore, this process consumes the battery power of wireless devices. A neighbor discovery optimization problem considers and focuses on how to minimize power consumption as much as possible.
In this paper, we propose an energy-efficient neighbor discovery algorithm for symmetric and asymmetric operations. We formulated the neighbor discovery optimization problem with a mathematical term,
borrowed the concept of block design in combinatorics for effective neighbor discovery, and analyzed the performance of certain representative neighbor discovery mechanisms numerically.
This paper is configured as follows:
Section 2
provides literature reviews of asynchronous neighbor discovery.
Section 3
specifies a neighbor discovery optimization process.
Section 4
introduces a neighbor discovery problem (NDP), few neighbor discovery schedules, the main features of block designs, and a practical challenge of existing block designs.
Section 5
elaborates the primary idea of merging two block designs and focuses on our key contributions. In
Section 6
, we consider an asymmetric NDP.
Section 7
evaluates and measures the performance of the proposed algorithm and compares the proposed scheme with that of other representative neighbor discovery protocols. Finally, we conclude the paper in
Section 8
2. Related Work
Several researchers have conducted studies pertaining to the NDP since this problem was first introduced. Consequently, there are a number of representative outcomes of NDP, referred to as neighbor
discovery protocol, in the sensor network research fields. In IoT communication networks, a network topology is constantly updated owing to nodes joining and leaving the network frequently.
Consequently, the neighbor discovery process should happen continually to ensure network connectivity. In this section, we present several related works of major neighbor discovery protocols.
The primary idea behind the
protocol [
] follows the birthday paradox in which the birthday of two individuals out of a set of
randomly selected people is the same. This paradox is applied to develop a neighbor discovery protocol in ad hoc networks. Nodes can wake up and sleep with their own probabilities in the network.
They merely pursue their probabilities and meet each other at a certain time owing to the paradox. This mechanism is likely to sound advantageous for neighbor discovery. However, it is known that a
randomness approach, such as the
protocol, performs well in average cases, but it cannot guarantee neighbor discovery at a worst case. Neighbor discovery should be assured in both average and worst cases with minimum energy
Each wireless node can randomly choose one row and one column of a two-dimensional array
in a
-based neighbor discovery protocol [
] for ad hoc networks. Every node activates all the slots from the selected rows and columns and the remaining slots are inactive for neighbor discovery. Consequently,
$2 n − 1 n 2$
slots remain awake and attempt to locate their neighbors at the awakening time. The
-based neighbor discovery protocol can guarantee that two neighbor nodes can identify each other quickly if the two nodes choose the same row or column. Even if this aspect demonstrates the advantage
of the
-based protocol, the performance of the protocol depends on which row and column are chosen. Furthermore, we cannot ensure that two different nodes will always select the same row or column.
There are two similar approaches to neighbor discovery that adopts prime numbers:
] and
]. In
, every node chooses two different prime numbers and activates the nodes that are multiples of these numbers to identify if there are neighboring nodes in the communication boundary. The main concept
ensures that there is at least one common time slot between two neighboring nodes at the time of the multiplication of the chosen prime numbers. When compared to previous protocols for neighbor
has its own distinct feature where two nodes are able to locate each other at different duty cycles. The study emphasizes that
can support an asymmetric duty cycle. The performance of
depends on choosing balanced primes (the difference between two prime numbers of each node is minimal) at a given duty cycle. Choosing two well-balanced prime numbers may not be easy because there is
no proper algorithm or formula for identifying balanced prime numbers.
The other neighbor discovery protocol using a prime number is
]. Only one prime number is used in
instead of using two prime numbers. Unfortunately, there may be no overlap between two neighboring nodes with one prime number.
activates certain number of slots to overcome the problem of non-rendezvous slots. For instance,
$p + 1 2$
consecutive slots wake up in the beginning of the time slots per
slots when a prime number
is selected for neighbor discovery. One of valuable contributions of
is the performance metric of neighbor discovery protocols called the power-latency (PL) product.
offers an alternative approach to
by using one prime number; however, it still demonstrates a weakness where two nodes are not likely to meet each other quickly in case of a mismatch during the early time slots of a neighbor
discovery schedule. This shortcoming may result in a significant increase in the worst-case discovery latency.
The concept of a block design in combinatorial mathematics was used for inventing a neighbor discovery protocol by Zheng et al. [
]. The neighbor discovery schedule can be easily generated using a certain duty cycle from an existing and well-known block design. The combinatorial approach demonstrates the fastest discovery
latency and lowest energy consumption when compared to other existing neighbor discovery protocols based on the PL product. It is to be noted that the combinatorial scheme performs well in a
situation where all nodes follow the same duty cycle (a symmetric case). However, it may be impossible to apply this approach to an asymmetric case of neighbor discovery. In addition, it is
impossible to create a discovery schedule using the concept of the block design if there is no proper set of block designs at a given duty cycle.
The simplest solution of neighbor discovery is to maintain all slots as active until two neighboring nodes identify each other. However, this approach is not applicable to low-power and
resource-constrained network environments. Activating half of the total slots in each duty cycle could be a better solution than the previously mentioned extreme one. However, this solution is also
not sufficiently good for developing an energy-efficient neighbor discovery protocol.
] offers a clue for energy-efficient neighbor discovery by probing half of the total time slots to determine at least one common active slot. There are two active slots within
contiguous slots: an anchor slot and a probe slot. The anchor slot is the first slot in the discovery period and the probe slot changes its location to search for the anchor slot of the other node.
Intuitively, the probing method can reduce the number of active slots marginally; however, the discovery latency takes longer as the total slots in the neighbor discovery period is increased.
Two different types of slots, called static and dynamic slots, are used in
]. The static slot is approximately the same as the anchor slot in
. The concept of dynamic slot is similar to that of the probe slot of
; however, the moving directions constantly change from left to right or right to left to locate neighboring nodes quickly. This mechanism can decrease the worst-case discovery latency when compared
. However, the performance of
is only marginally different from that of
3. Problem Statement
There are generally two power-operating modes called
modes in most energy-saving mechanisms in wireless networks [
]. In the
mode, a wireless node turns off its radio interface and expends only a small amount of power, thereby saving energy when it turns on its radio for communication in the
One of the simplest ways to reduce the power consumption of wireless nodes in the network minimizes the number of awakening modes. If the nodes sleep for most of their network lifetime, then we could
achieve the optimization of power consumption in wireless sensor networks effectively. Unfortunately, this power management mechanism is ineffective because the entire network remains disconnected.
Hence, we are required to accomplish both network connectivity and power management.
As we mentioned before, the power-saving policy consists of the
modes. Each node might follow its own power management schedule based on its own policy. We can formulate the power management schedule to represent two power-saving modes using a binary number. A
binary number “zero” can be used to represent the
mode and the number “one” can express the
mode. All the sensor devices that follow this power-saving policy alternate between the
modes continuously. Therefore, the power management schedule,
, can be illustrated as zero and one. Furthermore, each wireless node has its own duty cycle of a certain length of schedule,
. The power management schedule of node
$S u$
, can be represented by a polynomial of order
$S u ( x ) = ∑ i = 0 ℒ − 1 a i x i ,$
is the length of the schedule,
$a i$
= 0 or 1 (
$0 ≤ i ≤ ℒ − 1 )$
, and
is a place holder.
By definition, two nodes, u and v, wake up in slot i if a[i] = 1. In this situation, they can communicate with each other if they are within the same communication range; thus, it can be noted that
nodes u and v identify their neighbors. Conversely, it may be difficult for them to talk to each other when u is awake and v sleeps at slot i, or vice versa. In the latter scenario, an awakening node
cannot avoid wasting its energy and the neighboring nodes u and v are unable to identify each other. For neighbor discovery optimization to achieve low-power, lost-cost communication networks, an
ultimate goal is to minimize the number of wasted awakening slots as much as possible during the neighbor discovery process.
By the above definition of
$S u ( 1 )$
is the total number of slots where node
should be awake for neighbor discovery in every
slot when
= 1. If we apply a bitwise XOR operation (^) to
$S u ( x )$
$S v ( x )$
, then it would be possible to determine the number of wasted awakening slots that exist between
$S u ( x )$
$S v ( x )$
. In addition, we can calculate the number of overlapping awakening slots that occur between
$S u ( x )$
$S v ( x )$
if we assign a bitwise AND operation (&) to
$S u ( x )$
$S v ( x )$
. Let
$S u ^ v ( 1 )$
be the total number of wasted awakening slots and
$S u & v ( 1 )$
be the total number of overlapping awakening slots between
$S u ( x )$
$S v ( x )$
. Therefore, we define a neighbor discovery optimization problem as follows:
$Minimize : S u ^ v ( 1 ) Subject to : S u & v ( 1 ) ≥ 1$
4. Block Design for Neighbor Discovery
It is crucial to minimize the total number of wasted awakening slots in the neighbor discovery optimization problem. However, fundamentally, two nodes, u and v should locate each other within $ℒ$
slots. Hence, in this section, we introduce a block design for neighbor discovery. The NDP can be applied to the block design in combinatorics theory. Initially, we define certain important
terminologies in neighbor discovery.
Definition 1.
A discovery schedule (DS) is a repetition of zeroes or ones that represents the sleeping and awakening modes, respectively. When a wireless node is in the awakening mode, it should be scheduled to
turn its radio on for communication and prepare to transmit or receive packets. If the node stays in the sleeping mode, then it should turn its radio off and obtain certain environmental information.
In a DS, the binary number zero represents the sleeping mode and one denotes the awakening mode.
Definition 2.
A DS consists of a certain number of place holders containing the binary numbers, zero or one. We call this place holder a “slot”. Therefore, each DS has its own number of slots representing sleeping
or awakening modes.
Definition 3.
A duty cycle (D) is the ratio of the number of active slots over the total number of slots for a given DS. Therefore, D can be expressed as:where A is the number of awakening slots and T is the total
number of slots in a DS. Hence, D illustrates the number of slots that should be awake in a DS.
It is possible to express DS graphically as illustrated in
Figure 1
. Based on Definition 1, there are only two modes in the DS. Therefore, it can be said that the DS is a sequence of zeroes and ones representing awakening or sleeping slots.
Figure 1
shows a typical example of a DS. In this example, the DS consists of 10 slots and the duty cycle of the DS is
$5 10 × 100 % = 50 %$
We adapted the basic idea of the block design to address the NDP. The block design is defined as follows [
Definition 4.
A design is a pair (X, A) such that the following criteria are satisfied:
X is a set of elements (points), and
A is a collection of non-empty subsets of X (blocks).
Definition 5.
Let v, k, and λ be positive integers such that v > k ≥ 2. A (v, k, λ)-balanced incomplete block design ((v, k, λ)-BIBD) satisfies the following properties:
|X| = v,
Each block has exactly k points, and
Every pair of distinct points is included in exactly λ blocks.
Thus, (7, 3, 1)-BIBD is one of the representative BIBDs when λ = 1. In the (7, 3, 1)-BIBD,
X = {1, 2, 3, 4, 5, 6, 7}, and
A = {{1,2,4}, {2,3,5}, {3,4,6},
{4,5,7}, {5,6,1}, {6,7,2}, {7,1,3}}.
In (7,3,1)-BIBD, each block is composed of three points and each pair of distinct points is exactly related to one block. For example, a block {1,2,4} contains elements 1, 2, and 4 and a pair (2,3)
or (3,5) only appears in a block {2,3,5}.
In the theory of block design, if the number of blocks is the same as the number of points, i.e., |X| = |A|, then they call this design a symmetric-BIBD. In Reference [
], based on Theorem 1.2.1, it was discussed that if a block design is a symmetric-BIBD, then the arbitrarily chosen two blocks have λ common points. This theorem involves a basic property where two
random DSs might have overlapping awakening slots when applying the symmetric-BIBD to the NDP. Therefore, we introduce this theorem here.
Theorem 1.
(Theorem 1.2.1 in [25]): If (X, A) is a symmetric-BIBD with parameters (v, k, λ), any two different blocks have exactly λ common points.
It is possible to rewrite all the blocks in (
. For instance, a block {
} can be replaced by (
) with binary numbers, zero and one. The (
can be demonstrated explicitly as shown in
Figure 2
using the notation of a matrix. This binary matrix of blocks completely corresponds with the design of
. Hence, we can allocate one of the blocks to each wireless node at random as its own
This ensures that the symmetric-BIBD guarantees at least one overlapping awakening slot between two DSs for neighbor discovery. Theorem 2 demonstrates that any arbitrary two DSs have λ overlapping
awakening slots when we employ symmetric-BIBD to the design of DSs.
Theorem 2.
If two distinct DSs, S[i] and S[j], are applied to the (X, A) symmetric-BIBD with parameters (v, k, λ), then the two DSs, S[i] and S[j], have λ overlapping awakening slots.
As (
) is a
with parameters (
can be represented as follows:
$A = { B i | B i ⊂ X , | B i | = k }$
Two schedules, S[i] and S[j], can be connected to two distinct blocks in A, respectively. Let two blocks in A be B[i] and B[j]. According to Theorem 1, there exists λ points in B[i] and B[j].
Consequently, it is possible to mention that | B[i] ∩ B[j] | = λ. It shows that S[i] and S[j], which are applied to B[i] and B[j], respectively, have λ overlapped active slots. □
It is already known that if
is a power of a prime, there exists a block design (
k^2 + k + 1
+ 1,
= 1; further, this kind of
is a
]. (
)-BIBD is a case where
= 2. By choosing an appropriate power of the prime
, we would be able to create several
s with different duty cycles. Although this idea seems as a possible method for designing
s, there is a practical challenge in adopting a (
, 1)-
The practical challenge of a block design is that there is no algorithmic and well-known block construction method. In a general WSN application, it is required to make a set of DSs with a low duty
cycle to save considerable energy of wireless sensors. For maintaining a low duty cycle continuously, a relatively big prime number should be selected to generate a set of blocks. However, a big
prime number results in a large number of blocks. Hence, big prime numbers may considerably influence the computational time for generating blocks. It might not be possible to search for a proper
BIBD within a constant time when very low duty cycles are required. If we cannot select a suitable block construction technique for a target low duty cycle, we must consider $( k 2 + k + 1 k + 1 ) ×
( k 2 + k + 1 2 )$ number of trials as the worst-case scenario to search all the blocks of a (k^2 + k + 1, k + 1, 1)-BIBD.
5. Block Construction Mechanism
As we stated in the previous section, a practical challenge results in a difficulty in creating a DS by adopting a symmetric-BIBD directly. A new approach is required for producing neighbor discovery
schedules within a small amount of computational time. We have introduced our proposed DS construction mechanism and its specific features in this section.
A fundamental idea of the proposed scheme is mixing two existing block designs and creating a discovery schedule. The symmetric-BIBD is only used to illustrate and explain how our technique is
Definition 6.
Let v, k, and λ be positive integers such that v > k ≥ 2. A (v, k, λ)-neighbor discovery design ((v, k, λ)-NDD) is a design (X, A) such that the following characteristics are valid:
|X| = v,
Each block has exactly k awakening slots, and
Every pair of different blocks includes at least λ common awakening slots.
Definition 7.
A sleep schedule is a n × n matrix with a value of all zeroes.
We only created (4,3,2)- and (3,2,1)-designs for the purpose of explaining how to combine two block designs. It is to be noted that these two block designs were constructed only for demonstrating the
process of making neighbor discovery schedules.
Step 1: Preparing two block designs.
First, we prepared two well-known block designs. We assumed that there are two block designs: A = (v
, k
, λ
)-BIBD and B = (v
, k
, λ
)-BIBD. (4,3,2)- and (3,2,1)-designs were created to demonstrate the operation of the proposed combining process. The (4,3,2)- and (3,2,1)-designs are shown in
Figure 3
. We defined the (4, 3, 2)-design as base and the (3,2,1)-design as replacement.
Step 2: Replacing each awakening slot in the base by the entire slot of the replacement.
Secondly, each awakening slot in the base was replaced by the total slots of the replacement. In addition, all sleeping slots were changed into a sleep schedule as per Definition 7. Each awakening
slot in the (4,3,2)-design was transformed to entire slots in the (3,2,1)-design.
Step 3: Constructing a new block design.
It is possible to create a (v
× v
, k
× k
, λ
× λ
)-NDD by implementing steps 1 and 2 in the final step. A new block design, (12,6,2)-NDD, is shown in
Figure 4
. It is finally produced by combining the (4,3,2) and (3,2,1)-designs.
It can be guaranteed that we can produce and create a number of diverse DSs with a target duty cycle by repeating the proposed construction method through steps 1 to 3. If the existing well-known
block designs are able to cover the given duty cycle, it could simply use it to construct a neighbor discovery schedule. However, it is not always possible to do that because of the lack of a general
algorithm for generating BIBDs. Consequently, the proposed idea can be applied for solving the NDP. As you have seen, it is relatively simple to create a new block design for neighbor discovery.
Additionally, our technique is reasonable with respect to the computational time.
It is required to prove that the newly created block design has the same features as specified in Definition 5. The process for proving this is really important and critical. If the new block design
does not have the same properties of the original block design, then the new one is inoperable and it cannot be applied to the NDP. The proposed scheme can guarantee that the new DS has the same
properties as the original block designs.
Definition 8.
We assumed that F is defined as a (v[1], k[1], 1)-BIBD and $ℛ$is declared as a (v[2], k[2], 1)-BIBD. F $⊗ ℛ$represents that all the awakening slots in F are transformed to $ℛ$and all the sleeping
slots are changed into a v[2] $×$v[2] size matrix of the sleep schedule.
Theorem 3.
We assume that F is a (v[1], k[1], 1)-BIBD and $ℛ$is a (v[2], k[2], 1)-BIBD. F $⊗$$ℛ$results in a (v[1] × v[2], k[1] × k[2], 1)-NDD.
Both F and
. Hence, it is possible to say that F = {
$F i$
$F i$is a schedule
} and
= {
T[i] is a schedule
}. We assumed that
= F
. We know that
= {
$Ψ i$
| 1
$Ψ i$
} is a (
, 1)-NDD, where
. Thus,
$Ψ j$$∈$$Ψ ,$
i $≠$j
$≤$h $≤$v[1]v[2]
such that
= 1 for any pair of schedules
$Ψ i$
. Every awakening slot in
is transformed to
. In addition, every sleeping slot in
is changed into a
sized matrix of all zeros. Hence, it is possible to express that the total number of points in
and each block in
has exactly
awakening slots. These two chrematistics perfectly correspond to the first and second properties in Definition 10. Consequently, we concluded that every pair of differing two blocks has at least λ
common awakening slots.
can be illustrated with the notation of the matrix as follows:
$Ψ = [ ψ ( 1 , 1 ) ψ ( 1 , 2 ) ⋯ ψ ( 1 , v 1 v 2 ) ψ ( 2 , 1 ) ψ ( 2 , 2 ) ⋯ ψ ( 2 , v 1 v 2 ) ⋮ ⋮ ⋱ ⋮ ψ ( v 1 v 2 , 1 ) ψ ( v 1 v 2 , 2 ) ⋯ ψ ( v 1 v 2 , v 1 v 2 ) ]$
is a
sized matrix and each component of
denotes an awakening or a sleeping slot. Schedules
$Ψ i$
$Ψ j$
represent one of the rows in
is a
. Therefore,
has a schedule
$F i$
. Each
$F i$
has at least one awakening slot,
, where 1 ≤
. According to Definition 8,
is changed into
Case I: For all indexes
i $≠$j
$i , j ∈ [ v 2 × ( α − 1 ) + 1 … v 2 × ( α − 1 ) + v 2 ]$
where 1
$Ψ i$
$Ψ j$
contain at least one common awakening slot such that
= 1 by
Case II: For all values of
$α ≠ β$
and indexes
i $≠$j
$i ∈ [ v 2 × ( α − 1 ) + 1 … v 2 × ( α − 1 ) + v 2 ] and$
$j ∈ [ v 2 × ( β − 1 ) + 1 … v 2 × ( β − 1 ) + v 2 ]$
where 1 ≤
$Ψ i$
$Ψ j$
include at least one common awakening slot such that
= 1 by F. According to Cases I and II, schedules
$Ψ i$
$Ψ j$
always contain at least λ common awakening slots. Finally,
is a (
, 1)-NDD. □
6. Asymmetric Neighbor Discovery Algorithm
We assumed that the DS of each node operated at the same duty cycle. Unfortunately, this hypothesis may not be considered in real network environments because it is not a general situation. This is
because participant nodes could follow different duty cycles in a distributed manner (asymmetric duty cycle). However, both the concept of original block design and the proposed block construction
mechanism are intrinsically unable to support asymmetric duty cycles because the basic concept of these two ideas is centered on symmetric-BIBD.
An asymmetric neighbor discovery algorithm should focus on constructing proper neighbor discovery schedules by supporting both symmetric and asymmetric cases. The traditional theory of block design
works well in a given symmetric duty cycle. However, it cannot be applied to nodes working with asymmetric duty cycles. Furthermore, if we cannot determine an appropriate block design, then it might
not be possible to consider solving NDPs. One of the main shortcomings of the original block designs is that it could not assist certain duty cycles. The reason is that there is no proper block
design for certain specific duty cycles. We can easily solve this problem by adopting our new approach of constructing neighbor DSs with a given target duty cycle. However, both techniques cannot be
applied to solve the asymmetric NDP. That is why we have primarily discussed the algorithm dealing with the asymmetric NDP in this section.
The fundamental idea of solving the asymmetric NDP is to incorporate the original block designs with multiples of k. The asymmetric NDP can be resolved by integrating the concept of multiples of k
with the original block designs. The primary concept of the multiples of k is that if a slot number is the same as a multiple of k then the proposed algorithm activates that slot. Thus, the multiples
try to cordinate the wake-up time of two nodes operating at different duty cycles and provide the two nodes a chance to locate each other at the same time.
There are two different neighbor discovery schedules: one is a (
)-design and the other is obtained from a (
)-design. The duty cycle of these two block designs is not symmetric (asymmetric duty cycle). Hence, two nodes cannot communicate with each other because the two block designs do not have at least
one overlapping awakening slot. As seen from
Figure 5
, there is no common awakening slot between nodes
. In
Figure 5
, node
follows the (
)-design and
utilizes the (
)-design. Therefore, it was possible to apply our proposed asymmetric neighbor discovery algorithm to this problem. Initially, node
may think that there is no neighbor within its communication range in the first round of the duty cycle. There are two options in this situation: First, there is no neighbor in practice. Second,
there are neighbors around node
; however,
and its neighbors cannot meet each other because of different duty cycles. We have an opportunity to enable them to talk to each other with the proposed algorithm. From the second cycle, the proposed
mechanism can begin to work. In
Figure 5
, for the first time,
has three awakening slots:
, and
. Our asymmetric algorithm starts waking up slot numbers 3 and 6 because these number are multiples of
. From this second cycle, the proposed mechanism turns on its process. Even if node
wasted six awakening slots in the first and second duty cycles, nodes
still cannot talk to each other. However, the node following a (
)-design also thinks that there is no neighbor for communication. Node
proceeds with the proposed mechanism.
Figure 6
illustrates that node
wakes up slot numbers
, which implies that
adopts multiples of
. The yellow color in
Figure 5
Figure 6
represents additional awakening slots that occur using the proposed mechanism. Both
finally talk to each other twice in the fourth cycle of
and the second cycle of
, respectively, in
Figure 6
. The proposed asymmetric neighbor discovery algorithm is shown in
Figure 7
7. Numerical Analysis
The study of sensor network protocols [
] shows that the most energy consumption in wireless networked devices occurs during radio communication and idle listening. Idle listening activates a radio interface and waits for the signal from
neighboring nodes to maintain network connectivity. If a node wakes up and remains idle, then it wastes energy without any activity. Therefore, the main goal of neighbor discovery optimization is to
minimize idle listening time while maintaining network connectivity. Both minimum discovery latency time and a small amount of energy are critical requirements to achieve the primary goal. In
Section 7
, we present the numerical analysis conducted by comparing and analyzing the performance of certain representative neighbor discovery protocols and ours. There are two performance metrics we
considered in numerical analysis: discovery latency time and energy consumption. The first metric is significant because it represents the worst-case neighbor discovery latency. The second one is
also crucial because it involves the energy usage of each node.
It is easy to calculate the worst-case discovery latency time once a target duty cycle is provided. There are different duty cycles in an asymmetric case: one is lower than the other. We only
considered a lower duty cycle that primarily influences the performance of neighbor discovery in an asymmetric scenario. Unfortunately, the node with the lower cycle cannot discover its neighbors
until the total number of slots are utilized in a worst-case scenario. For the numerical study, we first defined the main parameters such as duty cycle, discovery latency time, and the number of
active slots for the analysis of discovery latency.
Table 1
lists three main parameters considered among the representative neighbor discovery protocols and our protocol. The next step was to determine an appropriate parameter for each protocol to apply these
parameters to compute the worst-case discovery latency time.
Table 2
lists the parameter settings for calculating the discovery latency of each protocol based on 10%, 5%, 2%, and 1% of the duty cycles.
Table 3
depicts the worst-case discovery latency with respect to the total number of slots. As we can realize from
Table 3
, the proposed approach has the smallest number of slots when compared to the three neighbor discovery protocols (
). It proves that the two nodes adopting the proposed protocol might have a chance to discover their neighbors faster than when utilizing other protocols in an asymmetric situation.
Figure 8
shows that the total number of slots is different in several asymmetric scenarios.
It is guaranteed that the number of active slots is closely related to energy consumption of the wireless sensors. When the node is in the active state, it turns its radio on and is ready to transmit
and receive packets. Transmitting and receiving packets are some of the most expensive activities. That is why active slots represent energy consumption. From this perspective, the numerical analysis
of energy consumption usually computes the number of active slots.
Table 1
lists the number of active slots for each neighbor discovery protocol. In the proposed mechanism, the number of awakening slots might be different at given duty cycles; therefore, a variable
is borrowed in
Table 1
. The variable
can be decided based on the multiples of
is not fixed in our proposed algorithm. If
is a small value, then the number of active slots will be increased. Otherwise, the active slots will be decreased. For example, if the duty cycle is 10% and
value is 9, the variable
will be approximately 10.
The following scenario was considered for the analysis of energy consumption: one node has a higher duty cycle than the other. This node cannot discover its neighbor within the first round of its
duty cycle. For instance, node
uses 10% of the duty cycle (higher duty cycle) and the other node
uses 5% of the duty cycle (lower duty cycle).
cannot talk to
within the first round of its duty cycle. The next step is to compute the total number of active slots among the four neighbor discovery protocols.
Figure 9
shows the total number of active slots for the four different algorithms. In general, 1% of the duty cycle requires a bigger number of active slots than others.
, and
show a similar pattern. For example, the primary concept of
use a prime number to activate the node. They make each slot active according to the multiples of a given prime number. If the given prime number is getting bigger, similar to 1% of the duty cycle,
the offset between one active slot and the other is also increased. Therefore, the asymmetric case of (10%, 1%) has a larger value than the cases of (10%, 5%) and (10%, 2%) in
Figure 9
. However, the proposed algorithm represents a different shape when compared to other protocols because the waking-up pattern is asymmetrical. In general, the total number of active slots in the
proposed technique is the lowest when compared to that of other protocols. The discovery latency in
Figure 8
and the number of awakening slots in
Figure 9
illustrate a reverse trend from (10%, 1%) to (5%, 2%). This situation happens because it is based on the selection of variable
. A simulation or real experiment with mobile sensors will be required to verify that our numerical study is reasonable. Consequently, our method can discover neighbors much faster than others with
minimum energy consumption.
8. Conclusions
We analyzed neighbor discovery optimization for low-power, low-cost communication networks in this paper. As we stated in the introduction section, these networks consist of various heterogeneous
communication devices. Every time the network participants are required to communicate with their neighbors, they should first comprehend that there are existing neighbor nodes in their communication
range. After locating their neighbors, they try to connect with them and finally establish a network connection. The communication mechanism we considered in this paper focused on M2M communication
in a distributed manner. This communication is mostly similar to the communication method of WSNs.
In this paper, we developed an asynchronous and asymmetric neighbor discovery protocol by combining the traditional block designs and the multiples of k. The shortcoming of the block design is that
it might be possible to use it only for certain duty cycles. We recommended a block construction scheme by combining two block designs to support any desired duty cycle. Furthermore, the traditional
block design can assist only symmetric duty cycles and not asymmetric cases. We provided an asymmetric neighbor discovery algorithm by introducing the use of the multiples of k to overcome the
weakness of typical block designs.
We performed a numerical study by analyzing the performance of existing representative neighbor discovery protocols and comparing our proposed algorithm with other protocols. The numerical analysis
revealed that the proposed algorithm utilizes the minimum number of active slots, which implies that our protocol discovers neighbors faster and wastes less energy than other protocols.
Future research directions should attempt a simulation study or a real experiment on distributed network environment settings. This study can verify that the total number of slots is related to the
energy consumption and network lifetime of nodes in the network. In addition, selecting the parameter values using the concept of multiples might affect the performance of the neighbor discovery
protocol that we proposed with a variety of parameter settings.
In addition, one of the significant research directions of neighbor discovery is the IP version 6 (IPv6) NDD [
]. In WSN environments, there is no concept of IP address in each wireless sensor. In the future, IoT devices could be either IP address or ad hoc based. Therefore, we plan to focus on a neighbor
discovery study in IPv6-based mobile devices.
Author Contributions
All authors contributed equally to this paper.
This work was supported by the Dongguk University Research Fund of 2016, BK21 Plus project of the National Research Foundation of Korea Grant, and National Research Foundation of Korea(NRF) funded
(NRF-2016R1D1A1A09919318, NRF-2019R1F1A1064019).
Conflicts of Interest
The authors declare no conflict of interest.
Protocol DC L Active slots
Disco [12] $p 1 + p 2 p 1 · p 2$ $p 1 · p 2$ $p 1 + p 2$
U-Connect [13] $p + 1 p 2$ p^2 $3 p − 1 2$
Searchlight [15] $2 t$ $t 2 2$ t
Combinatorial + Multiples of k $( k + 1 ) + α k 2 + k + 1$ $k 2 + k + 1$ $( k + 1 ) + α$
Protocol DC
10% 5% 2% 1%
Disco p[1] = 13, p[1] = 29, p[1] = 97, p[1] = 191,
p[2] = 31 p[2] = 61 p[2] = 101 p[2] = 211
U-Connect p = 11 p = 23 p = 53 p = 101
Searchlight t = 20 t = 40 t = 100 t = 200
Combinatorial + Multiples of k k = 9 k = 19 k = 49 k = 97
Protocol DC
10% 5% 2% 1%
Disco 403 1769 9797 40,301
U-Connect 121 529 2809 10,201
Searchlight 200 800 5000 20,000
Combinatorial + Multiples of k 91 381 2451 9507
© 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http:/
Share and Cite
MDPI and ACS Style
Choi, S.; Yi, G. Neighbor Discovery Optimization for Big Data Analysis in Low-Power, Low-Cost Communication Networks. Symmetry 2019, 11, 836. https://doi.org/10.3390/sym11070836
AMA Style
Choi S, Yi G. Neighbor Discovery Optimization for Big Data Analysis in Low-Power, Low-Cost Communication Networks. Symmetry. 2019; 11(7):836. https://doi.org/10.3390/sym11070836
Chicago/Turabian Style
Choi, Sangil, and Gangman Yi. 2019. "Neighbor Discovery Optimization for Big Data Analysis in Low-Power, Low-Cost Communication Networks" Symmetry 11, no. 7: 836. https://doi.org/10.3390/sym11070836
Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details
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Theory Weekly Highlights for March 2010
March 26, 2010
Improvements have been made to the MATLAB code NC to calculate neoclassical ion thermal conductivity and parallel flow using the full Fokker Planck operator. The code now employs normalized velocity
space basis functions to allow more terms in the expansion. It also uses the Knorr model to better define aspect ratio dependence and block matrix notations to separate velocity space and poloidal
angle dependences, allowing easy extension to general magnetic geometry. The number of terms in the expansion increases as collisionality is reduced. Convergence to one percent accuracy has been
demonstrated for the case with inverse aspect ratio of 0.1 and the customary collisionality parameter nustar of 1 using 40 each of Legendre polynomials, generalized Laguerre polynomials, and
trigonometric functions.
March 19, 2010
GYRO simulations of energetic particle (EP) driven very long wavelength local TAE/EPM turbulence embedded in shorter wavelength ITG/TEM turbulence have shown that fixed gradient nonlinearly-saturated
transport states can exist in a narrow range of EP pressure gradients. EP transport is enhanced over the ITG/TEM driven levels but the background plasma (BGP) transport is unchanged. The narrow range
is bounded by the TAE/EPM linear threshold gradient and a critical gradient where runaway in time and unbounded EP transport onsets. However, ExB shear from the runaway long wavelength TAE/EPM modes
stabilized the ITG/TEM turbulence and the BGP transport decreases. Since EP transport is limited by the EP source rate, the EP pressure gradients will profile relax to intermittent EP transport at
the critical gradient. The critical gradient collapses to the linear threshold gradient if the BGP gradients are not large enough. The results suggest an important conclusion: The onset of local high
n TAE/EPM modes does not degrade the background plasma confinement.
March 12, 2010
NIMROD simulations of DIII-D rapid shutdown by Ar-pellet injection showed that the application of RMP fields with n=1,2, and 3 symmetry reduces the magnitude of the runaway “prompt loss” that occurs
at the start of the current quench. In these simulations, orbits for 1800 fast electrons were integrated during the rapid shutdown to examine confinement of supra-thermal and runaway electrons. The
prompt loss reduction relative to the case with no RMPs occurred in every simulation. However, the n=1 fields show the most promise for sustaining a continuous runaway electron loss rate that exceeds
the theoretical avalanche growth rate. In future simulations, the n=1 poloidal spectrum will be tailored to incorporate stronger resonant components. A toroidal resolution study has also found
runaway electron confinement to be converged with 11 toroidal modes (n=0,10); the change in confined runaway electrons when the number of modes is doubled is 1%.
March 05, 2010
The MDSplus data system has recently been expanded to include a new server which doubles the overall usable disk storage to 13.5Tb. This expansion was critical, as the estimated disk usage would have
reached over 95% by the end of this years run period. Data migration has been completed in order to spread the usage out between the different data servers.
These highlights are reports of research work in progress and are accordingly subject to change or modification | {"url":"https://fusion.gat.com/global/theory/weekly/0310","timestamp":"2024-11-12T16:48:11Z","content_type":"application/xhtml+xml","content_length":"16241","record_id":"<urn:uuid:737baef1-253e-4518-b315-b54005f111cf>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00662.warc.gz"} |
What does the coefficient of determination r tell you?
The coefficient of determination (denoted by R2) is a key output of regression analysis. It is interpreted as the proportion of the variance in the dependent variable that is predictable from the
independent variable. An R2 between 0 and 1 indicates the extent to which the dependent variable is predictable.
What does the coefficient of determination r2 tell us?
R2 is a measure of the goodness of fit of a model. In regression, the R2 coefficient of determination is a statistical measure of how well the regression predictions approximate the real data points.
An R2 of 1 indicates that the regression predictions perfectly fit the data.
How do you interpret the coefficient of determination?
The most common interpretation of the coefficient of determination is how well the regression model fits the observed data. For example, a coefficient of determination of 60% shows that 60% of the
data fit the regression model. Generally, a higher coefficient indicates a better fit for the model.
What does the coefficient of variation tell us?
The coefficient of variation (CV) is the ratio of the standard deviation to the mean. The higher the coefficient of variation, the greater the level of dispersion around the mean. It is generally
expressed as a percentage. The lower the value of the coefficient of variation, the more precise the estimate.
What is the difference between coefficient of determination and coefficient of correlation?
Coefficient of Determination is the R square value i.e. . R square is simply square of R i.e. R times R. Coefficient of Correlation: is the degree of relationship between two variables say x and y.
It can go between -1 and 1.
How do you interpret coefficient of non determination?
Inversely, the Coefficient of Non-Determination explains the amount of unexplained, or unaccounted for, variance between two variables, or between a set of variables (predictors) in an outcome
variable. Where the Coefficient of Non-Determination is simply 1 – R2.
Is coefficient of determination always positive?
will always be a positive value between 0 and 1.0. When going from to , in addition to computing , the direction of the relationship must also be taken into account. If the relationship is positive
then the correlation will be positive. If the relationship is negative then the correlation will be negative.
How is coefficient of determination related to correlation?
Coefficient of correlation is “R” value which is given in the summary table in the Regression output. R square is also called coefficient of determination. Multiply R times R to get the R square
value. In other words Coefficient of Determination is the square of Coefficeint of Correlation.
Does CV measure accuracy or precision?
The CV is a more accurate comparison than the standard deviation as the standard deviation typically increases as the concentration of the analyte increases. Comparing precision for two different
methods using only the standard deviation can be misleading.
Why coefficient of variation is better than standard deviation?
Comparison to standard deviation The coefficient of variation is useful because the standard deviation of data must always be understood in the context of the mean of the data. In contrast, the
actual value of the CV is independent of the unit in which the measurement has been taken, so it is a dimensionless number.
What is the difference between R² and R²?
Adjusted R-squared is a modified version of R-squared that has been adjusted for the number of predictors in the model. The adjusted R-squared increases when the new term improves the model more than
would be expected by chance. It is always lower than the R-squared.
What is the coefficient of determination in machine learning?
Coefficient of determination also called as R2 score is used to evaluate the performance of a linear regression model. It is the amount of the variation in the output dependent attribute which is
predictable from the input independent variable(s). | {"url":"https://ru-facts.com/what-does-the-coefficient-of-determination-r-tell-you/","timestamp":"2024-11-07T13:10:05Z","content_type":"text/html","content_length":"54208","record_id":"<urn:uuid:10a1ef82-d0d0-4f27-b34f-d34eb6c73d54>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00891.warc.gz"} |
Dependent and Independent Variables in Economics
Dependent and Independent Variables in Economics
Dependent Variable
In experimental analysis, researchers use independent variables to find out what impact they have on the dependent variable. The dependent variable doesn’t change during the course of the experiment
and is held constant.
The purpose of the experiment is to discover whether the dependent variable is linked to a number of independent variables. For example, an economist may look at what variables affect economic
growth. The economist may look at factors such as trade restrictions, labor laws, taxes, or other independent variables. The dependent variable will be economic growth as the purpose of the
experiment is to find out what variables impact it.
The dependent variable is used to identify what independent variables have a statistically significant impact on it. For example, cornflakes may be growing in popularity, so researchers may want to
find out why. The dependent variable is the sales of cornflakes. There are a number of independent variables which may or may not cause the sale of cornflakes to rise. The point of the research is to
identify which one is most significant.
Key Points
1. An independent variable is changed to see what impact or relationship it has with the dependent variable.
2. If the dependent variable changes as a result of a change in the independent variable, then this would suggest there is a cause-and-effect relationship.
Independent Variable
In an experiment, the independent variable is the one by which is changed in order to find out which has a significant statistical impact on the dependent variable. In other words, are the two
connected with each other? For example, sales of sneakers may rise as a result of increased spending on advertising. The dependent variable is the sale of sneakers, whilst increased spending on
advertising is an independent variable.
There are many independent variables which may have an impact on the dependent variable. For example, the sale of sneakers may increase due to advertising, but they may also increase due to seasonal
variations or because the price of other footwear has increased. These other factors are known as independent variables – they may all have an impact on the dependent variable.
Dependent vs Independent Variable
The dependent variable differs from an independent variable in that it is held constant. The dependent variable stays the same, whilst the independent variables will change constantly as researchers
aim to find a correlation between the two. For example, researchers may look to learn the most effective way to study. They may look at several variables such as time spent studying, the number of
hours of tutoring, or perhaps even diet and other in-direct variables.
The dependent variable will always be the variable that is held constant. It is the main point of focus in the experiment whereby researchers are looking to find a cause to. By contrast, the
independent variable is the potential answer as to what is causing the change in the dependent variable.
For example, economists may want to find out what is triggering inflation. As that is the issue being looked at, it is the dependent variable. So the independent variables are compared to the
dependent variable. In this case, it is inflation. The aim of which is to find out what is causing it. Potentially independent variables may include a boom in consumer spending or an increase in the
money supply.
Independent and Dependent Variable Examples
Some examples of independent and dependent variables include:
1. An economist is trying to analyze what caused a recent increase in the inflation rate. This will be the dependent variable. The independent variable will include potential factors such as the
exchange rate, the money supply, the unemployment rate, or consumer spending. All of these may have a potential impact on the dependent variable.
2. A patient visits a doctor to find out why they are getting pain in their leg. This is the dependent variable. The doctor may suggest a number of independent variables which may be the cause. For
example, it may be onset arthritis or an issue with the cartilage in the knee.
3. Researchers may want to find out how sugar free soda affects consumer’s hunger. The dependent variable will be how hungry the consumer is. The independent variable will be the type of soda that
is used. For instance, the hunger might be more affected by soda with higher levels of caffeine in.
4. Researchers want to find out the impact social media usage has on an individual’s mental health. The independent variable will be the individual’s mental health. The independent variable will be
the level of social media usage.
5. Psychotherapists are looking to find out what are the main causes of anxiety. An individual’s level of anxiety is the dependent variable. The independent variables may include factors such as
diet or quality of sleep.
6. Researchers are looking to find out when people are most alert during the day. The dependent variable will be the individual’s alertness. The independent variable will be the time of day.
FAQs on Dependent and Independent Variables
What are independent and dependent variables?
The independent variable is the potential cause and the dependent variable is the effect. In an experiment, researchers will use different independent variables to identify how significantly they
impact on the dependent variable – i.e. what effect it has.
How do you know if a variable is independent?
A variable is independent if it is one of many factors which could have an effect on the dependent variable. The dependent variable stays the same, whilst researchers use different independent
variables to find out what impact it has.
What is the difference between independent and dependent variables?
An independent variable is one which is changed by the researcher in order to find out what impact it has on the dependent variable – which is kept constant.
About Paul
Paul Boyce is an economics editor with over 10 years experience in the industry. Currently working as a consultant within the financial services sector, Paul is the CEO and chief editor of BoyceWire.
He has written publications for FEE, the Mises Institute, and many others.
Further Reading
Embezzlement - Embezzlement is the act of fraudulently misappropriating or stealing funds or assets entrusted to one's care, often done by an…
Embargo Definition and Examples - An embargo is a government-imposed restriction on trade or economic activity with a specific country or group of countries. | {"url":"https://boycewire.com/dependent-and-independent-variables/","timestamp":"2024-11-12T07:13:26Z","content_type":"text/html","content_length":"162933","record_id":"<urn:uuid:5fb05ed4-8e20-4982-8dfd-5daf43d46130>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00592.warc.gz"} |
Ebook A History Of Modern Britain 2009
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| Tool Preview
This tool compares two sets of data and find the differential expression. One very important component of the tool is the reference set. Actually, to use the tool, you need the two input sets of
data, of course, and the reference set. The reference set is a set of genomic coordinates and, for each interval, it will count the number of feature on each sample and compute the differential
expression. For each reference interval, it will output the direction of the regulation (up or down, with respect to the first input set), and a p-value from a Fisher exact test.
This reference set seems boring. Why not computing the differential expression without this set? The answer is: the differential expression of what? I cannot guess it. Actually, you might want to
compare the expression of genes, of small RNAs, of transposable elements, of anything... So the reference set can be a list of genes, and in this case, you can compute the differential expression of
genes. But you can also compute many other things.
Suppose that you cluster the data of your two input samples (you can do it with the clusterize and the mergeTranscriptLists tools). You now have a list of all the regions which are transcribed in at
least one of the input samples. This can be your reference set. This reference set is interesting since you can detect the differential expression of data which is outside any annotation.
Suppose now that you clusterize using a sliding window the two input samples (you can do it with the clusterizeBySlidingWindows and the mergeSlidingWindowsClusters tools). You can now select all the
regions of a given size which contain at least one read in one of the two input samples (do it with selectByTag and the tag nbElements). Again, this can be an other interesting reference set.
In most cases, the sizes of the two input samples will be different, so you should probably normalize the data, which is an available option. The ---rather crude--- normalization increases the number
of data in the least populated sample and decreases the number of data in the most populated sample to the average number of data. | {"url":"https://toolshed.g2.bx.psu.edu/repository/display_tool?repository_id=613e73b56070623c&tool_config=%2Fsrv%2Ftoolshed%2Fmain%2Fvar%2Fdata%2Frepos%2F000%2Frepo_307%2FSMART%2Fgalaxy%2FGetDifferentialExpression.xml&changeset_revision=d96f6c9a39e0&render_repository_actions_for=tool_shed","timestamp":"2024-11-13T19:25:30Z","content_type":"text/html","content_length":"9141","record_id":"<urn:uuid:32e39796-7155-42b5-90fa-81bdf6aace15>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00213.warc.gz"} |
Basics: Relative Velocity
**pre reqs:** [Vectors and Vector Addition](http://scienceblogs.com/dotphysics/2008/09/basics-vectors-and-vector-ad…)
This was sent in as a request. I try to please, so here it is. The topic is something that comes up in introductory physics - although I am not sure why. There are many more important things to worry
about. Let me start with an example. Suppose you are on a train that is moving 10 m/s to the right and you throw a ball at 5 m/s to the right. How fast would someone on the ground see this ball? You
can likely come up with an answer of 15 m/s - that wasn't so hard right? But let me draw a picture of this situation:
The important thing is: If the velocity of the ball is 5 m/s, that is the velocity with respect to what? In the diagram, I listed the velocity of the ball as *v[ball-train]* this indicates it is with
respect to the train. There are three velocities in this example.
• The velocity of the ball with respect to the train
• The velocity of the train with respect to the ground
• The velocity of the ball with respect to the ground
These three velocities are related by the following:
**note**: The way I always remember this is to arrange it so that the frames match up on the left side. That is to say v(a-b) + v(b-c) - you can think of this as the "b's" canceling and giving v
Clearly this works for the simple case above, but actually it works no matter which direction as long as the equation remains as a **vector** equation. In general, with two reference frames (Say A
and B) then you (or I) can say:
The most important thing is that these are vectors and must be treated as such. If you treat these vectors as scalars, you will likely get the problem wrong.
Ok. Fine, that makes some sense - but these darn physics problems are killing me (or you). How about an example, everyone loves those. However, if you don't feel comfortable with vectors, go look [at
my introduction to vectors](http://scienceblogs.com/dotphysics/2008/09/basics-vectors-and-vector-ad…).
**Problem** Suppose you have a boat that travels at a speed of 2 m/s on the water. This boat is to cross a river that is 500 meters wide and has a speed (of the water) of 0.5 m/s. What angle should
you aim your boat so that it travels straight across the river (without going downstream at all).
Here is a picture:
What is given in the problem? There is the magnitude of the velocity of the boat with respect to the water, the velocity of the water with respect to the ground. There is one other thing that the
problem gives. The velocity of the boat with respect to the ground is only in the y-direction. The goal of the problem is to find the angle ? that the boat has to aim. Here is what is given
Actually, there is one other piece of information that is important. The velocity of the boat with respect to the ground is ONLY in the y direction. I can write this as:
Now, let me put these velocities together:
Where this is a vector addition equation. To add vectors, I can just add the x-components and then just add the y-components. In this case, I can JUST look at the x-direction:
Note that the (m/s) units cancel. Solving for sin(?):
So, there you have it. Let me recap what is important:
• Start with the relative velocity equation
• Write down the velocities you know (as vectors)
• Treat the velocities as vectors
See. That is not too bad, is it?
**Final Note:** This is known as Galilean relativity. It works when the velocities of the frames and objects are much less than the velocity of light. (example: a jet going at twice the speed of
sound is way slower than light). If the objects are moving close to the speed of light, this stuff does not work.
More like this
**Pre Reqs:** [What is a Force](http://scienceblogs.com/dotphysics/2008/09/basics-what-is-a-force.php) [Previously, I talked about the momentum principle](http://scienceblogs.com/dotphysics/2008/10/
basics-forces-and-the-moment…). Very useful and very fundamental idea. The other big (and useful)…
**Pre Reqs:** [Work-Energy](http://scienceblogs.com/dotphysics/2008/10/basics-work-energy.php) You need to be familiar with work and energy to understand this. If you are not familiar, look at the
pre requisite link. Ok? Now, let's begin. Suppose a ball moves from point A (3 m, 3 m) to B (1 m, 1…
**pre-reqs:** trig Think of the following two things. Temperature and wind speed. These are two different things that you could measure, but there is one big difference. Wind speed has two parts to
it - how fast and which direction. Temperature is just one thing (no direction). Temperature is…
**Pre Reqs:** [Intro to Forces](http://scienceblogs.com/dotphysics/2008/09/basics-what-is-a-force.php), [Vectors](http://scienceblogs.com/dotphysics/2008/09/basics-vectors-and-vector-ad…) Hopefully
now you have an idea of what a force is and what it isn't. What do you do with them? The useful…
thanks for your prompt action.I was teaching relative velocity to someone and I myself was not convinced about my explanations. your exposition makes it crystal clear.
Thank you so much for your kind information.
With best regards
ashek ullah
Thanks, Me too was looking for a clarification of
idea of relative velocity. I got it here.
your example got some numbers switched. you said water speed of 0.5 m/s. in the equation though you say 1 m/s for Vwater/ground | {"url":"https://scienceblogs.com/dotphysics/2008/09/24/basics-relative-velocity","timestamp":"2024-11-09T06:31:41Z","content_type":"text/html","content_length":"47002","record_id":"<urn:uuid:bf5987e4-8458-4d1c-9e73-c11069db0b9c>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00580.warc.gz"} |
High precision series solutions of differential equations: Ordinary and regular singular points of second order ODEs
High precision series solutions of differential equations: Ordinary and regular singular points of second order ODEs
Published: 1 October 2012| Version 1 | DOI: 10.17632/k934x6hbtj.1
Amna Noreen, Kåre Olaussen
Abstract A subroutine for a very-high-precision numerical solution of a class of ordinary differential equations is provided. For a given evaluation point and equation parameters the memory
requirement scales linearly with precision P , and the number of algebraic operations scales roughly linearly with P when P becomes sufficiently large. We discuss results from extensive tests of the
code, and how one, for a given evaluation point and equation parameters, may estimate precision loss and computing ti... Title of program: seriesSolveOde1 Catalogue Id: AEMW_v1_0 Nature of problem
The differential equation -s 2 (d 2 /dz 2 + (1 - ν + - ν - )/z d/dz + (ν + ν - )/z 2 )ψ(z) + 1/z Σ N n-0 v n z n ψ(z) =0, is solved numerically to very high precision. The evaluation point z and some
or all of the equation parameters may be complex numbers; some or all of them may be represented exactly in terms of rational numbers. Versions of this program held in the CPC repository in Mendeley
Data AEMW_v1_0; seriesSolveOde1; 10.1016/j.cpc.2012.05.015 This program has been imported from the CPC Program Library held at Queen's University Belfast (1969-2018)
Atomic Physics, Computational Physics, Computational Method | {"url":"https://data.mendeley.com/datasets/k934x6hbtj/1","timestamp":"2024-11-14T17:55:07Z","content_type":"text/html","content_length":"105050","record_id":"<urn:uuid:eba51afe-0b2e-4c67-b5ce-e2e48ca6f531>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00816.warc.gz"} |
Write the decimal number represented by the points A, B, C, D on the given number line
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Write the decimal number represented by the points A, B, C, D on the given number line.
We use the concepts of decimals and number line to write the decimal number represented by the points A, B, C, D on the given number line.
The decimal numbers represented by the points A, B, C and D on the given number line are:
Point Decimal Value on Number Line
A 0.8
B 1.3
C 2.2
D 2.9
NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.1 Question 9
Write the decimal number represented by the points A, B, C, D on the given number line.
The decimal numbers represented by the points A, B, C and D on the given number line are 0.8, 1.3, 2.2 and 2.9 respectively.
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The Hamermesh umpire/race study revisited -- part IX
The Hamermesh umpire/race study revisited -- part IX
This is the ninth post in this series, on the Hamermesh racial-bias study. The previous posts are here.
There's not going to be anything new in this post – I'm just going to recap the various issues I've already talked about, with fewer numbers. You can consider this a condensed version of the other
eight posts.
I'll start by summarizing the study one more time.
The Hamermesh study analyzed three seasons worth of pitch calls. It attempted to predict whether a pitch would be called a ball or a strike based on a whole bunch of variables: who the umpire was,
who the pitcher was, the score of the game, the inning, the count, whether the home team was batting, and so forth.
But it added one extra variable: whether the race of the pitcher (black, white, or hispanic) matched the race of the umpire. That was called the "UPM" variable, for "umpire/pitcher match." If umpires
had no racial bias, the UPM variable would come out close to zero, meaning knowing the races wouldn't help you predict whether the pitch was a ball or a strike. But if the UPM came out significant
and positive, that would mean that umpires were biased -- that all else being equal, pitches were more likely to be strikes when the umpire was of the same race as the pitcher.
It turned out, that, when the authors looked at *all* the data, the UPM variable was not significant; there was only small evidence of racial bias. However, when the data were split, it turned out
that the UPM coefficient *was* significant, at 2.17 standard deviations, in parks in which the QuesTec system was not installed. Umpires appear to have called more strikes for pitchers of their own
race in parks in which their calls were not second-guessed by QuesTec.
An even stronger result was found when selecting only games in parks where attendance was sparse. In those games, the UPM coefficient was significant at 2.71 standard deviations. The authors
interpreted this to mean that when fewer people were there scrutinzing the calls, umpires felt freer to indulge their taste for same-race discrimination.
In this latter case, the UPM coefficient was 0.0084, meaning that the percentage of strikes called for same-race pitchers increased by 0.84 of a percentage point. That would suggest that about 1
pitch in 119 is influenced by race.
That's the study.
I do not agree with the authors that the results show of widespread existence of same-race bias. I have two separate sets of arguments. First, there are statistical reasons to suggest that the
significance levels might be overinflated. And, second, the model the authors chose have embedded assumptions which I don't think are valid.
1. The Significance Arguments
In their calculations, the authors calculated standard errors as if they were analyzing a random sample of pitches. But the sample is not random. Major League Baseball does not assign a random umpire
for each pitch. They do, roughly, assign a random umpire for each *game*, but that means that a given umpire will see a given pitcher for many consecutive pitches.
If the sample of pitches were large enough, this wouldn't be a big issue – umpires would still see close to a random sample of pitchers. But there are very few black pitchers, and only 7 of the 90
umpires are of minority race (2 hispanic, 5 black). This means that some of the samples are very small. For instance, there were only about 900 ptiches called by black umpires on black pitchers in
low-attendance situations. That situation is very influential in the results, but it's only about 11 games' worth. It seems reasonable to assume that these umpires saw only a very few starting
What difference does that make? It means that the pitches are not randomly distributed among all other conditions, because they're clustered into only a very few games. That means that if the study
didn't control for everything correctly, the errors will not necessarily cancel out, because they're not independent for each pitch.
For instance, the authors didn't control for whether it was a day game or a night game. Suppose (and I'm making this up) that strikes are much more prevalent in night games than day games because of
reduced visibility. If the sample were very large, it wouldn't matter, because if there were 1000 starts or so, the day/night situation would cancel out. But suppose there were only 12 black/black
starting pitcher games. If, just by chance, 8 of those 12 happened to be night games, that might explain all those extra strikes.
8 out of 12 is 66%. The chance of 66% of 12 random games being night games is reasonably high. But the chance of 66% of 900 random *pitches* being in night games is practically zero. And it's the
latter that the study incorrectly assumes. (Throughout the paper, the standard errors are based on the normal approximation to binomial, which assumes independence.)
(I emphasize that this is NOT an argument of the form, "you didn't control for day/night, and day/night might be important, so your conclusions aren't right." That argument wouldn't hold much weight.
In any research study, there's always some possible explanation, some factor the study didn't consider. But, if that factor is random among observations, the significance level takes it into account.
The argument "you didn't control for X" might suggest that X is a *cause* of the statistical significance, but it is not an argument that the statistical significance is overstated.
So my argument is not "you didn’t control for day/night." My argument is, "the observations are not sufficiently independent for your significance calculation to be accurate enough." The day/night
illustration is just to show *why* independence matters.)
Now, I don't have any evidence that day/night is an issue. But there's one thing that IS an issue, and that's how the study corrected for score. The study assumed that the bigger the lead, the more
strikes get thrown, and that every extra run by which you lead (or trail) causes the same positive increase in strikes. But that's not true. Yes, there are more strikes with a five-run lead, but
there are also more strikes with a five-run deficit, as mop-up men are willing to challenge the batter in those situations. So when the pitcher's team is way behind, the study gets it exactly
backwards: it predicts very few strikes, instead of very many strikes.
Again, if the sample were big enough, all that would likely cancel out – all three races would have the same level of error. But, again, the black/black subsample has only a few games. What if one of
those games was a 6-inning relief appearance by a (black) pitcher down by 10 runs? The model expects almost no strikes, the pitcher throws many, and it looks like racial bias. That isn't necessarily
that likely, but it's MUCH more likely than the model gives it credit for. And so the significance level is overstated.
So we have at least one known problem, of the score adjustment. And there were many adjustments that weren't made: day/night, runners on base, wind speed, days of rest ... and you can probably think
of more. All these won't be indpendent, and there's going to be some clustering. So if any of those other factors influence balls and strikes – which they probably do -- the significance levels will
be even wronger.
How wrong? I don't know. It could be that if you did all the calculations, they'd be only only slightly off. It could be that they're way off. But they're definitely too high.
Note that this argument only applies to the significance levels, and not the actual bias estimates. Even with all the corrections, the 0.0084 would remain. The question is only whether it would still
be statistically significant.
2. The Model
As I mentioned earlier, the study included only one race-related UPM variable, for whether or not the umpire and pitcher were of the same race. Because the variable came out significantly different
from zero, the study concluded that its value represents the effect of umpires being biased in favor of pitchers of their own race.
However, the choice of a single value for UPM is based on two hidden assumptions. The first one:
-- Umpire bias is the same for all races.
That is: the study assumes that a white umpire is exactly as likely to favor a white pitcher as a black umpire is to favor a black pitcher, and exactly as likely as a hispanic umpire is to favor a
hispanic pitcher.
Why is this a hidden assumption? Because there is only one UPM variable that applies to all races. But it's not hard to think of an example where you'd need to measure bias for each race separately.
Suppose umpires were generally unbiased, except that, for some reason, black umpires had a grudge against black pitchers, and absolutely refused to call strikes against them (if you like, you can
suppose that fact is well-known and documented). If that were the case, the analysis done in this study would NOT pick that up. It would find that there is indeed racial discrimination against
same-race pitchers, but it would be forced to assume that it's equally distributed among the three races of umpires.
That's a contrived example, of course. But, the real world, things are different. In real life, is it necessarily true that all races of umpires would have exactly the same level of bias?
It seems very unlikely to me. Historically, discrimination has gone mostly one way, mostly whites discriminating against minorities, mostly men discriminating against women. There are probably a fair
number of white men who wouldn't want to work for a black or female boss. Are there as many black women who wouldn't want to work for a white or male boss? I doubt it.
Why, then, assume that signficant bias must exist for all races? And, why assume, as the study did, that not only does it exist for all races, but that the effects are *exactly the same* regardless
of which race you're looking at?
If you remove the assumption, you wind up with a much weaker result. There still turns out to be a statistically significant bias, but you no longer know where it is. Take another hypothetical
example: there are white and black pitchers and umpires, and three of the four combinations result in 50% strikes. However, the fourth case is off -- white umpires call 60% strikes for white
Who's discriminating? You can't tell. It could be whites favoring whites. But it could be that white pitchers are just better pitchers, and it's the black umpires discriminating against whites,
calling only 50% strikes when they should be calling 60%. If you open up the possibility that one set of umpires might be more biased than the other – an assumption which seems completely reasonable
to me – you can find that there's discrimination, but not what's causing it.
Also, you can't even tell how many pitches are affected. If the white/white case had 350,000 pitches and the black umpire/white pitcher case had only 45,000 pitches, you could have as many as 35,000
pitches affected (if it's the white umpires discriminating), as few as 4,500 pitches (if it's the black umpires) or something in the middle (both sets of umpires discriminate, to varying extents).
And maybe it's not white umpires favoring white pitchers, or black umpires disfavoring white pitchers. Maybe it's black umpires favoring black pitchers. Maybe black umpires generally have a smaller
strike zone, but they enlarge it for black pitchers. Since there are so few black pitchers, maybe in this case only 800 pitches are affected.
The point is that without the different-races-have-identical-biases assumption, all the conclusions fall apart. The only thing you *can* conclude, if you get a statistically-significant UPM
coefficient, is that there is bias *somewhere*. You can reject the hypothesis that bias is zero everywhere, but that's it.
The other hidden assumption is:
-- All umpires have identical same-race bias.
The study treats all pitches the same, again with the same UPM coefficient, and assumes the errors are independent. This assumes that any umpire/pitcher matchup shows the same bias as any other
umpire/pitcher matchup – in other words, that all umpires are biased by the same amount.
To me, that makes no sense. In everyday life, attitudes towards race vary widely. There are white people who are anti-black, there are people who are race-neutral, and there are people who favor
affirmative action. Why would umpires be any different? Admittedly, we are talking about unconscious bias, and not political beliefs, but, still, wouldn't you expect that different personalities
would exhibit different quantities of miscalled pitches?
Put another way: remember when MLB did its clumsy investigations of umpires' personal lives, asking neighbors if the ump was, among other things, a KKK member? Well, suppose they had found one umpire
who, indeed *was* a KKK member. Would you immediately assume that *all* white umpires now must be KKK members? That would be silly, but that's what is implied by the idea that all umpires are
I argue that umpires are human, and different humans must exhibit different degrees of conscious and unconscious racial bias.
Once you admit the possibility that umpires are different, it no longer follows that bias must be widespread among umpires or races of umpires. It becomes possible that the entire observed effect
could be caused by one umpire! Of course, it's not necessarily true that it's one umpire – maybe it's several umpires, or many. Or, maybe it is indeed all umpires – even though they have different
levels of bias, they might all have *some*.
How can we tell? What we can do is, for all 90 umpires, see how much they favor their one race over another. Compare them to the MLB average, so that the mean umpire becomes zero relative to the
league. Look at the distribution of those 90 umpire scores.
Now, suppose there is no bias at all. In that case, the distribution of the individual umpires should be normally distributed exactly as predicted by the binomial distribution, based on the number of
What if only one or two umpires are biased? In that case, we probably can't tell that apart from the case where no umpire is biased – it's only a couple of datapoints out of 90. Unless the offending
umps are really, really, really biased, like 3 or 4 standard deviations, they'll just fit in with the distribution.
What if half the umpires are biased? Then we should get something that's more spread out than the normal distribution – perhaps even a "two hump" curve, with the biased umps in one hump, and the
unbiased ones in the other. (The two humps would probably overlap).
What if all the umpires are (differently) biased? Again we should get a curve more spread out than the normal distribution. Instead of only 5% of umpires outside 2 standard errors, we should get a
lot more.
So we should be able to estimate the extent of bias by looking at the distribution of individual umpires. I checked, using a dataset similar to the one in the study (details are in part 8).
What I found was that the result looked almost perfectly normal. (You would have expected the SD of the Z-scores to be exactly 1. It was 1.02.)
This means one of the following:
-- no biased umps
-- 1 or 2 biased umps
-- many biased umps with *exactly the same bias*.
As I said, I don't find the third option credible, and the statistical significance, if we accept it, contradicts the first option. So I think the evidence suggests that only a very few umpires are
biased, but at least one.
However: the white/white sample is so large that one or two biased white umpires wouldn't be enough to create statistical significance. So, if we must assume bias, we should assume we're dealing with
a very small number of biased *minority* umpires. Maybe even just one.
And as it turns out (part 7), if you take out one of the two hispanic umpires (who, it turns out, both called more strikes for hispanic umpires), the statistical significance disappears.
If you take out a certain black umpire, who had the highest increase in strike calls for black batters out of all 90 umps, the statistical significance again disappears. This doesn't necessarily mean
that any or all of those umps is biased. It *does* mean that the possibility explains the data just as well as the assumption of universal unconscious racial bias.
Based on all that, here's where the study and I disagree.
Study: there is statistically significant evidence of bias.
Me: there *may* be statistically significant evidence of bias, but you can't tell for sure because some of the critical observations aren't independent.
Study: the findings are the result of all umpires being biased.
Me: the findings are more likely the result of one or two minority umpires being biased.
Study: many pitches are affected by this bias.
Me: there is no way to tell how many pitches are affected, but, if the effect is caused by one or two minority umpires favoring their own race, the number of pitches would be small.
Study: overall, minority pitchers are disadvantaged by the bias.
Me: That would be true if all umpires were biased, because most umpires are white. But if only one or two minority umpires are biased, then minority pitchers would be the *beneficiaries* of the bias.
Study: the data show that bias exists.
Me: I don't think the data shows that bias exists beyond a reasonable doubt. For instance, suppose the results found are significant at a 5% level. And suppose your ex-ante belief is that umpire
racism is rare, and the chance at least one minority umpire is biased is only about 5%. Then you have equal probabilities of luck and bias.
I do not believe the evidence in is study is strong enough to lead to a conclusion of bias beyond a reasonable doubt. But it is strong enough to suggest further investigation, specifically of those
minority umpires who landed on the "same-race" side. My unscientific gut feeling is that if you did look closely, perhaps by watching game tapes and such -- most of the effect would disappear and the
umps would be cleared.
But that's just my gut. Your gut my vary. I will keep an open mind to new evidence.
Labels: baseball, Hamermesh update, race
3 Comments:
At Tuesday, June 10, 2008 5:37:00 PM, said...
Isn't there also a possibility that there is one (or a very small number of) pitcher that every ump despises, and if he's non-white, even though all umps call him equally (but worse than all
other pitchers) we'd see evidence of bias under the Hamermesh study's assumptions?
At Tuesday, June 10, 2008 9:22:00 PM, Phil Birnbaum said...
Hmmm ... I think that would affect every umpire equally, so it wouldn't affect the race interactions. That is, there's no statistical difference between a minority pitcher getting fewer strikes
because he's legitimately worse, versus him getting fewer strikes because everyone hates him.
At Sunday, July 10, 2011 7:44:00 PM, said...
Great site. Anyway, onto the discussion.
1. Hasn't Implicit Race Bias ALREADY been shown to exist in other broader and more comprehensive social (see wiki) as well as sports (NBA) studies? So, as to your doubt that there could be "many
biased umps with *exactly the same bias*", and you "don't find the...option credible", isn't this exactly what such human implicit race bias would expect to find? That is, roughly the same amount
of unconscious bias across all umpires, and all umpires falling within a normal expected range of variance?
2. Isn't pulling 2 or 3 extreme umpire cases OUT of this study just as questionable as making sure you include them IN when interpreting this or any significance level?
You could be right on #1, it's just that I wouldn't be so sure that such a scenario exists.
On #2, it seems anyone could prove or disprove a study with this technique. So, how does one avoid this pitfall for a fully qualified and/or academic study? It seems that these included outliers
are exactly what ends up proving all such studies in the first place. | {"url":"http://blog.philbirnbaum.com/2008/06/hamermesh-umpirerace-study-revisited.html","timestamp":"2024-11-02T12:13:03Z","content_type":"application/xhtml+xml","content_length":"44914","record_id":"<urn:uuid:3ad8c471-0f2d-4869-9611-d2aa2a5d142f>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00739.warc.gz"} |
Introduction to Time complexity in Python | ESS Institute
Every material thing in the universe is defined by space and time. The efficiency of an algorithm can also be determined by space and time complexity. Even if there are many methods to approach a
particular programming challenge, understanding how an algorithm operates could help us become better programmers. Time complexity is suggested to be a key metric for measuring how well data
structures and algorithm’s function. It gives us a sense of how a data structure or algorithm will function as the volume of the input data increases. Writing scalable and efficient code, which is
crucial in data-driven applications, requires an understanding of time complexity. We will examine time complexity in Python data structures in more detail in this blogpost. Let’s go through it.
What is Time Complexity in Python?
In computer science, the phrase “time complexity” refers to the computational complexity which we can use to characterize how long it takes for an algorithm to run. Counting how many basic operations
the algorithm does, given that each basic operation takes a specific amount of time to complete.
Generally we have 3 cases in time complexity
1. Best case: Minimum time required
2. Average case: Average time required
3. Worst case: Maximum time it need to compute
It’s is a standard method for estimating time complexity. It’s usually expressed in big-O notation, which represents the upper bound of the worst-case scenario. Big-O notation helps us to compare the
performance of different algorithms and data structures in terms of their scalability.
Python counts the number of operations a data structure or method needs to perform in order to take a given task to end. Data structures in Python can have a variety of time complexity, including:
1. O(1) – Constant Time Complexity: This denotes that the amount of time required to complete an operation is constant and independent of the amount of input data. For example: When we access an
element in an array by index, it has an O(1) time complexity. because the time required to access an element is constant regardless of the size of the array.
2. (log n) – Logarithmic Time Complexity: OThis describes how the time required to complete an operation grows exponentially with the size of the input data. For example, Since the search time
increases exponentially with the number of items in the tree, a binary search tree has a temporal complexity of O(log n) when looking for an element.
3. O(n) – Linear Time Complexity: This specifies that as the size of the input data increases, the time required to complete an operation climbs linearly. As an example, running algorithm over an
array or linked list is having time complexity of O(n) because the amount of time needed to run over each member grows linearly as the size of the array or linked list does. Due to which iterating
over them has an O(n) time complexity.
4. O (n log n) – Log-Linear Time Complexity: This describes how an operation’s time complexity is a blend of linear and logarithmic. As an example, the time complexity of sorting an array with a
merge sort is O(n log n). It is Because of the time needed to divide and merge the array grows logarithmically while the time needed to compare and merge each element grows linearly.
5. O(n2) – Quadratic Time Complexity: This refers to the fact that as the size of the input data increases, an operation’s execution time climbs quadratically. For instance, sorting an array using
bubble sort has an O(n^2) time complexity because it takes more time to compare and swap each element as the array gets larger.
6. O(2^n) – Exponential Time Complexity: This refers to the fact that as the size of the input data increases, same will gonna happen with the time required to complete an operation. For instance,
because the number of subsets grows exponentially with the size of the set, creating all conceivable subsets of a set using recursion has a temporal complexity of O(2n).
7. Factorial time (n!): When every possible combination of a collection is calculated during a single operation, an algorithm is said to have a factorial time complexity. As a result, the time
required to complete a single operation is factorial of the collection’s item size. The Traveling Salesman Problem and the Heap’s algorithm (which creates every combination of n objects imaginable)
are both O(n!) time tough problems. Its slowness is an issue.
Python data structures’ time complexity must be understood in order to write scalable and effective code, especially for applications that work with big volumes of data. We can enhance the
performance and efficiency of our code by selecting the proper data structure and algorithm with the best time complexity.
Why do we need to know time complexity of data structures?
We can optimize our code for effectiveness and scalability by knowing time complexity, which is crucial when working with large data sets. We can decide which strategy to use to solve an issue based
on the time complexity of a specific algorithm or data structure in Python. When we kook out\ for an element in a huge dataset, we might go for a hash table rather than an array because the time
complexity of a hash table search is O(1) as opposed to O(n) for an array. When processing a lot of data, this improvement can help us save a lot of time. Additionally, it helps us stay clear of
programming errors like nested loops and recursive functions, which can cause extremely slow execution.
Common Time Complexities in Data Structures
Time Complexities of Array in Python
Arrays are a basic data structure that let us efficiently store and access components of the same type. Python implements arrays as lists, and the following operations have the following temporal
1) O(1) for accessing an element by index
2) Adding an element at the start: O(n)
3) Adding a component at the end: O(1)
4)Removing a component at the start: O(n)
5) Removing a component at the conclusion: O(1)
Time Complexity of Linked Lists
Linked Lists are Another basic data structure which are made up of a series of nodes, each of which has information and a link to the node after it. The use of classes and objects allows us to
implement linked lists in Python. Following are some operations with a high temporal complexity.
1) When accessing an element via its index O(n)
2) Inserting an element at the beginning: O(1)
3) Adding a last element: O(n)
4) Removing an element at the beginning: O(1)
5) For removing a component at the conclusion O(n)
Time complexity of Stack
Linear data structures that adhere to the Last-In-First-Out (LIFO) concept include stacks. Lists can be used to implement stacks in Python. The following operations have a high temporal complexity:
1) O(1) for adding an element to the stack.
2) O(1) for popping an element from the stack
3) Getting to the first element: O(1)
Time complexity of Queues
A queue is a type of linear data structure that follows the First-In-First-Out (FIFO) principle. In Python, we can also implement queues using lists. The time complexity of common operations is as
1) Enqueuing an element: O(1)
2) Dequeuing an element: O(1)
3) Accessing the front element: O(1)
Time complexity of Hash Tables
A hash table is categorized as a type of data structure that enables key-value pair-based data storage and retrieval. Hash tables can be created in Python by using dictionaries. The following
operations have a high temporal complexity.
1) Accessing an element by key: O(1)
2) Inserting an element: O(1)
3) Removing an element: O(1)
For creating code that is effective and scalable, it is crucial to comprehend the time complexity in data structures. For typical tasks, Python offers us a variety of built-in data structures with
varying time complexity. If we Know the temporal complexity of each data structure, it will allow us to select the best one for our application and will going to speed up our code. The main
operations of Python data structures were described in this article using Big-O notation. In essence, the big-o notation is a technique to gauge how time-consuming an operation is. A number of
typical operations for a list, set, and dictionary were also shown in the article.
How the programs use space and time efficiently is the key difference between an average python programmaer and a good python programmer. A good Python programmer will always end up with better
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Revised Probabilistic Approach to Hardy-Littlewood Twin Prime Conjecture with Asymptotic Independence
This document presents an exploration of the Hardy-Littlewood Twin Prime Conjecture through a probabilistic lens, aiming to provide a more accessible understanding and offer an alternative path
towards its potential resolution.
Theorem: Probabilistic Density of Twin Primes
Let π2(x) denote the number of twin primes less than or equal to x. Then, under the assumption of asymptotic independence of primality events for numbers of the form 6k-1 and 6k+1, the following
asymptotic relationship holds:
π[]2(x) ~ 2C[]2 ∫[]2^x (1/ln(t))^2 dt
where C[]2 is a constant that can be empirically estimated.
Part 1: Laying the Foundation
1. Prime Number Theorem (PNT): The PNT states that for large x, the number of primes less than x, denoted by π(x), can be approximated by x/ln(x). This implies that the probability of a randomly
chosen number near x being prime is approximately 1/ln(x).
2. Twin Prime Structure: All twin prime pairs, except for (3, 5), can be expressed in the form (6k – 1, 6k + 1) where k is an integer. This observation restricts our analysis to these specific
arithmetic progressions.
Part 2: Establishing Asymptotic Independence
This section replaces the previous reliance on an unproven assumption.
1. Definitions:
□ Let d(X) denote the asymptotic density of a set X of integers, defined as d(X) = limn→∞ |{k ∈ X : |k| ≤ n}| / (2n + 1), if the limit exists.
□ Define Ak as the event that |6k – 1| is prime. Let d(Ak) be the asymptotic density of integers k for which Ak occurs.
□ Define Bk as the event that |6k + 1| is prime. Let d(Bk) be the asymptotic density of integers k for which Bk occurs.
□ Note: Asymptotic density is not a probability measure (it lacks countable additivity) but serves as a useful tool for our analysis.
2. Symmetry: Observe that |6k-1| = |6k+1| for all integers k. This symmetry is crucial as it implies d(Ak) = d(Bk).
3. Chinese Remainder Theorem and Mirror Images:
□ For a prime p > 2 and an integer a, define the “mirror image” function μ as μ(a mod p) = (-a mod p). This function maps a residue class modulo p to its additive inverse.
□ For a finite set of primes S = {p1, p2, …, pr}, define MS = ∏i=1r pi. The Chinese Remainder Theorem guarantees a bijection between residue classes modulo MS and tuples of residue classes
modulo each prime in S.
□ Crucially, for any prime p > 3, if |6k-1| ≡ a (mod p), then |6k+1| ≡ μ(a) (mod p). This establishes a connection between the residue classes occupied by |6k-1| and |6k+1| modulo each prime.
4. Conditional Sets and Independence:
□ Let ES(Ak) = {k : |6k-1| is not divisible by any prime in S}, and similarly define ES(Bk).
□ Using the CRT and the mirror image property, we can show that:
d(ES(Ak) ∩ ES(Bk)) = ∏pi ∈ S, pi > 2 [(pi – 1)/pi]2 · (1/2)
□ This factorization demonstrates that, conditioned on not being divisible by primes in S, the events Ak and Bk are independent across different primes.
5. Error Analysis:
□ Let εS(Ak) = |d(Ak) – d(ES(Ak))|. This represents the error introduced by considering only primes in S.
□ Using Mertens’ third theorem and partial summation, we can show that εS(Ak) = O(1/ln(pS)), where pS is the largest prime not in S.
□ As S approaches the set of all primes, pS → ∞, and consequently, εS(Ak) → 0. The same argument holds for εS(Bk).
6. Convergence to Independence:
□ Combining the PNT and the symmetry argument, we have for large |k|:
d(Ak) = 1/ln(|6k-1|) + O(1/ln2(|6k-1|)) and d(Bk) = 1/ln(|6k+1|) + O(1/ln2(|6k+1|)).
□ From the error analysis, we know that:
|d(Ak ∩ Bk) – d(Ak) · d(Bk)| ≤ εS(Ak) + εS(Bk) + εS(Ak)εS(Bk)
□ As |k| → ∞, the right-hand side tends to 0, demonstrating the asymptotic independence of Ak and Bk in terms of their asymptotic densities.
Part 3: Deriving the Conjectured Density
1. Probabilistic Heuristic: Assuming asymptotic independence, the probability of a pair (6k – 1, 6k + 1) being a twin prime pair is:
P(Ak ∩ Bk) ≈ P(Ak) * P(Bk) ≈ (1/ln(6k))2
2. Summing Probabilities: To estimate the total number of twin primes up to x, we sum over potential twin prime pairs:
π2(x) ≈ Σk=1 to x/6 (1/ln(6k))2
3. Integral Approximation: This sum can be approximated by an integral:
π2(x) ≈ ∫1x/6 (1/ln(6t))2 dt
4. Change of Variables and Constant Adjustment: Applying the substitution u = 6t and adjusting the integration limits introduces the constant C2:
π2(x) ~ 2C2 ∫2x (1/ln(t))2 dt
This probabilistic approach provides an alternative perspective on the Hardy-Littlewood Conjecture. We have rigorously established the asymptotic independence of events Ak and Bk, addressing a
crucial gap in previous probabilistic arguments. While not a complete proof of the conjecture (as C2’s value is derived empirically), this method offers valuable insight into the distribution of twin
primes and highlights the potential of probabilistic reasoning within number theory. | {"url":"https://n01r.com/revised-probabilistic-approach-to-hardy-littlewood-twin-prime-conjecture-with-asymptotic-independence/","timestamp":"2024-11-06T18:05:33Z","content_type":"text/html","content_length":"115543","record_id":"<urn:uuid:7b522698-4181-4768-9602-d07981120ab5>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00140.warc.gz"} |
Coupling between opposite-parity modes in parallel photonic crystal waveguides and its application in unidirectional light transmission
Directional coupling between the even- and odd-parity modes of two parallel dissimilar linear defect waveguides in a square photonic crystal of cylindrical air holes in dielectric background is
numerically demonstrated. Projected band-structure computations through the plane-wave expansion method reveal that high-efficiency coupling can be achieved in a frequency range of approximately 9 %
extent around the central frequency. Coupling occurs if one row of spacing is maintained between the waveguides supporting even and odd modes, which are composed of annular air holes with outer radii
equal to the photonic crystal's scatterer radii and inner radii of 0.19 and 0.44 periods, respectively. Extinction ratio for coupling from the even to odd mode at the central frequency is 4.0 dB.
Coupling length calculated through finite-difference time-domain simulations is approximately 25 periods at the central frequency, in agreement with the estimation through band diagram.
Unidirectional light transmission is also demonstrated through finite-difference time-domain simulations, provided that waveguide and coupling lengths are equal. Forward and reverse transmittances of
71 and 0.3 %, respectively, are achieved at the central operation frequency in a 25-period system.
Parmak izi
Coupling between opposite-parity modes in parallel photonic crystal waveguides and its application in unidirectional light transmission' araştırma başlıklarına git. Birlikte benzersiz bir parmak izi | {"url":"https://research.bau.edu.tr/tr/publications/coupling-between-opposite-parity-modes-in-parallel-photonic-cryst","timestamp":"2024-11-14T21:53:32Z","content_type":"text/html","content_length":"56907","record_id":"<urn:uuid:b1381c2e-8fd2-475d-b02f-2b5c52914435>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00285.warc.gz"} |
Suppose that a population of bacteria triples every hour and starts with 400 bacteria. Find an expression for the number n of bacteria after t hours. n (t) = Use it to estimate the rate of growth of the bacteria population after 3.5 hours.
Suppose that a population of bacteria triples every hour and starts with 400 bacteria. Find an expression for the number n of bacteria after t hours. n (t) = Use it to estimate the rate of growth of
the bacteria population after 3.5 hours. (Round your answer to the nearest whole number.)
Get an answer to your question ✅ “Suppose that a population of bacteria triples every hour and starts with 400 bacteria. Find an expression for the number n of bacteria ...” in 📙 Mathematics if there
is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers | {"url":"https://educationexpert.net/mathematics/2414980.html","timestamp":"2024-11-07T02:58:44Z","content_type":"text/html","content_length":"22993","record_id":"<urn:uuid:5f6d1414-cef3-47b2-bbe1-c61df939abae>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00724.warc.gz"} |
Classify Solutions to Linear Equations (One, No, or Infinite Solutions)
Learning Objectives
• Classify solutions to linear equations
□ Solve equations that have one solution, no solution, or an infinite number of solutions
There are three cases that can come up as we are solving linear equations. We have already seen one, where an equation has one solution. Sometimes we come across equations that don’t have any
solutions, and even some that have an infinite number of solutions. The case where an equation has no solution is illustrated in the next examples.
Equations with no solutions
Solve for x. [latex]12+2x–8=7x+5–5x[/latex]
Show Solution
This is not a solution! You did not find a value for x. Solving for x the way you know how, you arrive at the false statement [latex]4=5[/latex]. Surely 4 cannot be equal to 5!
This may make sense when you consider the second line in the solution where like terms were combined. If you multiply a number by 2 and add 4 you would never get the same answer as when you multiply
that same number by 2 and add 5. Since there is no value of x that will ever make this a true statement, the solution to the equation above is “no solution.”
Be careful that you do not confuse the solution [latex]x=0[/latex] with “no solution.” The solution [latex]x=0[/latex] means that the value 0 satisfies the equation, so there is a solution. “No
solution” means that there is no value, not even 0, which would satisfy the equation.
Also, be careful not to make the mistake of thinking that the equation [latex]4=5[/latex] means that 4 and 5 are values for x that are solutions. If you substitute these values into the original
equation, you’ll see that they do not satisfy the equation. This is because there is truly no solution—there are no values for x that will make the equation [latex]12+2x–8=7x+5–5x[/latex] true.
Think About It
Try solving these equations. How many steps do you need to take before you can tell whether the equation has no solution or one solution?
a) Solve [latex]8y=3(y+4)+y[/latex]
Use the textbox below to record how many steps you think it will take before you can tell whether there is no solution or one solution.
Show Solution
b) Solve [latex]2\left(3x-5\right)-4x=2x+7[/latex]
Use the textbox below to record how many steps you think it will take before you can tell whether there is no solution or one solution.
Show Solution
Algebraic Equations with an Infinite Number of Solutions
You have seen that if an equation has no solution, you end up with a false statement instead of a value for x. It is possible to have an equation where any value for x will provide a solution to the
equation. In the example below, notice how combining the terms [latex]5x[/latex] and [latex]-4x[/latex] on the left leaves us with an equation with exactly the same terms on both sides of the equal
Solve for x. [latex]5x+3–4x=3+x[/latex]
Show Solution
You arrive at the true statement “[latex]3=3[/latex].” When you end up with a true statement like this, it means that the solution to the equation is “all real numbers.” Try substituting [latex]x=0[/
latex] into the original equation—you will get a true statement! Try [latex]x=-\frac{3}{4}[/latex], and it also will check!
This equation happens to have an infinite number of solutions. Any value for x that you can think of will make this equation true. When you think about the context of the problem, this makes
sense—the equation [latex]x+3=3+x[/latex] means “some number plus 3 is equal to 3 plus that same number.” We know that this is always true—it’s the commutative property of addition!
In the following video, we show more examples of attempting to solve a linear equation with either no solution or many solutions.
Solve for x. [latex]3\left(2x-5\right)=6x-15[/latex]
Show Solution
In this video, we show more examples of solving linear equations with either no solutions or many solutions.
In the following video, we show more examples of solving linear equations with parentheses that have either no solution or many solutions.
We have seen that solutions to equations can fall into three categories:
• One solution
• No solution, DNE (does not exist)
• Many solutions, also called infinitely many solutions or All Real Numbers
And sometimes, we don’t need to do much algebra to see what the outcome will be. | {"url":"https://courses.lumenlearning.com/aacc-collegealgebrafoundations/chapter/read-classify-solutions-to-linear-equations/","timestamp":"2024-11-10T23:54:05Z","content_type":"text/html","content_length":"58709","record_id":"<urn:uuid:3be76e66-b937-4df0-b011-9cf87c8c7efd>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00255.warc.gz"} |
Array indexing refers to any use of the square brackets ([]) to index array values. There are many options to indexing, which give numpy indexing great power, but with power comes some complexity and
the potential for confusion. This section is just an overview of the various options and issues related to indexing. Aside from single element indexing, the details on most of these options are to be
found in related sections.
Assignment vs referencing¶
Most of the following examples show the use of indexing when referencing data in an array. The examples work just as well when assigning to an array. See the section at the end for specific examples
and explanations on how assignments work.
Single element indexing¶
Single element indexing for a 1-D array is what one expects. It work exactly like that for other standard Python sequences. It is 0-based, and accepts negative indices for indexing from the end of
the array.
>>> x = np.arange(10)
>>> x[2]
>>> x[-2]
Unlike lists and tuples, numpy arrays support multidimensional indexing for multidimensional arrays. That means that it is not necessary to separate each dimension’s index into its own set of square
>>> x.shape = (2,5) # now x is 2-dimensional
>>> x[1,3]
>>> x[1,-1]
Note that if one indexes a multidimensional array with fewer indices than dimensions, one gets a subdimensional array. For example:
>>> x[0]
array([0, 1, 2, 3, 4])
That is, each index specified selects the array corresponding to the rest of the dimensions selected. In the above example, choosing 0 means that the remaining dimension of length 5 is being left
unspecified, and that what is returned is an array of that dimensionality and size. It must be noted that the returned array is not a copy of the original, but points to the same values in memory as
does the original array. In this case, the 1-D array at the first position (0) is returned. So using a single index on the returned array, results in a single element being returned. That is:
So note that x[0,2] = x[0][2] though the second case is more inefficient as a new temporary array is created after the first index that is subsequently indexed by 2.
Note to those used to IDL or Fortran memory order as it relates to indexing. NumPy uses C-order indexing. That means that the last index usually represents the most rapidly changing memory location,
unlike Fortran or IDL, where the first index represents the most rapidly changing location in memory. This difference represents a great potential for confusion.
Other indexing options¶
It is possible to slice and stride arrays to extract arrays of the same number of dimensions, but of different sizes than the original. The slicing and striding works exactly the same way it does for
lists and tuples except that they can be applied to multiple dimensions as well. A few examples illustrates best:
>>> x = np.arange(10)
>>> x[2:5]
array([2, 3, 4])
>>> x[:-7]
array([0, 1, 2])
>>> x[1:7:2]
array([1, 3, 5])
>>> y = np.arange(35).reshape(5,7)
>>> y[1:5:2,::3]
array([[ 7, 10, 13],
[21, 24, 27]])
Note that slices of arrays do not copy the internal array data but also produce new views of the original data.
It is possible to index arrays with other arrays for the purposes of selecting lists of values out of arrays into new arrays. There are two different ways of accomplishing this. One uses one or more
arrays of index values. The other involves giving a boolean array of the proper shape to indicate the values to be selected. Index arrays are a very powerful tool that allow one to avoid looping over
individual elements in arrays and thus greatly improve performance.
It is possible to use special features to effectively increase the number of dimensions in an array through indexing so the resulting array aquires the shape needed for use in an expression or with a
specific function.
Index arrays¶
NumPy arrays may be indexed with other arrays (or any other sequence- like object that can be converted to an array, such as lists, with the exception of tuples; see the end of this document for why
this is). The use of index arrays ranges from simple, straightforward cases to complex, hard-to-understand cases. For all cases of index arrays, what is returned is a copy of the original data, not a
view as one gets for slices.
Index arrays must be of integer type. Each value in the array indicates which value in the array to use in place of the index. To illustrate:
>>> x = np.arange(10,1,-1)
>>> x
array([10, 9, 8, 7, 6, 5, 4, 3, 2])
>>> x[np.array([3, 3, 1, 8])]
array([7, 7, 9, 2])
The index array consisting of the values 3, 3, 1 and 8 correspondingly create an array of length 4 (same as the index array) where each index is replaced by the value the index array has in the array
being indexed.
Negative values are permitted and work as they do with single indices or slices:
>>> x[np.array([3,3,-3,8])]
array([7, 7, 4, 2])
It is an error to have index values out of bounds:
>>> x[np.array([3, 3, 20, 8])]
<type 'exceptions.IndexError'>: index 20 out of bounds 0<=index<9
Generally speaking, what is returned when index arrays are used is an array with the same shape as the index array, but with the type and values of the array being indexed. As an example, we can use
a multidimensional index array instead:
>>> x[np.array([[1,1],[2,3]])]
array([[9, 9],
[8, 7]])
Indexing Multi-dimensional arrays¶
Things become more complex when multidimensional arrays are indexed, particularly with multidimensional index arrays. These tend to be more unusual uses, but they are permitted, and they are useful
for some problems. We’ll start with the simplest multidimensional case (using the array y from the previous examples):
>>> y[np.array([0,2,4]), np.array([0,1,2])]
array([ 0, 15, 30])
In this case, if the index arrays have a matching shape, and there is an index array for each dimension of the array being indexed, the resultant array has the same shape as the index arrays, and the
values correspond to the index set for each position in the index arrays. In this example, the first index value is 0 for both index arrays, and thus the first value of the resultant array is y[0,0].
The next value is y[2,1], and the last is y[4,2].
If the index arrays do not have the same shape, there is an attempt to broadcast them to the same shape. If they cannot be broadcast to the same shape, an exception is raised:
>>> y[np.array([0,2,4]), np.array([0,1])]
<type 'exceptions.ValueError'>: shape mismatch: objects cannot be
broadcast to a single shape
The broadcasting mechanism permits index arrays to be combined with scalars for other indices. The effect is that the scalar value is used for all the corresponding values of the index arrays:
>>> y[np.array([0,2,4]), 1]
array([ 1, 15, 29])
Jumping to the next level of complexity, it is possible to only partially index an array with index arrays. It takes a bit of thought to understand what happens in such cases. For example if we just
use one index array with y:
>>> y[np.array([0,2,4])]
array([[ 0, 1, 2, 3, 4, 5, 6],
[14, 15, 16, 17, 18, 19, 20],
[28, 29, 30, 31, 32, 33, 34]])
What results is the construction of a new array where each value of the index array selects one row from the array being indexed and the resultant array has the resulting shape (number of index
elements, size of row).
An example of where this may be useful is for a color lookup table where we want to map the values of an image into RGB triples for display. The lookup table could have a shape (nlookup, 3). Indexing
such an array with an image with shape (ny, nx) with dtype=np.uint8 (or any integer type so long as values are with the bounds of the lookup table) will result in an array of shape (ny, nx, 3) where
a triple of RGB values is associated with each pixel location.
In general, the shape of the resultant array will be the concatenation of the shape of the index array (or the shape that all the index arrays were broadcast to) with the shape of any unused
dimensions (those not indexed) in the array being indexed.
Boolean or “mask” index arrays¶
Boolean arrays used as indices are treated in a different manner entirely than index arrays. Boolean arrays must be of the same shape as the initial dimensions of the array being indexed. In the most
straightforward case, the boolean array has the same shape:
>>> b = y>20
>>> y[b]
array([21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34])
Unlike in the case of integer index arrays, in the boolean case, the result is a 1-D array containing all the elements in the indexed array corresponding to all the true elements in the boolean
array. The elements in the indexed array are always iterated and returned in row-major (C-style) order. The result is also identical to y[np.nonzero(b)]. As with index arrays, what is returned is a
copy of the data, not a view as one gets with slices.
The result will be multidimensional if y has more dimensions than b. For example:
>>> b[:,5] # use a 1-D boolean whose first dim agrees with the first dim of y
array([False, False, False, True, True])
>>> y[b[:,5]]
array([[21, 22, 23, 24, 25, 26, 27],
[28, 29, 30, 31, 32, 33, 34]])
Here the 4th and 5th rows are selected from the indexed array and combined to make a 2-D array.
In general, when the boolean array has fewer dimensions than the array being indexed, this is equivalent to y[b, …], which means y is indexed by b followed by as many : as are needed to fill out the
rank of y. Thus the shape of the result is one dimension containing the number of True elements of the boolean array, followed by the remaining dimensions of the array being indexed.
For example, using a 2-D boolean array of shape (2,3) with four True elements to select rows from a 3-D array of shape (2,3,5) results in a 2-D result of shape (4,5):
>>> x = np.arange(30).reshape(2,3,5)
>>> x
array([[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]],
[[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24],
[25, 26, 27, 28, 29]]])
>>> b = np.array([[True, True, False], [False, True, True]])
>>> x[b]
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[20, 21, 22, 23, 24],
[25, 26, 27, 28, 29]])
For further details, consult the numpy reference documentation on array indexing.
Combining index arrays with slices¶
Index arrays may be combined with slices. For example:
>>> y[np.array([0,2,4]),1:3]
array([[ 1, 2],
[15, 16],
[29, 30]])
In effect, the slice is converted to an index array np.array([[1,2]]) (shape (1,2)) that is broadcast with the index array to produce a resultant array of shape (3,2).
Likewise, slicing can be combined with broadcasted boolean indices:
>>> y[b[:,5],1:3]
array([[22, 23],
[29, 30]])
Structural indexing tools¶
To facilitate easy matching of array shapes with expressions and in assignments, the np.newaxis object can be used within array indices to add new dimensions with a size of 1. For example:
>>> y.shape
(5, 7)
>>> y[:,np.newaxis,:].shape
(5, 1, 7)
Note that there are no new elements in the array, just that the dimensionality is increased. This can be handy to combine two arrays in a way that otherwise would require explicitly reshaping
operations. For example:
>>> x = np.arange(5)
>>> x[:,np.newaxis] + x[np.newaxis,:]
array([[0, 1, 2, 3, 4],
[1, 2, 3, 4, 5],
[2, 3, 4, 5, 6],
[3, 4, 5, 6, 7],
[4, 5, 6, 7, 8]])
The ellipsis syntax maybe used to indicate selecting in full any remaining unspecified dimensions. For example:
>>> z = np.arange(81).reshape(3,3,3,3)
>>> z[1,...,2]
array([[29, 32, 35],
[38, 41, 44],
[47, 50, 53]])
This is equivalent to:
>>> z[1,:,:,2]
array([[29, 32, 35],
[38, 41, 44],
[47, 50, 53]])
Assigning values to indexed arrays¶
As mentioned, one can select a subset of an array to assign to using a single index, slices, and index and mask arrays. The value being assigned to the indexed array must be shape consistent (the
same shape or broadcastable to the shape the index produces). For example, it is permitted to assign a constant to a slice:
>>> x = np.arange(10)
>>> x[2:7] = 1
or an array of the right size:
>>> x[2:7] = np.arange(5)
Note that assignments may result in changes if assigning higher types to lower types (like floats to ints) or even exceptions (assigning complex to floats or ints):
>>> x[1] = 1.2
>>> x[1]
>>> x[1] = 1.2j
<type 'exceptions.TypeError'>: can't convert complex to long; use
Unlike some of the references (such as array and mask indices) assignments are always made to the original data in the array (indeed, nothing else would make sense!). Note though, that some actions
may not work as one may naively expect. This particular example is often surprising to people:
>>> x = np.arange(0, 50, 10)
>>> x
array([ 0, 10, 20, 30, 40])
>>> x[np.array([1, 1, 3, 1])] += 1
>>> x
array([ 0, 11, 20, 31, 40])
Where people expect that the 1st location will be incremented by 3. In fact, it will only be incremented by 1. The reason is because a new array is extracted from the original (as a temporary)
containing the values at 1, 1, 3, 1, then the value 1 is added to the temporary, and then the temporary is assigned back to the original array. Thus the value of the array at x[1]+1 is assigned to x
[1] three times, rather than being incremented 3 times.
Dealing with variable numbers of indices within programs¶
The index syntax is very powerful but limiting when dealing with a variable number of indices. For example, if you want to write a function that can handle arguments with various numbers of
dimensions without having to write special case code for each number of possible dimensions, how can that be done? If one supplies to the index a tuple, the tuple will be interpreted as a list of
indices. For example (using the previous definition for the array z):
>>> indices = (1,1,1,1)
>>> z[indices]
So one can use code to construct tuples of any number of indices and then use these within an index.
Slices can be specified within programs by using the slice() function in Python. For example:
>>> indices = (1,1,1,slice(0,2)) # same as [1,1,1,0:2]
>>> z[indices]
array([39, 40])
Likewise, ellipsis can be specified by code by using the Ellipsis object:
>>> indices = (1, Ellipsis, 1) # same as [1,...,1]
>>> z[indices]
array([[28, 31, 34],
[37, 40, 43],
[46, 49, 52]])
For this reason it is possible to use the output from the np.nonzero() function directly as an index since it always returns a tuple of index arrays.
Because the special treatment of tuples, they are not automatically converted to an array as a list would be. As an example:
>>> z[[1,1,1,1]] # produces a large array
array([[[[27, 28, 29],
[30, 31, 32], ...
>>> z[(1,1,1,1)] # returns a single value | {"url":"https://numpy.org/doc/1.15/user/basics.indexing.html","timestamp":"2024-11-01T19:55:23Z","content_type":"text/html","content_length":"44016","record_id":"<urn:uuid:7dab9295-af73-4dba-8105-8cdea4907095>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00235.warc.gz"} |
Fahrenheit to Kelvin
Converting from Fahrenheit to Kelvin
Converting from Fahrenheit to Kelvin can be tricky since both units have offsets and their increments are different. Kelvin is an absolute scale whereas Fahrenheit is a relative scale.
To convert Fahrenheit to Kelvin, you need to follow a three-step process. Firstly you will need to remove the Fahrenheit offset by subtracting 32. Next, divide this answer by 1.8 to obtain the
Celsius value. Finally add 273.15, the Kelvin offset, to convert the Celsius value to Kelvin.
For example, suppose we have a Fahrenheit temperature of 68°F. First subtract 32 giving 36. Now divide 36 by 1.8 to obtain the Celsius value of 20°C. Finally, add the Kelvin offset value of 273.15 to
get the resulting Kelvin value of 293.15K.
Converting from Fahrenheit to Kelvin can be more difficult than similar conversions- when you use the calculator on this page you can see the calculation steps underneath the result to better
understand the process.
Why convert from Fahrenheit to Kelvin?
Fahrenheit is commonly used in the United States for general temperature measurments whereas Kelvin is preferred in some scientific applications. Kelvin is an absolute temperature scale that starts
at absolute zero which is where all molecular motion ceases.
Converting Fahrenheit to Kelvin allows the values to become independent of the 32°F reference point of freezing water. This is useful in science such as physics, chemistry and engineering. A Kelvin
value is always positive removing the complication of negative values.
About the Fahrenheit scale
Fahrenheit is a relative temperature scale created by the Polish-German physicist Daniel Gabriel Fahrenheit. It is mainly used in the United States and is less common in science compared to the
Celsius (or Centigrade) scale.
Fahrenheit is based on the freezing and boiling points of water at standard atmospheric pressure with 32°F at freezing point and 212°F at boiling point.
Whilst Fahrenheit is still commonly used in the United States it is important to note that most of the world relies on Celsius.
About Kelvin
Kelvin is an absolute temperature measurement defined in the International System of Units (SI). It is named after the Scottish physicist William Thomson (Lord Kelvin) who studied the field of
thermodynamics. The Kelvin scale is based on absolute zero; the point at which all molecular motion ceases.
Unlike most other temperature scales, Kelvin does not have degrees as it is not a relative scale. The Kelvin scale is often used in physics, chemistry, and cosmology.
One advantage of Kelvin is that it does not have negative values making some calculations easier. This is useful in science calculations involving gases as it relates to the kinetic energy of
Is there an absolute scale related to Fahrenheit like Kelvin is related to Celsius?
Unlike the Celsius and Kelvin scales, Fahrenheit does not have an absolute zero point. Absolute zero is the lowest possible temperature, at which all molecular motion ceases. In the Celsius scale,
absolute zero is defined as 0 degrees Celsius, while in the Kelvin scale, it is defined as 0 Kelvin.
The Fahrenheit scale, however, does not have an absolute zero point. Instead, it is based on the freezing and boiling points of water. On the Fahrenheit scale, the freezing point of water is defined
as 32 degrees Fahrenheit, and the boiling point is defined as 212 degrees Fahrenheit. This means that the Fahrenheit scale is not directly related to an absolute scale like Kelvin is related to
Celsius. While the Celsius and Kelvin scales are based on the properties of water and have a clear reference point at absolute zero, the Fahrenheit scale is based on arbitrary points related to the
behavior of water at atmospheric pressure.
Rankine is a unit of temperature measurement in the absolute temperature scale, commonly used in engineering and thermodynamics. It is closely related to the Fahrenheit scale, which is primarily used
in the United States for everyday temperature measurements. The Rankine scale is an absolute temperature scale, meaning it starts at absolute zero, where all molecular motion ceases.
The Rankine scale is based on the Fahrenheit scale, with the same size degree and zero point. However, the zero point on the Rankine scale is set at absolute zero, which is equivalent to -459.67
degrees Fahrenheit. Therefore, to convert a temperature from Fahrenheit to Rankine, one simply needs to add 459.67 to the Fahrenheit temperature. Conversely, to convert a temperature from Rankine to
Fahrenheit, one subtracts 459.67 from the Rankine temperature.
What happens at absolute zero (0K)?
At absolute zero, 0 Kelvin (0K) or -273.15 degrees Celsius, the temperature is at the lowest possible point anything can possibly be. At this temperature the kinetic energy of atoms and molecules is
zero causing them to come to a complete standstill. All molecular motion ceases and matter becomes still.
Several amazing phenomena occur here. As there is no molecular motion there is no heat energy and this has significant implications for the physical properties of the substance. For example,
materials become very brittle and their electrical resistance becomes zero. Gases and liquids freeze into solids.
Scientists have never cooled anything down to absolute zero. However they have been able to see the effects of approaching absolute zero. This has provided insights into the behavior of matter and
have led to the understanding of superconductors and Bose-Einstein condensates. | {"url":"https://live.metric-conversions.org/temperature/fahrenheit-to-kelvin.htm","timestamp":"2024-11-06T15:28:06Z","content_type":"text/html","content_length":"77806","record_id":"<urn:uuid:3f31643d-a126-4aaa-bce5-899740bcad32>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00697.warc.gz"} |
Comments on Questions?: Standard DeviationTeacher or Student? Thanks for stopping by; I'...I've been reading your blog for a while, but t...To be honest, I've never really gotten into th...Maybe you do this already with sequences, but what...
tag:blogger.com,1999:blog-5964889903484807623.post3145295017349384887..comments2023-12-18T04:44:25.358-08:00David Coxhttp://www.blogger.com/profile/
06277427735527075341noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-5964889903484807623.post-59458407613418763122010-05-20T08:32:46.825-07:002010-05-20T08:32:46.825-07:00<b>Teacher or
Student?</b> Thanks for stopping by; I'm glad you're finding this blog helpful. <br /><br />I'll check out your rule...thanks.David Coxhttps://www.blogger.com/profile/
06277427735527075341noreply@blogger.comtag:blogger.com,1999:blog-5964889903484807623.post-72982076200414665862010-05-17T22:05:18.684-07:002010-05-17T22:05:18.684-07:00I've been reading your blog
for a while, but this is my first post. Just wanted to let you know that I'm enjoying and learning.<br /><br />@Calc Dave: I think of "never do any math" as don't simplify. "
Simplifying" is actually one of my pet peeves and I try my best to never use the term (although I sometimes fail because it is quite ingrained). Instead I talk about different forms and the form
for the context that is most informative. So in your case, 2+1+2+3 is much more informative than 8. <br /><br />@David: You mentioned that your students were looking for a different pattern that
worked for your 3rd sequence. It's ugly, but if you consider 1 to be the 0th term, (1/3)(590x^3-1770x^2+1210x+3) works. I'd give you the "unsimplified" expression, but that would
take away all the fun of how I figured this out. :)Avery Pickfordhttps://www.blogger.com/profile/
10433339146333801163noreply@blogger.comtag:blogger.com,1999:blog-5964889903484807623.post-36736835123977128992010-04-23T17:57:14.403-07:002010-04-23T17:57:14.403-07:00To be honest, I've never
really gotten into the notation in sequences with my 7th graders. Most of what happend in this lesson just kinda happened. I saw where they were taking it, I liked it and we went with it. <br /><br
/>I like what you describe. Re-writing the terms using what they all have in common will make defining a variable much easier. Thanks.David Coxhttps://www.blogger.com/profile/
06277427735527075341noreply@blogger.comtag:blogger.com,1999:blog-5964889903484807623.post-9964501371938604842010-04-22T15:03:32.316-07:002010-04-22T15:03:32.316-07:00Maybe you do this already with
sequences, but what I tell the students that helps them "see the light" on writing formulas (recursive or not) is that they should NEVER do any math once they see the pattern. So, your
first sequence would end up being 2, 2+1, 2+1+2, 2+1+2+3, ... Once you break it down into the pattern, then it's much easier to connect the subscript to the current term, I think.<br /><br />I
also like the "Give two possible answers for the next number in the sequence given the pattern for the first four" problem. I end up telling them that if you are only given a finite number
of terms in a sequence, you can NEVER definitively say what the next term will be. Even 1, 2, 3, 4, ... could be a sequence that continues to just repeat 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, ... and
you just happened to stop at the worst time possible.CalcDavehttps://www.blogger.com/profile/14039458440867020542noreply@blogger.com | {"url":"https://coxmath.blogspot.com/feeds/3145295017349384887/comments/default","timestamp":"2024-11-06T09:24:39Z","content_type":"application/atom+xml","content_length":"11077","record_id":"<urn:uuid:20a08ea2-f7d4-488e-90bf-44990cdb27c1>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00260.warc.gz"} |
Bending of a Curved Beam Numerical Results | Ansys Courses
Now we will examine the simulation results from Ansys. The Ansys model uses the 2D plane stress approximation.
Before we dive in to the solution, let's take a look at the mesh used for the simulation. In the outline window, click Mesh to bring up the meshed geometry in the geometry window.
Only one-half of the geometry is modeled using symmetry constraints, which reduces the problem size. Look to the outline window under "Mesh". Notice that there are two types of meshing entities: a
"mapped face meshing" and a "face sizing". The "mapped face meshing" is used to generate a regular mesh of quadrilaterals. The face sizing controls the size of the element edges in the 2D "face".
Okay! Now we can check our solution. Let's start by examining how the beam deformed under the load. Before you start, make sure the software is working in the same units you are by looking to the
menu bar and selecting Units > US Customary (in, lbm, lbf, F, s, V, A). Now, look at the Outline window, and select Solution > Total Deformation.
The colored section refers to the magnitude of the deformation (in inches) while the black outline is the undeformed geometry superimposed over the deformed model. The redder a section is, the more
it has deformed while the bluer a section is, the less it has deformed. For this geometry, the bar is bending inward and the largest deformation occurs where the moment is applied, as one would
intuitively expect.
Click Solution > Sigma-theta in the outline window. This will bring up the distribution for the normal stress in the theta direction.
Sigma-theta, the bending stress, is a function of r only as expected from theory. It is tensile (positive) in the top part of the beam and compressive (negative) in the bottom part. There is a
neutral axis that separates the tensile and compressive regions. The bending stress, Sigma-theta, is zero on the neutral surface. We will use the probe to locate the region where the bending stress
changes from tensile to compressive. To find the neutral axis, let's first enlarge the geometry. Do this by clicking the Box Zoom tool
We will now look at Sigma-theta along the symmetry line. Click Solution > Sigma-theta along symmetry in the outline window to bring up the stress distribution at the middle of the bar.
Look at the color bar to see the maximum and minimum stresses. The maximum theta-stress is 1697.63 psi and the minimum theta-stress is -1916.2 psi.
In the outline window, click Solution > Sigma-r. This will bring up the distribution for the normal stress in the r-direction.
Looking at the distribution, we can see that the stress varies only as a function of r as expected. The magnitude of Sigma-r is much lower than Sigma-theta (this is why Winkler-Bach theory assumes
Sigma-r =0). Also, we can see that there is a stress concentration in the area where the moment is applied. In the theory, this effect is ignored. To further examine the Sigma-r, let's look at the
variation along the symmetry line. Click on Solution > Sigma-r along symmetry. This solution is the normal stress in the r-direction at the midsection of the beam.
Looking at the color bar again, we can see that the maximum r-stress is -.110 psi, and the minimum r-stress is -82.302 psi. At r=a and r=b, Sigma-r ~ 0 as one would expect for a free surface.
In the details window, click Solution > Tau-r-theta to bring up the stress distribution for shear stress
Hover the probe tool over points on the geometry far from the moment. You will notice that the stress is on the order of 10e-7. For a beam in pure bending, we assume that the shear stress is zero.
However, Ansys Mechanical does not make this assumption: it calculates a value for shear stress at every point on the beam. Therefore, it is reassuring that the shear stress is almost negligible,
which reinforces our assumption that it is zero.
Solution at r = 11.5 Inches
Now that we have a good idea about the stress distribution, we will look specifically at solving the problem in the problem specification. First, we will look at the stress in the r-direction at r =
11.5 inches. In the outline window, click Solution > Sigma-r at r =11.5. This will bring up the stress in the r-direction along the path at r = 11.5 inches (from the center of curvature of the bar).
In the window below, there is a table of the stress values along the path. To find the value of sigma-r at r = 11.5 in, we want to look far away from the stress concentration region due to the
moment. The path is defined in a counterclockwise direction, so looking at the last value of the table should tell us the stress at r = 11.5 inches at the midsection of the bar. This value of sigma-r
is -57.042 psi.
Now, we will do the same for the stress in theta direction to determine sigma-theta at r = 11.5 inches. In the outline window, click Solution > Sigma-theta at r =11.5. This will bring up the stress
in the theta-direction along the path at r=11.5 inches.
Look again at the table containing the stresses along the path. Look to the bottom of the table to find the stress in the theta-direction at the midpoint of the bar. We find that sigma-theta at this
point is 910.950 psi. Compare this to what you would expect from curved beam theory.
Finally, we will examine the shear stress at r = 11.5 in. In the outline window, click Solution > Tau-r-theta at r =11.5.
Again, look at the bottom of the table. You will find that the shear stress is very small at this point as we mentioned above.
Now that we have our results from the Ansys simulation, let's compare them to the theory calculations. Below is chart comparing the values found in Ansys Mechanical, and through calculations using
the Elasticity Theory, Winkler-Bach Theory, and Straight Beam Theory (Note: all stress values are in psi)
Elastic Theory Winkler Bach Theory Straight Beam Theory Ansys Percent Difference (Elastic Theory) Percent Difference (Winkler Bach Theory) Percent Difference (Beam Theory)
Maximum Sigma_r 0.00 N/A 0 -0.11 N/A N/A N/A
Minimum Sigma_r -82.21 N/A 0 -82.30 0.11% N/A N/A
Maximum Sigma_theta 1697.00 1696.40 1800 1697.63 0.04% 0.07% 5.85%
Minimum Sigma_theta -1917.00 -1915 -1800 -1916.20 0.04% 0.03% 6.25%
Sigma_r at r = 11.5 -56.93 N/A 0 -57.04 0.20% N/A N/A
Sigma_theta at r = 11.5 910.70 911.14 900 910.95 0.03% 0.02% 1.21%
Now, let's see how the stress distributions vary along the beam for each theory.
First, let's see how the Elastic Theory compared to the Ansys solution (you can right-click on the tabular data and export it in Excel or Text format to make the plots below):
From what we can see from the graphs, the Elastic Theory matched the Ansys solution very well. The same can be said for the Winkler-Bach theory:
When we approximate the beam as a straight beam, the analytical solution deviates slightly from the Ansys solution.
Now that we have gone through a simulation for bending of a curved beam, it is time to see if you can do the same on your own! | {"url":"https://innovationspace.ansys.com/courses/courses/bending-of-a-curved-beam-using-ansys-mechanical/lessons/numerical-results-lesson-3/","timestamp":"2024-11-02T05:35:41Z","content_type":"text/html","content_length":"183481","record_id":"<urn:uuid:8066f912-301e-4824-a672-657fc103473f>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00712.warc.gz"} |
Circles and stars
A thinking mathematically context for practise resource focused on developing flexible multiplicative strategies and renaming using place value knowledge.
Adapted from Marilyn Burns – About Teaching Mathematics, 2015.
Syllabus outcomes and content descriptors from Mathematics K–10 Syllabus (2022) © NSW Education Standards Authority (NESA) for and on behalf of the Crown in right of the State of New South Wales,
Collect resources
You will need:
• playing cards (we used 2, 5 and 10 only)
• a dice
• paper
• markers or pencils.
Watch Circles and stars video (12:09).
Hello Barbara.
[Screen shows 2 pieces of A4 paper, playing cards, and a marker.]
Hello Michelle. How are you?
I am great. How are you?
I'm very well.
We are going to play a game today from Marilyn Burns called How many stars?
So, we need to organize our game board to start with.
So, with our game board, we need to make eighths.
So, one way that I make eighths is, I halve my paper. You can halve it a different way if you want.
You don't have to do the same. And then from my half, if I halve it again and then halve it again, I'm now quartering each half and that will make me have eighths.
[Michelle folds the paper on the right length way in half, and Barbara folds hers on the left width way in half. They both now fold their paper in half and then in half again.]
Oh, maybe I could go this way?
Oh well, a bit too skinny.
Too skinny. I think I'll go this way then.
You know what's interesting about that, is that we folded them in a different order.
[They both unfold their papers and place then on the table, the folds have created 8 square boxes.]
Yes, but we still got the same.
And you know what else it looks like? An array.
It does.
Look, it looks like 4 twos or 2 fours. Yeah. Our game's called circles and stars so we're gonna write circles and stars.
[The 2 A4 pieces of paper are turned to a portrait orientation. In the top left hand corner they write ‘circles and stars’.]
Okay. Okay.
Okay, so we are using playing cards 2, 5 and 10 today and that's gonna tell us how many is in each of our groups.
[Michelle picks up the playing cards which only has 2, 5 and 10 cards in the pack.]
And we roll the dice to say, how many groups do we need? So here, you go first.
Roll the dice.
So, 2 groups.
[Barbara rolls the dice and rolls 2. Michelle shuffles the cards.]
Two groups.
So, draw your 2 groups.
Two circles.
Two groups.
In this square any square.
In your, or one of those.
Yep ok.
We don’t use that one.
We don’t use this one, oh ok. Ok.
Oh, one of those?
Yeah, we don't use that one
Okay and then pick a card and that tells you how many.
[Barbara draws 2 circles on her piece of paper. She draws a number 10 from the deck.]
Now if you know it, you don't have to draw it. You can just explain how many 2 tens would be.
Okay so 2 tens would be 20.
Mm, cause of place value
So just rename it?
Yes, so just put 20 here cuz you know this.
[Barbara writes 20 under the 2 circles she drew.]
And then if we don't know it, we can use it to help us.
[Michelle rolls the dice, and she rolls one.]
One group.
[Michelle draws a card from the deck and draws a 10.]
One 10.
So, so I actually know that one 10 is 10, so, and that's 10. So, I don't need to draw them.
[Michelle draws a circle on her paper and writes 10 in the middle and 10 under the circle.]
Should I label mine as 10 and 10?
[Barbara writes 10 in each of her circles.]
Yes, that's a good idea.
Okay, your go.
[Barbara rolls the dice and rolls a one. She draws a larger circle on her paper in the top right-hand corner.]
Oh rats!
So, I'm guessing by that, that's not a good thing, right? We want to have lots.
Well, it could be if you get a 10 because 10 is a good move. But imagine now you get one 2.
Okay, so I want as many?
You want as many stars as possible at the end.
Oh, so these are stars? So, I should put in 20 stars.
Oh, that's a good idea! I should put stars too.
10 stars, 10 stars is 20 stars.
Okay, so now I've got one 2. Okay so that's just 2 stars.
[Barbara draws 1 star each under the tens in the circle and one star next to the 20, she draws a card from the deck and draws 2. She now writes 2 in her circle and underneath she writes 2 again. She
draws a star next to both digits.]
Yes, it's a known fact so you don't have to draw it.
Okay, my go. Oh 3. Fives. Okay, so if I didn't know 3 fives, I could draw my 3 and then I could draw my 5 stars.
[Michelle rolls the dice and rolls a 3. She draws a 5 from the card deck. She draws 3 circles on her paper and draws 5 stars in the first circle.]
Two, 3, 4, 5 and what I might do is now imagine in my mind's eye that there's 5 here, 5 here, and 5 here.
And what I do know is it that 5 and 5 combines to make 10.
And one 10 and 5 more is 15.
Okay, yep.
So, I don't have to draw everything.
[Michelle points with her pen to the first circle, then the second and third circles. She then writes 15 under the circles.]
Just enough.
But I can if I need to. I just draw enough to help me, you know, work out how many stars it would be in total. So that would be 15.
Okay. 15 stars.
Where do we put that? At the bottom after you've used it.
[Barbara places Michelle’s card at the bottom of the deck and then rolls the dice. She rolls a 2 and draws a 10.]
Okay, so 2. Tens.
Oh, 2 tens is nice. Also, nice because you can just use renaming with place value.
Exactly, if you know place value, then you instantly 2 tens can be renamed as 20.
[Barbara draws 2 circles and writes 10 in each circle and draws a star underneath them. Underneath the circles she writes 20 and draws a star.]
Okay, my go! Oh 5 is good.
[Michelle rolls the dice and rolls a 5.]
Imagine if you get a 10.
That would be good, imagine if I got a 2.
Ok let's see.
Oh yes! Yes! So, I can draw one, 2, 3, 4, 5, like a dice.
[Michelle draws a card from the deck and draws a 10.]
And each one is worth 10, but I actually just know that that's renamed as 50, because of place value knowledge.
[Michelle draws 5 circles and draws a star in each circle and underneath she writes 50 and draws a star next to it.]
Okay now I want 6 tens. Re-roll or…?
You can re-roll because I saw what it was.
All right, 2 fives. Well, I know that I know 2 fives is 10.
[Barbara rolls the dice. She rolls 2 and she draws a 5 from the deck of cards. She draws 2 circles.]
It's just...
How do you know it? Is this a fact you know?
It's a fact I know, but it's also my 2 hands.
Yes, because 2 fives together.
[Michelle displays her hands open, indicating 5 fingers on each hand add up to 10.]
Yeah, and also the 2 rows of a 10 frame.
Because 5 here and 5 there. 10.
Yeah. Okay.
Okay so, 10 stars. Oh, I should do 5.
[Barbara writes 5 in each of her circles and draws a star next to the digit. Underneath she writes equals 10.]
Okay, oh 6 is nice! Imagine if I got 10.
[Michelle rolls the dice and rolls a 6.]
Imagine if you get one.
No, there's no one in there, there's only 2, 5 or 10.
Come on 2.
Oh, rats!
[Michelle draws a 2.]
So, um, here's 6 and there's 2 in each one. And actually, I know that's 12 because if this was a 10 frame, say that moved to there and that moved to there.
You know, my 10 frame is going in this orientation. What I know is that for each dot there, there's actually 2.
[Michelle draws 6 smaller circles in a dice pattern, she draws a ten frame around 3 of her circles and draws arrows to signify the 2 circles going into the 10 frame. She draws an additional circle to
show the one circle left over.]
And then I'd have one more left over and I know that it's 12.
Okay. I like how you explained that. It made sense. Okay, 6. Fives. Not bad.
[Barbara rolls a 6 and draws a 5 from the deck of cards.]
Not bad, not bad.
Oh, I know how you could work that out.
Cuz, you could say if you halved 6, that would be 3.
[Michelle places her finger over 3 of the dots on the dice, showing only 3 dots, and then points to the 5 of diamonds on the card.]
And doubled 5 you get 10.
And then you just rename it.
Three 10s.
Because you can use your fives to work out tens.
Yeah, and I like how you actually, cuz I do that sometimes, but I don't always half and double.
Sometimes I just I get the result and then I halve it. I like the way you did that.
So how would I draw it then? Would I draw as 6 fives, or would I draw it as 3 tens?
Good question.
Because that's how we worked it out?
Maybe, can I draw on your paper?
Yes please.
Maybe you could draw like this, and that. So, I had 5, but I thought of them together.
[Michelle draws a rectangle on Barbara’s paper. She then draws 2 circles in the rectangles with 5 dots each in each circle. She then draws another 2 rectangles the same.]
Okay, oh I like that.
And then, there's, yeah, you had them like this and this, and then you know I'm actually thinking about fives as tens, and I only need 3 of them.
Okay, I really like that.
Okay I don't have to draw them all cause I know it now. So, this, so then we said that was 30 stars.
So, each one here it was 10 stars, 10 stars and 10 stars.
[Barbara now writes 10 at the bottom of each rectangle, and 30 underneath all of them.]
Okay, you've got one go, I've got 2 goes left. We lose, we use the last box to help us calculate.
Ah, 2. 2 twos. Well, that's 4 because I know this. But it would look like this, 2 twos, which is 4 stars all together. Your go.
[Michelle draws 2 circles with 2 dots in the circle. Underneath she writes 4.]
Okay. Oh, you're writing the word stars, I'm just drawing a star! Six! Yes! Twos.
[Barbara rolls 6 and draws 2 from the deck of cards.]
Oh, you go from this heightened state of yes 6, oh twos.
Oh, and well we've done this one before.
Still better than 6 ones.
Exactly, well twice as good. Okay, so, I know this number fact.
But otherwise, I could just think of it, you know, the idea of double 5 is 10 and then 2 more. So, um, with 2 in each one is 12 stars.
[Barbara draws 6 circles and writes 2 in the first one. Underneath she writes equals 12.]
Okay alright, last go for me. I need a good roll.
You've got that 50. So, I think you're ok.
That's true. Three fives. So, you know how last time I drew it and I said I could work that out as 5 and then visualize?
[Michelle rolls the dice and rolls a 3. She draws a 5 from the deck of cards.]
This time what I could think of is double 5, which is 10 and 1 more 5.
Because when you do your threes, you can work out double plus, plus one more. So how would I draw that?
I could do the same idea here. Oh, I know. I could go 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, and then one more group of 5.
[Michelle draws 2 rows of 5 circles she then draws a rectangle around the circles. Another row of 5 circles is drawn underneath the rectangle.]
Oh good.
And then there's the threes, look.
[Michelle draws a circle around each of the 3 columns of threes. She writes 15 underneath.]
Yeah, that's good recording.
Hmm, so that's 15 altogether.
So, now Barbara, what we need to do is work out how many stars we had and the person with the most, number of stars is declared the winner!
[Michelle circles the text and drawings on the paper and taps her fingers.]
Okay, great.
So, so now the fun part comes, right because we can use some really cool strategies. But we can work together to help each other.
So, what are you thinking when you look at your numbers? Because I might, what I might do is write down all of mine.
So, I've got 10 joined with 4, with 12, with 15, with 15 and with 50. And that helps me start to look for things.
[Michelle removes the dice and playing cards. In the last square of her game board she writes, 10 plus 4 plus 12 plus 15 plus 15 plus 50.]
Oh, I found something.
[Barbara writes 2 plus 20 plus 20 plus 10 plus 30 plus 12.]
Oh yeah, I can see some stuff too.
Okay. I ran out of space, but I'll put it down here. Okay, so when I wasn't, when I was writing them out, I realized that 20 and 20 and 10 actually makes 50.
[Barbara circles the 20 plus 20 plus 10 and writes the number 50 on top. Underneath she writes 2 plus 50 plus 30 plus 12.]
Yeah. So, all of that together, that's 50 there.
Yeah, so you could rewrite that now as 2, plus 50, plus 30, plus 12, because of our equivalent. If that helps?
No, that helps me, because then it's a much, it's a shorter number sentence as well. Okay, so.
Oh, now I can see something else.
Well, what I'm thinking, do you want to tell me what you're thinking? Or should?
Yeah, because I was thinking like 5 tens and 3 tens is 8 tens.
So, then it's 2 plus 8 tens which is 80, plus 12.
And that's even nicer to work with.
Yeah, or even, even get that 10 from here.
Oh yeah.
So, 5 tens and 3 tens is 8 tens and then 9 tens.
So, I've got 9 tens. Plus, 2, plus 2! Plus 2.
[Barbara writes underneath 9 tens plus 2 plus 2 and underneath that she writes equals 94.]
Oh, that is nice!
Okay and then that's 9 tens and 4, which we would rename as 94.
I think you've won!
So, but let's have a look. But it is close.
Very close.
So, so what I know actually, is that double 15 is 30.
I just happened to; I don't know why I know that, but I do. So, I'm gonna go.
I think you've won you know.
Well let's see. 10, plus 4, plus 12 plus 30, oh now I feel more confident, plus 50.
[Michelle writes 10 plus 4 plus 12 plus 30 plus 50 underneath her previous row.
Yeah, because now what I can see is that there's another hidden 50.
In there. So, if I take the 10 and the 10.
Oh yeah.
So, one 10 and one 10. Is 2 tens. Plus 3 tens is 50. So that would be 50, plus 4, plus 2, plus 50.
[Michelle writes 50 plus 4 plus 2 plus 5 underneath her previous equation.]
You won.
And then 5 tens and 5 tens is 10 tens, which you call 100. Plus 4, plus 2 and that's 106.
Aww but it was close.
[Michelle writes 100 plus 4 plus 2 equals 106 underneath.]
It was pretty close!
Only 12 away.
This one was really good.
That was a good lucky go!
Over to you mathematicians to enjoy Marilyn Burns' circles and stars!
[End of transcript]
• Divide your paper into eighths.
• Roll a dice to determine how many circles (groups) you need to make.
• Turn over a playing card (or roll the dice again) to determine how many stars to add into each circle.
• Determine how many stars there are in total. You can draw all or some of the stars in each circle - you only need to draw what you need to help you work out the product.
• Continue taking turns until each player has had 6 turns each.
• Work together to work out who has the most starts altogether. | {"url":"https://education.nsw.gov.au/teaching-and-learning/curriculum/mathematics/mathematics-curriculum-resources-k-12/thinking-mathematically-resources/mathematics-s1-circles-and-stars","timestamp":"2024-11-09T02:34:49Z","content_type":"text/html","content_length":"202502","record_id":"<urn:uuid:49748c73-0aff-4a58-83a6-8b07b51641fa>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00440.warc.gz"} |
Lesson 7
Similar Polygons
Problem 1
Triangle \(DEF\) is a dilation of triangle \(ABC\) with scale factor 2. In triangle \(ABC\), the largest angle measures \(82^\circ\). What is the largest angle measure in triangle \(DEF\)?
Problem 2
Draw two polygons that are similar but could be mistaken for not being similar. Explain why they are similar.
Problem 3
Draw two polygons that are not similar but could be mistaken for being similar. Explain why they are not similar.
Problem 4
These two triangles are similar. Find side lengths \(a\) and \(b\). Note: the two figures are not drawn to scale.
Problem 5
Jada claims that \(B’C’D’\) is a dilation of \(BCD\) using \(A\) as the center of dilation.
What are some ways you can convince Jada that her claim is not true?
Problem 6
1. Draw a horizontal line segment \(AB\).
2. Rotate segment \(AB\) \(90^\circ\) counterclockwise around point \(A\). Label any new points.
3. Rotate segment \(AB\) \(90^\circ\) clockwise around point \(B\). Label any new points.
4. Describe a transformation on segment \(AB\) you could use to finish building a square. | {"url":"https://curriculum.illustrativemathematics.org/MS/teachers/3/2/7/practice.html","timestamp":"2024-11-14T10:46:27Z","content_type":"text/html","content_length":"77449","record_id":"<urn:uuid:748e9b25-593b-4aa4-ac3a-74ede95abcde>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00546.warc.gz"} |
I thoroughly enjoy teaching. In Spring 2023, I offered a 4-week course (8 lectures in total) on fair machine learning (within the Causal Inference II course taught by Elias Bareinboim). Below, you
can find the course outline and all the necessary materials (including slides, lecture videos, and vignettes for software examples).
Fairness Course at Columbia
Lectures 1-2 (Week 1)
(L1) Theory of decomposing variations within the total variation fairness measure TV[x₀, x₁](y). Explaining the Fundamental Problem of Causal Fairness Analysis. Introducing contrasts and the
structural basis expansion for causal fairness measures. Introducing the Explainability Plane. Introducing the simplified cluster causal diagram called the Standard Fairness Model.
(L2) Measures in the TV family. Using contrasts in practice to measure discrimination. Structure of the TV family. Organizing the existing causal fairness measures into the Fairness Map.
Slides: Lecture 1, Lecture 2
Video: Lecture 1, Lecture 2
Lectures 3-4 (Week 2)
(L3) Identification of causal fairness measures from observational data. Estimation of causal fairness measures based on doubly-robust methods and double debiased machine learning.
(L4) Relationship to key existing notions in the fairness literature. Understanding where counterfactual fairness falls in the Fairness Map. Implications of causal fairness for the Fairness Through
Awareness framework. Connecting notions of predictive parity and calibration with causal fairness.
Slides: Lectures 3+4
Video: Lectures 3+4
Lectures 5-6 (Week 3)
(L5) Introducing the three key tasks of causal fairness analysis: (1) bias detection; (2) fair prediction; (3) fair decision-making. Discussing Task 1 of bias detection in depth with applications,
including the United States Government Census 2018 dataset, COMPAS dataset & other synthetic examples.
(L6) Discussing Task 2 of fair prediction. Proving the Fair Prediction Theorem that demonstrates why statistical notions of fairness are not sufficient in general.
Slides: Lectures 5, Lectures 6
Video: Lecture 5, Lecture 6
Vignettes: Census Task 1 Vignette, COMPAS Task 1 Vignette, COMPAS Task 3 Vignette
Lectures 7-8 (Week 4)
(L7) Moving beyond the Standard Fairness Model. Discussing how to extend causal fairness analysis to arbitrary causal diagrams. Discussing variable-specific and path-specific notions of indirect
effects. Discussing identifiability and estimation of variable-specific indirect effects.
(L8) Discussing decompositions of spurious effects. Introducing the partial abduction and prediction procedure. Introducing partially abducted submodels. Proving variable-specific spurious
decomposition results for Markovian causal models. Proving variable-specific spurious decomposition results for Semi-Markovian causal models.
Slides: Lectures 7+8
Video: Lectures 7+8
ETH Zurich
I was also involved in teaching during my PhD at ETH Zurich. Below is the list of courses for which I was the course assistant: | {"url":"https://www.cs.columbia.edu/~dplecko/","timestamp":"2024-11-10T09:38:51Z","content_type":"text/html","content_length":"27064","record_id":"<urn:uuid:34888623-3098-4503-a584-6c51cc13533a>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00813.warc.gz"} |
Ch. 6 Introduction - Calculus Volume 3 | OpenStax
Chapter Outline
Hurricanes are huge storms that can produce tremendous amounts of damage to life and property, especially when they reach land. Predicting where and when they will strike and how strong the winds
will be is of great importance for preparing for protection or evacuation. Scientists rely on studies of rotational vector fields for their forecasts (see Example 6.3).
In this chapter, we learn to model new kinds of integrals over fields such as magnetic fields, gravitational fields, or velocity fields. We also learn how to calculate the work done on a charged
particle traveling through a magnetic field, the work done on a particle with mass traveling through a gravitational field, and the volume per unit time of water flowing through a net dropped in a
All these applications are based on the concept of a vector field, which we explore in this chapter. Vector fields have many applications because they can be used to model real fields such as
electromagnetic or gravitational fields. A deep understanding of physics or engineering is impossible without an understanding of vector fields. Furthermore, vector fields have mathematical
properties that are worthy of study in their own right. In particular, vector fields can be used to develop several higher-dimensional versions of the Fundamental Theorem of Calculus. | {"url":"https://openstax.org/books/calculus-volume-3/pages/6-introduction","timestamp":"2024-11-04T18:58:26Z","content_type":"text/html","content_length":"307658","record_id":"<urn:uuid:f585b623-9c18-4b8e-9f4c-5ebbf0946120>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00884.warc.gz"} |
Logical, Comparision, Assignment and Arithmetic Operators in R | The Data Hall
Logical, Comparision, Assignment and Arithmetic Operators in R
In R, operators allow you to perform various tasks, from basic arithmetic operations to data manipulation and comparison tasks. There are different kinds of operators in R, which include arithmetic,
logical, assignment, and comparison operators etc.
Assignment Operators in R
Assignment operators are used to assign values to variables in R. They include the assignment operator “<-” and “=” the right assignment operator ->. The assignment operator “<-” is simply used to
assign a value to a variable. For instance, if we have a variable named z, and we need to assign it a value 5, we use the following command
z<- 5
Similarly, the assignment operator = is an alternative to <- and is also used to assign values to variables.
z= 5
The right assignment operator assigns the value to a variable from right to left. It can be used in the following way in a command
5 -> z
Comparison Operators
We use comparison operators in R to compare values. These comparison operators include greater than “>” , less than “<“, greater than or equal to “>=”, less than or equal to “<=”, equal to “==” and
Not equal to “!=”.
Now, if we want to check whether certain numbers or greater or less than others, we can use the following commands.
less <- 5 < 9 greater <- 12 > 20
Similarly, the comparison operators equal to and NOT equal to can be checked from the following command.
13!= 13
Here, the “==” sign shows equal to and “!=” sign shows not equal. Remember to use the signs correctly, or R will show an error for the command.
Logical Operators
The logical operators allow us to perform logical operations on logical values. The most basic logical operators are TRUE or just (T) and FALSE or just (F). These two are often used when it is
required to check the validity of an object, i.e. if 10 is less than 3 or not. To check this in R, the following command is used
The output of the above command will be FALSE, because 10, by no doubt, is greater than 3.
Some of the other commonly used logical operators are Logical AND “&” Logical OR “|” Logical NOT “!”.
The logical operators including AND “&” OR “I”, and NOT “!” can be used to compare the values. The & operator used for logical AND operations returns TRUE if both conditions on the left and right are
true. The | operator is used for logical OR operations. It returns TRUE if at least one of the conditions on the left or right is true. Similarly, the ! operator is used for logical NOT operations.
It negates a logical value, turning TRUE into FALSE and vice versa.
and <- 10 > 6 & 2 < 3 or <- 3 < 5 | 5 > 1 not <- !(2==3)
The output for the above commands will be following
We have explored the usage of different logical operations using simple numbers. What if we have a data frame containing numbers, and characters too? Let’s say we have the information of three
students, including their student ID’s, age, GPA, and their pass or fail status, and we want to use logical operators on it for filtering and sorting data. To create this data set, we use the
following command
student_data <- data.frame( StudentID = 1:3, Age = c(18, 20, 19), GPA = c(3.5, 3.2, 2.8), Pass = c(TRUE, TRUE, FALSE))
Once the above command is executed, the data frame has been created.
Now we perform different kind of logical operations and comparison operations on this data frame. Let’s say we just want to select the students that are above a certain age, i.e. 19 or above. To
perform this operation, we use the following command
old_stud <- stu_data[stu_data$Age > 19, ] or old_stud <- stu_data[stu_data$Age >= 19, ]
The first command just selects the student having age above than 19, while the second command select students of the age equal to and above than 19. The output of the second command is shown as below
Similarly, if we want to filter out students based on their GPA, we can also use logical operators. For example, if only student of 3.5 GPA are required, we use the above data in a command in
following way.
Multiple logical operators can also be used in a command, to filter multiple options from the data set. Let’s say we want to select students who have higher GPA and have passed the exams. We can
filter these students using logical AND operator (&) to combine these conditions in a command, in the following way
students <- stu_data[stu_data$GPA >=3.5 & stu_data$Pass == TRUE, ]
By using the above command, student fulfilling the above requirement will be filtered only.
Arithmetic Operators
Arithmetic operators are used to perform basic mathematical operations in R. They include addition (+), subtraction (-), multiplication (*), division (/), exponentiation (^), modulus (%%), and
integer division (%/%). We try these operators by using different examples.
Let’s say we have two objects by the name of x and y. We create these objects in R using following commands
x <- 15
y <- 9
The addition operator + is used to add two numbers or concatenate vectors. So, for the above example, we add x and y by using following command
add <- x + y
Similarly, we carry out subtraction, multiplication, division, exponential and integer division by using the following commands.
subtract <- x - y multiple <- x * y division <- x / y exponent<- x ^ y integer<- x %/% y
The output for all these commands is given below, showing that arithmetic operations have been carried out correctly.
Miscellaneous Operators in R
There are other operators in R too, which are as helpful as arithmetic, assignment, comparison, and logical operators are. The colon or sequence operator “:” is used to generate a sequence of values
from a starting point to an ending point. For instance, if we want to generate values from 1 to 5, following command should be used
seq<- 1:5
The concatenation operator “c” is used to combine multiple values into a vector. It is often used when creating vectors or lists. it can be used in a command in the following way
vec_a<- c(1, 2, 3)
vec_b <- c("patrick", "henry", "jack")
Similarly, there is another operator named subset operator in R. The subset operator “[ ]” is used to extract or subset elements from a data frame, vector, or list. For instance, if we create a
vector and then want to extract certain elements from it as a subset, we can use the following command
vector <- c(10, 20, 30, 40, 50) subset <- vector[2:4]
First, the vector will be created having values as shown above, and then we use the index number of values that we want separately in a subset. The values from 20 to 40 will be selected in the
subset, as shown in image below.
What if we want to select only a certain value in the subset, or more than one values not present in the sequence? This ca be done using the following commands
subset1 <- vector[4]
subset2 <- vector[c(1, 3)]
The above command will create two new vectors by the name of subset 1 and 2, having only the values for which we specified their index numbers. The result of above commands is shown below
Similarly, there is another operator named matrix multiplication operator. The “%*%” operator is used for matrix multiplication. It allows us to perform matrix multiplication between two matrices or
between a matrix and a vector.
Let’s say we create two matrices, named X and Y by using the following command
X <- matrix(c(1, 2, 3, 4), nrow = 2) Y <- matrix(c(5, 6, 7, 8), nrow = 2)
And now we perform matrix multiplication by using the following command
Z<- X%*% Y
The following matrix is created by using the above command
In conclusion, R offers a diverse range of operators, from arithmetic and assignment to comparison, logical, and specialized operators. We have discussed the important and useful operators in R, and
by practicing them you can manipulate and analyze your data as per your requirements.
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One-dimensional profile inversion of a half-space bounded by a three-part impedance ground
A method which permits one to reveal the one-dimensional electromagnetic profile of a half-space over a three-part impedance ground is established. The method reduces the problem to the solution of
two functional equations. By using a special representation of functions from the space L^1 (-∞, ∞), one of these equations is first reduced to a modified Riemann-Hilbert problem and then solved
asymptotically. The asymptotic solution is valid when the wave used for part of the boundary is sufficiently large as compared to the wavelength of the wave used for measurements. The second
functional equation is reduced under the Born approximation to a Fredholm equation of the first kind whose kernel involves the solution to the first equation. Since this latter constitutes an
ill-posed problem, its regularized solution in the sense of Tikhonov is given. The accuracy of the asymptotic solution to the first equation requires the use of waves of high frequencies while the
Born approximation in the second equation is accurate for lower frequencies. A criterion to fix appropriate frequencies meeting these contradictory requirements is also given. An illustrative
application shows the applicability and the accuracy of the theory. The results may have applications in profiling the atmosphere over non-homogeneous terrains.
Parmak izi
One-dimensional profile inversion of a half-space bounded by a three-part impedance ground' araştırma başlıklarına git. Birlikte benzersiz bir parmak izi oluştururlar. | {"url":"https://research.itu.edu.tr/tr/publications/one-dimensional-profile-inversion-of-a-half-space-bounded-by-a-th","timestamp":"2024-11-12T08:59:42Z","content_type":"text/html","content_length":"57979","record_id":"<urn:uuid:654cb783-62a5-4808-9563-b21413474791>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00849.warc.gz"} |
Margin of Error in Surveys: What is it and How to Tackle it?
It’s not always easy to determine the ideal sample size that balances practicality with statistical precision.
When you conduct surveys, finding that sweet spot is key to getting results you can trust without overdoing it. And this is where the margin of error might help.
Ready to find out more about this intriguing aspect of survey research? We will discover effective strategies for optimizing your data collection and not only.
But firstβ ¦
What is margin of error in surveys?
Margin of error, often called the confidence interval, is a concept in survey research. It represents how closely the survey results mirror the opinions of the entire population.
Conducting surveys is a delicate balancing act, especially when comparing different population segments. The margin of error is a useful metric in assessing the reliability and precision of your
How to calculate margin of error?
You need to understand a few key terms and their roles in the margin of error calculator:
1. Sample Size (n)
Itβ s the number of survey respondents. A larger sample size generally leads to a smaller margin of error, as it’s more likely to represent the population accurately.
π Read also: Quota Sampling in Surveys: All You Need to Know.
2. Confidence Level (%)
This reflects how sure you can be that the survey results fall within the margin of error. Common confidence levels are 90%, 95%, and 99%.
3. Z-Score (z)
Corresponds to the chosen confidence level. For instance, a 95% confidence level typically has a z-score of 1.96.
4. Standard Error (SE)
This is calculated using the sample proportion (p) and the population size (N). It’s a measure of the variability in your sample data.
5. Population Proportion (p)
The percentage of respondents choosing a particular option in your survey.
The formula for calculating the margin of error is:
Margin of Error (MoE)=Z-ScoreΓ pΓ (1β p)nMargin of Error (MoE)=Z-ScoreΓ npΓ (1β p)β β
Here’s a breakdown of the margin of error formula:
• The z-score is multiplied by the square root of the sample proportion (p) times one minus the sample proportion, divided by the sample size (n).
• This calculation gives you the margin of error in percentage points, which you can then apply to your survey results.
For example, if you have a sample size of 1000 respondents, with 50% choosing a particular option, and you’re working with a 95% confidence level (z-score of 1.96), the margin of error would be
calculated as follows:
MoE=1.96Γ 0.50Γ (1β 0.50)1000MoE=1.96Γ 10000.50Γ (1β 0.50)β β
The formula helps survey researchers identify the range within which the true population parameter likely falls. It also shows the reliability of the survey data.
π π » Remember, a smaller sample size or a higher confidence level will result in a higher margin of error. It indicates less confidence in the survey results being representative of the general
π π » Conversely, a larger sample size or a lower confidence level leads to a smaller margin of error, and implies more confidence in the survey’s accuracy.
π π » An acceptable margin of error used by most survey researchers typically falls between 4% and 8% at the 95% confidence level.
What affects margin of error?
For a better understanding of margin error, take a look at a few aspects.
#1 Impact of sample size on margin of error
Think of sample size like a snapshot of a crowd. The bigger the crowd you capture in your photo (or sample), the clearer the picture of the entire population. A big sample means a tight, small margin
of error, because it’s like getting a fuller, clearer picture of what everyone thinks.
#2 Confidence levels and their influence
Confidence levels show the probability that the margin of error represents the true population value. Higher confidence levels, such as 95% or 99%, make a margin of error bigger. Why? A higher
confidence level needs a wider interval to capture the true population parameter.
#3 The role of population size in margin of error
When we look at total population size, it’s a bit like deciding if a small town needs the same survey tactics as a big city. Smaller groups might see more impact on margin of error than larger ones,
but it’s always good to know the total number to keep things in perspective.
#4 Random sampling and error reduction
The use of a random sample minimizes random sampling error. When every member of the target market has the same chance of being picked, the sample more accurately reflects the overall population.
And, in turn, leads to a more reliable margin of error.
#5 Standard deviation: understanding population variability
Standard deviation is a way of measuring how spread out everyone’s opinions are. If everyone has wildly different views, you can expect a bigger margin of error.
#6 Critical value: z-score’s effect on the margin of error
Z-score in the margin of error formula, directly affects the margin. This value changes with the desired confidence level, and a higher z-score (representing a higher confidence level) shifts up the
margin of error.
#7 Representative sample: ensuring accuracy
A representative sample is like a mini-version of the entire population. If the sample accurately reflects the demographics and characteristics of the target market, the margin of error may show the
true population parameter.
#8 Calculating margin of error: the formula’s significance
Understanding the margin of error formula is like having the right recipe for your favorite dish. It mixes sample size, population proportion, and z-score to tell you how far off your survey results
might be from what the whole population thinks.
This often involves calculations related to standard deviation, which can be complex. Utilize a standard deviation calculator to simplify these calculations, ensuring your survey analysis is both
robust and reliable.
How to increase your survey data reliability?
Check out if you know all the tips. β¬ οΈ
01 Choose the right number of people
To get trustworthy survey data, having enough people in your survey (sample size) is key. More people mean a smaller margin of error, making the survey’s results closer to what everyone really
thinks. Survey researchers need to choose a sample size that balances practicality with the need for precision, considering how sample size affects the margin.
02 Use a robust margin of error calculator
A good margin of error calculator is a big help. It uses the margin of error formula, which considers things like how many people are surveyed (sample size), the z-score, and population standard
deviation to figure out the margin of error. That’ll help you figure out how trustworthy your survey results are.
03 Pick the right confidence intervals
The confidence interval you choose is really important for your survey’s reliability. A higher confidence level (like 95% or 99%) makes the margin of error bigger but means you can be more sure the
survey results include the real opinions of the entire population. Pick a confidence level that gives you a good balance of certainty and useful information.
04 Get a mix of people from the total population
Your survey data is more reliable if your sample looks like the total population. The sample should have the same mix of people with all their different characteristics and opinions. The purpose of
this is to make sure the survey’s margin is a true reflection of the population’s views.
05 Collect and analyzing data carefully
You need enough completed responses of good quality to make your analysis. Look at standard deviation and other statistics to understand the variety in your data. All to make more accurate error
And here you need a reliable surveying tool like Surveylab.
It offers features that help you manage your surveys efficiently. You can trust it to handle everything from setting up questions to analyzing the results. The tool is straightforward and easy to
use, making your survey process smooth.
Here are some of its key features:
1. User-friendly interface makes setup and execution quick.
2. It supports various question types.
3. Real-time data collection allows immediate analysis.
4. Offers robust data security to protect respondent information.
5. Integrates easily with other tools for enhanced functionality.
06 Apply correct error calculation methods
Getting correct error calculation methods right means you can be more confident about understanding the true population parameter. So, make sure you are familiar with the error formula and what the
error means. There’s more to it than just math. You must know what the numbers mean for your survey and the people you’re asking.
Example situation
For most survey researchers, the focus typically falls on how sample size affects the margin of error and the accuracy of the collected data.
When dealing with statistical measurements, such as the mean or median, researchers often encounter the concept of ‘maximum margin’β a term that signifies the extent of potential deviation from the
true value in a normal distribution. The error tells us about the reliability of our data. It’s also a statistic expressing the level of certainty in the poll result.
For instance, in a scenario where two candidates are in a statistical dead heat, the highest margin of error could greatly influence the interpretation of the survey’s outcome. Understanding this
concept is critical, especially when dealing with a smaller group or a specific subset of the total number of respondents.
The z-value, a key component in the following formula for calculating the margin of error, helps researchers feel confident about the values they present.
An example question might be to determine which product features target audiences prefer. Knowing the margin of error helps you make solid conclusions.
Key Takeaways
• Think about your sample size β the number of people you survey really affects your margin of error and how accurate your results are.
• Choose a good confidence level β find the right balance between needing precise results and being practical. Higher confidence means a bigger margin of error but more trust in your results.
• Get the margin of error right β use the margin of error formula carefully, and consider how many people you survey, the z-score, and population proportion to get accurate and reliable results.
• Make sure your sample represents everyone β aim for a sample that reflects the whole population’s variety to reduce error margins and make your survey more valid.
• Reduce random sampling error β use random sampling so everyone has an equal chance of being included, making your survey data more reliable.
• Analyze your data well β Look at your data carefully, focus on standard deviation and other statistics to understand the variability, and improve your error calculations.
• Apply error calculation methods properly β understand and use the right methods for calculating error, and interpret what the numbers mean for your survey and the people you’re asking to
accurately determine what the whole population thinks.
Margin of error concluded
The margin of error is your reality check in survey research. It tells you how much you can trust your results.
Paying attention to your sample size, picking the right confidence level, and using proper error calculation methods are non-negotiable for credible surveys.
So take the tips, apply them rigorously, and let your surveys truly speak for the people! Sign up for Surveylab.
FAQ on the margin of error
Let’s find out what the most frequently asked questions are:
What is margin of error in surveys?
Margin of error shows how much you can expect survey results to reflect the true opinions of the entire population. It helps you understand the accuracy of your survey results. A smaller margin of
error means more reliable results.
How do you calculate margin of error?
To calculate the margin of error, you use a formula that includes the sample size, the confidence level, and the population proportion. This formula considers how confident you are in the survey
results and how many people answered your survey. The calculation tells you how close the survey results are likely to be to the true opinions of the whole population.
Why is choosing the right sample size important?
Choosing the right sample size is crucial because it affects the margin of error. A larger sample size usually gives a smaller margin of error, making your survey results more accurate. You need to
balance having enough people to reduce error while keeping the survey manageable. | {"url":"https://www.surveylab.com/blog/margin-of-error/","timestamp":"2024-11-12T00:36:51Z","content_type":"text/html","content_length":"122807","record_id":"<urn:uuid:f87939cd-ade1-4233-bcab-0e9400b2bd1d>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00595.warc.gz"} |
The financing limit/funding limit is the maximum amount that can be invested in the project. Project volume: € 3,200,000
The financing limit/funding limit is the maximum amount that can be invested in the project. Project volume: € 880,000
The financing limit/funding limit is the maximum amount that can be invested in the project. Project volume: € 2.300,000
The financing limit/funding limit is the maximum amount that can be invested in the project. Project volume: € 2,045,000
The financing limit/funding limit is the maximum amount that can be invested in the project. Project volume: € 715,000
The financing limit/funding limit is the maximum amount that can be invested in the project. Project volume: Already paid out to investors | {"url":"https://www.momentum.investments/en/projects/","timestamp":"2024-11-07T07:36:45Z","content_type":"text/html","content_length":"162028","record_id":"<urn:uuid:9fd4a39f-d1dc-4a2f-b49a-1c8b5bedc506>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00309.warc.gz"} |
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Linear Regression (1 of 4)
Learning Objectives
• For a linear relationship, use the least squares regression line to model the pattern in the data and to make predictions.
So far we have used a scatterplot to describe the relationship between two quantitative variables. We described the pattern in the data by describing the direction, form, and strength of the
relationship. We then focused on linear relationships. When the relationship is linear, we used correlation (r) as a measure of the direction and strength of the linear relationship.
Our focus on linear relationships continues here. We will
• use lines to make predictions.
• identify situations in which predictions can be misleading.
• develop a measurement for identifying the best line to summarize the data.
• use technology to find the best line.
• interpret the parts of the equation of a line to make our summary of the data more precise.
Making Predictions
Earlier, we examined the linear relationship between the age of a driver and the maximum distance at which the driver can read a highway sign. Suppose we want to predict the maximum distance that a
60-year-old driver can read a highway sign. In the original data set, we do not have a 60-year-old driver.
How could we make a prediction using the linear pattern in the data?
Here again is the scatterplot of driver ages and maximum reading distances . (Note: Sign Legibility Distance = Max distance to read sign.) We marked 60 on the x-axis.
Of course, different 60-year-olds will have different maximum reading distances . We expect variability among individuals. But here our goal is to make a single prediction that follows the general
pattern in the data. Our first step is to model the pattern in the data with a line. In the scatterplot, you see a red line that follows the pattern in the data.
To use this line to make a prediction, we find the point on the line with an x-value of 60. Simply trace from 60 directly up to the line. We use the y-value of this point as the predicted maximum
reading distance for a 60-year-old. Trace from this point across to the y-axis.
We predict that 60-year-old drivers can see the sign from a maximum distance of just under 400 feet.
We can also use the equation for the line to make a prediction. The equation for the red line is
Predicted distance = 576 − 3 * Age
To predict the maximum distance for a 60-year-old, substitute Age = 60 into the equation.
Predicted distance = 576 − 3 * (60) = 396 feet
Shortly, we develop a measurement for identifying the best line to summarize the data. We then use technology to find the equation of this line. Later, in “Assessing the Fit of a Line,” we develop a
method to measure the accuracy of the predictions from this “best” line. For now, just focus on how to use the line to make predictions.
Before we leave the idea of prediction, we end with the following cautionary note:
Avoid making predictions outside the range of the data.
Prediction for values of the explanatory variable that fall outside the range of the data is called extrapolation. These predictions are unreliable because we do not know if the pattern observed in
the data continues outside the range of the data. Here is an example.
Cricket Thermometers
Crickets chirp at a faster rate when the weather is warm. The scatterplot shows data presented in a 1995 issue of Outside magazine. Chirp rate is the number of chirps in 13 seconds. The temperature
is in degrees Fahrenheit.
There is a strong relationship between chirp rate and temperature when the chirp rate is between about 18 and 45. What form does the data have? This is harder to determine. A line appears to
summarize the data well, but we also see a curvilinear form, particularly when we pay attention to the first and last data points.
Both the curve and line are good summaries of the data. Both give similar predictions for temperature when the chirp rate is within the range of the data (between 18 and 45). But outside this range,
the curve and the line give very different predictions. For example, if the crickets are chirping at a rate of 60, the line predicts a temperature just above 95°F. The curve predicts a much lower
temperature of about 85°F.
Which is a better prediction? We do not know which is better because we do not know if the form is linear or curvilinear outside the range of the data.
If we use our model (the line or the curve) to make predictions outside the range of the data, this is an example of extrapolation. We see in this example that extrapolation can give unreliable | {"url":"https://courses.lumenlearning.com/suny-hccc-wm-concepts-statistics/chapter/linear-regression-1-of-4/","timestamp":"2024-11-07T23:50:06Z","content_type":"text/html","content_length":"54346","record_id":"<urn:uuid:dd34d484-8987-4eb7-a857-8f1da341a615>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00275.warc.gz"} |
Viewable by the world
Group Access to Chombo Individual Access to Chombo
Can VIEW the space: chombo-editors , chombo-admin , anonymous , Can VIEW the space: [email protected] , [email protected] ,
Can EDIT the space: chombo-admin , chombo-editors , Can EDIT the space: [email protected] , [email protected] ,
Can ADMINISTER the space: chombo-admin , Can ADMINISTER the space: [email protected] , [email protected] ,
As in the rest of Chombo, data are held on rectangular patches. In the mapped multi-block framework, each patch must be contained in only one block. In other words, no patch may straddle two or more
Advanced Tables - Table Plus
columnAttributes width=40%,width=40%
Domain in 2D mapped space: five squares Range in 2D real space: disk
Advanced Tables - Table Plus
columnAttributes width=40%,width=40%
Domain in 2D mapped space: eight squares Range in 2D real space: single-null edge plasma geometry
Advanced Tables - Table Plus
columnAttributes width=40%,width=40%
Domain in 2D mapped space: six squares Range in 3D real space: surface of sphere
Advection on a sphere, with one level of refinement
width 100%%%%%%%%
name AdvectCosineBell.polar_lim.10.abs.AMR.64.mpeg
width 100%%%%%%%%
height 380
Exchanging data between blocks
The ghost cells of a patch that lie outside the block containing the patch will be called extra-block ghost cells of the patch. Two layers of extra-block ghost cells of block 2 of the disk example
are outlined with dotted blue lines below, in both mapped space and real space. The centers of four of them are marked with blue *. The cells of the interpolation stencils of these four ghost cells
are shown with thicker outlines, each with the color of its block.
Advanced Tables - Table Plus
columnAttributes height=100%
In mapped space: four ghost cells with their interpolation stencils
In real space: four ghost cells with their interpolation stencils
The function value for each extra-block ghost cell is interpolated from function values for the valid cells in its stencil.
Around each ghost cell, we approximate the function by a Taylor polynomial in real coordinates, of degree P. This polynomial has different coefficients for each ghost cell. In 2D the polynomial is of
the form:
LaTeX Formatting
$$f(X, Y) = \sum_{p, q \geq 0: p + q \leq P} a_{pq} \left(\frac{X - X_{\boldsymbol{g}}}{R}\right)^p \left(\frac{Y - Y_{\boldsymbol{g}}}{R}\right)^q$$
where (X[g], Y[g]) is the center of the ghost cell in real coordinates, and R is the mean distance from (X[g], Y[g]) to the centers in real space of the stencil cells of g.
Starting with averaged values of f over each valid cell, then the coefficients a[pq] come from solving the overdetermined system of equations
LaTeX Formatting
$$\sum_{p, q \geq 0; p + q \leq P} a_{pq} \left\langle \left(\frac{X - X_{\boldsymbol{g}}}{R}\right)^p \left(\frac{Y - Y_{\boldsymbol{g}}}{R}\right)^q \right \rangle_{\boldsymbol{j}} = \left\langle f \right\rangle_{\boldsymbol{j}}$$
where there is an equation for every cell j in the selected neighborhood of valid cells of the ghost cell. The notation
LaTeX Formatting
$\langle \cdot \rangle_{\boldsymbol{j}}$
indicates an average over cell j.
The average value of f on the ghost cell is then obtained by
LaTeX Formatting
$$\left\langle f \right\rangle_{\boldsymbol{g}} = \sum_{p, q \geq 0; p + q \leq P} a_{pq} \left\langle \left(\frac{X - X_{\boldsymbol{g}}}{R}\right)^p \left(\frac{Y - Y_{\boldsymbol{g}}}{R}\right)^q \right \rangle_{\boldsymbol{g}}$$ | {"url":"https://commons.lbl.gov/pages/diffpagesbyversion.action?pageId=73471623&selectedPageVersions=32&selectedPageVersions=33","timestamp":"2024-11-07T15:52:07Z","content_type":"text/html","content_length":"81323","record_id":"<urn:uuid:1e0221b3-684b-436a-9858-33c0d11db648>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00540.warc.gz"} |
Using and Understanding the Discriminant | sofatutor.com
Using and Understanding the Discriminant
Basics on the topic Using and Understanding the Discriminant
A quadratic equation in standard form ax² + bx + c = 0 can be solved in various ways. These include factoring, completing the square, graphing, and using the quadratic formula. Solving for the roots
or solutions of a quadratic equation by using the quadratic formula is often the most convenient way. The solutions to a quadratic equation can be determined by this quadratic formula: x = [-b ± √
(b²-4ac)] / 2a. We can predict the number of solutions to a quadratic equation by evaluating the discriminant given by b²-4ac or the radicand expression in the quadratic formula. The value of the
discriminant determines the nature and number of solutions of a quadratic equation.
If b²-4ac is positive, there will be two distinct, real solutions.
For example, the quadratic equation x² - 7x + 12 = 0 has two distinct, real solutions since b²-4ac = (-7)² - 4(1)(12) = 49 - 48 = 1 is positive.
If b²-4ac is zero, there will be one distinct, real solution.
For example, the quadratic equation 4x² - 4x + 1 = 0 has one distinct, real solution since b²-4ac = (-4)² - 4(4)(1) = 16 - 16 = 0.
If b2-4ac is negative, there are no real solutions.
For example, the quadratic equation 5x² + 2x + 3 = 0 has no real solutions since b2-4ac = (2)² - 4(5)(3) = 4 - 60 = -56 is negative.
Solve quadratic equations with real coefficients that have complex solutions.
Transcript Using and Understanding the Discriminant
”Hello! It's that time again for your favorite quiz show THE WALL OF FAME!” ”The first contestant on today's show is Theresa! I hope you're ready to play THE WALL OF FAME!” Before we begin today's
game, let's quickly review our game rules. Theresa, you're about to see today's wall. On the wall are 5 secret doors with one equation written over each door. During the game, you will get to open 3
of those doors. For each door you pick, you will receive a prize equal to the number of solutions the equation has. If each door you pick reveals a prize, you get to come back on the show tomorrow
for a chance to win an even bigger prize!
Understanding the Discriminant
The secret to this game is understanding and using the discriminant. If Theresa can figure out how to use this tool, she just might have a chance to win all the prizes and come back tomorrow! Let's
take a look at today's WALL OF FAME! The equations we have today are:
Equation 1: x-squared minus 10x plus 34.
Equation 2: 3 x-squared minus 4x plus 10.
Equation 3: x-squared minus 3x plus 5.
Equation 4: x-squared plus 2 root 2x plus 2.
Equation 5: x-squared plus 6x minus 16.
With her first choice, Theresa picks door number 4. Remember, we said we could use the discriminant to quickly determine whether or not Theresa has won one of our fabulous prizes. The discriminant
comes from the quadratic formula, and it's formula is b-squared minus 4ac where a, b and c refer to the coefficients and constant of a quadratic equation in standard form.
When the discriminant is positive, our equation will have 2 solutions; when the discriminant is equal to 0, our equation has 1 solution and when our discriminant is negative, our equation will have
no solutions. Let's take a look at the equation Theresa has just selected. In the equation Theresa has just selected, "a" is equal to 1, "b" is equal to 2 times the square root of 2, and "c" is equal
to 2.
Let's calculate the discriminant! The quantity 2 times the square root of 2 squared will be 8. 8 minus 8 will equal 0. Because the discriminant is equal to 0, this equation will have 1 solution.
Theresa, you've just won 1 brand new unicorn! Doesn't she look happy?! That's about as good of a start as you can ask for!
Second Example
With her second selection, Theresa picks the fifth equation. Let's take a closer look! In this equation, "a" is equal to 1, "b" is equal to 6, and "c" is equal to negative 16. Let's calculate the
discriminant! Plugging in our values, we get 6 squared minus 4 times 1 times negative 16.
Using PEMDAS, we get 36 plus 64 which equals 100. Wow! Look at that! Theresa has just won a pair of Positron and Negatron figurines! 2 prizes for an equation with 2 solutions! Theresa's doing great
so far! If she can pick a door with one more prize she'll be invited back for tomorrow's show and a chance to win the Grand Prize!
And with her final selection, Theresa picks the equation on door number 2! Let's take a look at her equation! In this equation, "a" is equal to 3, "b" is equal to negative 4, and "c" is equal to 10.
Let's calculate the discriminant! Plugging in our values, we get negative 4 squared minus 4 times 3 times 10. Calculating our math, we get 16 minus 120 which equals negative 104. What do you think
this means? Not only for how many solutions our equation has but for Theresa's chances at coming back tomorrow?
Remember, when our discriminant is greater than 0 that means our equation will have 2 solutions. When our discriminant is equal to 0 our equation will have 1 solution. And when our discriminant is
negative, our equation will have no real solutions.
Well Folks! That also means there's no prize! Although Theresa is going to be disappointed, at least she got some action figures and a new Unicorn! Although that Unicorn looks a bit suspicious... And
the host's hair looks a bit suspicious too!
Using and Understanding the Discriminant exercise
Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Using and Understanding the Discriminant.
• Define the discriminant.
The discriminant could be found for this equation:
What type of equation is this, and what form is it in?
A quadratic equation can have zero, one, or two solutions.
The discriminant is a mathematical expression that can be calculated for quadratic equations in standard form. Therefore the equation will be in the form:
The discriminant is equal to $b^2-4ac$.
The discriminant is useful because it tells you the number of solutions, or roots, that a quadratic equation has. If the discriminant is positive, the equation has two solutions. If it is equal
to zero, the equation has one solution. If it is negative, the equation has zero solutions.
• Determine the number of solutions.
The discriminant is equal to:
If the discriminant is positive, the equation has two solutions. If it is equal to zero, the equation has one solution. If it is negative, the equation has zero solutions.
Theresa sees that she has a quadratic equation in standard form. She knows that she can use the discriminant to find the number of solutions to the equation.
She knows that the discriminant is equal to:
She substitutes in the values from the given equation, and gets:
$(2 \sqrt2)^2-4(1)(2)$
Using PEMDAS, she finds the discriminant to be equal to $0$.
She knows that if the discriminant is positive, the equation has two solutions. If it is equal to zero, the equation has one solution. If it is negative, the equation has zero solutions.
Since the discriminant is equal to zero, Theresa can see that this equation has one solution, and she should get one prize by opening this door.
• Find the discriminants.
Pay close attention to signs.
Remember that the standard form of a quadratic equation is:
$a x^2 + b x + c = 0$
For the standard form of a quadratic equation, the discriminant is equal to:
$b^2 - 4(a)(c)$
Each quadratic equation is given in the standard form:
$a x^2 + b x + c = 0$
For the standard form of a quadratic equation, the discriminant is equal to:
$b^2 - 4(a)(c)$
We can therefore determine the values of $a$, $b$, and $c$, and use these values to find the expression for the discriminant.
From the equation:
$2 x^2 -5 x + 6 = 0$
We can see that:
$a = 2$
$b = -5$
$c = 6$
And therefore we can determine that the discriminant is equal to:
$(-5)^2 - 4(2)(6)$
Similarly, the other quadratic equations can be paired with their corresponding discriminants.
For $4 x^2 + 4 x + 1 = 0$ the discriminant is: $4^2 - 4(4)(1)$
For $5 x^2 + 3 x -7 = 0$ the discriminant is: $3^2 - 4(5)(-7)$
For $3 x^2 + 8 x + 5 = 0$ the discriminant is: $8^2 - 4(3)(5)$
For $-2 x^2 -3 x -4 = 0$ the discriminant is: $(-3)^2 - 4(-2)(-4)$
• Determine which equation has two solutions.
You can use the discriminant to check how many solutions there are to each equation.
If the discriminant is positive, the equation has two solutions.
The discriminant is equal to $b^2-4ac$.
The equations above the doors are all standard form quadratic equations. Therefore John can use the discriminant to check how many solutions each equation has.
If the discriminant is positive, the equation has two solutions. If it is equal to zero, the equation has one solution. If it is negative, the equation has zero solutions. Therefore John must
find the equation whose determinant is positive.
John decides to evaluate the determinant for each quadratic equation. He begins with the equation:
$3 x^2 +2 x + 1 = 0$
He finds the determinant to be:
$(2)^2 - 4(3)(1) = -8$
The determinant for this equation is negative. Therefore this is not the right door.
Similarly, for the other equations:
Door 2
$2 x^2 -4 x + 2 = 0$
The determinant is:
$(-4)^2 - 4(2)(2) = 0$
The determinant for this equation is zero. Therefore this is not the right door.
Door 3
$2 x^2 + 3 x +5 = 0$
The determinant is:
$(3)^2 - 4(2)(5) = -31$
The determinant for this equation is negative. Therefore this is not the right door.
Door 4
$- x^2 + 2 x + -1 = 0$
The determinant is:
$(2)^2 - 4(-1)(-1) = 0$
The determinant for this equation is zero. Therefore this is not the right door.
Door 5
$2 x^2 -6 x +3 = 0$
The determinant is:
$(-6)^2 - 4(2)(3) = 12$
The determinant for this equation is positive. Therefore this is the right door.
• Explain what a discriminant equal to $100$ tells you.
A quadratic equation can have zero, one, or two solutions.
If the discriminant is equal to zero, the equation has one solution.
If the discriminant is negative, the equation has zero solutions.
The discriminant is useful because it tells you the number of solutions, or roots, that a quadratic equation has. If the discriminant is positive, the equation has two solutions. If it is equal
to zero, the equation has one solution. If it is negative, the equation has zero solutions.
In this case, the discriminant is equal to $100$, and is positive. Therefore this equation has two solutions.
• Calculate the discriminant.
Pay close attention to signs in your calculations.
Remember that you can determine how many solutions an equation has by comparing the discriminant to zero.
Remember that a quadratic equation can only have $0$, $1$, or $2$ solutions.
Iris knows that the discriminant can be found for the standard form of a quadratic equation using the equation:
She also knows that if the discriminant is positive, the equation has two solutions. If it is equal to zero, the equation has one solution. If it is negative, the equation has zero solutions.
For the first equation:
$-x^2 + 3x -4 =0$
She evaluates the discriminant:
$= 9 - 16$
$= -7$
Since the discriminant is negative, the equation has $0$ solutions.
Similarly, for the other equations:
For the equation $-2x^2 +x +1 =0$ she finds that the discriminant is $9$. Therefore the equation has $2$ solutions.
For the equation $8x^2 + 6x +2 =0$ she finds that the discriminant is $0$. Therefore the equation has $1$ solution.
For the equation $7x^2 + 5x -3 =0$ she finds that the discriminant is $109$. Therefore the equation has $2$ solutions. | {"url":"https://us.sofatutor.com/math/videos/using-and-understanding-the-discriminant","timestamp":"2024-11-12T23:53:36Z","content_type":"text/html","content_length":"162029","record_id":"<urn:uuid:343387a8-6754-471b-9fe3-afd356454b49>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00262.warc.gz"} |
Make A Table Of Equivalent Ratios Based On The Graph Below
hoped this helped:)
At the 50th place, triangle shape pattern is used.
What is a triangle?
A triangle is a geometric figure with three edges, three angles and three vertices. It is a basic figure in geometry.
The sum of the angles of a triangle is always 180°
The square, triangle, circle, and oval pattern rules are listed.
The formula for the number of shapes (n) is:
In order to calculate the 50th shape, the following absolute operator is used,
Divide 50 by 4 and record the result,
4 divided by 50 leaves, 2 as the remaining.
This implies that, the shape at the 50th position is the same shape at the 2nd position.
In the 2nd position, the shape is circle, triangle.
Hence, the shape is triangle.
To know more about Triangle on: | {"url":"https://learning.cadca.org/question/make-a-table-of-equivalent-ratios-based-on-the-graph-below-aftb","timestamp":"2024-11-09T11:07:39Z","content_type":"text/html","content_length":"72026","record_id":"<urn:uuid:0d35ea1d-5635-4cae-a551-fb37d9dc5363>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00848.warc.gz"} |
Lec5101 Question Mock Quiz 1
Hi, I have a question when I solve the general solution of ODE.
When I calculate the integral, should I add the absolute value to the result? (see the picture)
I am confused about this because I think the constant "C" can cancel the absolute value there.
Also, can anyone help me to check if my answer totally correct or anywhere I can improve?
Look forward to a kind reply.
Thanks a lot! | {"url":"http://forum.math.toronto.edu/index.php?PHPSESSID=c218vmafat7tio9e3h5hpgldt0&topic=2367.0","timestamp":"2024-11-08T17:04:55Z","content_type":"application/xhtml+xml","content_length":"32245","record_id":"<urn:uuid:9bec9b3d-3307-46ae-aea8-2302240165f6>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00792.warc.gz"} |
Matrix Diagram
What is a Matrix Diagram?
Quality Glossary Definition: Matrix
Also called: matrix, matrix chart
A matrix diagram is defined as a new management planning tool used for analyzing and displaying the relationship between data sets. The matrix diagram shows the relationship between two, three, or
four groups of information. It also can give information about the relationship, such as its strength, of the roles played by various individuals or measurements.
Six differently shaped matrices are possible: L, T, Y, X, C, and roof-shaped, depending on how many groups must be compared.
When to Use Each Matrix Diagram Shape
Table 1 summarizes when to use each type of matrix. Click on the links below to see an example of each type. In the examples, matrix axes have been shaded to emphasize the letter that gives each
matrix its name.
Table 1: When to use differently-shaped matrices
L-shaped 2 groups A
T-shaped 3 groups B
Y-shaped 3 groups A
C-shaped 3 groups All three simultaneously (3D)
X-shaped 4 groups A
Roof-shaped 1 group A
This L-shaped matrix summarizes customers’ requirements. The team placed numbers in the boxes to show numerical specifications and used check marks to show choice of packaging. The L-shaped matrix
actually forms an upside-down L. This is the most basic and most common matrix format.
L-shaped Matrix Diagram: Customer Requirements
Customer Customer Customer Customer
D M R T
Purity % > 99.2 > 99.2 > 99.4 > 99.0
Trace metals (ppm) < 5 — < 10 < 25
Water (ppm) < 10 < 5 < 10 —
Viscosity (cp) 20-35 20-30 10-50 15-35
Color < 10 < 10 < 15 < 10
This T-shaped matrix relates product models (group A) to their manufacturing locations (group B) and to their customers (group C).
Examining the matrix (below) in different ways reveals different information. For example, focusing on model A shows that it is produced in large volume at the Texas plant and in small volume at the
Alabama plant. Time Inc. is the major customer for model A, while Arlo Co. buys a small amount. Focusing on the customer rows shows that only one customer, Arlo Co., buys all four models. Zig Corp.
buys just one. Time Inc. makes large purchases of A and D, while Lyle Co. is a relatively minor customer.
T-shaped Matrix Diagram: Products—Manufacturing Locations—Customers
This Y-shaped matrix shows the relationships between customer requirements, internal process metrics, and the departments involved. Symbols show the strength of the relationships: primary
relationships, such as the manufacturing department’s responsibility for production capacity; secondary relationships, such as the link between product availability and inventory levels; minor
relationships, such as the distribution department’s responsibility for order lead time; and no relationship, such as between the purchasing department and on-time delivery.
Y-shaped Matrix Diagram: Responsibilities for Performance to Customer Requirements
Because this matrix is three-dimensional, it is difficult to draw and infrequently used. If it is important to compare three groups simultaneously, consider using a three-dimensional model or
computer software that can provide a clear visual image.
C-shaped Matrix Diagram
This figure extends the T-shaped matrix example into an X-shaped matrix by including the relationships of freight lines with the manufacturing sites they serve and the customers who use them. Each
axis of the matrix is related to the two adjacent ones, but not to the one across. Thus, the product models are related to the plant sites and to the customers, but not to the freight lines.
X-shaped Matrix Diagram: Manufacturing Sites—Products—Customers—Freight Lines
The roof-shaped matrix is used with an L- or T-shaped matrix to show one group of items relating to itself. It is most commonly used with a House of Quality, where it forms the "roof" of the "house."
In the figure below, the customer requirements are related to one another. For example, a strong relationship links color and trace metals, while viscosity is unrelated to any of the other
Roof-shaped Matrix Diagram
Frequently Used Matrix Diagram Symbols
Adapted from The Quality Toolbox, Second Edition, ASQ Quality Press. | {"url":"https://asq.org/quality-resources/matrix-diagram","timestamp":"2024-11-06T07:50:46Z","content_type":"text/html","content_length":"71326","record_id":"<urn:uuid:04f985df-8733-4645-85d5-193ccdf95f6f>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00021.warc.gz"} |
On learning hamiltonian systems from data
On learning hamiltonian systems from data
Concise, accurate descriptions of phys. systems through their conserved quantities abound in the natural sciences. In data science, however, current research often focuses on regression problems,
without routinely incorpo- rating addnl. assumptions about the system that generated the data. Here, we propose to explore a particular type of underlying structure in the data: Hamiltonian systems,
where an "energy" is conserved. Given a collection of observations of such a Hamiltonian system over time, we extract phase space coordinates and a Hamiltonian function of them that acts as the
generator of the system dynamics. The approach employs an auto-encoder neural network component to estimate the transformation from observations to the phase space of a Hamiltonian system. An addi-
tional neural network component is used to approx. the Hamiltonian function on this constructed space, and the two components are trained simultaneously. As an alternative approach, we also
demonstrate the use of Gaussian processes for the estimation of such a Hamiltonian. After two illustrative examples, we extract an underlying phase space as well as the generating Hamiltonian from a
collection of movies of a pendulum. The approach is fully data-driven, and does not assume a particular form of the Hamiltonian function.
arXiv.org, e-Print Arch., Phys.
CAplus AN 2019:1914826 (Preprint; Article) | {"url":"https://yannis-kevrekidis.scholar.princeton.edu/publications/learning-hamiltonian-systems-data","timestamp":"2024-11-05T18:20:54Z","content_type":"text/html","content_length":"27746","record_id":"<urn:uuid:e62ba714-23e0-4efa-884a-594a267e4aac>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00728.warc.gz"} |
function p = dglnpdf(n,alpha,beta)
%DGLNPDF Discrete generalized log-normal probability density function.
% P = DGLNPDF(N,ALPHA,BETA) returns the probabilities for a discrete
% version of the generalized log-normal probability density function. In
% this case, Prob(x) ~ exp(-(log(x)/alpha)^beta) for x = 1:N.
% See also DPLPDF, GENDEGDIST.
% Reference:
% * T. G. Kolda, A. Pinar, T. Plantenga and C. Seshadhri. A Scalable
% Generative Graph Model with Community Structure, arXiv:1302.6636,
% March 2013. (http://arxiv.org/abs/1302.6636)
% Tamara G. Kolda, Ali Pinar, and others, FEASTPACK v1.1, Sandia National
% Laboratories, SAND2013-4136W, http://www.sandia.gov/~tgkolda/feastpack/,
% January 2014
Copyright (c) 2014, Sandia National Laboratories All rights reserved.
Redistribution and use in source and binary forms, with or without modification, are permitted provided that the following conditions are met:
1. Redistributions of source code must retain the above copyright notice, this list of conditions and the following disclaimer.
2. Redistributions in binary form must reproduce the above copyright notice, this list of conditions and the following disclaimer in the documentation and/or other materials provided with the
Sandia National Laboratories is a multi-program laboratory managed and operated by Sandia Corporation, a wholly owned subsidiary of Lockheed Martin Corporation, for the U.S. Department of Energy's
National Nuclear Security Administration under contract DE-AC04-94AL85000.
p = exp(-((log((1:n)'))/alpha).^beta);
p = p / sum(p); | {"url":"http://www.kolda.net/feastpack/dglnpdf.html","timestamp":"2024-11-04T15:40:34Z","content_type":"text/html","content_length":"10075","record_id":"<urn:uuid:db6a16c9-9199-4a7f-bc0c-bb5d658e5730>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00654.warc.gz"} |
Setting Up Inequalities
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