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MAT 540 Week 5 Mid Term Quiz
Deterministic techniques assume that no uncertainty exists in model parameters.
An inspector correctly identifies defective products 90% of the time. For the next 10 products, the probability that he makes fewer than 2 incorrect inspections is 0.736.
A continuous random variable may assume only integer values within a given interval.
A decision tree is a diagram consisting of circles decision nodes, square probability nodes, and branches.
Starting conditions have no impact on the validity of a simulation model.
Excel can only be used to simulate systems that can be represented by continuous random variables.
The Delphi develops a consensus forecast about what will occur in the future.
Qualitative methods are the least common type of forecasting method for the long-term strategic planning process.
__________ is a measure of dispersion of random variable values about the expected value.
In Bayesian analysis, additional information is used to alter the __________ probability of the occurrence of an event.
The __________ is the maximum amount a decision maker would pay for additional information.
Two hundred simulation runs were completed using the probability of a machine breakdown from the table below. The average number of breakdowns from the simulation trials was 1.93 with a standard
deviation of 0.20.
No. of breakdowns per week Probability Cumulative probability
0 .10 .10
1 .25 .35
2 .36 .71
3 .22 .93
4 .07 1.00
What is the probability of 2 or fewer breakdowns?
Pseudorandom numbers exhibit __________ in order to be considered truly random.
Selected Answer:
a uniform distribution
Correct Answer:
a uniform distribution
A seed value is a(n)
Random numbers generated by a __________ process instead of a __________ process are pseudorandom numbers.
Consider the following demand and forecast.
PeriodDemand Forecast
If MAD = 2, what is the forecast for period 4?
Given the following data on the number of pints of ice cream sold at a local ice cream store for a 6-period time frame:
If the forecast for period 5 is equal to 275, use exponential smoothing with α = .40 to compute a forecast for period 7.
Which of the following possible values of alpha would cause exponential smoothing to respond the most slowly to sudden changes in forecast errors?
Coefficient of determination is the percentage of the variation in the __________ variable that results from the __________ variable.
Consider the following graph of sales.
Which of the following characteristics is exhibited by the data?
__________ is a measure of the strength of the relationship between independent and dependent variables.
Consider the following graph of sales.
Which of the following characteristics is exhibited by the data?
__________ is a category of statistical techniques that uses historical data to predict future behavior.
14, and 15)estion worth 2 points, 1 hour time limit (chapters 1,ue units EXCEPT:The U.S. Department of Agriculture estimates that the yearly yield of limes per acre is distributed as follows:
Yield, bushels per acre Probability
350 .10
400 .18
450 .50
500 .22
The estimated average price per bushel is $16.80.
What is the expected yield of the crop?
The drying rate in an industrial process is dependent on many factors and varies according to the following distribution.
Compute the mean drying time. Use two places after the decimal.
A loaf of bread is normally distributed with a mean of 22 oz and a standard deviation of 0.5 oz. What is the probability that a loaf is larger than 21 oz? Round your answer to four places after the
A fair die is rolled 8 times. What is the probability that an even number (2,4, 6) will occur between 2 and 4 times? Round your answer to four places after the decimal.
An automotive center keeps tracks of customer complaints received each week. The probability distribution for complaints can be represented as a table or a graph, both shown below. The random
variable xi represents the number of complaints, and p(xi) is the probability of receiving xi complaints.
xi 0 1 2 3 4 5 6
p(xi) .10 .15 .18 .20 .20 .10 .07
What is the average number of complaints received per week? Round your answer to two places after the decimal.
An investor is considering 4 different opportunities, A, B, C, or D. The payoff for each opportunity will depend on the economic conditions, represented in the payoff table below.
Economic Condition
Poor Average Good Excellent
Investment (S1) (S2) (S3) (S4)
A 50 75 20 30
B 80 15 40 50
C -100 300 -50 10
D 25 25 25 25
If the probabilities of each economic condition are 0.5, 0.1, 0.35, and 0.05 respectively, what is the highest expected payoff?
The local operations manager for the IRS must decide whether to hire 1, 2, or 3 temporary workers. He estimates that net revenues will vary with how well taxpayers comply with the new tax code. The
following payoff table is given in thousands of dollars (e.g. 50 = $50,000).
If he thinks the chances of low, medium, and high compliance are 20%, 30%, and 50% respectively, what is the expected value of perfect information? Note: Please express your answer as a whole number
in thousands of dollars (e.g. 50 = $50,000). Round to the nearest whole number, if necessary.
Given the following random number ranges and the following random number sequence: 62, 13, 25, 40, 86, 93, determine the average demand for the following distribution of demand.
Demand Random
Number Ranges
5 00-14
6 15-44
7 45-69
8 70-84
9 85-99
Consider the following annual sales data for 2001-2008.
Year Sales
Calculate the correlation coefficient . Use four significant digits after the decimal.
Given the following data on the number of pints of ice cream sold at a local ice cream store for a 6-period time frame:
Compute a 3-period moving average for period 6. Use two places after the decimal.
The following data summarizes the historical demand for a product.
MonthActual Demand
April 25
May 40
June 35
July 30
August 45
Use exponential smoothing with α = .2 and the smoothed forecast for July is 32. Determine the smoothed forecast for August.
The following sales data are available for 2003-2008 :
Year Sales Forecast
Calculate the absolute value of the average error. Use three significant digits after the decimal.
Daily highs in Sacramento for the past week (from least to most recent) were: 95, 102, 101, 96, 95, 90 and 92. Develop a forecast for today using a 2 day moving average.
Selected Answer: 91
Correct Answer: 91
The following sales data are available for 2003-2008.
Determine a 4-year weighted moving average forecast for 2009, where weights are W1 = 0.1, W2 = 0.2, W3 = 0.2 and W4 = 0.5.
Given the following data, compute the MAD for the forecast.
Year Demand Forecast
Daily highs in Sacramento for the past week (from least to most recent) were: 95, 102, 101, 96, 95, 90 and 92. Develop a forecast for today using a weighted moving average, with weights of .6, .3 and
.1, where the highest weights are applied to the most recent data.
The following data summarizes the historical demand for a product
MonthActual Demand
April 25
May 40
June 35
July 30
August 45
If the forecasted demand for June, July and August is 32, 38 and 42, respectively, what is MAPD? Write your answer in decimal form and not in percentages. For example, 15% should be written as 0.15.
Use three significant digits after the decimal.
MAT 540 Midterm Exam Version 2
MAT 540 Midterm Quiz
Deterministic techniques assume that no uncertainty exists in model parameters.
An inspector correctly identifies defective products 90% of the time. For the next 10 products, the probability that he makes fewer than 2 incorrect inspections is 0.736.
A continuous random variable may assume only integer values within a given interval.
A decision tree is a diagram consisting of circles decision nodes, square probability nodes, and branches.
Starting conditions have no impact on the validity of a simulation model.
Excel can only be used to simulate systems that can be represented by continuous random variables.
The Delphi develops a consensus forecast about what will occur in the future.
Data cannot exhibit both trend and cyclical patterns.
Assume that it takes a college student an average of 5 minutes to find a parking spot in the main parking lot. Assume also that this time is normally distributed with a standard deviation of 2
minutes. What time is exceeded by approximately 75% of the college students when trying to find a parking spot in the main parking lot?
A company markets educational software products, and is ready to place three new products on the market. Past experience has shown that for this particular software, the chance of “success” is 80%.
Assume that the probability of success is independent for each product. What is the probability that exactly 1 of the 3 products is successful?
The __________ is the expected value of the regret for each decision.
Random numbers generated by a __________ process instead of a __________ process are pseudorandom numbers.
A seed value is a(n)
Pseudorandom numbers exhibit __________ in order to be considered truly random.
In the Monte Carlo process, values for a random variable are generated by __________ a probability distribution.
__________ is a category of statistical techniques that uses historical data to predict future behavior.
__________ methods are the most common type of forecasting method for the long-term strategic planning process.
Consider the following demand and forecast.
PeriodDemand Forecast
If MAD = 2, what is the forecast for period 4?
Consider the following graph of sales.
Which of the following characteristics is exhibited by the data?
Consider the following graph of sales.
Which of the following characteristics is exhibited by the data?
Which of the following possible values of alpha would cause exponential smoothing to respond the most slowly to sudden changes in forecast errors?
In exponential smoothing, the closer alpha is to __________, the greater the reaction to the most recent demand.
Given the following data on the number of pints of ice cream sold at a local ice cream store for a 6-period time frame:
If the forecast for period 5 is equal to 275, use exponential smoothing with α = .40 to compute a forecast for period 7.
__________ is a linear regression model relating demand to time.
The drying rate in an industrial process is dependent on many factors and varies according to the following distribution.
Compute the mean drying time. Use two places after the decimal.
An automotive center keeps tracks of customer complaints received each week. The probability distribution for complaints can be represented as a table or a graph, both shown below. The random
variable xi represents the number of complaints, and p(xi) is the probability of receiving xi complaints.
xi 0 1 2 3 4 5 6
p(xi) .10 .15 .18 .20 .20 .10 .07
What is the average number of complaints received per week? Round your answer to two places after the decimal.
A life insurance company wants to estimate their annual payouts. Assume that the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68
years and a standard deviation of 4 years. What proportion of the plan recipients would receive payments beyond age 75? Round your answer to four places after the decimal.
A fair die is rolled 8 times. What is the probability that an even number (2,4, 6) will occur between 2 and 4 times? Round your answer to four places after the decimal.
The local operations manager for the IRS must decide whether to hire 1, 2, or 3 temporary workers. He estimates that net revenues will vary with how well taxpayers comply with the new tax code. The
following payoff table is given in thousands of dollars (e.g. 50 = $50,000).
If he thinks the chances of low, medium, and high compliance are 20%, 30%, and 50% respectively, what is the expected value of perfect information? Note: Please express your answer as a whole number
in thousands of dollars (e.g. 50 = $50,000). Round to the nearest whole number, if necessary.
An investor is considering 4 different opportunities, A, B, C, or D. The payoff for each opportunity will depend on the economic conditions, represented in the payoff table below.
Economic Condition
Poor Average Good Excellent
Investment (S1) (S2) (S3) (S4)
A 50 75 20 30
B 80 15 40 50
C -100 300 -50 10
D 25 25 25 25
If the probabilities of each economic condition are 0.5, 0.1, 0.35, and 0.05 respectively, what is the highest expected payoff?
Consider the following distribution and random numbers:
If a simulation begins with the first random number, what would the first simulation value would be __________.
The following data summarizes the historical demand for a product
MonthActual Demand
April 25
May 40
June 35
July 30
August 45
If the forecasted demand for June, July and August is 32, 38 and 42, respectively, what is MAPD? Write your answer in decimal form and not in percentages. For example, 15% should be written as 0.15.
Use three significant digits after the decimal.
Given the following data, compute the MAD for the forecast.
Year Demand Forecast
The following sales data are available for 2003-2008 :
Year Sales Forecast
Calculate the absolute value of the average error. Use three significant digits after the decimal.
The following sales data are available for 2003-2008.
Determine a 4-year weighted moving average forecast for 2009, where weights are W1 = 0.1, W2 = 0.2, W3 = 0.2 and W4 = 0.5.
This is the data from the last 4 weeks:
Use the equation of the regression line to forecast the increased sales for when the number of ads is 10.
Daily highs in Sacramento for the past week (from least to most recent) were: 95, 102, 101, 96, 95, 90 and 92. Develop a forecast for today using a weighted moving average, with weights of .6, .3 and
.1, where the highest weights are applied to the most recent data.
The following data summarizes the historical demand for a product.
MonthActual Demand
April 25
May 40
June 35
July 30
August 45
Use exponential smoothing with α = .2 and the smoothed forecast for July is 32. Determine the smoothed forecast for August.
Robert wants to know if there is a relation between money spent on gambling and winnings.
What is the coefficient of determination? Note: please report your answer with 2 places after the decimal point.
Consider the following annual sales data for 2001-2008.
Year Sales
Calculate the correlation coefficient . Use four significant digits after the decimal.
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Chapter 1—Introduction and Goals of the Firm
1. The form of economics most relevant to managerial decision-making within the firm is:
welfare economics
free-enterprise economics
none of the above
ANS: PTS: 1
2. If one defines incremental cost as the change in total cost resulting from a decision, and incremental revenue as the change in total revenue resulting from a decision, any business decision is
profitable if:
it increases revenue more than costs or reduces costs more than revenue
it decreases some costs more than it increases others (assuming revenues remain constant)
it increases some revenues more than it decreases others (assuming costs remain constant)
all of the above
b and c only
ANS: PTS: 1
3. In the shareholder wealth maximization model, the value of a firm’s stock is equal to the present value of all expected future ____ discounted at the stockholders’ required rate of return.
profits (cash flows)
ANS: PTS: 1
4. Which of the following statements concerning the shareholder wealth maximization model is (are) true?
The timing of future profits is explicitly considered.
The model provides a conceptual basis for evaluating differential levels of risk.
The model is only valid for dividend-paying firms.
a and b
a, b, and c
ANS: PTS: 1
5. According to the profit-maximization goal, the firm should attempt to maximize short-run profits since there is too much uncertainty associated with long-run profits.
ANS: PTS: 1
6. According to the innovation theory of profit, above-normal profits are necessary to compensate the owners of the firm for the risk they assume when making their investments.
ANS: PTS: 1
7. According to the managerial efficiency theory of profit, above-normal profits can arise because of high-quality managerial skills.
ANS: PTS: 1
8. Which of the following (if any) is not a factor affecting the profit performance of firms:
differential risk
managerial skills
existence of monopoly power
all of the above are factors
ANS: PTS: 1
9. Agency problems and costs are incurred whenever the owners of a firm delegate decision-making authority to management.
ANS: PTS: 1
10. Economic profit is defined as the difference between revenue and ____.
explicit cost
total economic cost
implicit cost
shareholder wealth
none of the above
ANS: PTS: 1
11. Income tax payments are an example of ____.
implicit costs
explicit costs
normal return on investment
shareholder wealth
none of the above
ANS: PTS: 1
12. Various executive compensation plans have been employed to motivate managers to make decisions that maximize shareholder wealth. These include:
cash bonuses based on length of service with the firm
bonuses for resisting hostile takeovers
requiring officers to own stock in the company
large corporate staffs
a, b, and c only
ANS: PTS: 1
13. The common factors that give rise to all principal-agent problems include the
unobservability of some manager-agent action
presence of random disturbances in team production
the greater number of agents relative to the number of principals
a and b only
none of the above
ANS: PTS: 1
4. The Saturn Corporation (once a division of GM) was permanently closed in 2009. What went wrong with Saturn?
5. Saturn’s cars sold at prices higher than rivals Honda or Toyota, so they could not sell many cars.
6. Saturn sold cars below the prices of Honda or Toyota, earning a low 3% rate of return.
7. Saturn found that young buyers of Saturn automobiles were very loyal to Saturn and GM.
8. Saturn implemented a change management view that helped make first time Saturn purchasers trade up to Buick or Cadillac.
9. all of the above
ANS: PTS: 1
5. A Real Option Value is:
a. An option that been deflated by the cost of living index makes it a “real” option.
b. An opportunity cost of capital.
c. An opportunity to implement a new cost savings or revenue expansion activity that arises from business plans that the managers adopt.
d. An objective function and a decision rule that comes from it.
e. Both a and b.
ANS: PTS: 1
6. Which of the following will increase (V[0]), the shareholder wealth maximization model of the firm:
V[0]∙(shares outstanding) = Ó[t=1] (π[t ]) / (1+k[e])^t + Real Option Value.
1. a. Decrease the required rate of return (k[e]).
2. b. Decrease the stream of profits (π[t]).
3. c. Decrease the number of periods from to 10 periods.
4. d. Decrease the real option value.
5. e. All of the above.
ANS: PTS: 1
7. 17. The primary objective of a for-profit firm is to ___________.
8. maximize agency costs
9. minimize average cost
10. maximize total revenue
11. set output where total revenue equals total cost
e maximize shareholder value
ANS: PTS: 1
7. 18.Possible goals of Not-For-Profit (NFP) enterprises include all of the following EXCEPT:
8. maximize total costs
9. maximize output, subject to a breakeven constraint
10. maximize the happiness of the administrators of the NFP enterprise
11. maximize the utility of the contributors
12. a. and c.
ANS: PTS: 1
9. The flat-screen plasma TVs are selling extremely well. The originators of this technology are earning higher profits. What theory of profit best reflects the performance of the plasma screen
10. a. risk-bearing theory of profit
11. b. dynamic equilibrium theory of profit
12. c. innovation theory of profit
13. d. managerial efficiency theory of profit
14. e. stochastic optimization theory of profit
ANS: PTS: 1
1. To reduce Agency Problems, executive compensation should be designed to:
a. create incentives so that managers act like owners of the firm.
b. avoid making the executives own shares in the company.
c. be an increasing function of the firm’s expenses.
d. be an increasing function of the sales revenue received by the firm.
e. all of the above
ANS: PTS: 1
1. Recently, the American Medical Association changed its recommendations on the frequency of pap-smear exams for women. The new frequency recommendation was designed to address the family histories
of the patients. The optimal frequency should be where the marginal benefit of an additional pap-test:
1. a. equals zero.
2. b. is greater than the marginal cost of the test
3. c. is lower than the marginal cost of an additional test
4. d. equals the marginal cost of the test
5. e. both a and b.
ANS: PTS: 1
Chapter 2—Fundamental Economic Concepts
1. A change in the level of an economic activity is desirable and should be undertaken as long as the marginal benefits exceed the ____.
marginal returns
total costs
marginal costs
average costs
average benefits
ANS: PTS: 1
2. The level of an economic activity should be increased to the point where the ____ is zero.
marginal cost
average cost
net marginal cost
net marginal benefit
none of the above
ANS: PTS: 1
3. The net present value of an investment represents
an index of the desirability of the investment
the expected contribution of that investment to the goal of shareholder wealth maximization
the rate of return expected from the investment
a and b only
a and c only
ANS: PTS: 1
4. Generally, investors expect that projects with high expected net present values also will be projects with
low risk
high risk
certain cash flows
short lives
none of the above
ANS: PTS: 1
5. An closest example of a risk-free security is
General Motors bonds
AT&T commercial paper
U.S. Government Treasury bills
San Francisco municipal bonds
an I.O.U. that your cousin promises to pay you $100 in 3 months
ANS: PTS: 1
6. The standard deviation is appropriate to compare the risk between two investments only if
the expected returns from the investments are approximately equal
the investments have similar life spans
objective estimates of each possible outcome is available
the coefficient of variation is equal to 1.0
none of the above
ANS: PTS: 1
7. The approximate probability of a value occurring that is greater than one standard deviation from the mean is approximately (assuming a normal distribution)
none of the above
ANS: PTS: 1
8. Based on risk-return tradeoffs observable in the financial marketplace, which of the following securities would you expect to offer higher expected returns than corporate bonds?
U.S. Government bonds
municipal bonds
common stock
commercial paper
none of the above
ANS: PTS: 1
9. The primary difference(s) between the standard deviation and the coefficient of variation as measures of risk are:
the coefficient of variation is easier to compute
the standard deviation is a measure of relative risk whereas the coefficient of variation is a measure of absolute risk
the coefficient of variation is a measure of relative risk whereas the standard deviation is a measure of absolute risk
the standard deviation is rarely used in practice whereas the coefficient of variation is widely used
c and d
ANS: PTS: 1
10. The ____ is the ratio of ____ to the ____.
standard deviation; covariance; expected value
coefficient of variation; expected value; standard deviation
correlation coefficient; standard deviation; expected value
coefficient of variation; standard deviation; expected value
none of the above
ANS: PTS: 1
11. Sources of positive net present value projects include
buyer preferences for established brand names
economies of large-scale production and distribution
patent control of superior product designs or production techniques
a and b only
a, b, and c
ANS: PTS: 1
2. Receiving $100 at the end of the next three years is worth more to me than receiving $260 right now, when my required interest rate is 10%.
a. True
b. False
ANS: PTS: 1
13. The number of standard deviations z that a particular value of r is from the mean ȓ can be computed as z = (r – ȓ)/ ó. Suppose that you work as a commission-only insurance agent earning $1,000
per week on average. Suppose that your standard deviation of weekly earnings is $500. What is the probability that you zero in a week? Use the following brief z-table to help with this problem.
Z value Probability
-3 .0013
-2 .0228
-1 .1587
0 .5000
1. a. 1.3% chance of earning nothing in a week
2. b. 2.28% chance of earning nothing in a week
3. c. 15.87% chance of earning nothing in a week
4. d. 50% chance of earning nothing in a week
5. e. none of the above
ANS: PTS: 1
4. Consider an investment with the following payoffs and probabilities:
State of the Economy Probability Return
Stability .50 1,000
Good Growth .50 2,000
Determine the expected return for this investment.
a. 1,300
b. 1,500
c. 1,700
d. 2,000
e. 3,000
ANS: PTS: 1
5. Consider an investment with the following payoffs and probabilities:
State of the Economy Probability Return
GDP grows slowly .70 1,000
GDP grow fast .30 2,000
Let the expected value in this example be 1,300. How do we find thestandard deviationof the investment?
á. ó = { (1000-1300)^2 + (2000-1300)^2 }
â. ó = { (1000-1300) + (2000-1300) }
÷. ó = { (.5)(1000-1300)^2 + (.5)(2000-1300)^2 }
ä. ó = { (.7)(1000-1300) + (.3)(2000-1300) }
å. ó = { (.7)(1000-1300)^2 + (.3)(2000-1300)^2 }
ANS: PTS: 1
6. An investment advisor plans a portfolio your 85 year old risk-averse grandmother. Her portfolio currently consists of 60% bonds and 40% blue chip stocks. This portfolio is estimated to have an
expected return of 6% and with a standard deviation 12%. What is the probability that she makes less than 0% in a year? [A portion of Appendix B1 is given below, where z = (x – ì)/ó , with ì as
the mean and ó as the standard deviation.]
Table B1 for Z
Z Prob.
-3 .0013
-2.5 .0062
-2. .0228
-1.5 .0668
-1 .1587
-.5 ..3085
0 .5000
ANS: PTS: 1
7. Two investments have the following expected returns (net present values) and standard deviations:
PROJECT Expected Value Standard Deviation
Q $100,000 $20,000
X $50,000 $16,000
Based on the Coefficient of Variation, where the C.V. is the standard deviation dividend by the expected value.
All coefficients of variation are always the same.
Project Q is riskier than Project X
Project X is riskier than Project Q
Both projects have the same relative risk profile
There is not enough information to find the coefficient of variation.
PTS: 1
1. Suppose that the firm’s cost function is given in the following schedule (where Q is the level of output):
Output Total
Q (units) Cost
Determine the (a) marginal cost and (b) average total cost schedules
2. Complete the following table.
Total Marginal Average
Output Profit Profit Profit
−48 ______
1 −26 ______ ______
2 −8 ______ ______
3 6 ______ ______
4 16 ______ ______
5 22 ______ ______
6 24 ______ ______
7 22 ______ ______
8 16 ______ ______
9 6 ______ ______
10 −8 ______ ______
3. A firm has decided to invest in a piece of land. Management has estimated that the land can be sold in 5 years for the following possible prices:
Price Probability
10,000 .20
15,000 .30
20,000 .40
25,000 .10
(a) Determine the expected selling price for the land.
(b) Determine the standard deviation of the possible sales prices.
(c) Determine the coefficient of variation.
Chapter 3—Demand Analysis
1. Suppose we estimate that the demand elasticity for fine leather jackets is ‑.7 at their current prices. Then we know that:
a. a 1% increase in price reduces quantity sold by .7%.
b. no one wants to buy leather jackets.
c. demand for leather jackets is elastic.
d. a cut in the prices will increase total revenue.
e. leather jackets are luxury items.
ANS: PTS: 1
2. If demand were inelastic, then we should immediately:
3. cut the price.
4. keep the price where it is.
c. go to the Nobel Prize Committee to show we were the first to find an upward sloping demand curve.
1. stop selling it since it is inelastic.
2. raise the price.
ANS: PTS: 1
3. In this problem, demonstrate your knowledge of percentage rates of change of an entire demand function (Hint: %ÄQ = E[P]•%ÄP + E[Y]•%ÄY). You have found that the price elasticity of motor control
devices at Allen-Bradley Corporation is -2, and that the income elasticity is a +1.5. You have been asked to predict sales of these devices for one year into the future. Economists from the
Conference Board predict that income will be rising 3% over the next year, and AB’s management is planning to raise prices 2%. You expect that the number of AB motor control devices sold in one
year will:
fall .5%.
not change.
rise 1%r.
rise 2%.
rise .5%.
ANS: PTS: 1
4 A linear demand for lake front cabins on a nearby lake is estimated to be: Q[D] = 900,000 – 2P. What is the pointprice elasticity for lake front cabins at a price of P = $300,000? [Hint: E[p ]= (Q/
E[P] = -3.0
E[P] = -2.0
E[P] = -1.0
E[P] = -0.5
E[P] = 0
ANS: PTS: 1
5. Property taxes are the product of the tax rate (T) and the assessed value (V). The total property tax collected in your city (P) is: P = T•V. If the value of properties rise 4% and if Mayor and
City Council reduces the property the tax rate by 2%, what happens to the total amount of property tax collected? [hint: the percentage rate of change of a product is approximately the sum of the
percentage rates of change.}
6. It rises 6 %.
7. It rises 4 %.
8. It rises 3 %.
9. It rises 2 %
10. If falls 2%.
ANS: PTS: 1
6. Demand is given by Q[D] = 620 ‑ 10·P and supply is given by Q[S] = 100 + 3·P. What is the price and quantity when the market is in equilibrium?
a. The price will be $30 and the quantity will be 132 units.
b. The price will be $11 and the quantity will be 122 units.
c. The price will be $40 and the quantity will be 220 units.
d. The price will be $35 and the quantity will be 137 units
e. The price will be $10 and the quantity will be 420 units.
ANS: PTS: 1
7. Which of the following would tend to make demand INELASTIC?
a. the amount of time analyzed is quite long
b. there are lots of substitutes available
c. the product is highly durable
d. the proportion of the budget spent on the item is very small
e. no one really wants the product at all
ANS: PTS: 1
8. Which of the following best represents management’s objective(s) in utilizing demand analysis?
it provides insights necessary for the effective manipulation of demand
it helps to measure the efficiency of the use of company resources
it aids in the forecasting of sales and revenues
a and b
a and c
ANS: PTS: 1
9. Identify the reasons why the quantity demanded of a product increases as the price of that product decreases.
as the price declines, the real income of the consumer increases
as the price of product A declines, it makes it more attractive than product B
as the price declines, the consumer will always demand more on each successive price reduction
a and b
a and c
ANS: PTS: 1
10. An increase in the quantity demanded could be caused by:
an increase in the price of substitute goods
a decrease in the price of complementary goods
an increase in consumer income levels
all of the above
none of the above
ANS: PTS: 1
1. Iron ore is an example of a:
durable good
producers’ good
nondurable good
consumer good
none of the above
ANS: PTS: 1
12. If the cross price elasticity measured between items A and B is positive, the two products are referred to as:
inelastic as compared to each other
both b and c
a, b, and c
ANS: PTS: 1
3. When demand is ____ a percentage change in ____ is exactly offset by the same percentage change in ____ demanded, the net result being a constant total consumer expenditure.
elastic; price; quantity
unit elastic; price; quantity
inelastic; quantity; price
inelastic; price; quantity
none of the above
ANS: PTS: 1
14. Marginal revenue (MR) is ____ when total revenue is maximized.
greater than one
equal to one
less than zero
equal to zero
equal to minus one
ANS: PTS: 1
15. The factor(s) which cause(s) a movement along the demand curve include(s):
increase in level of advertising
decrease in price of complementary goods
increase in consumer disposable income
decrease in price of the good demanded
all of the above
ANS: PTS: 1
16. An increase in each of the following factors would normally provide a subsequent increase in quantity demanded, except:
price of substitute goods
level of competitor advertising
consumer income level
consumer desires for goods and services
a and b
ANS: PTS: 1
17. Producers’ goods are:
consumers’ goods
raw materials combined to produce consumer goods
durable goods used by consumers
always more expensive when used by corporations
none of the above
ANS: PTS: 1
18. The demand for durable goods tends to be more price elastic than the demand for non-durables.
ANS: PTS: 1
19. A price elasticity (E[D]) of −1.50 indicates that for a ____ increase in price, quantity demanded will ____ by ____.
one percent; increase; 1.50 units
one unit; increase; 1.50 units
one percent; decrease; 1.50 percent
one unit; decrease; 1.50 percent
ten percent; increase; fifteen percent
ANS: PTS: 1
20. Those goods having a calculated income elasticity that is negative are called:
producers’ goods
durable goods
inferior goods
nondurable goods
none of the above
ANS: PTS: 1
21. An income elasticity (E[y]) of 2.0 indicates that for a ____ increase in income, ____ will increase by ____.
one percent; quantity supplied; two units
one unit; quantity supplied; two units
one percent; quantity demanded; two percent
one unit; quantity demanded; two units
ten percent; quantity supplied; two percent
ANS: PTS: 1
22. When demand elasticity is ____ in absolute value (or ____), an increase in price will result in a(n) ____ in total revenues.
less than 1; elastic; increase
more than 1; inelastic; decrease
less than 1; elastic; decrease
less than 1; inelastic; increase
none of the above
ANS: PTS: 1
23. Empirical estimates of the price elasticity of demand [in Table 3.4] suggest that the demand for household consumption of alcoholic beverages is:
highly price elastic
price inelastic
unitarily elastic
an inferior good
none of the above
ANS: PTS: 1
1. The manager of the Sell-Rite drug store accidentally mismarked a shipment of 20-pound bags of charcoal at $4.38 instead of the regular price of $5.18. At the end of a week, the store’s inventory
of 200 bags of charcoal was completely sold out. The store normally sells an average of 150 bags per week.
(a) What is the store’s arc elasticity of demand for charcoal?
(b) Give an economic interpretation of the numerical value obtained in part (a)
PTS: 1
2. The Future Flight Corporation manufactures a variety of Frisbees selling for $2.98 each. Sales have averaged 10,000 units per month during the last year. Recently Future Flight’s closest
competitor, Soaring Free Company, cut its prices on similar Frisbees from $3.49 to $2.59. Future Flight noticed that its sales declined to 8,000 units per month after the price cut.
(a) What is the arc cross elasticity of demand between Future Flight’s and Soaring Free’s Frisbees?
(b) If Future Flight knows the arc price elasticity of demand for its Frisbees is −2.2, what price would they have to charge in order to obtain the same level of sales as before Soaring Free’s price
PTS: 1
3. The British Automobile Company is introducing a brand new model called the “London Special.” Using the latest forecasting techniques, BAC economists have developed the following demand function
for the “London Special”:
Q[D] = 1,200,000 − 40P
What is the point price elasticity of demand at prices of (a) $8,000 and (b) $10,000?
PTS: 1
4. Hanna Corporation markets a compact microwave oven. In 2010 they sold 23,000 units at $375 each. Per capita disposable income in 2010 was $6,750. Hanna economists have determined that the arc
price elasticity for this microwave oven is −1.2.
(a) In 2011 Hanna is planning to lower the price of the microwave oven to $325. Forecast sales volume for 2011 assuming that all other things remain equal.
However, in checking with government economists, Hanna finds that per capita disposable income is expected to rise to $7,000 in 2011. In the past the company has observed an arc income elasticity
(b) of +2.5 for microwave ovens. Forecast 2011 sales given that the price is reduces to $325 and that per capita disposable income increases to $7,000. Assume that the price and income effects are
independent and additive.
Chapter 4—Estimating Demand
1. 1. Using a sample of 100 consumers, a double-log regression model was used to estimate demand for gasoline. Standard errors of the coefficients appear in the parentheses below the coefficients.
Ln Q = 2.45 -0.67 Ln P + . 45 Ln Y – .34 Ln P[cars]
(.20) (.10) (.25)
Where Q is gallons demanded, P is price per gallon, Y is disposable income, and P[cars] is a price index for cars. Based on this information, which is NOT correct?
8. Gasoline is inelastic.
9. Gasoline is a normal good.
10. Cars and gasoline appear to be mild complements.
11. The coefficient on the price of cars (P[cars]) is insignificant.
12. All of the coefficients are insignificant.
ANS: PTS: 1
8. 2. In a cross section regression of 48 states, the following linear demand for per-capita cans of soda was found: Cans = 159.17 – 102.56 Price + 1.00 Income + 3.94Temp
Coefficients Standard Error t Stat
Intercept 159.17 94.16 1.69
Price -102.56 33.25 -3.08
Income 1.00 1.77 0.57
Temperature 3.94 0.82 4.83
R-Sq = 54.1% R-Sq(adj) = 51.0%
From the linear regression results in the cans case above, we know that:
1. Price is insignificant
2. Income is significant
3. Temp is significant
4. As price rises for soda, people tend to drink less of it
5. All of the coefficients are significant
ANS: PTS: 1
3. A study of expenditures on food in cities resulting in the following equation:
Log E = 0.693 Log Y + 0.224 Log N
where E is Food Expenditures; Y is total expenditures on goods and services; and N is the size of the family. This evidence implies:
1. that as total expenditures on goods and services rises, food expenditures falls.
2. that a one-percent increase in family size increases food expenditures .693%.
3. that a one-percent increase in family size increases food expenditures .224%.
4. that a one-percent increase in total expenditures increases food expenditures 1%.
5. that as family size increases, food expenditures go down.
ANS: PTS: 1
4. All of the following are reasons why an association relationship may not imply a causal relationship except:
the association may be due to pure chance
the association may be the result of the influence of a third common factor
both variables may be the cause and the effect at the same time
the association may be hypothetical
both c and d
ANS: PTS: 1
5. In regression analysis, the existence of a significant pattern in successive values of the error term constitutes:
a simultaneous equation relationship
ANS: PTS: 1
6. In regression analysis, the existence of a high degree of intercorrelation among some or all of the explanatory variables in the regression equation constitutes:
a simultaneous equation relationship
ANS: PTS: 1
7. When using a multiplicative power function (Y = a X[1]^b1 X[2]^b2 X[3]^b3) to represent an economic relationship, estimates of the parameters (a, and the b’s) using linear regression analysis can
be obtained by first applying a ____ transformation to convert the function to a linear relationship.
ANS: PTS: 1
8. The correlation coefficient ranges in value between 0.0 and 1.0.
ANS: PTS: 1
9. The coefficient of determination ranges in value between 0.0 and 1.0.
ANS: PTS: 1
10. The coefficient of determination measures the proportion of the variation in the independent variable that is “explained” by the regression line.
ANS: PTS: 1
1. The presence of association between two variables does not necessarily imply causation for the following reason(s):
the association between two variables may result simply from pure chance
the association between two variables may be the result of the influence of a third common factor
both variables may be the cause and the effect at the same time
a and b
a, b, and c
ANS: PTS: 1
2. The estimated slope coefficient (b) of the regression equation (Ln Y = a + b Ln X) measures the ____ change in Y for a one ____ change in X.
percentage, unit
percentage, percent
unit, unit
unit, percent
none of the above
ANS: PTS: 1
3. The standard deviation of the error terms in an estimated regression equation is known as:
coefficient of determination
correlation coefficient
Durbin-Watson statistic
standard error of the estimate
none of the above
ANS: PTS: 1
4. In testing whether each individual independent variables (Xs) in a multiple regression equation is statistically significant in explaining the dependent variable (Y), one uses the:
Durbin-Watson test
none of the above
ANS: PTS: 1
15. One commonly used test in checking for the presence of autocorrelation when working with time series data is the ____.
Durbin-Watson test
none of the above
ANS: PTS: 1
16. The method which can give some information in estimating demand of a product that hasn’t yet come to market is:
the consumer survey
market experimentation
a statistical demand analysis
plotting the data
the barometric method
ANS: PTS: 1
17. Demand functions in the multiplicative form are most common for all of the following reasons except:
elasticities are constant over a range of data
ease of estimation of elasticities
exponents of parameters are the elasticities of those variables
marginal impact of a unit change in an individual variable is constant
c and d
ANS: PTS: 1
18. The Identification Problem in the development of a demand function is a result of:
the variance of the demand elasticity
the consistency of quantity demanded at any given point
the negative slope of the demand function
the simultaneous relationship between the demand and supply functions
none of the above
ANS: PTS: 1
19. Consider the following linear demand function where Q[D] = quantity demanded, P = selling price, and Y = disposable income:
Q[D] = −36 −2.1P + .24Y
The coefficient of P (i.e., −2.1) indicates that (all other things being held constant):
for a one percent increase in price, quantity demanded would decline by 2.1 percent
for a one unit increase in price, quantity demanded would decline by 2.1 units
for a one percent increase in price, quantity demanded would decline by 2.1 units
for a one unit increase in price, quantity demanded would decline by 2.1 percent
none of the above
ANS: PTS: 1
1. Consider the following multiplicative demand function where Q[D] = quantity demanded, P = selling price, and Y = disposable income:
The coefficient of Y (i.e., .2) indicates that (all other things being held constant):
for a one percent increase in disposable income, quantity demanded would increase by .2 percent
for a one unit increase in disposable income, quantity demanded would increase by .2 units
for a one percent increase in disposable income quantity demanded would increase by .2 units
for a one unit increase in disposable income, quantity demanded would increase by .2 percent
none of the above
ANS: PTS: 1
1. One shortcoming of the use of ____ in demand analysis is that the participants are generally aware that their actions are being observed and hence they may seek to act in a manner somewhat
different than normal.
market experiments
consumer clinics
statistical (econometric) methods
a and b
none of the above
ANS: PTS: 1
2. The constant or intercept term in a statistical demand study represents the quantity demanded when all independent variables are equal to:
their minimum values
their average values
none of the above
ANS: PTS: 1
3. Novo Nordisk A/S, a Danish firm, sells insulin and other drugs worldwide. Activella, an estrogen and progestin hormone replacement therapy sold by Novo-Nordisk, is examined using 33 quarters of
Y = -204 + . 34X[1] – .17X[2]
(17.0) (-1.71)
Where Y is quarterly sales of Activella, X[1] is the Novo’s advertising of the hormone therapy, and X[2] is advertising of a similar product by Eli Lilly and Company, Novo-Nordisk’s chief competitor.
The parentheses contain t-values. Addition information is: Durbin-Watson = 1.9 and R^2 = .89.
Using the data for Novo-Nordisk, which is correct?
1. Both X[1] and X[2 ]are statistically significant.
2. Neither X[1] nor X[2 ]are statistically significant.
3. X[1] is statistically significant but X[2 ]is not statistically significant.
4. X[1] is not statistically significant but X[2 ]is statistically significant.
5. The Durbin-Watson statistic shows significant problems with autocorrelation
ANS: PTS: 1
4. In which of the following econometric problems do we find Durbin-Watson statistic being far away from 2.0?
5. the identification problem
6. autocorrelation
7. multicollinearity
8. heteroscedasticity
9. agency problems
ANS: PTS: 1
5. When there is multicollinearity in an estimated regression equation,
a. the coefficients are likely to be small.
b. the t‑statistics are likely to be small even though the R^2 is large.
c. the coefficient of determination is likely to be small.
d. the problem of omitted variables is likely.
e. the error terms will tend to have a cyclical pattern.
6. When two or more “independent” variables are highly correlated, then we have:
7. a. the identification problem
8. b. multicollinearity
9. c. autocorrelation
10. d. heteroscedasticity
11. e. complementary products
ANS: PTS: 1
7. Which is NOT true about the coefficient of determination?
8. As you add more variables, the R-square generally rises.
9. As you add more variables, the adjusted R-square can fall.
c. If the R-square is above 50%, the regression is considered significant.
1. The R-square gives the percent of the variation in the dependent variable that is explained by the independent variables.
2. The higher is the R-square, the better is the fit.
ANS: PTS: 1
1. Phoenix Lumber Company uses the number of construction permits issued to help estimate demand (sales). The firm collected the following data on annual sales and number of construction permits
issued in its market area:
No. of Construction Sales
Year Permits Issued (000) (1,000,000)
2003 6.50 10.30
2004 6.20 10.10
2005 6.60 10.50
2006 7.30 10.80
2007 7.80 11.20
2008 8.20 11.40
2009 8.30 11.30
(a) Which variable is the dependent variable and which is the independent variable?
(b) Determine the estimated regression line.
(c) Test the hypothesis (at the .05 significance level) that there is no relationship between the variables.
(d) Calculate the coefficient of determination. Give an economic interpretation to the value obtained.
(e) Perform an analysis of variance on the regression including an F-test (at the .05 significance level) of the overall significance of the results.
(f) Suppose that 8,000 construction permits are expected to be issued in 2010. What would be the point estimate of Phoenix Lumber Company’s sales for 2010?
(a) Give the regression equation for predicting restaurant sales.
(b) Give an interpretation of each of the estimated regression coefficients.
(c) Which of the independent variables (if any) are statistically significant at the .05 level in “explaining” restaurant sales?
(d) What proportion of the variation in restaurant sales is “explained” by the regression equation?
(e) Perform an F-test (at the .05 significance level) of the overall explanatory power of the regression model.
PTS: 1 NOTE: This problem requires the use of statistical tables.
3. The following demand function has been estimated for Fantasy pinball machines:
Q[D] = 3,500 − 40P + 17.5P[x] + 670U + .0090A + 6,500N
where P = monthly rental price of Fantasy pinball machines
P[x] = monthly rental price of Old Chicago pinball machines (their largest competitor)
U = current unemployment rate in the 10 largest metropolitan areas
A = advertising expenditures for Fantasy pinball machines
N = fraction of the U.S. population between ages 10 and 30
(a) What is the point price elasticity of demand for Fantasy pinball machines when P = $150, P[x] = $100, U = .12, A = $200,000 and N = .35?
(b) What is the point cross elasticity of demand with respect to Old Chicago pinball machines for the values of the independent variables given in part (a)?
PTS: 1
4. Given the following demand function:
Q = 2.0 P^−^1.33 Y^2.0 A^.50
where Q = quantity demanded (thousands of units)
P = price ($/unit)
Y = disposable income per capita ($ thousands)
A = advertising expenditures ($ thousands)
determine the following when P = $2/unit, Y = $8 (i.e., $8000), and A = $25 (i.e., $25,000)
(a) Price elasticity of demand
(b) The approximate percentage increase in demand if disposable income percentage increases by 3%.
(c) The approximate percentage increase in demand if advertising expenditures are increased by 5 percent.
Chapter 5—Business and Economic Forecasting
1. Time-series forecasting models:
are useful whenever changes occur rapidly and wildly
are more effective in making long-run forecasts than short-run forecasts
are based solely on historical observations of the values of the variable being forecasted
attempt to explain the underlying causal relationships which produce the observed outcome
none of the above
ANS: PTS: 1
2. The forecasting technique which attempts to forecast short-run changes and makes use of economic indicators known as leading, coincident or lagging indicators is known as:
econometric technique
time-series forecasting
opinion polling
barometric technique
judgment forecasting
ANS: PTS: 1
3. The use of quarterly data to develop the forecasting model Y[t] = a +bY[t][−][1] is an example of which forecasting technique?
Barometric forecasting
Time-series forecasting
Survey and opinion
Econometric methods based on an understanding of the underlying economic variables involved
Input-output analysis
ANS: PTS: 1
4. Variations in a time-series forecast can be caused by:
cyclical variations
secular trends
seasonal effects
a and b only
a, b, and c
ANS: PTS: 1
5. The variation in an economic time-series which is caused by major expansions or contractions usually
of greater than a year in duration is known as:
secular trend
cyclical variation
seasonal effect
unpredictable random factor
none of the above
ANS: PTS: 1
6. The type of economic indicator that can best be used for business forecasting is the:
leading indicator
coincident indicator
lagging indicator
current business inventory indicator
optimism/pessimism indicator
ANS: PTS: 1
7. Consumer expenditure plans is an example of a forecasting method. Which of the general categories best described this example?
time-series forecasting techniques
barometric techniques
survey techniques and opinion polling
econometric techniques
input-output analysis
ANS: PTS: 1
8. In the first-order exponential smoothing model, the new forecast is equal to a weighted average of the old forecast and the actual value in the most recent period.
ANS: PTS: 1
9. Simplified trend models are generally appropriate for predicting the turning points in an economic time series.
ANS: PTS: 1
10. Smoothing techniques are a form of ____ techniques which assume that there is an underlying pattern to be found in the historical values of a variable that is being forecast.
opinion polling
barometric forecasting
econometric forecasting
time-series forecasting
none of the above
ANS: PTS: 1
11. Seasonal variations can be incorporated into a time-series model in a number of different ways, including:
ratio-to-trend method
use of dummy variables
root mean squared error method
a and b only
a, b, and c
ANS: PTS: 1
2. For studying demand relationships for a proposed new product that no one has ever used before, what would be the best method to use?
3. ordinary least squares regression on historical data
4. market experiments, where the price is set differently in two markets
5. consumer surveys, where potential customers hear about the product and are asked their opinions
6. double log functional form regression model
7. all of the above are equally useful in this case
ANS: PTS: 1
3. Which of the following barometric indicators would be the most helpful for forecasting future sales for an industry?
a. lagging economic indicators.
b. leading economic indicators.
c. coincident economic indicators.
d. wishful thinking
e. none of the above
ANS: PTS: 1
4. An example of a time series data set is one for which the:
a. data would be collected for a given firm for several consecutive periods (e.g., months).
b. data would be collected for several different firms at a single point in time.
c. regression analysis comes from data randomly taken from different points in time.
d. data is created from a random number generation program.
d. use of regression analysis would impossible in time series.
ANS: PTS: 1
5. Examine the plot of data.
It is likely that the best forecasting method for this plot would be:
1. a two-period moving average
2. a secular trend upward
3. a seasonal pattern that can be modeled using dummy variables or seasonal adjustments
4. a semi-log regression model
5. a cubic functional form
ANS: PTS: 1
16. Emma uses a linear model to forecast quarterly same-store sales at the local Garden Center. The results of her multiple regression is:
Sales = 2,800 + 200•T – 350•D
where T goes from 1 to 16 for each quarter of the year from the first quarter of 2006 (‘06I) through the fourth quarter of 2009 (‘09 IV). D is a dummy variable which is 1 if sales are in the cold and
dreary first quarter, and zero otherwise, because the months of January, February, and March generate few sales at the Garden Center. Use this model to estimate sales in a store for the first quarter
of 2010 in the 17^th month; that is: {2010 I}. Emma’s forecast should be:
3. a. 5,950
4. b. 6,200
5. c. 6,350
6. d. 6,000
7. e. 5,850
ANS: PTS: 1
17. Select the correctstatement.
a. Qualitative forecasts give the direction of change.
b. Quantitative forecasts give the exact amount or exact percentage change.
c. Diffusion forecasts use the proportion of the forecasts that are positive to forecast up or down.
d. Surveys are a form of qualitative forecasting.
e. all of the above are correct.
ANS: PTS: 1
8. If two alternative economic models are offered, other things equal, we would
a. tend to pick the one with the lowest R^2.
b. select the model that is the most expensive to estimate.
c. pick the model that was the most complex.
d. select the model that gave the most accurate forecasts
e. all of the above
ANS: PTS: 1
9. Mr. Geppettouses exponential smoothing to predict revenue in his wood carving business. He uses a weight of ù = .4 for the naïve forecast and (1-ù) = .6 for the past forecast. What revenue did he
predict for March using the data below? Select closet answer.
Nov 100 100
Dec 90 100
Jan 115 —-
Feb 110 —-
MARCH ? ?
1. 106.2
2. 104.7
3. 103.2
4. 102.1
5. 101.7
ANS: PTS: 1
1. Suppose a plot of sales data over time appears to follow an S-shape as illustrated below.
Which of the following is likely that the best forecasting functional form to use for sales data above?
1. A linear trend, Sales = a + b T
2. A quadratic shape in T, using T-squared as another variable, Sales = a + b T + cT^2.
3. A semi-log form as sales appear to be growing at a constant percentage rate, Ln Sales = a + bT
4. A cubic shape in T, using T-squared and T-cubed as variables, Sales = a + b T + cT^2 + d T^3.
5. A quadratic shape in T and T-squared as variables, Sales = a + b T + cT^2
ANS: PTS: 1
1. The Accuweather Corporation manufactures barometers and thermometers for weather forecasters. In an attempt to forecast its future needs for mercury, Accuweather’s chief economist estimated
average monthly mercury needs as:
N = 500 + 10X
where N = monthly mercury needs (units) and X = time period in months (January 2008= 0). The following monthly seasonal adjustment factors have been estimated using data from the past five years:
Month Adjustment Factor
January 15%
April 10%
July −20%
September 5%
December −10%
(a) Forecast Accuweather’s mercury needs for January, April, July, September, and December of 2010.
(b) The following actual and forecast values of mercury needs in the month of November have been recorded:
Year Actual Forecast
What seasonal adjustment factor should the firm use for November?
2. Milner Brewing Company experienced the following monthly sales (in thousands of barrels) during 2010:
(a) Develop 2-month moving average forecasts for March through July.
(b) Develop 4-month moving average forecasts for May through July.
(c) Develop forecasts for February through July using the exponential smoothing method (with w = .5). Begin by assuming .
Chapter 6—Managing Exports
1. Using demand and supply curves for the Japanese yen based on the $/¥ price for yen, an increase in US INFLATION RATES would
1. Decrease the demand for yen and decrease the supply of the yen.
2. Increase the demand for yen and decrease the supply of the yen.
3. Increase the demand and increase the supply of yen.
4. Decrease both the supply and the demand of yen.
5. Have no impact on the demand or supply of the yen.
ANS: PTS: 1
2. If the British pound (₤) appreciates by 10% against the dollar:
3. both the US importers from Britain and US exporters to Britain will be helped by the appreciating pound.
4. the US exporters will find it harder to sell to foreign customers in Britain.
5. the US importer of British goods will tend to find that their cost of goods rises, hurting its bottom line.
6. both US importers of British goods and exporters to Britain will be unaffected by changes in foreign exchange rates.
7. all of the above.
ANS: PTS: 1
3. Purchasing power parity or PPP says the ratios composed of:
a. interest rates explain the direction of exchange rates.
b. growth rates explain the direction of exchange rates.
c. inflation rates explain the direction of exchange rates.
d. services explain the direction exchange rates.
e. public opinion polls explain the direction of exchange rates.
ANS: PTS: 1
4. If Ben Bernanke, Chair of the Federal Reserve Board, begins to tighten monetary policy by raising US interest rates next year, what is the likely impact on the value of the dollar?
a. The value of the dollar falls when US interest rates rise.
b. The value of the dollar rises when US interest rates rise.
c. The value of the dollar is not related to US interest rates.
d. This is known as Purchasing Power Parity or PPP.
e. The Federal Reserve has no impact at all on interest rates.
ANS: PTS: 1
5. If the domestic prices for traded goods rises 5% in Japan and rises 7% the US over the same period, what would happened to the Yen/US dollar exchange rate? Hint: S[1]/S[0] = (1+π[h]) / (1+ π[f])
where S[0] is the direct quote of the yen at time 0, the current period.
6. The direct quote of the yen ($/¥) rises, and the value of the dollar falls.
7. The direct quote of the yen ($/¥) falls, and the value of the dollar rises.
8. The direct quote of the yen would remain the same.
9. Purchasing power parity does not apply to inflation rates.
10. Both a and d.
ANS: PTS: 1
6. US and Canada can both grow wheat and can do mining. Use the following table to look for which country has a comparative advantage in mining. (Hint: Find the cost of mining in terms of wheat in
each country.)
Absolute Cost in US Absolute Cost in Canada
Wheat $5 C$8
Mining $10 C$12
1. Canada has a comparative advantage in mining.
2. The US has a comparative advantage in mining.
3. No comparative advantage in mining exists for either nation.
4. We must first know the exchange rate to be able to answer this question.
5. Both a and b.
ANS: PTS: 1
7. The optimal currency area involves a trade-off of reducing transaction costs but the inability to use changes in exchange rates to help ailing regions. If the US, Canada, and Mexico had one
single currency (the Peso-Dollar) we would tend to see all of the following EXCEPT:
8. Even more intraregional trade of goods across the three countries.
9. Lower transaction costs of trading within North America.
10. A greater difficulty in helping Mexico as you can no longer deflate the Mexican peso.
11. Less migration of workers across the three countries.
12. An elimination of correlated macroeconomic shocks across the countries.
ANS: PTS: 1
8. If the value of the U.S. dollar rises from 1.0 per dollar to 1.3 per dollar,
imports of automobiles from Germany will decline
American inflation will increase
German exports of all traded goods will decline
American exports to Germany will decrease
sales by American manufacturers for the export markets will increase.
ANS: PTS: 1
9. An appreciation of the U.S. dollar has what impact on Harley-Davidson (HD), a U.S. manufacturer of motorcycles?
domestic sales of HD motorcycles increase and foreign sales of HD motorcycles increase
domestic sales of HD motorcycles decrease and foreign sales of HD motorcycles increase
domestic sales of HD motorcycles increase and foreign sales of HD motorcycles decrease
domestic sales of HD motorcycles decrease and foreign sales of HD motorcycles decrease
only manufacturers who produce traded goods are affected
ANS: PTS: 1
10. In the last twenty-five years, the Yen and German mark and now the Euro have
fluctuated widely against the dollar
appreciated against the dollar and then depreciated against the dollar
exchanged without restrictions
all of the above
none of the above
ANS: PTS: 1
1. In an open economy with few capital restrictions and substantial import-export trade, a rise in interest rates and a decline in the producer price index of inflation will
raise the value of the currency
lower the nominal interest rate
increase the volume of trading in the foreign exchange market
lower the trade-weighted exchange rate
increase consumer inflation.
ANS: PTS: 1
12. When a manufacturer’s home currency appreciates substantially,
domestic sales decline
foreign sales decline
company-owned foreign plant and equipment will increase
margins often decline
all of the above
ANS: PTS: 1
13. An increase in the exchange rate of the U.S. dollar relative to a trading partner can result from
higher anticipated costs of production in the U.S.
higher interest rates and higher inflation in the U.S.
higher growth rates in the trading partner’s economy
a change in the terms of trade
lower export industry productivity
ANS: PTS: 1
4. The purchasing power parity hypothesis implies that an increase in inflation in one country relative to another will over a long period of time
increase exports
reduce the competitive pressure on prices
lower the value of the currency in the country with the higher inflation rate
increase foreign aid
increase the speculative demand for the currency
ANS: PTS: 1
15. Trading partners should specialize in producing goods in accordance with comparative advantage, then trade and diversify in consumption because
out-of-pocket costs of production decline
free trade areas protect infant industries
economies of scale are present
manufacturers face diminishing returns
more goods are available for consumption
ANS: PTS: 1
16. European Union labor costs exceed U.S. and British labor costs primarily because
worker productivity is lower in the EU
union wages are higher in the EU
layoffs and plant closings are more restrictive in the U.S. and Britain
the amount of paid time off is higher in the EU
labor-management relations are better in the EU
ANS: PTS: 1
17. Companies that reduce their margins on export products in the face of appreciation of their home currency may be motivated by a desire to
sacrifice market share abroad but build market share at home
increase production volume to realize learning curve advantages
sell foreign plants and equipment to lower their debt
reduce the costs of transportation
all of the above
ANS: PTS: 1
18. In a recession, the trade balance often improves because
service exports exceed manufactured good exports
banks sell depressed assets
fewer households can afford luxury imports
direct investment abroad declines
the capital account exceeds the current account
ANS: PTS: 1
1. Suppose nominal interest rates in the U.S. rise from 4.6% to 5% and decline in Britain from 6% to 5.5%, while U.S. consumer inflation remains unchanged at 1.9% and British inflation declines from
4% to 3%. In addition suppose, real growth in the U.S. is forecasted for next year at 4% and in Britain real growth is forecasted at 5%. Finally, suppose producer price inflation in the U.S. is
declining from 2% to 1% while in Britain producer price inflation is rising from 2% to 3.2%. Explain what effect each of these factors would have on the long-term trend exchange rate ( per $) and
Chapter 7—Production Economics
1. 1. What’s true about both the short-run and long-run in terms of production and cost analysis?
2. a. In the short-run, one or more of the resources are fixed
3. b. In the long-run, all the factors are variable
4. c. The time horizon determines whether or not an input variable is fixed or not
5. d. The law of diminishing returns is based in part on some factors of production being fixed, as they are in the short run.
6. e. All of the above
ANS: PTS: 1
9. 2. The marginal product is defined as:
1. The ratio of total output to the amount of the variable input used in producing the output
2. The incremental change in total output that can be produced by the use of one more unit of the variable input in the production process
3. The percentage change in output resulting from a given percentage change in the amount
4. The amount of fixed cost involved.
5. None of the above
ANS: PTS: 1
3. Fill in the missing data to solve this problem.
Variable Total Average Marginal
Input Product Product Product
4 ? 70 —-
5 ? ? 40
6 350 ? ?
What is the total product for 5 units of input, and what is the marginal product for 6 units of input?
a. 320 and 30
b. 350 and 20
c. 360 and 15
d. 400 and 10
e. 430 and 8
ANS: PTS: 1
1. 4. The following is a Cobb-Douglas production function: Q = 1.75K^0.5∙L^0.5. What is correct here?
2. a. A one-percent change in L will cause Q to change by one percent
3. b. A one-percent change in K will cause Q to change by two percent
4. c. This production function displays increasing returns to scale
5. d. This production function displays constant returns to scale
6. e. This production function displays decreasing returns to scale
ANS: PTS: 1
5. Suppose you have a Cobb-Douglas function with a capital elasticity of output (α) of 0.28 and a labor elasticity of output (β) of 0.84. What statement is correct?
a. There are increasing returns to scale
b. If the amount of labor input (L) is increased by 1%, the output will increase by 0.84%
c. If the amount of capital input (K) is decreased by 1%, the output will decrease by 0.28%
d. The sum of the exponents in the Cobb-Douglas function is 1.12.
e. All of the above
ANS: PTS: 1
1. 6. The Cobb-Douglas production function is: Q = 1.4*L^0.6*K^0.5. What would be the percentage change in output (%∆Q) if labor grows by 3.0% and capital is cut by 5.0%?
[Hint: %∆Q = (E[L] * %∆L) + (E[K] * %∆K)]
1. a. %∆Q = + 3.0%
2. b. %∆Q = + 5.0%
3. c. %∆Q = – 0.70%
4. d. %∆Q = – 2.50%
5. e. %∆Q = – 5.0%
ANS: PTS: 1
1. 7. If the marginal product of labor is 100 and the price of labor is 10, while the marginal product of capital is 200 and the price of capital is $30, then what should the firm?
2. a. The firm should use relatively more capital
3. b. The firm should use relatively more labor
4. c. The firm should not make any changes – they are currently efficient
5. d. Using the Equimarginal Criterion, we can’t determine the firm’s efficiency level
6. e. Both c and d
ANS: PTS: 1
8. The marginal rate of technical substitution may be defined as all of the following except:
the rate at which one input may be substituted for another input in the production process, while total output remains constant
equal to the negative slope of the isoquant at any point on the isoquant
the rate at which all combinations of inputs have equal total costs
equal to the ratio of the marginal products of X and Y
b and c
ANS: PTS: 1
9. The law of diminishing marginal returns:
states that each and every increase in the amount of the variable factor employed in the production process will yield diminishing marginal returns
is a mathematical theorem that can be logically proved or disproved
is the rate at which one input may be substituted for another input in the production process
none of the above
ANS: PTS: 1
10. The combinations of inputs costing a constant C dollars is called:
an isocost line
an isoquant curve
the MRTS
an isorevenue line
none of the above
ANS: PTS: 1
11. In a relationship among total, average and marginal products, where TP is maximized:
AP is maximized
AP is equal to zero
MP is maximized
MP is equal to zero
none of the above
ANS: PTS: 1
12. Holding the total output constant, the rate at which one input X may be substituted for another input Y in a production process is:
the slope of the isoquant curve
the marginal rate of technical substitution (MRTS)
equal to MP[x]/MP[y]
all of the above
none of the above
ANS: PTS: 1
13. Which of the following is never negative?
marginal product
average product
production elasticity
marginal rate of technical substitution
slope of the isocost lines
ANS: PTS: 1
14. Concerning the maximization of output subject to a cost constraint, which of the following statements (if any) are true?
At the optimal input combination, the slope of the isoquant must equal the slope of the isocost line.
The optimal solution occurs at the boundary of the feasible region of input combinations.
The optimal solution occurs at the point where the isoquant is tangent to the isocost lines.
all of the above
none of the above
ANS: PTS: 1
15. In a production process, an excessive amount of the variable input relative to the fixed input is being used to produce the desired output. This statement is true for:
stage II
stages I and II
when Ep = 1
stage III
none of the above
ANS: PTS: 1
16. Marginal revenue product is:
defined as the amount that an additional unit of the variable input adds to the total revenue
equal to the marginal factor cost of the variable factor times the marginal revenue resulting from the increase in output obtained
equal to the marginal product of the variable factor times the marginal product resulting from the increase in output obtained
a and b
a and c
ANS: PTS: 1
17. The isoquants for inputs that are perfect substitutes for one another consist of a series of:
right angles
parallel lines
concentric circles
right triangles
none of the above
ANS: PTS: 1
8. In production and cost analysis, the short run is the period of time in which one (or more) of the resources employed in the production process is fixed or incapable of being varied.
ANS: PTS: 1
19. Marginal revenue product is defined as the amount that an additional unit of the variable input adds to ____.
marginal revenue
total output
total revenue
marginal product
none of the above
ANS: PTS: 1
20. Marginal factor cost is defined as the amount that an additional unit of the variable input adds to ____.
marginal cost
variable cost
marginal rate of technical substitution
total cost
none of the above
ANS: PTS: 1
21. The isoquants for inputs that are perfect complements for one another consist of a series of:
right angles
parallel lines
concentric circles
right triangles
none of the above
ANS: PTS: 1
22. Given a Cobb-Douglas production function estimate of Q = 1.19L^.72K^.18 for a given industry, this industry would have:
increasing returns to scale
constant returns to scale
decreasing returns to scale
negative returns to scale
none of the above
ANS: PTS: 1
23. The primary purpose of the Cobb-Douglas power function is to:
allow one to make estimates of cost-output relationships
allow one to make predictions about a resulting increase in output for a given increase in the inputs
aid one in gaining accurate empirical values for economic variables
calculate a short-run linear total cost function
a and b
ANS: PTS: 1
24. The original Cobb-Douglas function was given as. It was subsequently rewritten as. What benefit was derived in the revision?
the function becomes a non-linear relationship so it would fit to production curves having an “S” shape
returns to scale can be shown in the revision
returns to scale become constant
a and b only
a, b, and c
ANS: PTS: 1
25. The Cobb-Douglas production function has which of the following properties?
output is a linear increasing function of each of the inputs
it provides a good fit to the traditional S-shaped production function
the elasticity of production is constant and equal to 1 minus the exponent of the appropriate variable
all of the above
none of the above
ANS: PTS: 1
26. In the Cobb-Douglas production function ():
the marginal product of labor (L) is equal to â1
the average product of labor (L) is equal to â2
if the amount of labor input (L) is increased by 1 percent, the output will increase by â1 percent
a and b
a and c
ANS: PTS: 1
1. Emco Company has an assembly line of fixed size A. Total output is a function of the number of workers (crew size) as shown in the following schedule:
Crew Size Total Output
(No. of Workers) (No. of Units)
Determine the following schedules:
(a) marginal productivity of labor
(b) average productivity of labor
(c) elasticity of production with respect to labor
2. A certain production process employs two inputs–labor (L) and raw materials (R). Output (Q) is a function of these two inputs and is given by the following relationship:
Q = 6L^2 R^2 − .10L^3 R^3
Assume that raw materials (input R) are fixed at 10 units.
(a) Determine the total product function (TP[L]) for input L.
(b) Determine the marginal product function for input L.
(c) Determine the average product function for input L.
(d) Find the number of units of input L that maximizes the total product function.
(e) Find the number of units of input L that maximizes the marginal product function.
(f) Find the number of units of input L that maximizes the average product function.
(g) Determine the boundaries for the three stages of production.
3. An industry can be characterized by the following production function:
Q = 2.5L^.60 C^.40
(a) What is the algebraic expression for the marginal productivity of labor?
(b) What is the algebraic expression for the average productivity of labor?
(c) How would you characterize the returns-to-scale in the industry?
Chapter 8—Cost Analysis
1. 1. Economies of Scope refers to situations where per unit costs are:
1. Unaffected when two or more products are produced
2. Reduced when two or more products are produced
3. Increased when two or more products are produced
4. Demonstrating constant returns to scale
5. Demonstrating decreasing returns to scale
ANS: PTS: 1
2. Economies of scale exist whenever long-run average costs:
a. Increase as output is increased
b. Remain constant as output is increased
c. Decrease as output is increased
d. Decline and then rise as output is increased
e. None of the above
ANS: PTS: 1
3. Which of the following is true with regards to a long-run cost function?
a. The shape of the firm’s long-run cost function is important in decisions to expand the scale of operations
b. The long-run average cost curve is U-shaped
c. The long-run average cost curve is flatter than the short-run average cost curve.
d. The curve consists of the lower boundary of all the short-run cost curves
e. All of the above
ANS: PTS: 1
4. If TC = 321 + 55Q – 5Q^2, then average total cost at Q = 10 is:
a. 10.2
b. 102
c. 37.1
d. 371
e. 321
ANS: PTS: 1
1. 5. Suppose that total cost is cubic: TC = 200 + 5Q – 0.4Q^2 + 0.001Q^3
2. a. Fixed cost (FC) is $200
3. b. Variable cost (VC) is 5Q – 0.4Q^2 + 0.001Q^3
4. c. Average variable cost (AVC) is 5 – 0.4Q + 0.001Q^2
5. d. Marginal cost (MC) is 5 – 0.8Q +.003Q^2
6. e. All of the above are correct
ANS: PTS: 1
6. What method of inventory valuation should be used for economic decision-making problems?
book value
original cost
current replacement cost
cost or market, whichever is lower
historical cost
ANS: PTS: 1
7. According to the theory of cost, specialization in the use of variable resources in the short-run results initially in:
decreasing returns and declining average and marginal costs
increasing returns and declining average and marginal costs
increasing returns and increasing average and marginal costs
decreasing returns and increasing average and marginal costs
none of the above
ANS: PTS: 1
8. For a short-run cost function which of the following statements is (are) not true?
The average fixed cost function is monotonically decreasing.
The marginal cost function intersects the average fixed cost function where the average variable cost function is a minimum.
The marginal cost function intersects the average variable cost function where the average variable cost function is a minimum.
The marginal cost function intersects the average total cost function where the average total cost function is a minimum.
b and c
ANS: PTS: 1
9. The cost function is:
a means for expressing output as a function of cost
a schedule or mathematical relationship showing the total cost of producing various quantities of output
similar to a profit and loss statement
incapable in being developed from statistical regression analysis
none of the above
ANS: PTS: 1
10. Which of the following statements about cost functions is true?
Variable costs will always increase in direct proportion to the quantity of output produced.
The less capital equipment employed in the production process relative to labor and other inputs, the longer will be the period of time required to increase significantly the scale of operation.
The shape of the firm’s long-run cost function is important in decisions to expand the scale of operations.
none of the above
ANS: PTS: 1
11. Which of the following statements concerning the long-run average cost curve of economic theory is true?
It is L-shaped
It is -shaped
It is -shaped
It is -shaped
It is M-shaped
ANS: PTS: 1
12. Possible sources of economies of scale (size) within a production plant include:
specialization in the use of capital and labor
imperfections in the labor market
transportation costs
a and b
a and c
ANS: PTS: 1
13. The existence of diseconomies of scale (size) for the firm is hypothesized to result from:
transportation costs
imperfections in the labor market
imperfections in the capital markets
problems of coordination and control encountered by management
All of the above
ANS: PTS: 1
14. The relevant cost in economic decision-making is the opportunity cost of the resources rather than the outlay of funds required to obtain the resources.
ANS: PTS: 1
15. ____ are defined as costs which are incurred regardless of the alternative action chosen in a decision-making problem.
Opportunity costs
Marginal costs
Relevant costs
Sunk costs
None of the above
ANS: PTS: 1
16. ____ include the opportunity costs of time and capital that the entrepreneur has invested in the firm.
Implicit costs
Explicit costs
a and b
None of the above
ANS: PTS: 1
17. A cottage industry exists in the home-manufacture of ‘country crafts’. Especially treasured are handmade quilts. If the fourth completed quilt took 30 hours to make, and the eighth quilt took 28
hours. What is the percentage learning? Hint: Percentage learning = 100% – (c[2]/c[1])•100%.
1. 5%
2. 6.7%
3. 10%
4. 100%
5. 122%
ANS: PTS: 1
1. During the last few days the Superior Company has been running into problems with its computer system. The last run of the production cost schedule resulted in the incomplete listing shown below.
From your knowledge of cost theory, fill in the blanks.
Q TC TFC TVC ATC AFC AVC MC
40 _____ _____ x x x x
1 _____ _____ _____ 52 _____ _____ _____
2 _____ _____ 20 _____ _____ _____ _____
3 _____ _____ _____ 21.33 _____ _____ _____
4 _____ _____ _____ _____ _____ _____ 4
5 _____ _____ 40 _____ _____ _____ _____
6 _____ _____ _____ 15.67 _____ _____ _____
7 _____ _____ _____ _____ _____ 10 _____
8 _____ _____ 96 _____ _____ _____ _____
9 _____ _____ _____ _____ _____ 15 _____
10 _____ _____ _____ _____ _____ _____ 45
2. The Jones Company has the following cost schedule:
Output Total Cost
(Units) ($)
Prepare (a) average total cost and (b) marginal cost schedules for the firm.
3. A firm has determined that its variable costs are given by the following relationship:
VC = .05Q^3 − 5Q^2 + 500Q
where Q is the quantity of output produced.
(a) Determine the output level where average variable costs are minimized.
(b) Determine the output level where marginal costs are minimized.
Question 1
Validation of a simulation model occurs when the true steady state average results have been reached.
Question 2
Adjusted exponential smoothing is an exponential smoothing forecast adjusted for seasonality.
Question 3
In an unbalanced transportation model, supply does not equal demand and one set of constraints uses ≤ signs.
Question 4
Fractional relationships between variables are not permitted in the standard form of a linear program.
Question 5
In a 0-1 integer programming problem involving a capital budgeting application (where xj= 1, if project j is selected, xj= 0, otherwise) the constraint x1– x2≤ 0 implies that if project 2 is
selected, project 1 cannot be selected.
Question 6
Excel can be used to simulate systems that can be represented by both discrete and continuous random variables.
In a break-even model, if all of the costs are held constant, how does an increase in price affect the model?
Breakeven point decreases
Breakeven point increases
Breakeven point does not change
The revenue per unit goes down
Question 8
In linear programming problems, multiple optimal solutions occur
when constraint lines are parallel to each other.
when the objective function is parallel to a constraint line
every possible solution point violates at least one constraint
when the dual price for a particular resource is very small
Question 9
Events that cannot occur at the same time in any trial of an experiment are:
mutually exclusive
Question 10
A business owner is trying to decide whether to buy, rent, or lease office space and has constructed the following payoff table based on whether business is brisk or slow.
If the probability of brisk business is .40 and for slow business is .60, the expected value of perfect information is:
Question 11
A business owner is trying to decide whether to buy, rent, or lease office space and has constructed the following payoff table based on whether business is brisk or slow.
The conservative (maximin) strategy is:
Question 12
Steinmetz furniture buys 2 products for resale: big shelves (B) and medium shelves (M). Each big shelf costs $100 and requires 100 cubic feet of storage space, and each medium shelf costs $50 and
requires 80 cubic feet of storage space. The company has $25000 to invest in shelves this week, and the warehouse has 18000 cubic feet available for storage. Profit for each big shelf is $85 and for
each medium shelf is $75.In order to maximize profit, how many big shelves (B) and how many medium shelves (M) should be purchased?
B = 225, M = 0
B = 0, M = 225
B = 150, M = 75
B = 75, M = 150
Question 13
Steinmetz furniture buys 2 products for resale: big shelves (B) and medium shelves (M). Each big shelf costs $100 and requires 100 cubic feet of storage space, and each medium shelf costs $50 and
requires 80 cubic feet of storage space. The company has $25000 to invest in shelves this week, and the warehouse has 18000 cubic feet available for storage. Profit for each big shelf is $85 and for
each medium shelf is $75. What is the objective function?
Max Z = 75B + 85M
Max Z = 85B + 75M
100B + 50M ≤ 25000
100B + 50M ≥ 25000
100B + 80M ≤ 18000
100B + 80M ≥ 18000
Question 14
The following is an Excel “Answer” and “Sensitivity” reports of a linear programming problem:
The Answer Report:
The Sensitivity Report:
Which additional resources would you recommend to be increased?
paint and seal
Cannot tell from the information provided
Question 15
The production manager for Beer etc. produces 2 kinds of beer: light (L) and dark (D). Two resources used to produce beer are malt and wheat. He can obtain at most 4800 oz of malt per week and at
most 3200 oz of wheat per week respectively. Each bottle of light beer requires 12 oz of malt and 4 oz of wheat, while a bottle of dark beer uses 8 oz of malt and 8 oz of wheat. Profits for light
beer are $2 per bottle, and profits for dark beer are $1 per bottle. What is the optimal weekly profit?
Question 16
The owner of Black Angus Ranch is trying to determine the correct mix of two types of beef feed, A and B which cost 50 cents and 75 cents per pound, respectively. Five essential ingredients are
contained in the feed, shown in the table below. The table also shows the minimum daily requirements of each ingredient.
Ingredient Percent per pound in Feed A Percent per pound in Feed B Minimum daily requirement (pounds)
The constraint for ingredient 3 is:
.5A + .75B = 20
.3B = 20
.3 B≤ 20
.3B ≥ 20
Question 17
Let xij = gallons of component i used in gasoline j. Assume that we have two components and two types of gasoline. There are 8,000 gallons of component 1 available, and the demands for gasoline types
1 and 2 are 11,000 and 14,000 gallons respectively. Write the supply constraint for component 1.
x11+ x21≤ 8000
x12+ x21≥ 8000
x11+ x12≤ 8000
x11+ x12≥ 8000
Question 18
The Kirschner Company has a contract to produce garden hoses for a customer. Kirschner has 5 different machines that can produce this kind of hose. Write a constraint to ensure that if machine 4 is
used, machine 1 will not be used.
Y[1]+ Y[4]≤0
Y[1]+ Y[4]= 0
Y[1]+ Y[4]≤1
Y[1]+ Y[4]≥ 1
Question 19
If we are solving a 0-1 integer programming problem, the constraintx1=x2is a __________ constraint.
multiple choice
mutually exclusive
Question 20
A professor needs help from 3 student helpers to complete 4 tasks. The first task is grading; the second is scanning; the third is copying, and the fourth is organizing student portfolios. The
estimated time for each student to do each task is given in the matrix below.
Which of the following constraints represents the assignment for student A?
XA1 +XA2+ XA3 + XA4 = 0
XA1 +XA2+ XA3 + XA4 = 1
XA1 +XA2+ XA3 + XA4 ≥ 1
XA1 +XA2+ XA3 + XA4 ≥ 0
Question 21
The following table represents the cost to ship from Distribution Center 1, 2, or 3 to Customer A, B, or C.
The constraint that represents the quantity demanded by Customer B is:
6X1B + 2X2B+ 8X3B≤ 350
6X1B + 2X2B+ 8X3B= 350
X1B + X2B+ X3B≤ 350
X1B + X2B+ X3B= 350
Question 22
Professor Dewey would like to assign grades such that 15% of students receive As. If the exam average is 62 with a standard deviation of 13, what grade should be the cutoff for an A? (Round your
Question 23
The metropolitan airport commission is considering the establishment of limitations on noise pollution around a local airport. At the present time, the noise level per jet takeoff in one neighborhood
near the airport is approximately normally distributed with a mean of 100 decibels and a standard deviation of 3 decibels. What is the probability that a randomly selected jet will generate a noise
level of more than 105 decibels?Note: please provide your answer to 2 places past the decimal point, rounding as appropriate.
Question 24
A bakery is considering hiring another clerk to better serve customers. To help with this decision, records were kept to determine how many customers arrived in 10-minute intervals. Based on 100
ten-minute intervals, the following probability distribution and random number assignments developed.
Number of Arrivals Probability Random numbers
6 .1 .01 – .10
7 .3 .11 – .40
8 .3 .41 – .70
9 .2 .71 – .90
10 .1 .91 – .00
Suppose the next three random numbers were .18, .89 and .67. How many customers would have arrived during this 30-minute period?
Question 25
__________ moving averages react more slowly to recent demand changes than do __________ moving averages.
Longer-period, shorter-period
Shorter-period, longer-period
Longer-period, longer-period
Shorter-period, shorter-period
Question 26
For the following frequency distribution of demand, the random number 0.8177 would be interpreted as a demand of:
Question 27
Given an actual demand of 59, a previous forecast of 64, and an alpha of .3, what would the forecast for the next period be using simple exponential smoothing?
Question 28
Suppose that a production process requires a fixed cost of $50,000. The variable cost per unit is $10 and the revenue per unit is projected to be $50. Find the break-even point.
Question 29
Nixon’s Bed and Breakfast has a fixed cost of $5000 per month and the revenue they receive from each booked room is $200. The variable cost per room is $75. How many rooms do they have to sell each
month to break even?(Note: The answer is a whole number. Give the answer as a whole number, omitting the decimal point. For instance, use 12 for twelve rooms).
Answer: 40
Question 30
Students are organizing a “Battle of the Bands” contest. They know that at least 100 people will attend. The rental fee for the hall is $200 and the winning band will receive $500. In order to
guarantee that they break even, how much should they charge for each ticket?(Note: Write your answer with two significant places after the decimal and do not include the dollar “$” sign. For
instance, for five dollars, write your answer as 5.00).
Question 31
Consider the following linear program, which maximizes profit for two products, regular (R), and super (S):
50R + 75S
1.2R + 1.6 S ≤ 600 assembly (hours)
0.8R + 0.5 S ≤ 300 paint (hours)
.16R + 0.4 S ≤ 100 inspection (hours)
Sensitivity Report:
Final Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficient Increase Decrease
$B$7 Regular = 291.67 0.00 50 70 20
$C$7 Super = 133.33 0.00 75 50 43.75
Final Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease
$E$3 Assembly (hr/unit) 563.33 0.00 600 1E+30 36.67
$E$4 Paint (hr/unit) 300.00 33.33 300 39.29 175
$E$5 Inspect (hr/unit) 100.00 145.83 100 12.94 40
A change in the market has increased the profit on the super product by $5. Total profit will increase by __________. Write your answers with two significant places after the decimal and do not
include the dollar “$” sign.
Question 32
Consider the following linear program, which maximizes profit for two products, regular (R), and super (S):
50R + 75S
1.2R + 1.6 S ≤ 600 assembly (hours)
0.8R + 0.5 S ≤ 300 paint (hours)
.16R + 0.4 S ≤ 100 inspection (hours)
Sensitivity Report:
Final Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficient Increase Decrease
$B$7 Regular = 291.67 0.00 50 70 20
$C$7 Super = 133.33 0.00 75 50 43.75
Final Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease
$E$3 Assembly (hr/unit) 563.33 0.00 600 1E+30 36.67
$E$4 Paint (hr/unit) 300.00 33.33 300 39.29 175
$E$5 Inspect (hr/unit) 100.00 145.83 100 12.94 40
If downtime reduced the available capacity for painting by 40 hours (from 300 to 260 hours), profits would be reduced by __________. Write your answers with two significant places after the decimal
and do not include the dollar “$” sign.
Question 33
Kalamazoo Kennels provides overnight lodging for a variety of pets. An attractive feature is the quality of care the pets receive, including well balanced nutrition. The kennel’s cat food is made by
mixing two types of cat food to obtain the “nutritionally balanced cat diet.” The data for the two cat foods are as follows:
Cat Food Cost/oz protien (%) fat (%)
Pet’s Choice 0.35 40 15
Feline Chow 0.32 20 30
Kalamazoo Kennels wants to be sure that the cats receive at least 5 ounces of protein and at least 3 ounces of fat per day. What is the optimal cost of this plan?Note: Please write your answers with
two significant places after the decimal and do not include the dollar “$” sign. For instance, $9.45 (nine dollars and fortyfive cents) should be written as 9.45
Question 34
Find the optimal Z value for the following problem. Do not include the dollar “$” sign with your answer.
Max Z = x1+ 6×2
Subject to: 17×1+ 8×2≤ 136
3×1+ 4×2≤ 36
x1, x2≥ 0 and integer
Question 35
Let’s say that a life insurance company wants to update its actuarial tables. Assume that the probability distribution of the lifetimes of the participants is approximately a normal distribution with
a mean of 72 years and a standard deviation of 5 years. What proportion of the plan participants are expected to survive to see their 75^th
birthday?Note: Round your answer, if necessary, to two places after the decimal. Please express your answer with two places after the decimal.
Question 36
Ms. James is considering four different opportunities, A, B, C, or D. The payoff for each opportunity will depend on the economic conditions, represented in the payoff table below.
Economic Conditions
Investment Poor Average Good Excellent
(S1) (S2) (S3) (S4)
A 18 25 50 80
B 19 100 50 75
C 100 26 120 60
D 20 27 50 240
Suppose all states of the world are equally likely (each state has a probability of 0.25). What is the expected value of perfect information?Note: Report your answer as an integer, rounding to the
nearest integer, if applicable
Question 37
The local operations manager for the IRS must decide whether to hire 1, 2, or 3 temporary workers. He estimates that net revenues will vary with how well taxpayers comply with the new tax code. The
probabilities of low, medium, and high compliance are 0.20, 0.30, and 0.50 respectively. What are the expected net revenues for the number of workers he will decide to hire? The following payoff
table is given in thousands of dollars (e.g. 50 = $50,000).Note: Please express your answer as a whole number in thousands of dollars (e.g. 50 = $50,000). Round to the nearest whole number, if
Question 38
The local operations manager for the IRS must decide whether to hire 1, 2, or 3 temporary workers. He estimates that net revenues will vary with how well taxpayers comply with the new tax code. The
probabilities of low, medium, and high compliance are 0.20, 0.30, and 0.50 respectively. What is the expected value of perfect information? Do not include the dollar “$” sign with your answer. The
following payoff table is given in thousands of dollars (e.g. 50 = $50,000).Note: Please express your answer as a whole number in thousands of dollars (e.g. 50 = $50,000). Round to the nearest whole
number, if necessary.
Question 39
Recent past demand for product ABC is given in the following table.
Month Actual Demand
May 33
June 32
July 39
August 37
The forecasted demand for May, June, July and August were 25, 30, 33, and 38 respectively. Determine the value of MAD.Note: Please express the result as a number with 2 decimal places. If necessary,
round your result accordingly. For instance, 9.146, should be expressed as 9.15
Question 40
Consider the following decision tree. The objective is to choose the best decision among the two available decisions A and B. Find the expected value of the best decision. Do not include the dollar
“$” sign with your answer. | {"url":"https://ctsaferoutes.org/article/www-hwmojo-comsearch-through-our-website-for-exams-and-quizzes-solutions-and-ace-your-class","timestamp":"2024-11-08T22:19:36Z","content_type":"text/html","content_length":"279598","record_id":"<urn:uuid:9f32bd0b-47ac-4e67-924a-faa629b2e4cd>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00316.warc.gz"} |
Long question about attatching records, unattatching records.
What to do if FS is telling me to attatch a record but when I go there it says only detatatch and profile (sources) page looks like record is already there, It is too much on my home page not to
mention confusing and adding more unnecessary work.
Best Answer
• It sounds as if the record may already be attached to another profile of the same name. If you hover over the "detach" symbol, the PID (Person ID) to which the record is attached will appear.
If that number is not the same as the person you are working on, I would recommend checking to see if the original attachment is correct and if, perhaps, there are duplicates of the same person
that need to be merged.
If you want to share a specific example, perhaps someone can help you review the records and profiles.
• Many sources come from different indexing batches but end up with the same title. For example on a person I was working on recently, a section of her source page looks like this:
These four source look identical from the title but they really are four different sources created when two different parish registers, the priest's copy and the deacon's copy, were both indexed
So the copy of the source you see on your sources page is not the same as the source you see in the hint, even though the title is the same, or else it would not show up as a hint.
Since some of the older indexed records were used to create IGI records which subsequently became Family Tree entries. and those indexed records were attached to the corresponding Family Tree
person, you will very often find that hints which show up as already being attached to someone else, really are, as Áine stated above, attached to a duplicate in Family Tree. Which is great! you
now have the opportunity to help clean up the tree by combining duplicates. And you may even find that the duplicate has information and extended family that you are missing.
To illustrate, just to be perfectly clear, hover over the warning triangle icon or the detach link to see this:
You can actually right-click on the ID number to jump right to the other person's profile in a new tab and compare the records. Then merge the two records if appropriate or detach the source from
the other person if the source is actually attached to a wrong person.
The next question that comes up is usually, why attach so many duplicate source when they have the same information?
My main reason for doing so, and why I encourage everyone to, is because that marks the historical record as being attached to someone in Family Tree. To take my starting example, I want any
searches to lead to the correct person profile in Family Tree:
I want any and all of Sara Nilsina's historical records to lead to her in Family Tree, not just some of them.
• I am not new, believe it or not, for a long time I didn't have a computer or internet. Everything changes. I hate changes, I know it's necessary, I guess, but I still hate changes. I lost a lot
of family photos of all my mother's uncles because of changes. Never to be found. Thank you so much for your help. I really do appreciated it I think I got it if I can remember it.
This discussion has been closed. | {"url":"https://community.familysearch.org/en/discussion/104714/long-question-about-attatching-records-unattatching-records","timestamp":"2024-11-02T22:05:27Z","content_type":"text/html","content_length":"394216","record_id":"<urn:uuid:0673c256-cd7a-4aba-a6b9-a26227f99f25>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00775.warc.gz"} |
0.0.2 Nov 1, 2024
0.0.1 Aug 1, 2024
📐 int_math
int_math is a Rust crate providing mathematical abstractions for 2D vectors and rectangles. It includes:
• UVec2: A 2D vector with unsigned integer coordinates.
• Vec2: A 2D vector with signed integer coordinates.
• Vec3: A 3D vector with signed integer coordinates.
• URect: A rectangle with unsigned integer coordinates for position and size.
• Rect: A rectangle with signed integer coordinates for position and unsigned integer dimensions.
✨ Features
• Vector Operations: Supports basic arithmetic operations for UVec2 and Vec2.
• Rectangles: Provides methods to create and manipulate rectangles, including calculating centers and applying offsets.
📦 Installation
Add int_math to your Cargo.toml:
int_math = "0.0.2"
Then, use it in your code:
use int_math::{URect, UVec2};
let rect = URect::new(10, 20, 30, 40);
let center = rect.center();
println!("Center: {:?}", center);
Licensed under the MIT License. See the LICENSE file for details. | {"url":"https://lib.rs/crates/int_math","timestamp":"2024-11-10T21:40:39Z","content_type":"text/html","content_length":"15538","record_id":"<urn:uuid:ffb6e229-7909-4994-b838-9a359f4426fe>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00010.warc.gz"} |
positive semidefinite matrix completion
The Gram dimension $\gd(G)$ of a graph is the smallest integer $k \ge 1$ such that, for every assignment of unit vectors to the nodes of the graph, there exists another assignment of unit vectors
lying in $\oR^k$, having the same inner products on the edges of the graph. The class of graphs satisfying $\gd(G) … Read more | {"url":"https://optimization-online.org/tag/positive-semidefinite-matrix-completion/","timestamp":"2024-11-09T00:04:51Z","content_type":"text/html","content_length":"88571","record_id":"<urn:uuid:5c07a679-e049-478c-a5ba-2772cf1ea8f2>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00657.warc.gz"} |
What are boundary conditions in CFD? | SolidWorks Assignment Help
What are boundary conditions in CFD? Let $G=I_{n}$ $(n\in{\ensuremath{\mathbb{R}}})$ be a smooth quadrilateral regular surface $\mathbb{X}:=\mathbb{R}\big((k\, /\, \mathbb{p})\big)/\mathbb{C}_{k}$
and let $\widetilde{G}:=I_{n}/\mathbb{Q}_{r}(k)$ be its boundary. The dual metric of the CFD metric is denoted $\widetilde{G}$ (see Eq. \[eq:CFD\]), i.e., on $X^{k}$, $$\label{eq:Fm} \widetilde{G}
(x,y)=\widetilde{G}_{k}(x)+y\widetilde{G}'(x).$$ Such a metric space is known as the internal metric space with finite volume, or any other metric space with bounded volume metric, so it is sometimes
used to refer to a point in the internal metric space. [\[sec:Viscous\]]{} In this paper, for a given small mass, i.e. sufficiently small $p<\operatorname{\displaystyle \frac{\operatorname{Re}}{2}\
beta/\operatorname{Im}\textup{ }} a_{im}(x)$ with $a_{im}(x)\approx p/2$, the internal metric space can be defined as follows: $$\label{eq:CFD5.1} \widetilde{G}(x,y)=\frac{1}{2|x|}\int_{{\ensuremath
mathrm{d}}u'}}{2k},$$ i.e., $G'(x,y)=G(x,y)+y\widetilde{G}(x,y)$. Equivalently, $x,y\in\mathcal{V}_{q}$, and $\mathcal{V}_{q}$ is the set of all real coordinates where $x=x(t)=\lambda f(x,0)$, $y=-y
(t=\lambda f(x,0)$, and $u(t,x,y)=\lambda^2 f(u(x,y),t)$. The boundary metric is defined in terms of contours where one may also use the normal form theory on the surface $\widetilde{G}$. Consider on
$\widetilde{G}$ the image of the image of its boundary: ${\ensuremath{\mathrm{CT}}}(k)\otimes {\ensuremath{\mathrm{CT}}}^{-1}(k)=(-1)^{k}\varepsilon_\mathcal{H}$ which consists of an elementary line
from the origin to the boundary, whose surface structure is given by $$\label{eq:CFD4.1} \begin{array}{lll} {\ensuremath{\mathrm{CT}}}(k)\otimes {\ensuremath{\mathrm{CT}}}^{-1}(k)\cong\{(x,y)\in\
widetilde{G}\,|\,(x)_i :=\lambda^i (x,0),\,y=-y(t=y(t=\lambda f(x,0))-\sum_{i=1}^m\alpha_ix+\dots),\,\sum_{i=1}^m \alpha_i =1 \}\\ \end{array}$$ on which coincides a neighborhood of the origin due to
the action of Hodge filtration. The surface conditions form a surface given by $$\label{eq:CFD4.2} \sum_{\{i_1,\,\cdots,\,i_k\}}{\ensuremath{\mathrm{d}}x}{\ensuremath{\mathrm{d}}y}-\sum_{\{i_1,\,\
cdots,\What are boundary conditions in CFD? In a world of infinite data, I often find that boundary things like time are either no longer relevant or are not observed in the world at all. For
example, if you look at world examples from a library like Berkeley, Berkeley itself is telling you what the truth is. I don't think this is very useful or helpful.
Online Classwork
In fact the word boundary can lead to a lot of frustration. If I told you the truth for some reason it leads to a lot of frustration and maybe you are not actually interested in understanding its
existence. See also: CfDE: in the early 90’s Atmospheric boundary conditions (CFD) are fundamentally different than those in a physical world – with this difference some are more or less physically
uninteresting (cf), while others just for use, or a better description could be made of that? Another way to approach I’ve been doing a lot of thinking is to look for ways to get deeper and
understand physical boundary conditions. You basically need to make the distinction that it is necessary to see ground-based data and not just the physical world. By abstracting boundary conditions
(as in the following two paragraphs) you can get a better understanding of a simple set of conditions that is just not present in it and is in fact quite More Bonuses uninteresting. How do you check
this? I think that most of what you’re doing is abstracting data in terms that represent the physical world. For this to work I need to focus on visual-descriptive ways to do that. What happens if
the boundary conditions? Are there local changes that alter the way you can see them? If so these changes include the so called “infusional transformation”. If I are on the assumption that I used
already these previous conditions as control variables then I would prefer something like this. But what happens to these infusional transforms and what is the most important thing? Well, I say that
what I think is the most important thing, but I don’t like all those possibilities: “I am looking in the space we aren’t using for the data, not just the physical world. While this isn’t the case far
from it, the data is valid for the present world.” This would look something like this: Another way to look for possible boundary conditions is to use some simple examples. For example you could
learn my life, where I got money, what countries where I saw photos. Back to the question of the shape of the world, I know the shape of the world is defined by some physical representation (such as
a circle, otherwise not so great). For example it’s determined implicitly the shape of a single dimensional cube that can be viewed from different angles (i.e. the cube itself, a triangular shape –
something like a cube in the shape of a tetrahedron) in that same cube. We’ll look at that for example. What are boundary conditions in CFD? – Guido Díaz-Neto, Martin C. In the work of the committee
under this article I discussed the so-called boundary conditions in dynamic fractal analysis – the boundary condition at the origin (anisotropic), the boundary condition at infinity
(non-anisotropic), the form of the boundary conditions and the corresponding solutions of time-frequency integral equations in spatially extended domains (3D), by introducing boundary conditions at
one and several places in the problem.
Do My Accounting Homework For Me
Since the boundary conditions (12) seem not quite clear the so-called anisotropic boundary conditions [1.1.4] (11) – a rather unusual theory developed by an experiment [1.1.5] on a large square
lattice [2] – have been resolved by what we call theory of inelastic shock waves in a static and isotropic system. These waves are in anisotropic diffusive shock wave picture, so the condition of
fluid interface in the interface on a free surface is a condition of homogeneous matter. This is analogous to the incompressible and convex case in usual homogeneous and isotropic one-dimensional
problem. This paper focused some attention on the two classical boundary conditions on a static medium, namely (13) in the form of diffusive shock wave and (14) in the form of at least one diffusion
Clicking Here where the surface is not only uniform but also isotropic, which in our case has no homogeneous or homogenous distribution. On a general example this last assumption must be treated with
care. These techniques lead to two different possible conclusions and proofs. By (13) in the form of anisotropic shock wave, the result is a purely general result. The result is a result of one type
of shock wave equations, but with the inelastic shock waves not only included themselves. The results are: the conditions of the shock wave structure are identical for the dissimilar and homogeneous
case; that is, dissimilar shocks belong to the same configuration of the uniform distribution; that is, dissimilar shock waves belong to a different set of disrelevant types of waves; etc. These are
relevant cases of a very general type of shock wave. However, these sets of shock waves are not necessarily identical anymore, though they still belong to the same configuration of the form. A simple
physical insight (still a general fact by the same authors) is that, within a general definition, they can not be distinguished with the well-known fact that mass and charge of the bodies will
therefore coincide. Because this is not an Ising, Theorem must be applied with a correct theoretical attention to the situation where dissimilar shock waves appear both for uniform and homogeneous
distribution, but they are not just non-separable groups of shock waves; the conditions on them can be completely different. Despite this, the sound waves should not appear separately, but be
uniformly equiparty and isot | {"url":"https://solidworksaid.com/what-are-boundary-conditions-in-cfd-12942","timestamp":"2024-11-08T04:27:18Z","content_type":"text/html","content_length":"156316","record_id":"<urn:uuid:b876c083-2a89-4100-afb6-815071746292>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00843.warc.gz"} |
MathFiction: The Visiting Professor (Robert Littell)
Lemuel Falk, a ``randomnist'' from the Steklov Institute in Russia gets a visiting position at a chaos research institute in Upstate New York in this academic farce. He meets a drunkard who studies
the chaotic aspects of water droplets (especially tears), a Harvard MBA who runs the local supermarket, a sexy barber/drug dealer named ``Occasional Rain Morgan'', and a dope smoking rabbi from
Brooklyn who believes ``God is Randomness.''
Yo! The thing is, halfway through reading this book, I sort of fell in love with it, which isn't at all what I expected to do from the start. At the start, the characters were flat parodies of
researchers at a mathematics institute and the ``ordinary people'' that populate the town. The jokes were funny enough, but I didn't expect it to go anywhere. Moreover, the author's obvious ignorance
of some key results in the areas of mathematics that are frequently discussed in the book were a source of disappointment. So, I even went as far as to write a review of the book which I posted here
(prematurely) treating it as a funny but lightweight book that gets the math all wrong.
But, something unexpected happened. The characters get fleshed out, some serious subtexts are addressed (even as the characters discuss and denounce the whole notion of subtexts in novels) and an
interesting if somewhat flawed characterization of mathematicians and mathematics gets deeper than I initially expected.
This book is all over the place. In addition to being a broad farce, it is
• a somewhat touching (though, I must warn, also quite kinky) love story between the middle aged Russian researcher and the young barber
• a murder mystery in which Falk determines the identity of serial killer using game theory and his expertise in randomness
• a philosophical exploration of the deep questions of life (in this case: is the universe random or just chaotic? do we have free will? what is ``god''?)
• a psychological drama in which a man overcomes the scars inflicted on his psyche in his youth by his role in the death of his father
• and, of course, mathematical fiction!
Mathematically, there really are some interesting things to say about the relationship between chaos and randomness, and the author seems to have understood just a few of those things and based the
book around them. In particular, it is true that the mathematical field of chaos theory has provided us with numerous examples of very simple deterministic rules whose resulting output looks random
to a naive observer and which is unpredictable (in the very practical sense that even though the system is completely deterministic, you need to be able to measure the present state of the system
much more precisely than we can in the real world to be able to make accurate predictions about what will happen next). The corollary is that when we look at things that we really think of as random
(like a coin flip) we now have reason to wonder whether they are really random or whether they are simply chaotic systems.
Unfortunately, in discussing this, Littell makes a few errors that reveal his ignorance of some of the basic theorems of chaos theory. For instance, at one point, he uses the motion of planets in our
solar system as an example of a system which is not chaotic. I can see why one might think so. After all, the planets seem to be quite orderly and predictable. Ironically, however, it is precisely
this system (the motion of more than two massive objects in three-dimensional space under the influence of gravity) in which chaos was first identified. [The word chaos was not used, but the hallmark
of chaos which we call ``sensisitive dependence'' was noticed by the mathematician Poincare when he realized that his prize winning paper which purported to be able to predict the motion of the
planets from Newtonian principles was flawed!] On the same page, he has Falk define chaos as ``order without periodicity''. My objection here will be quite technical, but the fact is that this
definition is way off. For instance, the quasi-periodic solutions to integrable systems are good examples of order without periodicity, but they are certainly not chaotic systems. Moreover, one of
the most famous results in chaos theory is that ``period 3 implies chaos''. Under the Li-Yorke definition of chaos, in fact, chaos necessarily contains periodicity. [To tie this together, the orbit
of the planets is a (near) periodic submanifold in the chaotic system which is the many-body problem.]
The author's naivete also is revealed through things he fails to mention. It is odd to me that he talks about ``randomness'' and ``randomnists'' without mentioning probability. (Probability is the
area of mathematics that really studies randomness and the practitioners are called probabilists.) He also seems unaware of the results of mathematical physics. Any late 20th century scientist
interested in the question of whether randomness is real would have to bring up the topic of quantum theory. After all, it is the apparent unpredictability of the collapse of the quantum wave
function which is the most obvious candidate for true randomness in the real world. However, this topic is never even discussed. Nor does Falk seem to know General Relativity, since his description
of the number pi refers to the diameter and circumference of the circular path of a space ship travelling around the universe...which probably would not be pi because of the curvature of spacetime.
Unfortunately, Littell is also not very good at imagining impressive mathematical results. Supposedly, Falk has become famous for looking at finitely many digits in the decimal expansion of pi and
failing to find any order. This is hardly the sort of result that would generate as much attention as his result supposedly did. Also, his idea for using pi in cryptography seems a bit lame. Finally,
the ``message'' that Falk and Rain find hidden in pi is ridiculously weak compared to the one in Sagan's Contact.
One final complaint: although Littell seems to have done his homework on the Jewish angle of the story, a few strange aspects may grate on the nerves of a Jewish reader. For instance, the rabbi uses
``goys'' for the plural of ``goy'' (I'm sure he would have said ``goyim'') and also uses ``goy'' as an adjective (when ``goyishe'' is the adjectival form).
But, as I said, there are lots of things I love about this book. I love the way Falk collects American colloquialisms. I love the anecdote about the chaos professor in Leningrad who -- because of
problems with the Party -- lectures on subway trains for students who stuff rubles in his pockets. I think the author does capture some aspects of what it is like to be in a mathematics ``think
tank''. And, of course, he does a great job of parodying the current state of American culture. Perhaps my favorite scene was when he visits a math class at Backwater University as a guest lecturer
and speaks to the students about randomness and pi in the slang dialect he has picked up from Rain.
Highly recommended!
Contributed by Anonymous
Despite the flaws in maths, his brilliant writing ability, plus the overarching grandeur of the story he tells, makes this one of my favourites novels.
Contributed by John C. Konrath
There are parts of this novel that I enjoyed very much and other parts which I detested. Mr. Littell's remarks concerning the American college system were skillfully delivered. However, I found the
explicit sexual content pornographic and out of step with the general theme of the work. There are numerous quotable lines for those who are interested in such things.
Note: Littell is better known as an author of straight-up spy stories, such as The Once and Future Spy, which features a character, Huxstep, who impresses strangers with his lightning fast mental
arithmetic. This trait alone does not seem sufficient to give this book its own entry in the database, but it probably is worth mentioning here, and I would like to thank Michael Henle for bringing
it to my attention. | {"url":"https://kasmana.people.charleston.edu/MATHFICT/mfview.php?callnumber=mf243","timestamp":"2024-11-10T07:50:00Z","content_type":"text/html","content_length":"17551","record_id":"<urn:uuid:723f97c4-61cf-4035-9180-307dd358b0e4>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00145.warc.gz"} |
Resistor - (College Physics III – Thermodynamics, Electricity, and Magnetism) - Vocab, Definition, Explanations | Fiveable
from class:
College Physics III – Thermodynamics, Electricity, and Magnetism
A resistor is a passive electrical component that opposes the flow of electric current, resulting in a drop in voltage. It is characterized by its resistance, measured in ohms ($\Omega$).
congrats on reading the definition of resistor. now let's actually learn it.
5 Must Know Facts For Your Next Test
1. In AC circuits, resistors impede current equally at all frequencies.
2. The power dissipated by a resistor is given by $P = I^2 R$ where $P$ is power, $I$ is current, and $R$ is resistance.
3. Resistors do not change the phase difference between voltage and current in an AC circuit.
4. The total resistance in a series circuit is the sum of individual resistances: $R_{total} = R_1 + R_2 + ... + R_n$.
5. In a parallel circuit, the reciprocal of the total resistance is the sum of reciprocals of individual resistances: $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$.
Review Questions
• How does a resistor affect current and voltage in an AC circuit?
• What formula represents the power dissipated by a resistor?
• How do you calculate total resistance in a series circuit?
© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website. | {"url":"https://library.fiveable.me/key-terms/physics-t-e-m/resistor","timestamp":"2024-11-07T23:11:41Z","content_type":"text/html","content_length":"166599","record_id":"<urn:uuid:11f72154-e3e0-40b3-bdc5-542f0510049b>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00632.warc.gz"} |
Re: Longitudinal study - how to plot the outcome with CI against time
That is G-Side random effect with SUBJECT_ID variable.
If you can get model convergency you need some skill to adjust parameters of model . Check this:
in proc mixed, the random and repeated statements model G-Side and R-Side random effect separatedly.
Correspond to PROC GLIMMIX ,it should be:
G_Side effect:
random intercept / subject = Patient_ID;
R_Side effect:
random visit/ subject = Patient_ID residal ;
random _residual_/ subject = Patient_ID;
07-13-2024 11:14 PM | {"url":"https://communities.sas.com/t5/SAS-Procedures/Longitudinal-study-how-to-plot-the-outcome-with-CI-against-time/m-p/945745","timestamp":"2024-11-08T21:05:03Z","content_type":"text/html","content_length":"211782","record_id":"<urn:uuid:ec88604f-b17e-4392-939b-bd38478fb806>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00385.warc.gz"} |
Posts for the month of May 2014
The point of these simulations is to study cooling instabilities that may be present in Francisco's experiments of colliding jets. To simplify the problem, the simulations are just 2-D colliding
flows with the appropriate Al cooling. The initial parameters are as follows:
T = 10 eV
rho = 10 ug/cc
v = 60 km/s
Z = 6 (ne/ni the electron to ion number density ratio)
The grid is 2 x 2 cm and 64 x 64 cells with 6 levels of AMR. The total runtime is 100 ns.
In the experiments, the "ambient" is a vacuum. To get around this in astrobear, I initialized the flows such that they already fill the domain. So the flows immediately collide at the beginning of
the simulation, and they continue to flow in from the top and bottom boundaries. The left and right boundary conditions are periodic.
Below are images and movies of the temperature in eV, and also the density in ug/cc. They are enlarged to look at the region of interest.
• The new high resolution 2.5-D pulsed jets (5 models) are running on bluehive2 and stampede.
• Wrote a problem module to study cooling instabilities in Francisco's colliding jets experiments. Now working on implementing an aluminum cooling table.
• Need to analyze Mach stem simulation data, and possibly do some more runs to write HEDLA proceedings.
• Made a quick fix to the code (ticket #401)
• Need to look at Martin's jet/outflow module and combine my work with his.
• Julio should have the code now. I will give him some time to get set up and play around with it, and then I will contact him at the end of the week to see where he is at.
Talked with Jonathan on tasks and threading, will need to adjust the slurm script on Bluestreak to use threading
Made tickets for documentation, assigned each group member a ticket to begin working on — what are our names on the wiki?
Changed due dates for development tickets
Read a lot of colliding flows papers last week, am putting together an email for Fabian to meet next week.
See here:
The wind is initially Mach 10, it continues for 0.2 mln yrs then linearly slow down to 0 for another 0.2 mln yrs.
I have now finished a simulation that contains all the relevant physics, including cooling, MHD and the outflow object.
But first I want to recall two of my previous simulations, which do not include MHD: The first animation shows the surface density of the accretion disk's inner region. The inner black circle marks
the outflow object, the outer back circle marks the initial inner rim of the accretion disk. We see that, although no physical viscosity is present and the outflow is interacting with the disk,
material is accreted, i.e. the accretion disk's inner rim moves inwards until it reaches the outflow object.
The second animation shows the same but without outflow object, this is just to show that the accretion disk's inner rim finds a stable configuration and the gas inside the inner cavity has a density
of about 1000 cm^-3, about 30 times lower than the accretion disk's density.
The third animation now includes MHD and the outflow object. The disk has a toroidal initial field configuration with an initial field strength of 1 mG. There are still clumps and streams of matter
forming and moving inwards, but besides these features the inner rim seems to be stable with densities of the inner cavity of about 1000 cm^-3 as seen in the second animation. So magnetic fields seem
to play an important role in forming the inner cavity of the galactic center's accretion disk.
The fourth animation shows the corresponding face-on magnetic field strength.
First animation: surface density, without magn. field
Second animation: surface density, without magn. field and without outflow
Third animation: surface density, with magn. field and with outflow
Fourth animation: face-on magnetic field strength in Gauss
This leads to two important questions:
1. Why is the disk collapsing when the outflow is switched on?
2. Why and how do magnetic fields prevent this?
Some time ago we discussed a simple model for the extraction of angular momentum from the inner accretion disk:
Lets assume the accretion disk has an inner rim , and the inner part of the accretion disk (a ring with mass M) fully interacts with the outflow. The outflow has a massflow , so after a time the mass
has been added to the ring. That means that the mass of the ring increases, but its angular momentum does not change because the wind does not have any angular momentum. So the specitic angular
momentum is decreasing, leading to the accretion of the disk's material. However, there must be a kind of critical outflow rate, because although the wind does not add any angular momentum to the
disk, it does add radial momentum to the disk, and at some point the radial momentum will just win against the loss of angular momentum. To test this simple model I did the following very rough
estimate: Both, the ring with mass and the wind with mass , have a specific potential energy and a specific kinetic energy (the wind due to its radial momentum and the ring due to its azimuthal
velocity ). If we assume perfect mixing of the ring with the wind material, we can calculate the total energy of the "new" ring (with mass ), which is just the sum of the aforementioned:
When we further assume that the "new" ring finds a new orbital configuration (at radius ) we can use the virial theorem, where
to calculate its new orbit:
So if the wind velocity is smaller than the azimuthal velocity we would expect an inflow, otherwise an outflow of disk material. In the GC we have , , , so we would always expect that the disk is
pushed away from the central black hole, contrary to what we see in the simulations. So apparently this simple model does not account for the angular momentum loss.
I furthermore made a rough estimate for the code performance on XSEDE's gordon cluster, as shown in the following figure:
Here are some plots for the CollidingFlows shear 15 data (beta 1, 10: x, y, z fields, and Hydro). Each of these are column density maps for each .bov file along their respective axes.
Beta 1, Shear 15
Beta 10, Shear 15
Hydro, Shear 15
Did the best I could to orient them appropriately without the triade VisIt produces (which is misleading as it does not indicate the actual direction of the flows). The min/max is kept the same for
each plot.
• Science
1. 2D Ablative RT: tried lower tolerance ( according to blog:johannjc05132014) with max iteration 10000, Maximum # of iteration reached and negative temperature found at frame 316 — comparing
356 frame with tolerance and max iteration 1000.
2. Help Rui set up on bluehive2. He's analyzing the front rate using his code
3. 3D Ablative RT: still working on transfering the hdf4 data from Rui to text form.
Hi!~ so here is a link to my page http://astrobear.pas.rochester.edu/trac/wiki/u/madams
Previously I had experimented with making visualizations with Erica's CollidingFlows data on Bamboo during the semester. I created some slices with the Shear 15 data for Beta 1, 10 and Hydro. The
most notable effect is how the magnetic field confines the fluid the higher its strength. This is expected, but as I post more in the next few days it'll become more clear for these data sets.
Here I will post some density (rho) plots pertaining to the Beta 1, Shear 15 set with a magnetic field in the x-direction. Each of these movies are slices in the yz, yx, and xz plane.
min: 0.65, max: 250
• Attempted to make a three slice visualization with these slices. Visit started crashing on hay. Need to talk to Rich sometime soon about Visit issues.
• Visualizations for density (Beta 10, and Hydro).
• Make some velocity plots for these three data sets (Beta 1, 10 and Hydro).
BE paper
• Got all the proofs back to ApJ
• Sent you the form for funding the paper. Not sure if this is needed before they publish?
MHD Shear Flows
• Need to decide on field orientation for the 'perpendicular' case. The Beta=1 effects for a uniform field in y or z are too strong (see below). Talking with Jonathan on this, it seems a highly
unrealistic initial field orientation to begin with — how do you have this large scale perpendicular field and a tilted interface to begin with? If there was some large scale perpendicular field,
then by the time the flows collided, they would have no tilted interface. A more realistic perp. case might be a helical field inside of the colliding flows only (no ambient field)? This would
simulate a perpendicular field, but without concerning the broken symmetry played by the tilted interface. For now though, am running 3D low res sims of the y and z fields with a weaker field.
• The data set for these runs as they are now, is extremely large. I am at ~ 120 GB now, ¼ of the way through the production Shear15, Beta=1, x-field run. This is for ~ 1500^3 cell resolution (48 +
5 levels). I estimate each run will be ~ 1TB in size. Where to store all this data?
• Stuff still need to finalize for production runs: 1) Potentially increasing the box size to prevent back-flow into the outflow-boundary. This most likely will be a problem in the Shear 30 and 60
cases (see my page). Can do this either by adding code to enable a restart with a bigger box once the back-flow hits the wall, or just run sim with a longer box in x to start. 2) Decide on
parameters for the flow, i.e. the mach, ramp down times, etc. for production.
• Runs possible: Beta 1/10, 3 shear angles, 2 field orientations = 12 runs..
High Res, Shear 15, Bx, Beta = 1
• ¼ way through on BS — no nans, but did see source errors
Low Res, Different Field Strengths/Orientations
Here is my wiki page on this: http://astrobear.pas.rochester.edu/trac/wiki/u/erica/LowResMHDShearFlows
And last week's blog post talked about what I was seeing in the Beta=1 case: http://astrobear.pas.rochester.edu/trac/blog/erica05122014
Beta10, Beta1
Beta 10, Beta1
Beta 10, Beta 1
Column Densities
Beta 10, Beta 1
I installed visit on my laptop and tuned to be able to use visit on alfalfa remotely.
Plan to modify the outflow module. Since as the radiation force changes, velocity and density profile changes dramatically within the sonic radius and sonic radius also moves. So I probably need to
solve this equation and plug the velocity and density profile into outflow module.
So I ran hypre with varying tolerances for the radiation solve and found that the solution only converged when I set the tolerance to be 1e-10
So I looked more closely at what the 'tolerance' actually refers to in hypre. The hypre user guide is not much help…
Convergence can be controlled by the number of iterations, as well as various tolerances such as relative residual, preconditioned residual, etc. Like all parameters, reasonable defaults are used.
Users are free to change these, though care must be taken. For example, if an iterative method is used as a preconditioner for a Krylov method, a constant number of iterations is usually required.
I'm assuming that hypre is trying to solve Ax=b and instead ends up solving Ax'=b' where r stands for the residual (r=b'-b) and x' is the approximate solution to x. I'm not sure what the _C means or
what the <C*b,b> represents… presumably this is the inner product of some matrix 'C' times b with b.
Iters ||r||_C conv.rate ||r||_C/||b||_C
----- ------------ --------- ------------
1 1.667442e-07 265.427358 3.581554e-08
2 2.421703e-07 1.452347 5.201657e-08
3 2.037404e-07 0.841310 4.376208e-08
4 1.176484e-07 0.577442 2.527008e-08
5 7.646604e-08 0.649954 1.642440e-08
6 4.446094e-08 0.581447 9.549914e-09
7 2.173844e-08 0.488933 4.669272e-09
8 1.033716e-08 0.475525 2.220354e-09
9 5.190075e-09 0.502079 1.114794e-09
10 2.514604e-09 0.484502 5.401202e-10
11 1.291694e-09 0.513677 2.774473e-10
12 6.662719e-10 0.515812 1.431108e-10
13 4.688375e-10 0.703673 1.007032e-10
14 2.451621e-10 0.522915 5.265918e-11
15 1.654131e-10 0.674709 3.552962e-11
16 8.812279e-11 0.532744 1.892819e-11
17 5.478943e-11 0.621740 1.176841e-11
18 7.612170e-11 1.389350 1.635043e-11
19 1.416387e-10 1.860687 3.042304e-11
20 2.346052e-10 1.656364 5.039163e-11
21 4.389153e-10 1.870868 9.427609e-11
22 1.170443e-09 2.666672 2.514034e-10
23 3.069051e-09 2.622127 6.592117e-10
24 3.268667e-09 1.065042 7.020879e-10
25 3.349935e-09 1.024863 7.195436e-10
26 1.136404e-09 0.339232 2.440919e-10
27 2.175246e-10 0.191415 4.672284e-11
28 7.067671e-11 0.324914 1.518089e-11
29 2.139085e-11 0.302658 4.594611e-12
30 7.493659e-12 0.350321 1.609588e-12
31 3.051552e-12 0.407218 6.554531e-13
<C*b,b>: 2.167928e+01
So I looked for an explanation of convergence for linear system solves and found this document
The condition number of a matrix is related to how much a relative error in the rhs affects the relative error in the lhs. ie
which would imply that
In general we have
and indicates how close to singular a matrix is. (ie a singular matrix would have an infinite condition number)
Also because of machine precision, there is an unavoidable error related to the condition number
AstroBEAR solveres
Now we have two solvers available in the code… One is GMres and the other is PCG (preconditioned conjugate gradient). GMres does not converge unless the tolerance is set to ~ 1e-3 - but then it does
nothing. PCG apparently expects symmetric positive definite matrices - but the matrices are certainly not symmetric. Not sure how much that matters… I'm also not sure whether PCG is actually doing
any preconditioning automatically. I am not setting the preconditioner etc…
Also the actual solution vector 'x' is in terms of the Radiation Energy, and not radiation temperature - and since E ~ T^4, there and there is a 3 order range of temperatures - there will be a 12
order range in Energies - and the relative error in the vector norm will be dominated by the larger values… A better metric would be the maximum local relative error…
MHD Shear Flows
• Need to decide on field orientation for the 'perpendicular' case. The Beta=1 effects for a uniform field in y or z are too strong (see below). Talking with Jonathan on this, it seems a highly
unrealistic initial field orientation to begin with — how do you have this large scale perpendicular field and a tilted interface to begin with? If there was some large scale perpendicular field,
then by the time the flows collided, they would have no tilted interface. A more realistic perp. case might be a helical field inside of the colliding flows only (no ambient field)? This would
simulate a perpendicular field, but without concerning the broken symmetry played by the tilted interface. For now though, am running 3D low res sims of the y and z fields with a weaker field.
• The data set for these runs as they are now, is extremely large. I am at ~ 120 GB now, ¼ of the way through the production Shear15, Beta=1, x-field run. This is for ~ 1500^3 cell resolution (48 +
5 levels). I estimate each run will be ~ 1TB in size. Where to store all this data?
• Stuff still need to finalize for production runs: 1) Potentially increasing the box size to prevent back-flow into the outflow-boundary. This most likely will be a problem in the Shear 30 and 60
cases (see my page). Can do this either by adding code to enable a restart with a bigger box once the back-flow hits the wall, or just run sim with a longer box in x to start. 2) Decide on
parameters for the flow, i.e. the mach, ramp down times, etc. for production.
3D High Res Colliding Flows
I ran a Beta = 1, Shear = 15, field = parallel case on BS, 8000 cores. It got ~¼ of the way through. It ran into some bad cfl numbers (at first, finite, then cfl=infinity) and src errors, requested
restarts, and then made a few more frames until it died with the error:
"hyperbolic/sweep/stencil_control.cpp.f90, line 353: 1525-108 Error encountered while attempting to allocate a data object. The program will stop."
I'm attaching log files from machine to this blog post.
The global info allocations are at 300 GB at frame 57. The Info allocations didn't reach that size until frame 160 in the Hydro run on BS.
I'm curious how the src errors are mitigated
I also am curious about this - I saw that on BH (~64 cores), the info allocations in the standard out was 10x larger than the size of the chombo. This we talked about before as being related to not
counting ghost zones. When I moved to BS, the difference was a factor of 100. Is this likely ghost zones again, this time the larger factor being due to the many more smaller patches being
distributed over 8,000 cores?
Here are density and mesh images over time:
Everything looks good..
3D Low Res Perpendicular Field Cases
I ran both the y-field and z-field cases. This is a uniform field that runs through the ambient object and the CF object, perpendicular to the CFs. Here is a movie of the y-field case. This is
density with the B vector field. Note the flow is colliding in x. Beta = 1.
We see that the oncoming flows cause the field to begin bending strongly, especially near the x faces of the box. The incoming flow is then directed along these field lines, squashing the material
into a pancake in the xz plane. The field is pushed together by the flow, and piles up strongly in the collision region, where the magnetic pressure becomes so strong it vacates/resists gas from
collecting there. Hence, the vacated cavity in the collision region, traced out by billowing shell of material traveling in y, away from the colliding flows.
The field is so strong in this case, that no interesting features of the flow are coming through. The NTSI for example is completely wiped out.
I also made column density maps of this case and the 3D low res z-field case and posted them on my page here.
I am running these same simulations now with Beta = 10 and with extrapolating/outflow only boundary conditions on all faces of the box, since the problem with the Bfields we saw before only happens
when the Bfields are normal to the outflow boundary. The cases above had reflecting, Bnormal boundaries where the field was coming from, but in trying to avoid the bounce off of the boundary,
changing BC's now too. Directories:
They are ~¼ way through on bamboo and grass and take ~ 20 hours to complete.
Now that I fixed that typo in the cooling routines, the non-equilibrium cooling will be a bit stronger. This is good news for the 2.5-D sims as long as I am still able to resolve the cooling length,
and it will make my resolution more impressive when compared to other papers. I redid the cooling length calculations and updated the following table in my paper:
This will, however, be bad news for the 3-D sims. We've already seen that it is very difficult to get high resolution on these runs. The shorter cooling length means we will be that much less
resolved. We may have to reconsider the problem set up, and do something that the code can handle more easily.
New time stepping that is based on actual time scale for energy to change.
• Ablative RT
1. Rui is running 2D RT on LLE machines. He will check the growth rate with his matlab program when getting the data (Find the front by checking the slope and subtract the speed of the whole
body) Will send me the 3D results and let me try the 3D.
2. Still working on the hypre chocking issue. http://astrobear.pas.rochester.edu/trac/blog/bliu05012014
"In 3D, the vector potential should be averaged along each edge of the component parallel to that edge. For example Ax should be averaged along each edge that is parallel to the x-axis. The 2nd order
errors due to estimating the average value along an edge by the midpoint will not produce divergence in the resulting B-field." For what field geometry can one just set the B fields directly?
Where does maintainauxarrays get set? There is a bit on it in amr/amr_control.cpp.f90:
But it seems can set this just in the objects.. | {"url":"https://bluehound.circ.rochester.edu/astrobear/blog/2014/5","timestamp":"2024-11-09T23:52:04Z","content_type":"text/html","content_length":"107801","record_id":"<urn:uuid:426e077c-95e4-4c41-81ba-13587b8bfb33>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00023.warc.gz"} |
Purpose of standard deviation/definition/Measure of position and variability
Research Writing
Purpose of standard deviation/definition/Measure of position and variability
The term standard deviation refers to a measure that is used to quantify the variation or spread of numerical data in a random variable, statistical population, data set, or distribution of a
probability. Purpose of standard deviation
The world of research and statistics can seem complex and foreign to the general population, since it seems that mathematical calculations happen under our eyes without us being able to understand
their underlying mechanisms. Nothing is further from reality.
In this opportunity we are going to relate in a simple but at the same time exhaustive way the context, the foundation and the application of a term as essential as the standard deviation in the
field of statistics.
What is the standard deviation?
Statistics is a branch of mathematics that is responsible for recording variability, as well as the random process that generates it following the laws of probability . This is said soon, but within
the statistical processes are the answers to everything that today we consider as “dogmas” in the world of nature and physics.
For example, let’s say that when you toss a coin three times into the air, two of them come up heads and one tails. Simple coincidence, right? On the other hand, if we toss the same coin 700 times
and 660 of them land heads, perhaps there is a factor that promotes this phenomenon beyond randomness (imagine, for example, that you only have time to give a limited number of turns in the air,
which makes it almost always fall the same way). Thus, observing patterns beyond mere coincidence prompts us to think about the underlying motives for the trend.
What we want to show with this bizarre example is that statistics is an essential tool for any scientific process , since on the basis of it we are able to distinguish random realities from events
governed by natural laws. Purpose of standard deviation
Thus, we can throw a hasty definition of the standard deviation and say that it is a statistical measure product of the square root of its variance. This is like starting the house with the roof,
because for a person who does not dedicate himself entirely to the world of numbers, this definition and not knowing anything about the term differ little. So let’s take a moment to dissect the world
of basic statistical patterns .
What is the purpose of the standard deviation?
A standard deviation determines the distribution of values in a data set. standard deviation is the measure of the variability of any set of numerical values over its arithmetic mean and is
represented by the Greek letter sigma. It is found by taking the square root of the variance, which is the average of the squared differences from the mean.
standard deviation is a value that is frequently used in social science and statistics, especially when analyzing data printed in research papers or journals. The standard deviation can be helpful in
determining how to continue research or a course of action based on the amount of variance in the data. For example, a teacher who finds that there is a large value for the standard deviation of test
scores, indicating that there is a large variation, may choose to adjust their teaching method to accommodate students of diverse backgrounds and abilities. When test results indicate little
variation, represented by a small standard deviation, and when they are consistently high, There can be little concern about how to instruct the class or make up the lesson plans. There are two types
of standard deviations: population standard deviation and sample deviation. Purpose of standard deviation
Measures of position and variability
Position measures are indicators used to indicate what percentage of data within a frequency distribution exceed these expressions, whose value represents the value of the data that is in the center
of the frequency distribution . Do not despair, because we define them quickly:
• Mean: The numerical average of the sample.
• Median: represents the value of the central position variable in an ordered data set.
In a rudimentary way, we could say that the position measures are focused on dividing the data set into equal percentage parts, that is, “reaching the middle”.
On the other hand, the variability measures are responsible for determining the degree of approach or distance between the values of a distribution compared to its location average (that is,
compared to the mean). These are the following:
• Range: measures the breadth of the data, that is, from the minimum value to the maximum.
• Variance: the expectation (mean of the data series) of the square of the deviation of said variable with respect to its mean.
• standard deviation: numerical index of the dispersion of the data set.
Of course, we are moving in relatively complex terms for someone who is not fully dedicated to the world of mathematics. We do not want to go into other measures of variability, because knowing that
the greater the numerical products of these parameters, the less homogenized the data set will be.
“The average of the atypical”
Once we have established our knowledge of the variability measures and their importance in data analysis, it is time to refocus our attention on the standard deviation. Purpose of standard deviation
Without going into complex concepts (and perhaps oversimplifying things), we can say that this measure is the product of calculating the average of the “outliers” . Let’s take an example to clarify
this definition:
We have a sample of six pregnant bitches of the same breed and age who have just given birth to their litters of puppies simultaneously. Three of them have given birth to 2 cubs each, while another
three have given birth to 4 cubs per female. Naturally, the average value of offspring is 3 cubs per female (the sum of all cubs divided by the total number of females).
What would the standard deviation be in this example? First, we would have to subtract the mean from the values obtained and square this figure (since we don’t want negative numbers),
for example: 4-3 = 1 or 2-3 = (-1, squared, 1) .
The variance would be calculated as the mean of the deviations from the mean value (in this case, 3). Here we would be before the variance, and therefore, we have to take the square root of this
value to transform it into the same numerical scale as the mean. After this, we would obtain the standard deviation. Purpose of standard deviation
So what would be the standard deviation of our example? Well, a puppy. It is estimated that the average of the litters is three offspring, but it is within normality for the mother to give birth to
one less puppy or one more per litter.
Perhaps this example could sound a bit confusing as far as variance and deviation are concerned (since the square root of 1 is 1), but if the variance were 4 in it, the result of the standard
deviation would be 2 (remember , its square root).
What we wanted to show with this example is that the variance and the standard deviation are statistical measures that seek to obtain the mean of the values other than the average . Remember: the
higher the standard deviation, the greater the dispersion of the population.
Returning to the previous example, if all the bitches are of the same breed and have similar weights, it is normal for the deviation to be one puppy per litter. But for example, if we take a mouse
and an elephant, it is clear that the deviation in terms of the number of descendants would reach values much greater than one. Again, the less the two sample groups have in common, the larger the
deviations are to be expected.
Still, one thing is clear: using this parameter we are calculating the variance in the data of a sample, but by no means does this have to be representative of an entire population. In this example
we have taken six female dogs, but what if we monitored seven and the seventh had a litter of 9 puppies?
Of course, the pattern of the deviation would change. For this reason, taking the sample size into account is essential when interpreting any data set . The more individual numbers that are collected
and the more times an experiment is repeated, the closer we are to postulating a general truth.
As we have seen, the standard deviation is a measure of data dispersion. The greater the dispersion, the greater this value will be , because if we were dealing with a completely homogeneous set of
results (that is, that all were equal to the mean), this parameter would be equal to 0.
This value is of enormous importance in statistics, since not everything comes down to finding common bridges between figures and events, but it is also essential to record the variability between
sample groups in order to ask ourselves more questions and obtain more knowledge in the long term. | {"url":"https://englopedia.com/purpose-of-standard-deviation/","timestamp":"2024-11-12T12:00:22Z","content_type":"text/html","content_length":"147654","record_id":"<urn:uuid:040c3fac-eb49-4f2f-bb2c-a19b3d862cad>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00229.warc.gz"} |
Quadratic Equation - Formula, Examples | Quadratic Formula - [[company name]] [[target location]], [[stateabr]]
Quadratic Equation Formula, Examples
If this is your first try to work on quadratic equations, we are thrilled regarding your adventure in mathematics! This is indeed where the amusing part begins!
The information can appear enormous at first. However, give yourself some grace and space so there’s no pressure or stress while figuring out these problems. To be efficient at quadratic equations
like a pro, you will require patience, understanding, and a sense of humor.
Now, let’s begin learning!
What Is the Quadratic Equation?
At its core, a quadratic equation is a math equation that states distinct scenarios in which the rate of deviation is quadratic or relative to the square of few variable.
Though it might appear similar to an abstract idea, it is just an algebraic equation described like a linear equation. It ordinarily has two solutions and uses complicated roots to solve them, one
positive root and one negative, through the quadratic formula. Solving both the roots should equal zero.
Definition of a Quadratic Equation
Primarily, keep in mind that a quadratic expression is a polynomial equation that comprises of a quadratic function. It is a second-degree equation, and its standard form is:
ax2 + bx + c
Where “a,” “b,” and “c” are variables. We can employ this equation to work out x if we replace these variables into the quadratic formula! (We’ll look at it next.)
Any quadratic equations can be written like this, which makes working them out simply, comparatively speaking.
Example of a quadratic equation
Let’s contrast the given equation to the subsequent formula:
x2 + 5x + 6 = 0
As we can see, there are two variables and an independent term, and one of the variables is squared. Consequently, linked to the quadratic formula, we can confidently state this is a quadratic
Commonly, you can observe these types of formulas when measuring a parabola, that is a U-shaped curve that can be graphed on an XY axis with the data that a quadratic equation gives us.
Now that we learned what quadratic equations are and what they look like, let’s move on to solving them.
How to Work on a Quadratic Equation Using the Quadratic Formula
Even though quadratic equations might seem greatly complicated initially, they can be cut down into few easy steps utilizing a straightforward formula. The formula for solving quadratic equations
includes setting the equal terms and using rudimental algebraic functions like multiplication and division to get 2 answers.
Once all functions have been carried out, we can work out the numbers of the variable. The results take us one step nearer to discover solutions to our first question.
Steps to Figuring out a Quadratic Equation Using the Quadratic Formula
Let’s promptly plug in the common quadratic equation once more so we don’t forget what it looks like
ax2 + bx + c=0
Before figuring out anything, keep in mind to detach the variables on one side of the equation. Here are the 3 steps to figuring out a quadratic equation.
Step 1: Note the equation in standard mode.
If there are terms on either side of the equation, sum all similar terms on one side, so the left-hand side of the equation totals to zero, just like the conventional model of a quadratic equation.
Step 2: Factor the equation if workable
The standard equation you will conclude with must be factored, generally through the perfect square method. If it isn’t workable, put the terms in the quadratic formula, which will be your best
friend for figuring out quadratic equations. The quadratic formula appears something like this:
All the terms coincide to the equivalent terms in a conventional form of a quadratic equation. You’ll be utilizing this significantly, so it is wise to memorize it.
Step 3: Implement the zero product rule and work out the linear equation to remove possibilities.
Now once you possess two terms equal to zero, figure out them to get 2 answers for x. We have two results due to the fact that the solution for a square root can either be positive or negative.
Example 1
2x2 + 4x - x2 = 5
At the moment, let’s fragment down this equation. Primarily, simplify and put it in the conventional form.
x2 + 4x - 5 = 0
Next, let's determine the terms. If we compare these to a standard quadratic equation, we will get the coefficients of x as follows:
To solve quadratic equations, let's put this into the quadratic formula and solve for “+/-” to involve each square root.
We work on the second-degree equation to get:
Next, let’s streamline the square root to obtain two linear equations and work out:
x=-4+62 x=-4-62
x = 1 x = -5
Next, you have your solution! You can review your work by using these terms with the first equation.
12 + (4*1) - 5 = 0
1 + 4 - 5 = 0
-52 + (4*-5) - 5 = 0
25 - 20 - 5 = 0
That's it! You've solved your first quadratic equation using the quadratic formula! Congrats!
Example 2
Let's check out one more example.
3x2 + 13x = 10
First, put it in the standard form so it is equivalent zero.
3x2 + 13x - 10 = 0
To figure out this, we will substitute in the numbers like this:
a = 3
b = 13
c = -10
figure out x employing the quadratic formula!
Let’s clarify this as far as possible by working it out just like we executed in the prior example. Work out all simple equations step by step.
You can figure out x by taking the positive and negative square roots.
x=-13+176 x=-13-176
x=46 x=-306
x=23 x=-5
Now, you have your solution! You can revise your work utilizing substitution.
3*(2/3)2 + (13*2/3) - 10 = 0
4/3 + 26/3 - 10 = 0
30/3 - 10 = 0
10 - 10 = 0
3*-52 + (13*-5) - 10 = 0
75 - 65 - 10 =0
And that's it! You will work out quadratic equations like nobody’s business with little practice and patience!
Granted this summary of quadratic equations and their basic formula, kids can now go head on against this complex topic with confidence. By beginning with this easy definitions, kids gain a strong
understanding before moving on to further intricate ideas ahead in their studies.
Grade Potential Can Guide You with the Quadratic Equation
If you are fighting to understand these ideas, you may require a mathematics tutor to help you. It is better to ask for assistance before you lag behind.
With Grade Potential, you can learn all the tips and tricks to ace your subsequent math test. Become a confident quadratic equation solver so you are ready for the following big ideas in your math | {"url":"https://www.longbeachinhometutors.com/blog/quadratic-equation-formula-examples-quadratic-formula","timestamp":"2024-11-12T17:31:47Z","content_type":"text/html","content_length":"79312","record_id":"<urn:uuid:ca38a496-e74f-49a4-847f-712839171d6c>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00830.warc.gz"} |
Automatic Peak Selection by a Benjamini-Hochberg-Based Algorithm
A common issue in bioinformatics is that computational methods often generate a large number of predictions sorted according to certain confidence scores. A key problem is then determining how many
predictions must be selected to include most of the true predictions while maintaining reasonably high precision. In nuclear magnetic resonance (NMR)-based protein structure determination, for
instance, computational peak picking methods are becoming more and more common, although expert-knowledge remains the method of choice to determine how many peaks among thousands of candidate peaks
should be taken into consideration to capture the true peaks. Here, we propose a Benjamini-Hochberg (B-H)-based approach that automatically selects the number of peaks. We formulate the peak
selection problem as a multiple testing problem. Given a candidate peak list sorted by either volumes or intensities, we first convert the peaks into -values and then apply the B-H-based algorithm to
automatically select the number of peaks. The proposed approach is tested on the state-of-the-art peak picking methods, including WaVPeak [1] and PICKY [2]. Compared with the traditional fixed
number-based approach, our approach returns significantly more true peaks. For instance, by combining WaVPeak or PICKY with the proposed method, the missing peak rates are on average reduced by 20%
and 26%, respectively, in a benchmark set of 32 spectra extracted from eight proteins. The consensus of the B-H-selected peaks from both WaVPeak and PICKY achieves 88% recall and 83% precision, which
significantly outperforms each individual method and the consensus method without using the B-H algorithm. The proposed method can be used as a standard procedure for any peak picking method and
straightforwardly applied to some other prediction selection problems in bioinformatics. The source code, documentation and example data of the proposed method is available at http://sfb.kaust.edu.sa
Citation: Abbas A, Kong X-B, Liu Z, Jing B-Y, Gao X (2013) Automatic Peak Selection by a Benjamini-Hochberg-Based Algorithm. PLoS ONE 8(1): e53112. https://doi.org/10.1371/journal.pone.0053112
Editor: Anna Tramontano, University of Rome, Italy
Received: July 27, 2012; Accepted: November 26, 2012; Published: January 7, 2013
Copyright: © 2013 Abbas et al. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction
in any medium, provided the original author and source are credited.
Funding: This work was supported by Award No. GRP-CF-2011-19-P-Gao-Huang, a GMSV-OCRF award from King Abdullah University of Science and Technology, and Hong Kong Research Grants Council grants
HKUST6019/10P and HKUST6019/12P. The funders had no role in study design, data collection and analysis, decision to publish, or preparation of the manuscript.
Competing interests: The authors have declared that no competing interests exist.
Many computational bioinformatics methods generate a large number of predictions for the correct solution to a problem among which are both true and false predictions. Such predictions are usually
sorted according to certain confidence scores. For instance, ab initio protein structure prediction methods sample tens of thousands of three-dimensional models. The energy values are calculated for
each model based on a given energy function, where lower values likely indicate better models. Another example is the protein function annotation problem in which the amino acid sequence or the
domain architecture of a protein is given and the Gene Ontology (GO) terms selected from among some 30,000 are used to annotate the function.
In nuclear magnetic resonance (NMR)-based protein structure determination, thousands of peaks are routinely predicted from the input spectra in which there are usually tens to hundreds of true
signals. The peaks are sorted according to either their intensities or estimated volumes. Both means of sorting, based on computational methods, have common properties. First, a large number of
predictions are generated. Second, the predictions are scored by the scoring functions of the methods. However, the scoring functions are not powerful enough to distinguish true predictions from the
false ones. Third, it is important to discover most of the true predictions while maintaining a reasonably low false positive rate. Therefore, it is crucial to know how many predictions should be
selected in such scenarios.
Peak picking is one of the key problems in NMR protein structure determination process [3]–[5]. The problem is defined as follows: given any NMR spectrum or a set of spectra, select the true signals,
i.e., peaks, while filtering the false ones. Typically, true peaks are assumed to have Gaussian-like shapes and high intensities so that they can be easily differentiated from false ones. However,
there are two main factors that make the peak picking problem difficult. On the one hand, depending on the quality of the protein sample, the property of the target protein and local dynamics, there
can be a number of weak peaks, i.e., peaks with low intensities or volumes. That is, if we sort the predicted peaks by volumes or intensities, there is no clear cutoff threshold to distinguish true
peaks from false ones. These peaks are difficult to identify even by manual processes. This is why computational methods are useful. On the other hand, due to the various sources of noise in NMR
spectra, such as water bands and artifacts, false peaks can have high intensities or volumes. The group of sorted peaks is therefore comprised of a mixture of true peaks and false ones, where most of
the true peaks tend to be ranked higher with a few strong, false peaks also included. It is extremely difficult, if not impossible, to select only the true peaks and eliminate all the false ones. In
NMR structure determination, a missing true peak may cause all the follow-up procedures to fail, whereas a false peak can still be eliminated later [6]–[9]. Therefore, an ideal method should identify
almost all the true peaks while maintaining reasonably high precision.
The peak picking problem has been studied for more than two decades. A variety of computational methods have been proposed [1], [2], [10]–[19]. The existing methods can be classified into two
categories according to the de-noising method. Included in the first category are hard threshold-based approaches. For instance, PICKY [2] assumes that the noise is white Gaussian and estimates the
noise level in small regions that do not contain signals. The data points that have lower intensities than the estimated noise level are eliminated from the spectra. Singular value decomposition is
applied to the connected components of the remainder of the spectra to yield one-dimensional lineshapes. The peaks are identified in each lineshape and sorted according to the intensity values. The
higher the intensity is, the greater the confidence that it is a true peak. However, the hard threshold-based methods cannot detect weak peaks that are embedded in the noise. In the second category
are soft threshold-based approaches, which do not eliminate any data point from the spectra. We recently proposed WaVPeak [1] to overcome the bottleneck in the hard threshold-based methods. WaVPeak
applies the high-dimensional version of the Daubechies 3 wavelet [20] to smooth the given spectra. The shapes of true peaks become sharper and smoother. A brute-force method is used to identify all
the local maxima in the smoothed spectra. In contrast to PICKY, the peaks are sorted according to their estimated volumes by WaVPeak. We have found that volume significantly outperforms intensity in
distinguishing true peaks from false ones.
However, the existing peak picking methods are not able to determine automatically how many peaks among many to identify in order to include most of the true peaks. This number should be large enough
to include as many true peaks as possible, and in the meanwhile small enough to achieve relatively high precision. In PICKY, the default number of peaks to return is , where is the length of the
protein. In [1], WaVPeak is mainly compared with PICKY on the top peaks, where is the number of manually identified peaks, which is unknown for a new target protein. However, such fixed number-based
approaches do not take the distribution of peaks into consideration. For instance, if there is a spectrum that is very noisy or has a large number of artifacts, there can be many strong but false
peaks, which are identified along with the true ones. Many true peaks will not be selected if or is used. No matter how powerful the peak picking method is, it is crucial to cleverly determine the
number of peaks to be selected. Otherwise, true peaks will be eliminated even if they have been identified by the methods.
In this paper, we propose a Benjamini-Hochberg (B-H)-based approach for the peak picking problem. We first cast the peak selection problem into a multiple testing problem [21]. Because there is no
clear cutoff threshold for intensities or volumes, we calculate the -value for each peak. The number of peaks to be selected is then automatically determined by the B-H-based algorithm. We
demonstrate that the proposed method significantly outperforms the fixed number-based method on selecting the true peaks from the predictions by the state-of-the-art peak picking methods, including
WaVPeak and PICKY.
Our goal is to develop a method to help us to determine how many peaks to select among candidate peaks that number usually in the order of several hundreds. Each candidate peak can be considered as a
null hypothesis, where each false peak is a true null hypothesis and each true peak is a false null hypothesis. Therefore, the goal is to simultaneously test all the hypotheses and to reject as many
false null hypotheses as possible. This is a multiple testing problem, which has received much attention in the literature (see, e.g., [22]). One prominent solution to multiple testing problem was
proposed by Benjamini and Hochberg [23]. We first describe how to cast our problem into that framework.
A Quick Review of Benjamini-Hochberg Method
We wish to test null hypotheses:on the basis of a data set . We have some decision rule that rejects or accepts each of the above cases (e.g., decides if the th candidate peak is a true peak or a
false peak). The data set consists of
where are a random sample from the th population (e.g. intensities or volumes in a neighborhood of the th candidate peak). We assume that our decision rule, , produces a -value, , for each case, (we
will discuss several different ways of calculating such -values later). Therefore, has a uniform distribution if is correct,
Intuitively, if the -value, , is small enough, will be rejected. In fact, the usual Bonferroni procedure [24], [25] rejects wheneverwhere is the significance level. This is typically a very
conservative procedure, particularly when is large, because it does not reject as many null hypotheses as it should. In other words, it tends to have a low discovery rate.
To improve the discovery rate, Benjamini and Hochberg (1995) proposed an algorithm based on ordered -values:
The Benjamini and Hochberg (B-H) algorithm uses the following rule: for a fixed value of , referred to as the control rate, let be the largest index for whichand reject , the null hypothesis
corresponding to , if
accepting otherwise. Figure 1 illustrates how the B-H procedure works.
In this example, the number of hypotheses () is 10 and the false discovery proportion () is 0.2. The largest index of the hypotheses that is below the line is 6 (). Therefore, the first six
hypotheses are rejected as the predicted peaks.
Benjamini and Hochberg proved the following result [23], which justified their procedure.
For independent test statistics, the B-H algorithm controls the expected false discovery proportion (FDP) at :where , is the number of cases rejected, is the number of those that are actually null,
and is the number of true null hypotheses.
Clearly, the above FDP control attempts to keep the number of false discoveries under control, and in a sense to keep the precision above a certain level. A good procedure should have as high recall
rates as possible with prescribed high precision (or low FDP).
Applying the B-H Procedure to the Peak Picking Problem
We will cast the NMR peak picking problem into the multiple testing framework. In WaVPeak (or PICKY), after data cleaning at the first stage by wavelet smoothing (or by hard thresholding), potential
peaks are identified. We wish to test that, for each ,against
We can view each candidate peak and its surroundings as one population. We have a random sample of intensities, from the th population. The sample size depends on which method is adopted. For
WaVPeak, we have if we use a rectangular neighborhood of length 1 in 2D spectra, such as ^15N-HSQC; for PICKY, we have since we only use one intensity at each candidate peak.
We implement the B-H procedure below in two steps.
• Step I: calculating -values.
For WaVPeak and PICKY, we use volume () and intensity () around the th candidate peak as the test statistics, respectively. Our decision rule is to reject if or is large, respectively. The
corresponding -values arewhere and are observed values of and .
• Step II: applying the B-H procedure at .
Rank the -values obtained from Step I in ascending order, and denote the ordered -values as . We can then plot vs , and apply the B-H procedure.
Calculation of P-values
We now explain how to calculate p-values and in Step I above. We assume that the observations from different peaks are independent, and that true peaks and false peaks are from two different normal
distributions. Then we can rewrite the above testing problem asagainst
Typically, the mean intensity from false peaks is much smaller than the mean intensity from true peaks, usually written as . However, may not be zero, and can be estimated from weak intensities. For
variances, we typically have .
The reason why is small (compared with ) but not zero is due to how the candidate peaks are selected. In WaVPeak and PICKY, the volumes and intensities are calculated for a grid of points,
respectively, those below certain thresholds are discarded, and the remaining ones are retained as candidate peaks. Therefore, the volumes and intensities for those candidate peaks should all have
mean volumes and intensities above the thresholds.
To calculate and , we need to standardize by subtracting the mean, , and divide the standard deviation (s.d.), , under ’s. Due to the different data structures of WaVPeak (volumes) and PICKY
(intensities), they are considered separately below.
Calculation of
In WaVPeak, the test statistic is the approximate volume under the th candidate peak: , where and is some constant. Then, the -value is(1)where is the standard normal distribution. The mean/median, ,
and variance, , of the false peaks are unknown, which can be estimated by the sample median and sample variance of the false peaks, respectively. To do this, we need to have a rough idea of where
those false peaks are located. It has been observed that the number of true peaks of a protein, , is always less than , where is the length of that protein and is the expected number of peaks per
residue for the corresponding spectrum. For instance, for , ; for , . Almost all true peaks are ranked in the top candidate peaks by volume in WaVPeak, while the remaining candidate peaks are mostly
false peaks, from which we can estimate and .
To be more specific, for , let and denote the sample mean and variance for the th candidate peak; and and the ordered sample means and variances, respectively. Then and can be estimated by the
medians of the smallest and :(2)(3)
Calculation of
In PICKY, the test statistic is the intensity, , at the th single peak. Here, . Its -value can be calculated similarly to that in WaVPeak, giving(4)
Although we could use the same estimators of and as above, we will propose some different ones for PICKY due to its unique features. It has been demonstrated that the intensity of a single peak point
is a much less reliable confidence score than the volume of the peak [1]. It is thus expected that the intensity curves are mixed up by fewer true peaks and more false peaks. Therefore, the median of
may no longer be accurate because the median may very likely come from a true peak. On the other hand, replacing the median by the minimum in (2) and (3) should produce better estimators of and ,
respectively, which turns out to be true for less reliable confidence scores (data not shown). Based on these considerations, we propose to estimate and in PICKY respectively by
We evaluated the performance of the proposed methods on the peaks predicted by WaVPeak and PICKY. The same dataset as the one used by both [1] and [2] was used as the benchmark dataset, the most
comprehensive dataset available for the peak picking problem. The dataset covers a wide range of spectrum types, including 2D ^15N-HSQC, and 3D HNCO, HNCA, HNCACB and CBCA(CO)NH, which were extracted
from the spectrum sets of eight proteins (TM1112, YST0336, RP3384, ATC1776, CASKIN, HACS1, VRAR, and COILIN).
We first demonstrate how our method performed when a more reliable confidence score is available, i.e., the estimated volumes of the peaks predicted by WaVPeak. We then present the performance of the
method when a less reliable confidence score is available, i.e., the single intensity values of the peaks provided by PICKY. We finally demonstrate how to combine the results of our method with both
WaVPeak and PICKY, to further eliminate false positive peaks.
Selecting WaVPeak Peaks
The B-H algorithm is first compared with a fixed number-based method, i.e., , on selecting peaks predicted by WaVPeak. is set to . That is, the top peaks predicted by WaVPeak are considered. The
results are presented in Table 1, about which we make the following observations.
• The B-H algorithm significantly outperforms the -based method in terms of the average missing peak rates, i.e., the percentage of true peaks that are not selected. On six out of the 32 spectra,
the B-H algorithm reduces the -based method on the missing peak rate by more than 50%. One exception is HNCACB, where the B-H algorithm is slightly worse than the -based selection in the missing
peak rate (but better in precision); however, this can be easily rectified by increasing the FDR to , which is commonly adopted in practice. Overall, the B-H algorithm is much more sensitive and
stable than the fixed number-based method. It is noticeable that the improvement in the sensitivity is at the cost of the reduced precision. This is expected because the B-H algorithm does not
change the order of the sorted candidate peaks. Instead, it provides a good tradeoff that prefers higher sensitivity by selecting a cutting point in the list of the sorted peaks.
• As expected, the fixed number-based method is not stable. It performs well on some spectra (e.g. RP3384), but poorly on the others (e.g., TM1112). This is further verified by its larger standard
deviations. The reason is that such a method does not take the properties of the input spectra into consideration. For instance, for a very noisy spectrum with weak signals, there can be many
false peaks sorted amongst the true ones (e.g., Figures 2(a) and 2(c)). Thus, by taking a fixed number of peaks, there is no way one can ensure that the true peaks are included.
• Reduction on the missing peak rate of B-H over can reach as high as , indicated in column . These improvements mostly occur in the weak peaks, which are the most difficult to find. Since there
are not many weak peaks to start with, improvements measured by relative missing peak rates (i.e., weak signals found/all weak signals) are very high, even though those measured by absolute
missing peak rates may not always appear.
(a) and (d): sorted volume curve (a) and the corresponding p-value curve (d) of peaks predicted by WaVPeak on the 2D ^15N-HSQC spectrum of the protein ATC1776; (b) and (e): sorted volume curve (b)
and the corresponding p-value curve (e) of peaks predicted by WaVPeak on the 3D HNCO spectrum of the protein VRAR; (c) and (f): sorted volume curve (c) and the corresponding p-value curve (f) of
peaks predicted by WaVPeak on the 3D CBCA(CO)NH spectrum of the protein COILIN. In all figures, true peaks are shown in black and false ones are shown in cyan. In (d), (e) and (f), the decision
boundaries of and B-H procedure are shown in black and magenta, respectively.
It is noticeable that all the missing peak rates in Table 1 are the results by comparing to the “expected” peak lists of the spectra. The “expected” peak lists were generated by NMR labs by combining
information from large sets of spectra. It is thus likely that an expected peak does not exist in some spectra, especially the noisy ones, such as HNCACB and CBCA(CO)NH. In practice, higher recall
rates (lower missing peak rates) than those reported here can be expected.
Figure 2 shows several representative examples of how different selection methods work. We make several remarks.
• It can be difficult to set a cutoff point from the original volume curves in Figures 2(a)–2(c) to separate true peaks from false ones. The best thing the fixed number-based methods can do is to
take a random guess. For example, the -based selection method overestimates the number of peaks to be selected for a less noisy spectrum as shown in Figure 2(e), but significantly underestimates
the number of peaks to be selected for a noisier spectrum as shown in Figure 2(f).
• The B-H algorithm works consistently well on the -value curves. As shown in Figure 2, after converting the volumes to -values, strong true peaks with high volumes are dragged down to the -axis,
i.e., the -values are almost equal to zero. Most of the weak true peaks with low volumes are also dragged to the -axis, making it possible to identify them in the -value curves. For instance, two
of the three weak peaks with low volumes in Figure 2(a) are dragged down to the -axis, and thus selected by the B-H algorithm. Note that the -value does not change the volume order of the peaks.
Instead, it provides a much better curve so that the weak peaks can be possibly selected.
Selecting PICKY Peaks
We then evaluated the performance of the proposed method with a less reliable confidence score, i.e., the intensity value of PICKY. PICKY has a default noise level threshold [2], which sometimes
causes insufficient numbers of predicted peaks. For fair comparison purposes, we lowered the noise level threshold of PICKY until it generated more than 1.5 peaks.
Table 2 presents the performance of the proposed method on selecting peaks predicted by PICKY. Similar conclusions to those about WaVPeak can be made here. For instance, the B-H method consistently
and significantly outperforms the fixed number-based method. There are seven spectra on which the B-H algorithm reduces the missing peak rate of the -based method by at least 50%. Six of these
spectra have original recall rates that were already higher than 90%. There are two spectra, HNCO of COILIN and CBCA(CO)NH of RP3384, on which the absolute improvements are greater than 15% with
highest being 26%. As shown in Figures 3(b) and 3(c), the original intensity curves for these two spectra are continuous and smooth. It is difficult to identify a cutoff point between true peaks and
false ones on such curves. Many false peaks are sorted amongst the true ones. After converting the intensity values into p-values, most of the true peaks are dragged down to the -axis, i.e., they
have very small p-values. The 5% slope is then able to select most of the true peaks. In the two cases, fewer than three true peaks are not selected and true peaks are almost the last ones selected
by the B-H algorithm.
(a) and (d): sorted intensity curve (a) and the corresponding p-value curve (d) of peaks predicted by PICKY on the 2D ^15N-HSQC spectrum of the protein TM1112; (b) and (e): sorted intensity curve (b)
and the corresponding p-value curve (e) of peaks predicted by PICKY on the 3D HNCO spectrum of the protein COILIN; (c) and (f): sorted intensity curve (c) and the corresponding p-value curve (f) of
peaks predicted by PICKY on the 3D CBCA(CO)NH spectrum of the protein RP3384. In these figures, true peaks are shown in black and false ones are shown in cyan. In (d), (e) and (f), the decision
boundaries of and the B-H procedure are shown in black and magenta, respectively.
Eliminating False Peaks
The proposed B-H algorithm automatically determines how many peaks we should select from the candidate peak lists that are sorted according to the confidence scores of different methods. Therefore,
the more true peaks it includes, the greater the possibility that it also includes false ones. This possibility is verified by the relatively low precision values in Table 1. The selected false peaks
usually have larger volumes (or even much larger volumes) than the true ones. This can be caused by a variety of reasons, such as water bands, artifacts and side-chains. It is thus very difficult to
eliminate them from a single spectrum. An effective way to eliminate false peaks is to use spectra that share same atoms to “cross-reference” the peaks [2].
The goal of such cross-referencing is to eliminate as many false peaks as possible, while maintaining as many true peaks as possible. Among the commonly used NMR spectra, ^15N-HSQC is the most
sensitive and reliable one. It is often used as the root spectrum by NMR spectroscopists. If ^15N-HSQC is not available, HNCO is usually considered to be the root, especially in non-linear
acquisition mode. If other types of spectra are used to cross-reference ^15N-HSQC, the recall will be significantly decreased. Therefore, we used a consensus method to refine the peaks selected for ^
15N-HSQC. Both WaVPeak and PICKY were used to pick peaks for the ^15N-HSQC spectra of the eight proteins. The two candidate peak lists were then selected by the proposed B-H algorithm. Only the peaks
that appeared in both selected peak lists were kept as the consensus peak list for ^15N-HSQC. As shown in Table 3, the consensus method retained all the true peaks while increasing the precision by
13% on average. The consensus peak list was then used to refine all the other peak lists of WaVPeak that were selected by the proposed B-H algorithm. The reason we used the peak lists of WaVPeak was
that WaVPeak was shown to be more sensitive than PICKY on noisier spectra [1]. Table 3 shows that for all the spectra, most of the true peaks were maintained, and the precision values were
significantly improved. F-score, which is the harmonic mean of precision and recall, suggests that the BH-based consensus method gives the best overall accuracy comparing to other methods, including
PICKY, WaVPeak, B-H WaVPeak, and the consensus of PICKY and WaVPeak by simply considering the top peaks from each method. On average, the BH-based consensus method was able to identify more than 88%
of the expected true peaks, whereas less than 17% of the selected peaks were false ones.
Note that the performance of PICKY and WaVPeak in Table 3 was taken from that reported in [1], in which the top peaks were selected for comparison, where , the number of true peaks that exist in the
spectrum, was assumed to be known. The consensus method in Table 3 was done by considering the top peaks of both PICKY and WaVPeak, which is much larger than the number of peaks used in [1]. This
explains the significant drop of precision for the consensus method with respect to PICKY and WaVPeak.
Figures 4(a)–(e) show the precision-recall curves of the six different peak picking methods on the five types of spectra. These six methods are PICKY, B-H PICKY, WaVPeak, B-H WaVPeak, consensus and
B-H consensus. For the sake of clearance, only the important parts of the curves, i.e., when recall is at least 0.7, are drawn. It is clear that B-H consensus always outperforms the five other
methods. That is, at the same recall value, B-H consensus always has less proportion of false positive peaks. The consensus is the second best method. This makes sense because the consensus methods,
comparing to the other methods, combine information from different, relevant spectra. B-H WaVPeak and B-H PICKY consistently outperform WaVPeak and PICKY. Note that WaVPeak has been shown to be
better than PICKY [1]. Thus, the improvement of B-H PICKY over WaVPeak is due to the use of our B-H algorithm. In practice, we suggest the users to use the B-H WaVPeak if high sensitivity is required
or only one spectrum is available, and use the B-H consensus method if high tradeoff between precision and recall is needed or a set of relevant spectra is given.
(a)–(e): precision-recall curves for different methods on ^15N-HSQC, HNCO, HNCA, CBCA(CO)NH and NHCACB, respectively. The solid black curves are for B-H consensus method; the dashed black curves are
for the 1.5 consensus method; the solid cyan curves are for B-H WaVPeak; the dashed cyan curves are for the original WaVPeak; the solid magenta curves are for B-H PICKY; and the dashed magenta curves
are for the original PICKY. The relative area under curve (AUC) values are in legends, which are the area under curve over the total area of recall at least 0.7. (f): sensitivity analysis for
different number of peaks. The precision and recall values of B-H WaVPeak are shown when , , and top peaks are used to calculate the p-values.
We further studied the sensitivity of the B-H algorithm with respect to the parameter. In this paper, we have been using 1.5 as the parameter value in . As shown in Figure 4(f), when the parameter
value is changed to 2, 2.5 or 3, there is no significant change on both precision and recall.
A common issue in bioinformatics is that a large number of predictions are made by computational methods. These predictions contain both true predictions and false ones. In most problems, a fixed
number of predictions is selected according to a certain confidence score. The confidence score, however, is not accurate enough to differentiate true predictions from false ones. Therefore,
selecting a fixed number of predictions or thresholding by a fixed score usually sacrifices a lot of true predictions because it does not take the properties of the problem into consideration. We
propose a general approach to partially resolve this issue. The original confidence score is first converted into -values, which have been demonstrated to have a much stronger distinguishing
capability than the original confidence score. The Benjamini-Hochberg algorithm is then applied to select a self-adapted number of predictions according to the false discovery rate that we want to
control. This approach provides a systematic way of selecting predictions of computational methods. We further demonstrate that the false predictions can be further eliminated by using consensus or
cross-referencing approaches.
The proposed approach has a wide range of potential applications. For instance, in protein function annotation problems, the amino acid sequences or domain architectures of proteins are given, and
the GO terms selected from among some 30,000 are used to annotate the function. Most of the existing methods estimate the probability for each GO term to annotate the given protein [26]–[29].
However, the number of GO terms that annotate a certain protein is unknown. Our approach can be directly applied to the protein function annotation problem such that the correct number of GO terms is
Theoretically speaking, the sum of the false discovery rate and the precision should be one. However, the precision values of B-H WaVPeak and B-H PICKY are way below 0.95, as shown in Tables 1 and 2.
This is due to the fact that the volume and the intensity used in the original WaVPeak and PICKY are not perfect measures to rank peaks. That is, although such measures contain information about peak
properties, the information is far from complete or correct. As shown in Figures 2 and 3, many true peaks can have much lower volume or intensity than some false ones. In order to achieve the
theoretical precision level, better measurements have to be used by the original peak picking methods. For instance, the symmetry of peak shapes can be considered as additional information to rank
peaks [4].
We are currently incorporating the proposed method as a plug-in into the available NMR software, such as CCPN and NMRView [15]. The source code of the proposed method is available at http://
We have proposed a sensitive and robust approach to select peaks from automatic peak picking methods. The original peak confidence scores are first converted into -values. The Benjamini-Hochberg
algorithm is then applied to select the number of peaks. In this paper, we demonstrated that the proposed approach worked consistently well using state-of-the-art peak picking methods. Therefore,
this can be a potentially general approach to select a good number of candidates from a large set of predictions.
We are grateful to Ming Li for making PICKY publicly available. We thank Virginia Unkefer for editorial work on the manuscript. The spectra for TM1112, YST0336, RP3384 and ATC1776 were generated by
Cheryl Arrowsmith’s Lab at the University of Toronto. The spectra for COILIN, VRAR, HACS1 and CASKIN were provided by Logan Donaldson’s Lab at York University.
Author Contributions
Critical revision of the manuscript: XK ZL. Conceived and designed the experiments: BJ XG. Performed the experiments: AA XG. Analyzed the data: AA BJ XG. Contributed reagents/materials/analysis
tools: XK ZL XG. Wrote the paper: AA BJ XG. | {"url":"https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0053112","timestamp":"2024-11-12T05:52:07Z","content_type":"text/html","content_length":"213463","record_id":"<urn:uuid:ba2f0375-4058-4533-a5f4-fbb325e167fa>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00388.warc.gz"} |
d. none of the above 6. find the length pm from the circle x^(2)+y^(2)-20 x-24 y+195=0 , whose center is m from the point p(25,32) . a. | Question AI
D. None of the above 6. Find the length PM from the circle x^(2)+y^(2)-20 x-24 y+195=0 , whose center is M from the point P(25,32) . A. 15 units B. 35 units C. 25 units D. 10 units 7. If the distance
between the points (2,-2) and (-1,x) is 5 , one of the values of x is qquad A. 2 B. -1 C. -2 D. 1 8. What is the equation of the line parallel to the line passing through (5,7) and (2,3) and having x
intercept as -4 . A. 3y=4x-16 B. 4y=3x-16 C. 3y=4x+16 D. 4y=3x+16 9. Find the gradient m of the line that passes through (2,3) and the point of intersection of the lines 3x+2y=2 and 4x+3y=7 . A. m=-1
B. m=-2 C. m=1 D. m=2 10.The mid-point of the line segment joining the points A(-2,-8) and B(-6,4) is A. (4,-6) B. (4,-2) C. (-4,-2) D. (-4,-6) 11.The line y+mx=c passes through the point (3,7) is
perpendicular to the line joining the points A(3,3) and D(9,12) . What is the value c ? A. 5 B. 9 C. (5)/(2) D. (13)/(2) 12. Find the coordinates of the midpoint of the line joining the centre of the
circle defined by x^(2)+y^(2)-2x-3y=0 and the point (11,9) . A. (-2,-3) B. (0,0) C. (2,4) D. (5,6) 13.If the mid-point of the line segment joining the points A(3,4) and B(k,6) is P(x,y) and x+y-10=0
, find the value of k .
Not your question?Search it
Sandy MaysVeteran · Tutor for 12 years
<p></p><br />6.A. 15 units<br />7.C. -2<br />8.B. \(4 y=3 x-16\)<br />9.B. \(m=-2\)<br />10.D. \((-4,-6)\)<br />11.D.\(\frac{13}{2}\)<br />12.D.\((5,6)\)<br />13.D. 6
<p></p><br />For question 6, we need to calculate the distance between the center of the circle and point P(25,32). The center of the circle fits the formula (h,k) where h= -(-20)/2 = 10 and k= -
(-24)/2 = 12. Therefore, the center M of the circle is at coordinates M(10,12). Distance MP can be calculated using the distance formula √[(x2 - x1)² + (y2 - y1)²]. <br /><br />For question 7, the
formula for the distance between two points (x1,y1) and (x2,y2) = √[(x2 - x1)² + (y2 - y1)²] and substituting the given coordinates and solving the equation can provide the estimate of the possible
values of 'x' .<br /><br />For question 8 & 11, first find the slope of the line through provided coordinates respectively, and two lines are parallel when their slope 'm' is equal. Equation of a
line is given as *y = mx + c*. The line parallel will have the same slope & to find line with its x-intercept; the general equation of the line changes accordingly *x=c* since point on the line has x
coordinate -4 according to question for 8. For question 11, equation put horizontal demand which suggests the slope isn't equal.<br /><br />For question 9, to find a line's slope the equation *m=
(y2-y1)/(x2-x1)* where, (x1,y1) & (x2,y2) are coordinates, therefore substituting given coordiantes can result the required slope 'm'.<br /> <br />For question10 & 12, Midpoint M of a line segment
with endpoints (x1,y1) and (x2,y2) is M = [(x1+x2)/2 , (y1+y2)/2]. Knowing the coordinates of the centre of the circle and the endpoint (11,9), the coordinates of the midpoint can thus be determined.
<br /><br />Question 13, find the coordinate lie as midpoint and given equation to ascertain the value of 'k'.
Step-by-step video
D. None of the above 6. Find the length PM from the circle x^(2)+y^(2)-20 x-24 y+195=0 , whose center is M from the point P(25,32) . A. 15 units B. 35 units C. 25 units D. 10 units 7. If the distance
between the points (2,-2) and (-1,x) is 5 , one of the values of x is qquad A. 2 B. -1 C. -2 D. 1 8. What is the equation of the line parallel to the line passing through (5,7) and (2,3) and having x
intercept as -4 . A. 3y=4x-16 B. 4y=3x-16 C. 3y=4x+16 D. 4y=3x+16 9. Find the gradient m of the line that passes through (2,3) and the point of intersection of the lines 3x+2y=2 and 4x+3y=7 . A. m=-1
B. m=-2 C. m=1 D. m=2 10.The mid-point of the line segment joining the points A(-2,-8) and B(-6,4) is A. (4,-6) B. (4,-2) C. (-4,-2) D. (-4,-6) 11.The line y+mx=c passes through the point (3,7) is
perpendicular to the line joining the points A(3,3) and D(9,12) . What is the value c ? A. 5 B. 9 C. (5)/(2) D. (13)/(2) 12. Find the coordinates of the midpoint of the line joining the centre of the
circle defined by x^(2)+y^(2)-2x-3y=0 and the point (11,9) . A. (-2,-3) B. (0,0) C. (2,4) D. (5,6) 13.If the mid-point of the line segment joining the points A(3,4) and B(k,6) is P(x,y) and x+y-10=0
, find the value of k .
All Subjects Homework Helper | {"url":"https://www.questionai.com/questions-twa5eeVHEm/d-none-6-find-length-pm-circle-x2y220-x24-y1950-center-m","timestamp":"2024-11-08T22:23:46Z","content_type":"text/html","content_length":"156809","record_id":"<urn:uuid:94565847-6d10-4570-8f7f-9d493636bec9>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00657.warc.gz"} |
euclidean_distance_cvip() - calculates the Euclidean distance between two feature vectors.
d = euclidean_distance_cvip( vector1, vector2)
Input Parameters include :
• vector1 - Input feature vector 1.
• vector2 - Input feature vector 2.
output Parameter include :
• d - euclidean distance between the two vectors.
The Euclidean distance for two vector x, and y is defined as:
The functions calculates the euclidean distance between the two input feature vectors according to the above formula.
1. Scott E Umbaugh. DIGITAL IMAGE PROCESSING AND ANALYSIS: Applications with MATLAB and CVIPtools, 3rd Edition.
% input vectors
vector1 = [1 3 4 2];
vector2 = [-3 3.2 sqrt(2) pi];
% euclidean distance between two vectors.
d = euclidean_distance_cvip( vector1, vector2);
fprintf('Euclidean Distance: %d', d);
Euclidean Distance: 4.901992e+00
Author: Mehrdad Alvandipour, March 2017
Copyright © 2017-2018 Scott E Umbaugh
For updates visit CVIP Toolbox Website | {"url":"https://cviptools.ece.siue.edu/downloads/CVIPhtml/euclidean_distance_cvip.html","timestamp":"2024-11-03T18:58:52Z","content_type":"text/html","content_length":"8288","record_id":"<urn:uuid:7ee128f9-e29c-41bc-afa0-cb1af46bb218>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00159.warc.gz"} |
Looking at variance vs implied vol term structure
VARIANCE TERM STRUCTURE CAN INDENTIFY TRADES To determine if a term structure trade is needed, we could look at variance term structure rather than implied volatility term structure...An ATM term
structure will not be ATM as soon as the spot moves, so it is effectively strike dependent.
Variance term structure is similar to ATM term structure, despite variance being long skew and skew being greater for near-dated implieds. This is because the time value of an OTM option increases
with maturity. Hence, the increased weight associated with OTM options cancels the effect of smaller skew for longer maturities. | {"url":"https://moontowerquant.com/looking-at-variance-vs-implied-vol-term-structure","timestamp":"2024-11-05T15:58:18Z","content_type":"text/html","content_length":"33454","record_id":"<urn:uuid:4cafa067-988d-4fb9-bcbb-e00ec4fceb95>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00592.warc.gz"} |
Comparison with ALICE previous publications, for the INEL event class, at $\sqrt{s} = 0.9$ TeV (top left) and the INEL$>$0 event class at $\sqrt{s} = 7$ TeV (top right); in both cases, ratios
between ALICE new data and ALICE previous data are also shown. The total uncertainties are shown as error bars for the previous data and as a band for the present measurement. For the NSD event
class, comparison with ALICE previous publication and with CMS data at $\sqrt{s} = 0.9$ TeV (bottom left), and comparison with CMS data at $\sqrt{s} = 7$ TeV (bottom right); ratios of the NBD fits
of ALICE data taken without errors to CMS data, for the various $\eta$ intervals (indicated on the figure), are also shown. Error bars represent the contributions of the CMS errors to the ratios,
the bands represent the ALICE total uncertainty assigned to the ratio of 1. | {"url":"https://alice-publications.web.cern.ch/index.php/node/3478","timestamp":"2024-11-08T02:45:04Z","content_type":"text/html","content_length":"40593","record_id":"<urn:uuid:0c30a354-6d43-4ab5-96ad-7658a0f0b2f2>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00266.warc.gz"} |
Calculeaza Integrala
Nu:)).Am rezolvat-o.Vreau sa vedem daca reuseste careva
Ti-o rezolvam eu daca le invatam la liceu..but fuck..
Take the integral:
integral csc(x+1) csc(x+2) dx
Write csc(x+1) csc(x+2) as 2/(cos(1)-cos(2 x+3)):
= integral 2/(cos(1)-cos(2 x+3)) dx
Factor out constants:
= 2 integral 1/(cos(1)-cos(2 x+3)) dx
For the integrand 1/(cos(1)-cos(2 x+3)), substitute u = 2 x+3 and du = 2 dx:
= integral 1/(cos(1)-cos(u)) du
For the integrand 1/(cos(1)-cos(u)), substitute s = tan(u/2) and ds = 1/2 sec^2(u/2) du. Then transform the integrand using the substitutions sin(u) = (2 s)/(s^2+1), cos(u) = (1-s^2)/(s^2+1) and du =
(2 ds)/(s^2+1):
= integral 2/((s^2+1) (cos(1)-(1-s^2)/(s^2+1))) ds
Simplify the integrand 2/((s^2+1) (cos(1)-(1-s^2)/(s^2+1))) to get 2/(s^2+s^2 cos(1)-1+cos(1)):
= integral 2/(s^2+s^2 cos(1)-1+cos(1)) ds
Factor out constants:
= 2 integral 1/(s^2+s^2 cos(1)-1+cos(1)) ds
Factor cos(1)-1 from the denominator:
= 2 integral 1/((cos(1)-1) ((s^2 (1+cos(1)))/(cos(1)-1)+1)) ds
Factor out constants:
= 2/(cos(1)-1) integral 1/((s^2 (1+cos(1)))/(cos(1)-1)+1) ds
For the integrand 1/((s^2 (1+cos(1)))/(cos(1)-1)+1), substitute p = s cot(1/2) and dp = cot(1/2) ds:
= (2 tan(1/2))/(cos(1)-1) integral 1/(1-p^2) dp
The integral of 1/(1-p^2) is tanh^(-1)(p):
= (2 tan(1/2) tanh^(-1)(p))/(cos(1)-1)+constant
Substitute back for p = s cot(1/2):
= csc(1/2) sec(1/2) (-tanh^(-1)(s cot(1/2)))+constant
Substitute back for s = tan(u/2):
= csc(1/2) sec(1/2) (-tanh^(-1)(cot(1/2) tan(u/2)))+constant
Substitute back for u = 2 x+3:
= csc(1/2) sec(1/2) (-tanh^(-1)(cot(1/2) tan(x+3/2)))+constant
Which is equivalent for restricted x values to:
Answer: |
| = csc(1) (log(sin(x+1))-log(sin(x+2)))+constant
csc(1) log(sin(1+x))-csc(1) log(sin(2+x))+C
Scuze de offtopic, dar ce-i cu saracia aia de semnatura ?
Numitorul este produs, deci provine dintr-o adunare de 2 fractii => ne gandim la [ln(ceva)]' * (ceva)' (cazul fericit
Calculand cos(x+1)/sin(x+1) - cos(x+2)/sin(x+2) face, prin aducere la numitor comun, sin1/sin(x+1)sin(x+2). Revenid la integrala initiala, aceasta devine 1/sin1*integrala[cos(x+1)/sin(x+1) - cos(x+2)
/sin(x+2)]dx = 1/sin1 * {ln[sin(x+1)] - ln[sin(x+2)]}+ C.
Sper ca nu am greseli, am facut-o in graba.
Edited by Fame | {"url":"https://rstforums.com/forum/topic/73579-calculeaza-integrala/","timestamp":"2024-11-11T12:59:22Z","content_type":"text/html","content_length":"219260","record_id":"<urn:uuid:e3900476-0540-43b4-bb20-6c80417e1d7d>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00489.warc.gz"} |
Subtraction to 50 with subtrahends >10
I can subtract numbers greater than 10 from numbers to 50.
Students learn to subtract with numbers greater than ten up to 50. They also cross tens.
It is important that students are able to subtract numbers to 50 so that they can determine how many they have left with larger quantities.
Start with a subtraction problem in which students are asked to drag blocks into a box to show subtraction. Then they practice a subtraction problem to 30 on the number line with a minuend greater
than ten.
Discuss the importance of subtraction to 50. This learning goal will be taught in three forms, visual, abstract and in story form. You may use the blue menu in the bottom right to select which form
is best suited to your classroom needs. Each form starts with explanation, then practicing together as a class, and ends with students practicing on their own or in small groups to check their
learning. The first form is visual. Explain a subtraction problem by using the blocks on the interactive whiteboard. Do a problem as a class and then ask students to solve the next visual problem.
Next the abstract problems are given. Tell students that one way of subtracting is to first subtract the tens, then count back to the next ten, and then take away the rest. It is helpful to imagine a
number line while doing this. You can also decompose both numbers and subtract the tens and then subtract the ones. Practice a few examples and then have students do a few problems on their own.
Finally go through the steps of solving a story problem with the class and solve the given story problem. Ask students if they can solve the next story problem in pairs, and the final story problem
individually.To check that students understand subtraction to 50 with minuends greater than ten you can ask the following questions:- Why is it useful to be able to subtract numbers to 50?- How do
you subtract numbers that cross tens? For example 45-17.
Students are given subtraction problems in the visual, abstract and story problem forms.
Ask students to do one subtraction problem in each form, visual, abstract, and story problem. Then divide your class into groups. Each group should get 3 six-sided dice, or 2 10-sided dice. Each
player starts with 50 points. Throw the dice. Those numbers must be subtracted from their 50 points. Who reaches 0 first? If you subtract to "more" than zero, you must throw the dice again.
Students who have difficulty with this can be supported with the use of manipulatives or use of a number line.
3 six-sided dice or 2 ten-sided dice per groupOptional: blocks or other manipulatives
Gynzy is an online teaching platform for interactive whiteboards and displays in schools.
With a focus on elementary education, Gynzy’s Whiteboard, digital tools, and activities make it easy for teachers to save time building lessons, increase student engagement, and make classroom
management more efficient. | {"url":"https://www.gynzy.com/en-us/library/items/subtraction-to-50-with-subtrahends-greater10/2354","timestamp":"2024-11-06T01:58:56Z","content_type":"text/html","content_length":"553556","record_id":"<urn:uuid:47139a23-d635-4650-be5f-3b84efcc5d2a>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00872.warc.gz"} |
Moderation model produces weird results
Replied on Tue, 04/16/2019 - 15:34
I wouldn't be able to even guess without at least seeing the script you're working with. `summary()` output might help, too.
Replied on Wed, 04/17/2019 - 07:00
Thank you for the answer!
Here are the scripts that I use and the output from full moderation models and models without any moderation. When having lowsupport_s as moderator, rE and rP are not consistent across Purcell
moderation model and CF moderation model and are different from the main effects model (and from the observed phenotypic correlation). When removing D, they at least become consistent across Purcell
and CF models, but still not equal to the no moderation results (not to the observed rP). And only when removing all the moderation, rP estimated is in agreement with rP observed.
Thank you for looking at it!
File attachments
Replied on Wed, 04/17/2019 - 11:41
some input
In the Purcell model, have you tried removing the lower bounds on the unmoderated terms of the path coefficients? I notice there's an active bound at the solution:
3 aU MZ.a 2 2 -6.908945e-04 0!
It seems to me that the bounds shouldn't matter, since you mean-centered the moderator, and you identify the liability scale by placing a constraint on the variance at the moderator's zero point.
Still, I'm not 100% sure about that. Notice that the point estimate above is actually negative, meaning that the bound is slightly violated. If the bound weren't there, would the optimizer try to
push that parameter into the negative region?
Note that you don't need to put `var_constraint` into both the MZ and DZ MxModels. It suffices to put it into only one of them.
Have you tried using a different optimizer?
Consider replacing `mxRun()` with mxTryHardOrdinal().
In the correlated-factors model, why is 'Rd' fixed to -0.7?
Replied on Wed, 04/17/2019 - 16:09
In reply to some input by AdminRobK
I tried
I tried to remove lower bounds for the path estimates, and the value went just below zero, but very little:
aU MZ.a 2 2 -0.0004634254
I also tried another optimizer as you suggested, and SLSQP produced nearly identical results, whereas CSOLNP was just hanging for 10 minutes and I had to terminate the session.
As for rD=-0.7 in the CF model, this was based on the best model in terms of AIC when trying different rD values in the range from -1 to 0 with 0.1 step (previous analysis indicated negative rD).
Also thank you for note about putting the variance constraint just once into the model. I actually think that in some other scripts I had it inside the global model. Don't know why I changed it here.
Replied on Wed, 04/17/2019 - 16:37
In reply to I tried by Julia
mxTryHardOrdinal() ?
I tried to remove lower bounds for the path estimates, and the value went just below zero, but very little
Did the fit value appreciably improve?
I also tried another optimizer as you suggested, and SLSQP produced nearly identical results, whereas CSOLNP was just hanging for 10 minutes and I had to terminate the session.
It's encouraging that SLSQP's results agree with NPSOL's. Did you try CSOLNP with the MxConstraint in both the MZ and DZ models? There is a known issue with CSOLNP and redundant equality constraints.
In the next OpenMx release, CSOLNP will at least not freeze uninterruptibly when there are redundant equalities.
As for rD=-0.7 in the CF model, this was based on the best model in terms of AIC when trying different rD values in the range from -1 to 0 with 0.1 step (previous analysis indicated negative rD).
But why isn't it a free parameter?
Have you tried `mxTryHardOrdinal()`?
Replied on Thu, 04/18/2019 - 08:34
In reply to mxTryHardOrdinal() ? by AdminRobK
Have you tried mxTryHardOrdinal()?
Yes, I did. The the results are still the same with no improvement in the fit.
Did the fit value appreciably improve?
The fit value was left unchanged.
Did you try CSOLNP with the MxConstraint in both the MZ and DZ models?
Yes, I put the constraint just into the MZ model and tried all the optimizers. Here are the fit indices:
Cholesky moderation (Purcell)
NPSOL: -2LL=12907.98 AIC= -2100.015
SLSQP: -2LL=12907.98, AIC=-2100.016
CSONLP: -2LL=12907.98, AIC=-2100.016
CF moderation
NPSOL: -2LL =12910.81, AIC = -2105.186
SLSQP: -2LL=12910.84, AIC=-2105.156 (Mx Status Red)
CSONLP: -2LL=12910.84, AIC=-2105.156
But why isn't it a free parameter?
I thought that rA and rD (just like rA and rC) could not be estimated simultaneously, could they?
Replied on Thu, 04/18/2019 - 11:18
In reply to answers by Julia
results are probably right
Well, it looks as though the optimizers really are finding the solution. I agree that the results seem odd, but I guess there's something wrong with our intuition!
I thought that rA and rD (just like rA and rC) could not be estimated simultaneously, could they?
I'm not used to thinking of moderation in terms of the correlated-factors parameterization, so I could be mistaken here, but I don't see any reason why they couldn't be estimated simultaneously.
After all, you were able to estimate a cross-path for _D_, 'dC', in the Cholesky-parameterized model, right? It should be possible to get a correlated-factors solution equivalent to the Cholesky
solution. But, it's possible that the correlated-factors parameterization is harder to optimize.
Replied on Thu, 04/18/2019 - 14:37
In reply to results are probably right by AdminRobK
Yes, what is puzzling me is
Yes, what is puzzling me is that the rP estimated are so different from rP observed! And that estimates at M=0 are totally different from estimates with no moderation. What can be the explanation
here? Can we trust the results of moderation here.
The reason why we try moderation in terms of CF is the paper by Rathouz PJ et al:
Rathouz PJ, Van Hulle CA, Rodgers JL, Waldman ID, Lahey BB. Specification, testing, and interpretation of gene-by-measured-environment interaction models in the presence of gene-environment
correlation. Behav Genet. 2008;38(3):301–315. doi:10.1007/s10519-008-9193-4
There they say that CF model has more power to detect moderation because it has fewer parameters to estimate. Since our power is quite limited due to a low number of twin pairs and due to a not very
prevalent dichotomous outcome, we thought to give CF a try. It doesn't seem to provide any evidence of moderation either, but its fit is better, although the correlation estimates are as weird as in
the Purcell's model (for two out of three moderators that we tested).
Replied on Thu, 04/18/2019 - 15:55
In reply to Yes, what is puzzling me is by Julia
It's been a few years since I
It's been a few years since I read that Rathouz et al. paper, so there's a good chance I'm mistaken in what I posted about the correlated-factors parameterization.
Replied on Sun, 08/18/2019 - 10:14
I just run the model posted by Julia(bivChol_Moderation.txt), but here are something wrong when there is no moderation in the model:
Error: The job for model 'MainEffects' exited abnormally with the error message: fit is not finite (Ordinal covariance is not positive definite in data 'DZ.data' row 13703 (loc1))
In addition: Warning message:
In model 'MainEffects' Optimizer returned a non-zero status code 10. Starting values are not feasible. Consider mxTryHard()
However, I don't know the exzact reason, could you give any questions.
Replied on Sun, 08/18/2019 - 10:14
I just run the model posted by Julia(bivChol_Moderation.txt), but here are something wrong when there is no moderation in the model:
Error: The job for model 'MainEffects' exited abnormally with the error message: fit is not finite (Ordinal covariance is not positive definite in data 'DZ.data' row 13703 (loc1))
In addition: Warning message:
In model 'MainEffects' Optimizer returned a non-zero status code 10. Starting values are not feasible. Consider mxTryHard()
However, I don't know the exzact reason, could you give any questions.
Replied on Sun, 08/18/2019 - 16:25
In reply to Hello by xiyuesenlinyu
start values
Evidently, you need better start values for the free parameters. Did you modify the block of syntax that sets the start values? Values that worked for Julia might not work well with your dataset.
Search this website for "start values". You'll find plenty of advice and discussion about the topic, e.g., this thread. Another thing you could try is to replace `mxRun()` with `mxTryHardOrdinal()`
in your script, and see if that helps. I can't really offer any more-specific advice without more details from you.
Replied on Mon, 08/19/2019 - 10:00
In reply to start values by AdminRobK
I still can’t run my model.
I still can’t run my model.
Firstly, I run the following syntax, which induce some warning and error messages .
# PREPARE DATA #
# Load Data
mxOption(NULL, "Default optimizer", "NPSOL")
#mxOption( NULL, 'Default optimizer' ,'SLSQP' )
# mxOption(NULL, "Default optimizer", "CSOLNP")
# -----------------------------------------------------------------------
# PREPARE DATA
# Read Data
#twinData <- read.table("D:/FHI/Sosiale forhold og helse/Analyses/IBS/SFH_17_09_2018_subset_pss_discordance_strain_lowsupport_family_format.txt", header=T, sep='\t', dec=',')
twinData <-read.dta13("-25-34581pair-coronary-ssex.dta")
twinData <- twinData [!is.na(twinData $coronary1)&!is.na(twinData $coronary2),]
twinData <- twinData [!is.na(twinData $drink_cur1)&!is.na(twinData $drink_cur2),]
nobs = dim(twinData)[1]
twinData$age1 = scale(twinData$age1)
twinData$age2 = scale(twinData$age2)
#twinData$Bmi1 = scale(twinData$bmi1)
#twinData$Bmi2 = scale(twinData$bmi2)
# Select Ordinal Variables
nth <- 1 # number of thresholds
varso <- c('coronary') # list of ordinal variables names
nvo <- length(varso) # number of ordinal variables
ntvo <- nvo*2 # number of total ordinal variables
ordVars <- paste(varso,c(rep(1,nvo),rep(2,nvo)),sep="")
# Select Variables for Analysis
Vars <- c("drink_cur","coronary")
nv <- length(Vars) # number of variables
nind <- 2
ntv <- nv*nind # number of total variables
selVars <- paste(Vars,c(rep(1,nv),rep(2,nv)),sep="")
def = c("sex", "age")
ndef= length(def)
defVars = paste(def,c(rep(1,ndef),rep(2,ndef)),sep="")
# Select Data for Analysis
mzData <- subset(twinData, zy_sex1==1|2, c(selVars,defVars))
dzData <- subset(twinData, zy_sex1==3|4, c(selVars,defVars))
mzData = mzData[complete.cases(mzData[,c(defVars, "drink_cur1", "drink_cur2")]),]
dzData = dzData[complete.cases(dzData[,c(defVars, "drink_cur1", "drink_cur2")]),]
mzDataF = mzData
dzDataF = dzData
mzDataF[,ordVars] <- mxFactor( x=mzDataF[,ordVars], levels=c(0,1) )
dzDataF[,ordVars] <- mxFactor( x=dzDataF[,ordVars], levels=c(0,1) )
# Raw data in OpenMx format
dataMZ <- mxData(observed = mzDataF, type = "raw" )
dataDZ <- mxData(observed = dzDataF, type = "raw" )
# ---------------------Cholesky part!------------------------------------
# Set up Cholesky ADE decomposition, with RawData and Matrices Input
# -----------------------------------------------------------------------
## Labeling
aLabs <- c("aM","aC","aU")
cLabs <- c("cM","cC","cU")
eLabs <- c("eM","eC","eU")
meanLabs <- c("meanM","meanP")
aModLabs <- c("aMod11","aMod21","aMod22")
cModLabs <- c("cMod11","cMod21","cMod22")
eModLabs <- c("eMod11","eMod21","eMod22")
threshLabs <- paste(varso,"thresh",sep="_")
betaLabs_age <- paste("beta","Age",Vars, sep="_")
betaLabs_sex <- paste("beta","Sex",Vars, sep="_")
# Set Starting Values
frMV <- c(TRUE, FALSE) # free status for variables
frCvD <- diag(frMV,ntv,ntv) # lower bounds for diagonal of covariance matrix
frCvD[lower.tri(frCvD)] <- TRUE # lower bounds for below diagonal elements
frCvD[upper.tri(frCvD)] <- TRUE # lower bounds for above diagonal elements
frCv <- matrix(as.logical(frCvD),4)
svMe <- c(0,0) # start value for means
svPa <- .4 # start value for path coefficient
svPaD <- vech(diag(svPa,nv,nv)) # start values for diagonal of covariance matrix
svPe <- .8 # start value for path coefficient for e
svPeD <- vech(diag(svPe,nv,nv)) # start values for diagonal of covariance matrix
lbPa <- 0 # start value for lower bounds
lbPaD <- diag(lbPa,nv,nv) # lower bounds for diagonal of covariance matrix
lbPaD[lower.tri(lbPaD)] <- -10 # lower bounds for below diagonal elements
lbPaD[upper.tri(lbPaD)] <- NA # lower bounds for above diagonal elements
svTh <- 1.5 # start value for thresholds
pathModVal = c(0,0.1,0.1)
B_AgeVal = 0.5
B_SexVal = 0.5
## Modeling
# Matrices a, c, and e to store a, c, and e Path Coefficients
pathA <- mxMatrix(name = "a", type = "Lower", nrow = nv, ncol = nv, free=T, labels = aLabs, values=svPaD, lbound=lbPaD)
pathC <- mxMatrix(name = "c", type = "Lower", nrow = nv, ncol = nv, free=T, labels = cLabs, values=svPaD, lbound=lbPaD)
pathE <- mxMatrix(name = "e", type = "Lower", nrow = nv, ncol = nv, free=T, labels = eLabs, values=svPeD, lbound=lbPaD)
modPathA = mxMatrix( "Lower", nrow=nv, ncol=nv, free=c(F,T,T), values=pathModVal, labels=aModLabs, name="aMod" )
modPathC = mxMatrix( "Lower", nrow=nv, ncol=nv, free=c(F,T,T), values=pathModVal, labels=cModLabs, name="cMod" )
modPathE = mxMatrix( "Lower", nrow=nv, ncol=nv, free=c(F,T,T), values=pathModVal, labels=eModLabs, name="eMod" )
# Moderator
mod_tw1 <- mxMatrix( type="Full", nrow=1, ncol=1, free=FALSE, labels='data.drink_cur1', name="Mod1")
mod_tw2 <- mxMatrix( type="Full", nrow=1, ncol=1, free=FALSE, labels='data.drink_cur2', name="Mod2")
# Matrices generated to hold A, D, and E computed Variance Components
varA1 <- mxAlgebra(name = "A1", expression = (a + Mod1%x%aMod) %*% t(a+ Mod1%x%aMod))
varC1 <- mxAlgebra(name = "C1", expression = (c + Mod1%x%cMod) %*% t(c+ Mod1%x%cMod))
varE1 <- mxAlgebra(name = "E1", expression = (e + Mod1%x%eMod) %*% t(e+ Mod1%x%eMod))
covA12 = mxAlgebra(name = "A12", expression = (a + Mod1%x%aMod) %*% t(a+ Mod2%x%aMod))
covC12 = mxAlgebra(name = "C12", expression = (c + Mod1%x%cMod) %*% t(c+ Mod2%x%cMod))
covE12 = mxAlgebra(name = "E12", expression = (e + Mod1%x%eMod) %*% t(e+ Mod2%x%eMod))
covA21 = mxAlgebra(name = "A21", expression = (a + Mod2%x%aMod) %*% t(a+ Mod1%x%aMod))
covC21 = mxAlgebra(name = "C21", expression = (c + Mod2%x%cMod) %*% t(c+ Mod1%x%cMod))
covE21 = mxAlgebra(name = "E21", expression = (e + Mod2%x%eMod) %*% t(e+ Mod1%x%eMod))
varA2 <- mxAlgebra(name = "A2", expression = (a + Mod2%x%aMod) %*% t(a+ Mod2%x%aMod))
varC2 <- mxAlgebra(name = "C2", expression = (c + Mod2%x%cMod) %*% t(c+ Mod2%x%cMod))
varE2 <- mxAlgebra(name = "E2", expression = (e + Mod2%x%eMod) %*% t(e+ Mod2%x%eMod))
myVarA <- mxAlgebra(name = "A0", expression = a %*% t(a))
myVarC <- mxAlgebra(name = "C0", expression = c %*% t(c))
myVarE <- mxAlgebra(name = "E0", expression = e %*% t(e))
# Algebra to compute total variances and standard deviations (diagonal only) per twin
var1 <- mxAlgebra( A1+C1+E1, name="V1" )
var2 <- mxAlgebra( A2+C2+E2, name="V2" )
myVar <- mxAlgebra( A0+C0+E0, name="V0" )
# Constraint on variance of Binary variables
matUnv <- mxMatrix( type="Unit", nrow=nvo, ncol=1, name="Unv1" )
var_constraint <- mxConstraint(V0[2,2] == Unv1, name="Var1")
### Algebra for expected variance/covariance matrices
expCovMZ <- mxAlgebra(name = "expCovMZ", expression = rbind (cbind(A1+C1+E1, A12+C12),
cbind(A21+C21, A2+C2+E2)))
expCovDZ <- mxAlgebra(name = "expCovDZ", expression = rbind (cbind(A1+C1+E1, 0.5%x%A12+C12),
cbind(0.5%x%A21+C21, A2+C2+E2)))
# Matrices for expected Means for females and males
#setting up the regression
intercept <- mxMatrix( type="Full", nrow=1, ncol=ntv, free=frMV, values=svMe, labels=meanLabs, name="intercept" )
threshold <-mxMatrix( type="Full", nrow=1, ncol=ntvo, free=T, values=svTh, labels=threshLabs, name="Threshold" )
# Regression effects
B_Age <- mxMatrix( type="Full", nrow=1, ncol=nv, free=TRUE, values=.1, labels=betaLabs_age, name="bAge" )
defAge <- mxMatrix( type="Full", nrow=1, ncol=2, free=FALSE, labels=c("data.drink_cur1","data.drink_cur2"), name="Age")
B_Sex <- mxMatrix( type="Full", nrow=1, ncol=nv, free=TRUE, values=.1, labels=betaLabs_sex, name="bSex" )
defSex <- mxMatrix( type="Full", nrow=1, ncol=2, free=FALSE, labels=c("data.sex1","data.sex2"), name="Sex")
expMean <- mxAlgebra( intercept + (Age %x% bAge) + (Sex %x% bSex) , name="expMean")
inclusions <- list (B_Age, B_Sex, defAge, defSex, intercept, expMean, threshold)
# Objective objects for Multiple Groups
expMZ <- mxExpectationNormal( covariance="expCovMZ", means="expMean", dimnames=selVars, thresholds="Threshold", threshnames=ordVars )
expDZ <- mxExpectationNormal( covariance="expCovDZ", means="expMean", dimnames=selVars ,thresholds="Threshold", threshnames=ordVars )
funML <- mxFitFunctionML()
# MZ and DZ models
modelMZ <- mxModel(pathA, pathC, pathE, modPathA, modPathC, modPathE, mod_tw1, mod_tw2,
varA1, varC1, varE1, covA12, covC12, covE12, covA21, covC21, covE21, varA2, varC2, varE2, var1, var2,
myVarA, myVarC, myVarE, myVar, matUnv, var_constraint,
B_Age, B_Sex, defAge, defSex, intercept, expMean, threshold,
dataMZ, expCovMZ, expMZ, funML, name = "MZ")
modelDZ <- mxModel(pathA, pathC, pathE, modPathA, modPathC, modPathE, mod_tw1, mod_tw2,
varA1, varC1, varE1, covA12, covC12, covE12, covA21, covC21, covE21, varA2, varC2, varE2, var1, var2,
myVarA, myVarC, myVarE, myVar, matUnv, var_constraint,
B_Age, B_Sex, defAge, defSex, intercept, expMean, threshold,
dataDZ, expCovDZ, expDZ, funML, name = "DZ")
#plan <- omxDefaultComputePlan()
#plan$steps$GD <- mxComputeNelderMead()
multi <- mxFitFunctionMultigroup( c("MZ","DZ") )
ACEmodModel <- mxModel( "ACEmod", modelMZ, modelDZ, funML, multi )
#ACEmodFit <- mxAutoStart(ACEmodModel)
ACEmodFit <- mxRun(ACEmodModel)
ACEmodFit <- mxTryHardOrdinal(ACEmodModel)
model <- mxTryHard(ACEmodFit)
mxGetExpected(ACEmodModel, "covariance", "MZ")
MainEffectsModel <- mxModel (ACEmodFit, name='MainEffects')
MainEffectsModel <- omxSetParameters(MainEffectsModel, labels=c('aMod21','aMod22'), values=c(0,0), free=FALSE)
MainEffectsModel <- omxSetParameters(MainEffectsModel, labels=c('cMod21','cMod22'), values=c(0,0), free=FALSE)
MainEffectsModel <- omxSetParameters(MainEffectsModel, labels=c('eMod21','eMod22'), values=c(0,0), free=FALSE)
#ACEmodFit <- mxAutoStart(MainEffectsModel)
#ACEmodFit <- mxRun(MainEffectsModel)
MainEffectsFit <- mxRun(MainEffectsModel)
MainEffectsFit <- mxTryHardOrdinal(MainEffectsModel)
mxCompare(ACEmodFit, MainEffectsFit)
model <- mxTryHard(MainEffectsFit)
Many of Std.Error are NA in ACEmodModel .
ACEmodFit <- mxRun(ACEmodModel)
Running ACEmod with 22 parameters
Summary of ACEmod
free parameters:
name matrix row col Estimate Std.Error A lbound
1 aM MZ.a 1 1 1.926922e-02 1.902139e-04 ! 0!
2 aC MZ.a 2 1 -4.604157e-01 4.172067e-03 -10
3 aU MZ.a 2 2 1.801983e-02 1.806019e-02 ! 0!
4 cM MZ.c 1 1 1.954307e-02 4.009311e-04 ! 0!
5 cC MZ.c 2 1 -2.463518e-01 1.425125e-02 -10
6 cU MZ.c 2 2 -7.894769e-03 1.938597e-02 ! 0!
7 eM MZ.e 1 1 1.837021e-02 2.447180e-04 ! 0!
8 eC MZ.e 2 1 -8.526084e-01 2.113446e-03 -10
9 eU MZ.e 2 2 -3.436889e-05 1.613466e-02 ! 0!
10 aMod21 MZ.aMod 2 1 -5.222943e+02 NA !
11 aMod22 MZ.aMod 2 2 7.894969e+01 3.810561e+01 !
12 cMod21 MZ.cMod 2 1 -4.150706e+02 NA
13 cMod22 MZ.cMod 2 2 1.734169e+02 NA !
14 eMod21 MZ.eMod 2 1 -1.133761e+03 NA !
15 eMod22 MZ.eMod 2 2 2.499588e+02 NA !
16 beta_Age_drink_cur MZ.bAge 1 1 1.000291e+00 3.464405e-04
17 beta_Age_coronary MZ.bAge 1 2 -1.675897e+03 NA !
18 beta_Sex_drink_cur MZ.bSex 1 1 5.461896e-05 3.377409e-04
19 beta_Sex_coronary MZ.bSex 1 2 -2.344350e-02 1.639405e-02
20 meanM MZ.intercept 1 1 -6.592594e-03 5.280465e-04
21 coronary_thresh MZ.Threshold 1 coronary1 1.178388e+00 2.735566e-02
As for MainEffectsModel(without moderator),
MainEffectsFit <- mxRun(MainEffectsModel)
Running MainEffects with 15 parameters
Error: The job for model 'MainEffects' exited abnormally with the error message: fit is not finite (Ordinal covariance is not positive definite in data 'DZ.data' row 13703 (loc1))
In addition: Warning message:
In model 'MainEffects' Optimizer returned a non-zero status code 10. Starting values are not feasible. Consider mxTryHard()
Secondly, I run mxGetExpected(). However, I just don’t know how to reset the start values.
drink_cur1 coronary1 drink_cur2 coronary2
drink_cur1 0.96 0.00 0.32 0.00
coronary1 0.00 0.96 0.00 0.32
drink_cur2 0.32 0.00 0.96 0.00
coronary2 0.00 0.32 0.00 0.96
drink_cur1 coronary1 drink_cur2 coronary2
drink_cur1 0.96 0.00 0.24 0.00
coronary1 0.00 0.96 0.00 0.24
drink_cur2 0.24 0.00 0.96 0.00
coronary2 0.00 0.24 0.00 0.96
I also replace mxRun() with MainEffectsFit <- mxTryHardOrdinal(MainEffectsModel). Though the error of NA are removed, the model without moderator didn’t work.
ACEmodFit <- mxTryHardOrdinal(ACEmodModel)
Solution found! Final fit=-448097.41 (started at 171627.54) (11 attempt(s): 8 valid, 3 errors)
Warning message:
In mxTryHard(model = model, greenOK = greenOK, checkHess = checkHess, :
The final iterate satisfies the optimality conditions to the accuracy requested, but the sequence of iterates has not yet converged. Optimizer was terminated because no further improvement could be
made in the merit function (Mx status GREEN).
MainEffectsFit <- mxTryHardOrdinal(MainEffectsModel)
All fit attempts resulted in errors - check starting values or model specification.
I read the suggestions you posted, but I failed to use Nelder-Mead implementation. I might be a little stupid, so I need your help.
Replied on Mon, 08/19/2019 - 11:54
In reply to I still can’t run my model. by xiyuesenlinyu
some suggestions
First off, only put `var_constraint` in one of the MZ or DZ MxModels, not both. That won't matter if you're using NPSOL, but it will be a problem if you use CSOLNP (or Nelder-Mead).
Many of Std.Error are NA in ACEmodModel .
It looks like the optimizer is reaching a solution where the Hessian (as calculated) isn't positive-definite (status code 5). Your phenotype is a threshold trait, and due to the limited accuracy of
the algorithm for the multivariate-normal probability integral, sometimes code 5 can occur even when the optimizer has found a minimum. Therefore, you'll want to find the solution with the smallest
fitfunction value, even if it has status code 5. Try requesting more attempts from `mxTryHardOrdinal()` via argument `extraTries`, e.g. `extraTries=30`.
As for MainEffectsModel(without moderator),
I suggest running the main-effects model *before* the moderation model. Use `free=FALSE` when creating `modPathA`, `modPathC`, and `modPathE`. Then, the first MxModel you run will be the main effects
model. Then, create the moderation model from the fitted main-effects model, and use omxSetParameters() to free the moderation parameters.
Secondly, I run mxGetExpected(). However, I just don’t know how to reset the start values.
What were you trying to do? Use omxSetParameters() to change free parameter values.
I read the suggestions you posted, but I failed to use Nelder-Mead implementation. I might be a little stupid, so I need your help.
If you want to try it, let me suggest some syntax:
plan <- omxDefaultComputePlan()
plan$steps <- list(
Then, put `plan` into the `mxModel()` statement for `MainEffectsModel`, assuming you are creating and running `MainEffectsModel` first. To clear the custom compute plan from an MxModel and go back to
the default compute plan, do model@compute <- NULL.
Replied on Tue, 08/20/2019 - 00:43
In reply to some suggestions by AdminRobK
I change the order of models,
I change the order of models, and fit the main-effect model first. I also delete the var_constraint in MZ MxModel(of course try this in DZ MxModel). However, there still error massage:
> MainEffectsFit <- mxTryHardOrdinal(MainEffectsModel)
Solution found! Final fit=-444773.78 (started at 169287.04) (11 attempt(s): 5 valid, 6 errors)
It seems the model didn’t run. I want to Try requesting more attempts from mxTryHardOrdinal() via argument extra Tries, e.g. extra Tries=30.
Then, I put plan into the mxModel() statement for MainEffectsModel, errors
plan <- omxDefaultComputePlan()
plan$steps <- list(
multi <- mxFitFunctionMultigroup( c("MZ","DZ") )
#ACEmodModel <- mxModel( "ACEmod", modelMZ, modelDZ, funML, multi
MainEffectsModel<- mxModel( "MainEffects", modelMZ, modelDZ, funML, multi,plan )
MainEffectsFit <- mxTryHardOrdinal(MainEffectsModel)
All fit attempts resulted in errors - check starting values or model specification
As for Try requesting more attempts from mxTryHardOrdinal() via argument extraTries, e.g. extra Tries=30, even 90, errors still.
All fit attempts resulted in errors - check starting values or model specification.
In the main effect model, start values of moderator Path Coefficients are 0, isn’t it? Do I need to change the start values of a, c, and e Path Coefficients.
pathModVal = c(0,0.1,0.1)
B_AgeVal = 0.5
B_SexVal = 0.5
# Matrices a, c, and e to store a, c, and e Path Coefficients
pathA <- mxMatrix(name = "a", type = "Lower", nrow = nv, ncol = nv, free=T, labels = aLabs, values=svPaD, lbound=lbPaD)
pathC <- mxMatrix(name = "c", type = "Lower", nrow = nv, ncol = nv, free=T, labels = cLabs, values=svPaD, lbound=lbPaD)
pathE <- mxMatrix(name = "e", type = "Lower", nrow = nv, ncol = nv, free=T, labels = eLabs, values=svPeD, lbound=lbPaD)
modPathA = mxMatrix( "Lower", nrow=nv, ncol=nv, free=c(F,F,F), values=0, labels=aModLabs, name="aMod" )
modPathC = mxMatrix( "Lower", nrow=nv, ncol=nv, free=c(F,F,F), values=0, labels=cModLabs, name="cMod" )
modPathE = mxMatrix( "Lower", nrow=nv, ncol=nv, free=c(F,F,F), values=0, labels=eModLabs, name="eMod" )
Replied on Tue, 08/20/2019 - 14:27
In reply to I change the order of models, by xiyuesenlinyu
First of all,
First of all, `mxTryHardOrdinal()` uses random-number generation. You can get different results each time you run it, which is especially likely if your model is hard to optimize for whatever reason.
If you want reproducible results, precede each call to `mxTryHardOrdinal()` with something to set the random-number generator's seed, e.g. `set.seed(1)`.
I change the order of models, and fit the main-effect model first. I also delete the var_constraint in MZ MxModel(of course try this in DZ MxModel). However, there still error massage:
> MainEffectsFit <- mxTryHardOrdinal(MainEffectsModel)
Solution found! Final fit=-444773.78 (started at 169287.04) (11 attempt(s): 5 valid, 6 errors)
It seems the model didn’t run.
Huh? But it did! `mxTryHardOrdinal()` reported "Solution found!".
Then, I put plan into the mxModel() statement for MainEffectsModel, errors
plan <- omxDefaultComputePlan()
plan$steps <- list(
multi <- mxFitFunctionMultigroup( c("MZ","DZ") )
#ACEmodModel <- mxModel( "ACEmod", modelMZ, modelDZ, funML, multi
MainEffectsModel<- mxModel( "MainEffects", modelMZ, modelDZ, funML, multi,plan )
MainEffectsFit <- mxTryHardOrdinal(MainEffectsModel)
All fit attempts resulted in errors - check starting values or model specification
OK, maybe the custom compute plan was a bad idea.
In the main effect model, start values of moderator Path Coefficients are 0, isn’t it?
Yes, and since they're fixed, they will remain at 0 during optimization.
Do I need to change the start values of a, c, and e Path Coefficients.
You don't have to change them, but maybe there is a better choice of start values than what you're currently using. Sorry I don't have any specific advice about that.
Replied on Tue, 08/20/2019 - 14:59
Premise: A without moderation = A when moderated by zero?
Hi Julia!
A few comments:
I think that the expectation - that the estimate of A when the moderator value is zero should be the same as when no moderation is specified in the model - is incorrect. I've not read the thread in
detail, but basically the estimate of A with no moderation is an average over all levels of Ac + Am * M (A constant plus A moderated * M). Depending on the distribution of M (is it symmetric around
zero?), and the size of Am, you should not generally expect Ac = A.
For starting values in moderated models, I usually start moderator effect at zero. Also, if the moderator is something like age in years, it can help a lot to rescale to age in centuries to
approximate a 0-1 range for the moderator value.
Finally, as you may already know, with binary data some models are not identified. See Medland, S.E., Neale, M.C., Eaves, L.J., Neale, B.M. (2009). A note on the parameterization of purcell's G x E
model for ordinal and binary Data. Behavior Genetics, 39 (2): 220-229.
Replied on Tue, 08/20/2019 - 16:41
isolating the question
I've tried to re-phrase the opening message to understand what's going on.
> I am running a bivariate GxE model with a dichotomous DV, and a continuous but skewed moderator not shared by twins
Is that right? Below, you mention two moderators - that's a valiant script :-)
> Estimates at the zero level of moderator with the estimates when there is no moderation in the model differ.
Non-significant variables happily take (random, drop-able without significant loss of fit) values.
> I am running ADE model even though D can be dropped out of the model but power is low.
I'm going to ask below: what's the N here?
> For one of the moderators...
Hang on... so two moderators?
> For the other two moderators
I'd say "for the other moderator"
What's your n? With a DV, and two moderators, all loading on the DV via three components, and having three moderated loadings on the DV, and moderating three influences on the DV, this will be a hard
model to get stable estimates on.
Replied on Tue, 08/20/2019 - 18:01
user xiyuesenlinyu?
AdminNeale, tbates: Any advice for xiyuesenlinyu (not the OP)? They're the reason I emailed the list about this thread.
Replied on Wed, 08/21/2019 - 01:53
In reply to user xiyuesenlinyu? by AdminRobK
Now, the model runs well. Another problem, the p-value=1 between the main effect model and model with moderator.
base comparison ep minus2LL df AIC diffLL diffdf p
1 MainEffects 18 -412071.5 159463 -730997.5 NA NA NA
2 MainEffects MainEffects 15 -599113.0 159466 -918045.0 -187041.4 3 1
Replied on Wed, 08/21/2019 - 09:54
In reply to Thanks! by xiyuesenlinyu
that doesn't seem right
Another problem, the p-value=1 between the main effect model and model with moderator.
If I'm reading the table correctly, then I think the model with 18 free parameters didn't converge. A model with 18 free parameters should have a minus2LL no greater than a model with 15 parameters.
I don't personally find `mxCompare()` very useful. Would you mind posting the `summary()` output you get from both fitted models, as well as your output from mxVersion()?
Replied on Wed, 08/21/2019 - 21:42
In reply to that doesn't seem right by AdminRobK
If I'm reading the table
If I'm reading the table correctly, then I think the model with 18 free parameters didn't converge. A model with 18 free parameters should have a minus2LL no greater than a model with 15
There were no error messages when the extraTries = 10 in the ACEmodModel(the full model with moderator). However, the following warning message appeared when I set extraTries = 30. And the warning
messages disappeared when extraTries = 80. As for the output of mxCompare(ACEmodFit,MainEffectsFit) ,there were no change.
> set.seed(1)
> ACEmodFit <- mxTryHardOrdinal(ACEmodModel)
Solution found! Final fit=-476335.1 (started at 171627.54) (11 attempt(s): 7 valid, 4 errors)
> ACEmodFit <- mxTryHardOrdinal(ACEmodModel,extraTries = 30)
Solution found! Final fit=-491704.8 (started at 171627.54) (31 attempt(s): 14 valid, 17 errors)
Warning message:
In mxTryHard(model = model, greenOK = greenOK, checkHess = checkHess, :
The model does not satisfy the first-order optimality conditions to the required accuracy, and no improved point for the merit function could be found during the final linesearch (Mx status RED)
> set.seed(1)
> ACEmodFit <- mxTryHardOrdinal(ACEmodModel,extraTries = 80)
Solution found! Final fit=-476335.1 (started at 171627.54) (81 attempt(s): 35 valid, 46 errors)
OpenMx version: 2.13.2 [GIT v2.13.2]
R version: R version 3.6.0 (2019-04-26)
Platform: x86_64-w64-mingw32
Default optimizer: CSOLNP
NPSOL-enabled?: Yes
OpenMP-enabled?: No
MainEffectsFit <- mxTryHardOrdinal(MainEffectsModel,extraTries = 30)
Solution found! Final fit=-599112.99 (started at 169287.04) (31 attempt(s): 14 valid, 17 errors)
> summary(MainEffectsFit)
Summary of MainEffects
free parameters:
name matrix row col Estimate Std.Error A lbound
1 aM MZ.a 1 1 1.449491e-02 7.028758e-05 ! 0!
2 aC MZ.a 2 1 8.430153e-01 9.187338e-03 -10
3 aU MZ.a 2 2 8.874368e-06 2.635970e-02 ! 0!
4 cM MZ.c 1 1 1.727313e-02 1.369073e-04 ! 0!
5 cC MZ.c 2 1 3.672863e-01 1.056641e-02 -10
6 cU MZ.c 2 2 4.954755e-06 3.044826e-02 ! 0!
7 eM MZ.e 1 1 6.485712e-04 9.930890e-06 ! 0!
8 eC MZ.e 2 1 3.361104e-01 1.644506e-02 -10
9 eU MZ.e 2 2 2.036068e-01 1.104726e-02 0
10 beta_Age_drink_cur MZ.bAge 1 1 1.000008e+00 1.929002e-05
11 beta_Age_coronary MZ.bAge 1 2 6.779762e-03 1.721201e-02
12 beta_Sex_drink_cur MZ.bSex 1 1 -3.644746e-05 2.361072e-04
13 beta_Sex_coronary MZ.bSex 1 2 -2.464578e-02 1.912105e-02
14 meanM MZ.intercept 1 1 3.261617e-03 3.507051e-04
15 coronary_thresh MZ.Threshold 1 coronary1 1.340007e+00 3.094911e-02
Model Statistics:
| Parameters | Degrees of Freedom | Fit (-2lnL units)
Model: 15 159466 -599113
Saturated: NA NA NA
Independence: NA NA NA
Number of observations/statistics: 39870/159481
Constraint 'DZ.Var1' contributes 1 observed statistic.
Information Criteria:
| df Penalty | Parameters Penalty | Sample-Size Adjusted
AIC: -918045 -599083.0 -599083.0
BIC: -2288397 -598954.1 -599001.8
CFI: NA
TLI: 1 (also known as NNFI)
RMSEA: 0 [95% CI (NA, NA)]
Prob(RMSEA <= 0.05): NA
To get additional fit indices, see help(mxRefModels)
timestamp: 2019-08-22 08:17:54
Wall clock time: 4.304247 secs
OpenMx version number: 2.13.2
Need help? See help(mxSummary)
> set.seed(1)
> ACEmodFit <- mxTryHardOrdinal(ACEmodModel)
Solution found! Final fit=-476335.1 (started at 171627.54) (11 attempt(s): 7 valid, 4 errors)
> summary(ACEmodFit)
Summary of MainEffects
free parameters:
name matrix row col Estimate Std.Error A lbound
1 aM MZ.a 1 1 7.190948e-04 -7.407524e+00 ! 0!
2 aC MZ.a 2 1 -2.504145e-02 1.653835e-01 -10
3 aU MZ.a 2 2 -5.604461e-08 1.628159e-07 0!
4 cM MZ.c 1 1 1.919975e-04 -2.375641e+01 ! 0!
5 cC MZ.c 2 1 -1.971316e-01 6.797870e-01 -10
6 cU MZ.c 2 2 1.692647e-02 -5.648652e-02 ! 0!
7 eM MZ.e 1 1 1.960369e-02 -1.176272e+01 ! 0!
8 eC MZ.e 2 1 9.799110e-01 -8.964625e-01 -10
9 eU MZ.e 2 2 -1.775966e-08 3.622713e-08 0!
10 aMod21 MZ.aMod 2 1 -2.648286e+01 -2.601806e-06
11 aMod22 MZ.aMod 2 2 -4.271794e-04 -7.992276e-10
12 cMod21 MZ.cMod 2 1 -7.242559e+01 -4.082045e-06
13 cMod22 MZ.cMod 2 2 2.722510e-01 6.434716e-08 !
14 eMod21 MZ.eMod 2 1 4.277475e+02 4.941039e-06
15 eMod22 MZ.eMod 2 2 -1.684815e-04 -4.634743e-10
16 beta_Age_drink_cur MZ.bAge 1 1 1.000149e+00 -1.141624e-01
17 beta_Age_coronary MZ.bAge 1 2 -4.276868e+02 5.793846e-06
18 beta_Sex_drink_cur MZ.bSex 1 1 -1.002649e-04 8.014452e+01
19 beta_Sex_coronary MZ.bSex 1 2 -2.274997e-02 -1.474523e+00
20 meanM MZ.intercept 1 1 4.591315e-03 5.418001e+01
21 coronary_thresh MZ.Threshold 1 coronary1 9.840132e-01 1.179441e+00
Model Statistics:
| Parameters | Degrees of Freedom | Fit (-2lnL units)
Model: 21 159460 -476335.1
Saturated: NA NA NA
Independence: NA NA NA
Number of observations/statistics: 39870/159481
Constraint 'DZ.Var1' contributes 1 observed statistic.
Information Criteria:
| df Penalty | Parameters Penalty | Sample-Size Adjusted
AIC: -795255.1 -476293.1 -476293.1
BIC: -2165555.4 -476112.6 -476179.4
CFI: NA
TLI: 1 (also known as NNFI)
RMSEA: 0 [95% CI (NA, NA)]
Prob(RMSEA <= 0.05): NA
To get additional fit indices, see help(mxRefModels)
timestamp: 2019-08-22 08:59:39
Wall clock time: 6.842392 secs
OpenMx version number: 2.13.2
Need help? See help(mxSummary)
> mxCompare(ACEmodFit,MainEffectsFit)
base comparison ep minus2LL df AIC diffLL diffdf p
1 MainEffects 21 -476335.1 159460 -795255.1 NA NA NA
2 MainEffects MainEffects 15 -599113.0 159466 -918045.0 -122777.9 6 1
Replied on Fri, 08/23/2019 - 05:01
In reply to If I'm reading the table by xiyuesenlinyu
Mr Robert
Mr Robert
Could you give some suggestions for the output of P-value=1?
Replied on Fri, 08/23/2019 - 11:04
In reply to Mr Robert by xiyuesenlinyu
Please post your current script, preferably as an attachment.
Replied on Sat, 08/24/2019 - 10:37
In reply to script? by AdminRobK
ok, thank you!
ok, thank you!
File attachments
Replied on Mon, 08/26/2019 - 10:13
In reply to ok, thank you! by xiyuesenlinyu
Create moderation model from fitted main-effects model
Try replacing this (line 249),
ACEmodModel <- mxModel (MainEffectsModel, name='MainEffects')
, with this,
ACEmodModel <- mxModel (MainEffectsFit, name='ACEmoderation')
. That way, you'll be starting the moderation model from the solution for the main-effects model.
Replied on Sat, 08/24/2019 - 11:11
In reply to Mr Robert by xiyuesenlinyu
A trick to help sticky optimization
When a sub model with fewer parameters fits better than its supermodel, the fit of the supermodel has to be wrong, because it should fit at least as well as the sub model. In such situations, poor
starting values for the supermodel are often the culprit. I suggest that you take the fitted sub model and free up the parameters (or stop having two parameters equated, estimate them separately)
necessary to turn the sub model into the supermodel. Then start fitting the supermodel from the sub model’s solution. At the least, the fit of the supermodel should not get any worse than that of the
sub model (which is where the supermodel optimization is beginning).
Replied on Sun, 08/25/2019 - 03:47
In reply to A trick to help sticky optimization by AdminNeale
Thanks for your suggestions,
Thanks for your suggestions, but I can’t understand well!
“I suggest that you take the fitted sub model and free up the parameters (or stop having two parameters equated, estimate them separately) necessary to turn the sub model into the supermodel.”
Did you mean to reset the three parameters like those
modPathA = mxMatrix( "Lower", nrow=nv, ncol=nv, free=c(F,T,T), values=pathModVal, labels=aModLabs, name="aMod" )
modPathC = mxMatrix( "Lower", nrow=nv, ncol=nv, free=c(F,T,T), values=pathModVal, labels=cModLabs, name="cMod" )
modPathE = mxMatrix( "Lower", nrow=nv, ncol=nv, free=c(F,T,T), values=pathModVal, labels=eModLabs, name="eMod" )
As for “or stop having two parameters equated, estimate them separately”. Though I can estimate them separately(other submodels), I need to estimate all of them in full model.
Another question, the -2LL are negative in some models, are they ok?
Replied on Tue, 08/27/2019 - 11:00
Hello, the model can be run
Hello, the model can be run now. however, I got puzzling output: the differences of -2LL are significant between the two models, and an odd p-value(0) for mxcompare().
Could you give some suggestions to me?
Thank you!
> mxCompare(ACEmodFit,MainEffectsFit)
base comparison ep minus2LL df AIC diffLL diffdf p
1 ACEmoderation 21 40711.46 159460 -278208.5 NA NA NA
2 ACEmoderation MainEffects 15 63773.39 159466 -255158.6 23061.93 6 0
Replied on Tue, 08/27/2019 - 12:11
In reply to Hello, the model can be run by xiyuesenlinyu
A chi-square test statistic
A chi-square test statistic of 23061.93 on 6df has a p-value computationally equal to zero:
> pchisq(23061.93,df=6,lower.tail=F)
[1] 0
Replied on Tue, 08/27/2019 - 19:10
In reply to A chi-square test statistic by AdminRobK
I’m sorry, I’m new for OpenMx
I’m sorry, I’m new for OpenMx, even for R. I have no idea for the output. Could you tell me what does that mean with a p-value=0? An Inappropriate model or poor model fitting? Is the model reliable?
What can I do for improvement?
Replied on Wed, 08/28/2019 - 10:09
In reply to I’m sorry, I’m new for OpenMx by xiyuesenlinyu
You're comparing the
You're comparing the moderation model to the main-effects model. A nonempty proper subset of the free parameters in the moderation model is fixed in the main-effects model, thus, the main-effects
model is said to be a "nested submodel" of the moderation model. The p-value of 0 is telling you that the main-effects model provides much, much worse fit to the data than the moderation model. Or,
to put it another way, you can reject the null hypothesis that the six moderation parameters are all equal to zero.
Replied on Wed, 08/28/2019 - 10:27
In reply to You're comparing the by AdminRobK
Thanks for your help. I think
Thanks for your help. I think that P-value <0.05 between two models may be significant, which suggests the moderated effects. My concern is p=0(just right), which seems not quite right.
Replied on Wed, 08/28/2019 - 11:45
In reply to Thanks for your help. I think by xiyuesenlinyu
The output of `pchisq()` is underflowing to zero. For the sake of perspective, the 99.99th percentile of a chi-square distribution on 6df is about 27.9:
> qchisq(0.9999,df=6)
[1] 27.85634
Replied on Wed, 08/28/2019 - 12:08
Disconcerting to have p-values, fit value reported incorrectly
This issue is the same: p = 1 being reported as p =0) : https://github.com/OpenMx/OpenMx/issues/131
Also this issue where RMSEA and TLI are misreported) https://github.com/OpenMx/OpenMx/issues/221
Replied on Wed, 08/28/2019 - 13:38
In reply to Disconcerting to have p-values, fit value reported incorrectly by tbates
not "incorrectly"
I don't agree with your use of the word "incorrectly". The p = 0 being discussed in this thread is being reported computationally correctly. Mathematically, the p-value is nonzero, but that value is
too small to be represented as a double-precision floating-point value. For that matter--Regarding issue #131, p was also being reported computationally correctly. But there, we decided to sacrifice
computational correctness for user experience in an edge case. | {"url":"https://openmx.ssri.psu.edu/comment/8364","timestamp":"2024-11-08T08:03:02Z","content_type":"text/html","content_length":"144062","record_id":"<urn:uuid:bcda1dc6-6581-4360-b120-c3427af65cac>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00360.warc.gz"} |
s - James McVittie
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What is implied by the word "weird", is it just something unexpected that comes up or something that hasn't been discussed in the course? Thanks!
In Lecture Notes 7 - 1D Wave Equation: IBVP, there was code below equation (10) and (15) that didn't get turned into text. It was $Oct$ and (\ref{eq-4}), what do these stand for? Thank you!
(With all appropriate boundary conditions defined) When the wave equation is defined for x>0, we can find a solution to the wave equation for x < ct and 0 < x < ct. If the equation is defined for all
x, how do the solutions change?
For Part (c) of Problem 4, does it refer to the solution of (5) or (6) or both? | {"url":"https://forum.math.toronto.edu/index.php?PHPSESSID=btr89b3i91s4kcve19i8suggu4&action=profile;area=showposts;sa=topics;u=12","timestamp":"2024-11-04T07:46:05Z","content_type":"application/xhtml+xml","content_length":"18420","record_id":"<urn:uuid:20fd2c55-2291-4681-8814-b2e1a51cf36f>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00686.warc.gz"} |
Volume 58, pp. 402-431, 2023.
Fast computation of Sep$_\lambda$ via interpolation-based globality certificates
Tim Mitchell
Given two square matrices $A$ and $B$, we propose a new approach for computing the smallest value $\varepsilon \geq 0$ such that $A+E$ and $A+F$ share an eigenvalue, where $\|E\|=\|F\|=\varepsilon$.
In 2006, Gu and Overton proposed the first algorithm for computing this quantity, called $\mathrm{sep}_\lambda(A,B)$ (“sep-lambda”), using ideas inspired from an earlier algorithm of Gu for computing
the distance to uncontrollability. However, the algorithm of Gu and Overton is extremely expensive, which limits it to the tiniest of problems, and until now, no other algorithms have been known. Our
new algorithm can be orders of magnitude faster and can solve problems where $A$ and $B$ are of moderate size. Moreover, our method consists of many “embarrassingly parallel” computations, and so it
can be further accelerated on multi-core hardware. Finally, we also propose the first algorithm to compute an earlier version of sep-lambda where $\|E\| + \|F\|=\varepsilon$.
Full Text (PDF) [1.4 MB], BibTeX
Key words
sep-lambda, eigenvalue separation, eigenvalue perturbation, pseudospectra, Hamiltonian matrix
AMS subject classifications
15A18, 15A22, 15A42, 65F15, 65F30
Links to the cited ETNA articles
[2] Vol. 8 (1999), pp. 115-126 Peter Benner, Volker Mehrmann, and Hongguo Xu: A note on the numerical solution of complex Hamiltonian and skew-Hamiltonian eigenvalue problems
Additional resources for this document
< Back | {"url":"https://etna.ricam.oeaw.ac.at/volumes/2021-2030/vol58/abstract.php?vol=58&pages=402-431","timestamp":"2024-11-12T03:05:26Z","content_type":"application/xhtml+xml","content_length":"8185","record_id":"<urn:uuid:e8d2ecb8-e66d-4dcc-b577-ddbd7834d690>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00769.warc.gz"} |
Injection of neutral particles from negative ions - Consorzio RFX
Injection of neutral particles from negative ions
The injection of neutral particles from negative ions (Neutral Beam Injection) is one of the principle heating methods of the plasma fuel inside magnetic confinement fusion devices.
What is Neutral Beam Injection and how can we reach the super high temperatures needed for fusion reactions?
Let’s ask Vanno Toigo, engineer and principal CNR researcher at Consorzio RFX, Project Manager of the NBTF Project.
In fusion reactors, the ohmic effect due to the high currents circulating in the plasma is not sufficient to bring the gas to the temperatures necessary for fusion reactions to take place. It is
therefore necessary to use additional heating systems. Among them, the most important is the neutral particle injector.
“To give an idea, let’s think of a large hot-water tap capable of producing a fast stream of “hot” particles accelerated to the energy of 1 megaelectronvolt (MeV). This particle beam injected into
the plasma mixes with the “lukewarm” plasma particles, whose energy is about 100 times lower, initiating an enormous number of very intense collisions. In this way it transfers its kinetic heat
energy to the matter inside the reactor to reach the hottest temperatures ever measured in the known Universe”.
Additional heating systems in fusion reactors
How does Neutral Beam Injection from negative ions work?
To generate neutral particles at high energies, negative hydrogen or deuterium ions are used which are extracted from a “plasma source” and accelerated by means of electrostatic fields.
What is a plasma source?
The plasma source is a container into which the operational gas (hydrogen or deuterium) is injected, which, due to the effect of radiofrequency electric fields, is transformed into a low-energy
plasma from which negative ions are then extracted.
What are electrostatic field accelerators?
Electrostatic field accelerators consist of a system of perforated plates (grids) to which an electrical voltage is applied and located on the front side of the source. Through the holes, the
negatively charged ions are extracted and accelerated by the attraction caused by these very strong positive electric fields.
The accelerated charged particles are subsequently neutralized (i.e. have their negative charge removed) to allow the beam to pass freely through the magnetic field which, in fusion experiments,
confine the plasma inside the reactor.
What is the neutralizer?
The neutralizer is a compartment in which neutral gas of the same kind is present at low pressure. High-energy negative ions passing through this component collide with the gas molecules. Due to the
shocks, a portion of the ions lose their electrical charge and become neutral, without however losing their energy, and thus continue undisturbed until they penetrate the reactor plasma. The system
has an average efficiency of 50% which means that only 50% of the ions become neutral and can pass through to the plasma, while the rest remain ions and get deflected away by the magnetic fields.
The residual ions are subsequently eliminated from the beam by effect of transverse electric fields; they are deflected laterally until they intercept the side walls where they remain trapped. Given
the high associated energy, the walls are water-cooled. This system is called the “Residual Ion Dump“.
The last component in line with the beam is the calorimiter. It is made up of two walls of water cooled pipes which, with its articulation system, can assume the so-called “V” configuration and
intercept the neutral beam, or remain parallel to allow the beam exit undisturbed from the injector.
When the calorimeter is closed, it absorbes all the beam and allows for power measurements. When open, the injector is delivering its beam power.
In this case, the beam of neutral particles exiting the injector transfers its kinetic energy to the plasma particles inside the Tokamak, raising the temperature to the point where the fusion
reactions are able to start.
The NBI system for ITER
ITER is the first fusion reactor prototype and is now under construction in France as part of a worldwide collaboration between 7 partners: Europe, China, Korea, Japan, India, Russia and the United
States of America.
Injection of neutral particles from negative ions is a widely and long-used system for heating plasma in fusion experiments already in operation. However, the dimensions of ITER requires energy,
power and continuity at parameters never reached before.
ITER requires denser particle beams and much faster particles able to penetrate to the heart of the reactor and guarantee ITER the extreme temperature conditions, 150 million degrees, necessary to
ignite the fusion reactions.
Compared to the systems already developed, the scientific and technological leap is enormous.
The technological challenges to be overcome to achieve the performance levels required by ITER made it necessary to create the NBTF Project, a dedicated laboratory
for the development of the neutral injection system.
The NBTF plant was entrusted to Italy by the international scientific community
For this reason it was decided to build a research and development plant for the NBI system, called the Neutral Beam Test Facility – NBTF. The NBTF facility will offer scientists the opportunity to
investigate very challenging physics and engineering aspects and to validate the concepts behind the operation of the system, before it is installed on ITER.
On the left, SPIDER, the negative ion source prototype,
in the centre MITICA, the prototype injector
The design, construction and operation of the NBTF plant was entrusted to Italy, in collaboration with European, Japanese and Indian laboratories.
NBTF project agreements and partnerships
A first agreement, in force from 2012 to 2019, for the construction of the plant defined the partners: the ITER Organization (IO), Europe, through the European Agency Fusion for Energy (F4E) and
Italy, through Consorzio RFX.
During this phase, several European laboratories contributed: such as IPP-Garching, KIT-Karlsruhe, CCFE-Culham, CEA-Cadarache, Italian research bodies, through some of their institutes: CNR, ENEA,
INAIL, University of Padua and Milan Bicocca , and international laboratories: QST (Japan), IPR (India), NIFS (Japan).
During this first agreement, the components and scientific equipment of the plant were designed for which all the supply contracts were launched, and the construction of the plant was completed.
The buildings and infrastructures were financed by the Italian government for around €25 M, while the components and experimental plants were supplied by the three ITER partners: Europe, Japan and
India, through their Domestic agencies: F4E, JADA, INDA for approximately €250 M.
Since 2020 a new agreement is in place for the operation of the experiments, with a ten-year duration, between ITER and Consorzio RFX, for a total funding of approximately 150 million euros.
Europe has joined this agreement, this time through EUROfusion, making scientists available from various European laboratories. | {"url":"https://www.igi.cnr.it/en/our-research-activities-press-area/neutral-beam-facility/","timestamp":"2024-11-02T03:20:11Z","content_type":"text/html","content_length":"52959","record_id":"<urn:uuid:270613b7-b90a-4ec3-954b-313728a88265>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00570.warc.gz"} |
An incorrect information in the Wiki
Well, I'm not sure if this is the right place for this, but basically there is an incorrect information in the wiki. It's in the page "Tail Recursion and Tail Call Optimization" (
https://wiki.osdev.org/Tail_Recursion_a ... timization
In this page, it's written that a tail-call is when a call occurs in the end of a function. This is incorrect, a tail-call is when, in assembly, a CALL occurs before RET.
Code: Select all
unsigned long long Factorial(unsigned start) {
if (start > 1) {
return Factorial(start - 1) * start;
} // if
return 1;
} // Factorial(start)
Now, according to the page, there is no tail-call here (which is correct, but for a very different reason. I'll get back to this)
Now the page also writes:
Code: Select all
unsigned long long Factorial(unsigned start) {
if (start <= 1) {
return 1;
} // if
return start * Factorial(start - 1);
} // Factorial(start)
The page claims there is a tail-call here, which is incorrect, there is none. As I said a tail-call is when a CALL occurs before RET. I'll just use pseudo-assembly for simplicity (since CALL, RET and
JMP are more or less the same in most architectures it doesn't matter much):
Code: Select all
CMP start,1
JMP.HI _endofblock
MOV return_value, 1
SUB tmp, start, 1
PUSH tmp
CALL Factorial
MUL return_value, return_value, start
As you can see there is a MUL between CALL and RET. As a result, there is no tail-call. The same logic also applies to the above code, as that code will also result in a MUL between CALL and RET.
Now an example of a tail-call would include:
Code: Select all
function foo(data) {
return b(data);
In assembly, this is:
Code: Select all
PUSH data
CALL a
PUSH data #assuming this is needed
CALL b
Now this can be tail-call optimized, by combining the CALL and RET to simply JMP.
Code: Select all
PUSH data
CALL a
PUSH data #assuming this is needed
JMP b
The new code will behave in the exact same way, with a minor and desirable difference, it won't push IP/PC to the stack, which is a great thing if the said tail-call is also a recursive call (though
it isn't in this case)
Contrary to what the said page claims, a tail call doesn't need to be at the tail of the code:
Code: Select all
function bar(data) {
if ( a(data) ) {
return b(data);
return c(data);
In assembly:
Code: Select all
PUSH data
CALL a
TEST return_value
JMP.FALSE _endofblock
PUSH data
CALL b
PUSH data
CALL c
Here, both b and c are tail-calls even though only c is at the tail, because both result in a case where CALL and RET follow each other, and can be optimized to:
Code: Select all
PUSH data
CALL a
TEST return_value
JMP.FALSE _endofblock
PUSH data
JMP b
PUSH data
JMP c
Another example, where the call isn't at the tail, but nonetheless is a tail-call:
Code: Select all
function foo()
int myInteger = bar();
return myInteger;
In assembly:
Code: Select all
CALL bar
MOV myInteger, return_value
MOV return_value, myInteger
We can first optimize the unnecessary MOVs, getting rid of myInteger:
And then simply apply tail-call optimization:
On the contrary, even if the statement is at the tail, it may not be a tail-call, and the code in the said page is a great example for that:
Code: Select all
unsigned long long Factorial(unsigned start) {
if (start <= 1) {
return 1;
} // if
return start * Factorial(start - 1);
} // Factorial(start)
Code: Select all
CMP start,1
JMP.HI _endofblock
MOV return_value, 1
SUB tmp, start, 1
PUSH tmp
CALL Factorial
MUL return_value, return_value, start
As you can see, this code results in a MUL between CALL and RET, therefore there is no tail-call, even though the call is actually on the tail of the function.
Now of course, this also depends on the exact architecture, but as I said, the way the instructions CALL, RET and JMP work are usually the same, so most of the above examples with pseudo-assembly
would still work on most real architectures.
If you, the person reading this, are a moderator or anyone else with the privilege to modify the wiki, I ask of you to correct this page, please. Thanks in advance.
Last edited by Clover5411 on Sat Jul 27, 2019 11:22 am, edited 1 time in total.
Re: An incorrect information in the Wiki
Actually everybody can edit the wiki, so also you could do it if you want to. To get edit rights, you just need to go to the User Control Panel here in the forum, then to the tab User Groups, and
join the wiki group.
Of course, you're right - nicely spotted!
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Re: An incorrect information in the Wiki
Oh, didn't know everyone could do it. Thanks.
Re: An incorrect information in the Wiki
I wasn't even aware that there was a page on this here. According to the history, it was posted in January of this year, and I am guessing no one except the OP (Johnburger) had noticed it until now.
FlandreScarlet: You're right, the example as given is not a tail call at all. The OP seems to think that the position of the call on the line of source code is what is relevant, which is not at all
the case - it has to do with it being the last action in the generated code before the function exits and returns. The real relevant factor is whether the activation record (or stack frame, or local
environment - take your pick, those terms all amount to almost the same thing) on the call stack can be reused without losing any necessary information, and in this case, the answer is no.
There is a related optimization which is referred to as a 'tail recursion modulo cons' which could be applied here, but it is significantly more complex to for the compiler writers to implement.
While a discussion of it might be appropriate, as it is the example is entirely incorrect.
The funny thing is, in Scheme textbooks (which I am guessing where Johnburger got this, as it seems like a garbled version of a common example), the reverse is usually given, replacing this linear
Code: Select all
; yes, I am aware that factorial is only defined for positive integers,
; but I wanted to keep it simple
(define (factorial-1 n)
(if (<= n 1)
(* n (factorial-1 (- n 1)))))
with this 'linear iteration' (i.e., a recursion suitable for TCO, which in Scheme terms is considered iteration because it gets optimized into one):
Code: Select all
; note that the 'named let' basically creates a special-purpose internal function;
; it's that function 'fact' which is recursing in this case.
(define (factorial-2 n)
(let fact ((product 1)
(counter 1))
(if (> counter n)
(fact (* product counter) (+ 1 counter)))))
However, this approach to iteration isn't necessary, or even particularly applicable, in languages which have built-in iterative operators such as while() and for() (or even a standard iteration
macro, such as Common Lisp's (loop)). In fact, even Scheme has one, (do), though it's rather odd:
Code: Select all
(define (factorial-3 n)
(do ((counter 1 (+ 1 counter))
(product 1 (* product counter)))
((> counter n) product)
;; note that there is no loop body in this case,
;; as the iteration clauses do all the heavy lifting
In any case, actual idiomatic Scheme would really be do this (she said, tongue firmly in cheek) with an accumulator applied to a stream (a lazy list). Assuming one got that far in learning the
language, that is. Note that standardization of SRFI library names between implementations is... problematic.
Code: Select all
; at this point I have probably lost everyone anyway, so a detailed explanation is probably fruitless.
; I will mention that I've tested this in Guile, so at least one implementation can use it...
(import (srfi :41))
(define (factorial-4 n)
(if (<= n 0)
(let ((range (stream-range 1 (+ 1 n))))
(lambda (x y)
(* x y))
(stream-car range) (stream-cdr range)))))
Enough of that, I think.
Rev. First Speaker Schol-R-LEA;2 LCF ELF JAM POEE KoR KCO PPWMTF
Ordo OS Project
Lisp programmers tend to seem very odd to outsiders, just like anyone else who has had a religious experience they can't quite explain to others.
Re: An incorrect information in the Wiki
Interesting. Thank you for putting this information here.
Although factorial isn't strictly defined only for positive integers.
Usually, it is also defined for zero where:
0! = 1
There is also more generalized definition with the gamma function, which can also be applied to non-integers.
So it depends on the exact definition for factorial you are using.
Not really an important matter, but nerd instincts kicked in.
Re: An incorrect information in the Wiki
FlandreScarlet wrote:Interesting. Thank you for putting this information here.
Although factorial isn't strictly defined only for positive integers.
Usually, it is also defined for zero where:
0! = 1
There is also more generalized definition with the gamma function, which can also be applied to non-integers.
So it depends on the exact definition for factorial you are using.
Not really an important matter, but nerd instincts kicked in.
The really interesting thing is that if you use the definition factorial=gamma(x+1), the only complex numbers for which factorial is not defined are real integers (specifically, the negative
So strictly speaking, factorial is defined for other numbers than the natural numbers over any domain that is a superset of the real integers, but over the real integers is defined only for the
Re: An incorrect information in the Wiki
Okay, so this took me way longer than it should have. The reason being a mixture of laziness and fearing my own ineptitude. That being said I finally had both the motivation and confidence to put up
a rudimentary explanation of tail call optimization. Unfortunately, I couldn't get into details about prologues and epilogues, and how this would effect optimization. I do have some idea on how it
could work, but I don't want to add misinformation.
I suppose I should also ask, can I even delete stuff from the main page? I don't know how to do that, so I took what felt like the most reasonable action at the time; but in retrospect, it was a bit
childish. (EDIT: Nevermind, it was surprisingly simple... Truth be told, temptation to delete that page and this topic is strong...)
Re: An incorrect information in the Wiki
KineticManiac wrote:Okay, so this took me way longer than it should have. The reason being a mixture of laziness and fearing my own ineptitude.
I haven't had the confidence to change much on the wiki myself either. There's even this "if someone writes something and I make an edit to it, how will the person that initially wrote the text feel?
" feeling that I probably shouldn't have that much of but I do. I feel like I know of several mistakes in the interrupts tutorial, but even though I see them as mistakes (or in one case, just doing a
thing without properly explaining why you're doing it that way) I don't really feel like I have the confidence to change what someone else wrote.
Re: An incorrect information in the Wiki
Do, do, do edit! That's the lifeblood of a Wiki.
Over the years, many people who have written a lot in the Wiki drop out of the hobby, partially or completely. If you hold whatever they wrote sacred, sooner or later the Wiki will wither and die.
Those who wrote into the Wiki before you might have been before you, but that doesn't make them Elders. A Wiki is a cooperation thing.
In the beginning, it was one static HTML page with a couple of good hints. It grew to what it is today only because we got people to contribute. Please, continue in that vein.
By posting to the Wiki, every author has agreed to have his / her work modified later on. The Wiki holds a history, and if your edit is considered faulty, there are discussion pages as well as the
ability to revert changes, wholly or partially. You cannot "destroy" information in the Wiki.
So... edit. Please.
Every good solution is obvious once you've found it.
Re: An incorrect information in the Wiki
Solar wrote: By posting to the Wiki, every author has agreed to have his / her work modified later on. The Wiki holds a history, and if your edit is considered faulty, there are discussion pages
as well as the ability to revert changes, wholly or partially. You cannot "destroy" information in the Wiki.
This is also a very good reason to use citations profusely, so that information can be cross-referenced and validated, and if any source is incorrect, the corrections can be escalated upstream in
order to reduce misinformation.
But yes, wikis should be ever evolving with new information, new knowledge, corrections, adaptations to more contemporary situations and so forth.
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Veranstalter: Almudena Arcones, Jens Braun, Michael Buballa, Hans-Werner Hammer, Kai Hebeler, Gabriel Martínez-Pinedo, Daniel Mohler, Guy Moore, Robert Roth, Achim Schwenk, Jochen Wambach
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Ort: S2|11, Raum 10
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gravitational mass loss of approximately 0.01 solar masses suffices to produce an eccentricity of the order of 0.1 without the need of a secondary kick mechanism. We also show that in warm
supernova and merger matter, thermal twin stars can be formed, even when the mass-radius diagram of cold neutron stars has no twins. We speculate about a correlation of the thermal twin
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13:30 rejected in a Metropolis step using the entropy as a statistical weight. This procedure reproduces the expected strains of dissipative relativistic hydrodynamics in a specific (and
S2|11 10 non-covariant) hydrodynamic frame known as the density frame. Numerical results, both with and without noise, are presented and compared to relativistic kinetics and analytical
expectations. An all order resummation of the density frame gradient expansion reproduces the covariant dynamics in a specific model. In contrast to all other numerical approaches to
relativistic dissipative fluids, the dissipative fluid formalism presented here is strictly first order in gradients and has no non-hydrodynamic modes. We will also present the extension
to relativistic viscous hydrodynamics and its comparison with BDNK formalism.
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04.07.2024 Parametric Matrix Models
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S2|11 10 however, missing.
Additionally, the ab initio many-body methods developed over the past two decades encounter limitations when it comes to dealing with excited-state properties. In this perspective, I will
present a systematic ab initio predictions of (giant) monopole resonances. Ab initio Quasiparticle-RPA (QRPA) and Projected Generator Coordinate Method (PGCM) calculations of monopole
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more specifically talk about the impact of the proto neutron star's early cooling, with an accent on a cooling process through heavy-lepton neutrinos and potentially axions, on the final
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02.05.2024 Unnuclear physics at large charge
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will discuss the potential relevance of the large-charge formalism to the study of special nuclear reactions with many low-energy neutrons in the final state.
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30.11.2023 billion years of evolution of a galaxy and also to keep track of the elements synthesized and ejected from every astrophysical site of interest. In this talk, I will give a general
13:30 overview of the Chemical Evolution of Galaxies field describing both its aims and explaining which are the basic ingredients necessary to build a Chemical Evolution simulation. I will then
S2|11 10 focus on the study of the evolution of heavy elements abundances. The majority of elements beyond the Fe peak are produced by neutron capture processes which can be rapid (r-process) or
slow (s-process) with respect to the beta-decay in nuclei. Understanding which are the astrophysical formation sites of these two processes has become one of the major challenges in
chemical evolution. In this talk, I will first present the main steps done in chemical evolution simulations to understand the origin of neutron capture elements and then I will show
results from our latest work.
Dr. Johannes Weber (Humboldt-Universität Berlin)
Hard Probes of Hot Nuclear Matter
23.11.2023 The hot nuclear medium that permeated the early universe can be studied experimentally with heavy-ion collisions and through various theoretical approaches. New or upgraded experiments
13:30 turn our attention to hard processes and a more fine-grained resolution of this primordial state of matter. In this endeavor quarkonia, open heavy flavors, and jets turn out to be
S2|11 10 versatile probes, which are usually described through models based on resummed perturbative QCD, AdS, and effective field theories. The lattice provides nonperturbative input and
constraints to such models. In-medium bottomonia, the complex static quark-antiquark potential, as well as the heavy-quark momentum and the jet transverse momentum diffusion transport
coefficients are key quantities where lattice gauge theory has recently achieved significant progress with major impact for heavy-ion phenomenology. I review these lattice results, relate
them to phenomenological applications, and close with an outlook towards expectations for the next few years.
Dr. Theo F. Motta (TU Darmstadt and JLU Giessen)
A Stability Analysis of Inhomogeneous Phases in QCD
16.11.2023 Understanding the phase structure of Quantum Chromodynamics (QCD) is of paramount importance for nuclear and particle physics. At large densities and low temperatures, many complex phases
13:30 are expected to appear. This is where the lattice sign problem is unavoidable and extrapolation methods such as Taylor expansions are out-of-bounds. Alongside colour-superconductivity,
S2|11 10 quarkyonic matter, and so on, the possibility of a crystalline phase has been studied for over twenty years. In simplified models of QCD such as NJL or quark-meson models, these phases are
present. However, no unambiguous determination exists that they appear in QCD. In this talk, I will discuss our efforts to develop a method of stability analysis that is compatible with
full QCD via Dyson-Schwinger Equations.
Martin Obergaulinger (Valencia University)
Core-collapse supernovae with magnetic jets
09.11.2023 Magnetic fields are a common feature of both the progenitors of core-collapse supernovae, i.e., massive stars, and of their remnants, neutron stars. If combined with rapid rotation, they
13:30 can affect the explosion dynamics and eject a part of the gas in the form of jets along the rotational axis. Besides the high explosion energies, these events also differ from the majority
S2|11 10 of neutrino-driven supernovae by their nucleosynthetic yields and observables like the gravitational-wave signal. I will present recent simulations of a set of three-dimensional
simulations combining magnetohydrodynamics and neutrino transport in which explosions with different degree of magnetic influence occur and highlight some of the key processes that
determine the outcome.
Dr. Aman Abhishek (Institute of Mathematical Sciences, Chennai, India)
Towards a universal description of hadronic phase of QCD
Mean-field model quantum field theories of hadrons were traditionally developed to describe cold and dense nuclear matter and are by now very well constrained from the recent neutron star
10.08.2023 merger observations. We show that when augmented with additional known hadrons and resonances but not included earlier, these mean-field models can be extended beyond its regime of
14:00 applicability. Calculating some specific ratios of baryon number susceptibilities for finite temperature and moderate values of baryon densities within mean-field approximation, we show
S2|11 10 that these match consistently with the lattice QCD data available at lower densities, unlike the results obtained from a non-interacting hadron resonance gas model. We also estimate the
curvature of the line of constant energy density, fixed at its corresponding value at the chiral crossover transition in QCD, in the temperature-density plane. The number density at low
temperatures and high density is found to be about twice the nuclear saturation density along the line of constant energy density of 348 +- 41 MeV/fm^3. Moreover from this line we can
indirectly constrain the critical end-point of QCD to be beyond 597 MeV mu_B for temperature 125 MeV.
Prof. Dr. Mark Alford (Washington University, St. Louis)
28.06.2023 Is nuclear matter in neutron star mergers driven out of equilibrium?
14:00 In a neutron star merger, nuclear matter experiences dramatic changes in temperature and density that happen in milliseconds. Mergers therefore probe dynamical properties that may help us
S2|11 10 uncover the phase structure of ultra-dense matter. I will describe some of the relevant material properties, focusing on chemical (beta) equilibration and its consequences such as bulk
viscosity and damping of oscillations.
Prof. Dr. Dam T. Son (University of Chicago)
22.06.2023 Nonrelativistic conformal field theory and nuclear reactions
14:00 We develop a formalism of nonrelativistic conformal field theory, which is then used to describe neutrons at low energies. We show that the rates of nuclear reactions with emission of a
S2|11 10 few neutrons in the final state show a power-law behavior in the kinematic region where the emitted neutrons have almost the same momentum. We show how corrections to this power-law
behavior can be computed using conformal perturbation theory.
Assistant Prof. Dr. Luka Leskovec (Universität Ljubljana)
15.06.2023 Baryonic Resonances in Lattice QCD
14:00 In recent decades, lattice QCD has been crucial in understanding the Standard Model. As a new era of experiments centered around nuclear physics begins, we also focus on baryons and their
S2|11 10 phenomena. In this talk, I will motivate some reasons for a deeper understanding of baryons from first principles QCD. After a brief overview of the formalism and the intricacies involved
with baryons, I will present our lattice QCD calculation of the lightest baryonic resonance, the (1232) – mass and decay width.
PD Dr. Sara Collins (Universität Regensburg)
Properties of the baryon octet from lattice QCD
25.05.2023 Numerous ongoing experimental investigations into physics beyond the Standard Model focus on nucleons as fundamental probes, prompting extensive efforts to extract nucleon structure
14:00 observables on the lattice. This includes the determination of quantities like the weak charges and axial form factors. However, it is also interesting to extend such studies to hyperons,
S2|11 10 which have received less attention. Exploring the properties of hyperons provides valuable insights into the SU(3) flavor symmetry encoded in low-energy effective field theory
descriptions. Moreover, studying their weak decays offers alternative means of determining the elements of the CKM matrix. As a first step we determine the spectrum, weak charges and sigma
terms of the baryon octet controlling all sources of systematic uncertainty.
Frederic Noël (Uni Bern)
Mu -> e conversion, LFV pseudoscalar decays, and nuclear charge densities
24.04.2023 Mu -> e conversion in nuclei gives one of the leading limits on BSM lepton-flavor violating (LFV) processes. In this process a muon bound to a nucleus converts into an electron without
15:00 neutrinos. Upcoming measurements call for a more consistent theoretical description of mu -> e conversion, which can be done model independently using an effective field theory frame work
S2|11 207 in terms of effective BSM operators. As it turns out, the relevant operators for the spin dependent part of mu -> e conversion also mediates LFV pseudoscalar decays, which makes it
possible to relate these processes and their experimental limits. Furthermore for the treatment of the bound state physics appearing in mu -> e conversion, quantifiable knowledge on the
charge densities of the considered nuclei is needed. In this talk I will give an overview over our ongoing work on mu -> e conversion, including some recent results regarding LFV
pseudoscalar decays, as well as the determination of nuclear charge densities from electron scattering.
Dr. Saga Aurora Säppi (TU München)
Exploring extremely dense matter with perturbative QCD
20.04.2023 With LIGO and its friends observing colliding neutron stars, and astrophysicists measuring their radii and masses with unprecedented precision, understanding how dense QCD matter behaves
14:00 is a particularly timely goal. I will approach this from the side of (very-)high-density theory: How do first-principles calculations in dense QCD in the small-coupling limit work, and how
S|11 10 have they advanced in the last few years? Some of the advancements I will discuss in this talk include an efficient and simple way to incorporate the effects of quark masses in
perturbative calculations, particularly useful for near-future calculations of the bulk viscosity, as well as an ongoing computation of the next-to-next-to-next-to-leading order pressure
of cold dense QCD, where there are both concrete recent (and upcoming) results for the self-energy as well as improved theoretical methods necessary for finishing the full computation.
Dr. Zhonghao Sun (Oak Ridge National Laboratory)
26.01.2023 Ab-initio computation of exotic nuclei
14:00 Precise and predictive calculations of the atomic nuclei from realistic nuclear force help us to understand how the fundamental interaction leads to the emergence of various exotic
via Zoom phenomena. The advances in computational power, emerging machine learning technology, and the development of many-body methods make it possible to perform uncertainty quantification
and sensitivity analyses in the nuclear structure calculations. In this talk, I will report the progress of the ab-initio coupled-cluster method in describing spherical and deformed
atomic nuclei. I will also introduce the quantified predictions of the neutron skin thickness of 208Pb and the drip line of oxygen isotopes.
Dr. Agnieszka Sorensen (INT Seattle)
The speed of sound of dense nuclear matter from heavy-ion collisions
The equation of state (EOS) of dense nuclear matter has been the center of numerous research efforts over the years. While numerous studies indicate that the EOS is relatively soft
15.12.2022 around the saturation density of nuclear matter, recent analyses of neutron star data strongly suggest that in the cores of neutron stars, where densities may reach several times
15:00 that of normal nuclear matter, the EOS becomes very stiff – so stiff, in fact, that the speed of sound squared may substantially exceed the conformal limit of 1/3. This striking
via Zoom behavior inspires the research
I will present in this talk. I will discuss a novel way of using higher moments of the baryon number distribution, measured in experiments, to infer the speed of sound in dense
nuclear matter created in low-energy heavy-ion collisions. I will then present the framework I developed to enable comprehensive hadronic transport studies of the influence of the
dense nuclear matter EOS on experimental observables, and I will discuss implications for the speed of sound of dense nuclear matter based on a recent analysis using this framework.
Prof. Dr. Owe Philipsen (Goethe Universität Frankfurt)
Chiral spin symmetry and the QCD phase diagram
24.11.2022 Recently, an emerging chiral spin symmetry was discovered in the multiplets of lattice QCD hadron correlators for a temperature window above the chiral crossover. This symmetry is
14:00 larger than the expected chiral symmetry. It can only be approximately and dynamically realised when colour-electric quark-gluon interactions dominate the quantum effective action.
S2|11 10 This suggests the chiral spin symmetric regime to be of a hadron-like rather than partonic nature. After a brief review of the symmetry, I show independent evidence from meson
screening masses and the pion spectral function, which support this picture. Finally, I discuss how this chiral spin symmetric band may continue across the QCD phase diagram, where
it may smoothly connect to quarkyonic matter at low temperatures and high densities.
Dr. Marcel Schmidt (d-fine)
17.11.2022 How to save the financial system: My journey from Physics to Risk Management @ d-fine
14:00 For over 3 years I have worked at d-fine, a leading consultancy for analytically demanding topics from branches like finance, energy industry or manufacturing. Early in my career, I
S2|08 171 have specialized in market risk management. In our projects, we help banks to secure against price fluctuations, using state-of-the-art methods from mathematics, machine learning,
(Uhrturmhörsaal) and modern software development.
In this talk, I would like to provide an impression on my career path and show how we as physicists contribute to a more secure financial system.
Prof. Dr. Frithjof Karsch (Universität Bielefeld)
QCD Phase Diagram and the Equation of State of Strong-Interaction Matter
10.11.2022 Lattice QCD calculations at non-zero temperature and with non-vanishinq chemical potentials provide a powerful framework for the analysis of the phase structure of strongly
14:00 interacting matter. Such calculations allow the determination of the crossover transition region at QCD with physical quark masses as well as the determination of the true chiral
S2|11 10 phase transition in the limit of vanishing light quark masses.
We present results on the determination of the pseudo-critical and chiral phase transition temperatures as well as a new, high statistics determination of the QCD equation of state.
We point out their importance for constraining the location of a possible critical end point in the QCD phase diagram. We furthermore present a new, high statistics determination of
the QCD equation of state.
Dr. Renwick James Hudspith (GSI)
27.10.2022 A complete lattice QCD determination of the hadronic light-bylight scattering contribution to the muon g-2
14:00 The g-2 of the muon provides a high-precision test of the Standard Model of particle physics, and a possible window into beyond the Standard Model physics. Currently, there is some
S2|11 10 tension between the theoretical prediction of this quantity and experiment. As experimental precision continues to improve it is paramount for theoretical computations to do so also,
in hope of resolving this tension. One of the most poorly-known contributions to the theory calculation of the muon g-2 comes from hadronic light-bylight scattering. I will present
an overview of our measurement of this contribution using lattice QCD techniques, where we have obtained the most precise determination to date.
Prof. Dr. Baha Balantekin (University of Wisconsin)
Collective Neutrino Oscillations and Quantum Entanglement
23.09.2022 Entanglement of constituents of a many-body system is a recurrent feature of quantum
14:00 behavior. Quantum information science provides tools, such as the entanglement entropy, to
S2|11 10 & help assess the amount of entanglement in such systems. Many-neutrino systems are present in
Zoom core-collapse supernovae, neutron star mergers, and the Early Universe. Recent work in
applying the tools of quantum information science to the description of the entanglement in
astrophysical many-neutrino systems is presented, in particular the connection between
entropy and spectral splits in collective neutrino oscillations is elaborated.
Prof. Dr. Derek Teaney (Stony Brook)
14.07.2022 Dynamics of the O(4) critical point in QCD
14:00 To motivate the simulations I review Lattice data on the chiral phase transition in QCD. Then I discuss the hydrodynamics of the chiral phase transition, reviewing the appropriate
S2|11 10 & dynamical equations above, below, and during the phase transition. Then I present a simulation of the dynamics of the phase transition, which shows how goldstone modes appear dynamically.
Zoom Finally I discuss soft pions in heavy ion collisions, which are enhanced relative to normal hydrodynamic simulations of heavy ion collisions. I suggest that this reflects the fingerprints
of the O(4) critical point.
Prof. Dr. Lorenz von Smekal (Justus-Liebig-Universität Gießen)
07.07.2022 Real-time methods for spectral functions
14:00 The real-time methods discussed and compared in this talk include classical-statistical lattice simulations, the Gaussian state approximation (GSA), and the functional renormalization
S2|11 10 & group (FRG) formulated on the Keldysh closed-time path. The quartic anharmonic oscillator coupled to an external heat bath after Caldeira and Leggett thereby serves as an illustrative
Zoom example where a benchmark solution can be obtained from exact diagonalization with constant Ohmic damping. To extend the GSA to open systems, we solve the corresponding Heisenberg-Langevin
equations in the Gaussian approximation. For the real-time FRG, we introduce a novel general prescription to construct causal regulators based on introducing scale-dependent fictitious
heat baths. As first field theory applications we have used our real-time FRG framework to calculate dynamical critical exponents for different dynamics.
04.07.2022 Dr. Robert Pisarski (Brookhaven National Laboratory)
15:00 A potpourri in extreme QCD
S2|11 10 & I discuss some combination of topics in SU(N) gauge theories at nonzero temperature and density, including: the exact solution of the low energy excitations for cold, dense quarks in 1+1
Zoom dimensions (you'll learn what a Luttinger liquid is); how to represent timelike Wilson loops in Hamiltonian form (bit obvious after the fact); configurations with topological charge 1/N in
SU(N) gauge theories without dynamical quarks
Carolyn Raithel (Princeton Center for Theoretical Science)
09.06.2022 Probing the Dense-Matter Equation of State with Neutron Star Mergers
14:00 Binary neutron star mergers provide a unique probe of the dense-matter equation of state (EOS) across a wide range of parameter space, from the cold EOS during the inspiral to the
S2|11 10 & finite-temperature EOS following the merger. In this talk, I will discuss the influence of finite-temperature effects on the post-merger evolution of a neutron star coalescence. I will
Zoom present a new set of neutron star merger simulations, which use a phenomenological framework for calculating the EOS at arbitrary temperatures and compositions. I will show how varying the
properties of the particle effective mass affects the thermal profile of the post-merger remnant and how this, in turn, and influences the post-merger evolution. Finally, I will discuss
several ways in which a future measurement of the post-merger gravitational waves can be used to constrain the dense-matter EOS.
Dr. Joanna Sobcyk (Uni Mainz)
02.06.2022 Nuclear ab initio studies for neutrino oscillations
14:00 We are entering an era of high-precision neutrino oscillation experiments (T2HK, DUNE), which potentially hold answers to some of the most exciting questions in particle physics. Their
S2|11 10 & scientific program requires a precise knowledge of neutrino-nucleus interactions coming from fundamental nuclear studies. Ab initio many-body theory has made great advances in the last
Zoom years and is able to give relevant predictions for medium-mass nuclei important for the neutrino experiments. In my talk I will give an overview of the recent progress that has been made
in describing neutrino-nucleus scattering within the ab-initio coupled-cluster framework, combined with the Lorentz integral transform. These techniques open the door to obtaining nuclear
responses (and consequently cross-sections) for medium-mass nuclei starting from first principles.
Dr. Aleksas Mazeliauskas (CERN)
Many-body QCD phenomena in high-energy proton and nuclear collisions
31.05.2022 The emergence of macroscopic medium properties over distances much smaller than a single atom is a fascinating and non-trivial manifestation of the many-body physics of Quantum
16:00 Chromodynamics in high-energy nuclear collisions. The observation of collective particle behaviour in collisions of heavy-ions at the Relativistic Heavy Ion Collider at BNL and the Large
S2|11 10 & Hadron Collider at CERN is strong evidence that a new exotic phase of matter called the Quark-Gluon Plasma is created in these large collision systems. However, the striking discovery of
Zoom the very same collective phenomena in much smaller systems of proton-proton and proton-lead collisions at the LHC has confounded heavy-ion physics expectations and is not predicted by the
conventional high-energy physics picture of elementary collisions. One of my main research goals is to uncover the physical origins of this universal macroscopic behaviour. In this talk I
will review the recent progress and future plans in developing theoretical description and experimental tests of these effects within the non-Abelian quantum field theory of strong
12.05.2022 Prof. Dr. Thomas Schäfer (NC State University)
14:00 Stochastic fluid dynamics: Effective actions and new numerical tools
S2|11 10 & Recent interest in stochastic fluid dynamics is motivated by the search for a critical point in the QCD phase diagram. I will discuss old ideas about effective actions that have recently
Zoom received new interest, and some new ideas about how to implement stochastic fluid dynamics in numerical simulations.
Lotta Jokiniemi (University of Barcelona)
What Can We Learn from Double-Beta Decay and Ordinary Muon Capture?
Observing neutrinoless double-beta (0vbb) would undoubtedly be one of the most anticipated breakthroughs in modern-day neutrino and nuclear physics. This is highlighted by the number of
massive experiments worldwide trying to detect the phenomenom, as well as the efforts of numerous theory groups trying to probe the process from different theory frameworks. When observed,
10.02.2022 the lepton-number-violating process would provide unique vistas beyond the Standard model of particle physics. However, the half-life of the process depends on coupling constants whose
14:00 effective values are under debate, and nuclear matrix elements (NMEs) that have to be extracted from theory. Unfortunately, at present different many-body calculations probe matrix
Zoom elements whose values disagree by more than a factor of two. Hence, it is crucial to gain a better understanding on both the coupling constants and the NMEs in order to plan future
experiments and to extract the beyond-standard-model physics from the experiments.
In my seminar I will discuss how the theory predictions can be improved either directly by investigating corrections to the 0vbb decay matrix elements, or indirectly by studying
alternative processes that can be or have been measured. First, I will introduce our recent work on a new leading-order correction to the standard 0vbb-decay matrix elements in heavy
nuclei. Then, I will discuss the potential of ordinary muon capture as a probe of 0vbb decay, and discuss the results of our recent muon-capture studies.
Laura Sagunski (Goethe Universität Frankfurt)
Gravitational Waves from the Dark Side of the Universe
03.02.2022 The first ever direct detections of gravitational waves from merging black holes and neutron stars by the Laser Interferometer Gravitational-Wave Observatory (LIGO) and the Virgo detector
14:00 have opened a fundamentally new window into the Universe. Gravitational waves from binary mergers are high precision tests of orbital dynamics and provide an unprecedented tool to probe
Zoom fundamental physics. Not only do they allow to test gravity under extreme conditions, but also to address the very fundamental open questions in the evolution of our Universe, namely the
mysteries of dark matter and dark energy (or possible modifications of general relativity). In my talk, I will show how we can turn binary mergers into cosmic labs where we can test the
very foundations of general relativity and explore the existence of new interactions and particles, like axions, which could be the dark matter.
Felipe Attanasio (Uni Heidelberg)
QCD equation of state via the complex Langevin method
The equation of state of hadronic matter is of high importance for
16.12.2021 many fields, ranging from heavy-ion collisions to neutron stars. Non-
14:00 perturbative methods to simulate QCD encounter difficulties at finite
Zoom chemical potential mu due to the so-called sign problem. We employ the complex Langevin method to circumvent this problem and carry out
simulations at a variety of values for temperature and mu. We present
results on the pressure, energy and entropy equations of state, as well
as a numerical observation of the Silver Blaze phenomenon.
Vittorio Soma (CEA Saclay)
A novel many-body method for the ab initio description of doubly open-shell nuclei
Recent developments in many-body theory and in the modelling of nuclear Hamiltonians have enabled the ab initio description of a considerable fraction of atomic nuclei up to mass A~100. In
09.12.2021 this context, one of the main challenges consists in devising a method that can tackle doubly open-shell systems and at the same time scales gently with mass number. This would allow both
14:00 to access all systems below A~100 and to open up perspectives for extending ab initio calculations to the whole nuclear chart.
Zoom In this seminar I will present a recently proposed many-body approach that aims towards this objective. After introducing the formalism based on a multi-reference perturbation theory [1],
I will discuss the first numerical applications [2,3] together with considerations on the state of the art and future perspective in ab initio nuclear structure.
[1] M. Frosini et al., arXiv:2110.15737
[2] M. Frosini et al., arXiv:2111.00797
[3] M. Frosini et al., arXiv:2111.01461
Nicolas Wink (TU Darmstadt)
Elementary correlation functions and their applications in QCD
25.11.2021 In this talk we explore the calculation of elementary correlation functions in the context of QCD. We consider these correlation functions in Euclidean and Minkowski space-time. For the
14:00 latter we consider direct calculations based on dimensional regularization in Dyson-Schwinger equations in a scalar theory and Yang-Mills. Additionally, we present results from analytic
Zoom continuation of Euclidean lattice data based on Gaussian Process Regression in full QCD. Afterwards we turn our attention to the calculation of transport coefficients in Yang-Mills, based
on gluon spectral functions obtained previously. In the last part of the talk we consider field dependencies in functional Renormalization Group equations. We focus in particular on
technical challenges at high densities and how to overcome them.
Andreas Ipp (Vienna University of Technology)
Simulating the Glasma stage in heavy ion collisions
18.11.2021 The earliest stage right after the collision of ultrarelativistic heavy ions is known as the Glasma stage. It is characterized by strong anisotropic color fields and forms the precursor of
14:00 the quark-gluon plasma. In this talk, I present our approach to simulating the Glasma using a colored particle-in-cell simulation. With this method, we can access the full 3+1 dimensional
Zoom space-time picture of the collision process. These simulations are inherently plagued by numerical Cherenkov instability, and we show how an improved action can cure this instability using
a semi-implicit scheme. Simulation results can be checked in a dilute limit against analytic calculations. I will present results for observables such as the rapidity profile or momentum
broadening of jets within the Glasma stage.
Weiguang Jiang (Chalmers)
28.10.2021 Exploring non-implausible nuclear-matter predictions with delta-full chiral interactions
14:00 Advances in quantum many-body methods and computing allow us to study the finite nuclei and infinite nuclear matter with realistic interaction models based on chiral effective field
Zoom theory. We develop the nuclear matter emulators and introduce the robust statistical approach called history matching to explore the non-implausible nuclear-matter predictions with chiral
interactions. We studied 1.6*10^6 non-implausible interaction samples in a huge LECs domain and reveal the connection between the finite nuclei and nuclear matter saturation properties. | {"url":"https://www.ikp.tu-darmstadt.de/aktuelles_kernphysik/veranstaltungen_sammlung/theorie_seminar/index.de.jsp","timestamp":"2024-11-14T14:30:53Z","content_type":"text/html","content_length":"112801","record_id":"<urn:uuid:a1f6b5db-2f8c-4736-8cca-914bbe691639>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00227.warc.gz"} |
Journal of the European Optical Society-Rapid Publications
Issue J. Eur. Opt. Society-Rapid Publ.
Volume 20, Number 1, 2024
Article Number 21
Number of page(s) 4
DOI https://doi.org/10.1051/jeos/2024019
Published online 24 May 2024
J. Eur. Opt. Society-Rapid Publ. 2024,
, 21
Short Communication
Uncertainty analysis of spectral flux measurement using Monte Carlo simulation
^1 TÜBİTAK UME, National Metrology Institute of Türkiye Barış Mah., Dr. Zeki Acar Cad., Gebze, Kocaeli 41470, Türkiye
^2 Gebze Technical University, 2254. Sok., No: 2, Gebze, Kocaeli 41400, Türkiye
^3 TechnoTeam Bildverarbeitung GmbH, Werner-von-Siemens-Straße 5, DE-98693 Ilmenau, Germany
^4 PTB, Bundesallee 100, 38116 Braunschweig, Germany
^* Corresponding author: senel.yaran@tubitak.gov.tr
Received: 5 September 2023
Accepted: 9 April 2024
In photometry, spectrally integrated quantities are commonly used, and their uncertainties are calculated using classical approaches. Since determining correlations at different wavelengths and
effects on spectrally integrated quantities is rather complex, a noise modification of the spectral value using Monte Carlo simulation was applied to estimate unknown correlations in the uncertainty
of total luminous flux based on integrated spectral luminous flux values. For this aim, an LED with 6500 K correlated colour temperature was measured in an integrated sphere flux measurement system.
The correlations between the measurements at different wavelengths were analysed, and the uncertainty boundaries of the integrated quantity and total luminous flux were obtained.
Key words: Uncertainty / Monte Carlo / Base functions / Integrated quantities / Correlations between quantities at different wavelengths
© The Author(s), published by EDP Sciences, 2024
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1 Introduction
Photometric parameters like total luminous flux, colour coordinates, photometer responsivity, spectral mismatch correction factor, etc., are attained by spectrally integrating quantities. The guide
to the expression of uncertainty in measurement (GUM) uncertainty framework or Monte Carlo (MC) simulations [1–3] are commonly used to assign uncertainties of the parameters at different wavelengths
in spectral measurements. However, fully uncorrelated contributions will average out while integrating the quantities over wavelength, and the fully correlated contributions can be assigned to the
integrated value without any change. The question is: what happens with partly correlated quantities?
The correlation between the spectral values of photometric parameters at various wavelengths possesses considerable intricacy. To overcome the challenges involved in determining the correlation
between values at different wavelengths, Kärhä, in 2017 [4], introduced an innovative MC based technique.
In this study, the novel MC-based method is utilised for analysing the uncertainty of the total luminous flux, one of the photometric integrated quantities.
2 Theoretical framework
The luminous flux is one of the main quantities of the light sources. This quantity value is determined using several methods and systems based on spatial and/or spectral measurements. The total
luminous flux can be determined depending on the measurement system and the integration of spatial and/or spectral measurement values.
As the luminous flux measurement, spectral measurements are generally preferred for more accurate results. To obtain the total luminous flux value, the spectral measurement results are integrated
over the visible spectrum (typically from approximately 380 nm to 780 nm).
The relation between the spectral measurements at different wavelength positions is found to be rather complex. The fully uncorrelated contributions will average out while integrating the quantities
over wavelength, and the fully correlated contributions can be assigned to the integrated value without any change. But the calculation of the partial correlations is very difficult.
Since there is a great difficulty in calculating the correlation between values at different wavelengths, a noise modification of the spectral value using the novel MC simulation was applied and used
for analysing and estimating the possible effects of partly correlated uncertainties in the measurement of spectral flux data to spectrally integrated total luminous flux quantity. A series of
orthogonal base functions were used in this MC-based method to simulate potential systematic deviations.
The uncertainty of spectral radiant flux at each wavelength, expressed in equation (1) and the uncertainty of the total luminous flux, expressed in equation (2) were calculated using classical
approaches GUM framework without any correlation contribution.
The uncertainty sources of the spectral radiant flux are given in Table 1.$ϕ eT ( λ ) = ϕ eR ( λ ) ∙ ( y T ( λ ) y R ( λ ) ) ∙ ( y AR ( λ ) y AT ( λ ) )$(1)
Table 1
Uncertainty budget of the radiant flux of the lamp at several wavelengths.
Φ[eT](λ): spectral radiant flux values of the measured lamp,
Φ[eR](λ): spectral radiant flux values of the reference lamp,
y[T](λ): spectral measured values when the measured lamp is on,
y[R](λ): spectral measured values when the reference lamp is on,
y[AT](λ): spectral measured values when the measured lamp is pleased in the integrating sphere and only the auxiliary lamp is on,
y[AR](λ): spectral measured values when the reference lamp is pleased in the integrating sphere and only the auxiliary lamp is on,$ϕ T = K m ∙ ∫ 380 nm 780 nm V ( λ ) ∙ ϕ eT ( λ ) ∙ d λ$(2)
Φ[T]: total luminous flux value of measured lamp,
Φ[eT](λ): spectral radiant flux values of the measured lamp,
V(λ): luminous efficiency function,
K[m]: maximum spectral luminous efficacy for photopic vision.
Later, the novel MC-based method was applied to the spectral values. The novel MC-based method has been employed to account for the correlations at different wavelengths for the total luminous flux
uncertainty analysis.
For this aim, the spectral radiant flux values, Φ[eT](λ), undergo a modification process in equation (3) by adding a component based on an uncertainty-weighted random error function, given in
equation (4). The error function includes the number of base functions and the weight as given in equation (5) and equation (6), respectively.$ϕ eTn ( λ ) = ϕ eT ( λ ) ∙ ( 1 + δ ( λ ) ∙ u c ( λ ) )$
(3) $δ ( λ ) = ∑ i = 0 N γ i ∙ f i ( λ )$(4) $f i ( λ ) = 2 ∙ sin [ i ∙ ( 2 π ∙ λ - λ 1 λ 2 - λ 1 ) + ϕ i ]$(5) ${ γ 0 , γ 1 , … , γ N } = { Y 0 Y 0 2 + Y 1 2 + … + Y N 2 , Y 1 Y 0 2 + Y 1 2 + … + Y
N 2 , … , Y N Y 0 2 + Y 1 2 + … + Y N 2 }$(6)
Φ[eTn](λ): spectral radiant flux with the random error function,
u[c](λ): combined uncertainty of spectral radiant flux,
δ(λ): random error function,
γ[ i ]: weight function,
f[ i ](λ): base function,
Φ[ i ]: phase,
λ[1] and λ[2]: start and end wavelength of the range,
Y[ i ]: random variables,
N: number of weight and base functions
The Y[ i ] variables, which relate to weighting the base functions, and the Φ[ i ] phase terms, which correspond to base functions, are randomly generated with 20,000 iterations. Following each
random generation of these variables, the resultant modified radiant flux values are computed for every wavelength in equation (3). The f[0](λ) function is assumed to be equal to one as a special
case to have the full correlation. Several random error functions given in equation (4) are illustrated in Figure 1 to show their pattern.
Fig. 1
Error functions for some N values.
3 Modification and results
An LED light source was measured using the integrating sphere flux measurement system with a spectrometer. Using the substitution method, the spectral radiant flux values of the LED light source were
calculated in equation (1) for each wavelenth. Then integrating the spectral radiant flux values in equation (2), the total luminous flux was obtained. The uncertainty of total luminous flux,
expressed in equation (2), was calculated using classical approaches GUM framework as 0.69%.
For the novel MC-based method for uncertainty analysis, a series of codes were developed in Python and are mainly based on the modules shared in empir19nrm02 GitHub repository [5, 6].
To determine the unpredictable uncertainty boundaries of the total luminous flux in equation (2), the spectral radiant flux values were disturbed using the MC-based method [4]. Total luminous flux
was obtained by integrating the disturbed spectral radiant flux in equation (2).
The modification process was repeated for each of the N base function numbers. After each modification, the correlations between the values at different wavelengths were calculated, and the
covariance and correlation matrices for N = 0, N = 1, N = 2, N = 5, and N = 200 are given in Figure 2.
Fig. 2
Correlation (on the left) and covariance (on the right) matrices for N given in ascending order.
The case of full correlation, in a special case where N equals zero and the maximum effect of correlation where N equals one are seen in Figure 2.
The other matrices are displayed in Figure 2 for three different values of N, exemplary N = 2, N = 5, and N = 200, to facilitate a comparative analysis. The correlation matrices with N = 0 and N = 1
can be compared to the correlation matrices for these N values. Figure 2 illustrates a diminishing trend in correlations as the value of N increases.
The uncertainties associated with the total luminous flux were also depicted in Figure 3 versus N number after the modification.
Fig. 3
Uncertainty (k = 2) of the total luminous flux.
The boundaries of the possible uncertainties, which cannot be unpredicted and determined, are discernible in Figure 3. The maximum uncertainty is observed when N attains unity. The convergence of
total luminous flux uncertainty towards the calculated uncertainty value from the GUM framework is observed as N increases.
4 Conclusions
The total luminous flux of an LED light source with 6500 K CCT, a photometric integrating parameter, was attained by integrating the spectral radiant flux. The uncertainty calculation of the total
flux using GUM framework [1] resulted in a value of 0.69%, which reflects the combined effects of various sources of uncertainty. However, the GUM analysis did not take into consideration any
correlations that may exist between the different sources of uncertainty.
Since there is complexity in calculating the correlation between values at different wavelengths, a novel MC-based method was utilised for analysing and estimating the uncertainties of the spectrally
integrated total luminous flux in this study. And the possible uncertainty distributions were attained. The maximum uncertainty is obtained 8.76% while GUM is 0.69%.
This project 19NRM02 RevStdLED has received funding from the EMPIR programme co-financed by the Participating States and from the European Union’s Horizon 2020 research and innovation programme.
Conflicts of interest
The authors declare that they have no competing interests to report.
Data availability statement
The raw data supporting the conclusions of this article will be made available by the authors on request.
Author contribution statement
Şenel Yaran; writing – original draft preparation/writing – review and editing/investigation/methodology. Zühal Alpaslan Kösemen; writing – review and editing/software/investigation. Çağrı Kaan
Akkan; investigation. Hilal Fatmagül Nişancı; software. Udo Krüger; supervision/software/conceptualization/methodology. Armin Sperlin; supervision/methodology.
All Tables
Table 1
Uncertainty budget of the radiant flux of the lamp at several wavelengths.
All Figures
Fig. 1
Error functions for some N values.
In the text
Fig. 2
Correlation (on the left) and covariance (on the right) matrices for N given in ascending order.
In the text
Fig. 3
Uncertainty (k = 2) of the total luminous flux.
In the text
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Six-Spheres - Cylinder & Piston - Compressive Strength Test - 911Metallurgist
Six-Spheres test: The conventional test consisting of compressing the pellet between parallel flat platens does not simulate very well the environment actually encountered by the pellet. Each pellet
is surrounded by similar pellets, that exert a pressure on it whose magnitude is dependent on the height of the pile, among other parameters. The most compact ways in which spheres pack themselves
close-packed hexagonal and face-centered cubic; the closest-packed planes are stacked in sequences AB AB and ABC ABC, respectively. The latter stacking sequence was chosen to represent the packing of
pellets. Two testing platens were machined, three steel spheres were inserted in each. The pellets were placed between the two platens, shown in Figure 12. The spheres on the top platen were not
placed vertically above the ones in the bottom layer in order to better simulate the ABC stacking sequence , in which the pellet corresponds to B. The results of tests conducted on approximately
seventy pellets are shown in Figure 13; the load of fracture is plotted against the cross-sectional area. Both the slope of the least-squares line and its intercept with the load axis are higher than
those of the conventional flat-platen test (Figure 8). Nevertheless, there seems to be no improvement in the scatter of the fracture load over the latter test. Hence, the increase in the number of
contact points from two to six, with a more uniform concentration of stress did not improve the results. This showed that the variation in pellet strength was not connected to irregularities in the
contact regions, confirming thus the mathematical analysis and observations of Section 2. One conclusion can be drawn from the six-spheres test: that the strength of the pellet is somewhat higher
(-30 pct) in the actual environment than between flat platens.
Most of the tests using the six-spheres system were continued beyond the first crack propagated through the pellet. Actually, the first crack did not propagate through its center; failure had a more
progressive, continuous response, and the pellet retained a considerable load-bearing ability after the first crack propagated. The cracks produced something that can be described as “scaling” or
“peeling off” of successive layers of the pellet. For this reason, the tests were continued unitl complete desintegration of the pellet.
Figure 14 shows a typical fractured pellet. The load versus extension curves are shown in Figure 15. Since the fracture load — first peak in load vs. extension curve — varied so much, it was thought
that the energy absorbed in the crushing process might be a better measure of the strength of the pellet. This energy is given by the area under the load vs. extension curves in Figure 15; it can be
readily seen that the units are N.m. Three plots are given in Figure 15; they show the different responses exhibited by pellets. For type A pellets, the fracture load is higher than the remainder of
the curve. For type B pellets, the pellet recovers its strength after the first drop, and the load reaches a level roughly equal to the initial fracture load. Type C pellets, on the other hand,
exhibit a higher strength after some deformation than at the first fracture. These three types of response show that the fracture load is not representative of the load-bearing ability of the
Figure 16 shows the energy absorbed in the crushing process as a function of pellet cross-sectional area. No improvement is obtained over the plots of Figure 8 and 13. The scatter in the data is
still very substantial.
In spite of the fact that the six-spheres test exhibits the same limitations than the flat-platen test, it is fundamentally more significant, since it simulates better the loading environment
actually encountered by the pellets.
Cylinder & Piston Test
The realization that the individual variation in pellet strength was unavoidable and due to (a) existing internal flaws and (b) size differences led to the development of a different testing
procedure consisting of-a thick-walled cylinder and a loosely fitting piston. The system is schematically shown in Figure 17. Pellets are loaded into the cylinder up to a standard height. The system
used in the present investigation had a height of 30.5 cm an internal diameter of 6.35 cm, and the pellets were loaded up to a height of 12.7 cm. The system was compressed in a Baldwin compression
testing machine at a velocity of 2 cm/min. The extension was monitored by means of a transducer (L.V.D. T.) attached to the top portion of the piston. The load and extension were fed into an X-Y
recorder. The total duration of a test was of approximately fifteen minutes.
The main goal in developing this test was of obtaining a parameter that satisfactorily described the compressive strength of a batch of pellet in a simple and readily reproduceable test. Figure 18
shows the types of stress versus strain plots to be expected of batches of pellets with different strength. The stress is calculated by dividing the load by the cross-sectional area of the piston;
the strain is obtained dividing the extension dL by the initial length of the pellet load. As the piston descends, the pellets are crushed; a progressively higher load is required. Since a large
number of pellets is involved, it was thought, sensibly, that the individual fracture events would average out, resulting in a smooch curve. The stress versus strain curve should asymptotically
approach the elastic loading line for totally crushed pellets. The total strain can be calculated; one assumes that the totally crushed pellets have the theoretical specific weight of the pellet
(3.83 gf/cm³) as a first approximation. The apparent specific weight of the pellets (weight of a certain quantity divided by its volume) was found to be 2.95 gf/cm². Dividing this value by the
theoretical specific weight, one finds that it is 0.5. Hence, complete crushing should be achieved at a strain of 0.5. At a strain of 0.5 the curves should approach the line for the elastic modulus
of hematite. Since this value of strain is only obtained at very high values of stress, for stronger pellets, a certain amount of strain was arbitrarely chosen. The area under the different curves
expresses the energy required to crush the pellets to the arbitrarely chosen strain and is a satisfactory measure of the compressive strength of pellets.
The application of the above concepts to actual pellets is shown in Figure 19. Four individual tests are shown in the figure, and they are in fair agreement. The energies absorbed up to a strain of
0.25 were measured from the area under the curve: the mean is 3128 N.m and the standard deviation is 651 N.m. The variation in this parameter from test to test is far lower than that of the
individual pellets (either fracture load or energy absorbed). Hence, it is thought that such a test, upon appropriate standardization after optimization of the parameters, would provide a simpler and
superior description of the pellet compressive strength. Figure 20 shows the percentage of pellets fractured as a function of strain. One can see that the curve is S shaped and that the strain of 0.5
was set as 100 pct fractured. Actually, a distinction should be traced between fractured and crushed: 100 pct fractured should correspond to a strain slightly lower than 100 pct crushed.
Nevertheless, the results of Fig. 20 show that the stress level used in the test is sufficient to produce fracture of a considerable percentage of the pellets; the five tests grouped around 80 pct
fractured correspond to the plots of Fig. 19. | {"url":"https://www.911metallurgist.com/blog/six-spheres-cylinder-piston-compressive-strength-test/","timestamp":"2024-11-05T20:11:32Z","content_type":"text/html","content_length":"164928","record_id":"<urn:uuid:5acbd855-7a6b-4c61-ab20-c3fc2b1a3e8e>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00883.warc.gz"} |
egional TACs
Compartmental model analysis of regional TACs
This page handles the general compartmental models with arterial plasma input that are applicable to most tracers; tracer specific models are described elsewhere, as well as reference tissue input
Model fitting software
In TPC, Carimas is available for all researchers, and it can be used to fit one- and two-tissue compartmental models to regional data. ROIs can either be drawn and regional TACs computed inside
Carimas, or regional TACs can be imported to Carimas.
Certain research groups in TPC have also acquired PMOD licenses.
Alternatively, in-house developed command-line programs (Open Source) can be used directly in Windows computers that are connected to TPC network, or can be downloaded for Windows, Linux, and macOS
Nonlinear least-squares (NLLS) fitting
Programs fitk2, fitk3, fitk4, fitk5, and fitkloss can be used to fit two, three, or four compartment models to the PET TACs; fitkloss can be used to fit a three-compartment model where the last rate
constant k[4] or k[Loss] represents the efflux of labelled metabolite or radioligand directly to the venous plasma.
These programs do use the weighting information if that is included in the tissue datafile.
In many cases all of the model parameters of two-tissue model can not be reliably fitted, but some of them need to be constrained to a pre-determined value. A commonly used method is to constrain the
K[1]/k[2], representing the distribution volume of nonspecifically bound and free radioligand in the tissue, to a value that is derived from a reference region with no specific binding or uptake (k
[3]=0). Not only does this method allow lower variances of model parameters, but it also enables us to calculate the binding potential in case of receptor studies.
As in all models applying reference tissue, also here it is assumed that K[1]/k[2] is the same in all (brain) regions. First, two- or three-compartment model is fitted to the reference tissue curve,
providing the values for K[1]/k[2] or K[1]/k[2] and k[5]/k[6]. These are used as constraints for these parameters when fitting the compartmental model to the regions of interest.
Constraining K[1]/k[2] can be optionally done with programs fitk3, fitk4, and fitk5, when the name of reference region in the TAC file is given as command-line argument.
Constraining K[1]/k[2] to zero with program fitk2 actually constrains k[2]=0, allowing the use of an extremely simple irreversible model with only two parameters, K[1] and V[B].
Akaike information criteria (AIC)
Various compartmental models can be constructed and used to analyze PET data. The more complicated the model is, the better is the achieved fit to the data. However, at the same time, also the
variance of the fitted parameters is increased. To find the optimum model, the programs compute also Akaike information criteria values: the smaller the AIC values are, the better the model is,
considering the degrees of freedom of the fit. However, the physiological interpretation of the fitted parameters is on the responsibility of the user.
General linear least squares method
Most compartmental models can be transformed into general linear least squares functions (Blomqvist, 1984), which can be solved using very fast linear methods, and are therefore suitable for
computing parametric images. For regional data, program lhsol can be used to estimate the model parameters using Lawson-Hanson nonnegative least squares (NNLS) algorithm. The compartmental model can
be selected with options -k1, -k2, -k3, -k4 for models excluding vascular volume fraction, or options -vk1, -vk2, -vk3, -vk4 for models including vascular volume fraction.
Note that if V[A] is fitted using this methods, the vascular blood TAC is assumed to be similar to the model input, i.e. metabolite corrected arterial plasma curve. This is close to truth for a few
tracers only, e.g. [^18F]FDG. For other studies, a fixed amount of blood background can be subtracted before the model fit using taccbv.
The fitted parameters from these programs may have non-physiological values, because there is no other constraints than non-negativity.
V[T] and K[i] using NNLS method
In receptor binding studies distribution volume (V[T]) is usually the only model parameter of interest. Instead of solving separate model rate constants and calculating V[T] from those afterwards,
more reliable estimates of V[T] can be obtained by solving V[T] directly without division (Zhou et al., 2002; Hagelberg et al., 2004). Program lhsoldv uses Lawson-Hanson non-negative least squares (
NNLS) algorithm and one or two tissue compartment models (with options -1 and -2) to solve the V[T] without division. The noise in regional TACs does not cause bias when using this method. Two tissue
compartment model is recommended, since one tissue compartment model may lead to biases with more complex tissue kinetics. By default (or with option -A), the model can be selected automatically,
based on lower AIC. With option -0 the AIC weighted average (Turkheimer et al. 2002) of V[T] from 1- and 2-tissue compartment models is calculated.
For irreversible tracer uptake models, the influx rate constant K[i] can be calculated accordingly with program lhsolki. The two tissue model is applied by default.
In-house analysis programs, including Carimas, automatically convert the sample time and radioactivity concentrations units of plasma, blood and tissue data, if the units are specified in the data
files. This is not always the case, and therefore it would be safest that researcher verifies that the units are the same in all data files before proceeding to the modelling.
In TPC, the units of radioactivity concentrations in plasma and blood are by default given per volume (mL), not per mass (g). Therefore the unit of model parameter K[1] is (mL plasma)*(mL tissue)^
-1*min^-1, and the units of other rate constants k[2], ... k[6] are min^-1. The units of the vascular blood or arterial plasma volume fractions V[B] and V[A] are mL/mL.
Steps of calculation using command-line tools
All of of the following steps can be done in Linux terminal window or MS Windows command prompt window (preferably using scripts):
1. Adding weights to regional tissue TAC data
2. Computing the parameter estimates: execute one of the model fit programs with the command line parameters that are specified in the programs user help information (with option --help).
See also:
Blomqvist G. On the construction of functional maps in positron emission tomography. J Cereb Blood Flow Metab. 1984; 4:629-632. doi: 10.1038/jcbfm.1984.89.
Lawson CL, Hanson RJ. Solving least squares problems. Prentice-Hall, 1974.
Turkheimer FE, Hinz R, Cunningham VJ. On the undecidability among kinetic models: from model selection to model averaging. J Cereb Blood Flow Metab. 2003; 23: 490-498. doi: 10.1097/
Zhou Y, Brasic J, Endres CJ, Kuwabara H, Kimes A, Contoreggi C, Maini A, Ernst M, Wong DF. Binding potential image based statistical mapping for detection of dopamine release by [^11C]raclopride
dynamic PET. NeuroImage 2002; 16: S91.
Tags: Modeling, Compartmental model, Rate constant, Fitting, NLLS, NNLS, Analysis
Updated at: 2020-05-01
Created at: 2013-05-17
Written by: Vesa Oikonen | {"url":"http://www.turkupetcentre.net/petanalysis/model_cm_plasma.html","timestamp":"2024-11-07T02:44:01Z","content_type":"text/html","content_length":"15277","record_id":"<urn:uuid:7e8c60ac-7a1d-4005-8ad0-c0d43d811bd9>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00573.warc.gz"} |
Robust covariance and precision matrix estimators. Based on the review of P.-L. Loh and X. L. Tan. (2018)
To install:
There are in total 4 robust covariance and 3 correlation estimation implemented, namely:
• corSpearman: Spearman correlation
• corKendall: Kendall’s tau
• corQuadrant: Quadrant correlation coefficients
• covGKmat: Gnanadesikan-Kettenring estimator by Tarr et al. (2015) and Oellerer and Croux (2015)
• covSpearmanU: SpearmanU covariance estimator by P.-L. Loh and X. L. Tan. (2018), The pairwise covariance matrix estimator proposed in Oellerer and Croux (2015), where the MAD estimator is
combined with Spearman’s rho
• covOGK: Orthogonalized Gnanadesikan-Kettenring (OGK) estimator by Maronna, R. A. and Zamar, R. H. (2002)
• covNPD: Nearest Positive (semi)-Definite projection of the pairwise covariance matrix estimator considered in Tarr et al. (2015).
P.-L. Loh and X. L. Tan. (2018) then used these robust estimates in Graphical Lasso (package glasso) or Quadratic Approximation (package QUIC) to obtain sparse solutions to precision matrix
With glasso, a function robglasso stand for robust graphical LASSO is implemented. It has build in cross validation described in P.-L. Loh and X. L. Tan. (2018), for instance, to use the method with
cross validation:
Where data should be a matrix and covest should be a function that estimate the covariance e.g. anyone mentioned above. The result list contains everything from glasso output with the optimal tuning
parameter found by cross validation. One can also decide fold by setting fold in robglasso. For more details see ?robglasso. | {"url":"https://pbil.univ-lyon1.fr/CRAN/web/packages/robustcov/readme/README.html","timestamp":"2024-11-14T12:21:24Z","content_type":"application/xhtml+xml","content_length":"7305","record_id":"<urn:uuid:838d73cf-d110-4ddc-b149-f95a0b3e694b>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00176.warc.gz"} |
12.4 Testing the Significance of the Correlation Coefficient (Optional)
The correlation coefficient, r, tells us about the strength and direction of the linear relationship between x and y. However, the reliability of the linear model also depends on how many observed
data points are in the sample. We need to look at both the correlation coefficient r and the sample size n, together.
We perform a hypothesis test of the significance of the correlation coefficient to decide whether the linear relationship in the sample data is strong enough to use to model the relationship in the
The sample data are used to compute r, the correlation coefficient for the sample. If we had data for the entire population, we could find the population correlation coefficient. But, because we have
only sample data, we cannot calculate the population correlation coefficient. The sample correlation coefficient, r, is our estimate of the unknown population correlation coefficient.
• The symbol for the population correlation coefficient is ρ, the Greek letter rho.
• ρ = population correlation coefficient (unknown).
• r = sample correlation coefficient (known; calculated from sample data).
The hypothesis test lets us decide whether the value of the population correlation coefficient ρ is close to zero or significantly different from zero. We decide this based on the sample correlation
coefficient r and the sample size n.
If the test concludes the correlation coefficient is significantly different from zero, we say the correlation coefficient is significant.
• Conclusion: There is sufficient evidence to conclude there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero.
• What the conclusion means: There is a significant linear relationship between x and y. We can use the regression line to model the linear relationship between x and y in the population.
If the test concludes the correlation coefficient is not significantly different from zero (it is close to zero), we say the correlation coefficient is not significant.
• Conclusion: There is insufficient evidence to conclude there is a significant linear relationship between x and y because the correlation coefficient is not significantly different from zero.
• What the conclusion means: There is not a significant linear relationship between x and y. Therefore, we cannot use the regression line to model a linear relationship between x and y in the
• If r is significant and the scatter plot shows a linear trend, the line can be used to predict the value of y for values of x that are within the domain of observed x values.
• If r is not significant or if the scatter plot does not show a linear trend, the line should not be used for prediction.
• If r is significant and the scatter plot shows a linear trend, the line may not be appropriate or reliable for prediction outside the domain of observed x values in the data.
Performing the Hypothesis Test
Performing the Hypothesis Test
• Null hypothesis: H[0]: ρ = 0
• Alternate hypothesis: H[a]: ρ ≠ 0
What the Hypothesis Means in Words:
• Null hypothesis H[0]: The population correlation coefficient is not significantly different from zero. There is not a significant linear relationship (correlation) between x and y in the
• Alternate hypothesis H[a]: The population correlation coefficient is significantly different from zero. There is a significant linear relationship (correlation) between x and y in the population.
Drawing a Conclusion:There are two methods to make a conclusion. The two methods are equivalent and give the same result.
• Method 1: Use the p-value.
• Method 2: Use a table of critical values.
In this chapter, we will always use a significance level of 5 percent, α = 0.05
Using the p-value method, you could choose any appropriate significance level you want; you are not limited to using α = 0.05. But, the table of critical values provided in this textbook assumes we
are using a significance level of 5 percent, α = 0.05. If we wanted to use a significance level different from 5 percent with the critical value method, we would need different tables of critical
values that are not provided in this textbook.
METHOD 1: Using a p-Value to Make a Decision
Using the TI-83, 83+, 84, 84+ Calculator
To calculate the p-value using LinRegTTEST:
1. Complete the same steps as the LinRegTTest performed previously in this chapter, making sure on the line prompt for β or σ, ≠ 0 is highlighted.
2. When looking at the output screen, the p-value is on the line that reads p =.
If the p-value is less than the significance level (α = 0.05),
• Decision: Reject the null hypothesis.
• Conclusion: There is sufficient evidence to conclude there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero.
If the p-value is not less than the significance level (α = 0.05),
• Decision: Do not reject the null hypothesis.
• Conclusion: There is insufficient evidence to conclude there is a significant linear relationship between x and y because the correlation coefficient is not significantly different from zero.
You will use technology to calculate the p-value, but it is useful to know that the p-value is calculated using a t distribution with n – 2 degrees of freedom and that the p-value is the combined
area in both tails.
An alternative way to calculate the p-value (p) given by LinRegTTest is the command 2*tcdf(abs(t),10^99, n-2) in 2nd DISTR.
Third Exam vs. Final Exam Example: p-Value Method
• Consider the third exam/final exam example.
• The line of best fit is ŷ = –173.51 + 4.83x, with r = 0.6631, and there are n = 11 data points.
• Can the regression line be used for prediction? Given a third exam score (x value), can we use the line to predict the final exam score (predicted y value)?
H[0]: ρ = 0
H[a]: ρ ≠ 0
α = 0.05
• The p-value is 0.026 (from LinRegTTest on a calculator or from computer software).
• The p-value, 0.026, is less than the significance level of α = 0.05.
• Decision: Reject the null hypothesis H[0].
• Conclusion: There is sufficient evidence to conclude there is a significant linear relationship between the third exam score (x) and the final exam score (y) because the correlation coefficient
is significantly different from zero.
Because r is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores.
METHOD 2: Using a Table of Critical Values to Make a Decision
The 95 Percent Critical Values of the Sample Correlation Coefficient Table (Table 12.11) can be used to give you a good idea of whether the computed value of r is significant. Use it to find the
critical values using the degrees of freedom, df = n – 2. The table has already been calculated with α = 0.05. The table tells you the positive critical value, but you should also make that number
negative to have two critical values. If r is not between the positive and negative critical values, then the correlation coefficient is significant. If r is significant, then you may use the line
for prediction. If r is not significant (between the critical values), you should not use the line to make predictions.
Example 12.7
Suppose you computed r = .801 using n = 10 data points. The degrees of freedom would be 8 (df = n – 2 = 10 – 2 = 8). Using Table 12.11 with df = 8, we find that the critical value is 0.632. This
means the critical values are really ±0.632. Since r = .801 and .801 > 0.632, r is significant and the line may be used for prediction. If you view this example on a number line, it will help you to
see that r is not between the two critical values.
Try It 12.7
For a given line of best fit, you computed that r = .6501 using n = 12 data points, and the critical value found on the table is 0.576. Can the line be used for prediction? Why or why not?
Example 12.8
Suppose you computed r = –0.624 with 14 data points, where df = 14 – 2 = 12. The critical values are –0.532 and 0.532. Since –0.624 –0.532, r is significant and the line can be used for prediction
Try It 12.8
For a given line of best fit, you compute that r = .5204 using n = 9 data points, and the critical values are ±0.666. Can the line be used for prediction? Why or why not?
Example 12.9
Suppose you computed r = .776 and n = 6, with df = 6 – 2 = 4. The critical values are – 0.811 and 0.811. Since 0.776 is between the two critical values, r is not significant. The line should not be
used for prediction.
Try It 12.9
For a given line of best fit, you compute that r = –.7204 using n = 8 data points, and the critical value is 0.707. Can the line be used for prediction? Why or why not?
Third Exam vs. Final Exam Example: Critical Value Method
Third Exam vs. Final Exam Example: Critical Value Method
Consider the third exam/final exam example. The line of best fit is: ŷ = –173.51 + 4.83x, with r = .6631, and there are n = 11 data points. Can the regression line be used for prediction? Given a
third exam score (x value), can we use the line to predict the final exam score (predicted y value)?
• H[0]: ρ = 0
• H[a]: ρ ≠ 0
• α = 0.05
• Use the 95 Percent Critical Values table for r with df = n – 2 = 11 – 2 = 9.
• Using the table with df = 9, we find that the critical value listed is 0.602. Therefore, the critical values are ±0.602.
• Since 0.6631 > 0.602, r is significant.
• Decision: Reject the null hypothesis.
• Conclusion: There is sufficient evidence to conclude there is a significant linear relationship between the third exam score (x) and the final exam score (y) because the correlation coefficient
is significantly different from zero.
Because r is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores.
Example 12.10
Suppose you computed the following correlation coefficients. Using the table at the end of the chapter, determine whether r is significant and whether the line of best fit associated with each
correlation coefficient can be used to predict a y value. If it helps, draw a number line.
a. r = – 0.567 and the sample size, n, is 19.
To solve this problem, first find the degrees of freedom. df = n – 2 = 17
Then, using the table, the critical values are ±0.456.
–0.567 < –0.456, or you may say that –0.567 is not between the two critical values.
r is significant and may be used for predictions.
b. r = 0.708 and the sample size, n, is 9.
df = n – 2 = 7
The critical values are ±0.666.
0.708 > 0.666
r is significant and may be used for predictions.
c. r = 0.134 and the sample size, n, is 14.
df = 14 – 2 = 12.
The critical values are ±0.532.
0.134 is between –0.532 and 0.532
r is not significant and may not be used for predictions.
d. r = 0 and the sample size, n, is 5.
It doesn’t matter what the degrees of freedom are because r = 0 will always be between the two critical values, so r is not significant and may not be used for predictions.
Try It 12.10
For a given line of best fit, you compute that r = 0 using n = 100 data points. Can the line be used for prediction? Why or why not?
Assumptions in Testing the Significance of the Correlation Coefficient
Assumptions in Testing the Significance of the Correlation Coefficient
Testing the significance of the correlation coefficient requires that certain assumptions about the data be satisfied. The premise of this test is that the data are a sample of observed points taken
from a larger population. We have not examined the entire population because it is not possible or feasible to do so. We are examining the sample to draw a conclusion about whether the linear
relationship that we see between x and y in the sample data provides strong enough evidence that we can conclude there is a linear relationship between x and y in the population.
The regression line equation that we calculate from the sample data gives the best-fit line for our particular sample. We want to use this best-fit line for the sample as an estimate of the best-fit
line for the population. Examining the scatter plot and testing the significance of the correlation coefficient helps us determine whether it is appropriate to do this.
The assumptions underlying the test of significance are as follows:
• There is a linear relationship in the population that models the sample data. Our regression line from the sample is our best estimate of this line in the population.
• The y values for any particular x value are normally distributed about the line. This implies there are more y values scattered closer to the line than are scattered farther away. Assumption 1
implies that these normal distributions are centered on the line; the means of these normal distributions of y values lie on the line.
• Normal distributions of all the y values have the same shape and spread about the line.
• The residual errors are mutually independent (no pattern).
• The data are produced from a well-designed, random sample or randomized experiment. | {"url":"https://texasgateway.org/resource/124-testing-significance-correlation-coefficient-optional?book=79081&binder_id=78271","timestamp":"2024-11-11T07:33:02Z","content_type":"text/html","content_length":"70552","record_id":"<urn:uuid:c193d45e-3096-4861-a566-bcacd31bbd1f>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00107.warc.gz"} |
Constructing “ex ante” real interest rates on FRED
Interest rates are some of the most popular series on FRED. Almost all the interest rates on FRED are nominal interest rates, which reflect the annual cost of borrowing money. A nominal interest rate
doesn’t account for the effects of inflation, though. For example, if a lender lends $100 for a year at 5% interest, the borrower repays the lender with $105 at the end of the year. But, if inflation
has been 10% over that same year, the lender is actually able to buy less with the $105 repayment at the end of that year than they could have bought with the $100 originally loaned at the beginning
of that year.
A real interest rate is an inflation-adjusted interest rate. You might think of a real interest rate as the price of borrowing in goods, not money. Because people and firms make decisions based on
real quantities, not nominal quantities, real interest rates are more useful than nominal interest rates. For example, real interest rates are much more informative than nominal interest rates about
the stance of monetary policy.
Technically, a gross real interest rate (1+r) is calculated as the ratio of gross nominal rates (1+i) to the gross inflation rate (1+π):
(1+r) = (1+i) / (1+π)
Suppose that candy bars cost $1 on January 1, 2022. The lender could use the $100 to buy 100 candy bars, but forgoes the purchase to make a loan of $100 instead. When the borrower repays the loan at
5% interest on January 1, 2023, the lender receives $105 dollars. If inflation has raised the price of candy bars by 10% by January 1, 2023, then each candy bar costs $1.10 and the lender can buy
only 95 candy bars: 105/1.1 = 95.4545. The gross real rate of return equals the real goods one can buy with the payoff from the loan (95.4545 candy bars) over the initial real value of the loan (100
candy bars). So, the gross real rate of interest is 95.4545/100 = 1.05/1.10 = (1+i)/(1+π).
This is often approximated as the interest rate minus the inflation rate.
r ≅ i – π
This approximation is generally useful for relatively low rates of interest and inflation. With the example above, it would be -5% = 5% – 10%. And yes, real interest rates can be negative.
To calculate historical real interest rates, one can either use a forecast of inflation or the average rate of inflation that actually occurred over the period of the loan/bond. When one uses a
forecast of inflation to construct a real rate, that measure is called an “ex ante” real rate, while using realized inflation produces “ex post” real rates. Because forecasts of inflation will
generally differ from each other and from the average rate of inflation realized over a period, estimates of real interest rates for the same date and same horizon can differ from each other.
Despite the usefulness of real interest rates, FRED only has a few real interest rates: 1-month, 1-year, and 10-year real rates, all at the monthly frequency, constructed by the Cleveland Fed with a
variety of data to estimate the expected rate of inflation.
FRED users can also construct daily historical series for real rates of interest with market-implied forecasts of inflation, called “breakeven” inflation rates derived from options prices. There are
breakeven inflation rates on FRED for 5-, 7-, 10-, 20-, and 30-year horizons.
The FRED graph at the top compares the monthly Cleveland Fed 10-year real interest rate with a daily 10-year real rate derived from breakeven inflation. The two series track each other reasonably
well for most of the sample, but diverge at times when the breakeven inflation rate is particularly volatile, such as during the Financial Crisis of 2008 and the COVID-19 pandemic of 2020-2021.
Using the methods to construct the above graph, FRED users can investigate real interest rates in several ways.
• It would be easy to compare the exact formula for a real interest rate r = ((1+i)/(1+π)-1) with the approximation (r ≅ i – π) by using the “add line” and “formula” functions to create another
series. You will see that the lines are difficult to distinguish.
• One could also compare the 10-year real interest rate above with the implied 5-year real interest rate from the 5-year constant maturity Treasury yield and 5-year breakeven inflation rate.
• One could download yield and inflation data to construct “ex post” real interest rates in Excel or another application.
How this graph was created: Search for and select “10-year constant maturity Treasury yield” and choose “Market Yield on U.S. Treasury Securities at 10-Year Constant Maturity (DGS10).” From the “Edit
Graph” panel, use the “Customize data” field to search for “10-year” and select “10-year breakeven inflation rate.” The 10-year yield (i.e., nominal interest rate) will be series “a” and the 10-year
break-even inflation rate will be series “b”. From the formula bar, type in the following formula for a real interest rate: 100*((1+a/100)/(1+b/100) – 1) and click “Apply.” To compare this series
with the Cleveland Fed 10-year real rate, use the “Add Line” tab at the top of the editing box to search for and select “10-year real interest rate” and click “Add data series.” The two series should
now be displayed from 1982, but there will be no values for the constructed real rate until 2003. To see them over a common sample, set the sample to start on January 1, 2003.
Suggested by Christopher Neely. | {"url":"https://fredblog.stlouisfed.org/2022/05/constructing-ex-ante-real-interest-rates-on-fred/?utm_source=series_page&utm_medium=related_content&utm_term=related_resources&utm_campaign=fredblog","timestamp":"2024-11-02T18:32:04Z","content_type":"text/html","content_length":"82815","record_id":"<urn:uuid:8a8dd3d2-6fd7-40bd-9bc6-07ce67056a21>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00091.warc.gz"} |
How To Calculate Quartiles
digitalgenetics/iStock/Getty Images
When ranking numbers, such as test scores or the length of elephant tusks, it can be helpful to conceptualize one rank in relation to another. For example, you might want to know if you scored higher
or lower than the rest of your class or if your pet elephant has longer or shorter tusks than most of the other pet elephants on your block. One way to conceptualize a ranking system is through the
use of quartiles, which represent three divides within your data that splits the data into four equal parts.
Aquir/iStock/Getty Images
Step 1
Rank your values in order from lowest to highest; you will use this ranked value order in all of the different methods for computing quartiles. The first method for computing quartiles is to divide
your newly ordered dataset into two halves at the median.
Step 2
Find the median, or middle value, of your dataset. For example, if your dataset is (1, 2, 5, 5, 6, 8, 9), the median is 5 because that is the middle value. This middle value represents your second
quartile, or 50th percentile. Fifty percent of your values are higher than this value, and 50 percent are lower.
Step 3
Draw a line at the median to separate the lower half of your data, which is now (1, 2, 5), and the upper half of your data, which is (6, 8, 9). The first quartile value, or 25th percentile, is the
median of the lower half, which is 2. The third quartile, or 75th percentile, is the median of the upper half, which is 8. So you know that about 25 percent of your numbers are lower than 2, half of
your numbers are 5 or lower, and about three-quarters of your values are lower than 8.
Step 4
Find the difference between your upper quartile, or 75th percentile, and your lower quartile, or 25th percentile. Using the dataset (1, 2, 5, 5, 6, 8, 9), your interquartile range is the difference
between 8 and 2, so your interquartile range is 6.
Cite This Article
Reid, Ari. "How To Calculate Quartiles" sciencing.com, https://www.sciencing.com/calculate-quartiles-5162215/. 24 April 2017.
Reid, Ari. (2017, April 24). How To Calculate Quartiles. sciencing.com. Retrieved from https://www.sciencing.com/calculate-quartiles-5162215/
Reid, Ari. How To Calculate Quartiles last modified March 24, 2022. https://www.sciencing.com/calculate-quartiles-5162215/ | {"url":"https://www.sciencing.com:443/calculate-quartiles-5162215/","timestamp":"2024-11-06T20:05:10Z","content_type":"application/xhtml+xml","content_length":"70612","record_id":"<urn:uuid:3c7988b0-6544-402f-a6ba-edd09c10bba0>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00810.warc.gz"} |
Sixty Degrees and Other Angles in Geometry
In geometry, angles can be measured in two ways. The first way is in terms of degrees, with a full circle being 360 degrees. The second way is in terms of radians, with a full circle being 2π
radians. An angle of 60 degrees is therefore equal to π/3 radians.
There are 90 degrees in a right angle, 180 degrees in a straight angle, and 360 degrees in a full angle. Other angles are named by the number of degrees they contain, with the symbol ° after the
number, such as 45°. A right angle is an angle that measures exactly 90°. It is called a right angle because it forms a square corner, like the corner on a right-angled triangle. A right angle is
denoted by the small raised square symbol shown here: ?
Angles that are less than 90° are acute angles. Angles that are more than 90° but less than 180° are obtuse angles. An angle of 180° is called a straight angle because it appears as a straight line
when drawn on paper. When two lines intersect at a point and form an angle greater than 180° but less than 360°, this is called a reflex angle. Angles that are greater than or equal to 360° (for
example, 540°) but less than 720° are called exteriors angles on the same side as the interior angles.
In geometry, angles can be measured in two ways. The first way is in terms of degrees, with a full circle being 360 degrees. The second way is in terms of radians, with a full circle being 2π
radians. An angle of 60 degrees is therefore equal to π/3 radians.
There are 90 degrees in a right angle, 180 degrees in a straight angle, and 360 degrees in a full angle. Other angles are named by the number of degrees they contain, with the symbol ° after the
number, such as 45°. A right angle is an angle that measures exactly 90°. It is called a right angle because it forms a square corner, like the corner on a right-angled triangle. A right angle is
denoted by the small raised square symbol. Angles that are less than 90° are acute angles. Angles that are more than 90° but less than 180° are obtuse angles. An angle of 180° is called a straight
angle because it appears as a straight line when drawn on paper. When two lines intersect at a point and form an angle greater than 180° but less than 360°, this is called a reflex angle. Angles that
are greater than or equal to 360° (for example 540°) but less then 720° exterior angles on the same side as the interior angles."
What does 60 mean in geometry?
In geometry, angles can be measured in two ways. The first way is in terms of degrees, with a full circle being 360 degrees. The second way is in terms of radians, with a full circle being 2π
radians. An angle of 60 degrees is therefore equal to π/3 radians.
How do you make a 60 degree angle?
There are 90 degrees in a right angle, 180 degrees in a straight angle, and 360 degrees in a full angle. Other angles are named by the number of degrees they contain, with the symbol ° after the
number, such as 45°. A right angle is an angle that measures exactly 90°. It is called a right angle because it
What is the answer of 60 degree?
The answer to 60 degrees is 3.14 radians. | {"url":"https://www.intmath.com/functions-and-graphs/sixty-degrees-and-other-angles-in-geometry.php","timestamp":"2024-11-12T17:32:12Z","content_type":"text/html","content_length":"101338","record_id":"<urn:uuid:7bfdac41-2d33-4ebf-a1ea-ede8fac8486f>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00608.warc.gz"} |
III Quantum Computation - Some quantum algorithms
3Some quantum algorithms
III Quantum Computation
3.4 Search problems and Grover’s algorithm
We are now going to turn our attention to search problems. These are very
important problems in computing, as we can formulate almost all problems as
some sort of search problems.
One important example is simultaneous constraint satisfaction. Here we have
a large configuration space of options, and we want to find some configuration
that satisfies some constraints. For example, when designing a lecture timetable
for Part III courses, we need to schedule the courses so that we don’t clash
two popular courses in the same area, and the courses need to have big enough
lecture halls, and we have to make sure a lecturer doesn’t have to simultaneously
lecture two courses at the same time. This is very complicated.
In general, search problems have some common features:
Given any instance of solution attempt, it is easy to check if it is good or
(ii) There are exponentially many possible instances to try out.
One example is the boolean satisfiability problem, which we have already
seen before.
(Boolean satisfiability problem)
The boolean satisfiability problem
(SAT ) is as follows — given a Boolean formula
→ B
, we want to know if
there is a “satisfying argument”, i.e. if there is an x with f(x) = 1.
This has complexity class
, standing for non-deterministic polynomial
time. There are many ways to define
, and here we will provide two. The
first definition of NP will involve the notion of a verifier:
Definition (Verifier). Suppose we have a language L ⊆ B
, where
is the set of all bit strings.
A verifier for L is a computation V (w, c) with two inputs w, c such that
(i) V halts on all inputs.
(ii) If w ∈ L, then for some c, V (w, c) halts with “accept”.
(iii) If w 6∈ L, then for all c, V (w, c) halts with “reject”.
A polynomial time verifier is a
that runs in polynomial time in
|w| + |c|!).
We can think of
as “certificate of membership”. So if you are a member,
you can exhibit a certificate of membership that you are in there, and we can
check if the certification is valid. However, if you are not a member, you cannot
“fake” a certificate.
(Non-deterministic polynomial time problem)
. NP
is the class of
languages that have polynomial time verifiers.
The SAT problem is in
. Here
is the satisfying argument, and
V (f, c) just computes f(c) and checks whether it is 1.
Determining if a number is composite is in
, where a certificate
is a factor of the number.
However, it is not immediately obvious that testing if a numb er is prime is
. It is an old result that it indeed is, and recent progress shows that it is
in fact in P.
It is rather clear that
P ⊆ NP
. Indeed, if we can check membership in
polynomial time, then we can also construct a verifier in polynomial time that
just throws the certificate away and check directly.
There is another model of
, via non-deterministic computation. Recall
that in probabilistic computation, in some steps, we had to pick a random
number, and picking a different number would lead to a different “branch”. In
the case of non-deterministic computation, we are allowed to take all paths at
the same time. If some of the paths end up being accepting, then we accept the
input. If all paths reject, then we reject the input. Then we can alternatively
say a problem is in
if there is a p olynomial-time non-deterministic machine
that checks if the string is in the language.
It is not difficult to see that these definitions of
are equivalent. Suppose
we have a non-deterministic machine that checks if a string is in the language.
Then we can construct a verifier whose certificate is a prescription of which
particular branch we should follow. Then the verifier just takes the prescription,
follows the path described and see if we end up being accepted.
Conversely, if we have a verifier, we can construct a non-deterministic machine
by testing a string on all possible certificates, and check if any of them accepts.
Unfortunately, we don’t know anything about how these different complexity
classes compare. We clearly have
P ⊆ BPP ⊆ BQP
P ⊆ NP
. However,
we do not know if these inclusions are strict, or how
compares to the others.
Unstructured search problem and Grover’s algorithm
Usually, when we want to search something, the search space we have is structured
in some way, and this greatly helps our searching problem.
For example, if we have a phone book, then the names are ordered alpha-
betically. If we want to find someone’s phone number, we don’t have to look
through the whole b ook. We just open to the middle of the book, and see if the
person’s name is before or after the names on the page. By one lookup like this,
we have already eliminated half of the phone book we have to search through,
and we can usually very quickly locate the name.
However, if we know someone’s phone number and want to figure out their
name, it is pretty much hopeless! This is the problem with unstructured data!
So the problem is as follows: we are given an unstructured database with
= 2
items and a unique good item (or no good items). We can query any
item for good or bad-ness. The problem is to find the good item, or determine if
one exists.
) queries are necessary and sufficient. Even if we are asking
for a right result with fixed probability
, if we pick items randomly to check,
then the probability of seeing the “good” one in
queries is given by
. So
we still need O(N ) queries for any fixed probability.
Quantumly, we have Grover’s algorithm. This needs
) queries, and
this is both necessary and sufficient.
The database of
= 2
items will be considered as an oracle
→ B
it is promised that there is a unique
∈ B
) = 1. The problem is to
find x
. Again, we have the quantum version
|xi|yi = |xi|y ⊗ f(x)i.
However, we’ll use instead I
on n qubits given by
|xi =
|xi x 6= x
−|xi x 6= x
This can be constructed from
as we’ve done before, and one use of
be done with one use of U
|si I
We can write I
= I − 2 |x
where I is the identity operator.
We are now going to state the Grover’s algorithm, and then later prove that
it works.
For convenience, we write
= H ⊗ ··· ⊗ H
| {z }
n times
We start with a uniform sup erposition
i = H
|0 ···0i =
all x
We consider the Grover iteration operator on n qubits given by
Q = −H
Here running
requires one query (whereas
is “free” because it is just
I − 2 |0ih0|).
Note that all these operators are all real. So we can pretend we are living
in the real world and have nice geometric pictures of what is going on. We let
) be the (real) plane spanned by |x
i and |ψ
i. We claim that
(i) In this plane P(x
), this operator Q is a rotation by 2α, where
sin α =
= hx
(ii) In the orthogonal complement P(x
, we have Q = −I.
We will prove these later on. But if we know this, then we can repeatedly apply
to rotate it near to
, and then measure. Then we will obtain
with very high probability:
The initial angle is
cos β = hx
i =
So the number of iterations needed is
2 sin
In general, this is not an integer, but applying a good integer approximation to it
will bring us to very close to
, and thus we measure
with high probability.
For large n, the number of iterations is approximately
Example. Let’s do a boring example with N = 4. The initial angle satisfies
cos β =
So we know
β =
Similarly, we have
2α = 2 sin
So 1 iteration of
will rotate
exactly to
, so we can find it with certainty
with 1 lookup.
Now we prove that this thing actually works. In general, for any unitary
and I
= I − 2 |ψihψ|, we have
= U IU
− 2U |ψihψ|U
= I
In particular, since
is self-adjoint, i.e.
, and that by definition
i, we know
Q = −H
= −I
Next we note that for any |ψi and |ξi, we know by definition
|ξi = |ξi−2 |ψihψ|ξi.
So this modifies |ξi by some multiple of |ψi. So we know our operator
Q|ψi = −I
first by some multiple of
, then by some multiple of
. So if
|ξi ∈ P(x
), then Q|ψi ∈ P(x
) too! So Q preserves P(x
We know that
is a unitary, and it is “real”. So it must be a rotation or
a reflection, since these are the only things in O(2). We can explicitly figure
out what it is. In the plane
), we know
is reflection in the mirror line
perpendicular to
. Similarly,
is reflection in the mirror line perpendicular
to |ψ
We now use the following facts about 2D Euclidean geometry:
is a reflection in mirror
, then
is reflection in mirror
To see this, we know any vector can be written as a |Mi + b
. Then
sends this to
a |Mi − b
, while
sends it to
−a |Mi
and this is reflection in
(ii) Suppose we have mirrors M
and M
making an angle of θ:
Then reflection in
then reflection in
is the same as rotating coun-
terclockwise by 2θ.
So we know
Q = −I
is reflection in
then reflection in
. So this is a rotation by 2
where α is the angle between
and |ψ
i, i.e.
sin α = cos β = hx
To prove our second claim that
acts as
, we simply note that if
|ξi ∈ P(x
, then |ξi ⊥ |ψi
and ξ ⊥ |x
i. So both I
and I
fix |ξi.
In fact, Grover’s algorithm is the best algorithm we can achieve.
be any quantum algorithm that solves the unique search
problem with probability 1
− ε
(for any constant
), with
queries. Then
at least O(
N). In fact, we have
T ≥
(1 − ε)
So Grover’s algorithm is not only optimal in the growth rate, but in the
constant as well, asymptotically.
Proof is omitted.
Further generalizations
Suppose we have multiple good items instead, say
of them. We then replace
with I
, where
|xi =
−|xi x good
|xi x bad
We run the same algorithm as before. We let
i =
x good
Then now
is a rotation through 2
in the plane spanned by
sin α = hψ
i =
So for large N, we need
i.e. we have a
reduction over the unique case. We will prove that these
numbers are right later when we prove a much more general result.
What if we don’t know what
is? The above algorithm would not work,
because we will not know when to stop the rotation. However, there are some
tricks we can do to fix it. This involves cleverly picking angles of rotation at
random, and we will not go into the details. | {"url":"https://dec41.user.srcf.net/h/III_M/quantum_computation/3_4","timestamp":"2024-11-11T10:10:11Z","content_type":"text/html","content_length":"266898","record_id":"<urn:uuid:c3876718-8a00-434f-b97f-e6a58cfb1fbe>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00527.warc.gz"} |
Matrix Equations: Linear Combinations
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This lecture introduces the concept of matrix equations as linear combinations of columns, illustrating how to find solutions and interpret them geometrically. It covers the definition of vector
spaces, span, and linear combinations, providing insights into the geometric interpretation of vector spaces. The lecture also explains the rule of 'row-column' for matrix-vector multiplication and
the geometric interpretation of vector spaces. Additionally, it discusses the properties of matrix equations, including the existence of solutions and the geometric interpretation of solutions. The
lecture concludes with examples demonstrating how to find the equation of a plane spanned by given vectors.
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Item #1 of 10
The low-amplitude and high frequency (8-12 cps) waveforms produced during the stage of transition between waking and sleeping are called _________waves.
INCORRECT - Delta waves are much slower (< 4 cps) and characteristic of stage 3 & 4 sleep.
[Review Part 1 of the tutorial] [Review Part 2 (Matching) of the tutorial] | {"url":"https://psych.athabascau.ca/html/Psych289/Biotutorials/20/quiz.cgi?ques=1&c=1","timestamp":"2024-11-13T05:52:15Z","content_type":"text/html","content_length":"4505","record_id":"<urn:uuid:fca1c26d-625e-4cbb-b86d-d00076d0f11e>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00767.warc.gz"} |
Work, Energy
Mechanics: Work, Energy and Power
Work, Energy and Power: Problem Set Overview
There are 20 ready-to-use problem sets on the topic of Work, Energy, and Power. The problems target your ability to use equations related to work and power, to calculate the kinetic, potential and
total mechanical energy, and to use the work-energy relationship in order to determine the final speed, stopping distance or final height of an object.
Work results when a force acts upon an object to cause a displacement (or a motion) or, in some instances, to hinder a motion of a moving object. Three variables are of importance in this definition
- force, displacement, and the extent to which the force causes or hinders the displacement. Each of these three variables find their way into the equation for work. That equation is:
Work = Force • Displacement • Cosine(theta)
W = F • d • cos(Θ)
Since the standard metric unit of force is the Newton and the standard meteric unit of displacement is the meter, then the standard metric unit of work is a Newton•meter, defined as a Joule and
abbreviated with a J.
The most complicated part of the work equation and work calculations is the meaning of the angle theta (Θ) in the above equation. The angle is not just any stated angle in the problem; it is the
angle between the F and the d vectors. In solving work problems, one must always be aware of this definition - theta is the angle between the force and the displacement which it causes. If the force
is in the same direction as the displacement, then the angle is 0 degrees. If the force is in the opposite direction as the displacement, then the angle is 180 degrees. If the force is up and the
displacement is to the right, then the angle is 90 degrees. This is summarized in the graphic below.
Power is defined as the rate at which work is done upon an object. Like all rate quantities, power is a time-based quantity. Power is related to how fast a job is done. Two identical jobs or tasks
can be done at different rates - one slowly or and one rapidly. The work is the same in each case (since they are identical jobs) but the power is different. The equation for power shows the
importance of time:
Power = Work / time
P = W / t
The unit for standard metric work is the Joule and the standard metric unit for time is the second, so the standard metric unit for power is a Joule / second, defined as a Watt and abbreviated W.
Special attention should be taken so as not to confuse the unit Watt, abbreviated W, with the quantity work, also abbreviated by the letter W.
Combining the equations for power and work can lead to a second equation for power. Power is W/t and work is F•d•cos(theta). Substituting the expression for work into the power equation yields P =
F•d•cos(theta)/t. If this equation is re-written as
P = F • cos(Θ) • (d/t)
one notices a simplification which could be made. The d/t ratio is the speed value for a constant speed motion or the average speed for an accelerated motion. Thus, the equation can be re-written as
P = F • v • cos(Θ)
where v is the constant speed or the average speed value. A few of the problems in these problem sets will utilize this derived equation for power.
Mechanical, Kinetic and Potential Energies
There are two forms of mechanical energy - potential energy and kinetic energy.
Potential energy is the stored energy of position. In this set of problems, we will be most concerned with the stored energy due to the vertical position of an object within Earth's gravitational
field. Such energy is known as the gravitational potential energy (PE[grav]) and is calculated using the equation
PE[grav] = m•g•h
where m is the mass of the object (with standard units of kilograms), g is the gravitational field constant (9.8 N/kg) and h is the height of the object (with standard units of meters) above some
arbitraily defined zero level (such as the ground or the top of a lab table in a physics room).
Potential energy can also be stored in a mass-spring system. This is known as elastic potential energy (PE[elastic]). As a spring is stretched some distance x from its equilibrium position, potential
energy is stored in the spring-mass system. This type of potential energy is calculated using the equation
PE[elastic] = 1/2• k • x^2
where x is the amount of stretch from the equilibrium position and k is the spring constant of the spring.
Kinetic energy is defined as the energy possessed by an object due to its motion. An object must be moving to possess kinetic energy. The amount of kinetic energy (KE) possessed by a moving object is
dependent upon mass and speed. The equation for kinetic energy is
KE = 0.5 • m • v^2
where m is the mass of the object (with standard units of kilograms) and v is the speed of the object (with standard units of m/s).
The total mechanical energy possessed by an object is the sum of its kinetic and potential energies.
Work-Energy Connection
There is a relationship between work and total mechanical energy. The relationship is best expressed by the equation
TME[i] + W[nc] = TME[f]
In words, this equation says that the initial amount of total mechanical energy (TME[i]) of a system is altered by the work which is done to it by non-conservative forces (W[nc]). The final amount of
total mechanical energy (TME[f]) possessed by the system is equivalent to the initial amount of energy (TME[i]) plus the total work done by these non-conservative forces (W[nc]).
The mechanical energy possessed by a system is the sum of the kinetic energy and the potential energy. Thus the above equation can be re-arranged to the form of
KE[i] + PE[i] + W[nc] = KE[f] + PE[f]
0.5 • m • v[i]^2 + m • g • h[i] + F • d • cos(Θ) = 0.5 • m • v[f]^2 + m • g • h[f]
The work done to a system by non-conservative forces (W[nc]) can be described as either positive work or negative work. Positive work is done on a system when the force doing the work acts in the
direction of the motion of the object. Negative work is done when the force doing the work opposes the motion of the object. When a positive value for work is substituted into the work-energy
equation above, the final amount of energy will be greater than the initial amount of energy; the system is said to have gained mechanical energy. When a negative value for work is substituted into
the work-energy equation above, the final amount of energy will be less than the initial amount of energy; the system is said to have lost mechanical energy. There are occasions in which the only
forces doing work are conservative forces (sometimes referred to as internal forces). Typically, such conservative forces include gravitational forces, elastic or spring forces, electrical forces and
magnetic forces. When the only forces doing work are conservative forces, then the W[nc] term in the equation above is zero. In such instances, the system is said to have conserved its mechanical
The proper approach to a work-energy problem involves carefully reading the problem description and substituting values from it into the work-energy equation listed above. Inferences about certain
terms will have to be made based on a conceptual understanding of kinetic and potential energy. For instance, if the object is initially on the ground, then it can be inferred that the PE[i] is 0 and
that term can be canceled from the work-energy equation. In other instances, the height of the object is the same in the initial state as in the final state, so the PE[i] and the PE[f] terms are the
same. As such, they can be mathematically canceled from each side of the equation. In other instances, the speed is constant during the motion, so the KE[i] and KE[f] terms are the same and can thus
be mathematically canceled from each side of the equation. Finally, there are instances in which the KE and or the PE terms are not stated; rather, the mass (m), speed (v), and height (h) is given.
In such instances, the KE and PE terms can be determined using their respective equations. Make it your habit from the beginning to simply start with the work and energy equation, to cancel terms
which are zero or unchanging, to substitute values of energy and work into the equation and to solve for the stated unknown.
Habits of an Effective Problem-Solver
An effective problem solver by habit approaches a physics problem in a manner that reflects a collection of disciplined habits. While not every effective problem solver employs the same approach,
they all have habits which they share in common. These habits are described briefly here. An effective problem-solver...
• ...reads the problem carefully and develops a mental picture of the physical situation. If needed, they sketch a simple diagram of the physical situation to help visualize it.
• ...identifies the known and unknown quantities in an organized manner, often times recording them on the diagram itself. They equate given values to the symbols used to represent the
corresponding quantity (e.g., m = 1.50 kg, v[i] = 2.68 m/s, F = 4.98 N, t = 0.133 s, v[f] = ???).
• ...plots a strategy for solving for the unknown quantity. The strategy will typically center saround the use of physics equations and is heavily dependent upon an understanding of physics
• ...identifies the appropriate formula(s) to use, often times writing them down. Where needed, they perform the needed conversion of quantities into the proper unit.
• ...performs substitutions and algebraic manipulations in order to solve for the unknown quantity.
Additional Readings/Study Aids:
The following pages from The Physics Classroom tutorial may serve to be useful in assisting you in the understanding of the concepts and mathematics associated with these problems.
Watch a Video
We have developed and continue to develop Video Tutorials on introductory physics topics. You can find these videos on our YouTube channel. We have an entire Playlist on the topic of Work and Energy. | {"url":"https://staging.physicsclassroom.com/calcpad/energy/Equation-Overview","timestamp":"2024-11-02T22:25:45Z","content_type":"application/xhtml+xml","content_length":"210747","record_id":"<urn:uuid:f4989306-b6f5-4779-a708-f4395630c2ed>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00825.warc.gz"} |
How about a closet Tool?
My closets are almost always 24" deep. There are a multiple, but finite number of combinations of shelves, drawers, single/double hanging etc. A simple tool that inserts a 3d BIM closet in-fill
object would be really useful to me.
11 answers to this question
• 0
@Andrew Pollock What about making a 3D symbol that employs the cabinet plug-ins? This is what I do for my Doctor Office Exam Rooms. Two basic symbols that I customize for each suite of exam rooms
depending on the project.
• 0
Guest Frank Brault
• 0
In a large home, there will be many closets all of slightly different sizes. Very few exact repeats. Symbols wouldn't be efficient. I wish there was an un-styled closets symbol that worked like an
un-styled window type.
• 0
Have a look at the InteriorCAD XS plugin - infinite possibilities and you can save a standard design as a symbol and it remains parametric when reused from your library. It’s awesome for kitchens,
bathrooms, wardrobe, built in units and office cabinetry.
Edited by Aspect_Design
• 0
7 hours ago, Andrew Pollock said:
In a large home, there will be many closets all of slightly different sizes. Very few exact repeats. Symbols wouldn't be efficient. I wish there was an un-styled closets symbol that worked like
an un-styled window type.
That's where plug-in style bridge the gap well.
• 0
@Aspect_Design I showed IC to a client a few years ago & for some reason, (god only knows why) they bought the software that their CAM dealer suggests (Bessiework maybe?) Anywho, I was very
impressed with IC, both how it works with VW & what it adds. It would be a real improvement if VW came with IC as a module like RW or Landscape.
@Andrew Pollock I get that every closet is different; I worked on a home a few years back that had his & hers walk-ins that rival the main floor of my house in area. I also get that a PIO would
likely make things easier, but if @Aspect_Design 's suggestion of using IC is overkill, you may want to consider a symbol or three. For example, a closet with interior dims of 600x1850 (2'x6') with
generic walls and a generic door that includes several PIO's such as the Closet Rod, Utility Cabinet, Wall & Base Cabinet. In each room one could place the closet with all the classes turned on then
as the symbol requires amending turn the symbol on the page to a Group to edit. Not as elegant as IC but, may take some of the drudgery out of the process.
• 0
@Jim SmithThe NZ distro has taken VW+IC a step further with QuantumCad which does away with the need for Biesseworks, microvellum etc...
• 0
@Aspect_Design Sheesh! between that & Windor it seems you have a better VW Experience than we do in the Great White North! I spent days of un-billable hours attempting to build a pathway to
Biesseworks for this client including working with the Biesseworks people.
Biesseworks is just plain evil.
• 0
@Jim Smith The good thing about Quantumcad is it isn't region locked like - might be worth an email 😉
• 0
• 0
@Jim Smith / @Aspect_Design
Would be interested to hear how you get on. It looks really interesting but there isn't much information or videos showing how it actually works. | {"url":"https://forum.vectorworks.net/index.php?/topic/71514-how-about-a-closet-tool/","timestamp":"2024-11-05T00:51:34Z","content_type":"text/html","content_length":"181077","record_id":"<urn:uuid:81077580-4eb5-4bca-991c-c1c9828dbbb9>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00032.warc.gz"} |
The Essentials of Geometry (solid)
Webster Wells
From inside the book
Results 1-5 of 18
Page 319 Webster Wells. BOOK VIII . THE CYLINDER , CONE , AND SPHERE . DEFINITIONS . 540. A cylindrical ... circular cylinder is a cylinder whose base is a circle . A plane is said to be tangent to a
cylinder 319 THE CYLINDER, Cone, and SPHERE.
Page 324 ... CONE . DEFINITIONS . 553. A conical surface is a surface generated by a moving straight line , which constantly ... circular cone is a cone whose base is a circle . The axis of a
circular cone is a straight line 324 SOLID GEOMETRY ...
Page 325 ... circular cone is a circular cone whose axis is per- pendicular to its base . A frustum of a cone is a portion of a cone included between the base and a plane parallel to the base . The
base of the cone is called the lower base , and ...
Page 326 ... circular cone made by a plane parallel to the base is a circle . B Given A'B'C ' a section of circular cone S - ABC , made by a plane to the base . To Prove A'B'C ' a O. Proof . Draw
axis OS , intersecting plane A'B'C ' 326 SOLID ...
Page 327 ... circular cone passes through the centre of every section parallel to the base . PROP . VIII . THEOREM . 559. A plane drawn through an element of the lateral sur- face of a circular cone
and a tangent ... cone . ( Prove THE CONE . 327.
Bibliographic information | {"url":"https://books.google.com.jm/books?id=6XkwAQAAMAAJ&q=circular+cone&dq=editions:HARVARD32044097046064&output=html_text&source=gbs_word_cloud_r&cad=5","timestamp":"2024-11-04T07:19:50Z","content_type":"text/html","content_length":"61462","record_id":"<urn:uuid:fa7b317e-baea-4f80-9d1b-d6a246b105e5>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00354.warc.gz"} |
Programming Assignment # 1 Heuristic Search
On page 103 of your textbook, you will find a diagram of the 8-puzzle. Your
work for this assignment is to implement several search techniques to find the shortest
path between the start state and the goal state.
Programming Assignment
Here is the goal state and four start states for the 8-puzzle.
Goal: Easy: Medium: Hard: Worst:
Implement the following search algorithms and test them on the start states and goal
state shown above. Only A* needs to detect duplicate states and eliminate them from the
remainder of the search.
1. A* search using the heuristic function f*(n) = g(n) + h*(n), where h*(n) is the
number of tiles out of place (not counting the blank).
2. A* search using the Manhattan heuristic function.
3. Iterative deepening A* with the Manhattan heuristic function.
4. Depth-first Branch and Bound with the Manhattan heuristic function.
When defining the successor function, it is helpful to define the actions in terms of
the empty tile, that is, the four actions are moving the empty tile right, down, left, and
up. Consider the actions in this order in your implementation.
If your algorithms take too long or too much memory to find any solution, your
implementation may be inefficient. Consider optimizing it. If you still can’t find solutions
within a reasonable amount of time, enforce a time limit, say 30 minutes, on your
algorithm and specify it in your report.
Problem Analysis:
For all the algorithms, include in your report a table the number of nodes expanded,
the total time required to solve the puzzle, and the sequence of moves in the optimal
solution. For Depth-first Branch and Bound, also record the time the optimal solution is
found (typically not the same as the finish time).
Besides, answer the following questions:
1. What is the number of possible states of the board?
2. What is the average number of possible moves from a given position of the board?
3. Estimate how many moves might be required for an optimal (minimum number of
moves) solution to a “worst-case” problem (maximum distance between starting and
goal states). Explain how you made your estimate (Note this is an open-ended
question; any logical answer may suffice).
4. Assuming the answer to question #2 is the “average branching factor” and a depth as
in the answer to question #3, estimate the number of nodes that would have to be
examined to find an answer by the brute force breadth-first search.
5. Assuming that your computer can examine one move per millisecond, would such a
blind-search solution to the problem terminate before the end of the semester?
6. The “worst” example problem given above is actually one of the easiest for humans to
solve. Why do you think that is the case? What lessons for AI programs are suggested
by this difference between human performance and performance of your search
7. Compare A*, DFBnB, and IDA* and discuss their advantages and disadvantages.
Send a single .zip file named LastName.FirstInitial.HW# containing the following
to the instructor:
1) Well commented code;
2) A readme file explaining how to compile and run your code;
3) A written report explaining your experimental results and analysis.
Also submit the report in hardcopy before the class.
Academic honesty: Please do your own work; do not give or receive any assistance in
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Find the Number 360 among 368 in 10 Secs - EduViet Corporation
Find the Number 360 among 368 in 10 Secs
Find the number 360 out of 368 numbers in 10 seconds
You can test your brain with a variety of puzzles of varying difficulty, battle in your knowledge arena, and win by finding the answers to start your fun-filled path. Embark on an enjoyable journey
to test your brain power with a variety of puzzles of varying difficulty, compete in the arena of your knowledge, and win by uncovering the answers. For your convenience, the answer is at the end.
To make it easier for you to understand, the answers are provided at the end of each puzzle. Now, focus the image a little; your incredible question is shown below. Take a few seconds to jot down
this question and think about it. You can keep a pen and paper nearby, or use an iPad if you prefer to solve this puzzle. The answer might be a little challenging, but if you can figure out the trick
in the question, you can solve it in under a minute. Now, focus on the image; your brain activation question can be found in the image below.
You are watching: Find the Number 360 among 368 in 10 Secs
Find the number 360 out of 368 numbers in 10 seconds – solution
How is the question stated in the above paragraph? For some of them, this may be difficult. But if you discover the trick in the above question, your eyes will not see the difficulty. If you haven’t
found the answer yet, try the solution to the above question again. oh! It’s okay.. If you still can’t find the answer after trying a lot, don’t worry. Calm down; the answer is given at the end of
this article.
Now you can understand the essence of this answer. You are smart enough and smart enough to have figured out the answer. Then, cross-check your answers with the solutions we have given in this
passage. If you find the correct answer, you can clap once.
IQ test: If 9+1=91, 8+2=75, then 7+3=?
The sequence begins: 9 plus 1 gives 91, 8 plus 2 gives 75. Now, the mystery deepens: When 7 plus 3 follows this pattern, what is the result?
Breaking down this pattern: In the first equation, 9 plus 1 neatly works out to 9 squared plus 1 plus 9, which gives us 91. Continuing this pattern, for 7 plus 3, the calculation is 7 squared plus 3
plus 9, which results in an answer of 61.
Math IQ Test: Solve 63÷3×5+9-2
Start the brainteaser math IQ test using the equation 63 ÷ 3 x 5 + 9 – 2. Your task is to carefully follow the order of operations and calculate the final result.
See more : Brain Teaser: What Comes Next in this Series 48:122::168:?
To solve this equation, follow the order of operations. Start with division: 63 ÷ 3 equals 21. Then, continue the multiplication: 21 x 5 equals 105. Add 9 to 105 to get 114, and finally subtract 2
from 114 to get 112. Therefore, equation 63 ÷ 3 x 5 + 9 – 2 equals 112.
Speed math test: 55÷5x(2+7)=?
Dive into the world of brainteaser math speed tests using the equation 55 ÷ 5 x (2 + 7). Your task is to quickly perform calculations and determine the final result.
To solve this equation efficiently, start with the parentheses: 2 + 7 equals 9. Then divide: 55 ÷ 5 equals 11. Finally, multiply 11 by 9 to get the result 99. Therefore, the equation 55 ÷ 5 x (2 + 7)
equals 99.
IQ test: What if 3+4=7, 5+3=15, 2+6=?
The sequence begins: 3 added to 4 to get 7, then the pattern continues: 7 added to 5, 3 to 15. Now, here comes the challenge: when 2 plus 6 are combined into this sequence, what is the result?
In a sequence, the value of each number is the sum of the previous result, the number itself, and the following numbers. In the first equation, 3 + 4 equals 7. Following the pattern, for 5 + 3, the
result is 15. Extending this rule, for 2 + 6, the result is 23.
Math IQ Test: Solving 49÷7×6+2-5
Take a fun brainteaser math IQ test using the equation 49 ÷ 7 x 6 + 2 – 5. Your task is to decipher the order of operations and calculate the final result. To solve this equation, follow the order of
operations. Start with division: 49 ÷ 7 equals 7. Then, continue the multiplication: 7 x 6 equals 42. Add 2 to 42 to get 44, and finally subtract 5 from 44 to get 39. Therefore, equation 49 ÷ 7 x 6 +
2 – 5 equals 39.
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Light, Shadow, and a World of Nails
Rendering 3D Worlds
Making a computer make images isn’t easy at the best of times, but 3D graphics are a substantial step up in difficulty. Balancing speed with quality is a challenge, especially when the objects we
want to render are complex and detailed.
Raster Graphics
The leading realtime rendering method (i.e. for video games) is raster graphics, which stores 3D objects as a collection of triangles, then projects those triangles onto a 2D screen. We can use the
relative depths of triangles to cull any obscured triangles, greatly speeding up rendering time, and we can use the normal vectors of the triangles to shade the resulting image.
However, effects like soft shadows, reflections, and antialiasing (smoothing jagged edges) are prohibitively difficult to implement with raster graphics, and the complexity of the objects we can
render is limited by the number of triangles that make them up.
Ray Tracing
An alternative to raster graphics is ray tracing, which more closely matches our physical world. We emit “light” rays from the camera and compute their intersection with the entire scene. We can then
continue with the reflected ray, intersecting and bouncing it until it reaches a light source. In effect, we simulate only the light rays that eventually reach the camera.
Unfortunately, computing those intersections is extraordinarily taxing, and although fancy effects are possible, they are also quite expensive to implement. Outside of limited use in modern video
games with cutting-edge hardware, ray tracing is still constrained to non-realtime rendering (e.g. 3D movies).
Distance Estimators
How can we get the effects of ray tracing without having to compute those intersections?
We could avoid having to calculate the geometry of an object if we instead knew a distance estimator, or DE, for it: a function that takes in a point in $\mathbb{R}^3$ and outputs (no more than) the
minimum distance to the surface of the object, with negative outputs indicating the point is inside.
Sphere of radius $r$ at the origin: $d(p) = |p| - r$
Plane $z = 0$: $d(p) = |p_z|$
Cube of side length $2r$ at the origin: $d(p) = \max\{|p_x| - r, |p_y| - r, |p_z| - r\}$
Union of objects with DEs $d_1$ and $d_2$: $d(p) = \min\{d_1(p), d_2(p)\}$
Intersection of objects with DEs $d_1$ and $d_2$: $d(p) = \max\{d_1(p), d_2(p)\}$
Set difference of objects with DEs $d_1$ and $d_2$: $d(p) = d_1(p) - d_2(p)$
Ray Marching
Given a camera position $c \in \mathbb{R}^3$ and a unit vector $v \in T_c\mathbb{R}^3$, we can “march” a ray from $c$ in the direction of $v$.
Since there are no objects within $d(c)$ of the camera, we can move to $p_1 = c + d(c)v$ without marching inside an object.
We can then march safely to $p_1 + d(p_1)v$, and repeat until the DE is below some threshold $\varepsilon$ or above some clip distance — lower values of $\varepsilon$ will result in a more accurate
Repeating this process for every pixel in a grid, using a different direction for each, we can render an image by coloring pixels based on whether they reach the object.
What else can we do with DEs? Everything might look a nail to our hammer, but this world might genuinely be made of nails.
Once we hit an object, we can find the normal vector to the surface by computing the gradient of the DE numerically. By taking the normal vector’s dot product with a vector pointing to a light
source, we can approximate shading.
We can also slightly darken the color based on the number of marches taken to simulate ambient occlusion, and by blending the final color with the sky color based on distance from the camera, we can
simulate fog.
Shadows and Reflections
Once we hit an object, we can use the surface normal to bump the position back outside of the object, then turn and march straight toward the light source. If we hit something else along the way, we
darken the pixel.
Amazingly, we also get soft shadows for free with this method. If we don’t hit anything on the way to the light source, we darken the pixel based on how close it came to hitting something.
Reflections are even simpler: by reflecting the direction through the surface normal, we can start a new march and mix it with our current color.
Folding Space
The real power of ray marching comes from the ways in which we can manipulate DEs to effectively cheat on computation time while the ray is still traveling.
For example, if we modulo the position by a constant in all three directions at every step, the effect is to render infinitely many spheres at no additional time complexity, and we can invert the
sphere DE to render a room-like space instead.
Let’s render a cube fractal that’s reminiscent of the free group on three elements. We’ll begin with a DE for a cube, and based on which of the six sides we’re closest to, we’ll union it with a
smaller cube on that side. Repeating this produces $O(6^n)$ cubes in $O(n)$ time.
By carving out cubes instead of adding them, we can also easily render the Menger Sponge.
Curved Space
By changing the paths along which light travels, we can simulate a different curvature of space.
With (a lot) more work, we can render scenes in the eight Thurston geometries on 3-manifolds.
Ray marching is an excellent tool for rendering anything in 3D, particularly when it involves complicated mathematical objects.
More broadly, it’s an excellent example of extracting an incredible amount of detail and quality from a limited amount of information.
Thank You! | {"url":"https://cruzgodar.com/slides/light-shadow-world-of-nails/","timestamp":"2024-11-09T02:37:05Z","content_type":"text/html","content_length":"10448","record_id":"<urn:uuid:344952d3-ae46-41a3-8ffd-16654c576708>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00709.warc.gz"} |
Renaud Raquépas
I am a Courant Instructor in the Mathematics Department of the Courant Institute at New York University hosted by Professor Lai-Sang Young. Prior to this, I was a postdoctoral reasearcher in
mathematics at CY Cergy Paris Université (Laboratoire AGM), working with Professor Armen Shirikyan.
I completed my PhD in 2020 at McGill University (Department of mathematics and statistics) and Université Grenoble Alpes (Institut Fourier), under the joint supervision of Professors Vojkan Jakši&
cacute; and Alain Joye.
Find more about me on the About page or by visiting my ORCID, arXiv and Research Gate profiles.
Research Interests
I study mathematical physics, with emphasis on time-dependent aspects of statistical mechanics and entropy production, in both quantum and classical systems. Relevant mathematical tools to study such
problems include:
• probability theory (large deviations, stochastic differential equations);
• dynamical systems and ergodic theory (recurrence, mixing, theory of C*-algebras, random dynamical systems);
• operator theory (spectra, resolvents, perturbation theory, one-parameter semigroups).
Courant Dynamical Systems Seminar
I am co-organizing the Courant Dynamical Systems Seminar, which takes place on Wednesday afternoon at the Courant Institute.
Preprints and publications
Links to archived PDFs and more details can be found on the Research page.
• N. Barnfield, R. Grondin, G. Pozzoli and R. Raquépas. On the Ziv–Merhav theorem beyond Markovianity II.
• Z. Wu, R. Raquépas, J. Xin and Z. Zhang Computing large deviation rate functions of entropy production for diffusion processes in the vanishing-noise limit and high dimensions by an interacting
particle method.
• N. Barnfield, R. Grondin, G. Pozzoli and R. Raquépas. On the Ziv–Merhav theorem beyond Markovianity I. To appear in Can. J. Math.
• N. Cuneo and R. Raquépas. Large deviations of return times and related entropy estimators on shift spaces. Commun. Math. Phys. 405, Article 135 (2024).
• R. Raquépas. The large-time and vanishing-noise limits for entropy production in nondegenerate diffusions. Ann. Inst. Henri Poincaré Probab. Stat. 60, 431–462 (2024).
• G. Cristadoro, M. Degli Esposti, V. Jakšić and R. Raquépas. On a waiting-time result of Kontoyiannis: mixing or decoupling? Stoch. Proc. Appl. 166, 104222 (2023).
• R. Raquépas. A gapped generalization of Kingman's subadditive ergodic theorem. J. Math. Phys. 64, 062702 (2023).
• G. Cristadoro, M. Degli Esposti, V. Jakšić and R. Raquépas. Recurrence times, waiting times and universal entropy production estimators. Lett. Math. Phys. 113, Article 19 (2023).
• S. Andréys, A. Joye and R. Raquépas. Fermionic walkers driven out of equilibrium. J. Stat. Phys. 184, Article 14 (2021).
• V. Nersesyan and R. Raquépas. Exponential mixing under controllability conditions for SDEs driven by a degenerate Poisson noise. Stoch. Proc. Appl. 138, 26–55 (2021).
• R. Raquépas. On Fermionic walkers interacting with a correlated structured environment. Lett. Math. Phys. 110, 121–145 (2020).
• T. Benoist, A. Panati and R. Raquépas. Control of fluctuations and heavy tails for heat variation in the two-time measurement framework. Ann. Henri Poincaré 20, 631–674 (2019).
• R. Raquépas. A note on Harris’ ergodic theorem, controllability and perturbations of harmonic networks. Ann. Henri Poincaré 20, 605–629 (2019).
• E. P. Hanson, A. Joye, Y. Pautrat and R. Raquépas. Landauer’s principle for trajectories of repeated interaction systems. Ann. Henri Poincaré 19, 1939–1991 (2018).
• E. P. Hanson, A. Joye, Y. Pautrat and R. Raquépas. Landauer’s principle in repeated interaction systems. Commun. Math. Phys. 349, 285–327 (2017). | {"url":"https://renaudraquepas.github.io/","timestamp":"2024-11-05T01:23:20Z","content_type":"text/html","content_length":"7614","record_id":"<urn:uuid:45de2a56-3015-4adc-bb55-a63cd2e4e506>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00247.warc.gz"} |
How to do Big O with Recursion | Joe McCann
A Victim of my Own Design
Prior, we have been working on very nice and fun programs that have been super straightforward to do counting on, but many algorithms that are taught in classes are self referential and recursive.
How can we count operations when our operations refer to themselves?? In case my late night rambling isn’t clear, consider this simple code to calculate Fibonacci numbers, which are given by the
formula $$ F_n = F_{n-1}+F_{n-2} $$ with initial conditions $F_0=0, F_1=1$
def fib(n):
# return the nth fibbonacci number
if n <= 1:
# base cases n=0, n=1
return n
return fib(n-1)+fib(n-2)
This code will give us the sequence we know and love $0,1,1,2,3,5,8,13,\ldots$. Now lets count some operations to see the runtime of this code. In the event that $n=0$ or $n=1$ we get that $T(n)=2$
by our base case. However, if $n>1$, we find that our operations function has $1$ comparison, $3$ arithmetic, $1$ return, and $2$ function calls. But the two function calls are to the function itself
which creates this bizzare situation of $$ T(n) = T(n-1)+T(n-2)+7. $$ How can we try to classify this into some $\mathcal{O}$? Can we come up with a clean expression for it?
The answer is yes, but we need to learn a new technique.
Recurrance Relations
Whenever we have a function that relates back to itself in some previous case, we call this a recurrance relation. For example, the Fibonacci sequence itself is a recurrance relation, as we say that
$F_n=F_{n-1}+F_{n-2}$, which shows us that in order to solve for a value of $F_n$ we need to know what comes before it. These equations also come with initial conditions that actually change how the
sequence unfolds over time.
Now lets actually show how to solve these sorts of equations.
First Order Linear Equations
Before we get into the actual example that we want, lets consider some easier cases of recurrance relations that we can easily solve by patter recognition. First, lets consider the example
Example: Let $f_n = 2f_{n-1}$ with $f_0=1$. Find an equation in terms of $n$ for $f_n$
Well let’s plug in some values and see what happens $$ f_0 &= 1 \\ f_1 &= 2f_0 = 2 \\ f_2 &= 2f_1 = 4 \\ f_3 &= 2f_2 = 8 \\ &\vdots $$ Hopefully you can see the pattern here in that every iteration
we are adding a factor of $2$ which gives us that $f_n=2^n$. But what if we weren’t multiplying by $2$, and what if we had a different initial condition?
This type of recurrance relation is called a first order relation because it only goes back one step into $n-1$, and since we are continually multiplying by the same number over and over again, we
can assume that our solution probably should look something like $f_n=Cr^n$ where $C$ is some constant that depends on the initial condition. Why we can make this assumption should be fairly obvious
in this first order case (multiply by the same number $n$ times gives $r^n$), but I will justify it later on with more advanced, abliet uneccessary for CS, math.
Let’s consider a more general example
Example: Let $f_n = kf_{n-1}$ with initial condition $f_0$. Find an equation in terms of $n$ for $f_n$
Well lets assume that $f_n = Cr^n$ where $C, r$ are some constants that solve the problem. Lets plug these in to the equations, and notice that we can say that $$ f_n = Cr^n $$ and that $$ f_{n-1} =
Cr^{n-1}. $$ Plugging these into our recurrance relation we can solve for $r$. $$ f_n &= kf_{n-1} \\ \implies Cr^n &= kCr^{n-1} \\ \implies \frac{Cr^n}{Cr^{n-1}} &= \frac{kCr^{n-1}}{Cr^{n-1}} \\ \
implies r &= k. $$ And just like that we know that $f_n = Ck^n$ by solving for $r$! Now in order to find $C$ we just plug in our initial conditions $$ f_0 = Ck^0 = C, $$ which gives us our final
solution of $$ f_n = f_0\cdot k^n. $$
Ok thats cool, but also seems pretty simple, lets move on to something more “challenging”
Second Order Linear Equations
The vast majority of the time, we will not have only one term in our recurrance relation, so we need to have a method to solve higher order equations that go back more steps than just one at every
point. Lets start by working with the Fibonacci sequence
Example: Define the Fibonacci sequence as $F_n = F_{n-1} + F_{n-2}$ with initial conditions $F_0=0, F_1=1$. Find an equation for $F_n$ in terms of $n$.
To begin, we are going to make the same assumption we did in the first order case in which we assume that $F_n = Cr^n$. This might seem like a weird assumption but I promise it will work. When we
plug this in to our equation with $F_{n-1}=Cr^{n-1}$ and $F_{n-2}=Cr^{n-2}$ then we get $$ Cr^n &= Cr^{n-1} + Cr^{n-2} \\ \implies r^2 &= r + 1 \\ \implies r^2 - r - 1 &= 0 $$ Woah now it turns our
that we have a second degree polynomial for $r$ instead of just a number, which means that we actually will have two solutions to find! In fact, when we apply this process to an $n$ order equation,
its exactly the same except we will end up with an $n$ degree polnomial!
To solve for $r$, plug it in to the quadratic formula (or wolfram alpha) to find that $$ r_1 = \frac{1+\sqrt{5}}{2}, r_2 = \frac{1-\sqrt{5}}{2}. $$ In fact, $r_1$ is normally notated as $\phi$ and
has a special name, the golden ratio! Also known as the most overrated thing in all of mathematics. Anyway, lets solve for constants, we know that $$ F_0 = 0 = C_1 + C_2 \implies C_2 = -C_1 $$ and
that $$ F_1 = 1 = C_1r_1 + C_2r_2 = C_1r_1-C_1r_2= C_1(r_1-r_2). $$ Solving for $C_1$ we get that $$ C_1 = \frac{1}{r_1-r_2} = \frac{1}{\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}} = \frac{1}{\sqrt
{5}}. $$ As such we also know that $C_2 = -\frac{1}{\sqrt{5}}$. Adding these into our final equation we get that we can express the Fibonacci numbers as $$ F_n = \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt
{5}}{2}\right)^n-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n. $$
This is so cool because (in theory) it means that if you want to express any particular Fibonacci number, you can just plug in that number into this equation and you’ll get the corresponding $F_n$.
Also crazy because every one of these terms are irrational numbers and yet when you add them together the irrational parts cancel out and you always end up with an integer!
Note that when you are dealing with an recurrance relation of order $n$, you are guaranteed $n$ linearly independent solutions^1. Linearly independent means that the solutions are not the same with
respect to a constant, so like $3\cdot 2^n$ and $2^n$ are not independent. This will be important later.
Why Can We Assume $r^n$?
This is going to be a semi-technical discussion into why even in higher order equations we can assume $r^n$ and then get a corresponding solution. Also for the purposes of discussion, we will be
using our previous example of the Fibonnachi numbers, but this also generalizes to higher order linear equations.
Take note of how we specifically said linear equations. The linear here actually matters a lot, as linear means that for any two recurrance terms such as $F_{n-1}, F_{n-2}$, we are only adding them
or multiplying them by a constant. So something like $$ f_n = 5f_{n-1}-2f_{n-2}+\frac{1}{2}f_{n-6} $$ would be considered a valid linear equation that we can solve using the methods described in this
page, however something like $$ f_n = f_{n-1}f_{n-2} $$ would be non-linear and thus full of pain when trying to describe its behavior over time.
Not only this, but notice that we don’t just have one equation, we actually have two! In order to see this, let me show a function in Python that will calculate Fibonacci numbers
def fib(n):
if n == 0 or n == 1:
return n
old = 0
new = 1
for i in range(n-1):
temp = new
new = new + old
old = temp
return new
Yes, we have the normal, but we also have an equation to set new to the sum of the previous new and old, and the old to the previous new. We can refer to these in terms of $f_n, f_{n-1}, f_{n-2}$. So
writing this out we could say $$ f_n &= f_{n-1}+f_{n-2} \\ f_{n-1} &= f_{n-1}. $$ Heres the crazy thing, when we have a system of coupled linear equations, we can use the tools of linear algebra.
Converting these equations into a matrix we get the system $$ \begin{bmatrix} f_n \\ f_{n-1}\end{bmatrix} = \begin{bmatrix} 1&1\\ 1&0 \end{bmatrix}\begin{bmatrix} f_{n-1} \\ f_{n-2}\end{bmatrix}. $$
If we notate this in matrix equation form, then we get $$ \vec{f}n = A\vec{f}{n-1} $$ where $\vec{f}n$ is the vector containing $f_n, f{n-1}$ and $A$ is a matrix containing information on the
recurrance relation. Well at this point this looks literally identical to the first order equation, so we know that our final solution would end up being $$ \vec{f}_n = A^n\vec{f}_0. $$
Ok, like but why does this mean that we can solve using $r^n$? Well remember from linear algebra when we could diagonalize a matrix into a $PDP^{-1}$, where $D$ is a diagonal matrix of the
eigenvalues of $A$? Well if not its ok but if you do that you get that $$ \vec{f}_n = P\begin{bmatrix}r_1^n & 0 \\ 0 & r_2^n\end{bmatrix}P^{-1}\vec{f}_0, $$ and if you expand this out for the
particular values of $P,P^{-1}$ you’ll get an equation for $f_n$ in terms of powers of $r_1,r_2$!!
Now this assumes that all the roots are unique and the matrix is diagonalizable though…
Repeated Roots in Recurrance Relations
Consider the following piece of code for computing a factorial
def factorial(n):
if n == 0:
return 1
return factorial(n-1)
Cutting to the chase, this function has operations $T(n) = T(n-1) + 4$ with $T(0)=2$. How do we solve this equation with that constant term inside of there? What we are going to do is employ a super
clever trick, and subtract the equation by the other equation $T(n-1)=T(n-2)+4$ to cancel out the terms. We get $$ T(n)-T(n-1) &= \left(T(n-1)+4\right)-\left(T(n-2)+4\right)\\ \implies T(n)-T(n-1) &=
T(n-1) - T(n-2) \\ \implies T(n) &= 2T(n-1) - T(n-2) $$ and this is a second order recurrance relation that we can solve. Note that we now need a second initial condition which we get as $T(1)=6$ by
plugging in for $n=1$. Since its a second order equation we expect $2$ linearly independent solutions, and solving this recurrance relation we get to assume $T(n)=Cr^n$ giving us $$ Cr^n &= 2Cr^{n-1}
- Cr^{n-2} \\ \implies r^2 &= 2r - 1 \\ \implies (r-1)^2 &= 0 $$ which gives us that $r=1$. We can’t just have this though cause we need two independent solutions, also $C\cdot 1^n$ clearly is not
the answer. As such, in order to get a second independent solution, we multiply our repeated root by $n$^2 to give us the roots $$ T(n) = C_1\cdot 1^n + C_2 n\cdot 1^n = C_1 + C_2 n. $$ In general,
if a root is repeated $k$ times, you would multiply it by $1,n,n^2,n^3,\ldots, n^{k-1}$. Anyway, the fact that this is our answer actually is pretty obvious when you plug values in to our initial
recurrance relation, as we’re just adding $4$ every single time. Solving for our constants we get $$ T(n) = 4n+2, $$ which is $T(n)\in\Theta(n)$ as should be expected for this algorithm.
Note that you can even have complex roots depending on the values of your recurrance relation! Nothing is off the table 😈
Solving for $T(n)$ of fib(n)
Now we have enough to solve for the recurrance relation shown at the start of this page for Fibonacci numbers. $$ T(n) = T(n-1)+T(n-2)+7, $$ which we subtract both sides by $T(n-1)=T(n-2)+T(n-3)+7$
to get rid of the constant to get $$ T(n)-T(n-1) = \left(T(n-1)+T(n-2)+7\right) - \left(T(n-2)+T(n-3)+7\right) $$ which simplified down gives us $$ T(n) = 2T(n-1) - T(n-3) $$ with additional initial
condition $T(2)=0+1+7=8$. Assuming that $T(n)=Cr^n$ and simplifying we get that $$ r^3 = 2r^2 - 1 \implies r^3-2r^2+1=0. $$ For this polynomial we get our solution is $r_1=1$, $r_2=\frac{1+\sqrt{5}}
{2}$, and $r_3=\frac{1-\sqrt{5}}{2}$. This means that our solution is $$ T(n)=C_1 + C_2\left(\frac{1+\sqrt{5}}{2}\right)^n + C_3\left(\frac{1-\sqrt{5}}{2}\right)^n. $$ We could very much solve for
the coefficients at this point, but it would be soooooo annoying, so we’ll skip it and analyze the equation. Since $C_1$ is a constant and $-1 < \frac{1-\sqrt{5}}{2} < 0$, we can see that as $n$ gets
very very large, the only term that matters is $C_2\left(\frac{1+\sqrt{5}}{2}\right)^n=C_1\phi^n$ since $\phi$ is the golden ratio. This means that $$ T(n)\in\Theta\left(\phi^n\right). $$
This time complexity is exponential which is terrible! But on this same page we also use a loop to calculate Fibonacci numbers, and if you do the calculations, you would find it is $\Theta(n)$. This
is interesting because calculating this out is the key idea behind dynamic programming, which is applied strong induction 😄.
Divide and Conquer Algorithms
Wait a second, there’s another type of recursive. What happenes when we don’t go back some fixed amount, but instead divide our input by some amount? This is called a Divide and Conquer algorithm.
The best example of this is to observe what happens when we do a merge sort. Lets look at some code for a merge sort so we can show whats going on^3.
def merge_sort(arr):
n = len(arr)
if n == 0 or n == 1:
return arr
left = merge_sort(arr[0:n//2])
right = merge_sort(arr[n//2:n])
i = 0
j = 0
n_left = len(left)
n_right = len(right)
while (i < n_left) and (j < n_right):
if left[i] < right[j]:
arr[i+j] = left[i]
i += 1
arr[i+j] = right[j]
j += 1
while (i < n_left):
arr[i+j] = left[i]
i += 1
while (j < n_right):
arr[i+j] = right[j]
j += 1
return arr
Honestly I’m proud of myself for being able to write a merge sort from scratch when this exhausted, currently I’m in a hotel with no internet in the middle of nowhere Pennsylvania for a family
reunion. Normally I wouldn’t write something like this in the notes, but I don’t think any of my students are going to actually read this page and bring it up in class lmfao.
Anyway, outside of the recursive calls, the function performs $\Theta(n)$ work, you can count yourself for best and worst possible cases to very for yourself if you’d like. Now, the function
recursively calls itself twice, but both times require only to sort a list of half the size, to give us the recurrance relation $$ T(n) = 2T\left(\frac{n}{2}\right) + \Theta(n). $$ Notice that if we
were being very careful we would be careful about integer division n//2, but this is good enough for our purposes.
What sort of techniques can we use for this? Can we assume $r^n$? Unfortunately that will not work, as there isn’t a super easy way to come up with a formula for these sorts of equations, even if we
had a explicity formula there for the merge operations. Rather, we will need to use a theorem that will give us a good way of stating the growth rate.
The Master Theorem
What I am about to show you is the most insane, balls to the walls theorem that you will ever see. The cases behind it are freaking crazy, and the proof is even more nuts. This theorem, called the
Master Theorem, tells us how function asymptotics behave depending on their structure.
This is kinda more of a theorem to look at to explain, but I’ll provide some examples
The Master Theorem: Let $$T(n) = aT\left(\frac{n}{b}\right) + f(n)$$ where $a$ is the number of divisions per call, $b$ is how much you are dividing the input, and $f(n)$ is the work per call.
Let $\varepsilon = \log_b(a)$, then
1. If $f(n)\in\mathcal{O}(n^k)$, with $k < \varepsilon$ then $T(n)\in\Theta(n^\varepsilon)$
2. If $f(n)\in\Theta(n^k)$, with $k = \varepsilon$ then $T(n)\in\Theta(n^\varepsilon\log(n))$
3. If $f(n)\in\Omega(n^k)$, with $k > \varepsilon$ and $$0\leq\lim_{n\rightarrow\infty}\frac{af\left(\frac{n}{b}\right)}{f(n)} < 1,$$ then $T(n)\in\Theta(f(n))$
Note that this theorem applies for both our practical and our true version of big $O$, but you just need to make sure that you are consistent.
The whole idea of this theorem hinges on the value $\log_b(a)$ which represents the ratio of how many times we are splitting compared to how small we are dividing up the problem. Based on this and
the amount of work we do every iteration, we can come up with our bounds. Note that the proof is based off the idea that you have an exact expression for the sum, and then find the asymptotics.
“Exact” is a little loosey goosey here, but there are more exact methods for coming up with the Big $O$ of these, this is just the simplest. Lets run through examples of the cases.
Example: What are the asymptotics of $T(n)=9T\left(\frac{n}{3}\right)+n$
Proof Since $a=9$ and $b=3$ then $\log_3(9)=2$. $f(n)\in\Theta(n)$ which means that $k=1$. Since $k < \varepsilon$ then we are in Case $1$, which means that $T(n)\in\Theta(n^2)$QED
Case $1$ represents the situation in which there is just so much splitting that it completely overpowers all the work that actually gets done. If you represent a recursion tree, then its too many
Example: What are the asymptotics of merge sort with $T(n)=2T\left(\frac{n}{2}\right)+f(n)$ with $f(n)\in\Theta(n)$?
Proof Since $a=2$ and $b=2$ then $\log_2(2)=1$. $f(n)\in\Theta(n)$ which means that $k=1$. Since $k = \varepsilon$ then we are in Case $2$, which means that $T(n)\in\Theta(n\log(n))$QED
Case $2$ represents the situation in which the splitting and the work done in the calls perfectly balances out to add just a little bit of extra work. As you can see, this shows us how merge sort is
$\mathcal{O}(n\log n)$!!
Example: What are the asymptotics of $T(n)=2T\left(\frac{n}{2}\right)+n^4$
Proof Since $a=2$ and $b=2$ then $\log_2(2)=1$. $f(n)\in\Theta(n^4)$ which means that $k=4$. Since $k > \varepsilon$ then we are in Case $3$, which means that $T(n)\in\Theta(n^4)$ since the
additional constraint of $$ \lim_{n\rightarrow\infty} \frac{2\left(\frac{n}{2}\right)^4}{n^4} = \frac{1}{8} < 1 $$QED
Case $3$ represents the case in which there is so much work every iteration that it dominates the splitting. Also, we need that extra condition to make sure that we don’t completely blow up our
asymptotics from above.
The Master’s Master
In this section I will give the Akra-Brazzi method which is a more general example of the Master Theorem.
Strange Multiplication
Here I will give a weird example of multiplication that uses master theorem
1. This is the first piece of witchcraft on this page ↩︎
2. This is the second piece of witchcraft on this page ↩︎
3. Apologies if this section is not coherent, I’m pretty tired and theres some true crime murder mystery thing that keeps distracting me ↩︎ | {"url":"https://wjmccann.com/course/computationalcomplexity/sections/bigorecursion/","timestamp":"2024-11-10T03:17:56Z","content_type":"text/html","content_length":"42316","record_id":"<urn:uuid:0cff88e2-65f2-4dd3-b88e-1943e960ccd3>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00111.warc.gz"} |
Studying Islam | Forums
UNITED ARAB EMIRATES Topic initiated on Wednesday, February 14, 2007 - 11:51 AM
Dilemma of Applying Reason
Dilemma of Applying Reason
Almost all of us have been faced with the questioning of a child by repeating one word over and over. He can be very frustrating to us as he asks "Why?" If you put a knife beyond his reach, he wants
to know, "Why?" When you explain it is sharp, he asks "Why?" And so you explain, "in order to cut fruit," and he asks, "Why?" And so it goes.
It illustrates the dilemma of applying reason. What we have to do when we apply reason is first to set standards of proof. We decide for ourselves, "What will be satisfied with if I find such and
such and so and so that constitutes for me a final proof?". We have to decide on that first.
What happens though, is that on the really important issues, the philosophical matters, thinkers set standards of proof and they take a look at their subjects and eventually they may arrive at their
standards. They may arrive at the point which they say would constitute a proof. But then they ask for a proof of the proof.
Setting Standards
The key to avoiding an endless dissatisfaction is to satisfy ourselves about standards first; to satisfy ourselves that such and such are a list of criteria that constitute proof, satisfying proof,
and then we test the subjects that we examine.
Taking a Stand
Everyone must be committed to something. You have to put your foot down some place. It is impossible to be neutral at all times. There has to be a point of reference in the life of any thinking
individual. You have to take a stand somewhere. The question, of course, is to put your foot down in the right place. Since there is no such thing as a proof of a proof of a proof and so on, in order
to find the right place to put one’s foot down, to take a stand, we have to search and find that place and it is by a method that I hope to illustrate here.
It is a question of finding a point of convergence. You see, we search for truth in many places and we begin to know that we are succeeding in finding the truth if all our different paths start to
converge; they start to come together at the same point.
If we are examining a book, looking for evidence of divine origin, and we are led to Islam, this is one path. If at the same time, we are examining the words of all those who were called prophets and
we find ourselves led to Islam, we have a firmly grounded basis for belief We started looking for truth in two different places and found ourselves going down the path headed for the same
No one ever proves all things. We have to stop at some point being satisfied with our standards as I have mentioned earlier. The point is, in order to take a stand and to be sure it is in the right
place, we want to examine all the evidence around us and see where does it lead us and anticipate this point of convergence; to say it looks like all things are pointing to this place. We go to that
place and then look at the data around us to see if it fits into place. Does it now make sense? Are we standing is on right place? | {"url":"http://studying-islam.org/forum/topic.aspx?topicid=2349&lang=&forumid=1","timestamp":"2024-11-03T23:10:18Z","content_type":"text/html","content_length":"34539","record_id":"<urn:uuid:e87f5bc5-fc91-4f29-8e7a-0f517ed2e5e1>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00586.warc.gz"} |
7.03 Thevenin's theorem.
Thevenin's theorem states, that the equivalent circuit will simply be an e.m.f. in series with an internal resistance , (exactly like the power supply models introduced in the previous section).
The values of the e.m.f and internal resistance are calculated as follows.
• The value for the e.m.f., is simply the voltage across the output terminals of the original circuit, when no external components are connected.
• The value for the internal resistance of the equivalent circuit, is found by shorting out the e.m.f. in the original circuit and then calculating the resistance between the output terminals
The formal statement of Thevenin's theorem is: "The linear network behind a pair of output terminals, can be replaced with an e.m.f. in series with an internal resistance." (The term linear network
,just refers to a circuit containing components like resistors to which Ohm's law applies).
Thevenin's theorem is best illustrated with the example on the following page. This example only uses a very simple circuit to illustrate how to apply the theorem. In practice the circuits we use
Thevenin's theorem with, can be much more complicated. | {"url":"https://science-campus.com/engineering/electrical/dc_theory/chapter7/dctheory_7_3.html","timestamp":"2024-11-02T07:29:07Z","content_type":"text/html","content_length":"8486","record_id":"<urn:uuid:1c731cec-5be6-44d1-89c1-057806c074aa>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00689.warc.gz"} |
EViews Help: Pooled Time Series, Cross-Section Data
Pooled Time Series, Cross-Section Data
Data often contain information on a relatively small number of cross-sectional units observed over time. For example, you may have time series data on GDP for a number of European nations. Or perhaps
you have state level data on unemployment observed over time. We term such data pooled time series, cross-section data.
EViews provides a number of specialized tools to help you work with pooled data. EViews will help you manage your data, perform operations in either the time series or the cross-section dimension,
and apply estimation methods that account for the pooled structure of your data.
The EViews object that manages time series/cross-section data is called a pool. The remainder of this chapter will describe how to set up your data to work with pools, and how to define and work with
pool objects.
Note that the data structures described in this chapter should be distinguished from data where there are large numbers of cross-sectional units. This type of data is typically termed
data. Working with panel structured data in EViews is described in
“Working with Panel Data”
“Panel Estimation” | {"url":"https://help.eviews.com/content/pool-Pooled_Time_Series,_Cross-Section_Data.html","timestamp":"2024-11-05T07:38:28Z","content_type":"application/xhtml+xml","content_length":"7161","record_id":"<urn:uuid:1d24316c-4a0a-4b04-906f-544448f78ee2>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00119.warc.gz"} |
Analytic torsion for graphs - Quantum Calculus
Analytic torsion for graphs
A bit of a personal health scare this summer (I’m completely fine again) reminded me that we don’t live for ever. This prompted me to rummage through some past initiated work which has never seen the
light. One of them was the work on natural groups which still needs to be wrapped up. One of the (nice) dilemmas one has when working independently without a research group pushing for deadlines or
polishing things is that math is an infinite playground where one can get lost easily. While floating around in various areas one often gets distracted by new things. For me, when looking at
unfinished things, there is one topic I definitely want to make a bit more progress. This is the concept of analytic torsion for graphs. Unlike in a classical setting which relies on (standard) but
rather heavy analysis, the case of graphs is completely combinatorial. The topic can be understood and studied by any student who has seen a fewweeks of linear algebra.
Unlike for manifolds or varieties, one can study analytic torsion for general graphs. Every finite simple graph comes with a natural simplicial complex and so also with a natural differential
complex. Almost all notions from the continuum can be studied now for general finite simple graphs. One of them is analytic torsion. I got dragged into this in 2012, when working on Cauchy-Binet for
pseudo determinants. One of the questions was to understand the pseudo determinant Det of the Dirac operator $D=d+d^*$ for graphs where d is the exterior derivative. I noticed (but still can up to
now not prove) that for contractible graphs $G=(V,E)$, the quantity $Det(D)/|V|$ is a square suggesting that the Dirac operator itself is a square in that case. For general cases, like for example if
the fundamental group of the graph is non-trivial, then this number theoretical oddity disappears. This is mentioned in section 8 of my Cauchy-Binet paper. I still believe that the Dirac pseudo
determinant somehow is related to Analytic torsion and this hunch has gotten stronger even in that for spheres, there are analogue coincidences for the pseudo determinant of the Dirac operator of
spheres. It suggests that there is an even deeper symmetry hidden.
In the summer of 2015, while traveling in Europe, I made experiments with analytic torsion $A(G)=\prod_{k=0}^d Det(L_k)^{k(-1)^{k+1}}$, where $L_k$ are the form Laplacians, the blocks of the Hodge
Laplacian $L=D^2$. I noticed A(G)/|V|=1 for contractible graphs (and more generally for graphs homotopic to 1 like the dunce hat) and looked also at various other cases like discrete manifolds. For
spheres, one has either A(G)=|V|/|G| or A(G)=|V| |G| depending on whether the dimension of the sphere is even or odd. [Classically, one takes $\sqrt{1/A(G)}$. I like to get integer values if possible
in our combinatorial setting and taking the square root does not make any sense here. Note that $A(G)=\prod_{k=0}^d Det(L_k)^{k (-1)^{k+1}}$ by McKean Singer which tells $\prod_{k=0}^d Det(L_k)^{(-1)
^k}=1$.] The quantity |G| is the volume of the graph, the number of maximal complete subgraphs. The youtube video reports a bit about this. I worked around Christmas of this year (2021) a bit more on
it. It turns out that the case of 2-dimensional spheres (which was still not done while recording the video) is quite easy. Remember that a graph is a 2-dimensional sphere, if every unit sphere is a
1-sphere, that is a circular graph with 4 or more elements. Examples are the icosahedron graph or octahedron graph.
So, here is my Christmas theorem (not mentioned in the movie which was done a week before on December 18th):
Theorem: If G is a 2-sphere, then A(G)=|V|/|G|.
Even so, it is small progress only, I like this result very much because it taps into very classical and very beautiful (completely non-technical) mathematics ranging from the mid 19th century
(Kirchhoff, Von Staudt or Cauchy) to mid 20th century (like Dirac andMcKean_Singer). So, here is my proof. Of course, I hope to extend this to all d-spheres still. But that requires some more work,
as I rely here on situations which only work in 2-dimensions.
Proof: In two dimensions, $A(G)=Det(L_1)^2/(Det(L_0) Det(L_2)^3/)$. By McKean-Singer super symmetry, $Det(L_0) Det(L_2)^2=Det(L_1)$, so that $A(G)=Det(L_1)/Det(L_2)^2 = Det(L_2)/Det(L_0)$. Now recall
that $Det(L_0)/|V|$ is by the famous Matrix Tree theorem the number of spanning trees in the graph (the matrix tree theory is a direct consequence of Cauchy-Binet). Now comes a beautiful insight from
the first part of the 19th century: in modern form it tells that the number of spanning trees in a planar graph is the same than the number of spanning trees in the dual graph. I call it the maze
theorem because a two-dimensional maze illustrates that every spanning tree (the walls of a maze) comes with a complementary tree. [Van Staute has first noticed in the first few pages of his book
from 1847 that this immediately implies the Euler Gem Formula in two dimensions: the set of edges E of a graph can be partitioned into two sets of size |V|-1 and |F|-1 and |E|=|V|-1 + |G|-1 is just
the Euler Gem formula. ] Since $Det(L_2)/|G|$ is the number of spanning trees in the dual graph, this implies that $A(G) = Det(L_2)/Det(L_0) = |V|/|G|$. QED.
Example: If that is not beautiful, I don’t know what is. We combine duality insight from the 19th century to super symmetry insight from the 20th century. To illustrate this, let us just take the
Icosahedron graph. We have $(Det(L_0),Det(L_1),Det(L_2)) = (62208000, 6449725440000000, 103680000)$ and $(|V|,|E|,|G|) = (12, 30, 20)$. Now, $A(G) = 12/20 = 3/5$. We can check McKean-Singer:
$62208000*103680000/6449725440000000=1$ and $6449725440000000^2/(62208000*103680000^3)=3/5$.
To show an other example, lets just take a random 2-sphere and compute everything. Of course with the computer. You see how large these determinants can become. The Dirac matrix is a 410 x 410 | {"url":"https://www.quantumcalculus.org/analytic-torsion-for-graphs/","timestamp":"2024-11-02T08:46:43Z","content_type":"text/html","content_length":"106320","record_id":"<urn:uuid:dbc0024e-e6ab-4ab2-9280-5a12702d7040>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00600.warc.gz"} |
Key Words For Word Problems Printable
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Related Post: | {"url":"https://sop.edu.pl/printable/key-words-for-word-problems-printable.html","timestamp":"2024-11-09T07:07:57Z","content_type":"text/html","content_length":"26254","record_id":"<urn:uuid:e337679e-64ca-438e-a78d-3ac676be2f38>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00411.warc.gz"} |
Vector3 - Unity 스크립팅 API
This structure is used throughout Unity to pass 3D positions and directions around. It also contains functions for doing common vector operations.Besides the functions listed below, other classes can
be used to manipulate vectors and points as well. For example the Quaternion and the Matrix4x4 classes are useful for rotating or transforming vectors and points.
back Shorthand for writing Vector3(0, 0, -1).
down Shorthand for writing Vector3(0, -1, 0).
forward Shorthand for writing Vector3(0, 0, 1).
left Shorthand for writing Vector3(-1, 0, 0).
negativeInfinity Shorthand for writing Vector3(float.NegativeInfinity, float.NegativeInfinity, float.NegativeInfinity).
one Shorthand for writing Vector3(1, 1, 1).
positiveInfinity Shorthand for writing Vector3(float.PositiveInfinity, float.PositiveInfinity, float.PositiveInfinity).
right Shorthand for writing Vector3(1, 0, 0).
up Shorthand for writing Vector3(0, 1, 0).
zero Shorthand for writing Vector3(0, 0, 0).
magnitude Returns the length of this vector (Read Only).
normalized Returns this vector with a magnitude of 1 (Read Only).
sqrMagnitude Returns the squared length of this vector (Read Only).
this[int] Access the x, y, z components using [0], [1], [2] respectively.
x X component of the vector.
y Y component of the vector.
z Z component of the vector.
Vector3 Creates a new vector with given x, y, z components.
Equals Returns true if the given vector is exactly equal to this vector.
Set Set x, y and z components of an existing Vector3.
ToString Returns a formatted string for this vector.
Angle Returns the angle in degrees between from and to.
ClampMagnitude Returns a copy of vector with its magnitude clamped to maxLength.
Cross Cross Product of two vectors.
Distance Returns the distance between a and b.
Dot Dot Product of two vectors.
Lerp Linearly interpolates between two points.
LerpUnclamped Linearly interpolates between two vectors.
Max Returns a vector that is made from the largest components of two vectors.
Min Returns a vector that is made from the smallest components of two vectors.
MoveTowards Calculate a position between the points specified by current and target, moving no farther than the distance specified by maxDistanceDelta.
Normalize Makes this vector have a magnitude of 1.
OrthoNormalize Makes vectors normalized and orthogonal to each other.
Project Projects a vector onto another vector.
ProjectOnPlane Projects a vector onto a plane defined by a normal orthogonal to the plane.
Reflect Reflects a vector off the plane defined by a normal.
RotateTowards Rotates a vector current towards target.
Scale Multiplies two vectors component-wise.
SignedAngle Returns the signed angle in degrees between from and to.
Slerp Spherically interpolates between two vectors.
SlerpUnclamped Spherically interpolates between two vectors.
SmoothDamp Gradually changes a vector towards a desired goal over time.
operator - Subtracts one vector from another.
operator != Returns true if vectors different.
operator * Multiplies a vector by a number.
operator / Divides a vector by a number.
operator + Adds two vectors.
operator == Returns true if two vectors are approximately equal. | {"url":"https://docs.unity3d.com/kr/2020.2/ScriptReference/Vector3.html","timestamp":"2024-11-07T20:21:08Z","content_type":"text/html","content_length":"31864","record_id":"<urn:uuid:4f066318-5414-4b97-892d-03bcb5350802>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00648.warc.gz"} |
Learning Alloy the hard way
Reading Software Abstractions was a blast. It is complete, very insightful in first order logic, and makes Alloy an intuitive tool. That was until page 171 and the chapter "Leader Election In A
Ring". This chapter gave me a serious headache, and I needed to write about it here so I can clear my head out.
This article is not meant to be a tutorial on Alloy and I won't explain the logic or syntax of the language here. Sorry if you're not familiar to the language, this post won't be very easy on you.
How the book states the problem
Let's imagine you have a ring of processes, the goal is to elect the leader of the group of processes. Each process will be given a unique identifier (say a MAC ID, or something) and the leader will
be the process with the largest identifier. To achieve that, we'll use the Chang and Roberts algorithm, a well known approach to solve the problem. The given Alloy code is explained first:
open util/ordering[Time]
open util/ordering[Process]
sig Time {}
sig Process {
succ: Process,
toSend: Process -> Time,
elected: set Time
fact Ring { all p: Process | Process in p.^succ }
As a complete stranger to this kind of algorithm, it took me quite some time to understand the ordering of the processes: it simulates the unique identifiers of processes.
I have to point out something that bothers me at this point. The book reads "[...] a token can be taken from the pool of one process and moved to the pool of its successor in the ring" (I put
emphasis on the words myself). Out of 4 concepts, two of them are completely dropped from the Alloy specification, token and pool does never appear, and it seems is replaced by toSend, which kind of
feel an arbitrary name.
Then the core of the algorithm is presented:
pred step (t, t': Time, p: Process) {
let from = p.toSend, to = p.succ.toSend |
some id: from.t {
from.t' = from.t - id
to.t' = to.t + (id - p.succ.prevs)
fact DefineElected {
no elected.first
all t: Time - first |
elected.t = { p: Process | p in p.toSend.t - p.toSend.(t.prev) }
And this is where I lost myself, not only are these steps explained in a paragraph or less, but there's no more explanation to relate the choices of writing this way compared to the original
algorithm. Where are the pools, the tokens, and above all why toSend?!
I was angry and I could not wrap my head around the problem. Plus it was the first time graphs presented by the examples did not help me. I did not understand the distance between the original
algorithm and the solution in Alloy. I doubted Alloy for the first time.
Revamping the specifications
My hubris took place: I'm smarter than Daniel Jackson, am I not? I will revamp his example into a more faithful example. Let's look at the description of the algorithm in Wikipedia:
1. Initially each process in the ring is marked as non-participant.
2. A process that notices a lack of leader starts an election. It creates an election message containing its UID. It then sends this message clockwise to its neighbour.
3. Every time a process sends or forwards an election message, the process also marks itself as a participant.
4. When a process receives an election message it compares the UID in the message with its own UID.
1. If the UID in the election message is larger, the process unconditionally forwards the election message in a clockwise direction.
2. If the UID in the election message is smaller, and the process is not yet a participant, the process replaces the UID in the message with its own UID, sends the updated election message
in a clockwise direction.
3. If the UID in the election message is smaller, and the process is already a participant (i.e., the process has already sent out an election message with a UID at least as large as its own
UID), the process discards the election message.
4. If the UID in the incoming election message is the same as the UID of the process, that process starts acting as the leader.
This is the part the specification treats, the second phase of the algorithm is not modeled here.
What do we read? Firstly, there is a notion of participant that is not in the specification of the book. Secondly, the message being carried out is not present in the original specification. So here
is the new Process definition:
sig Time {}
sig Process {
neighbor: Process,
token: Process -> Time,
inbox: Process -> Time,
participant: set Time,
elected: set Time
Our goals are:
• make the message passing more obvious, so we name an inbox
• the succ being too close from next, we rename it neighbor
• instead of having a sendTo, rename it token just as the description says
• introduce the concept of participant that is in the description as well
Since we renamed all of these concepts, I feel more confident in refactoring the step method:
pred startsElection (t, t': Time, p: Process) {
p not in participant.t implies // (1)
p in participant.t' // (2)
and p.neighbor.inbox.t' = p.token.t // (3)
and messageReception [t', p.neighbor] // (4)
Whenever a process p is not a participant (1), it becomes one (2) and sends a message in it's neighbor's inbox (3), and the neighbor will act the message reception as it should (4). The message
reception logic follows the one from the description:
pred messageReception (t: Time, p: Process) {
p.inbox.t = p implies p in elected.t and p not in participant.t // (4.4)
p.inbox.t in p.^next implies messageReception [t.next, p.neighbor] // (4.1)
p.inbox.t in p.^prev and p not in participant.t implies // (4.2)
p in participant.t.next
and p.neighbor.inbox.t.next = p.token.t
and messageReception [t.next, p.neighbor]
For each proposition, the comment links to the rule in the algorithm description above. Although this code is not perfect, I was confident it was an improvement compared to the example in the book.
Let's run the code...
This is what happens when you are not attentive enough
Hubris is not a good advisor
Alright, this approach is uselessly aggressive and full of pride. I admit my anger blinded me and hubris brought me to think I could do better! First, I don't know a penny about concurrent algorithms
(apart from distant lessons during my studies). Two, I am still a beginner in Alloy. I only used the language for its boosted graphs drawing capabilities and not for its formal logic powers. I forgot
the one and only rule the author has being repeating again and again in the book: Alloy is a first order logic langage. It means that recursions, variables bindings, etc. are not part of the tools
available to express ideas.
Therefore, Alloy is definitely not the intuitive approach to specifications like a typical language would be. One must bind their mind to the first order logic (pun intended).
What now?
I want to try two things: first rename variables in the algorithm from the book to understand it better, then fix my own implementation to check my understanding of the approach and to formulate the
limitations of the algorithm. Alright, so let's rename the variables and understand how an election can be generated by the algorithm:
Let's analyse the traces carefully
We'll try to understand this example of an election.
0. At the initial state, no process is elected. We know Process2 should become elected, as it has the higher id. Each process has a token to itself. This is the initial state of every simulation
1. Process0 has lost its token while others have their own token. This can be explained for a call of step [Time0, Time1, Process0]. Let's see how by replacing the terms in the predicate:
step [Time0, Time1, Process0] iff {
let from = Process0.token, to = Process0.neighbor.token |
some id: from.Time0 {
from.Time1 = from.Time0 - id
to.Time1 = to.Time0 + (id - Process0.neighbor.prevs)
This substitution unravels the following predicate:
some id: Process0.token.Time0 {
Process0.token.Time1 = Process0.token.Time0 - id
Process2.token.Time1 = Process2.token.Time0 + (id - Process2.prevs)
Since we know that Process0.token.Time0 is a scalar, id must be Process0.token.Time0 and since we know Process0.token.Time0 = Process0 and Process2.token.Time0 = Process2 we can keep on reducing:
Process0.token.Time1 = Process0 - Process0
no Process0.token.Time1
Process2.token.Time1 = Process2 + (Process0 - Process2.prevs)
We confirmed that the Process0.token is empty on the second step, but what about Process2.token.Time1? For that we need to explain Process2.prevs: it's all the processes with a smaller id than the
current one. This term is the translation of the rule 4.3: the bigger process will discard messages of smaller ids. So:
Process2.token.Time1 = Process2 + (Process0 - (Process1 + Process0))
Process2.token.Time1 = Process2
That's it, the reduction is valid, we proved the predicate step[Time0, Time1, Process0]. Let's be a bit quicker for other steps
2. Here token points from Process0 to Process1, that would happen if Process1 passed its token to Process0, this is step[Time1, Time2, Process1]:
let from = Process1.token, to = Process1.neighbor.token |
some id: from.Time1 {
from.Time2 = from.Time1 - id
to.Time2 = to.Time1 + (id - Process1.neighbor.prevs)
Again Process1.token.Time1 = Process1, therefor id = Process1, Process1.neighbor = Process0 and Process1.neighbor.prevs = {} the empty set. We have our final reduction:
no Process1.token.Time2
Process0.token.Time2 = Process1
3. We can use the same proof: this time Process2 sent a message to Process1.
4. This time, it seems that Process1 sent the message to Process0, since Process0 already had a token for Process1, it also has a token to Process2 now.
5. Okay, we arrived at the crux of the algorithm, the two next step are the most important. This one is trickier, because so far, we only solved the step predicate with a single possible value for
id. Since Process0 has two tokens now, there is a choice to make. We'll focus on the three options:
// id = Process1
Process0.token.Time6 = Process2
Process2.token.Time6 = Process1 - (Process0 + Process1)
no Process.token.Time6
// id = Process2
Process0.token.Time6 = Process1
Process2.token.Time6 = Process2
// id = Process2 + Process1
no Process0.token.Time6
Process2.token.Time6 = Process2
Alright, so we can see clearly that this time, it's the last option that is right. We'll keep the second option in our mind, since we can wonder if it can imply the election of Process2 in the next
6. Finally last step, this time, we need to understand the election predicate:
fact DefineElected {
all t: Time - first |
elected.t = { p: Process | p in p.token.t - p.token.(t.prev) }
// t = Time6 and p = Process2
elected.Time6 = Process2.token.Time6 - Process.token.Time5
It means that the elected process will be the one having received his own token. The fact it just received it is garanteed by the fact that the system only evolves with the step predicate. The
condition - p.token.(t.prev) also prevents processes that did not change to elect themselves. The process being elected is the process that has it's own token and received it from its left neighbor
with the step predicate.
Now that I've understood the version of the book, I'm confident that I won't be able to do better than renaming the variables. I understand now how Alloy can help in this kind of approach but I can't
stop thinking that a temporal logic tool like TLA+ is a much better fit. As Hillel Wayne (yes him again!) puts it:
the more “timelike” the problem gets, the worse Alloy becomes at handling it
Don't be discouraged by this post to learn Alloy. I'm still convinced it is a very powerful tool to understand constraints in systems. I think that the first order logic means that you cannot use
familiar techniques like recursion.
Also traces can be misleading, the first read might make you feel confident, but you need to decypher and fully understand the underlying specification. And I think this is what Daniel Jackson meant
at the beginning of his book when he says Alloy focuses on the deep concepts behind your design and not intricacies of transient technology. | {"url":"https://sadraskol.com/posts/learning-alloy-the-hard-way/","timestamp":"2024-11-11T07:51:33Z","content_type":"text/html","content_length":"23073","record_id":"<urn:uuid:798a50e3-f4fb-4b2e-a5b2-a2da1b8fc70c>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00879.warc.gz"} |
CWG Issue 614
This is an unofficial snapshot of the ISO/IEC JTC1 SC22 WG21 Core Issues List revision 115d. See http://www.open-std.org/jtc1/sc22/wg21/ for the official list.
614. Results of integer / and %
Section: 7.6.5 [expr.mul] Status: CD1 Submitter: Gabriel Dos Reis Date: 15 January 2007
[Voted into the WP at the September, 2008 meeting as part of paper N2757.]
The current Standard leaves it implementation-defined whether integer division rounds the result toward 0 or toward negative infinity and thus whether the result of % may be negative. C99, apparently
reflecting (nearly?) unanimous hardware practice, has adopted the rule that integer division rounds toward 0, thus requiring that the result of -1 % 5 be -1. Should the C++ Standard follow suit?
On a related note, does INT_MIN % -1 invoke undefined behavior? The % operator is defined in terms of the / operator, and INT_MIN / -1 overflows, which by Clause 7 [expr] paragraph 5 causes undefined
behavior; however, that is not the “result” of the % operation, so it's not clear. The wording of 7.6.5 [expr.mul] paragraph 4 appears to allow % to cause undefined behavior only when the second
operand is 0.
Proposed resolution (August, 2008):
Change 7.6.5 [expr.mul] paragraph 4 as follows:
The binary / operator yields the quotient, and the binary % operator yields the remainder from the division of the first expression by the second. If the second operand of / or % is zero the
behavior is undefined[DEL:; otherwise (a/b)*b + a%b is equal to a. If both operands are nonnegative then the remainder is nonnegative; if not, the sign of the remainder is implementation-defined.
[Footnote: According to work underway toward the revision of ISO C, the preferred algorithm for integer division follows the rules defined in the ISO Fortran standard, ISO/IEC 1539:1991, in which
the quotient is always rounded toward zero. —end footnote]:DEL]
[Drafting note: see C99 6.5.5 paragraph 6.] | {"url":"https://cplusplus.github.io/CWG/issues/614.html","timestamp":"2024-11-09T23:03:33Z","content_type":"text/html","content_length":"4163","record_id":"<urn:uuid:e34184fe-bcd5-474a-bce2-786144a3c454>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00072.warc.gz"} |
Problem of Separating point sets with a circle
Separating point sets with a circle
Let there be given 2n + 3 points (n > 0) such that no 4 lie on the same circle. Prove that it's possible to select 3 points so that the circle passing through these points contains in its interior
exactly n points from the set whereas the remaining n points lie in its exterior.
Let's use the notion of convex hull that served us so well so many times before. Thus, let's enclose our set into its convex hull. We should start somewhere looking for our three points. Choose any
side of the convex hull. This gives two points. Fix these two. Selecting every time a new point in addition to these two, we can draw (2n+1) circles because no 4 points belong to the same circle.
These circles are naturally ordered. Can't we use this order?
But what is it, to start with? What can we say about it? Do all circles have different radii? No, it must not necessarily be true. (To see this, start with two intersecting circles of equal radius.
Besides the two points of intersection, we may choose three additional points: one on each of the circles and one elsewhere. This a counterexample.) Recollect now that, in addition to the two fixed
points, every circle contains exactly one point from the given set. Connecting this point with the fixed two determines an angle. The angle under which the fixed segment is seen from the selected
point. For a fixed circle, this angle does not depend on the exact location of the point on the circle. Therefore, it's rather a circle related than a point related. Might it be a candidate to
determine the order of the circles?
Yes, indeed. As the angles become smaller so the circles (or rather their part above the two fixed points) expand. The circle with the largest angle contains no points inside it. The next one
contains the point that defined the previous circle. The next one contains this point too but, in addition, it also contains the point that defines the second circle, and so on. The circle number
(n+1) contains exactly n points inside and n points outside itself.
Note that the solution may not be unique because it depends on the original selection of two fixed points.
Richard Beigel sent me the following simple proof:
Let there be a set S of N points. For each pair of points in S, draw its perpendicular bisector. Let p be any point in the plane that is not on any of those perpendicular bisectors. Then no two
points in S are equidistant from p. Let the distances from p to each point in S be d[1], d[1], ..., d[N]. Relabel the points so that 0 = d[1] < d[2] < ... < d[N]. Then, for any k > 0, the circle with
center p and radius (d[k] + d[k+1])/2 separates some k from the remaining N - k points.
Richard's result allows 4 or more points to lie on the same circle. Returning to the original problem, N = 2n + 3, we first find a circle that separates n points from the remaining n + 3. Then we
start expanding it until it first touches at least one of the n + 3 points. If 3 points lie on the circle, t's all finished. If only 1 or 2 points lie on the circle we continue extending it keeping
the "acquired" points on the circle until there are exactly three points there (it can't be more by the conditions of the problem.)
Richard also sent another proof that made use of less elementary mathematics and the following variation on the first proof.
Consider pairwise distances of the given 2n + 3 points and select two points A and B such that dist(A,B) is the shortest. Consider a pencil of circles (i.e., all circles) through these two points.
Start with a circle whose diameter is AB. Move the center along the perpendicular bisector of AB, starting from the midpoint, out to infinity in one direction, and back in from infinity in the other
direction. As the circle crosses points in S, the number of points inside/outside the circle progresses as follows.
inside 0 1 2 ... k ... (2n + 1)-(k + 1) (2n + 1) - (k + 2) ... 0
outside 2n+1 2n 2n - 1 ... (2n + 1) - k ... k + 1 k + 2 ... 2n + 1
Assume there are k+1 points on one side of the line AB, (2n + 1)-k on the other. The three stars indicate marks where the center goes to infinity and back around the other side. Interchanging inside/
outside after infinity is reached, you see the progression (0, 2n + 1), ..., (2n + 1, 0) so at some point there were n points on each side of the circle with one addtional point on the circle.
|Contact| |Front page| |Contents| |Geometry| |Up|
Copyright © 1996-2018 Alexander Bogomolny | {"url":"https://www.cut-the-knot.org/Generalization/scircles.shtml","timestamp":"2024-11-10T10:39:51Z","content_type":"text/html","content_length":"15782","record_id":"<urn:uuid:d29b9a5c-4628-4e6c-98a9-0ed873264c22>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00248.warc.gz"} |
Why study Mathematics? - The Culture SGWhy study Mathematics?
So the A’levels is coming to an end. Good job to all the JC2s! I did receive a few questions on studying Mathematics at undergraduate level. So I thought the following article might just answer it.
A few (living) mathematicians you can google are, Jim Simons, Edward Thorp and James Stewart too.
The article is a fairly interesting read since many students have come up to me and asked why do I study and like Mathematics so much, and what can I do in my life. Of course, there are some students
who got their A’levels grade and took an A for H2 Math as a calling to study Mathematics. Well, tread carefully.
Personally, having a Mathematics background do not just lead to one being a teacher or tutor. At the end of the day, Mathematics boils down to problem solving (It is the core of Mathematics) and from
the article, they draw a few real world problems to pique some interests. 🙂
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JOI Spring Training 2024 Online Contest - Codeforces
Hello, everyone!
Japanese Olympiad in Informatics (JOI) Spring Training 2024 will be held from March 21 to March 24. There are 4 days in the contest.
The duration of the contest is 5 hours, but you can start any time in 22-hour window. Note that if you start competing after 19:00 GMT, the contest ends at 24:00 GMT and you cannot compete 5 full
hours. For example, if you start competing at 21:00 GMT, the contest will be shortened to 3 hours.
The contest information is as follows. Details are available in contest page.
• Number of problems for each contest: 3-4 problems
• Max Score for each task: 100 points
• Style: IOI-Style (There may be some partial scores)
• Problem Statement: Japanese & English
• There may be some unusual task (e.g. output-only task, communication task) like IOI
Note that from this year, you can submit with C++20.
The registration page and live scoreboard will be announced 10 minutes before the contest.
We welcome all participants. Good luck and have fun!
UPD1: The contest will start in 10 hours.
UPD2: Contest Day1 is started.
UPD3: Contest Day1 is finished. Now you can discuss Day1 problems.
UPD4: Due to technical issue, the contest Day2 will be delayed by about 1 hour.
UPD5: Contest Day2 is started.
UPD6: Contest Day2 is finished. Now you can discuss Day2 problems.
UPD7: Contest Day3 is started.
UPD8: Contest Day3 is finished. Now you can discuss Day3 problems.
UPD9: Contest Day4 is started.
UPD10: Contest Day4 is finished and now the entire series of contests is ended. Thank you for your participation.
8 months ago, # |
» -29
I_love_Leyeli_Meredova hello!
Why can't I register for the competition?
• » 8 months ago, # ^ |
» +11
Abito Read the blog please.
□ » 8 months ago, # ^ |
» 0
• » 8 months ago, # ^ |
» +3
Vectrizz there you can register 10 minutes before the start of the contest
Good luck have fun everybody! I always have fun with the problems from JOI Spring Camp.
Is it like Div2 or does JOI has same difficulty as Div1 ?
• » 8 months ago, # ^ |
» +20
SebR As far as I know it's not comparable with a Codeforces round, it's more like a contest with 3-4 3000ish problems
□ » 8 months ago, # ^ |
» +3
I believe that not all problems in the contest are 3000ish. Last year there were several problems that might be easy enough for solving (Currencies, Council, Tourism, Bitaro's
minhcool travel) (you can also throw in Passport and Conveyor Belt). But yeah, focus on solving the subtasks might be a better strategy.
When will the analysis mode be opened?
• » 8 months ago, # ^ |
» +11
qiuzx After the 4-day contest ends.
7 months ago, # |
← Rev. 6 → +75
I believe I have a solution for day 1 P2 using $$$1363 \approx k \log (q+1) + (q-1) \log (k)$$$ queries after contest ended.
The organizers probably purposely set the limits close to $$$k \log (2q)$$$ to purposely troll people to believe that it is close to optimal (and I got trolled during contest to do
errorgorn disgusting optimizations, I mind solved up to 96 points, but ran out of time and only got 94 :V).
How to solve Q=1
1363 solution
96 point solution (for reference)
• » 7 months ago, # ^ |
» +23
ksun48 Nice approach! I believe it is possible to optimize the 1500+? solution to 100 points though by sending the compressed tree more efficiently
» 7 months ago, # |
How to solve d1p3 ?
» 7 months ago, # |
A critical error has occurred :-(
• » 7 months ago, # ^ |
» +15
andreystefanov I have the same problem :-(
• »
» Does someone know why this might be happening? It seems to work when I use a VPN
□ »
» It is probably because of an IP ban. It seems to work with mobile data/hotspot
☆ »
» 7 months ago, # ^ |
» +13
Probably my result on Day 1 was too suspicious :D
» 7 months ago, # |
how did Japanese design these communication task perfectly to let me get unbelievably low score and simply can not notice the key to a great solution in 2+ hours?
• » 7 months ago, # ^ |
» +14
qiuzx Agreed, I couldn't think of anything helpful at these communication tasks :(
• » 7 months ago, # ^ |
» +12
Dominater069 Being good isnt an excuse to not follow the rules. Kindly do not discuss about the problems. You spoiled atleast 2 things about the contest in this comment.
How to solve d2p3? I passed it in n*log(1e9)*log(n) with some careful implementations and dirty optimizations, so I think someone will have a better solution.
• »
» After the binary search, each element in $$$a$$$ makes it impossible to choose a position within an interval as a starting point. The process of labelling these intervals that cannot be
chosen can easily be done in $$$O(n)$$$ time, and the process of finding these intervals can be done in $$$O(n)$$$ time using many monotonically moving pointers.
□ »
» Can you explain in more detail? I don’t understand what "each element in a makes it impossible to choose a position within an interval as a starting point" means.
☆ 7 months ago, # ^ |
← Rev. 2 → +10
First, when we want to match two arrays $$$A$$$ and $$$B$$$, the best strategy is to match the $$$i$$$-th largest element in $$$A$$$ with the $$$i$$$-th largest element in $$$B$$$.
» Wlog assume that $$$b$$$ forms an interval when matching $$$a$$$, and let's focus only on the starting point $$$l$$$ of this interval $$$[l,l+n)$$$.
» Consider an element $$$a_i\ (n+1\le i\le 2n)$$$, when this element is included in the interval (that means $$$l+n>i$$$), we've known that $$$a_{n+1}>a_{n+2}>\dots>a_{i-1}>a_i$$$,
and we will have $$$\min(c,n-l+1)$$$ elements in $$$a_l,a_{l+1},\dots,a_n$$$ to be greater than $$$a_i$$$, where $$$c$$$ is the total number of elements in $$$a_1,a_2,\dots,a_n$$$
lldar which is greater than $$$a_i$$$. $$$c$$$ can be found in linear time.
We can know that we want to have $$$[L,R]$$$ elements in the interval to be greater than $$$a_i$$$, which $$$L,R$$$ can be found in linear time. So when $$$l+n>i$$$, $$$l$$$ should
satisfy $$$i-n-1+\min(c,n-l+1)\in [L,R]$$$, and this restriction means that when $$$l$$$ is in some ($$$O(1)$$$) intervals, we cannot choose that $$$l$$$.
The rest part of $$$a$$$ can be treated in similar ways, resulting to 4 similar processes.
○ »
» Nice solution.
7 months ago, # |
» +11
qiuzx Day 3 Problem 1 and 3 were really amazing. Spent a long time getting the 71-point subtask right and made stupid mistakes when rushing out problem 3 in the last minutes XD. Though I think the
71-point subtask of problem 1 is probably harder to think and implement than the full solution of problem 3, so I guess it deserves more points.
• »
» How to get 71-point subtask of Day 3 Problem 1?
□ » 7 months ago, # ^ |
» +1
The idea is to think of the ones $$$>x$$$ as $$$1$$$ and the ones $$$<x$$$ as $$$-1$$$ when the given query is $$$x$$$. Then solve the problem for three values. Here some careful
qiuzx classifications are required. I have written an explanation of the whole solution in Chinese right here, but it's too much work translating it in English.
Are these problems available in Codeforces gym After the end of contest?
• » 7 months ago, # ^ |
» +17
libobil No, but they will certainly be available in oj.uz
» 7 months ago, # |
How can I achieve 50 points in Problem 2 Day 2? I've only managed to score 15 points so far and couldn't think of anything helpful to proceed further in the task.
7 months ago, # |
Contest Day4 is Over
The contest Day4 (final day) is finished. The overall ranking is as follows. Thank you for your participation, and let's discuss the problem.
The overall ranking is as follows.
Place Username Day1 Day2 Day3 Day4 Total
1st tricyzhkx 300 215 271 300 1086
2nd TheContingency 296 250 200 300 1046
3rd PersistentLife 296 176 211 300 983
4th SuddeNDeath 291 182 197 300 970
» 5th dXqwq 239 210 211 300 960
6th JoesSR 254 205 211 272 942
E869120 7th zhoukangyang 200 182 257 300 939
8th ChenyaoZhong 268 167 200 300 935
9th qiuzx 223 210 201 300 934
9th smwtcat 292 99 271 272 934
9th zh0uhuanyi 293 130 211 300 934
12th sjc061031 159 220 211 300 900
13th tute7627 200 210 271 204 885
14th Little09 200 136 211 300 847
15th Alan_Zhao 208 230 130 272 840
16th AFewSuns 242 117 131 300 790
17th map 200 217 100 272 789
18th goujingyu 186 136 179 277 778
19th zhouhuanyi 243 130 100 300 773
20th abcsadeveryhour 185 116 163 300 764
20th koosaga 295 168 136 165 764
• » 7 months ago, # ^ |
» +5
wxhtzdy Is Japanese IOI team selected?
7 months ago, # |
» +11
khaledkunbargi Will the problems be available for upsolving ?
if so then on which website ?
• 7 months ago, # ^ |
» ← Rev. 2 → +3
We can upsolve problems in the Contest Site, which is supposed to be on Analysis Mode now. It will end 7 days after the contest (if not started yet, it will start soon).
Also, we can upsolve past JOI problems in AtCoder. This year's one is not prepared yet, but I think that the judge will be prepared within some weeks.
□ »
» Is official solution code available?
☆ »
» I think it will be prepared soon (in a few days or a week).
» Like all solutions and test data in JOI Spring Camp 2023 can be obtained in the link https://www2.ioi-jp.org/camp/2023/2023-sp-tasks/index.html, a new page with the link just
replacing 2023 to 2024 will soon be created.
○ »
» 6 months ago, # ^ |
» +11
» Any updates on this?
7 months ago, # |
I've decided to share my solutions to the problems I've managed to fully solve. If you can add solutions to the rest of the problems/provide a better solution/ask a question, please, feel
free to do so.
$$$\mathbf{Day\ 1.\ Problem\ A}$$$
I'll begin with a little transformation of the statement — instead of adding $$$1$$$ or $$$D$$$, I will subtract from the numbers and aim to achieve subarrays of zeroes. My first observation
was that one's goal should be to use operation $$$A$$$ to sort the numbers in the queried range and then continue with operation $$$B$$$, as you can achieve any sorted sequence with it. We
will answer the queries offline. If we are willing to answer the queries ending on some position $$$r$$$, we should seek to find the optimal usage of operations to sort the sequence for
$$$1$$$ to $$$pos$$$, as then if $$$b_i$$$ is the number of usages of operation $$$A$$$ on $$$c_i$$$, the answer for the query from $$$l$$$ to $$$r$$$ will be $$$\sum_{i=l}^r b_i$$$. This is
true because adding a number to the left of an interval will only result in a potential addition of operation $$$A$$$-s to it. Imagine we've found $$$b_i$$$ for some $$$r-1$$$ and now we want
to see how $$$b_i$$$-s change with the appendment of $$$c_r$$$. Let's denote the changed heights with $$$h_1, h_2, h_3, ..., h_{r-1}$$$. You can see, that after sorting, the sequence is split
into blocks of "dominoes" — subarrays $$$[l,r]$$$, for which if you decrease $$$h_r$$$ by $$$D$$$, then $$$h_l, h_{l+1}, ..., h_{r-1}$$$ will also decrease by $$$D$$$, as if you push the last
one domino, every other will fall as well. The blocks of dominoes are separated by "barriers" between the end $$$r_{prev}$$$ and the beginning $$$l_{next}$$$ of such blocks, which if broken
will result in the merge of the two ranges. Each of these barriers takes some number of $$$D$$$-s before breaking. So, now imagine the following two scenarios — $$$c_r \geq h_{r-1}$$$ and
$$$c_r < h_{r-1}$$$. If $$$c_r \geq h_{r-1}$$$, then it's easy — $$$h_r = c_r$$$ and only possibly add a barrier between $$$r-1$$$ and $$$r$$$. The other case is much more interesting, as you
then must decrease the height of $$$c_r$$$. Let $$$x$$$ be the required number of times you need to perform operation $$$A$$$ on $$$c_r$$$. Then, every element in the domino-range of
$$$r-1$$$ will each decrease by $$$x \times D$$$. Then, when you meet the barrier between the current and previous range, you should check whether you should remove the barrier and decrease
$$$x$$$, or stop the procedure. Let's say that you need to decrease $$$h_{l_{next}}$$$ by $$$y \times D$$$ in order to merge the ranges. If $$$x \geq y$$$, then every element in the previous
range should be decreased by $$$(x-y) \times D$$$ units. Then you repeat the procedure until you run out of "decreases" or of barriers. If implemented with the correct data structure for
range updates, this algorithm will be fast, as with every addition of an element, you add at most $$$1$$$ barrier, remove some number of barriers, then stop at $$$1$$$ barrier. This will
result in $$$O(N)$$$ amortized traversal of barriers. In order to implement the range updates, you can keep a segment tree with lazy propagation. The answer will be, as mentioned, the sum of
$$$b_i$$$-s and the check whether an interval is valid is to see if $$$h_l \geq 0$$$. The final complexity is $$$O((N+Q)\ log_{\,2}\ N)$$$.
$$$\mathbf{Day\ 3.\ Problem\ B}$$$
As it gets messy very quickly, I will only mention the key ideas of my solution. Its main idea is to use a centroid decomposition of the tree and keep online updates. By keeping a centroid
decomposition, our goal is to calculate the number of triplets in each centroid's subtree, whose path passes trough the current centroid. Let's denote $$$u$$$-s centroid subtree, where
$$$u$$$ is root, by $$$T_u$$$. The "subtrees" in $$$T_u$$$ are the subtrees of the children of $$$u$$$ in $$$T_u$$$. The paths are split into the following categories:
$$$(1)$$$ Triplets, including the centroid
• If $$$f_{centroid} = 0$$$, their number is the count of paths of the type $$$1-2$$$ in each subtree.
• If $$$f_{centroid} = 1$$$, their number is the ways of choosing $$$0$$$ and $$$2$$$ from separate subtrees in $$$T_u$$$.
• If $$$f_{centroid} = 2$$$, their number is the count of paths of the type $$$1-0$$$ in each subtree.
$$$(2)$$$ Paths, not including the centroid
• $$$0-1-centroid-2$$$, equal to the number of ways of selecting paths of the type $$$1-0$$$ and $$$2$$$ from different subtrees.
• $$$0-centroid-1-2$$$, equal to the number of ways of selecting paths of the type $$$0$$$ and $$$1-2$$$ from different subtrees.
The whole solution is based on keeping the mentioned pairwise products in a sane way. You should be able to track down how an update is changing the count of $$$1-0$$$ and $$$1-2$$$ paths for
every subtree. For every node, you keep in which subtree is situated for every of its parent centroids, and keep three Fenwicks on the dfs order for each $$$T_u$$$, one for $$$0$$$, $$$1$$$,
and $$$2$$$ updates, which helps calculating the change in the count of triplets. Instead of keeping $$$3 \times N$$$ separate fenwicks, I glued them together for better memory and time
management. With very careful implementation of $$$356$$$ lines, I managed to provide an $$$O(N\ log_{\,2}^{\,2} N)$$$ solution with a big constant, and it passed for $$$2.3$$$ seconds out of
» the provided $$$3 seconds$$$.
libobil $$$\mathbf{Day\ 3.\ Problem\ C}$$$
The first thing I did was to solve the third subtask. It only requires to be able to determine whether you can reach a certain position or not. It could be solved by finding $$$[0,+\infty] \
setminus [L_1, R_1] \setminus [L_2, R_2] \setminus [L_3, R_3]\ \setminus \ ... \setminus \ [L_N, R_N]$$$. Then, you sort the remaining intervals and keep a vector with the reachable
positions. Imagine you have determined the reachable positions of the first $$$x-1$$$ intervals and now you seek to find the available positions of the $$$x$$$-th interval, which is $$$[l,r]
$$$. If you can reach the $$$pos$$$-th position $$$(l \leq pos \leq r)$$$, then you can reach every position in the range $$$[pos + 1, r]$$$. Now, we need to find that position $$$pos$$$. As
the intervals are uninteresecting, then we can't reach the interval $$$[l,r]$$$ by doing a simple step, so we need to find the farthest reachable position down, from which we can jump to the
current interval. This can be done with a binary search, because what you really search for is the first interval, which intersects with $$$[l-D, r-D]$$$. To answer the queries you can do a
binary search on the intervals of available positions. Now, after solving the third subtask, we can notice that the optimal path of all paths to a certain $$$X_j$$$ is one using the least or
the most $$$D$$$-jumps possible. Finding the least amount of $$$D$$$-jumps to a certain position is easy, the count is equal to the minimum $$$D$$$-jumps to reach the interval as a whole, so
if the jumps for every interval $$$i$$$ is $$$minJumps_i$$$, and the interval, which firstly intersects with $$$[l-D,r-D]$$$ is with index $$$parent$$$, then $$$minJumps_i = minJumps_{parent}
+ 1$$$. Now, the more interesting part is to count the path with the most $$$D$$$ jumps. I tried the following greedy strategy — if you want to find the path with the most $$$D$$$ jumps,
which ends at position $$$X$$$, check whether position $$$X-D$$$ is reachable, and if it is, then jump to it. If it isn't, go to $$$X-1$$$. It passed the tests for the first subtask, so I
decided to optimize it, hoping for the $$$O(NQ)$$$ subtask, by doing the following:
1) Have a memoization, so I don't end up calculating the same answer twice.
2) Instead of always checking whether $$$X-D$$$ is reachable, find the interval you're in and do all the jumps possible at once, so you stay in the interval.
3) If $$$X-D$$$ is outside the current interval, check if it's in another. If it is, directly jump to it.
4) Otherwise find the first interval, before $$$X-D$$$. Directly jump to its right border.
By doing the described optimizations, the solution passed for $$$100$$$ points. I haven't proved it yet, so if you could provide proof/antitest to it it wouldn't be left unappreciated. Still,
I believe that the tests are strong enough to not allow such a solution to pass, if it has bad complexity.
$$$\mathbf{Day\ 4.\ Problem\ A}$$$
We can follow this given greedy strategy: for every query, catch the flight which arrives at the next destination the earliest. Then you can choose which flight you'll be taking first and
then stick to the greedy. By doing this, you will score $$$14$$$ points.
Then, instead of doing this procedure over and over again, you can precalculate for every flight $$$i$$$ the transferring one $$$par_i$$$ from the next position, where $$$i$$$ and $$$par_i$$$
follow some kind of numeration. After doing this, you can build a tree, where the edges are between $$$par_i$$$ and $$$i$$$. Every edge has $$$2$$$ fixated costs — one when transferring and
one when ending your jouyney with the current flight. Every path between $$$(l,r)$$$ can be broken down to $$$r-l-1$$$ flights of the first type and one of the second type. Now, every query
is depicted as follows: for a set of nodes (the flights, coming from $$$l$$$), find the path with edge count $$$r-l$$$, which is with the lowest length. For the subtasks with $$$M_i \leq
5$$$, this can be implemented directly with binary lifting for a complexity of $$$O(N\ log_{\,2}\ N + Q\ M_i\ log_{\,2} \ N)$$$, but there is a better approach. Instead of using binary
lifting, you can notice, that the set of the mentioned $$$l$$$-flights is all the nodes of a certain depth $$$x$$$ in the tree, and all the destination nodes ($$$r-1$$$-flights) are at
another depth $$$y$$$. You answer the aforementioned queries offline by doing a dfs traversal of the tree, writing down the parents at all depths, and when reaching a node with the dfs, you
can access all the parents you want with $$$O(1)$$$ complexity. This will remove the $$$log$$$ from the complexity, giving you $$$O(N\ log_{\,2}\ N + QM_i)$$$, still giving you $$$54$$$
Now, the only thing what remains is what with do with big $$$M_i$$$-s. Let's split the $$$l$$$-s into two types — big and small. The small $$$l$$$-s will be with $$$M_i \leq \sqrt{\sum M_i}
$$$ and the big ones will be with $$$M_i > \sqrt{\sum M_i}$$$. We can answer the queries with the small $$$M_i$$$-s for a complexity of $$$O(N\ log_{\,2}\ N + Q \sqrt{\sum M_i})$$$, which is
fast enough. The only thing what remains is to find a way to answer all the queries for the big $$$M_i$$$-s. One can notice, that there will be at most $$$O(\sqrt{\sum M_i})$$$ big
$$$M_i$$$-s, and if we answer the queries for a big $$$M_i$$$ with a complexity of $$$O(N+Q)$$$, our solution would be finished. For doing this, we can do a dfs, similar to the one mentioned
before, as we keep the best children at the depth of $$$l$$$-s flights for every node and chkmin the answer for it's $$$r$$$. We now have a solution with complexity $$$O(N\ log_{\,2}\ N + Q \
sqrt{\sum M_i})$$$, which passes for $$$100$$$ points.
$$$\mathbf{Closing\ remarks}$$$
This analysis of mine isn't meant to be professional and I'm sure I've made mistakes in it. I hope it still is useful for solving the tasks. I very much hope to receive as comments for the
rest of the tasks. Also, If you have any questions, again, feel free to ask :). And also, sorry if my English isn't the best, but I think it's understandable enough :).
• 7 months ago, # ^ |
» I haven't proved it yet, so if you could provide proof/antitest to it it wouldn't be left unappreciated.
Really depends on what "Have a memoization" means, but if I am not mistaken, it seems what you are effectively doing is precalculating the maximum jumps for each right end, and
histrictchki seeing to which of those the current query rounds down to? Or on second thought, I feel that a testcase where the availible segments are something akin to $$$[i * (D + 1), i * (D +
1) + D - 1]$$$ might fail your code (because memoization might fail on such cases where you succesively apply step 3)-- could you link your code somewhere and maybe I can figure out
some counter-testcase :))?
Maybe a more «sure» solution which looks to be the same?
□ » Well, yeah, if it works, it would be because the solution mostly relies on the precalced answers for the right ends. However, I think a test in the form of blocks of $$$D$$$ free
» spaces and $$$1$$$ occupied after them should get me a TL because I will be randomly traversing through the tower. Still, here is my code: https://pastebin.com/mnmePxZc
And yeah, I think your idea should be correct, probably by doing amortized stuff with data structures, which could even be with a set.
7 months ago, # |
← Rev. 3 → +87
JOI Spring Camp is certainly a great contest series in terms of its OI-style format, problem quality, difficulty, and the total length of 20 hours. I always solved most camp problems at some
point in the year (usually during our national IOI camp training), but I could never participate in all seriousness. This year (right after participating in Potyczki Algorytmiczne 2024,
another great contest spanning six days), I reserved my 10 pm to 3 am time to focus solely on virtual participation. It was a rough ride (I still suffer from headaches from 5 am sleep), but
I had a lot of fun; thank you as always!
Here are my very honest impressions of each problem:
Day 1
Day 2
Day 3
Day 4
My ranking for the past 13 years of JOISC: 2017 > 2023 > 2020 > 2014 > 2012 > 2022 > 2019 > 2016 > 2015 > 2021 > 2024 > 2013 > 2018. I think this year was slightly less than a "great"
contest, but still an interesting problemset anyway.
• 7 months ago, # ^ |
» ← Rev. 2 → +30
Ok, here's my implementation for everything besides tricolor and collection: Link
Btw, seems like Mr. JOI has a sneaky plot for invading the KOI plateau...
» ojuz Any plans on adding them to your website?
BunBunChocolat Or is there another place where they are already added with English statements? I saw Atcoder has them but the translation is pretty bad IMO.
• »
» 7 months ago, # |
When will you upload test data?
» 6 months ago, # |
When will test data be published? Usually it is published within a week and problems are also then added to atcoder/oj.uz for upsolving.
6 months ago, # |
» ← Rev. 2 → +20
ojuz You can solve all problems here: https://oj.uz/problems/source/647
Why I couldn't upload day4 tennis (now uploaded)
• 6 months ago, # ^ |
» +3
Writing the checker for this problem is really easy assuming you've solved it, but in case you don't want to be spoiled I'll hide it below:
small tennis spoiler
□ » 6 months ago, # ^ |
» 0
Thanks, done
6 months ago, # |
← Rev. 3 → +25
» Contest Data
E869120 We updated the contest page of JOI Spring Camp. Test Data, Problem Statement & Sample Solutions are added.
• https://www2.ioi-jp.org/camp/2024/2024-sp-tasks/index.html
We are very sorry for late upload. All of the responsibility are belong to us. | {"url":"https://mirror.codeforces.com/blog/entry/127315","timestamp":"2024-11-05T05:50:33Z","content_type":"text/html","content_length":"294519","record_id":"<urn:uuid:16faa6fd-eb57-4d2e-b38b-cd5d3d5f6ef9>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00807.warc.gz"} |
Compressed Trie
A Compressed Trie compress the paths of nodes with only one child.
• Each node stores an index, corresponding to the depth in the uncompressed trie
□ This gives the next bit to be tested during a search
• A compressed trie with keys has at most internal nodes | {"url":"https://stevengong.co/notes/Compressed-Trie","timestamp":"2024-11-14T23:58:30Z","content_type":"text/html","content_length":"12885","record_id":"<urn:uuid:61b732e0-f28d-4ca8-8992-0b22d08e32da>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00047.warc.gz"} |
Talk Keyword Index
This page contains an index consisting of author-provided keywords.
Abstract machines
Algorithmic game theory
Applied pi-calculus
asynchronous automata
Asynchronous hyperproperties
Asynchronous transition systems
Boolean Büchi objectives
bounded linear logic
burst automata
Büchi automata
Büchi non-emptiness problem
cascade product
Closed languages
concurrency theory
Concurrent games
control flow
Controller Synthesis
Decidability issues
Depth-bounded processes
Dynamical Systems
edge clique cover
Energy games
Equational basis
Event-clock automata
Expressiveness issues
Failure Handling
fair simulation
fair termination
Fast-growing complexity classes
finite-memory strategies
fixed points
Game theory
Hidden Markov Model
higher-dimensional automata
Infinite-state systems
interface graph
interval posets
invariant checking
Kleene algebra
Kleene theorem
Krohn Rhodes theorem
labelled precube categories
language equivalence
language inclusion
linear logic
linear π-calculus
local construction
Local-time semantics
Lyapunov exponent
Markov chains
markov decision process
Markov decision processes
Mazurkiewicz traces
memoryless optimal strategies
Model checking
Nash Equilibrium
Networks of timed automata
Nondeterministic Moore machines
One-Counter Net
Parallel composition
Parameter Synthesis
parameterized verification
Parametric Markov chains
Parity objectives
Partial information
Petri box
Petri net
probabilistic polynomial time
process calculi
Process Calculus (pi-calculus)
propositional dynamic logic
randomised strategies
Rational verification
reconfigurable systems
regular model-checking
regular trace languages
regular transition systems
resource logics
separation logic
Sequential Probability Ratio Test
Session types
static construction
stochastic games
Strong linearizability
Strong observational refinement
Structural operational semantics
Temporal logics for hyperproperties
Timed automata
tree automata
two-player games on graphs
Type System
Vector Addition System
vector addition systems
Weak progressive forward simulation
Weak semantics
Weighted timed games
Zones and abstractions | {"url":"https://easychair.org/smart-program/CONCUR2022/talk_keyword_index.html","timestamp":"2024-11-03T09:04:40Z","content_type":"application/xhtml+xml","content_length":"55591","record_id":"<urn:uuid:03f7b726-e160-4485-8fe9-abe89010d479>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00498.warc.gz"} |
Working Capital Is Current Assets
Working capital is the difference between a company's current assets, such as cash, accounts receivable (customers' unpaid bills) and inventories of raw. Related to this Question · Working capital
is: A. · The working capital ratio is: a. · The current ratio is: a. · The current ratio is computed as: a. · Working. Working capital measures how effectively a business can pay down its debts. It's
calculated by subtracting your current liabilities from your current assets. A. It includes both operating assets and liabilities, such as accounts receivable, accounts payable, and inventory, as
well as financial assets and liabilities. Broadly defined, working capital is the excess of current assets over current liabilities. It is cash and other assets expected to be consumed or converted
Working Capital Metrics · Net Working Capital (NWC) is figured by subtracting the total current liabilities from the total current assets. · Current Ratio is. Working capital is calculated by
subtracting current liabilities from current assets. Due to differences in businesses and the fact that working capital is. Working capital is equal to current assets minus current liabilities.
Working capital is the difference between a company's current assets and current. Net working capital (also known as working capital) is the overall result of all the assets obtained by a company
minus the operating current liabilities. This. Working capital (also known as net working capital) is defined as current assets minus current liabilities. Therefore, a company with $, of current
assets. A company's working capital is defined as the difference between a company's current assets (such as cash, accounts receivable, and inventory) and its current. Simply put, Net Working Capital
(NWC) is the difference between a company's current assets and current liabilities on its balance sheet. It is a measure of a. The net working capital formula is current assets minus current
liabilities. Current is short-term, meaning conversion to cash within twelve months or the. Working capital is equal to current assets minus current liabilities. Changes in this account are crucial
to translating net income into cash because when. A positive working capital indicates that a company has more current assets than current liabilities. This is typically a healthy sign as it shows
that the.
Net working capital may not always provide an accurate measure of liquidity because some current assets can't be easily converted to cash. Excessive NWC may. In financial accounting, working capital
is a specific subset of balance sheet items and is calculated by subtracting current liabilities from current assets. Operating Current Assets → Accounts Receivables (A/R), Inventory, Prepaid
Expenses; Operating Current Liabilities → Accounts Payable (A/P), Accrued Expense. Unlike inventory, accounts receivable and other current assets, cash then earns a fair return and should not be
included in measures of working capital. Are. Working capital ratio is a measurement that shows a business's current assets as a proportion of its liabilities. It's a metric that provides an overview
of. Current assets include items such as cash, accounts receivable, and inventory items. · Current liabilities refer to outstanding debts like accounts payable and. Generally, current assets and
current liabilities are expected to generate or use cash within a short-term period, typically 12 months or less. NWC is a measure. Changes in working capital is included in cash flow from operations
because companies typically increase and decrease their current assets and current. Net working capital (also known as working capital) is the overall result of all the assets obtained by a company
minus the operating current liabilities. This.
The working capital ratio is calculated by subtracting current liabilities from current assets. Working capital formula. Working capital = current assets –. Current assets can include cash, accounts
receivable, inventory, cash equivalent in checking and savings accounts, prepaid expenses, inventory, and raw. The working capital is the difference between a company's current assets, such as cash,
accounts receivable (unpaid invoices from customers) and inventories of. Net Working Capital Ratio - A firm's current assets less its current liabilities divided by its total assets. It shows the
amount of additional funds. With the current ratio calculation, a business's current assets are divided by its current liabilities. A ratio below one means that its net working capital is.
Shark Tank Curtain Rod Hangers | What Is P And L Statement | {"url":"https://cazinobitcoin1.site/gainers-losers/working-capital-is-current-assets.php","timestamp":"2024-11-06T05:29:35Z","content_type":"text/html","content_length":"12317","record_id":"<urn:uuid:f77d6167-bb66-4ccb-ac09-209058898d56>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00747.warc.gz"} |
Section: Research Program
Functional Inference, semi- and non-parametric methods
Participants : El-Hadji Deme, Jonathan El-Methni, Stéphane Girard, Gildas Mazo, Farida Enikeeva, Seydou-Nourou Sylla.
Key-words: dimension reduction, extreme value analysis, functional estimation.
We also consider methods which do not assume a parametric model. The approaches are non-parametric in the sense that they do not require the assumption of a prior model on the unknown quantities.
This property is important since, for image applications for instance, it is very difficult to introduce sufficiently general parametric models because of the wide variety of image contents.
Projection methods are then a way to decompose the unknown quantity on a set of functions (e.g. wavelets). Kernel methods which rely on smoothing the data using a set of kernels (usually probability
distributions) are other examples. Relationships exist between these methods and learning techniques using Support Vector Machine (SVM) as this appears in the context of level-sets estimation (see
section 3.3.2 ). Such non-parametric methods have become the cornerstone when dealing with functional data [66] . This is the case, for instance, when observations are curves. They enable us to
model the data without a discretization step. More generally, these techniques are of great use for dimension reduction purposes (section 3.3.3 ). They enable reduction of the dimension of the
functional or multivariate data without assumptions on the observations distribution. Semi-parametric methods refer to methods that include both parametric and non-parametric aspects. Examples
include the Sliced Inverse Regression (SIR) method [71] which combines non-parametric regression techniques with parametric dimension reduction aspects. This is also the case in extreme value
analysis [65] , which is based on the modelling of distribution tails (see section 3.3.1 ). It differs from traditional statistics which focuses on the central part of distributions, i.e. on the
most probable events. Extreme value theory shows that distribution tails can be modelled by both a functional part and a real parameter, the extreme value index.
Modelling extremal events
Extreme value theory is a branch of statistics dealing with the extreme deviations from the bulk of probability distributions. More specifically, it focuses on the limiting distributions for the
minimum or the maximum of a large collection of random observations from the same arbitrary distribution. Let ${X}_{1,n}\le ...\le {X}_{n,n}$ denote $n$ ordered observations from a random variable
$X$ representing some quantity of interest. A ${p}_{n}$-quantile of $X$ is the value ${x}_{{p}_{n}}$ such that the probability that $X$ is greater than ${x}_{{p}_{n}}$ is ${p}_{n}$, i.e. $P\left(X>
{x}_{{p}_{n}}\right)={p}_{n}$. When ${p}_{n}<1/n$, such a quantile is said to be extreme since it is usually greater than the maximum observation ${X}_{n,n}$ (see Figure 1 ).
Figure 1. The curve represents the survival function $x\to P\left(X>x\right)$. The $1/n$-quantile is estimated by the maximum observation so that ${\stackrel{^}{x}}_{1/n}={X}_{n,n}$. As illustrated
in the figure, to estimate ${p}_{n}$-quantiles with ${p}_{n}<1/n$, it is necessary to extrapolate beyond the maximum observation.
To estimate such quantiles therefore requires dedicated methods to extrapolate information beyond the observed values of $X$. Those methods are based on Extreme value theory. This kind of issue
appeared in hydrology. One objective was to assess risk for highly unusual events, such as 100-year floods, starting from flows measured over 50 years. To this end, semi-parametric models of the tail
are considered:
$P\left(X>x\right)={x}^{-1/\theta }\ell \left(x\right),\phantom{\rule{0.277778em}{0ex}}x>{x}_{0}>0,$ (2)
where both the extreme-value index $\theta >0$ and the function $\ell \left(x\right)$ are unknown. The function $\ell$ is a slowly varying function i.e. such that
$\frac{\ell \left(tx\right)}{\ell x}\to 1\phantom{\rule{4.pt}{0ex}}\text{as}\phantom{\rule{4.pt}{0ex}}x\to \infty$ (3)
for all $t>0$. The function $\ell \left(x\right)$ acts as a nuisance parameter which yields a bias in the classical extreme-value estimators developed so far. Such models are often referred to as
heavy-tail models since the probability of extreme events decreases at a polynomial rate to zero. It may be necessary to refine the model (2 ,3 ) by specifying a precise rate of convergence in (3 ).
To this end, a second order condition is introduced involving an additional parameter $\rho \le 0$. The larger $\rho$ is, the slower the convergence in (3 ) and the more difficult the estimation of
extreme quantiles.
More generally, the problems that we address are part of the risk management theory. For instance, in reliability, the distributions of interest are included in a semi-parametric family whose tails
are decreasing exponentially fast. These so-called Weibull-tail distributions [9] are defined by their survival distribution function:
$P\left(X>x\right)=exp\left\{-{x}^{\theta }\ell \left(x\right)\right\},\phantom{\rule{0.277778em}{0ex}}x>{x}_{0}>0.$ (4)
Gaussian, gamma, exponential and Weibull distributions, among others, are included in this family. An important part of our work consists in establishing links between models (2 ) and (4 ) in order
to propose new estimation methods. We also consider the case where the observations were recorded with a covariate information. In this case, the extreme-value index and the ${p}_{n}$-quantile are
functions of the covariate. We propose estimators of these functions by using moving window approaches, nearest neighbor methods, or kernel estimators.
Level sets estimation
Level sets estimation is a recurrent problem in statistics which is linked to outlier detection. In biology, one is interested in estimating reference curves, that is to say curves which bound $90%$
(for example) of the population. Points outside this bound are considered as outliers compared to the reference population. Level sets estimation can be looked at as a conditional quantile estimation
problem which benefits from a non-parametric statistical framework. In particular, boundary estimation, arising in image segmentation as well as in supervised learning, is interpreted as an extreme
level set estimation problem. Level sets estimation can also be formulated as a linear programming problem. In this context, estimates are sparse since they involve only a small fraction of the
dataset, called the set of support vectors.
Dimension reduction
Our work on high dimensional data requires that we face the curse of dimensionality phenomenon. Indeed, the modelling of high dimensional data requires complex models and thus the estimation of high
number of parameters compared to the sample size. In this framework, dimension reduction methods aim at replacing the original variables by a small number of linear combinations with as small as a
possible loss of information. Principal Component Analysis (PCA) is the most widely used method to reduce dimension in data. However, standard linear PCA can be quite inefficient on image data where
even simple image distorsions can lead to highly non-linear data. Two directions are investigated. First, non-linear PCAs can be proposed, leading to semi-parametric dimension reduction methods [67]
. Another field of investigation is to take into account the application goal in the dimension reduction step. One of our approaches is therefore to develop new Gaussian models of high dimensional
data for parametric inference [64] . Such models can then be used in a Mixtures or Markov framework for classification purposes. Another approach consists in combining dimension reduction,
regularization techniques, and regression techniques to improve the Sliced Inverse Regression method [71] . | {"url":"https://radar.inria.fr/report/2013/mistis/uid11.html","timestamp":"2024-11-05T03:27:23Z","content_type":"text/html","content_length":"57120","record_id":"<urn:uuid:45d812a5-ebdd-4609-8616-04a5f6119183>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00039.warc.gz"} |
AS and A Level Further Mathematics B (MEI): Do I have to study A Level Maths in order to study A Level Further Maths?
No, GCE qualifications are now linear and decoupled so there is no requirement to be studying A Level Maths in order to take Further Maths.
A Level Maths is, however, assumed knowledge for A Level Further Maths so if you previously studied the legacy unitised A Level Maths or an international equivalent to A Level Maths then you will
need to be familiar with the content of the new A Level Maths. Similarly, AS Maths is assumed knowledge for AS Further Maths.
Each qualification is a separate entity and there is no restriction on sitting Maths and Further Maths with different awarding organisations.
Stay connected
If you have any queries or questions, email us via maths@ocr.org.uk, call us on 01223 553998 or Tweet us @OCR_Maths. You can also sign up to subject updates and receive up-to-date email information
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0 comments
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Interaction of short waves in a barotropic atmosphere
The Cauchy problem for barotropic atmosphere equations is analyzed for low Mach numbers. An asymptotic solution with respect to the Mach number (M much less than 1) is obtained which describes the
interaction and generation of rapidly oscillating vortices. A new model equation is obtained which describes the formation and evolution of a two-phase vortex; the equation is a nonlinear
pseudodifferential equation on a torus.
Akademiia Nauk SSSR Doklady
Pub Date:
□ Atmospheric Physics;
□ Barotropic Flow;
□ Cauchy Problem;
□ Short Wave Radiation;
□ Wave Interaction;
□ Asymptotic Methods;
□ Existence Theorems;
□ Mach Number;
□ Theorem Proving;
□ Two Phase Flow;
□ Vortices;
□ Geophysics | {"url":"https://ui.adsabs.harvard.edu/abs/1989DoSSR.307..844O/abstract","timestamp":"2024-11-11T22:32:50Z","content_type":"text/html","content_length":"34039","record_id":"<urn:uuid:0694a0cf-cdc4-42a1-8678-7fc98dd3c66e>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00588.warc.gz"} |
Define 'Pearson correlation coefficient'. - Asksia.ai
Answer from Sia
Posted 7 months ago
Solution by Steps
step 1
Define the Pearson correlation coefficient: The Pearson correlation coefficient, denoted as $r$, is a measure of the linear correlation between two variables $X$ and $Y$. It is the ratio of the
covariance of the two variables to the product of their standard deviations
step 2
The formula for the Pearson correlation coefficient is: $r = \frac{\sum_{i=1}^{n}(X_i - \bar{X})(Y_i - \bar{Y})}{\sqrt{\sum_{i=1}^{n}(X_i - \bar{X})^2}\sqrt{\sum_{i=1}^{n}(Y_i - \bar{Y})^2}}$ where $
\bar{X}$ and $\bar{Y}$ are the means of the $X$ and $Y$ variables, respectively, and $n$ is the number of data points
step 3
Interpretation of $r$: The value of $r$ ranges from -1 to 1. A value of 1 implies a perfect positive linear relationship, -1 implies a perfect negative linear relationship, and 0 implies no linear
relationship between the variables
The Pearson correlation coefficient $r$ is a measure of the linear relationship between two variables, calculated using the formula provided.
Key Concept
Pearson correlation coefficient
The Pearson correlation coefficient quantifies the degree of linear relationship between two variables, with values ranging from -1 to 1.
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Heath Jarrow and Morton Example Three: Modeling Interest Rates with Two Factors and Rate and Maturity-Dependent Volatility - SAS Risk Data and Analytics
In the first two blogs in this series, we provided worked examples of how to use the yield curve simulation framework of Heath, Jarrow and Morton using two different assumptions about the volatility
of forward rates. The first assumption was that volatility was dependent on the maturity of the forward rate and nothing else. The second assumption was that volatility of forward rates was
dependent on both the level of rates and the maturity of forward rates being modeled. Both of these models were one factor models, implying that random rate shifts are either all positive, all
negative, or zero. This kind of yield curve movement is not consistent with the yield curve twists that are extremely common in the U.S. Treasury market. In this blog we generalize the model to
include two risk factors in order to increase the realism of the simulated yield curve.
The author wishes to thank Robert A. Jarrow for his encouragement and advice on this series of worked examples of the HJM approach. What follows is based heavily on Prof. Jarrow’s Modeling Fixed
Income Securities and Interest Rate Options (second edition, 2002), particularly chapters 4, 6, 8, 9 and 15.
The first two blogs in this series implemented the work of Heath Jarrow and Morton (1990, 1990, and 1992) using one random risk factor to drive interest rate movements under two different volatility
van Deventer, Donald R. “Heath Jarrow and Morton Example One:
Modeling Interest Rates with One Factor and Maturity-Dependent Volatility,” Kamakura blog, www.kamakuraco.com, March 2, 2012.
van Deventer, Donald R. “Heath Jarrow and Morton Example Two:
Modeling Interest Rates with One Factor and Rate and Maturity-Dependent Volatility,” Kamakura blog, www.kamakuraco.com, March 6, 2012.
The volatility assumptions used so far include (a) interest rate volatility that is a function of the maturity of the forward rate and (b) interest rate volatility that depends both on the maturity
of the forward rate and the level of the one year spot rate of interest. Both assumptions are much more general assumption that that used by Ho and Lee [1986] (constant volatility) or by Vasicek
[1977] and Hull and White [1990] (declining volatility).
In this blog series, we use data from the Federal Reserve statistical release H15 published on April 1, 2011. U.S. Treasury yield curve data was smoothed using Kamakura Risk Manager version 7.3 to
create zero coupon bonds via the maximum smoothness forward rate technique of Adams and van Deventer as documented in these two recent blog issues:
van Deventer, Donald R. “Basic Building Blocks of Yield Curve Smoothing, Part 10: Maximum Smoothness Forward Rates and Related Yields versus Nelson-Siegel,” Kamakura blog, www.kamakuraco.com,
January 5, 2010. Redistributed on www.riskcenter.com on January 7, 2010.
van Deventer, Donald R. “Basic Building Blocks of Yield Curve Smoothing, Part 12: Smoothing with Bond Prices as Inputs,” Kamakura blog, www.kamakuraco.com, January 20, 2010. Redistributed on
The smoothed U.S. Treasury yield curve and the implied forward yield curves monthly for ten years looks like this:
The continuous forward rate curve and zero coupon bond yield curve that prevailed as of the close of business on March 31, 2011 were as follows:
Probability of Yield Curve Twists in the U.S. Treasury Market
The table below is taken from the blog entry
van Deventer, Donald R. “Pitfalls in Asset and Liability Management: One Factor Term Structure Models,” Kamakura blog, www.kamakuraco.com, November 7, 2011. Reprinted in Bank Asset and Liability
Management Newsletter, January, 2012.
It shows that, for 12,386 days of movements in U.S. Treasury forward rates, yield curve twists occurred on 94.3% of the observations.
In order to incorporate yield curve twists in interest rate simulations under the Heath Jarrow and Morton framework, we introduce a second risk factor driving interest rates in this blog. The single
factor yield curve models of Ho and Lee, Vasicek, Hull and White, Black Derman and Toy, and Black and Karasinski are unable to model yield curve twists and therefore they do not provide a sufficient
basis for interest rate risk management.
Objectives of the Example and Key Input Data
Following Jarrow (2002), we make the same modeling assumptions for our worked example as in the first blog in this series:
• Zero coupon bond prices for the U.S. Treasury curve on March 31, 2011 are the basic inputs.
• Interest rate volatility assumptions are based on the Dickler, Jarrow and van Deventer blog series on daily U.S. Treasury yields and forward rates from 1962 to 2011. In this blog, we retain the
volatility assumptions used in the second blog but expand the number of random risk factors driving interest rates to two factors.
• The modeling period is 4 equal length periods of one year each.
• The HJM implementation is that of a “bushy tree” which we describe below
The HJM framework is usually implemented using Monte Carlo simulation or a “bushy tree” approach where a lattice of interest rates and forward rates is constructed. This lattice, in the general
case, does not “recombine” like the popular “binomial” or “trinomial” trees used to replicate Black-Scholes options valuation or simple 1 factor term structure models. In general, the bushy tree
does not recombine because the interest rate volatility assumptions imply a path-dependent interest rate model, not one that is path independent like the simplest one factor term structure model
implementations. In the first two blogs in this series, the bushy tree consisted solely of “up shifts” and “down shifts” because we were modeling as if only one random factor was driving interest
rates. In this blog, with two random factors, we move to a bushy tree that has “up shifts,” “mid shifts,” and “down shifts” at each node in the tree. Please consider these terms as labels only,
since forward rates and zero coupon bond prices, of course, move in opposite directions:
At each of the points in time on the lattice (time 0, 1, 2, 3 and 4) there are sets of zero coupon bond prices and forward rates. At time 0, there is one set of data. At time one, there are three
sets of data, the “up set,” “the mid set” and the “down set.” At time two, there are nine sets of data (up up, up mid, up down, mid up, mid mid, mid down, down up, down mid, and down down), and at
time three there are 27=3^3 sets of data.
The volatilities of the U.S. Treasury one year spot rate and 1 year forward rates with maturities in years 2, 3, … 10 depend dramatically on the starting level of the one year U.S. Treasury spot
rate. The graph below shows the standard deviation in the annual changes in the one year U.S. Treasury forward rates maturing in years 2, 3 and 4 as a function of the starting U.S. Treasury 1 year
spot rate. Volatilities are reported for spot rates between 0.2498% and 0.50%, 0.50% and 0.75%, 0.75% and 1.00%, and then in single percent increments up to 10%. Spot rates over 10% made up the
final grouping:
Forward rate volatility rises in a smooth but complex way as the level of interest rates rises.
The results also confirm that the level of interest rates does impact the level of forward rate volatility but in a more complex way than the lognormal interest rate movements assumed by Fischer
Black with co-authors Derman and Toy (1990), and Karasinski (1991).
The table below shows the actual volatilities for the 1 year changes in continuously compounded forward rates from 1963 to 2011. There were no observations at the time this data was compiled for
starting 1 year U.S. Treasury yields below 0.002499%, so we have set those volatilities by assumption. With the passage of time, these assumptions can be replaced with facts or with data from other
low rate counties like Japan.
We use this table later to divide total volatility between two uncorrelated risk factors.
We will use the zero coupon bond prices prevailing on March 31, 2011 as our other inputs:
Introducing a Second Risk Factor Driving Interest Rates
In the first two worked examples of the Heath Jarrow and Morton approach, the nature of the single factor shocking 1 year spot and forward rates was not specified. In this blog, we take a cue from
the popular academic models of the term structure of interest rates and postulate that one of the two factors driving changes in forward rates is the change in the short run rate of interest. In the
context of our annual model, the “short rate” is the one year U.S. Treasury yield. For each of the 1 year U.S. Treasury forward rates f[k](t), we run the regression
where the change in continuously compounded yields is measured over annual intervals from 1963 to 2011. The coefficients of the regressions for the 1 year forward rates maturing in years 2, 3, …, 10
are as follows:
Graphically, it is easy to see that the historical response of forward rates to the spot rate of interest is neither constant (as assumed by Ho and Lee [1986]) nor declining, as assumed by Vasicek
[1977] and Hull and White [1990]. Indeed, the response of the 1 year forwards maturing in years 8, 9, and 10 year to changes in the 1 year spot rate is larger than the response of the 1 year forward
maturing in year 7:
We use these regression coefficients to separate the total volatility (shown in the table above) of each forward rate between two risk factors. The first risk factor is changes in the one year spot
U.S. Treasury. The second factor represents all other sources of shocks to forward rates.
Because of the nature of the linear regression of changes in 1 year forward rates on the first risk factor, changes in the 1 year spot U.S. Treasury rate, we know that risk factor 1 and risk factor 2
are uncorrelated. Therefore total volatility for the forward rate maturing in T-t=k years can be divided as follows between the two risk factors:
We also know that the risk contribution of the first risk factor, the change in the spot 1 year U.S. treasury rate, is proportional to the regression coefficient αk of changes in forward rate
maturing in year k on changes in the 1 year spot rate. We denote the total volatility of the 1 year U.S. Treasury spot rate by the subscript “1,total”:
This allows us to solve for the volatility of risk factor 2 using this equation:
Because the total volatility for each forward rate varies by the level of the 1 year U.S. Treasury spot rate, so will the values of σ[k,1] and σ[k,2]. In one case (data group 6 for the forward rate
maturing at time T=3), we set the sigma for the second risk factor to zero and ascribed all of the total volatility to risk factor one because the calculated contribution of risk factor 1 was greater
than the total volatility. In the worked example, we will select the volatilities from look up tables for the appropriate risk factor volatility. Using the equations above, the look up table for
risk factor 1 (changes in the 1 year spot rate) volatility is given here:
The look up table for the general “all other” risk factor 2 is as follows:
Key Implications and Notation of the HJM Approach
The Heath Jarrow and Morton conclusions are very complex to derive but their implications are very straightforward. Once the zero coupon bond prices and volatility assumptions are made, the mean of
the distribution of forward rates (in a Monte Carlo simulation) and the structure of a bushy tree are completely determined by the constraints that there be no arbitrage in the economy. Modelers who
are unaware of this insight would choose means of the distributions for forward rates such that valuation would provide different prices for the zero coupon bonds on March 31, 2011 than those used as
input. This would create the appearance of an arbitrage opportunity, but it is in fact a big error that calls into question the validity of the calculation, as it should.
We show in this example that the zero coupon bond valuations are 100% consistent with the inputs. We now introduce the same notation used in the first blog in this series:
Note that this is a slightly different definition of K than we used in the first two blogs in this series.
We will also see the rare appearance of a trigonometric function in finance, one found in common spreadsheet software:
Note that the current times that will be relevant in building a bushy tree of zero coupon bond prices are current times t=0, 1, 2, and 3. We’ll be interested in maturity dates T=2, 3, and 4. We
know that at time zero, there are 4 zero coupon bonds outstanding. At time 1, only the bonds maturing at T = 2, 3, and 4 will remain outstanding. At time 2, only the bonds maturing at times T = 3
and 4 will remain, and at time 3 only the bond maturing at time 4 will remain. For each of the boxes below, we need to fill in the relevant bushy tree (one for each of the four zero coupon bonds)
with each up shift and down shift of the zero coupon bond price as we step forward one more period (by Δ = 1) on the tree. In the interests of saving space, we’ll again arrange the tree to look like
a table by stretching the bushy tree as follows:
A completely populated zero coupon bond price tree would then be summarized like this; prices shown are for the zero coupon bond price maturing at time T=4 at times 0, 1, 2, and 3:
The mapping of the sequence of up and down states is shown here, consistent with the stretched tree above:
In order to populate the trees with zero coupon bond prices and forward rates, there is one more piece of information which we need to supply.
Pseudo Probabilities
In Chapter 7 of Jarrow (2002), Prof. Jarrow shows that a necessary and sufficient condition for no arbitrage is that, at every node in the tree, the one period return on a zero coupon bond neither
dominates nor is dominated by a one period investment in the risk free rate. As explained in the two previous blogs in this series, If the computed probabilities of an up shift, a “mid shift” and a
down shift are between 0 and 1 everywhere on the bushy tree, then the tree is arbitrage free. Without loss of generality, we set the probability of an up shift to ¼, the probability of a mid shift to
¼, and the probability of a down shift to ½. The no arbitrage restrictions that stem from this set of pseudo probabilities are given below.
Prof. Jarrow goes on to explain on page 129 that “risk neutral valuation” is computed by “taking the expected cash flow, using the pseudo probabilities, and discounting at the spot rate of interest.”
He adds “this is called risk neutral valuation because it is the value that the random cash flow ‘x’ would have in an economy populated by risk-neutral investors, having the pseudo probabilities as
their beliefs.”
We now demonstrate how to construct the bushy tree and use it for risk-neutral valuation.
The Formula for Zero Coupon Bond Price Shifts with Two Risk Factors
In the first two blogs in this series, we used equation 15.17 in Jarrow (2002, page 286) to calculate the no arbitrage shifts in zero coupon bond prices. Alternatively, when there is one risk
factor, we could have used equation 15.19 in Jarrow (2002, page 287) to shift forward rates and derive zero coupon bond prices from the forward rates.
Now that we have two risk factors, it is convenient to calculate the forward rates first. We do this using equations 15.32, 15.39a, and 15.39b in Jarrow (2002, pages 293 and 296). We use this
equation for the shift in forward rates:
The values for the pseudo probabilities and Index(1) and Index (2) are set as follows:
Building the Bushy Tree for Zero Coupon Bonds Maturing at Time T=2
We now populate the bushy tree for the 2 year zero coupon bond. We calculate each element of equation (1). When t=0 and T=2, we know Δ=1 and
P(0,2,s[t]) = 0.98411015.
The one period risk free rate is again
The 1 period spot rate for U.S. Treasuries is r(0, s[t]) =R(0,s[t])-1=0.3003206%. At this level of the spot rate for U.S. Treasuries, volatilities for risk factor 1 are selected from data group 3 in
the look up table above. The volatilities for risk factor 1 for the 1 year forward rates maturing in years 2, 3 and 4 are 0.000492746, 0.000313424, and 0.00016918. The volatilities for risk factor
2 for the 1 year forward rates maturing in years 2, 3 and 4 are 0.003810177, 0.006694414, and 0.008636852.
and therefore K(1,0,T,s[t]) =(√1)( 0.000492746) = 0.000492746. Similarly, K(2,0,T, s[t])= 0.003810177. We also can calculate that [μ(t,t+Δ)Δ]Δ = 0.00000738.
Using formula 1 with these inputs and the fact that the variable Index(1)=-1 and Index(2)=-1 for an up shift gives us the forward returns for an up shift, mid shift and down shift as follows:
1.007170561, 1.018083343, and 1.013610665. From these we calculate the zero coupon bond prices:
P(1,2,s[t] = up) = 0.992880 =1/F(1,2,s[t] = up)
For a mid shift, we set Index(1)=-1 and Index(2)=+1 and calculate
P(1,2,s[t] = mid) = 0.982238=1/F(1,2,s[t] = mid)
For a down shift we set Index(1) = 1 and Index(2)= 0 and recalculate formula 1 to get
P(1,2,s[t] = down) = 0.986572 =1/F(1,2,s[t] = down)
We have fully populated the bushy tree for the zero coupon bond maturing at T=2 (note values have been rounded to six decimal places above for display only), since all of the up, mid and down states
at time t=2 result in a riskless pay-off of the zero coupon bond at its face value of 1.
Building the Bushy Tree for Zero Coupon Bonds Maturing at Time T=3
For the zero coupon bonds and 1 period forward returns ( = 1 + forward rate) maturing at time T=3, we use the same volatilities listed above for risk factors 1 and 2 to calculate
K(1,0,3,s[t]) = 0.00080617
K(2,0,3,s[t]) = 0.010504592
[μ(t,T)Δ]Δ = 0.00004816
The resulting forward returns for an up shift, mid shift and down shift are 1.013189027, 1.032556197, and 1.023468132. Zero coupon bond prices are calculated from the 1 period forward returns, so
P(1,3,s[t] = up) =1/[F(1,2,s[t] = up)F(1,3,s[t] = up)]
The zero coupon bond prices for an up shift, mid shift, and down shift are 0.979956, 0.951268, and 0.963950. To eight decimal places, we have populated the second column of the zero coupon bond
price table for the zero coupon bond maturing at T=3.
Building the Bushy Tree for Zero Coupon Bonds Maturing at Time T = 4
We now populate the bushy tree for the zero coupon bond maturing at T=4. Using the same volatilities as before for both risk factors 1 and 2, we find that
K(1,0,4,s[t]) = 0.000975351
K(2,0,4,s[t]) = 0.019141444
[μ(t,T)Δ]Δ = 0.00012830
Using formula (1) with the correct values for Index(1) and Index (2) leads to the following forward returns for an up shift, mid shift and down shift: 1.020756042, 1.045998862, and 1.033650059. The
zero coupon bond price is calculated as follows:
P(1,4,s[t] = up) =1/[F(1,2,s[t] = up)F(1,3,s[t] = up) F(1,4,s[t] = up)]
This gives us the three zero coupon bond prices of the column labeled 1 in this table for up, mid and down shifts: 0.960029, 0.909435, and 0.932569.
Now we move to the third column, which displays the outcome of the T=4 zero coupon bond price after 9 scenarios: up-up, up-mid, up-down, mid-up, mid-mid, mid-down, down-up, down-mid, and down-down.
We calculate P(2,4,s[t] = up), P(2,4,s[t]=mid) and P(2,4,s[t] = down) after the initial “down” state as follows. When t=1, T=4, and Δ=1 then the volatilities for the two remaining 1 period forward
rates that are relevant are taken from the lookup table for data group 6 for risk factor 1: 0.007871303 , and 0.006312739. For risk factor 2, the volatilities for the two remaining 1 period forward
rates are also chosen from data group 6: 0.003086931 , and 0. The zero value was described above-the implied volatility for risk factor 1 was greater than measured total volatility for the forward
rate maturing at T=3 in data group 6, so the risk factor 1 volatility was set to total volatility and risk factor 2 volatility was set to zero.
At time 1 in the down state, the zero coupon bond prices for maturities at T=2, 3 and 4 are 0.986572, 0.963950, and 0.932569. We make the intermediate calculations as above for the zero coupon
bond maturing at T=4:
K(1,1,4,s[t]) = 0.014184042
K(2,1,4,s[t]) = 0.003086931
[μ(t,T)Δ]Δ = 0.00006964
We can calculate the values of the 1 period forward return maturing at time T=4 in the up state, mid state, and down state as follows: 1.027216983, 1.027216983, and 1.040268304. Similarly, using the
appropriate intermediate calculations, we can calculate the forward returns for maturity at T=3: 1.011056564, 1.01992291, and 1.031592858. Since
P(2,4,s[t] = down up) =1/[F(1,3,s[t] = down up) F(1,4,s[t] = down up)]
the zero coupon bond prices for maturity at T=4 in the down up, down mid, and down down states are as follows:
We have correctly populated the seventh, eighth and ninth rows of column 3 (t=2) of the bushy tree above for the zero coupon bond maturing at T=4 (note values have been rounded to six decimal places
for display only). The remaining calculations are left to the reader.
If we combine all of these tables, we can create a table of the term structure of zero coupon bond prices in each scenario as in examples one and two. The shading highlights two nodes of the bushy
tree where values are identical because of the occurrence of σ[2] = 0 at one point on the bushy tree:
At any point in time t, the continuously compounded yield to maturity at time T can be calculated as y(T-t)=-ln[P(t,T)]/(T-t). Note that we have no negative rates on this bushy tree and that yield
curve shifts are much more complex than in the two prior examples using one risk factor:
We can graph yield curve movements as shown below at time t=1. We plot yield curves for the up, mid and down shifts. These shifts are relative to the 1 period forward rates prevailing at time zero
for maturity at time T=2 and T=3. Because these 1 period forward rates were much higher than yields as of time t=0, all three shifts produce yields higher than time zero yields.
When we add 9 yield curves prevailing at time t=3 and 27 “single point” yield curves prevailing at time t=4, two things are very clear. First, yield curve movements in a two factor model are much
more complex and much more realistic than what we saw in the two one-factor examples. Second, in the low yield environment prevailing as of March 31, 2011, no arbitrage yield curve simulation shows
“there is nowhere to go but up” from a yield curve perspective.
Finally, we can display the 1 year U.S. Treasury spot rates and the associated term structure of 1 year forward rates in each scenario.
Valuation in the Heath Jarrow and Morton Framework
Prof. Jarrow in a quote above described valuation as the expected value of cash flows using the risk neutral probabilities. Note that column 1 denotes the riskless 1 period interest rate in each
scenario. For the scenario number 39 (three consecutive down shifts in zero coupon bond prices), cash flows at time T=4 would be discounted by the one year spot rates at time t=0, by the one year
spot rate at time t=1 in scenario 3 (“down”), by the one year spot rate in scenario 12 (“down down”) at time t=2, and by the one year spot rate at time t=3 in scenario 39 (“down down down”). The
discount factor is
Discount Factor (0,4, down down down) =1/(1.003003)(1.013611)(1.031593)(1.050231)
These discount factors are displayed here for each potential cash flow date:
When taking expected values, we can calculate the probability of each scenario coming about since the probabilities of an up shift, mid shift, and down shift are ¼, ¼ and 1/2:
It is convenient to calculate the “probability weighted discount factors” for use in calculating the expected present value of cash flows:
We now use the HJM bushy trees we have generated to value representative securities.
Valuation of a Zero Coupon Bond Maturing at Time T=4
A riskless zero coupon bond pays $1 in each of the 8 nodes of the bushy tree that prevail at time T=4:
When we multiply this vector of 1s times the probability weighted discount factors in the time T=4 column in the previous table and add them, we get a zero coupon bond price of 0.93085510, which is
the value we should get in a no-arbitrage economy, the value observable in the market and used as an input to create the tree.
Valuation of a Coupon-Bearing Bond Paying Annual Interest
Next we value a bond with no credit risk that pays $3 in interest at every scenario at times T=1, 2, 3, and 4 plus principal of 100 at time T=4. The valuation is calculated by multiplying each cash
flow by the matching probability weighted discount factor, to get a value of 104.70709974. It will surprise many that this is the same value that we arrived at in examples one and two, even though
the volatilities used and number of risk factors used are different. The values are the same because, by construction, our valuations for the zero coupon bond prices at time zero for maturities at T
= 1, 2, 3, and 4 continue to match the inputs. Multiplying these zero coupon bond prices times 3, 3, 3, and 103 also leads to a value of 104.70709974 as it should.
Valuation of a Digital Option on the 1 Year U.S. Treasury Rate
Now we value a digital option that pays $1 at time T=3 if (at that time) the one year U.S. Treasury rate (for maturity at T=4) is over 4%. If we look at the table of the term structure of one year
spot rates over time, this happens at the seven shaded scenarios out of 27 possibilities at time t=3.
The evolution of the spot rate can be displayed graphically as follows:
The cash flow payoffs in the 7 relevant scenarios can be input in the table below and multiplied by the probability weighted discount factors to find that this option has a value of 0.29701554:
Replication of HJM Example 3 in Excel
Kamakura Risk Manager and Kamakura Risk Information Services clients may request a copy of the Excel spreadsheet supporting this blog after signing a supplemental confidentiality agreement. Please
request a copy of the spreadsheet from your Kamakura representative or from info@kamakuraco.com.
The Dickler, Jarrow and van Deventer studies of movements in U.S. Treasury yields and forward rates from 1962 to 2011 confirm that 5-10 factors are needed to accurately model interest rate
movements. Popular one factor models (Ho and Lee, Vasicek, Hull and White, Black Derman and Toy) cannot replicate the actual movements in yields that have occurred. The interest rate volatility
assumptions in these models (constant, constant proportion, declining, etc.) are also inconsistent with observed volatility.
In order to handle a large number of driving factors and complex interest rate volatility structures, the Heath Jarrow and Morton framework is necessary. This blog, the third in a series, shows how
to simulate zero coupon bond prices, forward rates and zero coupon bond yields in an HJM framework with two risk factors and rate-dependent and maturity-dependent interest rate volatility. The
results show a rich twist in simulated yield curves and a pull of rates upward from a very low rate environment. Monte Carlo simulation, an alternative to the bushy tree framework, can be done in a
fully consistent way.
In the next blog in this series, we introduce a third risk factor to further advance the realism of the model.
Other References
Dicker, Daniel T., Robert A. Jarrow and Donald R. van Deventer, “Inside the Kamakura Book of Yields: A Pictorial History of 50 Years of U.S. Treasury Forward Rates,” Kamakura Corporation memorandum,
September 13, 2011.
Dickler, Daniel T. and Donald R. van Deventer, “Inside the Kamakura Book of Yields: An Analysis of 50 Years of Daily U.S. Treasury Forward Rates,” Kamakura blog, www.kamakuraco.com, September 14,
Dicker, Daniel T., Robert A. Jarrow and Donald R. van Deventer, “Inside the Kamakura Book of Yields: A Pictorial History of 50 Years of U.S. Treasury Zero Coupon Bond Yields,” Kamakura Corporation
memorandum, September 26, 2011.
Dickler, Daniel T. and Donald R. van Deventer, “Inside the Kamakura Book of Yields: An Analysis of 50 Years of Daily U.S. Treasury Zero Coupon Bond Yields,” Kamakura blog, www.kamakuraco.com,
September 26, 2011.
Dicker, Daniel T., Robert A. Jarrow and Donald R. van Deventer, “Inside the Kamakura Book of Yields: A Pictorial History of 50 Years of U.S. Treasury Par Coupon Bond Yields,” Kamakura Corporation
memorandum, October 5, 2011.
Dickler, Daniel T. and Donald R. van Deventer, “Inside the Kamakura Book of Yields: An Analysis of 50 Years of Daily U.S. Par Coupon Bond Yields,” Kamakura blog, www.kamakuraco.com, October 6, 2011.
Heath, David, Robert A. Jarrow and Andrew Morton, “Contingent Claims Valuation with a Random Evolution of Interest Rates,” The Review of Futures Markets, 9 (1), 1990.
Heath, David, Robert A. Jarrow and Andrew Morton, “Bond Pricing and the Term Structure of Interest Rates: A Discrete Time Approximation,” Journal of Financial and Quantitative Analysis, December
Heath, David, Robert A. Jarrow and Andrew Morton, “Bond Pricing and the Term Structure of Interest Rates: A New Methodology for Contingent Claims Valuation,” Econometrica, 60(1), January 1992.
Jarrow, Robert A. Modeling Fixed Income Securities and Interest Rate Options, second edition, Stanford Economics and Finance, Stanford University Press, Stanford, California, 2002.
Jarrow, Robert A. and Stuart Turnbull. Derivative Securities, 1996, Southwestern Publishing Co., second edition, fall 2000.
van Deventer, Donald R. “Pitfalls in Asset and Liability Management: One Factor Term Structure Models,” Kamakura blog, www.kamakuraco.com, November 7, 2011. Reprinted in Bank Asset and Liability
Management Newsletter, January, 2012.
van Deventer, Donald R. “Pitfalls in Asset and Liability Management: One Factor Term Structure Models and the Libor-Swap Curve,” Kamakura blog, www.kamakuraco.com, November 23, 2011. Reprinted in
Bank Asset and Liability Management Newsletter, February, 2012.
Slattery, Mark and Donald R. van Deventer, “Model Risk in Mortgage Servicing Rights,” Kamakura blog, www.kamakuraco.com, December 5, 2011.
van Deventer, Donald R., Kenji Imai, and Mark Mesler, Advanced Financial Risk Management, John Wiley & Sons, 2004. Translated into modern Chinese and published by China Renmin University Press,
Beijing, 2007. Second edition forthcoming in 2012.
Donald R. van Deventer
Kamakura Corporation
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system, mathematical models have been widely used. In this talk, I will introduce the typical process of mathematical modelling including mathematical representation, model fitting to data, analysis
and simulations, and experimental validation. Across each step of modelling process, I will also describe our efforts to develop a new mathematical tools and point to the parts of current toolbox of
mathematical biology that need further mathematical development. Finally, I will present how mathematical modelling reveals the key mechanism underlying circadian clocks. | {"url":"http://www.math.snu.ac.kr/board/index.php?mid=colloquia&sort_index=speaker&order_type=desc&l=en&page=7&document_srl=765016","timestamp":"2024-11-14T01:18:57Z","content_type":"text/html","content_length":"43868","record_id":"<urn:uuid:4944c7a6-53ca-4346-810b-fff51d46d172>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00806.warc.gz"} |
Best Strategies to Solve Math Word Problems l Introduction To Math
We Grow People
15 Oct 202103:13
TLDRThis video introduces a two-pronged strategy for solving math word problems effectively. First, determine if the answer should be larger or smaller than the initial parts, guiding you to choose
between addition/multiplication or subtraction/division. Second, identify signal words that suggest specific operations: 'all together' for addition, 'decrease' for subtraction, 'per' for
multiplication, and 'half' for division. By recognizing these clues, viewers can confidently select the correct operation to tackle real-life math challenges.
• 🔍 Identify if the word problem suggests the answer is larger or smaller than the initial parts given; this guides whether to use addition/multiplication or subtraction/division.
• 📚 Look for signal words that indicate specific operations: 'took' and 'how many are left' imply subtraction.
• 🍕 Use subtraction in problems where something is taken away from a total, like subtracting slices of pizza taken by Susie from the total.
• 📒 For problems involving multiplication, look for words like 'for each' and 'how many total', as in calculating the total number of notebooks needed for students.
• 📈 Multiplication is often used when dealing with multiples of a quantity, such as notebooks for each student.
• 📊 Signal words for addition include 'all together', 'combined', 'increase', 'sum', 'total', and 'how many more'.
• ➖ Signal words for subtraction are 'decrease', 'difference', 'how many left', 'remain', 'take away', and 'how many more'.
• ✖️ Signal words for multiplication are 'per', 'each', 'times', 'twice', 'triple', 'etc.', and 'total'.
• 🔄 Signal words for division include 'half', 'third', 'quarter', 'how many each', 'out of', 'percent', and 'quotient'.
• 🧠 Knowing the right operation to choose is crucial for solving word problems and applying math in real-life situations.
• 🚀 Memorizing these strategies and signal words will empower you to tackle any word problem effectively.
Q & A
• What is the main focus of the video transcript provided?
-The video transcript focuses on strategies for solving math word problems by identifying key words and phrases that indicate the appropriate mathematical operation to use.
• What is the first strategy mentioned for choosing the right operation in a word problem?
-The first strategy is to determine if the answer to the word problem is likely to be larger or smaller than the initial parts given, which can suggest whether to use addition/multiplication or
• How does the word 'took' in the pizza example suggest the operation to be used?
-The word 'took' implies that something has been removed or subtracted from the initial amount, indicating that subtraction is the appropriate operation.
• In the notebooks example, what signal words indicate that multiplication is the correct operation?
-The words 'for each' and 'how many total' suggest that the operation involves scaling up the initial quantity (24 students) by a factor (2 notebooks per student), indicating multiplication.
• What are some signal words associated with addition in math word problems?
-Signal words for addition include 'all together', 'combined', 'increase', 'sum', 'total', and 'how many more'.
• What signal words are indicative of the need for subtraction in a word problem?
-Signal words for subtraction include 'decrease', 'difference', 'how many left', 'remain', 'take away', and 'how many more'.
• Can you provide some examples of signal words for multiplication in word problems?
-Examples of signal words for multiplication are 'per', 'each', 'times', 'twice', 'triple', etc., and 'total'.
• What signal words should you look for when division is the required operation in a word problem?
-Signal words for division include 'half', 'third', 'quarter', 'how many each', 'out of', 'percent', and 'quotient'.
• Why is knowing the right operation crucial for solving word problems?
-Knowing the right operation is crucial because it directly affects the solution to the word problem, ensuring accurate results in practical, real-life math situations.
• How can the strategies discussed in the transcript help in real-life situations involving math?
-The strategies help by providing a systematic approach to identifying the necessary operation for solving word problems, which is essential for making correct calculations in everyday scenarios.
• What is the significance of signal words in determining the operation needed to solve a word problem?
-Signal words are significant as they act as cues that guide the problem solver towards the correct mathematical operation, thus helping to avoid errors and solve the problem efficiently.
📚 Mastering Word Problems with Clues
This paragraph introduces the concept of solving word problems by identifying key words and phrases that indicate the mathematical operation needed. It explains a two-pronged strategy: assessing
whether the answer is likely larger or smaller than the initial parts and looking for signal words associated with specific operations. The paragraph uses the example of Susie taking slices from a
pizza to illustrate subtraction and Professor Salmon needing notebooks for her students to demonstrate multiplication. It also lists signal words for addition, subtraction, multiplication, and
division, emphasizing their importance in solving word problems effectively.
💡Word Problems
Word problems are a type of mathematical problem that requires the solver to understand the context and apply mathematical operations based on the given situation. In the video, word problems are the
central theme, as they are presented as a common challenge in real-life scenarios. The script discusses strategies for solving these problems by identifying clues within the text that suggest the
appropriate mathematical operation to use.
Operations in mathematics refer to the basic arithmetic actions such as addition, subtraction, multiplication, and division. The video emphasizes the importance of choosing the correct operation to
solve word problems. It suggests that understanding whether the answer should be larger or smaller than the given parts can guide the selection of the operation, such as addition or multiplication
for larger answers, and subtraction or division for smaller ones.
💡Signal Words
Signal words are specific terms that indicate the mathematical operation needed to solve a word problem. The script provides examples of such words, like 'for each' and 'how many total', which
suggest multiplication, and 'how many left', which implies subtraction. Recognizing these words is crucial for determining the correct approach to solving the problem.
Clues in the context of word problems are the words or phrases that hint at the mathematical operation required to find the solution. The video script uses examples like 'took' and 'how many are
left' to illustrate how these clues point towards subtraction in a pizza-sharing scenario.
Addition is the mathematical operation of combining two or more numbers to find their total or sum. The script mentions words like 'all together' and 'combined' as indicators that addition is the
appropriate operation to use in certain word problems.
Subtraction is the operation of finding the difference between two numbers. In the script, the word 'took' is a clue for subtraction, as it indicates that a certain amount has been removed from the
initial quantity, as seen in the pizza example where two slices were taken from a group of ten.
Multiplication is used to calculate the total when a number is to be added to itself repeatedly. The script uses 'for each' as a signal word for multiplication, as in the example where Professor
Salmon needs to order two notebooks for each of her 24 students, totaling 48 notebooks.
Division is the process of splitting a number into equal parts. The script suggests looking for words like 'half' and 'third' as indicators that division might be the necessary operation in a word
The term 'total' in the script is used in the context of finding the sum or the overall quantity after performing an operation, such as multiplying the number of notebooks needed for each student to
get the total number Professor Salmon must order.
💡Real-life Situations
Real-life situations are practical scenarios where mathematical concepts are applied. The video emphasizes that understanding how to solve word problems is essential for dealing with math in everyday
contexts, such as ordering supplies or sharing food.
Strategies in the context of the video refer to the methods or approaches one can use to solve word problems effectively. The script outlines a two-pronged strategy involving the evaluation of
whether the answer should be larger or smaller and the identification of signal words to determine the correct mathematical operation.
Word problems often require you to infer the solution method from the text rather than being explicitly told.
A two-pronged strategy is introduced to determine the correct mathematical operation for word problems.
Assess if the answer to the problem is likely larger or smaller than the initial values given to choose between addition/multiplication or subtraction/division.
Look for signal words that suggest specific operations such as 'took' and 'how many are left' indicating subtraction.
In the pizza example, 'took' and 'how many are left' confirm the need for subtraction to find remaining slices.
For the notebooks example, 'for each' and 'how many total' suggest multiplication to calculate the total number needed.
Signal words for addition include 'all together', 'combined', 'increase', 'sum', 'total', and 'how many more'.
Subtraction problems often contain words like 'decrease', 'difference', 'how many left', 'remain', 'take away', and 'how many more'.
Multiplication is indicated by words such as 'per', 'each', 'times', 'twice', 'triple', and 'total'.
Division problems may include signal words like 'half', 'third', 'quarter', 'how many each', 'out of', 'percent', and 'quotient'.
Understanding signal words is crucial for choosing the correct operation to solve word problems effectively.
Practical real-life situations often involve word problems, making these strategies essential for everyday math.
The video provides a structured approach to identifying operations needed for word problems, enhancing problem-solving skills.
Learning to solve word problems is not just about academics but also about navigating everyday math challenges.
Mastering these strategies can help tackle any word problem encountered, promoting confidence in mathematical abilities.
The video concludes with the importance of applying these strategies to real-life scenarios for practical math skills. | {"url":"https://math.bot/blog-Best-Strategies-to-Solve-Math-Word-Problems-l-Introduction-To-Math-38517","timestamp":"2024-11-07T00:34:10Z","content_type":"text/html","content_length":"118157","record_id":"<urn:uuid:106b8029-7825-46c0-a9ab-4f271fe1ff2a>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00708.warc.gz"} |
Invariant Factor -- from Wolfram MathWorld
Invariant Factor
The polynomials in the diagonal of the Smith normal form or rational canonical form of a matrix are called its invariant factors.
See also
Rational Canonical Form
Smith Normal Form
Explore with Wolfram|Alpha
Ayres, F. Jr. "Smith Normal Form." Ch. 24 in
Schaum's Outline of Theory and Problems of Matrices.
New York: Schaum, pp. 188-195, 1962.Dummit, D. S. and Foote, R. M.
Abstract Algebra, 2nd ed.
Englewood Cliffs, NJ: Prentice-Hall, 1998.
Referenced on Wolfram|Alpha
Invariant Factor
Cite this as:
Weisstein, Eric W. "Invariant Factor." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/InvariantFactor.html
Subject classifications | {"url":"https://mathworld.wolfram.com/InvariantFactor.html","timestamp":"2024-11-13T07:59:55Z","content_type":"text/html","content_length":"50471","record_id":"<urn:uuid:c84d73c0-ac58-4606-b750-7b2fb4d91f57>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00066.warc.gz"} |
[Solved] Which technical spec can be most easily m | SolutionInn
Which technical spec can be most easily modified without changing current choices for the other two technical
Which technical spec can be most easily modified without changing current choices for the other two technical specs?
Transcribed Image Text:
Quick Start Quick Start QFD Matrix 1 = Strong positive correlation = Some positive correlation = Strong negative correlation X = Some negative correlation Relative Importance (Out of 100) Technical
Spec 1 O Technical Spec 2 Technical Spec 3 value Feature A 25 value Feature B 75
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Plus Two Physics
Friday, April 30, 2010
New Delhi: The Prime Minister's Office, according to sources, is rethinking their earlier decision for letting 14 renowned innovation universities to be set up in places that were selected during
former Human Resource Development (HRD) Minister Arjun Singh's tenure.The decision was put on hold becau...
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Wednesday, April 28, 2010
Follow the link below to read on
• Various types of capacitor
• Factors Affecting Capacitance
• Capacitor Networks
• Charging and Discharging Capacitors
Click Here to Visit the details
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Capacitance is the property of an electric conductor that characterizes its ability to store an electric charge. An electronic device called a capacitor is designed to provide capacitance in an
electric circuit by providing a means for storing energy in an electric field between two conducting bodies.
Read more and watch the interactive JAVA tutorial
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History of the Nobel Prize in Physics
Physics Laureates and Their Work
Laureates' Workplace
Other Articles
Visit the site
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A good collection of java simulations for teaching and learning Physics
Visit the link below
Java Simulations in Physics
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Physics Problem solving strategy
Two factors can help make you a better physics problem solver. First of all, you must know and understand the principles of physics. Secondly, you must have a strategy for applying these principles
to new situations in which physics can be helpful. We call these situations problems. Many students say, “I understand the material, I just can’t do the problems.” If this is true of you as a physics
student, then maybe you need to develop your problem-solving skills. Having a strategy to organize those skills can help you.
Read more Here
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Visit the the website giving the previous question Papers and solutions of all the previous International Physics Olympiads and the solutions.
The site also give details of the past and the coming Olypiads.
Click the Link below to visit the site
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The "four horsemen" of bad health -- poor diet, inactivity, smoking, and excessive drinking -- may indeed add up to a personal apocalypse, researchers have found.
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Tuesday, April 27, 2010
We have started a new
Physics FAQ
section for publishing
frequently asked questions
and answers in Physics by various Boards and schools. The
section will also include questions from visitors and their answers.
Visit the FAQ section
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Monday, April 26, 2010
Bangalore: The All India Engineering Entrance Examination (AIEEE) which was held on Sunday had a 'moderate' difficulty level despite of the paper being set on the expected lines of students, leaving
them baffled with the differential marking scheme.Although the AIEEE had given it a skip in its 2009 ...
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Saturday, April 24, 2010
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Taking another step towards reforming the education system by making it more student-centric, the Central Board of Secondary Education has "advised" teachers in affiliated schools to act more as
facilitators than instructors by cutting down on teacher-talking time and instead enabling more peer-group learning.
In a recent set of guidelines spelling out, in no uncertain terms, the curriculum to be followed in Classes IX and X in CBSE schools, the board has asked teachers "to reduce teacher-talking time to
the minimum; encourage classroom interaction among peers, students and teachers; take up questions for discussion to encourage pupils to participate, and to marshal their ideas and express and defend
their views."
Academics say it is important that teachers speak less in class in order to adopt a constructive approach towards a student's learning and development. "The child must discover things for itself.
Unfortunately, in our country teacher training schools still advocate the teacher talk method. Such guidelines will allow the teacher to look at other teaching methodologies, some of which have been
prescribed," says Neera Chopra, an educational consultant based in Delhi who has worked with several schools in Delhi and abroad, and is an expert on the continuous and comprehensive evaluation
suggested by the CBSE in all affiliated schools.
"In the schools that I visit, I often come across children who say they are able to understand 60% from the textbooks, and that teachers make the whole learning experience boring by trying to explain
things to them. Instead, teachers should focus on guiding children to the right places to look for information, not by explaining everything in textbooks," she says.
While the guidelines break several barriers by making rote learning infeasible and instead focussing on learning by doing, there are fears that such precise instructions could stifle a teacher's
spontaneity and tendency to innovate in the classroom.
A facilitator makes learning happen by demystifying the process of learning and making students autonomous learners.
It is believed by many educators now that facilitation is a never ending function of a teacher who continually develops and supports the learning potential of students.
As a teacher, what is your opinion? Let's interact and discuss. Post your comments and ideas.
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Thursday, April 22, 2010
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Consider the shown arrangement in which a thin rope ‘A’ with a linear density
is connected to a thick rope ‘B’ with linear mass density
= 4 kg/m,
= 0.4 kg/m, M = 1 kg and L = 4l = 1 m. Under these conditions, find:
a. The least possible frequency of vibration
b. The total energy of vibration if the amplitude for both the string is A = 1 mm and the string vibrates at the frequency obtained in (a).
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New Delhi: The Union Public Service Commission (UPSC) has announced the results of the Engineering Services Examination, 2009. The examination for these services was held in June 2009 and the
interviews for Personality Test held in February - March 2009. A total of 469 candidates have been recomm...
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Theoretical problem one
, and
answer form
Theoretical problem two
, and
answer form
Theoretical problem three
, and
answer form
Instructions for Experimental exam
Experimental problem one
, and
answer form
Experimental problem two
, and
answer form
Links to the websites of some past International Physics Olympiads26^th International Physics Olympiad -199527^th International Physics Olympiad - 199628^th International Physics Olympiad - 199729^th
International Physics Olympiad - 199830^th International Physics Olympiad - 199931^st International Physics Olympiad - 200032^nd International Physics Olympiad - 2001
33^rd International Physics Olympiad - 200234^th International Physics Olympiad - 200335^th International Physics Olympiad - 200436^th International Physics Olympiad - 200537^th International Physics
Olympiad - 200638^th International Physics Olympiad - 200738^th International Physics Olympiad - 200839th International Physics Olympiad - 2009, Mexico
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Tuesday, April 20, 2010
New Delhi: Eligible Scheduled Caste (SC) students are being offered scholarships for pursuing higher studies abroad under 'National Overseas Scholarship for SC etc. Candidates' scheme, Minister of
State for Social Justice and Empowerment, D. Napoleon said today. The scholarship includes annual maint...
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Monday, April 19, 2010
1. State the principle of conservation of angular momentum and explain its applications.
2. Define radius of gyration
3. Establish equation of continuity of an ideal liquid flow
4. Explain the variation of acceleration due to gravity with (a) altitude (b) depth
5. State Kepler’s laws of planetary motion
6. Define escape velocity and derive an expression for the escape velocity of a body on the surface of earth.
7. State theorem of perpendicular axis and the theorem of parallel axes.
8. Show that the oscillations of a simple pendulum are simple harmonic.
9. What are beats? How are they produced?
10. What is Doppler Effect?
11. State Carnot’s theorem.
12. Describe Carnot’s cycle. (Draw graph)
13. State the three laws of thermo dynamics (Zeroth law, first law and second law)
14. Apply I law of thermodynamics to (a) isothermal process (b) adiabatic process
15. Derive an expression for the work done in an isothermal process
16. Find the angle of projection for a projectile motion whose range R is 4 times the maximum height H.
17. The airplane shown is in level flight at an altitude of 1 km and a speed of 200 km/h. At what distance ‘S’ should it release a heavy bomb to hit the target X? Take g = 10 m/s2.
18. At what temperature is the Fahrenheit scale reading equal to half that of the Celsius scale?
19. What is a simple harmonic motion? State its characteristics.
20. What is capillarity? Derive an expression for capillary ascent.
21. Define surface tension.
22. Derive an expression for velocities after collision when two bodies undergo perfectly elastic collision in one dimension.
23. State triangle law of vector addition. Find out the angle between the resultant vector and the vector A when two vectors A =3m is acting towards east and vector B=4m is acting towards the north
24. Prove work-energy theorem for a variable force.
25. The moments of inertia of two rotating bodies A and B are IA and IB (IA > IB) and their angular momenta are equal. Which one has greater kinetic energy?
26. Derive an equation for the work done in an isothermal process and draw the P-V indicator diagram.
27. Deduce an expression for the total energy of a system executing S.H.M. Draw the energy curve.’
28. What do you mean by standing wave? Deduce an expression for the nodes and antinodes of a standing wave. Write down the positions of nodes and antinodes.
29. Friction is a necessary evil, comment. Mention some methods of reducing friction.
30. State Bernoulli’s principle. Deduce an expression for the volume of the liquid flowing per second through the wider tube of a venturi meter.
31. What do mean by damped oscillation?
32. What is resonance in oscillations?
33. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer?
34. A light body and a heavy body have the same momentum which of the two bodies will have greater kinetic energy?
35. When is work done by the force is negative? Give condition for work to be positive?
36. Define one newton.
37. Distinguish between free, forced, and resonant oscillation with illustration.
38. Explain displacement, velocity, acceleration, and time period of a simple harmonic motion. Find relation between them.
39. State and explain superposition of waves.
40. Discuss the characteristics of stationary waves.
41. Explain the need of banking of tracks.
42. Explain why a gas has two specific heats?
43. The momentum of a body increases by 20%. What is the percentage increase in kinetic energy?
44. Calculate the energy spent in spraying a drop of mercury of 1cm radius into 106 droplets, all of same size. Surface tension of mercury is 55´10-2 Nm-1.
45. From kinetic theory of gas, find an expression for pressure exerted by gas on the walls of container.
46. Define orbital velocity of a satellite and derive an expression for orbital velocity.
47. Derive Stoke’s law by the method of dimensions.
48. Why does a freely falling body experience weightlessness?
49. Derive an expression for the apparent weight of a person in a lift when (a) the lift is moving up with acceleration (b) moving down with acceleration (c) moving up with acceleration (d) moving
down with deceleration (e) moving up or down with constant velocity.
50. State Hooke’s law of elasticity. Draw stress vs strain for a wire subjected to gradually increasing tension and explain the various points of the curve. (Elastic region, proportional limit,
elastic limit, plastic region, yield point and breaking point)
51. State Torricelli’s theorem and prove it. (Derive an expression for velocity of efflux)
52. Define thermal conductivity.
53. Explain the principle and working of carnot refrigerator.
54. Show that Cp is greater than Cv
55. Distinguish centre of mass and centre of gravity.
56. What are geostationary satellites?
57. Define Young’s modulus, bulk modulus and rigidity modulus.
58. State Pascal’s law and explain any one application. (Hydraulic brakes, hydraulic lift, hydraulic press)
59. Derive the relation L=I for a rigid body
60. Show that it is easier to pull a lawn roller than to push it.
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Friday, April 16, 2010
In a single fixed pulley the velocity ratio is always more than mechanical advantage. Why?
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We notice that there are many comments with unrelated links in it and we directly delete them without posting on to the site; however good the comments are.
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New Delhi: The Minister of State for Tourism, Sultan Ahmed today gave away the National Awards for Excellence in Hospitality Education for the year 2008-09.Speaking on the occasion he said, "The
Ministry of Tourism is making all efforts to expand the institutional infrastructure for training in the ...
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Wednesday, April 14, 2010
New Delhi: Reiterating that the India will return to a nine percent growth soon, Finance Minister Pranab Mukherjee on Tuesday urged the student community to serve at home and not leave the country
for greener pastures."While abundant opportunities await the graduating students, as you begin your car...
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Sunday, April 11, 2010
CBSE Results will be published in the following website
CBSE Results Link 1CBSE Results Link 2
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ELIGIBILITY FOR JEE-2010
Candidates must make sure that they satisfy all the eligibility conditions given below for appearing in JEE-2010:
Date of Birth
The date of birth of candidates belonging to GE,
categories should be on or after October 1,1985. Whereas the date of birth of those belonging to
categories should be on or after October 1,1980.
The date of birth as recorded in the high school/first Board/ Pre-University certificate will be accepted. If the certificate does not mention the date of birth, a candidate must submit along with
the application, an authenticated document indicating the date of birth.
Year of passing Qualifying Examination (QE)
A candidate must have passed the QE for the first time, after October 1, 2008 or in the year 2009 or will be appearing in 2010.
Those who are going to appear in the QE later than October 1, 2010 are not eligible to apply for JEE-2010.
The qualifying examinations (QE) are listed below:
i) The final examination of the 10+2 system, conducted by any recognized central / state Board, such as Central Board of Secondary Education, New Delhi; Council for Indian School Certificate
Examination, New Delhi; etc.
ii) Intermediate or two-year Pre-University examination conducted by a recognized Board / University.
iii) Final examination of the two-year course of the Joint Services Wing of the National Defence Academy.
iv) General Certificate Education (GCE) examination (London/Cambridge/Sri Lanka) at the Advanced (A) level.
v) High School Certificate Examination of the Cambridge University or International Baccalaureate Diploma of the International Baccalaureate Office, Geneva.
vi) Any Public School/Board/University examination in India or in any foreign country recognized as equivalent to the 10+2 system by the Association of Indian Universities (AIU).
vii) H.S.C. vocational examination.
viii) Senior Secondary School Examination conducted by the National Institute of Open Schooling with a minimum of five subjects.
ix) 3 or 4 year Diploma recognized by AICTE or a state Board of technical education. In case the relevant qualifying examination is not a public examination, the candidate must have passed at least
one public (Board or Pre-University) examination at an earlier level.
Minimum Percentage of Marks in QE
Candidates belonging to GE, OBC and DS categories must secure at least 60% marks in aggregate in their QE. Whereas, those belonging to SC, ST and PD categories must secure at least 55% marks in
aggregate in the QE.
The percentage of marks awarded by the Board will be treated as final. If the Board does not award the percentage of marks, it will be calculated based on the marks obtained in all subjects listed in
the mark sheet. If any Board awards only letter grades without providing an equivalent percentage of marks on the grade sheet, the candidate should obtain a certificate from the Board specifying the
equivalent marks, and submit it at the time of counselling/ admission. In case such a certificate is not provided then the final decision rests with the Joint Implementation Committee of JEE-2010.
4. Important Points to note
(i) One can attempt JEE only twice, in consecutive years. That means one should have attempted JEE for the first time in 2009 or will be appearing in 2010.
(ii) Those who have accepted admission after qualifying in JEE in earlier years by paying full fees at any of the IITs, IT-BHU, Varanasi or ISM, Dhanbad, are NOT ELIGIBLE to write JEE at all
irrespective of whether or not they joined in any of the programmes.
(iii) The year of passing the Qualifying Examination is the year in which the candidate has passed, for the first time, any of the examinations listed above, irrespective of the minimum percentage
marks secured.
(iv) The offer of admission is subject to verification of original certificates/ documents at the time of admission. If any candidate is found ineligible at a later date even after admission to an
Institute, his/ her admission will be cancelled automatically.
(iv) If a candidate is expecting the results of the QE in 2010, his/her admission will only be provisional until he/she submits the relevant documents. The admission stands cancelled if the documents
are not submitted in original to the concerned institute before September 30,2010.
(v) If a candidate has passed any of the examinations, listed in Sub-section III.2, before October 1,2008, he/she is not eligible to appear in JEE-2010.
(vi) If a Board invariably declares the results of the QE late (only after September 30, every year), the candidate is advised to attempt JEE in 2011 or later.
(vii) The decision of the Joint Admission Board of JEE-201 0 regarding the eligibility of any applicant shall be final.
Posted by AskPhysics at 3:20:00PM 1 comment:
General: Units and dimensions, dimensional analysis; least count, significant figures; Methods of measurement and error analysis for physical quantities pertaining to the following experiments:
Experiments based on using Vernier calipers and screw gauge (micrometer), Determination of g using simple pendulum, Young’s modulus by Searle’s method, Specific heat of a liquid using calorimeter,
focal length of a concave mirror and a convex lens using u-v method, Speed of sound using resonance column, Verification of Ohm’s law using voltmeter and ammeter, and specific resistance of the
material of a wire using meter bridge and post office box.
Mechanics: Kinematics in one and two dimensions (Cartesian coordinates only), projectiles; Uniform Circular motion; Relative velocity.
Newton’s laws of motion; Inertial and uniformly accelerated frames of reference; Static and dynamic friction; Kinetic and potential energy; Work and power; Conservation of linear momentum and
mechanical energy.
Systems of particles; Centre of mass and its motion; Impulse; Elastic and inelastic collisions.
Law of gravitation; Gravitational potential and field; Acceleration due to gravity; Motion of planets and satellites in circular orbits; Escape velocity.
Rigid body, moment of inertia, parallel and perpendicular axes theorems, moment of inertia of uniform bodies with simple geometrical shapes; Angular momentum; Torque; Conservation of angular
momentum; Dynamics of rigid bodies with fixed axis of rotation; Rolling without slipping of rings, cylinders and spheres; Equilibrium of rigid bodies; Collision of point masses with rigid bodies.
Linear and angular simple harmonic motions.
Hooke’s law, Young’s modulus.
Pressure in a fluid; Pascal’s law; Buoyancy; Surface energy and surface tension, capillary rise; Viscosity (Poiseuille’s equation excluded), Stoke’s law; Terminal velocity, Streamline flow, equation
of continuity, Bernoulli’s theorem and its applications.
Wave motion (plane waves only), longitudinal and transverse waves, superposition of waves; Progressive and stationary waves; Vibration of strings and air columns;Resonance; Beats; Speed of sound in
gases; Doppler effect (in sound).
Thermal physics: Thermal expansion of solids, liquids and gases; Calorimetry, latent heat; Heat conduction in one dimension; Elementary concepts of convection and radiation; Newton’s law of cooling;
Ideal gas laws; Specific heats (Cv and Cp for monoatomic and diatomic gases); Isothermal and adiabatic processes, bulk modulus of gases; Equivalence of heat and work; First law of thermodynamics and
its applications (only for ideal gases); Blackbody radiation: absorptive and emissive powers; Kirchhoff’s law; Wien’s displacement law, Stefan’s law.
Electricity and magnetism: Coulomb’s law; Electric field and potential; Electrical potential energy of a system of point charges and of electrical dipoles in a uniform electrostatic field; Electric
field lines; Flux of electric field; Gauss’s law and its application in simple cases, such as, to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and
uniformly charged thin spherical shell.
Capacitance; Parallel plate capacitor with and without dielectrics; Capacitors in series and parallel; Energy stored in a capacitor.
Electric current; Ohm’s law; Series and parallel arrangements of resistances and cells; Kirchhoff’s laws and simple applications; Heating effect of current.
Biot–Savart’s law and Ampere’s law; Magnetic field near a current-carrying straight wire, along the axis of a circular coil and inside a long straight solenoid; Force on a moving charge and on a
current-carrying wire in a uniform magnetic field.
Magnetic moment of a current loop; Effect of a uniform magnetic field on a current loop; Moving coil galvanometer, voltmeter, ammeter and their conversions.
Electromagnetic induction: Faraday’s law, Lenz’s law; Self and mutual inductance; RC, LR and LC circuits with d.c. and a.c. sources.
Optics: Rectilinear propagation of light; Reflection and refraction at plane and spherical surfaces; Total internal reflection; Deviation and dispersion of light by a prism; Thin lenses;
Combinations of mirrors and thin lenses; Magnification.
Wave nature of light: Huygen’s principle, interference limited to Young’s double-slit experiment.
Modern physics: Atomic nucleus; Alpha, beta and gamma radiations; Law of radioactive decay; Decay constant; Half-life and mean life; Binding energy and its calculation; Fission and fusion processes;
Energy calculation in these processes.
Photoelectric effect; Bohr’s theory of hydrogen-like atoms; Characteristic and continuous X-rays, Moseley’s law; de Broglie wavelength of matter waves.
Posted by AskPhysics at 3:14:00PM 2 comments:
Only those candidates who attempted both Paper-I and Paper-II will be considered for the ranking. Marks in Chemistry in JEE will be equal to marks in Chemistry section of Paper-I plus marks in
Chemistry section of Paper-II. Similar procedure will be followed for Mathematics and Physics. The sum of the marks obtained in the individual subjects in JEE will be the aggregate mark for the
The average of the marks scored by all such candidates will be computed for each of the three subjects. These will be the Minimum Qualifying Marks for Ranking (MQMR) in the individual subjects.
Based on the MQMR in the individual subjects as well as the aggregate marks in the examination, a Common Merit List (CML) will be prepared without any relaxed criteria, such that the number of
candidates in this list is equal to the total number of seats available in all the participating institutes put together. The aggregate marks scored by the last candidate in the CML will be the CML
cut-off score (CCS).
Next, the merit list of the OBC candidates will be prepared. If the number of OBC candidates in the CML is equal to or more than 1.4 times the number of available OBC seats, then the OBC merit list
will contain all these candidates.
In case the number of OBC candidates qualified in the CML is less than 1.4 times the number of available OBC seats, then relaxation (maximum of 10%) to the individual MQMR as well as to the CCS will
be applied, and an OBC merit list will be prepared, in which the number of candidates will be at most 1.4 times the number of available OBC seats.
By applying 50% relaxation to the individual MQMR as well as to the CCS, separate merit list for SC, ST and PD candidates will be prepared. The number of candidates in each of these lists will be, at
most 1.4 times the number of available seats in the respective categories.
While preparing the merit lists, if a candidate belongs to more than one category/ subcategory of relaxed norms, then he/she for the purpose of ranking shall be considered in all the categories in
which he/she qualifies.
There will be no separate waiting list for candidates.
Posted by AskPhysics at 3:13:00PM 3 comments:
Vijayawada: The Vijayawada Book Festival Society (VBFS) is organizing a two-week long course in Book Publishing from May 17 in association with the National Book Trust (NBT).The full time course is
only open for candidates who have already worked in the publication industry and its fee is to be Rs.1...
Posted by AskPhysics at 1:03:00AM No comments:
Saturday, April 10, 2010
New Delhi: According to an official press release from the Ministry of Personnel, Public Grievances & Pensions, the candidates short-listed for the personality test for Civil Services (Main)
Examination, 2009, will be taking the test on April 14, 2010 despite of the government declaring it a hol...
Posted by AskPhysics at 1:43:00AM No comments:
Friday, April 9, 2010
New Delhi: The Cabinet Committee on Economic Affairs today approved the proposal to provide financial support of Rs.282.25 crore to Indian Maritime University (IMU), Chennai to meet capital
expenditure as well as recurring deficit. The IMU has been established as Central University with all India Ju...
Posted by AskPhysics at 2:42:00AM No comments:
Thursday, April 8, 2010
Posted by AskPhysics at 10:02:00PM No comments:
Posted by AskPhysics at 5:14:00PM No comments:
Click on the names to know more on the Nobel Laureates in Physics
Posted by AskPhysics at 5:05:00PM 3 comments:
Friday, April 2, 2010
New Delhi: The Union Public Service Commission (UPSC) will conduct the National Defence Academy (NDA) and Naval Academy Examination (I) 2010 at 745 venues located in 41 centres throughout the country
on April 18, 2010 (Sunday).According to an official press release, "Admission certificates to candid...
Posted by AskPhysics at 2:08:00AM No comments: | {"url":"http://blog.plustwophysics.com/2010/04/","timestamp":"2024-11-13T08:46:14Z","content_type":"application/xhtml+xml","content_length":"547955","record_id":"<urn:uuid:a7f29f53-fee0-49c0-8b4e-62658eba93c1>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00185.warc.gz"} |
Brain Teaser Math IQ Test: Solve 34÷2x1+6-9 - EduViet Corporation
Brain Teaser Math IQ Test: Solve 34÷2×1+6-9
Brain teasers are puzzles that require lateral thinking to solve. If you like solving complex puzzles and their solutions, then you should certainly try Brain Teasers. Playing these brain teasers not
only keeps your mind active but also helps reduce stress levels and combat fatigue. There are tons of fantasy, brain teasers, and puzzles out there. Give your brain a good workout by focusing on
solving problems that align with your interests. Take a moment to explore the puzzles below.
Brain Teasers Math IQ Test: Solve 34÷2×1+6-9
Math puzzles engage readers by presenting scenarios that require active application of problem-solving skills. The puzzles are carefully designed to encourage individuals to think critically, analyze
the information provided, and creatively apply mathematical principles to arrive at solutions.
You are watching: Brain Teaser Math IQ Test: Solve 34÷2×1+6-9
The image provided above has an equation and the key to solving it is to discern the hidden patterns that control its elements. However, this task comes with a sense of urgency, as you must quickly
uncover the logic underpinning the pattern. This challenge requires quick cognitive responses and skilled analytical skills, all within a short time frame. Mastering this puzzle requires meticulous
attention to detail and keen observation of the components of the image.
See more : Observation Brain Test: If you have Eagle Eyes Find the Number 623 among 923 in 12 Secs
This is a challenge of moderate complexity that individuals with sharp intelligence and a keen eye for detail have the ability to overcome quickly. The ticking of the clock marks the beginning of the
countdown, adding to the anticipation.
The puzzle at hand provides a unique opportunity to improve your cognitive abilities and ultimately improve your IQ. This enhancement has profound implications for your future endeavors, providing
you with valuable skills that are sure to positively impact every aspect of your life. Research even shows that participating in brainteasers like this plays an important role in maintaining
cognitive health. Improving your intelligence through puzzles like these not only improves your ability to solve immediate problems, but also develops broader mental agility that benefits academic,
professional, and personal pursuits.
Although the puzzle may seem daunting, the solver is tasked with finding a solution that seamlessly matches the specified conditions to effectively decipher the code. The following sections will
provide insight into the exact nature of this mathematical puzzle and the satisfying solution that awaits.
Brain Teasers Math IQ Test: Solve 34÷2×1+6-9 Problem
This particular mathematical puzzle presents a great challenge and you are warmly invited to accept this task and try to solve it.
To solve the expression 34÷2×1+6-9, you should follow the order of operations, usually remembered using the abbreviation PEMDAS: parentheses, exponents, multiplication and division (from left to
right) and addition and subtraction (from left to right) to right).
In this expression, there are no parentheses or exponents, so we do multiplication and division from left to right. First, we divide 34 by 2, which equals 17. Then we multiply 17 by 1 to get 17.
Therefore, the expression becomes 17 + 6 – 9. Now, we perform addition and subtraction from left to right. First, add 17 and 6, which equals 23. Then, subtract 9 from 23 to get the final result of
34÷2×1+6-9 = 14
And this arrangement is indeed accurate.
Disclaimer: The above information is for general information purposes only. All information on this website is provided in good faith, but we make no representations or warranties, express or
implied, as to the accuracy, adequacy, validity, reliability, availability or completeness of any information on this website.
Source: https://truongnguyenbinhkhiem.edu.vn
Category: Brain Teaser
Leave a Comment | {"url":"https://truongnguyenbinhkhiem.edu.vn/brain-teaser-math-iq-test-solve-34%C3%B72x16-9","timestamp":"2024-11-03T03:35:22Z","content_type":"text/html","content_length":"121313","record_id":"<urn:uuid:539556f8-c5de-4d68-a8aa-b23bd3103bb8>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00509.warc.gz"} |
Why the linearized, binding equation has a zero shadow price?
Hello all,
I have a simple quadratic model that requires NLP solvers to solve (this is a simple version of my actual model). However, I have linearized this model to solve it with MIP solvers. The linearized
model works well and gives the exact same solution as the quadratic model. My problem is that one of the linearized equations (constraint BALANCE in my model) that seems to be binding has a zero
shadow price, level, and Upper values. I am wondering what is the explanation for this and how I can obtain a shadow price for this constraint. My guess is that the zero level and shadow price are
due to grid points values that I have used for linearization. I have attached the model.
I will appreciate any help,
OPTION LIMCOL = 0;
OPTION LIMROW = 0;
SETS CURVEPARM CURVE PARAMETERS /INTERCEPT,SLOPE/
CURVES TYPES OF CURVES /DEMAND,SUPPLY/
SETS DK 'SEGMENTS FOR DEMAND' /DK1*DK21/
SETS SK 'SEGMENTS FOR SUPPLY' /SK1*SK21/
TABLE DATA(CURVES,CURVEPARM) SUPPLY DEMAND DATA
INTERCEPT SLOPE
DEMAND 6 -0.30
SUPPLY 1 0.20
/SUPPLY -1, DEMAND 1/
PARAMETER DHVALUE(DK) VALUES OF GRID POINTS FOR DEMAND
SHVALUE(SK) VALUES OF GRID POINTS FOR SUPPLY ;
DHVALUE(DK) = 10 * (ORD(DK) - 1)/10;
SHVALUE(SK) = 10 * (ORD(SK) - 1)/10;
SOS2 VARIABLES LAMH1(DK), LAMH2(SK);
VARIABLES OBJ NUMBER TO BE MAXIMIZED
EQUATIONS OBJJ OBJECTIVE FUNCTION
BALANCE COMMODITY BALANCE
LAML1CONVEX WEIGHTS OF GRID POINTS USED IN LINEARIZATION OF DEMAND
LAML2CONVEX WEIGHTS OF GRID POINTS USED IN LINEARIZATION OF SUPPLY
OBJJ.. OBJ =E= SUM(CURVES, SIGN(CURVES)*
+0.5*DATA(CURVES,"SLOPE")*QUANTITY(CURVES)**2)) ;
BALANCE.. SUM(CURVES, SIGN(CURVES)*QUANTITY(CURVES)) =L= 0 ;
OBJ =E=
SUM(DK,(DATA("DEMAND","INTERCEPT") + 0.5 * DATA("DEMAND","SLOPE") * DHVALUE(DK)) * DHVALUE(DK) * LAMH1(DK))
- (SUM(SK,(DATA("SUPPLY","INTERCEPT") + 0.5 * DATA("SUPPLY","SLOPE") * SHVALUE(SK)) * SHVALUE(SK) * LAMH2(SK)))
*** DEMAND AND SUPPLY BALANCE
BALANCE.. SUM(DK,DHVALUE(DK) * LAMH1(DK)) =L= SUM(SK,SHVALUE(SK) * LAMH2(SK));
LAML1CONVEX.. SUM(DK,LAMH1(DK)) =E= 1;
LAML2CONVEX.. SUM(SK,LAMH2(SK)) =E= 1;
MODEL PRICEEND /ALL/ ;
DIFFERENCE = SUM(DK,DHVALUE(DK) * LAMH1.L(DK)) - SUM(SK,SHVALUE(SK) * LAMH2.L(SK));
The duals produced by a MIP are the duals of the “fixed” problem. So the MIP solver link fixes your SOS variables and then solves the resulting LP. These marginals might have little to do with your
original model. The fixed SOS variables LAMH1 and LAMH2 make the BALANCE equation a fixed equation (nothing can move), so increasing the rhs by 1 will not allow us to change the objective and hence
the dual of this constraint is 0. A reformulation often works in the primal space but not so much in the dual… You can try to get a primal/dual pair from your solution returned by the mip solver by
calling conopt on the rmip and set the iterlim to 4.
Moreover, don’t write **2 use sqr() then you can solve as qcp and since your qp is convex you can also solve with cplex, xpress, gurobi, mosek, …
Thank you very much. | {"url":"https://forum.gams.com/t/why-the-linearized-binding-equation-has-a-zero-shadow-price/3835","timestamp":"2024-11-02T17:30:45Z","content_type":"text/html","content_length":"20809","record_id":"<urn:uuid:e121f6d0-a672-4469-bae6-0541fb4fed2f>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00004.warc.gz"} |
C# – Operators
An operator is a symbol that tells the compiler to perform specific mathematical or logical manipulations. C# has rich set of built – in operators and provide the following type of operators –
• Arithmetic Operators
• Relational Operators
• Logical and Bitwise Operators
• Assignment Operators
• Other Operators
Arithmetic Operators
Several common arithmetic operators are allowed in C#.
Operand Description
+ Add
– Subtract
* Multiply
/ Divide
% Remainder or modulo
++ Increment by 1
— Decrement by 1
Relational Operators
Relational Operators are used for comparison purpose in conditional statements. Common relational operators in c# are :
Operand Description
== Equality check
!= Un – equality check
> Greater than
< Less than
>= Greater than or equal to
<= Less than or equal to
Relational operators always result in a Boolean statement; either true or false.
Logical and Bitwise Operators
These operators are used for logical and bitwise calculations. Common logical and bitwise operators in C# are:
Operand Description
& Bitwise AND
| Bitwise OR
^ Bitwise XOR
! Bitwise NOT
&& “Logical ” or ” short circuit ” AND
|| ” Logical” or ” short circuit ” OR
The operators &, | and ^ are rarely used in usual programming practice. The NOT operator is used to negate a Boolean or bitwise expression.
Assignment Operators
Assignment Operators are used to assign values to variables. Common assignment operators in C# are :
Operand Description
= Simple assignment
+= Additive assignment
-= subtractive assignment
*= Multiplicative assignment
/= Division assignment
%= Modulo assignment
The equal (=) operator is used to assign value to an object.
Other Operators
There are some other operators present in c#. A short description of these is given below:-
Operand Description
<< Left shift bitwise operator
>> Right shift bitwise operator
. Member access for objects
[] Index operator used in arrays and collections
() Cast operator
?: Ternary operator | {"url":"https://studymam.com/c-sharp/c-sharp-operators/","timestamp":"2024-11-02T21:32:19Z","content_type":"text/html","content_length":"27900","record_id":"<urn:uuid:f77429c6-d923-4b61-8a99-0cd5e4e7be1a>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00260.warc.gz"} |
SAE-AISI 4820 (G48200) Nickel Steel
SAE-AISI 4820 steel is an alloy steel formulated for primary forming into wrought products. Cited properties are appropriate for the annealed condition. 4820 is the designation in both the SAE and
AISI systems for this material. G48200 is the UNS number.
It has a moderately high base cost among SAE-AISI wrought steels. In addition, it has a moderately high embodied energy and a moderately high electrical conductivity.
The graph bars on the material properties cards below compare SAE-AISI 4820 steel to: SAE-AISI wrought steels (top), all iron alloys (middle), and the entire database (bottom). A full bar means this
is the highest value in the relevant set. A half-full bar means it's 50% of the highest, and so on.
Mechanical Properties
Elastic (Young's, Tensile) Modulus
190 GPa 27 x 10^6 psi
Fatigue Strength
260 MPa 38 x 10^3 psi
Shear Modulus
73 GPa 11 x 10^6 psi
Shear Strength
360 MPa 52 x 10^3 psi
Tensile Strength: Ultimate (UTS)
570 MPa 83 x 10^3 psi
Tensile Strength: Yield (Proof)
370 MPa 54 x 10^3 psi
Thermal Properties
Latent Heat of Fusion
250 J/g
Maximum Temperature: Mechanical
410 °C 780 °F
Melting Completion (Liquidus)
1460 °C 2660 °F
Melting Onset (Solidus)
1420 °C 2580 °F
Specific Heat Capacity
470 J/kg-K 0.11 BTU/lb-°F
Thermal Conductivity
51 W/m-K 29 BTU/h-ft-°F
Thermal Expansion
11 µm/m-K
Electrical Properties
Electrical Conductivity: Equal Volume
7.6 % IACS
Electrical Conductivity: Equal Weight (Specific)
8.7 % IACS
Otherwise Unclassified Properties
Base Metal Price
4.3 % relative
7.9 g/cm^3 490 lb/ft^3
Embodied Carbon
1.8 kg CO[2]/kg material
Embodied Energy
24 MJ/kg 10 x 10^3 BTU/lb
Embodied Water
53 L/kg 6.3 gal/lb
Common Calculations
Resilience: Ultimate (Unit Rupture Work)
110 MJ/m^3
Resilience: Unit (Modulus of Resilience)
370 kJ/m^3
Stiffness to Weight: Axial
13 points
Stiffness to Weight: Bending
24 points
Strength to Weight: Axial
20 points
Strength to Weight: Bending
19 points
Thermal Diffusivity
14 mm^2/s
Thermal Shock Resistance
19 points
Alloy Composition
Among alloy steels, the composition of SAE-AISI 4820 steel is notable for including nickel (Ni). Nickel is used to improve mechanical properties, and to make the alloy easier to heat treat.
Fe 94.6 to 95.7
Ni 3.3 to 3.8
Mn 0.5 to 0.7
Si 0.15 to 0.35
Mo 0.2 to 0.3
C 0.18 to 0.23
S 0 to 0.040
P 0 to 0.035
All values are % weight. Ranges represent what is permitted under applicable standards.
Further Reading
ASTM A752: Standard Specification for General Requirements for Wire Rods and Coarse Round Wire, Alloy Steel
ASTM A322: Standard Specification for Steel Bars, Alloy, Standard Grades
ASM Specialty Handbook: Carbon and Alloy Steels, J. R. Davis (editor), 1996
Manufacture and Uses of Alloy Steels, Henry D. Hibbard, 2005
Steels: Processing, Structure, and Performance, 2nd ed., George Krauss, 2015 | {"url":"https://www.makeitfrom.com/material-properties/SAE-AISI-4820-G48200-Nickel-Steel","timestamp":"2024-11-05T12:16:31Z","content_type":"text/html","content_length":"25673","record_id":"<urn:uuid:4ed14e67-02d7-4823-83a4-aef60c86bc7f>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00604.warc.gz"} |
Beyond Outerplanarity
We study straight-line drawings of graphs where the vertices are placed in convex position in the plane, i.e., convex drawings. We consider two families of graph classes with nice convex drawings:
outer k-planar graphs, where each edge is crossed by at most k other edges; and, outer k-quasi-planar graphs where no k edges can mutually cross. We show that the outer k-planar graphs are (√(4k+1)
+1)-degenerate, and consequently that every outer k-planar graph can be (√(4k+1)+2)-colored, and this bound is tight. We further show that every outer k-planar graph has a balanced separator of size
O(k). This implies that every outer k-planar graph has treewidth O(k). For fixed k, these small balanced separators allow us to obtain a simple quasi-polynomial time algorithm to test whether a given
graph is outer k-planar, i.e., none of these recognition problems are NP-complete unless ETH fails. For the outer k-quasi-planar graphs we prove that, unlike other beyond-planar graph classes, every
edge-maximal n-vertex outer k-quasi planar graph has the same number of edges, namely 2(k-1)n - 2k-12. We also construct planar 3-trees that are not outer 3-quasi-planar. Finally, we restrict outer
k-planar and outer k-quasi-planar drawings to closed drawings, where the vertex sequence on the boundary is a cycle in the graph. For each k, we express closed outer k-planarity and closed outer
k-quasi-planarity in extended monadic second-order logic. Thus, closed outer k-planarity is linear-time testable by Courcelle's Theorem. | {"url":"https://cdnjs.deepai.org/publication/beyond-outerplanarity","timestamp":"2024-11-04T01:42:00Z","content_type":"text/html","content_length":"154963","record_id":"<urn:uuid:96d9331d-7e22-4a60-a0c4-0914e4eddce3>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00095.warc.gz"} |
Model sometimes works, sometimes hangs indefinitely
Hi all,
I have been using Gurobi (both in the cloud and, when that seems to fail, on my local machine) to perform optimizations for months. For the last three weeks, I have been running a high volume of
optimizations in the cloud (eg, I have performed several hundred runs successfully and saved the outputs). These optimizations generally take 6-7 minutes to complete, though some take 20-25 for no
apparent reason. However, sometimes, a version of my model with a very similar formulation to all the others that have run successfully will hang indefinitely--the optimizer will seem to start (ie,
the line in my code just before the .solve prints 'optimizer starting', and the cloud machine turns green and launches), and no error is thrown, but no result is returned. It will hang in this state
for >5 hours if I don't catch it.
This is expensive, but (more importantly) baffling--I can't figure out why certain runs won't complete. To me it seems totally arbitrary which runs cause this hangup; I'm not yet sure what's similar
between them / different from the runs that do successfully complete. Does anyone have suggestions for troubleshooting this?
Thank you,
• Do you have a Gurobi log for one of the runs that doesn't complete?
• I don't, unfortunately. It doesn't seem to have been saving log files. I'm coding in Python and have been saving .ilp files, but not .log files. Can you tell me if the following looks like the
right way to save a log file?
solver_parameters = "ResultFile=model.ilp, LogFile=model_log.log"
results = opt.solve(model, options_string=solver_parameters, symbolic_solver_labels=True)
If that should work, at what point in a non-completing run would the log file be generated? I've started a run just now but have not seen a log file appear in my working directory.
• Eli,
Looks like I've saved a log file. It's not clear that this run will *never* complete, but it's been going for almost an hour and a half, whereas the previous runs that completed in this session
took 5-10 mines. Here's the head; please let me know if you need more information.
Gurobi 9.0.1 (mac64) logging started Wed Nov 18 05:13:18 2020
Changed value of parameter LogFile to model_log.log
Prev: Default:
Gurobi Optimizer version 9.0.1 build v9.0.1rc0 (mac64)
Optimize a model with 98554 rows, 24625 columns and 10736065 nonzeros
Model fingerprint: 0xa341d6d1
Model has 2114415 quadratic objective terms
Variable types: 1 continuous, 24624 integer (0 binary)
Coefficient statistics:
Matrix range [1e+00, 1e+00]
Objective range [3e-11, 1e+09]
QObjective range [2e+00, 8e+00]
Bounds range [0e+00, 0e+00]
RHS range [1e+00, 1e+06]
Warning: Model contains large objective coefficients
Consider reformulating model or setting NumericFocus parameter
to avoid numerical issues.
Presolve removed 73786 rows and 6697 columns (presolve time = 8s) ...
Presolve removed 73955 rows and 6697 columns (presolve time = 10s) ...
Presolve removed 75351 rows and 6697 columns (presolve time = 15s) ...
Presolve removed 75375 rows and 6646 columns
Presolve time: 18.87s
Presolved: 23179 rows, 17979 columns, 4558096 nonzeros
Presolved model has 1145648 quadratic objective terms
Variable types: 0 continuous, 17979 integer (0 binary)
Found heuristic solution: objective 5.188266e+12
Root simplex log...
Iteration Objective Primal Inf. Dual Inf. Time
0 2.7172068e+10 0.000000e+00 6.130384e+07 21s
7731 6.9597478e+06 0.000000e+00 1.206543e+04 25s
14727 -3.9400031e+05 0.000000e+00 3.662503e+02 30s
24701 -1.6692974e+01 2.499054e+09 0.000000e+00 35s
33853 -6.5574627e+02 1.697637e+09 0.000000e+00 40s
42493 -6.7434083e+02 1.138674e+09 0.000000e+00 45s
51183 -6.7160092e+02 8.935296e+08 0.000000e+00 50s
Warning: 1 variables dropped from basis
62668 8.6028767e-01 8.708114e+08 0.000000e+00 55s
72410 -4.2136698e+02 2.849482e+08 0.000000e+00 60s
82202 -1.5684586e+03 6.906356e+08 0.000000e+00 65s
Warning: 1 variables dropped from basis
91520 -1.0398743e+03 1.386282e+09 0.000000e+00 70s
Warning: 4 variables dropped from basis
• For what it's worth, the log file of the forever-hanging run looks similar (to my eye) to the log file of a run that completed in 10 minutes:
Gurobi 9.0.1 (mac64) logging started Tue Nov 17 22:08:52 2020
Changed value of parameter LogFile to model_log.log
Prev: Default:
Gurobi Optimizer version 9.0.1 build v9.0.1rc0 (mac64)
Optimize a model with 69181 rows, 17281 columns and 5045761 nonzeros
Model fingerprint: 0x6d78f2ea
Model has 1560240 quadratic objective terms
Variable types: 1 continuous, 17280 integer (0 binary)
Coefficient statistics:
Matrix range [1e+00, 1e+00]
Objective range [2e-13, 1e+09]
QObjective range [2e+00, 8e+00]
Bounds range [0e+00, 0e+00]
RHS range [1e+00, 7e+05]
Warning: Model contains large objective coefficients
Consider reformulating model or setting NumericFocus parameter
to avoid numerical issues.
Presolve removed 49485 rows and 3651 columns (presolve time = 5s) ...
Presolve removed 50099 rows and 3593 columns
Presolve time: 9.91s
Presolved: 19082 rows, 13688 columns, 2518175 nonzeros
Presolved model has 983254 quadratic objective terms
Variable types: 0 continuous, 13688 integer (0 binary)
Found heuristic solution: objective 1.277261e+13
Root simplex log...
Iteration Objective Primal Inf. Dual Inf. Time
0 3.3086991e+10 0.000000e+00 6.642891e+07 11s
9595 6.1536134e+07 0.000000e+00 2.757745e+04 15s
23130 6.6995621e-04 2.782785e+09 0.000000e+00 20s
36022 -1.2855606e+03 8.941475e+08 0.000000e+00 25s
48326 -1.2955765e+03 6.308836e+08 0.000000e+00 30s
60050 -1.3420210e+03 6.067081e+08 0.000000e+00 35s
Warning: 1 variables dropped from basis
73902 3.3387456e+00 9.911898e+08 0.000000e+00 40s
86910 -1.3406267e+01 2.250933e+09 0.000000e+00 45s
100862 -7.6920727e+02 2.880118e+09 0.000000e+00 50s
Warning: 1 variables dropped from basis
113222 -1.9792591e+01 1.951922e+09 0.000000e+00 55s
124891 -2.3535904e+02 8.992859e+08 0.000000e+00 60s
• Do you have a link to the full log files? These logs only show the first minute or so of solve time. Also, have you tried Gurobi 9.1?
The objective coefficient range is pretty suspicious:
Coefficient statistics:
Matrix range [1e+00, 1e+00]
Objective range [3e-11, 1e+09]
QObjective range [2e+00, 8e+00]
Bounds range [0e+00, 0e+00]
RHS range [1e+00, 1e+06]
Warning: Model contains large objective coefficients
Consider reformulating model or setting NumericFocus parameter
to avoid numerical issues.
Large objective coefficients like \( 10^9 \) can make it difficult for Gurobi to determine if a solution is truly optimal. Additionally, I'm curious where the very small objective coefficients
like \( 3 \cdot 10^{-11} \) come from. If any of the objective coefficients represent penalty terms, perhaps a hierarchical multi-objective approach would be more appropriate.
It is best to reformulate the problem yourself to remove these very large and very small objective coefficients. You could alternatively try using the ObjScale parameter to scale the objective. I
can't say for certain if rescaling the objective function will help, but it's the first place I would look.
Other parameters to try are PreQLinearize and NumericFocus.
Note there's no guarantee that if Gurobi solves a model in five minutes, it will solve a similar problem in five minutes (see Is Gurobi Optimizer deterministic?). And the models are pretty large
- they have 5-10 million nonzeros and 1-2 million quadratic objective terms.
• Thanks, Eli. This is a really helpful answer. For context, I have fairly limited optimization experience--I'm running a model I've inherited from someone else. Whoever takes over my job in a few
months will likely have more optimization experience and will be able to implement some of your suggestions. In the meantime, do you have any stopgap measures you could recommend to let me wrap
up these last few optimizations without making major changes to the code? ie, adjusting my cloud pool to have more or more powerful machines? (I'm currently running on a single c5.9xlarge
And I don't think I've tried Gurobi 9.1--would that help if I'm running on Gurobi Cloud? And here's a link to the log file--the last run logged there should be the one that failed.
• The failed run gets stuck in the root relaxation solve:
Root simplex log...
Iteration Objective Primal Inf. Dual Inf. Time
0 1.2863960e+11 0.000000e+00 1.765539e+08 138s
2299 1.3210152e+10 0.000000e+00 8.395725e+06 140s
5989 5.2098164e+09 0.000000e+00 2.544591e+06 145s
9132 2.1776792e+09 0.000000e+00 1.106070e+06 150s
12131 8.8095846e+08 0.000000e+00 4.704857e+05 155s
1894891 4.1334679e+08 1.716058e+08 0.000000e+00 12261s
1895489 4.1334657e+08 1.328328e+08 0.000000e+00 12266s
1896068 4.1334160e+08 1.004277e+08 0.000000e+00 12271s
1896676 4.1334140e+08 1.814993e+08 0.000000e+00 12276s
1897224 4.1334134e+08 2.409831e+08 0.000000e+00 12281s
Warning: 1 variables dropped from basis
I don't think a more powerful machine would help, since the problem is the model itself. To avoid long runs, you could set the TimeLimit parameter. But the solution returned by Gurobi could be
quite bad. From the above problem:
Found heuristic solution: objective 4.469094e+13
I would try testing different values of ObjScale, PreQLinearize, and NumericFocus on the problematic model. AggFill=0 or Aggregate=0 might also help.
To use Gurobi 9.1 on the Cloud, you only have to update the Gurobi installation on your local machine. When you optimize, Gurobi Cloud recognizes which version of Gurobi you used to build the
model and uses that version to solve the model. There's a chance Gurobi 9.1 performs a better on these models.
If you switch to Gurobi 9.1, you could try the new NoRelHeurTime parameter. This controls a heuristic that runs before the root relaxation is even solved. If the heuristic works well and Gurobi
hits a time limit while solving the root relaxation, you could get a decent solution to the problem. The tradeoff is (i) the heuristic might not always work on your problems, and (ii) models that
don't get stuck in the root relaxation might solve faster without spending time in this heuristic.
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Bayesian Algorithms for Mobile Terminal Positioning in Outdoor Wireless Environments
Bayesian Algorithms for Mobile Terminal Positioning in Outdoor Wireless Environments (2008)
Bayesian Signal Processing Techniques for GNSS Receivers: from multipath mitigation to positioning
This dissertation deals with the design of satellite-based navigation receivers. The term Global Navigation Satellite Systems (GNSS) refers to those navigation systems based on a constellation of
satellites, which emit ranging signals useful for positioning. Although the american GPS is probably the most popular, the european contribution (Galileo) will be operative soon. Other global and
regional systems exist, all with the same objective: aid user's positioning. Initially, the thesis provides the state-of-the-art in GNSS: navigation signals structure and receiver architecture. The
design of a GNSS receiver consists of a number of functional blocks. From the antenna to the fi nal position calculation, the design poses challenges in many research areas. Although the Radio
Frequency chain of the receiver is commented in the thesis, the main objective of the dissertation is on the signal processing algorithms applied after signal digitation. These ...
Closas, Pau — Universitat Politecnica de Catalunya
Exploiting Sparse Structures in Source Localization and Tracking
This thesis deals with the modeling of structured signals under different sparsity constraints. Many phenomena exhibit an inherent structure that may be exploited when setting up models, examples
include audio waves, radar, sonar, and image objects. These structures allow us to model, identify, and classify the processes, enabling parameter estimation for, e.g., identification, localisation,
and tracking. In this work, such structures are exploited, with the goal to achieve efficient localisation and tracking of a structured source signal. Specifically, two scenarios are considered. In
papers A and B, the aim is to find a sparse subset of a structured signal such that the signal parameters and source locations may be estimated in an optimal way. For the sparse subset selection, a
combinatorial optimization problem is approximately solved by means of convex relaxation, with the results of allowing for different types of ...
Juhlin, Maria — Lund University
Robust Signal Processing in Distributed Sensor Networks
Statistical robustness and collaborative inference in a distributed sensor network are two challenging requirements posed on many modern signal processing applications. This dissertation aims at
solving these tasks jointly by providing generic algorithms that are applicable to a wide variety of real-world problems. The first part of the thesis is concerned with sequential detection---a
branch of detection theory that is focused on decision-making based on as few measurements as possible. After reviewing some fundamental concepts of statistical hypothesis testing, a general
formulation of the Consensus+Innovations Sequential Probability Ratio Test for sequential binary hypothesis testing in distributed networks is derived. In a next step, multiple robust versions of the
algorithm based on two different robustification paradigms are developed. The functionality of the proposed detectors is verified in simulations, and their performance is examined under different
network conditions and outlier concentrations. Subsequently, ...
Leonard, Mark Ryan — Technische Universität Darmstadt
Performance Analysis of Bistatic Radar and Optimization methodology in Multistatic Radar System
This work deals with the problem of calculating the Cramer-Rao lower bounds (CRLBs) for bistatic radar channels. To this purpose we exploited the relation between the Ambiguity Function (AF) and the
CRLB. The bistatic CRLBs are analyzed and compared to the monostatic counterparts as a function of the bistatic geometric parameters. In the bistatic case both geometry factors and transmitted
waveforms play an important role in the shape of the AF, and therefore in the estimation accuracy of the target range and velocity. In particular, the CRLBs depend on the target direction of arrival,
the bistatic baseline length, and the distance between the target and the receiver. The CRLBs are then used to select the optimum bistatic channel (or set of channels) for the tracking of a radar
target moving along a trajectory in a multistatic scenario and for design ...
Stinco, Pietro — Universita di Pisa
Theoretical aspects and real issues in an integrated multiradar system
In the last few years Homeland Security (HS) has gained a considerable interest in the research community. From a scientific point of view, it is a difficult task to provide a definition of this
research area and to exactly draw up its boundaries. In fact, when we talk about the security and the surveillance, several problems and aspects must be considered. In particular, the following
factors play a crucial role and define the complexity level of the considered application field: the number of potential threats can be high and uncertain; the threat detection and identification can
be made more complicated by the use of camouflaging techniques; the monitored area is typically wide and it requires a large and heterogeneous sensor network; the surveillance operation is strongly
related to the operational scenario, so that it is not possible to define a ...
Fortunati Stefano — University of Pisa
A Contribution to Efficient Direction Finding using Antenna Arrays
It is save to say that there is no such thing as the direction finding (DF) algorithm. Rather, there are algorithms that are tuned to resolve hundreds of paths, algorithms that are designed for
uniform linear arrays or uniform circular arrays, and algorithms that strive for efficiency. The doctoral thesis at hand deals with the latter type of algorithms. However, the approach taken does not
only incorporate the actual DF algorithm but approaches the problem from different perspectives. The first perspective concerns the description of the array manifold. Current interpolation schemes
have no notion of polarization. Hence, the array manifold interpolation is performed separately for each state of polarization. In this thesis, we adopted the idea of interpolation via a 2-D discrete
Fourier transform. However, we transform the problem into the quaternionic domain. Here, a 2-D discrete quaternionic Fourier transform ...
Neudert-Schulz, Dominik — Technische Universität Ilmenau
Exploiting Prior Information in Parametric Estimation Problems for Multi-Channel Signal Processing Applications
This thesis addresses a number of problems all related to parameter estimation in sensor array processing. The unifying theme is that some of these parameters are known before the measurements are
acquired. We thus study how to improve the estimation of the unknown parameters by incorporating the knowledge of the known parameters; exploiting this knowledge successfully has the potential to
dramatically improve the accuracy of the estimates. For covariance matrix estimation, we exploit that the true covariance matrix is Kronecker and Toeplitz structured. We then devise a method to
ascertain that the estimates possess this structure. Additionally, we can show that our proposed estimator has better performance than the state-of-art when the number of samples is low, and that it
is also efficient in the sense that the estimates have Cramér-Rao lower Bound (CRB) equivalent variance. In the direction of ...
Wirfält, Petter — KTH Royal Institute of Technology
Robust Wireless Localization in Harsh Mixed Line-of-Sight/Non-Line-of-Sight Environments
This PhD thesis considers the problem of locating some target nodes in different wireless infrastructures such as wireless cellular radio networks and wireless sensor networks. To be as realistic as
possible, mixed line-of-sight and non-line-of-sight (LOS/NLOS) localization environment is introduced. Both the conventional non-cooperative localization and the new emerging cooperative localization
have been studied thoroughly. Owing to the random nature of the measurements, probabilistic methods are more advanced as compared to the old-fashioned geometric methods. The gist behind the
probabilistic methods is to infer the unknown positions of the target nodes in an estimation process, given a set of noisy position related measurements, a probabilistic measurement model, and a few
known reference positions. In contrast to the majority of the existing methods, harsh but practical constraints are taken into account: neither offline calibration nor non-line-of-sight state
identification is equipped in ...
Yin, Feng — Technische Universität Darmstadt
Bayesian methods for sparse and low-rank matrix problems
Many scientific and engineering problems require us to process measurements and data in order to extract information. Since we base decisions on information, it is important to design accurate and
efficient processing algorithms. This is often done by modeling the signal of interest and the noise in the problem. One type of modeling is Compressed Sensing, where the signal has a sparse or
low-rank representation. In this thesis we study different approaches to designing algorithms for sparse and low-rank problems. Greedy methods are fast methods for sparse problems which iteratively
detects and estimates the non-zero components. By modeling the detection problem as an array processing problem and a Bayesian filtering problem, we improve the detection accuracy. Bayesian methods
approximate the sparsity by probability distributions which are iteratively modified. We show one approach to making the Bayesian method the Relevance Vector ...
Sundin, Martin — Department of Signal Processing, Royal Institute of Technology KTH
Robust Signal Processing with Applications to Positioning and Imaging
This dissertation investigates robust signal processing and machine learning techniques, with the objective of improving the robustness of two applications against various threats, namely Global
Navigation Satellite System (GNSS) based positioning and satellite imaging. GNSS technology is widely used in different fields, such as autonomous navigation, asset tracking, or smartphone
positioning, while the satellite imaging plays a central role in monitoring, detecting and estimating the intensity of key natural phenomena, such as flooding prediction and earthquake detection.
Considering the use of both GNSS positioning and satellite imaging in critical and safety-of-life applications, it is necessary to protect those two technologies from either intentional or
unintentional threats. In the real world, the common threats to GNSS technology include multipath propagation and intentional/unintentional interferences. This thesis investigates methods to mitigate
the influence of such sources of error, with the final objective of ...
Li, Haoqing — Northeastern University
Bayesian data fusion for distributed learning
This dissertation explores the intersection of data fusion, federated learning, and Bayesian methods, with a focus on their applications in indoor localization, GNSS, and image processing. Data
fusion involves integrating data and knowledge from multiple sources. It becomes essential when data is only available in a distributed fashion or when different sensors are used to infer a quantity
of interest. Data fusion typically includes raw data fusion, feature fusion, and decision fusion. In this thesis, we will concentrate on feature fusion. Distributed data fusion involves merging
sensor data from different sources to estimate an unknown process. Bayesian framework is often used because it can provide an optimal and explainable feature by preserving the full distribution of
the unknown given the data, called posterior, over the estimated process at each agent. This allows for easy and recursive merging of sensor data ...
Peng Wu — Northeastern University
Sensor Fusion for Automotive Applications
Mapping stationary objects and tracking moving targets are essential for many autonomous functions in vehicles. In order to compute the map and track estimates, sensor measurements from radar, laser
and camera are used together with the standard proprioceptive sensors present in a car. By fusing information from different types of sensors, the accuracy and robustness of the estimates can be
increased. Different types of maps are discussed and compared in the thesis. In particular, road maps make use of the fact that roads are highly structured, which allows relatively simple and
powerful models to be employed. It is shown how the information of the lane markings, obtained by a front looking camera, can be fused with inertial measurement of the vehicle motion and radar
measurements of vehicles ahead to compute a more accurate and robust road geometry estimate. Further, it ...
Lundquist, Christian — Linköping University
Wireless Network Localization via Cooperation
This dissertation details two classes of cooperative localization methods for wireless networks in mixed line-of-sight and non-line-of-sight (LOS/NLOS) environments. The classes of methods depend on
the amount of prior knowledge available. The methods used for both classes are based on the assumptions in practical localization environments that neither NLOS identification nor experimental
campaigns are affordable. Two major contributions are, first, in methods that provide satisfactory localization accuracy whilst relaxing the requirement on statistical knowledge about the measurement
model. Second, in methods that provide significantly improved localization performance without the requirement of good initialization. In the first half of the dissertation, cooperative localization
using received signal strength (RSS) measurements in homogeneous mixed LOS/NLOS environments is considered for the case where the key model parameter, the path loss exponent, is unknown. The approach
taken is to model the positions and the path ...
Jin, Di — Signal Processing Group, Technische Universität Darmstadt
Direct Pore-based Identification For Fingerprint Matching Process
Fingerprint, is considered one of the most crucial scientific tools in solving criminal cases. This biometric feature is composed of unique and distinctive patterns found on the fingertips of each
individual. With advancing technology and progress in forensic sciences, fingerprint analysis plays a vital role in forensic investigations and the analysis of evidence at crime scenes. The
fingerprint patterns of each individual start to develop in early stagesof life and never change thereafter. This fact makes fingerprints an exceptional means of identification. In criminal cases,
fingerprint analysis is used to decipher traces, evidence, and clues at crime scenes. These analyses not only provide insights into how a crime was committed but also assist in identifying the
culprits or individuals involved. Computer-based fingerprint identification systems yield faster and more accurate results compared to traditional methods, making fingerprint comparisons in large
databases ...
Vedat DELICAN, PhD — Istanbul Technical University
Tracking and Planning for Surveillance Applications
Vision and infrared sensors are very common in surveillance and security applications, and there are numerous examples where a critical infrastructure, e.g. a harbor, an airport, or a military camp,
is monitored by video surveillance systems. There is a need for automatic processing of sensor data and intelligent control of the sensor in order to obtain efficient and high performance solutions
that can support a human operator. This thesis considers two subparts of the complex sensor fusion system; namely target tracking and sensor control.The multiple target tracking problem using
particle filtering is studied. In particular, applications where road constrained targets are tracked with an airborne video or infrared camera are considered. By utilizing the information about the
road network map it is possible to enhance the target tracking and prediction performance. A dynamic model suitable for on-road target tracking with ...
Skoglar, Per — Linköping University, Department of Electrical Engineering
The current layout is optimized for mobile phones. Page previews, thumbnails, and full abstracts will remain hidden until the browser window grows in width.
The current layout is optimized for tablet devices. Page previews and some thumbnails will remain hidden until the browser window grows in width. | {"url":"https://theses.eurasip.org/theses/172/bayesian-algorithms-for-mobile-terminal/similar/","timestamp":"2024-11-07T19:19:25Z","content_type":"text/html","content_length":"31464","record_id":"<urn:uuid:896d7a2b-b931-4101-ae4c-c47bfeec0ff0>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00404.warc.gz"} |
National Curriculum (Vocational) Mathematics Level 3
Data handling: Calculate, represent and interpret measures of central tendency and dispersion in univariate numerical ungrouped data
Subject outcome
Subject outcome 4.1: Calculate, represent and interpret measures of central tendency and dispersion in univariate numerical ungrouped data.
Learning outcomes
• Work out the five-number summary by:
□ Calculating the maximum, minimum and quartiles
□ Determining the fences
□ Constructing the box and whisker diagram
□ Indicating any outliers.
• Interpret the meaning of the representation of the box and whisker diagram with its outliers.
Unit 1 outcomes
By the end of this unit you will be able to:
• Find the mean for ungrouped data.
• Find the median for ungrouped data.
• Find the mode for ungrouped data.
• Find the range and interquartile range.
Unit 2 outcomes
By the end of this unit you will be able to:
• Combine measures of central tendency and dispersion to work out the five-number summary.
• Construct the box-and-whisker plot.
• Interpret the box-and-whisker plot. | {"url":"http://ncvm3.books.nba.co.za/part/data-handling-calculate-represent-and-interpret-measures-of-central-tendency-and-dispersion-in-univariate-numerical-ungrouped-data/","timestamp":"2024-11-12T01:56:37Z","content_type":"text/html","content_length":"77020","record_id":"<urn:uuid:1dab9385-7123-4a4f-ab99-5901e891c1eb>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00189.warc.gz"} |
09-21-2021, 11:32 AM (This post was last modified: 09-21-2021, 11:32 AM by CUwindows00.)
Of course, we have to admit that your algorithm is the most accurate at present. It is estimated that no one has a better algorithm than yours, at least as far as I know.
Because I started from the hccap hccapx algorithm to 22000, and have been following up the test, it turns out that your algorithm is currently the most accurate | {"url":"https://hashcat.net/forum/showthread.php?tid=10253&pid=53614&mode=threaded","timestamp":"2024-11-12T05:53:09Z","content_type":"application/xhtml+xml","content_length":"58422","record_id":"<urn:uuid:3afcaaa0-e291-4b26-9e01-059748f373f4>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00679.warc.gz"} |
1.1 Points, Lines and Planes
1.1 Points, Lines and Planes
In this lesson I will introduce you to points, lines and planes. The building blocks of geometry are points, lines and planes so it's important that you understand how we think of these
concepts. Geometry is very different than algebra so you will be learning an entirely new mathematical language. As such, you need to take excellent and organized notes because I will be teaching you
a lot of new symbols, notations and theorems. In this lesson my goal is to get you to understand how points, lines and planes relate. Now, this may sound strange but points, lines and planes can not
be defined in geometry- I will explain why in the video. I'm sure you have a good sense of what a "point" is or a "line" is but as you will see in the lesson these terms can not be strictly
defined. However, we still can learn a lot about the properties of points, lines and planes and apply this knowledge to the wonderful world of geometry. Welcome to course I know you will learn a lot!
1. Watch The Lesson Video First - Take Good Notes.
2. Next, Scroll All The Way Down The Page To View The Practice Problems - Try Them On Your Own.
3. Check The Solutions To The Practice Problems By Looking At The Answer Key At The End Of The Worksheet.
4. However, YOU MUST Still Watch The Video Solutions To The Practice Problems; These Are The Videos Labeled EX A, EX B, etc. - They Are Located Next To The Lesson Video.
5. After You Did All Of The Practice Problems - Complete The Section and Advance To The Next Topic.
1.1 GeoFoundationsPointsLinesPreview.pdf
Complete and Continue | {"url":"https://tabletclass-academy.teachable.com/courses/tabletclass-math-geometry1/lectures/9269117","timestamp":"2024-11-02T18:13:47Z","content_type":"text/html","content_length":"197765","record_id":"<urn:uuid:25af12c6-0203-44d5-a4d9-c1022c3c38ce>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00410.warc.gz"} |
nucleation Class Reference | OpenFOAM
Mix-in interface for nucleation models. Provides access to properties of the nucleation process, such as diameter and rate of production of nuclei. More...
Mix-in interface for nucleation models. Provides access to properties of the nucleation process, such as diameter and rate of production of nuclei.
Source files
Definition at line 55 of file nucleation.H. | {"url":"https://cpp.openfoam.org/dev/classFoam_1_1fv_1_1nucleation.html","timestamp":"2024-11-06T05:20:17Z","content_type":"application/xhtml+xml","content_length":"19881","record_id":"<urn:uuid:c8569b09-ba59-4b10-8184-4aaa1e0c1f07>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00297.warc.gz"} |
LinBox is a C++ template library for exact, high-performance linear algebra computation with dense, sparse, and structured matrices over the integers and over finite fields. LinBox has the following
top-level functions: solve linear system, matrix rank, determinant, minimal polynomial, characteristic polynomial, Smith normal form and trace. A good collection of finite field and ring
implementations is provided, for use with numerous black box matrix storage schemes.
Available via
Operating Systems
Programming Languages | {"url":"https://orms.mfo.de/project@all_keywords=1&terms=linear+least-squares+problem&id=261.html","timestamp":"2024-11-11T06:35:51Z","content_type":"application/xhtml+xml","content_length":"10552","record_id":"<urn:uuid:ae99bf3f-92d8-4ce2-b56d-311ad5e3e898>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00208.warc.gz"} |
Latent Class Analysis | Mplus Data Analysis Examples
Hypothetical Scenarios
Example 1
You are interested in studying drinking behavior among adults. Rather than conceptualizing drinking behavior as a continuous variable, you conceptualize it as forming distinct categories or
typologies. For example, you think that people fall into one of three different types: abstainers, social drinkers and alcoholics. Since you cannot directly measure what category someone falls into,
this is a latent variable (a variable that cannot be directly measured). However, you do have a number of indicators that you believe are useful for categorizing people into these different
categories. Using these indicators, you would like to:
1. Create a model that permits you to categorize these people into three different types of drinkers, hopefully fitting your conceptualization that there are abstainers, social drinkers and
2. Be able to categorize people as to what kind of drinker they are.
3. Count how many people would be considered abstainers, social drinkers and alcoholics.
4. Determine whether three latent classes is the right number of classes (i.e., are there only two types of drinkers or perhaps are there as many as four types of drinkers).
Example 2
High school students vary in their success in school. This might be indicated by the grades one gets, the number of absences one has, the number of truancies one has, and so forth. A traditional way
to conceptualize this might be to view “degree of success in high school” as a latent variable (one that you cannot directly measure) that is normally distributed. However, you might conceptualize
some students who are struggling and having trouble as forming a different category, perhaps a group you would call “at risk” (or in older days they would be called “juvenile delinquents”). Using
indicators like grades, absences, truancies, tardies, suspensions, etc., you might try to identify latent class memberships based on high school success.
Data Description
Let’s pursue Example 1 from above. We have a hypothetical data file that we created that contains 9 fictional measures of drinking behavior. For each measure, the person would be asked whether the
description applies to him/herself (yes or no). The 9 measures are
1. I like to drink
2. I drink hard liquor
3. I have drank in the morning
4. I have drank at work
5. I drink to get drunk
6. I like the taste of alcohol
7. I drink help me sleep
8. Drinking interferes with my relationships
9. I frequently visit bars
We have made up data for 1000 respondents and stored the data in a file called https://stats.idre.ucla.edu/wp-content/uploads/2016/02/lca1.dat, which is a comma-separated file with the subject id
followed by the responses to the 9 questions, coded 1 for yes and 0 for no. Using Stata, here is what the first 10 cases look like
list id item1-item9 in 1/10
| id item1 item2 item3 item4 item5 item6 item7 item8 item9 |
1. | 1 1 0 0 0 0 0 0 0 0 |
2. | 2 1 1 0 1 1 1 1 0 0 |
3. | 3 1 0 0 0 0 0 0 0 0 |
4. | 4 1 0 0 0 0 1 1 0 0 |
5. | 5 1 0 0 0 1 0 0 0 1 |
6. | 6 0 1 0 0 0 1 0 0 0 |
7. | 7 1 1 0 0 0 0 0 0 1 |
8. | 8 1 0 1 0 0 0 0 0 0 |
9. | 9 1 0 0 0 0 0 0 1 0 |
10. | 10 0 0 0 0 0 1 0 0 0 |
Some Strategies You Might Try
Before we show how you can analyze this with Latent Class Analysis, let’s consider some other methods that you might use:
• Cluster Analysis – You could use cluster analysis for data like these. However, cluster analysis is not based on a statistical model. It can tell you how the cases are clustered into groups, but
it does not provide information such as the probability that a given person is an alcoholic or abstainer. Also, cluster analysis would not provide information such as: given that someone said
“yes” to drinking at work, what is the probability that they are an alcoholic.
• Factor Analysis – Because the term “latent variable” is used, you might be tempted to use factor analysis since that is a technique used with latent variables. However, factor analysis is used
for continuous and usually normally distributed latent variables, where this latent variable, e.g., alcoholism, is categorical.
Mplus Results Using Latent Class Analysis
Note that I am showing you results before showing you the program. I will show you the program later.
Conditional Probabilities
First, the probability of answering “yes” to each question is shown for each type of drinker (latent class). For example, consider the question “I have drank at work”. The probability of answering
“yes” to this might be 70% for the first class, 10% for the second class, and 9% for the third class. This would be consistent with the first class being alcoholics. Looking at the pattern of
responses for all classes gives you an overall picture of the meaning of the three classes that are identified and helps us create descriptive labels for the classes. We are hoping to find three
classes that correspond to abstainers, social drinkers, and alcoholics. Abstainers would have a pattern that they generally avoid drinking, social drinkers would show a pattern of drinking but
generally in moderation and seldom in self-destructive ways, while alcoholics would show a pattern of drinking frequently and in very self-destructive ways.
The Mplus output shows a section labeled
which contains the conditional probabilities as describe above, but it is hard to read. I have reformatted that output to make it easier to read, shown below. Each row represents a different item,
and the three columns of numbers are the probabilities of answering “yes” to the item given that you belonged to that class. So, if you belong to Class 1, you have a 90.8% probability of saying “yes,
I like to drink”. By contrast, if you belong to Class 2, you have a 31.2% chance of saying “yes, I like to drink”.
Class 1 Class 2 Class 3 Item Label
ITEM1 0.908 0.312 0.923 I like to drink
ITEM2 0.337 0.164 0.546 I drink hard liquor
ITEM3 0.067 0.036 0.426 I have drank in the morning
ITEM4 0.065 0.056 0.418 I have drank at work
ITEM5 0.219 0.044 0.765 I drink to get drunk
ITEM6 0.320 0.183 0.471 I like the taste of alcohol
ITEM7 0.113 0.098 0.512 I drink help me sleep
ITEM8 0.140 0.110 0.619 Drinking interferes with my relationships
ITEM9 0.325 0.188 0.349 I frequently visit bars
Looking at item1, those in Class 1 and Class 3 really like to drink (with 90.8% and 92.3% saying yes) while those in Class 2 are not so fond of drinking (they have only a 31.2% probability of saying
they like to drink). Jumping to item5, 76.5% of those in Class 3 say they drink to get drunk, while 21.9% of those in Class 1 agreed to that, and only 4.4% of those in Class 2 say that.
I am starting to believe that Class 3 may be labeled as “alcoholics”. Focusing just on Class 3 (looking at that column), they really like to drink (92%), drink hard liquor (54.6%), a pretty large
number say they have drank in the morning and at work (42.6% and 41.8%), and well over half say drinking interferes with their relationships (61.9%).
It seems that those in Class 2 are the “abstainers” we were hoping to find. Not many of them like to drink (31.2%), few like the taste of alcohol (18.3%), few frequently visit bars (18.8%), and for
the rest of the questions they rarely answered “yes”.
This leaves Class 1; might they fit the idea of the “social drinker”? They like to drink (90.8%), but they don’t drink hard liquor as often as Class 3 (33.7% versus 54.6%). They rarely drink in the
morning or at work (6.7% and 6.5%) and rarely say that drinking interferes with their relationships (14%). They say they frequently visit bars similar to Class 3 (32.5% versus 34.9%), but that might
make sense. Both the social drinkers and alcoholics are similar in how much they like to drink and how frequently they go to bars, but differ in key ways such as drinking at work, drinking in the
morning, and the impact of drinking on their relationships.
While we should study these conditional probabilities some more, I think we can start to assign labels to these classes. As I hypothesized, the classes seem to make sense to be labeled “social
drinkers” (which is Class 1), “abstainers” (which is Class 2), and “alcoholics” (which is Class 3).
We can also take the results from the above table and express it as a graph. The X axis represents the item number and the Y axis represents the probability of answering “yes” to the given item,
given that you belong to a particular drinking class. The three drinking classes are represented as the three different lines.
Class Membership
For each person, Mplus will estimate what class the person belongs to (i.e., what type of drinker the person is). For a given person, Mplus estimates the probability that the person belongs to the
first, second, or third class. For example, for subject 1 these probabilities might be 15% that the person belongs to the first class, 80% probability of belonging to the second class, and 5% of
belonging to the third class. For such a person I would say that I think the person belongs to the second class since that class was the most likely. Mplus will also categorize people into a single
class using the same kind of rule.
Mplus creates an output file which contains the original data used in the analysis (i.e., item1 to item9) followed by the probability that Mplus estimates that the observation belongs to Class 1,
Class2, and Class 3. Next, the class with the highest probability (the modal class) is shown. I have taken a snippet of the output and labeled it to make it easier to read.
Items 1 - 9
----------------- P(c1) P(c2) P(c3) Class
1 0 0 0 0 0 0 0 0 0.645 0.354 0.001 1
1 1 0 1 1 1 1 0 0 0.098 0.001 0.901 3
1 0 0 0 0 0 0 0 0 0.645 0.354 0.001 1
1 0 0 0 0 1 1 0 0 0.797 0.177 0.026 1
1 0 0 0 1 0 0 0 1 0.934 0.041 0.025 1
0 1 0 0 0 1 0 0 0 0.312 0.686 0.002 2
1 1 0 0 0 0 0 0 1 0.903 0.092 0.005 1
1 0 1 0 0 0 0 0 0 0.766 0.218 0.017 1
1 0 0 0 0 0 0 1 0 0.696 0.290 0.014 1
0 0 0 0 0 1 0 0 0 0.149 0.850 0.000 2
For the first observation, the pattern of responses to the items suggests that the person has a 64.5% chance of being in Class 1 (which we called social drinkers), a 35.4% chance of being in Class 2
(abstainer), and a 0.1% chance of being in Class 3 (alcoholic). Note that these sum to 100% (since a person has to be in one of these classes). For this person, Class 1 is the most likely class, and
Mplus indicates that in the last column. It is interesting to note that for this person, the pattern of results made it almost certain that s/he was not alcoholic, but it was less clear whether s/he
was a social drinker or an abstainer (perhaps because the person said “yes” to item 1 (I like to drink). Note how the third row of data has the same pattern of responses for the items and has the
same predicted class probabilities. Consider row 2 of the data. This person has a 90.1% chance of being an alcoholic, a 9.8% chance of being a social drinker, and a 0.1% chance of being an abstainer.
One important point to note here is that for some subjects, the class membership is pretty well determined (like subject 2), while it is a bit more ambiguous (like subjects 1 and 3) where there is no
single class that they certainly belong to.
Size of Classes
Once we have come up with a descriptive label for each of the classes, we can look at the number of people who are categorized into each of the classes. I predict that about 20% of people are
abstainers, 70% are social drinkers, and about 10% are alcoholics. I can compare my predictions to the results that Mplus produces.
How many alcoholics are there? How many abstainers are there? How many social drinkers are there? One simple way we could determine this is by taking the information from the Class Membership above
and doing a simple tabulation on the last column. In fact, the Mplus output provides this to you like this.
Class Counts and Proportions
1 646 0.64600
2 288 0.28800
3 66 0.06600
Out of the 1,000 subjects we had, 646 (64.6%) are categorized as Class 1 (which we label as social drinkers), 66 (6.6%) are categorized as Class 3 (alcoholics), and 288 (28.8%) are categorized as
Class 2 (abstainers). This is consistent with my hunches that most people are social drinkers, a very small portion are alcoholics, and a moderate portion are abstainers.
There is a second way we could compute the size of the classes. Consider subject 1 from the above output on class membership. Rather than considering this person as entirely belonging to class 1, we
could allocate membership to the classes in proportion to the probability of being in each class. So, subject 1 has fractional memberships in each class, 0.645 to Class 1, 0.001 to Class 3, and 0.354
to Class 2. Mplus also computes the class sizes in this manner, as shown below.
1 557.56836 0.55757
2 363.13989 0.36314
3 79.29175 0.07929
These two methods yield largely similar results, but this second method suggests that there are somewhat more abstainers (36.3%) compared to the previous method (28.8%) and slightly fewer social
drinkers (55.7% compared to 64.6%), but these differences are not very troublesome to me.
Number of Classes
So far we have been assuming that we have chosen the right number of latent classes. Perhaps, however, there are only two types of drinkers, or perhaps there are four or more types of drinkers. So
far we have liked the three class model, both based on our theoretical expectations and based on how interpretable our results have been. We can further assess whether we have chosen the right number
of classes using the Vuong-Lo-Mendell-Rubin test (requested using TECH11, see Mplus program below) and the bootstrapped parametric likelihood ratio test (requested using TECH 14, see Mplus program
below). This test compares the model with K classes (in our case 3) to a model with (K-1) classes (in our case, K – 1 = 2 classes). The results are shown below.
TECHNICAL 11 OUTPUT
VUONG-LO-MENDELL-RUBIN LIKELIHOOD RATIO TEST FOR 2 (H0) VERSUS 3 CLASSES
H0 Loglikelihood Value -4251.208
2 Times the Loglikelihood Difference 39.025
Difference in the Number of Parameters 10
Mean 20.255
Standard Deviation 22.224
P-Value 0.1457
LO-MENDELL-RUBIN ADJUSTED LRT TEST
Value 38.468
P-Value 0.1500
TECHNICAL 14 OUTPUT
BOOTSTRAPPED PARAMETRIC LIKELIHOOD RATIO TEST FOR 2 (H0) VERSUS 3 CLASSES
H0 Loglikelihood Value -4251.208
2 Times the Loglikelihood Difference 39.025
Difference in the Number of Parameters 10
Approximate P-Value 0.0000
The Vuong-Lo-Mendell-Rubin test has a p-value of .1457 and the Lo-Mendell-Rubin adjusted LRT test has a p-value of .1500. Those tests suggest that two classes are sufficient and that three classes
are not really needed. However, the bootstrapped parametric likelihood ratio test has a p value of 0.0000, so this test suggests that three classes are indeed better than two classes. Because we have
seen unpublished results that suggest that the bootstrap method may be more reliable, and the three class model fits our theoretical expectations, we will go with the three class model.
The Mplus Program
Here is the whole Mplus program
Fictitious Latent Class Analysis.
File is lca1.dat ;
names = id item1 item2 item3 item4 item5 item6 item7 item8 item9;
usevariables = item1 item2 item3 item4 item5 item6 item7 item8 item9;
categorical = item1 item2 item3 item4 item5 item6 item7 item8 item9;
classes = c(3);
type is plot3;
series is item1 (1) item2 (2) item3 (3) item4 (4) item5 (5)
item6 (6) item7 (7) item8 (8) item9 (9);
file is lca1_save.txt ;
save is cprob;
format is free;
tech11 tech14;
• Title gives a title to the program.
• Data tells Mplus about the data file, in particular that the data are stored in the file lca1.dat which is the in the same folder or path as the command INP file.
• Variable tells Mplus more about the variables in the data file. The names statement lists the names of the variables (in the order they appear in the data file). The usevariables statement
indicates which variables we will use for this analysis. The categorical statement indicates that the specified variables are categorical variables. The classes statement indicates that there is
one categorical latent variable (which we will call c), and it has 3 levels.
• Analysis specifies the type of analysis as a mixture model, which is how you request a latent class analysis.
• Plot is used to make the plot we created above. The type was plot3, and the series statement is used to associate the items with the X axis, with item1 labeled as 1, item2 labeled as 2 … and
item9 labeled as 9 on the X axis.
• Savedata saves the original data file along with additional statistics in it. The data file will be named lca1_save.txt and the save statement indicates we want to save the class probabilities,
i.e., for each case the probability that case belongs to class 1, class 2 and class 3. This also saves the modal class membership. The format of the file is free, space separated.
• Output requests additional output. In this case, the tech11 option requests the Vuong-Lo-Mendell-Rubin test for assessing the number of classes and the tech14 option requests the bootstrapped
parametric likelihood ratio test.
Cautions, Flies in the Ointment
We have focused on a very simple example here just to get you started. Here are some problems to watch out for.
1. Have you specified the right number of latent classes? Perhaps you have specified too many classes (i.e., people largely fall into 2 classes) or you may have specified too few classes (i.e.,
people really fall into 4 or more classes).
2. Are some of your measures/indicators lousy? All of our measures were really useful in distinguishing what type of drinker the person was. However, say we had a measure that was “Do you like
broccoli?”. This would be a poor indicator, and each type of drinker would probably answer in a similar way, so this question would be a good candidate to discard.
3. Having developed this model to identify the different types of drinkers, we might be interested in trying to predict why someone is an alcoholic, or why someone is an abstainer. For example, we
might be interested in whether parental drinking predicts being an alcoholic. Such analyses are possible, but not discussed here. (references forthcoming)
See Also | {"url":"https://stats.oarc.ucla.edu/mplus/dae/latent-class-analysis/","timestamp":"2024-11-10T08:41:43Z","content_type":"text/html","content_length":"57710","record_id":"<urn:uuid:bee77ca6-1a60-4dd3-a3af-9c95de6b6e0b>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00560.warc.gz"} |
Is Math Truly Forever?
Posted on September 8, 2016 by Matthew Simonson
Guest Author: Andrea McNally
Anyone involved in the discipline of math can most likely recall one, if not multiple, instances of being questioned on the usefulness of math. Eduardo Saenz de Cabezon addresses this question in his
TED talk “Math is fore ver” (which can be found here). He claims there are three types of responses. First, the attacking one, which states math has a meanings all its own without the need for
application. Next is the defensive one, which replies math is behind everything from bridge building to credit card numbers. The third response is where Eduardo claims math’s utility stems from its
ability to control intuition, thus making it eternal.
Is math forever? Eduardo seems to think so stating diamonds aren’t forever, a theorem is. Mathematicians spend their lives generating conjectures and searching for ways to prove them. Once a
conjecture is proven true though, it becomes a theorem, which is a truth that will remain so forever. Therefore, concepts such as the Pythagorean Theorem and the Honeycomb Theorem will forever be
true, regardless of whether or not we are here to acknowledge it. This idea is rooted from Platonism, which is the philosophical view that there are abstract math objects that exist independently
from our thoughts. Thus, all math truths are waiting to be discovered and not invented.
There are two main contributors in the world of mathematical philosophy. The first is German mathematician David Hilbert (pictured to the left), creator of Hilbert’s Program. He claimed that all math
is formulized in axiomatic form with a proof to accompany it; it is done so by using finitary methods only which gives proper justification for classical mathematic problems. Hilbert believed
theories could be developed without the need for intuition and would generate a set of rules and axioms that are consistent so one cannot prove an assertion as well as its opposite. Hilbert, like
Eduardo, believed the capabilities of math were limitless.
Hilbert’s work, in turn, inspired the work of Kurt Gӧdel (pictured right) and his Incompleteness Theorems. Gӧdel proved that Hilbert’s concept of a decision procedure that generates axioms cannot be
possible; there will always be conjectures that need a proof that may not actually exist. Gӧdel’s first incompleteness theorem proved that math knowledge cannot be specifically summed up and
identified. Even the soundest basic rules will have statements about numbers that can’t be verified. It is important to note however, that Gӧdel never had the intention of disproving Hilbert’s
program but rather to offer a new view.
So this leaves the math community open to explore if math is created or exists regardless of human recognition. If a tree falls in the woods when no one is around, does it make a sound? If no one has
been able to prove a conjecture, does that theorem still exist? Like many schools of thought, there is ambiguity and uncertainty. As an individual in the math community, we all are responsible for
looking into the information and opinions and coming to our own conclusions. Yet one thing remains certain, intuition and creativity
are absolutely essential in mathematics.
de Cabezon, Eduardo Saenz. “Math is Forever.” TED. TED, Oct. 2014. Web. 05 Apr. 2016.
Elwes, Richard. “Ultimate Logic. (Cover Story).” New Scientist. 211.2823 (2011): 30-33. Academic Search Complete. Web. 4 Apr. 2016
Linnebo, Oystein. “Platonism in the Philosophy of Mathematics.” Stanford University. Stanford University, 18 July 2009. Web. 05 Apr. 2016.
Peterson, Ivars. “The Limits of Mathematics.” Science News. Society for Science & the Public, 2 Mar. 2006. Web. 5 Apr. 2016.
Zach, Richard. “Hilbert’s Program.” Stanford University. Stanford University, 31 July 2003. Web. 05 Apr. 2016.
Image 1 retrieved from: https://www.bing.com/images/search?q=Incompleteness+Theorems
Image 2 retrieved from: https://www.bing.com/images/search?q=david+hilbert
Image 3 retrieved from: https://www.bing.com/images/search?q=Incompleteness+Theorems
About Matthew Simonson
I am a second-year Network Science doctoral student at Northeastern University in Boston. I model homophily and time-varying dynamics on social networks.
This entry was posted in Math, Math History and tagged math history, nature of proof, philosophy of math. Bookmark the permalink. | {"url":"https://blogs.ams.org/mathgradblog/2016/09/08/math-forever/","timestamp":"2024-11-09T06:53:17Z","content_type":"text/html","content_length":"58711","record_id":"<urn:uuid:edc05360-b233-4f2a-af9a-d7f600c4ed80>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00872.warc.gz"} |
Exercise 8.3 class 8 solution
Table of Contents
Exercise 8.3 class 8 solution-Introduction
Welcome to our website, your one-stop destination for free mathematics solutions for Class 8!students can easily download Exercise 8.3 class 8 solution pdf here as well as other chapter solution
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Exercise 8.3 class 8 solution – Multiplying a monomial by polynomial
Exercise 8.3 class 8 solution-Multiplying a monomial by binomial –
When you multiply a monomial (a single term) by a binomial (an expression with two terms), you can use the distributive property to distribute the monomial to both terms in the binomial. Here’s how
to do it:
Step 1: Distribute the monomial to each term in the binomial.
Let’s say you want to multiply the monomial 3x by the binomial 2x+5
Distribute the monomial to both terms in the binomial:
Step 2: Simplify each term.
Now, perform the multiplications for each term:
Step 3: Combine the terms.
Now, combine the simplified terms.
Exercise 8.3 class 8 solution-multiplying a monomial by trinomial
When you multiply a monomial (a single term) by a trinomial (an expression with three terms), you can use the distributive property to distribute the monomial to each term in the trinomial. Here’s
how to do it:
Exercise 8.3 class 8 solution- exercise preview
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Exercise 8.3 class 8 solution- solution pdf
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These downloadable resources are designed to simplify complex concepts, provide clear step-by-step explanations, and boost your confidence in solving math problems. We believe that learning
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So, get ready to unlock the world of mathematics and explore Exercise 8.3 class 8 solutions that will help you not only in your exams but also in building a strong foundation for future mathematical
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Exercise 10.1 class 8 solution
Exercise 10.2 class 8 solution
Exercise 11.1 class 8 solution
Exercise 11.2 class 8 solution
Exercise 12.1 class 8 solution
Exercise 12.2 class 8 solution
Exercise 12.3 class 8 solution
Exercise 13.1 class 8 solution
Exercise 13.2 class 8 solution
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Check if Value is NaN in Python - EnableGeek
About NaN value
NaN stands for “Not a Number”. It is a special floating-point value that is used to represent the result of an undefined or unrepresentable mathematical operation.
NaN can arise in several ways, such as when dividing a number by zero, taking the square root of a negative number, or performing operations with infinity.
NaN is often used to indicate missing or undefined data in data analysis and scientific computing. In Python, NaN is represented by the special value ‘float('nan')‘ or ‘numpy.nan‘ when using the
NumPy library.
It’s important to note that NaN values do not compare equal to any other value, including other NaN values. This means that a comparison such as ‘NaN == NaN‘ will always return ‘False‘. In Python,
you can use the ‘math.isnan()‘ function to check whether a value is NaN or not.
Discover more Python tips and elevate your coding skills with our comprehensive guide, “Python Beginner to Advanced.” Whether you’re a beginner or an experienced developer, this book covers
everything from basic concepts to advanced techniques. And if you’re interested in expanding your programming collection, don’t miss out on our “Java Beginner to Advanced” guide as well!
Check for NaN Values in Python
In Python, you can check for NaN values using the math.isnan() function or the NumPy library’s ‘numpy.isnan()‘ function. Here are some examples:
import math
import numpy as np
# Check if a value is NaN using math.isnan()
x = float('nan')
if math.isnan(x):
print('x is NaN')
print('x is not NaN')
# Check if a value is NaN using numpy.isnan()
arr = np.array([1.0, float('nan'), 2.0, np.nan])
nan_indices = np.isnan(arr)
In the first example, math.isnan() is used to check whether the value of ‘x‘ is NaN. If ‘x‘ is NaN, the function returns ‘True‘ and the program prints 'x is NaN'. Otherwise, the function returns
‘False” and the program prints 'x is not NaN'.
In the second example, a NumPy array containing some NaN values is created. The np.isnan() function is used to check which elements of the array are NaN, returning a boolean array where ‘True‘
indicates a NaN value. This boolean array can be used to mask or filter the original array to work with only the non-NaN values.
Note that the ‘==‘ operator should not be used to check for NaN values, as it will always return ‘False‘ even when comparing NaN to itself.
Handling NaN in Data Analysi
In the realm of data analysis, the presence of NaN (Not a Number) values poses a significant challenge. These missing or undefined values, if not handled judiciously, can cast shadows over the
integrity and reliability of analyses. Common challenges emerge as data scientists navigate the intricate waters of NaN, seeking ways to mitigate its impact on statistical insights, visualizations,
and algorithmic models.
One of the primary challenges lies in data integrity. NaN values scattered throughout a dataset can act as silent saboteurs, subtly compromising the reliability of analyses. Identifying and
addressing these gaps in data become paramount to fortify the foundation upon which subsequent insights are built.
Moreover, the specter of statistical bias looms large. NaN values, if not properly handled, can skew statistical measures such as mean, variance, and correlations. The consequences of such biases can
ripple through analyses, potentially leading to flawed interpretations of data patterns and trends.
Visualizations, as powerful storytelling tools in data analysis, face their own set of challenges in the presence of NaN. Gaps in data, if not appropriately managed, can distort visual
representations, rendering them misleading. Tackling these visualization challenges involves crafting visual narratives that transparently reflect the true nature of the underlying data.
Algorithmic models, the backbone of many data analyses, confront hurdles when NaN values are in play. Certain machine learning algorithms struggle with missing data, raising the stakes in terms of
model accuracy and performance. Successfully addressing algorithmic impact requires strategic handling of NaN values to ensure the robustness of models.
However, the journey through NaN-laden datasets is not without its nuances. The decision to handle NaN often involves striking a balance between data cleaning overhead and information preservation.
Deleting rows or columns with NaN values might streamline the process, but at the cost of potential data loss.
In the arsenal of techniques available for handling NaN values, one approach involves dropping NaN values selectively. This method, while effective, necessitates a careful consideration of the
trade-offs involved, weighing the benefits against the potential loss of valuable information.
Alternatively, data analysts can turn to imputation techniques. Imputing missing values involves filling the NaN gaps with estimated or predicted values. This may include simple methods like mean or
median imputation, or more sophisticated approaches such as regression imputation, where relationships between variables are taken into account.
For time-series data, the strategy may involve forward and backward filling, wherein NaN values are replaced with preceding or succeeding values in the sequence. This approach aligns with the logical
progression of data over time, ensuring a coherent representation.
Delving deeper, interpolation techniques offer another avenue. These methods estimate missing values based on the surrounding data points, with linear interpolation being a common choice. More
complex techniques, such as cubic spline interpolation, provide smoother estimates for nuanced datasets.
Data scientists may also turn to advanced imputation models, employing machine learning algorithms like K-Nearest Neighbors (KNN) or Decision Trees to predict and impute missing values based on
observed patterns in the data. These models add a layer of sophistication, capturing intricate relationships that simpler imputation methods might overlook.
Libraries like Pandas and NumPy come equipped with functions such as fillna() or interpolate() that provide efficient tools for handling NaN values in datasets. Leveraging these functions streamlines
the data cleaning process, offering a practical solution for analysts.
Ultimately, analysts might find solace in establishing customized business rules for handling NaN values. Depending on the nature of the data and the domain, tailored imputation strategies or
flagging mechanisms can be devised to align with specific business requirements.
In navigating the seas of NaN in data analysis, data scientists wield a diverse toolkit of techniques. The effectiveness of these tools hinges on a nuanced understanding of the data at hand, the
specific challenges posed by NaN, and the broader goals of the analysis. By mastering the art of NaN handling, analysts pave the way for robust, accurate, and reliable insights in the ever-evolving
landscape of data science.
Fix NaN Value Problem
The approach to fixing NaN values depends on the specific problem and the nature of the data. However, here are some common techniques that can be used to address NaN values in Python:
1. Remove NaN values: If the NaN values are in a small proportion of the dataset and do not significantly affect the analysis, you can simply remove the rows or columns that contain NaN values. You
can use the pandas.DataFrame.dropna() function to remove NaN values from a pandas DataFrame.
2. Fill NaN values with a constant: If the NaN values represent missing data, you can fill them with a constant value that is representative of the data. For example, you can fill NaN values with
the mean, median, or mode of the non-NaN values in the column. You can use the ‘pandas.DataFrame.fillna()‘ function to fill NaN values in a pandas DataFrame.
3. Interpolate NaN values: If the NaN values represent missing data that has some level of predictability or correlation with the other data, you can interpolate the NaN values based on the adjacent
non-NaN values. For example, you can use linear or polynomial interpolation to estimate the NaN values based on the surrounding data. You can use the ‘pandas.DataFrame.interpolate()‘ function to
interpolate NaN values in a pandas DataFrame.
4. Use machine learning techniques: If the NaN values are part of a predictive modeling problem, you can use machine learning techniques to impute the missing values. For example, you can use
regression models or neural networks to predict the missing values based on the other data in the dataset.
It’s important to note that filling or interpolating NaN values can potentially introduce bias or noise into the data, and should be done with caution. It’s also a good practice to carefully examine
the data to understand the reasons for the NaN values and to choose an appropriate approach for handling them. | {"url":"https://www.enablegeek.com/tutorial/check-if-value-is-nan-python/","timestamp":"2024-11-04T10:31:03Z","content_type":"text/html","content_length":"250789","record_id":"<urn:uuid:e4994066-2f01-4388-b57d-97a6117776c5>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00238.warc.gz"} |
Data Science | College Coding
Introduction to Data Science
Data science is a multidisciplinary field that employs various techniques to extract insights and knowledge from data. A well-structured data science course module covers the essential aspects of
data collection, cleaning, analysis, and visualization. This guide aims to provide an overview of what you can expect from a comprehensive data science course module.
• Overview of data science
• Importance and applications of data science
• Data science lifecycle
• Setting up the Python environment (Anaconda, Jupyter Notebook, PyCharm)
• Python syntax and structure
• Basic data types and variables
• Control flow (if statements, loops)
• Functions and modules
• Lists, tuples, and sets
• Dictionaries
• Comprehensions (list, dictionary, set)
• Reading data from CSV, Excel, JSON, and SQL databases
• Web scraping with BeautifulSoup and Scrapy
• Accessing APIs with requests
□ Introduction to relational databases
□ SQL basics (SELECT, INSERT, UPDATE, DELETE)
□ Connecting Python to SQL databases using SQLAlchemy
• Handling missing values
• Removing duplicates
• Data transformation (scaling, normalization)
• DataFrames and Series
• Indexing, slicing, and filtering
• Aggregation and grouping
• Objectives of EDA
• Tools and libraries (Pandas, NumPy, Matplotlib, Seaborn)
• Creating plots and charts with Matplotlib
• Advanced visualizations with Seaborn
• Interactive visualizations with Plotly
• Summary statistics (mean, median, mode)
• Measures of dispersion (variance, standard deviation)
• Correlation and covariance
• Basic probability concepts
• Probability distributions (normal, binomial, Poisson)
• Hypothesis testing
• Confidence intervals
• t-tests, chi-square tests, ANOVA
• Simple linear regression
• Multiple linear regression
• Logistic regression
• Overview of machine learning
• Supervised vs. unsupervised learning
• Model evaluation and selection
□ Classification algorithms (decision trees, random forest, k-nearest neighbors)
□ Regression algorithms (linear regression, polynomial regression)
• Clustering algorithms (k-means, hierarchical clustering)
• Dimensionality reduction (PCA, t-SNE)
□ Basics of neural networks
□ Building neural networks with TensorFlow and Keras
• Image classification and processing
• Building and training CNN models
• Overview of big data technologies
• Working with Hadoop and Spark
• Generating reports with Jupyter Notebook
• Using BI tools (Tableau, Power BI)
• Designing and implementing a data science project
• Collecting, cleaning, and analyzing data
• Building predictive models
• Visualizing and presenting results | {"url":"https://collegecoding.com/data-science/","timestamp":"2024-11-04T07:11:15Z","content_type":"text/html","content_length":"200708","record_id":"<urn:uuid:0c9717fc-1687-4d42-822a-74c152e44491>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00423.warc.gz"} |
d/dx sin(ax) rule | formula
$\dfrac{d}{dx}{\,\sin{(ax)}}$ rule
$\dfrac{d}{dx}{\,\sin{(ax)}}$ $\,=\,$ $a\cos{(ax)}$
Let $a$ and $x$ be a constant and a variable respectively but the variable $x$ represents an angle in this case. The product of $a$ and $x$ is $ax$, which represents a multiple angle in mathematical
form. The sine of a multiple angle $ax$ is written as $\sin{(ax)}$ mathematically.
The derivative of sine of a multiple angle $ax$ with respect to $x$ is written in the following mathematical form in calculus.
The derivative of sine of multiple angle $ax$ with respect to $x$ is equal to the product of multiple constant $a$ and the cosine of multiple angle $ax$.
$\implies$ $\dfrac{d}{dx}{\,\sin{(ax)}}$ $\,=\,$ $a \times \cos{(ax)}$
$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{d}{dx}{\,\sin{(ax)}}$ $\,=\,$ $a\cos{(ax)}$
It is called the derivative rule for the sine of a multiple angle.
It is used as a formula to find the derivative of a sine function in which multiple or submultiple angle is involved.
Find $\dfrac{d}{dx}{\,\sin{(2x)}}$
In this example, $a \,=\, 2$, so, substitute it in the derivative of sine of multiple angle formula to find the derivative of $\sin{(2x)}$ with respect to $x$.
$\implies$ $\dfrac{d}{dx}{\,\sin{(2x)}}$ $\,=\,$ $2 \times \cos{(2x)}$
$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{d}{dx}{\,\sin{(2x)}}$ $\,=\,$ $2\cos{(2x)}$
Learn how to prove the differentiation formula for finding the derivative of the sine of a multiple angle with respect to a variable. | {"url":"https://www.mathdoubts.com/derivative-of-sin-multiple-angle-rule/","timestamp":"2024-11-11T01:43:03Z","content_type":"text/html","content_length":"27945","record_id":"<urn:uuid:e154ae2f-df45-404d-bab3-ca666559b1a4>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00615.warc.gz"} |
The Oldest Preserved Mathematical Work of Slavs
1. Lemma
2. Najstariji sacuvani matematicki rukopis Slovena
3. Serbian
4. 12-11-2018
5. Dejic, Mirko [Author]. The oldest Slav (Russian) Mathematical Manuscript, the Work of the Monk Kirk. 79–88
6. Mathematics - Slavs
1. The paper traces the correlation between religion and science in the oldest preserved mathematical work of Slavs, in Russian. It was written by the monk Kirk of Antonije Monastery in
Novograd. There are four units in the manuscript and twenty-seven paragraphs. There are different chronological calculations, as well as calendar and astronomic questions.
Kirk shows a great degree of numeric and astronomic abilities and shows a high degree of development of mathematical and astronomical sciences in Russia in the 12^th century.
The paper enlightens the oldest preserved mathematical manuscript of the Russians and Slavic people. The authors show that Kirk used mathematical and astronomic achievements, which were known
in the 12^th century and wider from Russia, and note of big figures may be seen as the original achievement of the Slavs, and Kirk’s work reveals the ways they were noted.
The paper shows that science and religion attract not only local research attention, but their intersection is significant in global terms. | {"url":"http://k2.altsol.gr/archive/item/5000?lang=en","timestamp":"2024-11-01T19:28:27Z","content_type":"application/xhtml+xml","content_length":"16011","record_id":"<urn:uuid:929a5991-bc20-4740-a58b-e995dbbf795c>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00140.warc.gz"} |
(iii) ∠ABD=∠BAC
3. AD and BC are equal and perpendiculars to a ... | Filo
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(iii) 3. and are equal and perpendiculars to a line segment . Show that bisects .
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Question Text (iii) 3. and are equal and perpendiculars to a line segment . Show that bisects .
Updated On Dec 19, 2022
Topic All topics
Subject Mathematics
Class Class 9
Answer Type Video solution: 2
Upvotes 193
Avg. Video Duration 4 min | {"url":"https://askfilo.com/user-question-answers-mathematics/iii-3-and-are-equal-and-perpendiculars-to-a-line-segment-33343431323432","timestamp":"2024-11-10T14:43:48Z","content_type":"text/html","content_length":"188910","record_id":"<urn:uuid:d42a1b26-65c5-44ac-bcb7-d5b3740d1fba>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00341.warc.gz"} |
Assumptions in Linear Regression: A Comprehensive Guide
You will learn the fundamentals of the assumptions in linear regression and how to validate them using real-world examples for practical data analysis.
• Linear regression is a widely used predictive modeling technique for understanding relationships between variables.
• The normality of residuals helps ensure unbiased predictions and trustworthy confidence intervals in linear regression.
• Homoscedasticity guarantees that the model’s predictions have consistent precision across different values.
• Identifying and addressing multicollinearity improves the stability and interpretability of your regression model.
• Data preprocessing and transformation techniques, such as scaling and normalization, can mitigate potential issues in linear regression.
Linear regression is a technique to model and predict the relationship between a target variable and one or more input variables.
It helps us understand how a change in the input variables affects the target variable.
Linear regression assumes that a straight line can represent this relationship.
For example, let’s say you want to estimate the cost of a property considering its size (measured in square footage) and age (in years).
In this case, the price of the house is the target variable, and the size and age are the input variables.
Using linear regression, you can estimate the effect of size and age on the price of the house.
Assumptions in Linear Regression
Six main assumptions in linear regression need to be satisfied for the model to be reliable and valid. These assumptions are:
1. Linearity
This assumption states a linear relationship exists between the dependent and independent variables. In other words, the change in the dependent variable should be proportional to the change in the
independent variables. Linearity can be assessed using scatterplots or by examining the residuals.
2. Normality of errors
The residuals should follow a normal distribution with a mean of zero. This assumption is essential for proper hypothesis testing and constructing confidence intervals. The normality of errors can be
assessed using visual methods, such as a histogram or a Q-Q plot, or through statistical tests, like the Shapiro-Wilk test or the Kolmogorov-Smirnov test.
3. Homoscedasticity
This assumption states that the residuals’ variance should be constant across all independent variable levels. In other words, the residuals’ spread should be similar for all values of the
independent variables. Heteroscedasticity, violating this assumption, can be identified using scatterplots of the residuals or formal tests like the Breusch-Pagan test.
4. Independence of errors
This assumption states that the dataset observations should be independent of each other. Observations may depend on each other when working with time series or spatial data due to their temporal or
spatial proximity. Violating this assumption can lead to biased estimates and unreliable predictions. Specialized models like time series or spatial models may be more appropriate in such cases.
5. Absence of multicollinearity (Multiple Linear Regression)
Multicollinearity takes place when two or more independent variables in the linear regression model are highly correlated, making it challenging to establish the precise effect of each variable on
the dependent variable. Multicollinearity can lead to unstable estimates, inflated standard errors, and difficulty interpreting coefficients. You can use the variance inflation factor (VIF) or
correlation matrix to detect multicollinearity. If multicollinearity is present, consider dropping one of the correlated variables, combining the correlated variables, or using techniques like
principal component analysis (PCA) or ridge regression.
6. Independence of observations
This assumption states that the dataset observations should be independent of each other. Observations may depend on each other when working with time series or spatial data due to their temporal or
spatial proximity. Violating this assumption can lead to biased estimates and unreliable predictions. Specialized models like time series or spatial models may be more appropriate in such cases.
By ensuring that these assumptions are met, you can increase your linear regression models’ accuracy, reliability, and interpretability. If any assumptions are violated, it may be necessary to apply
data transformations, use alternative modeling techniques, or consider other approaches to address the issues.
Assumptions Description
Linearity Linear relationship between dependent and independent variables, checked using scatterplots
Normality Normal distribution of residuals, assessed using Shapiro-Wilk test
Homoscedasticity Constant variance in error terms, evaluated using Breusch-Pagan test
Independence of errors Independent error terms, verified using Durbin-Watson test
Independence of observations Independently collected data points without autocorrelation
Absence of multicollinearity No multicollinearity among independent variables, determined using VIF and Tolerance measures
Here is a demonstration of a linear regression model problem with two independent variables and one dependent variable.
In this example, we will model the relationship between a house’s square footage and age with its selling price.
The dataset contains the square footage, age, and selling price of 40 houses.
We will use multiple linear regression to estimate the effects of square footage and age on selling price.
Here is a table with the data that you can copy and paste:
Copy Data to Clipboard
House SquareFootage Age Price
1 1500 10 250000.50
2 2000 5 300000.75
3 1200 15 200500.25
4 2500 2 400100.80
5 1800 8 270500.55
6 1600 12 220800.60
7 2200 4 320200.10
8 2400 1 420300.90
9 1000 18 180100.15
10 2000 7 290700.40
11 1450 11 240900.65
12 2050 6 315600.20
13 1150 16 190800.75
14 2600 3 410500.50
15 1750 9 260200.55
16 1550 13 210700.85
17 2300 3 330400.45
18 2450 2 415200.90
19 1100 17 185300.65
20 1900 8 275900.80
21 1400 12 235800.55
22 2100 6 305300.40
23 1300 14 195400.25
24 2700 3 410200.75
25 1700 10 255600.20
26 1650 11 215400.60
27 2150 5 325500.50
28 1250 15 205700.85
29 2550 4 395900.90
30 1850 9 265100.65
31 1350 13 225900.40
32 1950 7 285800.15
33 1100 16 195900.80
34 2800 3 430700.55
35 1750 10 245500.20
36 1600 12 225300.10
37 2000 7 310700.50
37 2000 7 310700.50
38 1200 15 201200.90
39 2600 4 380800.65
40 1800 8 279500.25
Evaluate the normality assumption by conducting the Shapiro-Wilk test, which assesses the residuals’ distribution for significant deviations from a normal distribution.
In the Shapiro-Wilk test, a high p-value (typically above 0.05) indicates that the residuals’ distribution does not significantly differ from a normal distribution.
4. Independence of errors
A Durbin-Watson statistic close to 2 suggests that the errors are independent, with minimal autocorrelation present.
Values below or above 2 indicate positive or negative autocorrelation, respectively.
The p-value signifies that the DW statistic is not significantly different from 2.
5. Absence of multicollinearity
Assess the absence of multicollinearity using Variance Inflation Factor (VIF) and Tolerance measures. Low VIF values (typically below 10) and high Tolerance values (above 0.1) indicate that
multicollinearity is not a significant concern in the regression model.
Our data indicate the presence of multicollinearity between the variables age and square footage. We will need to remove one of them. The variable to be removed can be determined in various ways,
such as testing with simple linear regressions to see which fits the model better or deciding based on the underlying theory.
6. Independence of observations
To avoid violating the independence of observations assumption, ensure that your data points are collected independently and do not exhibit autocorrelation, which can be assessed using the
Durbin-Watson test.
It is crucial to examine and address these assumptions when building a linear regression model to ensure validity, reliability, and interpretability.
By understanding and verifying the six assumptions — linearity, independence of errors, homoscedasticity, normality of errors, independence of observations, and absence of multicollinearity — you can
build more accurate and reliable models, leading to better decision-making and improved understanding of the relationships between variables in your data.
Seize the opportunity to access FREE samples from our newly released digital book and unleash your potential.
Dive deep into mastering advanced data analysis methods, determining the perfect sample size, and communicating results effectively, clearly, and concisely.
Click the link to uncover a wealth of knowledge: Applied Statistics: Data Analysis.
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Assumptions in Linear Regression
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Modus ponens
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In propositional logic, modus ponens (/ˈmoʊdəs ˈpoʊnɛnz/; MP; also modus ponendo ponens (Latin for "mode that affirms by affirming")^[1] or implication elimination) is a rule of inference.^[2] It can
be summarized as "P implies Q and P is asserted to be true, therefore Q must be true."
Modus ponens is closely related to another valid form of argument, modus tollens. Both have apparently similar but invalid forms such as affirming the consequent, denying the antecedent, and evidence
of absence. Constructive dilemma is the disjunctive version of modus ponens. Hypothetical syllogism is closely related to modus ponens and sometimes thought of as "double modus ponens."
The history of modus ponens goes back to antiquity.^[3] The first to explicitly describe the argument form modus ponens was Theophrastus.^[4]
Formal notation[edit]
The modus ponens rule may be written in sequent notation as
${\displaystyle P\to Q,\;P\;\;\vdash \;\;Q}$
where P, Q and P → Q are statements (or propositions) in a formal language and ⊢ is a metalogical symbol meaning that Q is a syntactic consequence of P and P → Q in some logical system.
The argument form has two premises (hypothesis). The first premise is the "if–then" or conditional claim, namely that P implies Q. The second premise is that P, the antecedent of the conditional
claim, is true. From these two premises it can be logically concluded that Q, the consequent of the conditional claim, must be true as well. In artificial intelligence, modus ponens is often called
forward chaining.
An example of an argument that fits the form modus ponens:
If today is Tuesday, then John will go to work.
Today is Tuesday.
Therefore, John will go to work.
This argument is valid, but this has no bearing on whether any of the statements in the argument are true; for modus ponens to be a sound argument, the premises must be true for any true instances of
the conclusion. An argument can be valid but nonetheless unsound if one or more premises are false; if an argument is valid and all the premises are true, then the argument is sound. For example,
John might be going to work on Wednesday. In this case, the reasoning for John's going to work (because it is Wednesday) is unsound. The argument is not only sound on Tuesdays (when John goes to
work), but valid on every day of the week. A propositional argument using modus ponens is said to be deductive.
In single-conclusion sequent calculi, modus ponens is the Cut rule. The cut-elimination theorem for a calculus says that every proof involving Cut can be transformed (generally, by a constructive
method) into a proof without Cut, and hence that Cut is admissible.
The Curry–Howard correspondence between proofs and programs relates modus ponens to function application: if f is a function of type P → Q and x is of type P, then f x is of type Q.
Justification via truth table[edit]
The validity of modus ponens in classical two-valued logic can be clearly demonstrated by use of a truth table.
│ p │ q │ p → q │
│ T │ T │ T │
│ T │ F │ F │
│ F │ T │ T │
│ F │ F │ T │
In instances of modus ponens we assume as premises that p → q is true and p is true. Only one line of the truth table—the first—satisfies these two conditions (p and p → q). On this line, q is also
true. Therefore, whenever p → q is true and p is true, q must also be true.
While modus ponens is one of the most commonly used argument forms in logic it must not be mistaken for a logical law; rather, it is one of the accepted mechanisms for the construction of deductive
proofs that includes the "rule of definition" and the "rule of substitution".^[5] Modus ponens allows one to eliminate a conditional statement from a logical proof or argument (the antecedents) and
thereby not carry these antecedents forward in an ever-lengthening string of symbols; for this reason modus ponens is sometimes called the rule of detachment^[6] or the law of detachment.^[7]
Enderton, for example, observes that "modus ponens can produce shorter formulas from longer ones",^[8] and Russell observes that "the process of the inference cannot be reduced to symbols. Its sole
record is the occurrence of ⊦q [the consequent] . . . an inference is the dropping of a true premise; it is the dissolution of an implication".^[9]
A justification for the "trust in inference is the belief that if the two former assertions [the antecedents] are not in error, the final assertion [the consequent] is not in error".^[9] In other
words: if one statement or proposition implies a second one, and the first statement or proposition is true, then the second one is also true. If P implies Q and P is true, then Q is true.^[10]
Correspondence to other mathematical frameworks[edit]
Probability calculus[edit]
Modus ponens represents an instance of the Law of total probability which for a binary variable is expressed as:
${\displaystyle \Pr(Q)=\Pr(Q\mid P)\Pr(P)+\Pr(Q\mid \lnot P)\Pr(\lnot P)\,}$,
where e.g. ${\displaystyle \Pr(Q)}$ denotes the probability of ${\displaystyle Q}$ and the conditional probability ${\displaystyle \Pr(Q\mid P)}$ generalizes the logical implication ${\displaystyle P
\to Q}$. Assume that ${\displaystyle \Pr(Q)=1}$ is equivalent to ${\displaystyle Q}$ being TRUE, and that ${\displaystyle \Pr(Q)=0}$ is equivalent to ${\displaystyle Q}$ being FALSE. It is then easy
to see that ${\displaystyle \Pr(Q)=1}$ when ${\displaystyle \Pr(Q\mid P)=1}$ and ${\displaystyle \Pr(P)=1}$. Hence, the law of total probability represents a generalization of modus ponens ^[11].
Subjective logic[edit]
Modus ponens represents an instance of the binomial deduction operator in subjective logic expressed as:
${\displaystyle \omega _{Q\|P}^{A}=(\omega _{Q|P}^{A},\omega _{Q|\lnot P}^{A})\circledcirc \omega _{P}^{A}\,}$,
where ${\displaystyle \omega _{P}^{A}}$ denotes the subjective opinion about ${\displaystyle P}$ as expressed by source ${\displaystyle A}$, and the conditional opinion ${\displaystyle \omega _{Q|P}^
{A}}$ generalizes the logical implication ${\displaystyle P\to Q}$. The deduced marginal opinion about ${\displaystyle Q}$ is denoted by ${\displaystyle \omega _{Q\|P}^{A}}$. The case where ${\
displaystyle \omega _{P}^{A}}$ is an absolute TRUE opinion about ${\displaystyle P}$ is equivalent to source ${\displaystyle A}$ saying that ${\displaystyle P}$ is TRUE, and the case where ${\
displaystyle \omega _{P}^{A}}$ is an absolute FALSE opinion about ${\displaystyle P}$ is equivalent to source ${\displaystyle A}$ saying that ${\displaystyle P}$ is FALSE. The deduction operator ${\
displaystyle \circledcirc }$ of subjective logic produces an absolute TRUE deduced opinion ${\displaystyle \omega _{Q\|P}^{A}}$ when the conditional opinion ${\displaystyle \omega _{Q|P}^{A}}$ is
absolute TRUE and the antecedent opinion ${\displaystyle \omega _{P}^{A}}$ is absolute TRUE. Hence, subjective logic deduction represents a generalization of both modus ponens and the Law of total
probability ^[12].
Alleged cases of failure[edit]
The philosopher and logician Vann McGee has argued that modus ponens can fail to be valid when the consequent is itself a conditional sentence.^[13] Here is an example:
Either Shakespeare or Hobbes wrote Hamlet.
If either Shakespeare or Hobbes wrote Hamlet, then if Shakespeare didn't do it, Hobbes did.
Therefore, if Shakespeare didn't write Hamlet, Hobbes did it.
The first premise seems reasonable enough, because Shakespeare is generally credited with writing Hamlet. The second premise seems reasonable, as well, because with the range of Hamlet 's possible
authors limited to just Shakespeare and Hobbes, eliminating one leaves only the other. But the conclusion is dubious, because if Shakespeare is ruled out as Hamlet's author, there are many more
plausible alternatives than Hobbes.
The general form of McGee-type counterexamples to modus ponens is simply ${\displaystyle P,P\rightarrow (Q\rightarrow R)}$, therefore ${\displaystyle Q\rightarrow R}$; it is not essential that ${\
displaystyle P}$ have the form of a disjunction, as in the example given. That these kinds of cases constitute failures of modus ponens remains a minority view among logicians, but there is no
consensus on how the cases should be disposed of.
Possible fallacies[edit]
The fallacy of affirming the consequent is a common misinterpretation of the modus ponens.
See also[edit]
External links[edit] | {"url":"https://static.hlt.bme.hu/semantics/external/pages/kett%C5%91s_tagad%C3%A1s/en.wikipedia.org/wiki/Implication_elimination.html","timestamp":"2024-11-09T16:54:15Z","content_type":"text/html","content_length":"117642","record_id":"<urn:uuid:d35383b6-944b-42e6-a4d6-12df162039ea>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00400.warc.gz"} |
How Domo calculate the average of a column
In my data set, I have a column that has a score of 1-7. I want to use the average of this column. However, some of the rows in this column is blank. Is Domo just going to ignore these blank when
calculate the average? Or I need to add anything in the blank rows to make the calculation accurate?
Best Answers
• Hi @sky00221155
Domo will ignore those blanks when calculating the average. It depends on your use case and what a blank value represents. NULLs typically mean there is no data so there's no certainty as to what
that value represents and is typically ignored when calculating averages. If you want the NULLs to represent something then you'd need to set the NULLs to be a specific value.
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**Did this solve your problem? Accept it as a solution!**
• @sky00221155 the NULL handling on aggregate functions (like AVG) is the same as in Excel. If the column contains a NULL, it will not impact the numerator or denominator while calculating average.
If, for your data, NULL is synonymous with 0, then you have to replace the NULLs with 0 and then they will impact the average.
Jae Wilson
Check out my 🎥
Domo Training YouTube Channel
**Say "Thanks" by clicking the ❤️ in the post that helped you.
**Please mark the post that solves your problem by clicking on "Accept as Solution"
• Hi @sky00221155
Domo will ignore those blanks when calculating the average. It depends on your use case and what a blank value represents. NULLs typically mean there is no data so there's no certainty as to what
that value represents and is typically ignored when calculating averages. If you want the NULLs to represent something then you'd need to set the NULLs to be a specific value.
**Was this post helpful? Click Agree or Like below**
**Did this solve your problem? Accept it as a solution!**
• @sky00221155 the NULL handling on aggregate functions (like AVG) is the same as in Excel. If the column contains a NULL, it will not impact the numerator or denominator while calculating average.
If, for your data, NULL is synonymous with 0, then you have to replace the NULLs with 0 and then they will impact the average.
Jae Wilson
Check out my 🎥
Domo Training YouTube Channel
**Say "Thanks" by clicking the ❤️ in the post that helped you.
**Please mark the post that solves your problem by clicking on "Accept as Solution"
• Thank you! That helps a lot.
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The Basics of Backtracking | CodingDrills
The Basics of Backtracking
Introduction to Backtracking: The Basics of Backtracking
Welcome to this comprehensive tutorial on backtracking algorithms! In this post, we will delve into the fundamentals of backtracking, a technique used for solving complex computational problems by
exploring all possible solutions.
What is Backtracking?
Backtracking is a systematic approach used to find the optimal solution for a given problem by traversing a search space using a trial-and-error method. It involves incrementally building a solution
and subsequently abandoning it if it is determined to be invalid, thus "backtracking" to explore other possibilities.
The general idea is to incrementally construct potential solutions, evaluating them against a set of constraints. If a partial solution fails to meet the constraints, we backtrack to a previous
decision point and try other options, continuing until a valid solution is found, or all possibilities have been exhausted.
Key Components of Backtracking
To effectively utilize backtracking, we need to understand its key components. They are:
1. State Space: The state space represents the problem domain and consists of all possible candidate solutions.
2. Constraints: These are the conditions or rules that define whether a particular solution is viable or not.
3. Decision Space: The decision space comprises the choices or decisions that we make at every step during the backtracking process.
4. Backtracking Algorithm: This is the algorithmic framework that guides the entire process, determining the order in which decisions are made and searched.
Pseudocode for Backtracking
To better comprehend backtracking, let's examine a common pseudocode template that can be applied to a wide range of problems:
solveProblem(state, ...):
if state is a valid solution:
return state
for decision in decisionSpace:
if decision is feasible:
apply decision to state
result = solveProblem(state, ...)
if result is a valid solution:
return result
undo decision on state
return failure
This generic pseudocode illustrates the general control flow of a backtracking algorithm. By applying this template to specific problems, we can develop personalized solutions.
Example: Solving the N-Queens Problem
Let's explore a classic example to illustrate the power of backtracking: the N-Queens problem. In this problem, we aim to place N queens on an N x N chessboard in a way that no two queens threaten
each other.
To solve this problem using backtracking, we can follow the pseudocode mentioned earlier with some specific modifications. Here's an implementation in Python:
def solveNQueens(n):
board = [['.' for _ in range(n)] for _ in range(n)]
def isSafe(row, col):
# Check if the current position is threatened by any other queen
for i in range(n):
if board[i][col] == 'Q' or board[row][i] == 'Q':
return False
if row + i < n and col + i < n and board[row + i][col + i] == 'Q':
return False
if row - i >= 0 and col - i >= 0 and board[row - i][col - i] == 'Q':
return False
if row + i < n and col - i >= 0 and board[row + i][col - i] == 'Q':
return False
if row - i >= 0 and col + i < n and board[row - i][col + i] == 'Q':
return False
return True
def backtrack(row):
if row == n:
# Solution found
return ["".join(row) for row in board]
for col in range(n):
if isSafe(row, col):
board[row][col] = 'Q'
result = backtrack(row + 1)
if result:
return result
board[row][col] = '.'
return None
return backtrack(0)
In this Python implementation, we use a 2D board to represent the chessboard, where 'Q' represents a queen and '.' denotes an empty cell. The recursive backtrack function places queens on the board
row by row, adhering to the four constraints mentioned earlier.
Backtracking algorithms provide a powerful approach for solving complex problems by systematically exploring all possible solutions. By backtracking, we incrementally construct potential solutions,
evaluating them against constraints and refining them until an optimal solution is found.
In this tutorial, we introduced the basics of backtracking and provided an example of solving the N-Queens problem using Python. Understanding the fundamental components and having a clear
understanding of the backtracking process is essential for successfully applying this technique to a wide range of computational problems.
Now that you have a solid foundation in backtracking, feel free to experiment and apply this technique to solve challenging programming puzzles and optimization tasks. Happy coding!
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The brms package provides an interface to fit Bayesian generalized (non-)linear multivariate multilevel models using Stan, which is a C++ package for performing full Bayesian inference (see https://
mc-stan.org/). The formula syntax is very similar to that of the package lme4 to provide a familiar and simple interface for performing regression analyses. A wide range of response distributions are
supported, allowing users to fit – among others – linear, robust linear, count data, survival, response times, ordinal, zero-inflated, and even self-defined mixture models all in a multilevel
context. Further modeling options include non-linear and smooth terms, auto-correlation structures, censored data, missing value imputation, and quite a few more. In addition, all parameters of the
response distribution can be predicted in order to perform distributional regression. Multivariate models (i.e., models with multiple response variables) can be fit, as well. Prior specifications are
flexible and explicitly encourage users to apply prior distributions that actually reflect their beliefs. Model fit can easily be assessed and compared with posterior predictive checks,
cross-validation, and Bayes factors.
How to use brms
As a simple example, we use poisson regression to model the seizure counts in epileptic patients to investigate whether the treatment (represented by variable Trt) can reduce the seizure counts and
whether the effect of the treatment varies with the (standardized) baseline number of seizures a person had before treatment (variable zBase). As we have multiple observations per person, a
group-level intercept is incorporated to account for the resulting dependency in the data.
The results (i.e., posterior draws) can be investigated using
#> Family: poisson
#> Links: mu = log
#> Formula: count ~ zAge + zBase * Trt + (1 | patient)
#> Data: epilepsy (Number of observations: 236)
#> Draws: 4 chains, each with iter = 2000; warmup = 1000; thin = 1;
#> total post-warmup draws = 4000
#> Multilevel Hyperparameters:
#> ~patient (Number of levels: 59)
#> Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> sd(Intercept) 0.59 0.07 0.46 0.74 1.01 566 1356
#> Regression Coefficients:
#> Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> Intercept 1.78 0.12 1.55 2.01 1.00 771 1595
#> zAge 0.09 0.09 -0.08 0.27 1.00 590 1302
#> zBase 0.71 0.12 0.47 0.96 1.00 848 1258
#> Trt1 -0.27 0.16 -0.60 0.05 1.01 749 1172
#> zBase:Trt1 0.05 0.17 -0.30 0.38 1.00 833 1335
#> Draws were sampled using sampling(NUTS). For each parameter, Bulk_ESS
#> and Tail_ESS are effective sample size measures, and Rhat is the potential
#> scale reduction factor on split chains (at convergence, Rhat = 1).
On the top of the output, some general information on the model is given, such as family, formula, number of iterations and chains. Next, group-level effects are displayed separately for each
grouping factor in terms of standard deviations and (in case of more than one group-level effect per grouping factor; not displayed here) correlations between group-level effects. On the bottom of
the output, population-level effects (i.e. regression coefficients) are displayed. If incorporated, autocorrelation effects and family specific parameters (e.g., the residual standard deviation
‘sigma’ in normal models) are also given.
In general, every parameter is summarized using the mean (‘Estimate’) and the standard deviation (‘Est.Error’) of the posterior distribution as well as two-sided 95% credible intervals (‘l-95% CI’
and ‘u-95% CI’) based on quantiles. We see that the coefficient of Trt is negative with a zero overlapping 95%-CI. This indicates that, on average, the treatment may reduce seizure counts by some
amount but the evidence based on the data and applied model is not very strong and still insufficient by standard decision rules. Further, we find little evidence that the treatment effect varies
with the baseline number of seizures.
The last three values (‘ESS_bulk’, ‘ESS_tail’, and ‘Rhat’) provide information on how well the algorithm could estimate the posterior distribution of this parameter. If ‘Rhat’ is considerably greater
than 1, the algorithm has not yet converged and it is necessary to run more iterations and / or set stronger priors.
To visually investigate the chains as well as the posterior distributions, we can use the plot method. If we just want to see results of the regression coefficients of Trt and zBase, we go for
A more detailed investigation can be performed by running launch_shinystan(fit1). To better understand the relationship of the predictors with the response, I recommend the conditional_effects
This method uses some prediction functionality behind the scenes, which can also be called directly. Suppose that we want to predict responses (i.e. seizure counts) of a person in the treatment group
(Trt = 1) and in the control group (Trt = 0) with average age and average number of previous seizures. Than we can use
newdata <- data.frame(Trt = c(0, 1), zAge = 0, zBase = 0)
predict(fit1, newdata = newdata, re_formula = NA)
#> Estimate Est.Error Q2.5 Q97.5
#> [1,] 5.91200 2.494857 2 11
#> [2,] 4.57325 2.166058 1 9
We need to set re_formula = NA in order not to condition of the group-level effects. While the predict method returns predictions of the responses, the fitted method returns predictions of the
regression line.
fitted(fit1, newdata = newdata, re_formula = NA)
#> Estimate Est.Error Q2.5 Q97.5
#> [1,] 5.945276 0.7075160 4.696257 7.450011
#> [2,] 4.540081 0.5343471 3.579757 5.665132
Both methods return the same estimate (up to random error), while the latter has smaller variance, because the uncertainty in the regression line is smaller than the uncertainty in each response. If
we want to predict values of the original data, we can just leave the newdata argument empty.
Suppose, we want to investigate whether there is overdispersion in the model, that is residual variation not accounted for by the response distribution. For this purpose, we include a second
group-level intercept that captures possible overdispersion.
fit2 <- brm(count ~ zAge + zBase * Trt + (1|patient) + (1|obs),
data = epilepsy, family = poisson())
We can then go ahead and compare both models via approximate leave-one-out (LOO) cross-validation.
loo(fit1, fit2)
#> Output of model 'fit1':
#> Computed from 4000 by 236 log-likelihood matrix.
#> Estimate SE
#> elpd_loo -671.7 36.6
#> p_loo 94.3 14.2
#> looic 1343.4 73.2
#> ------
#> MCSE of elpd_loo is NA.
#> MCSE and ESS estimates assume MCMC draws (r_eff in [0.4, 2.0]).
#> Pareto k diagnostic values:
#> Count Pct. Min. ESS
#> (-Inf, 0.7] (good) 228 96.6% 157
#> (0.7, 1] (bad) 7 3.0% <NA>
#> (1, Inf) (very bad) 1 0.4% <NA>
#> See help('pareto-k-diagnostic') for details.
#> Output of model 'fit2':
#> Computed from 4000 by 236 log-likelihood matrix.
#> Estimate SE
#> elpd_loo -596.8 14.0
#> p_loo 109.7 7.2
#> looic 1193.6 28.1
#> ------
#> MCSE of elpd_loo is NA.
#> MCSE and ESS estimates assume MCMC draws (r_eff in [0.4, 1.7]).
#> Pareto k diagnostic values:
#> Count Pct. Min. ESS
#> (-Inf, 0.7] (good) 172 72.9% 83
#> (0.7, 1] (bad) 56 23.7% <NA>
#> (1, Inf) (very bad) 8 3.4% <NA>
#> See help('pareto-k-diagnostic') for details.
#> Model comparisons:
#> elpd_diff se_diff
#> fit2 0.0 0.0
#> fit1 -74.9 27.2
The loo output when comparing models is a little verbose. We first see the individual LOO summaries of the two models and then the comparison between them. Since higher elpd (i.e., expected log
posterior density) values indicate better fit, we see that the model accounting for overdispersion (i.e., fit2) fits substantially better. However, we also see in the individual LOO outputs that
there are several problematic observations for which the approximations may have not have been very accurate. To deal with this appropriately, we need to fall back to other methods such as reloo or
kfold but this requires the model to be refit several times which takes too long for the purpose of a quick example. The post-processing methods we have shown above are just the tip of the iceberg.
For a full list of methods to apply on fitted model objects, type methods(class = "brmsfit").
Developing and maintaining open source software is an important yet often underappreciated contribution to scientific progress. Thus, whenever you are using open source software (or software in
general), please make sure to cite it appropriately so that developers get credit for their work.
When using brms, please cite one or more of the following publications:
• Bürkner P. C. (2017). brms: An R Package for Bayesian Multilevel Models using Stan. Journal of Statistical Software. 80(1), 1-28. doi.org/10.18637/jss.v080.i01
• Bürkner P. C. (2018). Advanced Bayesian Multilevel Modeling with the R Package brms. The R Journal. 10(1), 395-411. doi.org/10.32614/RJ-2018-017
• Bürkner P. C. (2021). Bayesian Item Response Modeling in R with brms and Stan. Journal of Statistical Software, 100(5), 1-54. doi.org/10.18637/jss.v100.i05
As brms is a high-level interface to Stan, please additionally cite Stan (see also https://mc-stan.org/users/citations/):
• Stan Development Team. YEAR. Stan Modeling Language Users Guide and Reference Manual, VERSION. https://mc-stan.org
• Carpenter B., Gelman A., Hoffman M. D., Lee D., Goodrich B., Betancourt M., Brubaker M., Guo J., Li P., and Riddell A. (2017). Stan: A probabilistic programming language. Journal of Statistical
Software. 76(1). doi.org/10.18637/jss.v076.i01
Further, brms relies on several other R packages and, of course, on R itself. To find out how to cite R and its packages, use the citation function. There are some features of brms which specifically
rely on certain packages. The rstan package together with Rcpp makes Stan conveniently accessible in R. Visualizations and posterior-predictive checks are based on bayesplot and ggplot2. Approximate
leave-one-out cross-validation using loo and related methods is done via the loo package. Marginal likelihood based methods such as bayes_factor are realized by means of the bridgesampling package.
Splines specified via the s and t2 functions rely on mgcv. If you use some of these features, please also consider citing the related packages.
How do I install brms?
To install the latest release version from CRAN use
The current developmental version can be downloaded from GitHub via
if (!requireNamespace("remotes")) {
Because brms is based on Stan, a C++ compiler is required. The program Rtools (available on https://cran.r-project.org/bin/windows/Rtools/) comes with a C++ compiler for Windows. On Mac, you should
install Xcode. For further instructions on how to get the compilers running, see the prerequisites section on https://github.com/stan-dev/rstan/wiki/RStan-Getting-Started.
I am new to brms. Where can I start?
Detailed instructions and case studies are given in the package’s extensive vignettes. See vignette(package = "brms") for an overview. For documentation on formula syntax, families, and prior
distributions see help("brm").
Where do I ask questions, propose a new feature, or report a bug?
Questions can be asked on the Stan forums on Discourse. To propose a new feature or report a bug, please open an issue on GitHub.
If you have already fitted a model, apply the stancode method on the fitted model object. If you just want to generate the Stan code without any model fitting, use the stancode method on your model
Can I avoid compiling models?
When you fit your model for the first time with brms, there is currently no way to avoid compilation. However, if you have already fitted your model and want to run it again, for instance with more
draws, you can do this without recompilation by using the update method. For more details see help("update.brmsfit").
What is the difference between brms and rstanarm?
The rstanarm package is similar to brms in that it also allows to fit regression models using Stan for the backend estimation. Contrary to brms, rstanarm comes with precompiled code to save the
compilation time (and the need for a C++ compiler) when fitting a model. However, as brms generates its Stan code on the fly, it offers much more flexibility in model specification than rstanarm.
Also, multilevel models are currently fitted a bit more efficiently in brms. For detailed comparisons of brms with other common R packages implementing multilevel models, see vignette
("brms_multilevel") and vignette("brms_overview"). | {"url":"https://cran.itam.mx/web/packages/brms/readme/README.html","timestamp":"2024-11-13T04:59:31Z","content_type":"application/xhtml+xml","content_length":"33343","record_id":"<urn:uuid:6e285c20-64bc-4d11-973b-41f104e3ac1b>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00543.warc.gz"} |
Debt to Income Ratio Formula | Calculator (Excel template) (2024)
Updated November 23, 2023
Debt to Income Ratio Formula (Table of Contents)
• Debt to Income Ratio Formula
• Debt to Income Ratio Calculator
• Debt to Income Ratio Formula in Excel(With Excel Template)
Debt to Income Ratio Formula
The debt to income ratio is the measure of estimating an individual’s capacity to repay the debt by comparing his recurring monthly debt to gross monthly income.
Examples of Debt to Income Ratio Formula
Example #1
You can download this Debt to Income Ratio Template here –Debt to Income Ratio Template
Let’s take an example for Jim, whose Gross Monthly Income is $10000. Jim has a housing mortgage payment of $3000 per month. Jim has also taken a car loan with a monthly payment of $1000. He also has
other smaller monthly debt payments, which amount to $500.
• Overall Recurring Monthly Debt for Jim = $4500
• Gross Monthly Income = $10000
Using the Debt to Income Ratio Formula, We get –
• Debt to Income Ratio =Overall Recurring Monthly Debt for Jim/Gross Monthly Income
• Debt to Income Ratio = $4500/$10000
• Debt to Income Ratio = 0.45 or 45%
Example #2
Generally, Debt to Income Ratios are used by lenders to determine whether the borrower will be able to repay the loan. It is assumed that the highest debt-to-income ratio is 43%, beyond which the
borrower has a diminishing ability to return the loan.
Suppose John has a gross monthly income of $20000 while Alan has a gross monthly income of $15000. John has a recurring monthly debt of $ 10,000, while Alan has a recurring monthly debt of $ 5,000.
The debt to Income Ratio of John is Calculated as:
• Debt to Income Ratio of John =Recurring Monthly Debt/Gross Monthly Income
• Debt to Income Ratio of John = $10000/$20000
• Debt to Income Ratio of John = 0.5 or 50%
Debt to debt-to-income ratio of Alan is Calculated as follows:
• Debt to Income Ratio of Alan =Recurring Monthly Debt/Gross Monthly Income
• Debt to Income Ratio of Alan = $5000/$15000
• Debt to Income Ratio of Alan = 0.33 or 33%
Hence, lenders will be more inclined to lend money to Alan as his debt-to-income ratio is lower.
Example #3
There are two types of Debt to Income ratios: the Front-end debt to income ratio and the Back-end debt to income ratio. The front-end debt to income ratio generally indicates the percentage of income
that goes towards housing costs, whether rent or payment towards a mortgage, which includes both principal and interest. The back-end debt to income ratio encompasses all other recurring debt
payments such as car loans, credit card payments, education loans, etc.
Lenders use a debt to income ratio of 28/36 to determine whether the borrower should be lent money. 28/36 norm indicates that 28% of the gross income can be expensed for housing costs, while 36% can
be used to expense all other recurring debt payments.
For example,
• If the gross monthly income = $10000.
• The amount allowed for housing expenses = 0.28*10000
• The amount allowed for housing expenses = $2800
• The amount allowed for housing expenses and recurring debt = 0.36*10000
• The amount allowed for housing expenses and recurring debt = $3600
Therefore, the amount allowed for housing expenses is $2800, and the amount allowed for housing expenses and recurring debt is $3600
Explanation of Debt to Income Ratio Formula
Lenders use the debt to income ratio to determine whether a further loan could be issued to the borrower and whether the borrower can return the loan payments. It is generally preferred that the
borrower should have a low Debt to Income Ratio. A ratio of 28% is usually preferable, while 43% is the highest that the Debt to Income ratio could be. A debt ratio to income higher than 43% signals
that the borrower might not be able to return the loan taken.
As can be understood from the formula, there are two ways of lowering one’s debt to income ratio. One can reduce their recurring monthly debt or increase their gross monthly income. Lowering
recurring debt payments can be achieved by prepaying some of the loans.
Significanceand Use ofDebt to Income Ratio Formula
As stated above, lenders use debt to income ratio to determine whether borrowers should be issued new loans or not. There are two types of Debt to Income ratios: the Front-end debt to income ratio
and the Back-end debt to income ratio. The front-end debt to income ratio generally indicates the percentage of income that goes towards housing costs, whether rent or payment towards a mortgage,
which includes both principal and interest. The back-end debt to income ratio encompasses all other recurring debt payments such as car loans, credit card payments, education loans, etc.
Lenders usually use a figure such as 28/36 to determine the amount of the expense that a borrower can afford for it to be eligible to give loans.
Numerator 28 indicates the Front-end debt to income ratio should be 28% of the overall gross monthly income. In contrast, denominator 36 indicates that the back-end debt to income ratio should be 36%
of the overall gross monthly income.
Debt to Income Ratio Formula Calculator
You can use the following Debt to Income Ratio Formula Calculator
Recurring Monthly Debt
Gross Monthly Income
Debt to Income Ratio Formula
Debt to Income Ratio Formula = Recurring Monthly Debt =
Gross Monthly Income
Debt to Income Ratio Formula in Excel (With Excel Template)
Here, we will do an Excel example of the Debt to Income Ratio Formula. It is very easy and simple. You need to provide the two inputs, i.e., Recurring Monthly Debt andGross Monthly Income
You can easily calculate the Debt to Income Ratio Formula in thetemplate provided.
Debt income ratio is one of the essential criteria, along with the credit score, that creditors use to determine whether further debt can be given to the borrowers. The historical limit of 28/36 has
been extended since. Currently, housing prices are higher all over the world. Even if borrowers have DTI ratios as high as 50%, they are given loans at a higher interest rate than others.
Recommended Articles
This has guided the Debt to Income Ratio Formula; here, we discuss its uses and practical examples. We also provide a Debt to Income Ratio calculator and a downloadable Excel template.
1. Price to Book Value Formula
2. DuPont Formula
3. Return on Assets (ROA) Formula
4. net working capital formula
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I'm an expert in financial analysis and debt management, with a deep understanding of the Debt to Income Ratio (DTI) formula. My expertise stems from practical experience and a thorough knowledge of
financial concepts. Let's delve into the information presented in the article by Madhuri Thakur regarding the Debt to Income Ratio Formula.
The Debt to Income Ratio is a crucial metric used to assess an individual's ability to repay debt by comparing their recurring monthly debt to gross monthly income. The formula for DTI is expressed
as follows:
[ \text{Debt to Income Ratio} = \frac{\text{Overall Recurring Monthly Debt}}{\text{Gross Monthly Income}} ]
Now, let's break down the examples provided in the article:
Example #1
• Jim's Gross Monthly Income: $10,000
• Housing Mortgage Payment: $3,000
• Car Loan Payment: $1,000
• Other Monthly Debt Payments: $500
[ \text{Overall Recurring Monthly Debt for Jim} = $3,000 + $1,000 + $500 = $4,500 ]
[ \text{Debt to Income Ratio for Jim} = \frac{$4,500}{$10,000} = 0.45 \text{ or } 45\% ]
Example #2
• John's Gross Monthly Income: $20,000
• Alan's Gross Monthly Income: $15,000
• John's Recurring Monthly Debt: $10,000
• Alan's Recurring Monthly Debt: $5,000
[ \text{Debt to Income Ratio for John} = \frac{$10,000}{$20,000} = 0.5 \text{ or } 50\% ]
[ \text{Debt to Income Ratio for Alan} = \frac{$5,000}{$15,000} = 0.33 \text{ or } 33\% ]
Hence, lenders would be more inclined to lend money to Alan due to his lower DTI ratio.
Example #3
• Front-end Debt to Income Ratio: Indicates housing costs (e.g., rent or mortgage)
• Back-end Debt to Income Ratio: Encompasses all other recurring debt payments
Lenders often use a 28/36 norm, where 28% of gross income can be allocated to housing costs, and 36% for all other recurring debt payments.
These examples illustrate how lenders use DTI ratios to determine a borrower's eligibility for loans. A lower DTI ratio is preferable, with 43% being the upper limit. Borrowers can improve their DTI
by reducing recurring debt or increasing gross monthly income.
The article also highlights the significance of the Debt to Income Ratio in the lending process and provides a calculator for easy computation.
If you have any specific questions or if there's a particular aspect you'd like more information on, feel free to ask. | {"url":"https://julalikariarts.com/article/debt-to-income-ratio-formula-calculator-excel-template","timestamp":"2024-11-04T17:28:05Z","content_type":"text/html","content_length":"73825","record_id":"<urn:uuid:76377837-dc0d-47b0-a880-32f04215d2a3>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00061.warc.gz"} |
How to Grasp Parallel Vectors
Parallel vectors are vectors that have the same or opposite direction. In simpler terms, they point along the same straight line in space. It's essential to note that the magnitude (or length) of the
vectors doesn't have to be the same for them to be parallel; only their direction must be consistent.
Step-by-step Guide to Understand Parallel Vectors
Here is a step-by-step guide to understanding parallel vectors:
Step 1: Setting the Stage – The Alluring World of Vectors
Venturing into the realm of vectors, we encounter entities with both magnitude and direction. These celestial arrows float in space, guiding us through mathematical challenges.
1. The Birth of a Vector:
□ Born from necessity, vectors are the heart and soul of linear motions. Represented visually as directed arrows, they possess both direction and magnitude.
□ E.g., \(\overrightarrow{V}=V_xi+V_yj\) (in \(2D\) space).
Step 2: The Crux – What Does ‘Parallel’ Really Mean?
Parallel vectors, in the abstract landscape of mathematics, are vectors that flow harmoniously in the same or opposite directions, never deviating from their shared path.
1. The Geometrical Insight:
□ Two vectors are parallel if they lie along the same line or their corresponding lines are equidistant at all points. In simpler terms, they point in the same or exactly opposite directions.
□ The angle between parallel vectors is either \(0^\circ\) (same direction) or \(180^\circ\) (opposite direction).
2. Algebraic Aura:
□ Vectors are parallel if one is a scalar multiple of the other.
□ If \(\overrightarrow{A}\) and \(\overrightarrow{B}\) are parallel, there exists a scalar \(k\) such that: \(\overrightarrow{A}=k\overrightarrow{B}\) or \(\overrightarrow{B}=k\overrightarrow
Step 3: The Dot Product Divulgence
The dot product can offer a secret passage to discern if two vectors are parallel.
1. Recollection:
□ For vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\), the dot product is: \(\overrightarrow{A}⋅\overrightarrow{B}=\overrightarrow{∣A∣} \ \overrightarrow{∣B∣}cos(θ)\)
2. Decoding Parallelism:
□ If vectors are parallel, \(cos(θ)\) is either \(1\) or \(-1\).
□ Thus, if \(\overrightarrow{A}⋅\overrightarrow{B}=\overrightarrow{∣A∣} \ \overrightarrow{∣B∣}\) or \(\overrightarrow{A}⋅\overrightarrow{B}=−\overrightarrow{∣A∣}\ \overrightarrow{∣B∣}\), then \
(\overrightarrow{A}\) and \(\overrightarrow{B}\) are parallel.
Step 4: The Cross-Product Conundrum
In a \(3\)-dimensional space, the cross-product provides another tantalizing technique.
1. Recollection:
□ For vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\), the cross-product results in a vector perpendicular to both.
□ Its magnitude is ∣\( \overrightarrow{A}×\overrightarrow{B}\)∣\(=\overrightarrow{∣A∣} \ \overrightarrow{∣B∣}sin(θ)\).
2. Deciphering Parallelism:
□ If vectors are parallel, \(sin(θ)\) is \(0\).
□ Hence, if \( \overrightarrow{A}× \overrightarrow{B}= \overrightarrow{0}\), then vectors \( \overrightarrow{A}\) and \( \overrightarrow{B}\) are parallel.
Step 5: The Elegant Dance of Parallel Vectors in Applications
1. Physics: Parallel vectors often represent forces acting in sync or collinear displacements.
2. Computer Graphics: Vectors play a role in transformations, where preserving parallelism ensures objects don’t get distorted.
3. Geometry: Recognizing parallel vectors helps in proofs and constructions, ensuring congruency and similarity.
Step 6: Conclusion & Reflection
In the vast cosmos of vectors, parallelism is a poetic notion of harmony, a ballet of vectors moving in tandem. Recognizing this dance requires a blend of geometry, algebra, and intuition. It’s a
journey worth undertaking, for the elegance of parallel vectors illuminates many a mathematical path. Embrace the dance, and let the vectors guide your way!
Example 1:
Given vectors: \( \overrightarrow{M}=(3,6)\), and \( \overrightarrow{N}=(1,2)\), determine if the vectors are parallel.
We must see if one is a scalar multiple of the other. Specifically, for some scalar \(k\), all components of \(k \overrightarrow{N}\) should match those of \( \overrightarrow{M}\).
For the \(x\)-components:
For the \(y\)-components:
Since for both components, the scalar \(k\) is consistent, the vectors are indeed parallel.
Example 2:
Given vectors: \( \overrightarrow{P}=(4,8)\), and \( \overrightarrow{Q}=(−2,−4)\), determine if the vectors are parallel.
Again, we’ll check if one is a scalar multiple of the other:
For the \(x\)-components:
For the \(y\)-components:
Once more, the scalar \(k\) remains consistent across the components. Therefore, the vectors \( \overrightarrow{P}\) and \( \overrightarrow{Q}\) are parallel, albeit pointing in opposite directions.
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Bar Graph Horizontal: Learn Definition, Types, Construction & Examples
Have you heard the term Bar graph horizontal or horizontal bar graph? A horizontal bar graph or bar graph horizontal is nothing but a way to represent data horizontally in a graph using bars. In
horizontal bar graphs, we represent data categories on the x-axis whereas data values are on the y-axis. Horizontal bar graphs are generally used to compare different observations. Read the article
below to have detailed information on Bar Graph Horizontal with interesting examples.
What is a Bar Graph?
Let us first know what a bar graph is, before knowing the horizontal bar graph in detail.
A bar graph, also known as a bar chart, is a graphical representation of data using bars of different heights.
Imagine you did a survey of 145 people to know which type of fruits are most liked by people.
On the bar graph, we can show this information as:
Image: Bar graph of fruits
A bar graph is an optimum way to show relative size. For example, in the above bar graph, we can say blueberries are most liked by the people whereas grapes are least liked by the people.
Types of Bar Graph
There are two different types of bar graphs namely:
• Horizontal bar graph and
• Vertical bar graph
In this article, we will discuss horizontal bar graphs in detail.
What Does Horizontal Bar Graph Mean?
A horizontal bar graph or bar graph horizontal is a way to represent data horizontally in a graph using bars. In horizontal bar graphs, data categories are represented on the x-axis whereas data
values are represented on the y-axis. Horizontal bar graphs are widely used for easy and quick comparison among various observations based on certain parameters. In horizontal bar graphs, the length
of the rectangular bars is proportional to their values. Also, all bars in the horizontal bar graph go from left to right.
How to Construct a Horizontal Bar Graph?
Following are the steps to construct a horizontal bar graph:
Step 1: On a graph paper, draw two perpendicular lines intersecting at O.
Step 2: The horizontal line on the bar graph is the x-axis whereas the vertical line on the bar graph is the y-axis.
Step 3: Along a horizontal axis, choose a suitable scale in order to determine the height of the bar for the given values (frequency is represented along the y-axis).
Step 4: Along a vertical axis, choose the uniform width of the bar, and uniform gap between the bars. Also, write the name of the data items whose values are to be marked.
Step 5: Calculate the height of the bar according to the value chosen for each graph and draw the horizontal bars accordingly.
Step 6: Give a suitable title to the graph.
The horizontal bar graph given below gives the information about temperature in celsius on different days of the week.
Image: Horizontal bar graph showing maximum temperature in a week
How to Read a Horizontal Bar Graph?
The horizontal bar graph is ead in the following manner:
• The title of the bar graph tells about the data being represented by the graph.
• The vertical axis of the horizontal bar graph represents the data categories. The data categories in the horizontal bar graph given below are “colours”.
• The horizontal axis of the horizontal bar graph represents the values corresponding to each data value. The data values in the below horizontal bar graph represent the number of students who like
a particular colour represented on the vertical axis.
• The scale is showing the value of 1 unit on the horizontal axis.
Image: Reading horizontal bar graph
From the above horizontal bar graph, we can infer that 1 unit represents 5 students. Accordingly, 10 students like the red colour, 5 students like the yellow colour, 20 students like blue colour, and
15 students like the green colour.
Horizontal Bar Graph Example
Let us understand the horizontal bar graph with an example:
Draw a horizontal bar graph on the basis of the following data.
Image: A bar graph showing subjects and the number of students who passed
The horizontal bar graph gives the following information:
• The total number of students passed in different subjects.
• The maximum number of students passed in Music.
• The minimum number of students passed in English and Maths.
• Students scored highest marks in Music but scored worst in English and Maths.
Hope you have understood what a horizontal bar graph is, and how to read and represent a horizontal bar graph. Now, you can visit Vedantu’s official website and try different questions based on this
to understand the concept better.
FAQs on Bar Graph Horizontal: How to Construct Horizontal Bar?
1. When is Horizontal Bar Graph Mostly Preferred?
The horizontal bar graph is a preferred option when larger quantities of data are involved as the positioning of labels on the vertical axis offers better readability.
2. What values are mostly represented in horizontal bar graphs?
Ensure to create a horizontal bar graph to represent a value that is nominal/categorical. The example of nominal data includes the city of residence, blood type, political party, favourite food, etc.
Also discussed above, horizontal bar graphs are ideal for larger data sets or those with longer labels.
3. What does the title of the horizontal bar graph represent?
The tile of the horizontal bar graph provides precise information on what is there in your graph. This enables readers to know what they are about to look at. The title of the bar graph should be
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Glossary of Scientific Terms - The Elegant Universe: Superstrings, Hidden Dimensions, and the Quest for the Ultimate Theory - Brian Greene
The Elegant Universe: Superstrings, Hidden Dimensions, and the Quest for the Ultimate Theory - Brian Greene (2010)
Glossary of Scientific Terms
Absolute zero. The lowest possible temperature, about -273 degrees Celsius, or 0 on the Kelvin scale.
Acceleration. A change in an object's speed or direction. See also velocity.
Accelerator. See particle accelerator.
Amplitude. The maximum height of a wave peak or the maximum depth of a wave trough.
Anthropic principle. Doctrine that one explanation for why the universe has the properties we observe is that, were the properties different, it is likely that life would not form and therefore we
would not be here to observe the changes.
Antimatter. Matter that has the same gravitational properties as ordinary matter, but that has an opposite electric charge as well as opposite nuclear force charges.
Antiparticle. A particle of antimatter.
ATB. Acronym for "after the bang"; usually used in reference to time elapsed since the big bang.
Atom. Fundamental building block of matter, consisting of a nucleus (comprising protons and neutrons) and an orbiting swarm of electrons.
Big bang. Currently accepted theory that the expanding universe began some 15 billion years ago from a state of enormous energy, density, and compression.
Big crunch. One hypothesized future for the universe in which the current expansion stops, reverses, and results in all space and all matter collapsing together; a reversal of the big bang.
Black hole. An object whose immense gravitational field entraps anything, even light, that gets too close (closer than the black hole's event horizon).
Black-hole entropy The entropy embodied within a black hole.
Boson. A particle, or pattern of string vibration, with a whole number amount of spin; typically a messenger particle.
Bosonic string theory. First known string theory; contains vibrational patterns that are all bosons.
BPS states. Configurations in a supersymmetric theory whose properties can be determined exactly by arguments rooted in symmetry.
Brane. Any of the extended objects that arise in string theory. A one-brane is a string, a two-brane is a membrane, a three-brane has three extended dimensions, etc. More generally, a p-brane has p
spatial dimensions.
Calabi-Yau space, Calabi-Yau shape. A space (shape) into which the extra spatial dimensions required by string theory can be curled up, consistent with the equations of the theory.
Charge. See force charge.
Chiral, Chirality. Feature of fundamental particle physics that distinguishes left-from right-handed, showing that the universe is not fully left-right symmetric.
Closed string. A type of string that is in the shape of a loop.
Conifold transition. Evolution of the Calabi-Yau portion of space in which its fabric rips and repairs itself, yet with mild and acceptable physical consequences in the context of string theory. The
tears involved are more severe than those in a flop transition.
Cosmic microwave background radiation. Microwave radiation suffusing the universe, produced during the big bang and subsequently thinned and cooled as the universe expanded.
Cosmological constant. A modification of general relativity's original equations, allowing for a static universe; interpretable as a constant energy density of the vacuum.
Coupling constant. See string coupling constant.
Curled-up dimension. A spatial dimension that does not have an observably large spatial extent; a spatial dimension that is crumpled, wrapped, or curled up into a tiny size, thereby evading direct
Curvature. The deviation of an object or of space or of spacetime from a flat form and therefore from the rules of geometry codified by Euclid.
Dimension. An independent axis or direction in space or spacetime. The familiar space around us has three dimensions (left-right, back-forth, up-down) and the familiar spacetime has four (the
previous three axes plus the past-future axis). Superstring theory requires the universe to have additional spatial dimensions.
Dual, Duality, Duality symmetries. Situation in which two or more theories appear to be completely different, yet actually give rise to identical physical consequences.
Electromagnetic field. Force field of the electromagnetic force, consisting of electric and magnetic lines of force at each point in space.
Electromagnetic force. One of the four fundamental forces, a union of the electric and magnetic forces.
Electromagnetic gauge symmetry. Gauge symmetry underlying quantum electrodynamics.
Electromagnetic radiation. The energy carried by an electromagnetic wave.
Electromagnetic wave. A wavelike disturbance in an electromagnetic field; all such waves travel at the speed of light. Visible light, X rays, microwaves, and infrared radiation are examples.
Electron. Negatively charged particle, typically found orbiting the nucleus of an atom.
Electroweak theory. Relativistic quantum field theory describing the weak force and the electromagnetic force in one unified framework.
Eleven-dimensional supergravity. Promising higher-dimensional supergravity theory developed in the 1970s, subsequently ignored, and more recently shown to be an important part of string theory.
Entropy. A measure of the disorder of a physical system; the number of rearrangements of the ingredients of a system that leave its overall appearance intact.
Equivalence principle. See principle of equivalence.
Event horizon. The one-way surface of a black hole; once penetrated, the laws of gravity ensure that there is no turning back, no escaping the powerful gravitational grip of the black hole.
Extended dimension. A space (and spacetime) dimension that is large and directly apparent; a dimension with which we are ordinarily familiar, as opposed to a curled-up dimension.
Extremal black holes. Black holes endowed with the maximal amount of force charge possible for a given total mass.
Families. Organization of matter particles into three groups, with each group being known as a family. The particles in each successive family differ from those in the previous by being heavier, but
carry the same electric and nuclear force charges.
Fermion. A particle, or pattern of string vibration, with half a whole odd number amount of spin; typically a matter particle.
Feynman sum-over-paths. See sum-over-paths.
Field, Force field. From a macroscopic perspective, the means by which a force communicates its influence; described by a collection of numbers at each point in space that reflect the strength and
direction of the force at that point.
Flat. Subject to the rules of geometry codified by Euclid; a shape, like the surface of a perfectly smooth tabletop, and its higher-dimensional generalizations.
Flop transition. Evolution of the Calabi-Yau portion of space in which its fabric rips and repairs itself, yet with mild and acceptable physical consequences in the context of string theory.
Foam. See spacetime foam.
Force charge. A property of a particle that determines how it responds to a particular force. For instance, the electric charge of a particle determines how it responds to the electromagnetic force.
Frequency. The number of complete wave cycles a wave completes each second.
Gauge symmetry. Symmetry principle underlying the quantum-mechanical description of the three nongravitational forces; the symmetry involves the invariance of a physical system under various shifts
in the values of force charges, shifts that can change from place to place and from moment to moment.
General relativity. Einstein's formulation of gravity, which shows that space and time communicate the gravitational force through their curvature.
Gluon. Smallest bundle of the strong force field; messenger particle of the strong force.
Grand unification. Class of theories that merge all three nongravitational forces into a single theoretical framework.
Gravitational force. The weakest of the four fundamental forces of nature. Described by Newton's universal theory of gravity, and subsequently by Einstein's general relativity.
Graviton. Smallest bundle of the gravitational force field; messenger particle for the gravitational force.
Heterotic-E string theory (Heterotic E[8] × E[8] string theory). One of the five superstring theories; involves closed strings whose right-moving vibrations resemble those of the Type II string and
whose left-moving vibrations involve those of the bosonic string. Differs in important but subtle ways from the Heterotic-O string theory.
Heterotic-O string theory (Heterotic O(32) string theory). One of the five superstring theories; involves closed strings whose right-moving vibrations resemble those of the Type II string and whose
left-moving vibrations involve those of the bosonic string. Differs in important but subtle ways from the Heterotic-E string theory.
Higher-dimensional supergravity. Class of supergravity theories in more than four spacetime dimensions.
Horizon problem. Cosmological puzzle associated with the fact that regions of the universe that are separated by vast distances nevertheless have nearly identical properties such as temperature.
Inflationary cosmology offers a solution.
Infinities. Typical nonsensical answer emerging from calculations that involve general relativity and quantum mechanics in a point-particle framework.
Inflation, Inflationary cosmology. Modification to the earliest moments of the standard big bang cosmology in which universe undergoes a brief burst of enormous expansion.
Initial conditions. Data describing the beginning state of a physical system.
Interference pattern. Wave pattern that emerges from the overlap and the intermingling of waves emitted from different locations.
Kaluza-Klein theory. Class of theories incorporating extra curled-up dimensions, together with quantum mechanics.
Kelvin. A temperature scale in which temperatures are quoted relative to absolute zero.
Klein-Gordon equation. A fundamental equation of relativistic quantum field theory.
Laplacian determinism. Clockwork conception of the universe in which complete knowledge of the state of the universe at one moment completely determines its state at all future and past moments.
Light clock. A hypothetical clock that measures elapsed time by counting the number of round-trip journeys completed by a single photon between two mirrors.
Lorentz contraction. Feature emerging from special relativity, in which a moving object appears shortened along its direction of motion.
Macroscopic. Refers to scales typically encountered in the everyday world and larger; roughly the opposite of microscopic.
Massless black hole. In string theory, a particular kind of black hole that may have large mass initially, but that becomes ever lighter as a piece of the Calabi-Yau portion of space shrinks. When
the portion of space has shrunk down to a point, the initially massive black hole has no remaining mass—it is massless. In this state, it no longer manifests such usual black hole properties as an
event horizon.
Maxwell's theory, Maxwell's electromagnetic theory. Theory uniting electricity and magnetism, based on the concept of the electromagnetic field, devised by Maxwell in the 1880s; shows that visible
light is an example of an electromagnetic wave.
Messenger particle. Smallest bundle of a force field; microscopic conveyer of a force.
Mirror symmetry. In the context of string theory, a symmetry showing that two different Calabi-Yau shapes, known as a mirror pair, give rise to identical physics when chosen for the curled-up
dimensions of string theory.
M-theory. Theory emerging from the second superstring revolution that unites the previous five superstring theories within a single overarching framework. M-theory appears to be a theory involving
eleven spacetime dimensions, although many of its detailed properties have yet to be understood.
Multidimensional hole. A generalization of the hole found in a doughnut to higher-dimensional versions.
Multi-doughnut, Multi-handled doughnut. A generalization of a doughnut shape (a torus) that has more than one hole.
Multiverse. Hypothetical enlargement of the cosmos in which our universe is but one of an enormous number of separate and distinct universes.
Neutrino. Electrically neutral particle, subject only to the weak force.
Neutron. Electrically neutral particle, typically found in the nucleus of an atom, consisting of three quarks (two down-quarks, one up-quark).
Newton's laws of motion. Laws describing the motion of bodies based on the conception of an absolute and immutable space and time; these laws held sway until Einstein's discovery of special
Newton's universal theory of gravity. Theory of gravity declaring that the force of attraction between two bodies is proportional to the product of their masses and inversely proportional to the
square of the distance between them. Subsequently supplanted by Einstein's general relativity.
Nonperturbative. Feature of a theory whose validity is not dependent on approximate, perturbative calculations; an exact feature of a theory.
Nucleus. The core of an atom, consisting of protons and neutrons.
Observer. Idealized person or piece of equipment, often hypothetical, that measures relevant properties of a physical system.
One-loop process. Contribution to a calculation in perturbation theory in which one virtual pair of strings (or particles in a point-particle theory) is involved.
Open string. A type of string with two free ends.
Oscillatory pattern. See vibrational pattern.
Particle accelerator. Machine for boosting particles to nearly light speed and slamming them together in order to probe the structure of matter.
Perturbation theory. Framework for simplifying a difficult problem by finding an approximate solution that is subsequently refined as more details, initially ignored, are systematically included.
Perturbative approach, Perturbative method. See perturbation theory.
Phase. When used in reference to matter, describes its possible states: solid phase, liquid phase, gas phase. More generally, refers to the possible descriptions of a physical system as features on
which it depends (temperature, string coupling constant values, form of spacetime, etc.) are varied.
Phase transition. Evolution of a physical system from one phase to another.
Photoelectric effect. Phenomenon in which electrons are ejected from a metallic surface when light is shone upon it.
Photon. Smallest packet of the electromagnetic force field; messenger particle of the electromagnetic force; smallest bundle of light.
Planck energy. About 1,000 kilowatt hours. The energy necessary to probe to distances as small as the Planck length. The typical energy of a vibrating string in string theory.
Planck length. About 10^-33 centimeters. The scale below which quantum fluctuations in the fabric of spacetime would become enormous. The size of a typical string in string theory.
Planck mass. About ten billion billion times the mass of a proton; about one-hundredth of a thousandth of a gram; about the mass of a small grain of dust. The typical mass equivalent of a vibrating
string in string theory.
Planck's constant. Denoted by the symbol h, Planck's constant is a fundamental parameter in quantum mechanics. It determines the size of the discrete units of energy, mass, spin, etc. into which the
microscopic world is partitioned. Its value is 1.05 × 10^-27 grams-cm/sec.
Planck tension. About 10^39 tons. The tension on a typical string in string theory.
Planck time. About 10^-43 seconds. Time at which the size of the universe was roughly the Planck length; more precisely, time it takes light to travel the Planck length.
Primordial nucleosynthesis. Production of atomic nuclei occurring during the first three minutes after the big bang.
Principle of equivalence. Core principle of general relativity declaring the indistinguishability of accelerated motion and immersion in a gravitational field (over small enough regions of
observation). Generalizes the principle of relativity by showing that all observers, regardless of their state of motion, can claim to be at rest, so long as they acknowledge the presence of a
suitable gravitational field.
Principle of relativity. Core principle of special relativity declaring that all constant-velocity observers are subject to an identical set of physical laws and that, therefore, every
constant-velocity observer is justified in claiming that he or she is at rest. This principle is generalized by the principle of equivalence.
Product. The result of multiplying two numbers.
Proton. Positively charged particle, typically found in the nucleus of an atom, consisting of three quarks (two up-quarks and one down-quark).
Quanta. The smallest physical units into which something can be partitioned, according to the laws of quantum mechanics. For instance, photons are the quanta of the electromagnetic field.
Quantum chromodynamics (QCD). Relativistic quantum field theory of the strong force and quarks, incorporating special relativity.
Quantum claustrophobia. See quantum fluctuations.
Quantum determinism. Property of quantum mechanics that knowledge of the quantum state of a system at one moment completely determines its quantum state at future and past moments. Knowledge of the
quantum state, however, determines only the probability that one or another future will actually ensue.
Quantum electrodynamics (QED). Relativistic quantum field theory of the electromagnetic force and electrons, incorporating special relativity.
Quantum electroweak theory. See electroweak theory.
Quantum field theory. See relativistic quantum field theory.
Quantum fluctuation. Turbulent behavior of a system on microscopic scales due to the uncertainty principle.
Quantum foam. See spacetime foam.
Quantum geometry. Modification of Riemannian geometry required to describe accurately the physics of space on ultramicroscopic scales, where quantum effects become important.
Quantum gravity. A theory that successfully mergers quantum mechanics and general relativity, possibly involving modifications of one or both. String theory is an example of a theory of quantum
Quantum mechanics. Framework of laws governing the universe whose unfamiliar features such as uncertainty, quantum fluctuations, and wave-particle duality become most apparent on the microscopic
scales of atoms and subnuclear particles.
Quantum tunneling. Feature of quantum mechanics showing that objects can pass through barriers that should be impenetrable according to Newton's classical laws of physics.
Quark. A particle that is acted upon by the strong force. Quarks exist in six varieties (up, down, charm, strange, top, bottom) and three "colors" (red, green, blue).
Radiation. The energy carried by waves or particles.
Reciprocal. The inverse of a number; for example, the reciprocal of 3 is 1/3, the reciprocal of 1/2 is 2.
Relativistic quantum field theory. Quantum-mechanical theory of fields, such as the electromagnetic field, that incorporates special relativity.
Resonance. One of the natural states of oscillation of a physical system.
Riemannian geometry. Mathematical framework for describing curved shapes of any dimension. Plays a central role in Einstein's description of spacetime in general relativity.
Schrödinger equation. Equation governing the evolution of probability waves in quantum mechanics.
Schwarzschild solution. Solution to the equations of general relativity for a spherical distribution of matter; one implication of this solution is the possible existence of black holes.
Second law of thermodynamics. Law stating that total entropy always increases.
Second superstring revolution. Period in the development of string theory beginning around 1995 in which some nonperturbative aspects of the theory began to be understood.
Singularity. Location where the fabric of space or spacetime suffers a devastating rupture.
Smooth, Smooth space. A spatial region in which the fabric of space is flat or gently curved, with no pinches, ruptures, or creases of any kind.
Space-tearing flop transition. See flop transition.
Spacetime. A union of space and time originally emerging from special relativity. Can be viewed as the "fabric" out of which the universe is fashioned; it constitutes the dynamical arena within which
the events of the universe take place.
Spacetime foam. Frothy, writhing, tumultuous character of the spacetime fabric on ultramicroscopic scales, according to a conventional point-particle perspective. An essential reason for the
incompatibility of quantum mechanics and general relativity prior to string theory.
Special relativity. Einstein's laws of space and time in the absence of gravity (see also general relativity).
Sphere. The outer surface of a ball. The surface of a familiar three-dimensional ball has two dimensions (which can be labeled by two numbers such as "latitude" and "longitude," as on the surface of
the earth). The concept of a sphere, though, applies more generally to balls and hence their surfaces, in any number of dimensions. A one-dimensional sphere is a fancy name for a circle; a
zero-dimensional sphere is two points (as explained in the text). A three-dimensional sphere is harder to picture; it is the surface of a four-dimensional ball.
Spin. A quantum-mechanical version of the familiar notion of the same name; particles have an intrinsic amount of spin that is either a whole number or half a whole number (in multiples of Planck's
constant), and which never changes.
Standard model of cosmology. Big bang theory together with an understanding of the three nongravitational forces as summarized by the standard model of particle physics.
Standard model of particle physics, Standard model, Standard theory. An enormously successful theory of the three nongravitational forces and their action on matter. Effectively the union of quantum
chromodynamics and the electroweak theory.
String. Fundamental one-dimensional object that is the essential ingredient in string theory.
String coupling constant. A (positive) number that governs how likely it is for a given string to split apart into two strings or for two strings to join together into one—the basic processes in
string theory. Each string theory has its own string coupling constant, the value of which should be determined by an equation; currently such equations are not understood well enough to yield any
useful information. Coupling constants less than 1 imply that perturbative methods are valid.
String mode. A possible configuration (vibrational pattern, winding configuration) that a string can assume.
String theory. Unified theory of the universe postulating that fundamental ingredients of nature are not zero-dimensional point particles but tiny one-dimensional filaments called strings. String
theory harmoniously unites quantum mechanics and general relativity, the previously known laws of the small and the large, that are otherwise incompatible. Often short for superstring theory.
Strong force, Strong nuclear force. Strongest of the four fundamental forces, responsible for keeping quarks locked inside protons and neutrons and for keeping protons and neutrons crammed inside of
atomic nuclei.
Strong force symmetry. Gauge symmetry underlying the strong force, associated with invariance of a physical system under shifts in the color charges of quarks.
Strongly coupled. Theory whose string coupling constant is larger than 1.
Strong-weak duality. Situation in which a strongly coupled theory is dual—physically identical—to a different, weakly coupled theory.
Sum-over-paths. Formulation of quantum mechanics in which particles are envisioned to travel from one point to another along all possible paths between them.
Supergravity. Class of point-particle theories combining general relativity and supersymmetry.
Superpartners. Particles whose spins differ by 1/2 unit and that are paired by supersymmetry.
Superstring theory. String theory that incorporates supersymmetry.
Supersymmetric quantum field theory. Quantum field theory incorporating supersymmetry.
Supersymmetric standard model. Generalization of the standard model of particle physics to incorporate supersymmetry. Entails a doubling of the known elementary particle species.
Supersymmetry. A symmetry principle that relates the properties of particles with a whole number amount of spin (bosons) to those with half a whole (odd) number amount of spin (fermions).
Symmetry. A property of a physical system that does not change when the system is transformed in some manner. For instance, a sphere is rotationally symmetrical since its appearance does not change
if it is rotated.
Symmetry breaking. A reduction in the amount of symmetry a system appears to have, usually associated with a phase transition.
Tachyon. Particle whose mass (squared) is negative; its presence in a theory generally yields inconsistencies.
Thermodynamics. Laws developed in the nineteenth century to describe aspects of heat, work, energy, entropy, and their mutual evolution in a physical system.
Three-brane. See brane.
Three-dimensional sphere. See sphere.
Time dilation. Feature emerging from special relativity, in which the flow of time slows down for an observer in motion.
T.O.E. (Theory of Everything). A quantum-mechanical theory that encompasses all forces and all matter.
Topologically distinct. Two shapes that cannot be deformed into one another without tearing their structure in some manner.
Topology. Classification of shapes into groups that can be deformed into one another without ripping or tearing their structure in any way.
Topology-changing transition. Evolution of spatial fabric that involves rips or tears, thereby changing the topology of space.
Torus. The two-dimensional surface of a doughnut.
Two-brane. See brane.
Two-dimensional sphere. See sphere.
Type I string theory. One of the five superstring theories; involves both open and closed strings.
Type IIA string theory. One of the five superstring theories; involves closed strings with left-right symmetric vibrational patterns.
Type IIB string theory. One of the five superstring theories; involves closed strings with left-right asymmetric vibrational patterns.
Ultramicroscopic. Length scales shorter than the Planck length (and also time scales shorter than the Planck time).
Uncertainty principle. Principle of quantum mechanics, discovered by Heisenberg, that there are features of the universe, like the position and velocity of a particle, that cannot be known with
complete precision. Such uncertain aspects of the microscopic world become ever more severe as the distance and time scales on which they are considered become ever smaller. Particles and fields
undulate and jump between all possible values consistent with the quantum uncertainty. This implies that the microscopic realm is a roiling frenzy, awash in a violent sea of quantum fluctuations.
Unified theory, Unified field theory. Any theory that describes all four forces and all of matter within a single, all-encompassing framework.
Uniform vibration. The overall motion of a string in which it moves without changes in shape.
Velocity. The speed and the direction of an object's motion.
Vibrational mode. See vibrational pattern.
Vibrational pattern. The precise number of peaks and troughs as well as their amplitude as a string oscillates.
Vibration number. Whole number describing the energy in the uniform vibrational motion of a string; the energy in its overall motion as opposed to that associated with changes in its shape.
Virtual particles. Particles that erupt from the vacuum momentarily; they exist on borrowed energy, consistent with the uncertainty principle, and rapidly annihilate, thereby repaying the energy
Wave function. Probability waves upon which quantum mechanics is founded.
Wavelength. The distance between successive peaks or troughs of a wave.
Wave-particle duality. Basic feature of quantum mechanics that objects manifest both wavelike and particle-like properties.
W bosons. See weak gauge boson.
Weak force, Weak nuclear force. One of the four fundamental forces, best known for mediating radioactive decay.
Weak gauge boson. Smallest bundle of the weak force field; messenger particle of the weak force; called W or Z boson.
Weak gauge symmetry. Gauge symmetry underlying the weak force.
Weakly coupled. Theory whose string coupling constant is less than 1.
Winding energy. The energy embodied by a string wound around a circular dimension of space.
Winding mode. A string configuration that wraps around a circular spatial dimension.
Winding number. The number of times a string is wound around a circular spatial dimension.
World-sheet. Two-dimensional surface swept out by a string as it moves.
Wormhole. A tube-like region of space connecting one region of the universe to another.
Z boson. See weak gauge boson.
Zero-dimensional sphere. See sphere. | {"url":"https://publicism.info/science/elegant/18.html","timestamp":"2024-11-04T01:02:50Z","content_type":"text/html","content_length":"46237","record_id":"<urn:uuid:be1522b4-e444-47a0-9073-dfe0e2c99cb8>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00817.warc.gz"} |
Probability (Statistics) Class X PDF | Quizgecko
Probability (Statistics) Class X PDF
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Uploaded by LightHeartedSard3046
Ryan International School
probability statistics mathematics class X
This document contains probability and statistics questions for class X. The questions cover various topics in probability, including calculating probabilities of events involving coins, dice, and
Full Transcript
Assignment Probability (Statistics) Class X Q1. A coin is tossed. Find the probability that a head is obtained. Q2. Find probability of throwing 5 with an ordinary dice. Q3. Probability of winning a
game is 0.4. What is the probability of loosin...
Assignment Probability (Statistics) Class X Q1. A coin is tossed. Find the probability that a head is obtained. Q2. Find probability of throwing 5 with an ordinary dice. Q3. Probability of winning a
game is 0.4. What is the probability of loosing the game? Q4. A person is known to hit the target in 3 shots out of 4 shots. Find the probability that the target is not hit. Q5. Tickets numbered from
1 to 20 are mixed together and a ticket is drawn at random. What is the probability that the ticket has a number which is multiple of 3 or 7? Q6. A bag contains 100 identical tokens, on which numbers
1 to 100 are marked. A token is drawn at random. What is the probability that the number on the token is: (a) an even number (b) an odd number (c) a multiple of 3 (d) a multiple of 5 (f) a multiple
of 3 and 5 (g) a multiple of 3 or 5 (h) a number less than 20 (i) a number greater than 70 (j) a perfect square number (k) a prime number less than 20. Q7. A card is drawn from a well-shuffled pack
of cards. Find the probability that the card drawn is: (a) a queen (b) a king bearing diamond sign (c) a black card (d) a jack (e) black and a queen (f) either black or a queen (g) a red card (h) a
face card (i) a diamond or a club (j) neither heart nor a jack (k) a 2 of diamond (l) an ace of hearts (m) a face card of red color (n) 10 of a black “suit” Q8. In a simultaneous toss of two coins,
find: (a) P(2 tails) (b) P(exactly one tail) (c) P(no tails) (d) P(at most one head) (e) P(one head) Q9. A coin is tossed successively three times. Find probability of getting exactly one head or two
heads. Q10. Three coins are tossed once. Find probability of: (a) 3 heads (b) exactly 2 heads (c) atleast 2 heads (d) atmost 2 heads (e) no tails (f) head and tail appear alternatively (g) atleast
one head and one tail Q11. A dice is thrown once. Find: (a) P(number 5) (b) P(number 7) (c) P(an even number) (d) P( a number greater than 4) (e) P( a number less than or equal to 4) (f) P(a prime
number) Q12. A bag contains 10 white, 6 black and 4 red balls. Find probability of getting: (a) a white ball (b) a black ball (c) not a red ball (d) a white or a red ball Q13. Two dice are thrown
simultaneously. Find: (a) P(an odd number as a sum) (b) P(sum as a prime number) (c) P(a doublet of odd numbers) (d) P(a total of atleast 9) (e) P( a multiple of 2 on one die and a multiple of 3 on
other die) (f) P(a doublet) (g) P(a multiple of 2 as sum) (h) P(getting the sum 9) (i) P(getting a sum greater than 12) (j) P( a prime number on each die) (k) P( a multiple of 5 as a sum) Q14. Find
the probability that a leap year at random contains 53 Sundays. Q15. Two black kings and two black jacks are removed from a pack of 52 cards. Find the probability of getting: (a) a card of hearts (b)
a black card (c) either a red card or a king (d) a red king (e) neither an ace nor a king (f) a jack, queen or a king *NOTE: A pack of playing cards consists of 52 cards, which are divided into 4
suits of 13 cards each. Each suit consists of one ace, one king, one queen, one jack and 9 other cards numbered from 2 to 10. Four suits are named as spades(),clubs(), hearts() and diamonds().
(spades & clubs are black. hearts & diamonds are red) ANSWERS Ans(1) 1/2 Ans(2) 1/6 Ans(3) 0.6 Ans(4) 1/4 Ans(5) 2/5 Ans(6) (a) 1/2 (b) 1/2 (c) 33/100 (d) 1/5 (e) 3/50 (f) 47/100 (g) 19/100 (h) 3/10
Ans(7) (a) 1/13 (b) 1/52 (c) 1/2 (d) 1/13 (e) 1/26 (f) 7/13 (g) 1/2 (h) 4/13 (i) 1/2 (j) 9/13 (k) 1/52 (l) 1/52 (m) 3/26 (n) 1/26 Ans(8) (a) 1/4 (b) 1/2 (c) 1/4 (d) 3/5 (e) 1/2 Ans(9) 3/4 Ans(10) (a)
1/8 (b) 3/8 (c) 1/2 (d) 7/8 (e) 1/8 (f) 1/4 (g) 3/4 Ans(11) (a) 1/6 (b) 0 (c) 1/2 (d) 1/3 (e) 2/3 (f) 1/2 Ans(12) (a) 1/2 (b) 3/10 (c) 4/5 (d) 7/10 Ans(13) (a) 1/2 (b) 5/12 (c) 1/12 (d) 5/18 (e) 11/
36 (f) 1/6 (g) 1/2 (h) 1/9 (i) 0 (j) 1/12 (k) 7/36 Ans(14) 2/7 Ans(15) (a) 13/48 (b) 11/24 (c) 13/24 (d) 1/24 (e) 7/8 (f) 1/6 | {"url":"https://quizgecko.com/uploads/ch15-probability-mainpdf-YlBjJT","timestamp":"2024-11-11T07:50:01Z","content_type":"text/html","content_length":"161038","record_id":"<urn:uuid:99c4f0fe-8737-4496-9d89-274f84db2c34>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00232.warc.gz"} |
Placement Papers
Technicalsymposium.com - New Updates Alerts-Subscribe
Download Free Placement Materials with Answers
1. A man engaged a servant on a condn that he’ll pay Rs 40 and also give him a bag at the end of the yr. He served for 9 months and was given a turban and Rs 55. So the price of turban is
i. Rs 10 / 29 / 0 / none
2.How many 4 digit no. can b formed wit digits 1, 2, 3,4,5 which r divisible by 4 and digits not repeated
144 / 168 / 182 / none
3.. If 1= (3/4)(1+ (y/x) ) then
i. x=3y
ii. x=y/3
iii. x=(2/3)y
iv. none
4.There is a rectangular Garden whose length and width are 60m X 20m.There is a walkway of uniform width around garden. Area of walkway is 516m^2. Find width of walkway
5. In a race from pt. X to pt Y and back, Jack averages 0 miles/hr to pt Y and 10 miles/hr back to pr X.Sandy averages 20 miles/hr in both directions. If Jack and Sandy start race at same tym, who’ll
finish 1st
Jack/Sandy/they tie/Impossible to tell
6 Fresh Grapes contain 90% water by wt. Dried grapes contain 20% water by %age. What will b wt of dried grapes when we begin with 20 kg fresh grapes?
2kg / 2.4kg / 2.5kg /none
7. Three wheels make 36, 24, 60 rev/min. Each has a black mark on it. It is aligned at the start of the qn. When does it align again for the first tym?
14/20/22/5 sec
8. Asish was given Rs. 158 in denominations of Rs 1 each. He distributes these in diff bags, such that ne sum of money of denomination betn 1 and 158 can be given in bags. The min no. of such bags
10 / 17 / 15 / none
9. The sum of six consecutive odd nos. is 888. What is the average of the nos.?
i. 147
ii. 148
iii. 149
iv. 146
10. 1010/104*102=10?
i. 8
ii. 6
iii. 4
iv. None
11). A is 4 yrs old and B is thrice A>when A is 14 yrs, how old will B be?
26 28 24 none
12) Find min value of fn:
|-6-x| + |4-x|+|5-x|+10-x|; where x is an integer
10 /17 /23 /none
13) units digit in expansion os 4 raised to 51 is:
2 /4 /6 /8
14) 2 men at same tym start walking towards each other from A n B 72 kms apart. sp of A is 4kmph.Sp of B is 2 kmph in 1st hr,2.5 in 2nd, 3 in rd. n so on…when will they meet
i in 7 hrs ii at 35 kms from A iii in 10 hrs iv midway
15) (9*76+10*?-60) / (?*5*12+3-52)=1
7 9 3 none
16) 45 grinders brought @ 2215/-.transpot expense 2190/-.2760/-on octroi . Find SP/piece to make profit of 20%
17) in a 2 digit no unit’s place is halved and tens place is doubled.diff bet the nos is 37.digit in unit’s place is 2 more than tens place.
24 46 42 none
18) if x-y + z = 29 , y + z =30 , x-z=3 , find d value of x+4y-5z
22 38 17 none
19) Find approx value of 59.987/0.2102+1.187*18.02
52 16 86 none
20) If the ratio of prod of 3 diff comp’s A B & C is 4:7:5 and of overall prod last yr was 4lac tones and if each comp had an increase of 20% in prod level this yr what is the prod of Comp B this yr?
2.1L 22.1L 4.1L none
21). If 70% of a no. is subtracted from itself it reduces to 81.what is two fifth of that no.?
108 54 210 none
22). If a certain sum of money at SI doubles itself in 5 yrs then what is d rate?
5% 20% 25% 14.8%
23). If radius of cylinder and sphere r same and vol of sphere and cylinder r same what is d ratio betn the radius and height of the cylinder
i. R= H ii. R= (3/4)H iii. R = (4/3)H iv. R=2/3H
one question was from conversion of hectare to kilametreReasonings were like this.These qns are based on situations given below:
7 Uni crick players are to be honored at a special luncheon. The players will be seated on a dais along one side of a single rectangular table.
A and G have to leave the luncheon early and must be seated at the extreme right end of table, which is closest to exit.
B will receive Man of the Match and must be in the centre chair
C and D who are bitter rivals for the position of Wicket keeper dislike one another and should be seated as far apart as possible
E and F are best friends and want to seat together.
13.Which of the foll may not be seated at both end of the table?
i. C &D
ii. D&F
iii. C&G
iv. C&F
14.Which of the foll pairs may be seated together?
i. E & A
ii. B & D
iii. C & F
iv. NONE
An employee has to allocate offices to 6 staff members. The offices are no. 1-6. the offices are arranged in a row and they are separated from each other by dividers>hence voices, sounds and
cigarette smoke flow easily from one office to another
Miss R needs to use the telephone quite often throughout the day. Mr. M and Mr. B need adjacent offices as they need to consult each other often while working. Miss H is a senior employee and his to
be allotted the office no. 5, having the biggest window.
Mr D requires silence in office next to his. Mr. T, Mr M and Mr. D are all smokers. Miss H finds tobacco smoke allergic and consecutively the offices next to hers are occupied by non-smokers. Unless
specifically stated all the employees maintain an atmosphere of silence during office hrs.
15.The ideal candidate to occupy office farthest from Mr. B will be
i. Miss H
ii. Mr. M
iii. Mr. T
iv. Mr. D
16.The three employees who are smokers should be seated in the offices
i. 1 2 4
ii. 2 3 6
iii. 1 2 3
iv. 1 2 3
17.The ideal office for Mr. M would be
i. 2
ii. 6
iii. 1
iv. 3
A robot moves on a graph sheet with x-y axes. The robot is moved by feeding it with a sequence of instructions. The different instructions that can be used in moving it, and their meanings are:
Instruction Meaning
GOTO(x,y) move to pt with co-ord (x,y) no matter where u are currently
WALKX(P) move parallel to x-axis through a distance of p, in the +ve direction if p is +ve and in –ve if p is –ve
WALKY(P) move parallel to y-axis through a distance of p, in the +ve direction if p is +ve and in –ve if p is –ve
19.The robot reaches point (5,6) when a sequence of 3 instr. Is executed, the first of which is GOTO(x,y) , WALKY(2), WALKY(4). What are the values of x and y??
i. 2,4
ii. 0,0
iii. 3,2
iv. 2,3
20. The robot is initially at (x.y), x>0 and y<0. The min no. of Instructions needed to be executed to bring it to origin (0,0) if you are prohibited from using GOTO instr. Is:
i. 2
ii. 1
iii. x + y
iv. 0
Ten coins are distr. Among 4 people P, Q, R, S such that one of them gets a coin, another gets 2 coins,3rd gets 3 coins, and 4th gets 4 coins. It is known that Q gets more coins than P, and S gets
fewer coins than R
21.If the no. of coins distr. To Q is twice the no. distr. to P then which one of the foll. is necessarily true?
i. R gets even no. of coins
ii. R gets odd no. of coins
iii. S gets even no. of coins
iv. S gets odd no. of coins
22. If R gets at least two more coins than S which one of the foll is necessarily true?
i. Q gets at least 2 more coins than S
ii. Q gets more coins than P
iii. P gets more coins than S
iv. P and Q together get at least five coins
23.If Q gets fewer coins than R, then which one of the foll. is not necessarily true?
i. P and Q together get at least 4 coins
ii. Q and S together get at least 4 coins
iii. R and S together get at least 5 coins
iv. P and R together get at least 5 coins
Elle is 3 times older than Zaheer. Zaheer is ½ as old as Waheeda. Yogesh is elder than Zaheer.
24.What is sufficient to estimate Elle’s age?
i. Zaheer is 10 yrs old
ii. Yogesh and Waheeda are both older than Zaheer by the same no of yrs.
iii. Both of the above
iv. None of the above
25. Which one of the foll. statements can be inferred from the info above
i. Yogesh is elder than Waheeda
ii. Elle is older than Waheeda
iii. Elle’s age may be less than that of Waheeda
iv. None of the above
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Basic Operations on Stacks | CodingDrills
Basic Operations on Stacks
Introduction to Stacks
A stack is a linear data structure that follows the Last-In-First-Out (LIFO) principle. It can be visualized as a stack of plates, where the last plate placed is the first one to be removed. Stacks
are widely used in programming and have various applications, such as function call stacks, undo/redo operations, and expression evaluation.
Basic Operations on Stacks
Push Operation
The push operation adds an element to the top of the stack. It requires two things: the stack itself and the element to be pushed. Let's see how this operation can be implemented in Python:
class Stack:
def __init__(self):
self.stack = []
def push(self, element):
In the above code snippet, we define a Stack class with an empty list as the underlying data structure. The push method appends the given element to the end of the list, effectively adding it to the
top of the stack.
Pop Operation
The pop operation removes and returns the topmost element from the stack. It requires only the stack itself. Here's the implementation of the pop operation in Python:
class Stack:
# previous code...
def pop(self):
if not self.is_empty():
return self.stack.pop()
raise Exception("Stack is empty!")
In the above code snippet, we add the pop method to the Stack class. It checks if the stack is empty using the is_empty method (which we will define shortly). If the stack is not empty, it uses the
pop method of the list to remove and return the topmost element. Otherwise, it raises an exception indicating that the stack is empty.
Peek Operation
The peek operation returns the topmost element from the stack without removing it. It is useful when you want to access the top element without modifying the stack. Here's the implementation of the
peek operation:
class Stack:
# previous code...
def peek(self):
if not self.is_empty():
return self.stack[-1]
raise Exception("Stack is empty!")
In the above code snippet, we define the peek method, which returns the last element of the list (i.e., the topmost element of the stack) if the stack is not empty. Otherwise, it raises an exception.
Is Empty Operation
The is empty operation checks whether the stack is empty or not. It returns a boolean value indicating the result. Here's the implementation:
class Stack:
# previous code...
def is_empty(self):
return len(self.stack) == 0
In the above code snippet, we define the is_empty method, which checks if the length of the list (i.e., the number of elements in the stack) is zero. If it is zero, the stack is empty, and the method
returns True; otherwise, it returns False.
In this tutorial, we covered the basic operations on stacks, including push, pop, peek, and is empty. Stacks are fundamental data structures in programming, and understanding their operations is
crucial for efficient problem-solving. By implementing these operations, you can utilize stacks in various applications and optimize your code.
Remember to practice and experiment with stacks to solidify your understanding. Happy coding!
Ada AI
Hi, I'm Ada, your personal AI tutor. I can help you with any coding tutorial. Go ahead and ask me anything.
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