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If you add several normally distributed random numbers, the result is normally distributed, where the mean of the sum is the sum of the individual means, and the variance of the sum is the sum of the individual variances. (Remember that variance is the square of standard deviation.) This is a difficult result to prove mathematically, but it is easy to demonstrate with simulation. To do so, run a simulation where you add three normally distributed random numbers, each with mean 100 and standard deviation 10. Your single output variable should be the sum of these three numbers. Verify with @RISK that the distribution of this output is approximately normal with mean 300 and variance 300 (hence, standard deviation
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Chapter 10 Solutions
Practical Management Science
- When you use @RISKs correlation feature to generate correlated random numbers, how can you verify that they are correlated? Try the following. Use the RISKCORRMAT function to generate two normally distributed random numbers, each with mean 100 and standard deviation 10, and with correlation 0.7. To run a simulation, you need an output variable, so sum these two numbers and designate the sum as an output variable. Run the simulation with 1000 iterations and then click the Browse Results button to view the histogram of the output or either of the inputs. Then click the Scatterplot button below the histogram and choose another variable (an input or the output) for the scatterplot. Using this method, are the two inputs correlated as expected? Are the two inputs correlated with the output? If so, how?arrow_forwardIn Example 11.2, the gamma distribution was used to model the skewness to the right of the lifetime distribution. Experiment to see whether the triangular distribution could have been used instead. Let its minimum value be 0, and choose its most likely and maximum values so that this triangular distribution has approximately the same mean and standard deviation as the gamma distribution in the example. (Use @RISKs Define Distributions window and trial and error to do this.) Then run the simulation and comment on similarities or differences between your outputs and the outputs in the example.arrow_forwardSimulation can be used to illustrate a number of results from statistics that are difficult to understand with nonsimulation arguments. One is the famous central limit theorem, which says that if you sample enough values from any population distribution and then average these values, the resulting average will be approximately normally distributed. Confirm this by using @ RISK with the following population distributions (run a separate simulation for each): (a) discrete with possible values 1 and 2 and probabilities 0.2 and 0.8; (b) exponential with mean 1 (use the RISKEXPON function with the single argument 1); (c) triangular with minimum, most likely, and maximum values equal to 1,9, and 10. (Note that each of these distributions is very skewed.) Run each simulation with 10 values in each average, and run 1000 iterations to simulate 1000 averages. Create a histogram of the averages to see whether it is indeed bell-shaped. Then repeat, using 30 values in each average. Are the histograms based on 10 values qualitatively different from those, based on 30?arrow_forward
- Suppose you simulate a gambling situation where you place many bets. On each bet, the distribution of your net winnings (loss if negative) is highly skewed to the left because there are some possibilities of really large losses but not much upside potential. Your only simulation output is the average of the results of all the bets. If you run @RISK with many iterations and look at the resulting histogram of this output, what will it look like? Why?arrow_forwardAssume that all of a companys job applicants must take a test, and that the scores on this test are normally distributed. The selection ratio is the cutoff point used by the company in its hiring process. For example, a selection ratio of 25% means that the company will accept applicants for jobs who rank in the top 25% of all applicants. If the company chooses a selection ratio of 25%, the average test score of those selected will be 1.27 standard deviations above average. Use simulation to verify this fact, proceeding as follows. a. Show that if the company wants to accept only the top 25% of all applicants, it should accept applicants whose test scores are at least 0.674 standard deviation above average. (No simulation is required here. Just use the appropriate Excel normal function.) b. Now generate 1000 test scores from a normal distribution with mean 0 and standard deviation 1. The average test score of those selected is the average of the scores that are at least 0.674. To determine this, use Excels DAVERAGE function. To do so, put the heading Score in cell A3, generate the 1000 test scores in the range A4:A1003, and name the range A3:A1003 Data. In cells C3 and C4, enter the labels Score and 0.674. (The range C3:C4 is called the criterion range.) Then calculate the average of all applicants who will be hired by entering the formula =DAVERAGE(Data, "Score", C3:C4) in any cell. This average should be close to the theoretical average, 1.27. This formula works as follows. Excel finds all observations in the Data range that satisfy the criterion described in the range C3:C4 (Score0.674). Then it averages the values in the Score column (the second argument of DAVERAGE) corresponding to these entries. See online help for more about Excels database D functions. c. What information would the company need to determine an optimal selection ratio? How could it determine the optimal selection ratio?arrow_forwardSuppose you have invested 25% of your portfolio in four different stocks. The mean and standard deviation of the annual return on each stock are shown in the file P11_46.xlsx. The correlations between the annual returns on the four stocks are also shown in this file. a. What is the probability that your portfolios annual return will exceed 30%? b. What is the probability that your portfolio will lose money during the year?arrow_forward
- The annual demand for Prizdol, a prescription drug manufactured and marketed by the NuFeel Company, is normally distributed with mean 50,000 and standard deviation 12,000. Assume that demand during each of the next 10 years is an independent random number from this distribution. NuFeel needs to determine how large a Prizdol plant to build to maximize its expected profit over the next 10 years. If the company builds a plant that can produce x units of Prizdol per year, it will cost 16 for each of these x units. NuFeel will produce only the amount demanded each year, and each unit of Prizdol produced will sell for 3.70. Each unit of Prizdol produced incurs a variable production cost of 0.20. It costs 0.40 per year to operate a unit of capacity. a. Among the capacity levels of 30,000, 35,000, 40,000, 45,000, 50,000, 55,000, and 60,000 units per year, which level maximizes expected profit? Use simulation to answer this question. b. Using the capacity from your answer to part a, NuFeel can be 95% certain that actual profit for the 10-year period will be between what two values?arrow_forwardIf you want to replicate the results of a simulation model with Excel functions only, not @RISK, you can build a data table and let the column input cell be any blank cell. Explain why this works.arrow_forwardUse @RISK to draw a binomial distribution that results from 50 trials with probability of success 0.3 on each trial, and use it to answer the following questions. a. What are the mean and standard deviation of this distribution? b. You have to be more careful in interpreting @RISK probabilities with a discrete distribution such as this binomial. For example, if you move the left slider to 11, you find a probability of 0.139 to the left of it. But is this the probability of less than 11 or less than or equal to 11? One way to check is to use Excels BINOM.DIST function. Use this function to interpret the 0.139 value from @RISK. c. Using part b to guide you, use @RISK to find the probability that a random number from this distribution will be greater than 17. Check your answer by using the BINOM.DIST function appropriately in Excel.arrow_forward
- You are considering a 10-year investment project. At present, the expected cash flow each year is 10,000. Suppose, however, that each years cash flow is normally distributed with mean equal to last years actual cash flow and standard deviation 1000. For example, suppose that the actual cash flow in year 1 is 12,000. Then year 2 cash flow is normal with mean 12,000 and standard deviation 1000. Also, at the end of year 1, your best guess is that each later years expected cash flow will be 12,000. a. Estimate the mean and standard deviation of the NPV of this project. Assume that cash flows are discounted at a rate of 10% per year. b. Now assume that the project has an abandonment option. At the end of each year you can abandon the project for the value given in the file P11_60.xlsx. For example, suppose that year 1 cash flow is 4000. Then at the end of year 1, you expect cash flow for each remaining year to be 4000. This has an NPV of less than 62,000, so you should abandon the project and collect 62,000 at the end of year 1. Estimate the mean and standard deviation of the project with the abandonment option. How much would you pay for the abandonment option? (Hint: You can abandon a project at most once. So in year 5, for example, you abandon only if the sum of future expected NPVs is less than the year 5 abandonment value and the project has not yet been abandoned. Also, once you abandon the project, the actual cash flows for future years are zero. So in this case the future cash flows after abandonment should be zero in your model.)arrow_forwardBased on Babich (1992). Suppose that each week each of 300 families buys a gallon of orange juice from company A, B, or C. Let pA denote the probability that a gallon produced by company A is of unsatisfactory quality, and define pB and pC similarly for companies B and C. If the last gallon of juice purchased by a family is satisfactory, the next week they will purchase a gallon of juice from the same company. If the last gallon of juice purchased by a family is not satisfactory, the family will purchase a gallon from a competitor. Consider a week in which A families have purchased juice A, B families have purchased juice B, and C families have purchased juice C. Assume that families that switch brands during a period are allocated to the remaining brands in a manner that is proportional to the current market shares of the other brands. For example, if a customer switches from brand A, there is probability B/(B + C) that he will switch to brand B and probability C/(B + C) that he will switch to brand C. Suppose that the market is currently divided equally: 10,000 families for each of the three brands. a. After a year, what will the market share for each firm be? Assume pA = 0.10, pB = 0.15, and pC = 0.20. (Hint: You will need to use the RISKBINOMLAL function to see how many people switch from A and then use the RISKBENOMIAL function again to see how many switch from A to B and from A to C. However, if your model requires more RISKBINOMIAL functions than the number allowed in the academic version of @RISK, remember that you can instead use the BENOM.INV (or the old CRITBENOM) function to generate binomially distributed random numbers. This takes the form =BINOM.INV (ntrials, psuccess, RAND()).) b. Suppose a 1% increase in market share is worth 10,000 per week to company A. Company A believes that for a cost of 1 million per year it can cut the percentage of unsatisfactory juice cartons in half. Is this worthwhile? (Use the same values of pA, pB, and pC as in part a.)arrow_forwardContinuing the previous problem, assume, as in Problem 11, that the damage amount is normally distributed with mean 3000 and standard deviation 750. Run @RISK with 5000 iterations to simulate the amount you pay for damage. Compare your results with those in the previous problem. Does it appear to matter whether you assume a triangular distribution or a normal distribution for damage amounts? Why isnt this a totally fair comparison? (Hint: Use @RISKs Define Distributions tool to find the standard deviation for the triangular distribution.)arrow_forward
- Practical Management ScienceOperations ManagementISBN:9781337406659Author:WINSTON, Wayne L.Publisher:Cengage,