Practical Management Science
6th Edition
ISBN: 9781337406659
Author: WINSTON, Wayne L.
Publisher: Cengage,
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Chapter 8.6, Problem 12P
Summary Introduction
To determine: The evolutionary solver to fit a pearl and Gompertz curve.
Introduction: The variation between the present value of the
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Chapter 8 Solutions
Practical Management Science
Ch. 8.3 - Prob. 1PCh. 8.3 - Prob. 2PCh. 8.4 - Prob. 3PCh. 8.4 - Prob. 4PCh. 8.4 - Prob. 5PCh. 8.5 - Prob. 6PCh. 8.5 - Prob. 7PCh. 8.5 - In the lawn mower production problem in Example...Ch. 8.6 - Prob. 9PCh. 8.6 - Prob. 10P
Ch. 8.6 - Prob. 11PCh. 8.6 - Prob. 12PCh. 8.7 - Prob. 13PCh. 8.7 - Prob. 14PCh. 8.8 - Prob. 15PCh. 8.8 - Prob. 16PCh. 8.8 - Prob. 17PCh. 8.8 - Prob. 18PCh. 8.8 - Prob. 19PCh. 8 - Prob. 20PCh. 8 - Prob. 21PCh. 8 - Prob. 22PCh. 8 - Prob. 23PCh. 8 - Prob. 24PCh. 8 - Prob. 25PCh. 8 - Prob. 26PCh. 8 - Prob. 27PCh. 8 - Prob. 28PCh. 8 - Prob. 29PCh. 8 - Prob. 30PCh. 8 - Prob. 31PCh. 8 - Prob. 32PCh. 8 - Prob. 33PCh. 8 - Prob. 34PCh. 8 - Prob. 35PCh. 8 - Prob. 36PCh. 8 - Prob. 37PCh. 8 - Prob. 38PCh. 8 - Prob. 39PCh. 8 - Prob. 1CCh. 8 - Prob. 2C
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- A company manufacturers a product in the United States and sells it in England. The unit cost of manufacturing is 50. The current exchange rate (dollars per pound) is 1.221. The demand function, which indicates how many units the company can sell in England as a function of price (in pounds) is of the power type, with constant 27556759 and exponent 2.4. a. Develop a model for the companys profit (in dollars) as a function of the price it charges (in pounds). Then use a data table to find the profit-maximizing price to the nearest pound. b. If the exchange rate varies from its current value, does the profit-maximizing price increase or decrease? Does the maximum profit increase or decrease?arrow_forwardYou 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_forwardIn the financial world, there are many types of complex instruments called derivatives that derive their value from the value of an underlying asset. Consider the following simple derivative. A stocks current price is 80 per share. You purchase a derivative whose value to you becomes known a month from now. Specifically, let P be the price of the stock in a month. If P is between 75 and 85, the derivative is worth nothing to you. If P is less than 75, the derivative results in a loss of 100(75-P) dollars to you. (The factor of 100 is because many derivatives involve 100 shares.) If P is greater than 85, the derivative results in a gain of 100(P-85) dollars to you. Assume that the distribution of the change in the stock price from now to a month from now is normally distributed with mean 1 and standard deviation 8. Let EMV be the expected gain/loss from this derivative. It is a weighted average of all the possible losses and gains, weighted by their likelihoods. (Of course, any loss should be expressed as a negative number. For example, a loss of 1500 should be expressed as -1500.) Unfortunately, this is a difficult probability calculation, but EMV can be estimated by an @RISK simulation. Perform this simulation with at least 1000 iterations. What is your best estimate of EMV?arrow_forward
- Assume the demand for a companys drug Wozac during the current year is 50,000, and assume demand will grow at 5% a year. If the company builds a plant that can produce x units of Wozac per year, it will cost 16x. Each unit of Wozac is sold for 3. Each unit of Wozac produced incurs a variable production cost of 0.20. It costs 0.40 per year to operate a unit of capacity. Determine how large a Wozac plant the company should build to maximize its expected profit over the next 10 years.arrow_forwardThe IRR is the discount rate r that makes a project have an NPV of 0. You can find IRR in Excel with the built-in IRR function, using the syntax =IRR(range of cash flows). However, it can be tricky. In fact, if the IRR is not near 10%, this function might not find an answer, and you would get an error message. Then you must try the syntax =IRR(range of cash flows, guess), where guess" is your best guess for the IRR. It is best to try a range of guesses (say, 90% to 100%). Find the IRR of the project described in Problem 34. 34. Consider a project with the following cash flows: year 1, 400; year 2, 200; year 3, 600; year 4, 900; year 5, 1000; year 6, 250; year 7, 230. Assume a discount rate of 15% per year. a. Find the projects NPV if cash flows occur at the ends of the respective years. b. Find the projects NPV if cash flows occur at the beginnings of the respective years. c. Find the projects NPV if cash flows occur at the middles of the respective years.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_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_forwardSometimes curvature in a scatterplot can be fit adequately (especially to the naked eye) by several trend lines. We discussed the exponential trend line, and the power trend line is discussed in the previous problem. Still another fairly simple trend line is the parabola, a polynomial of order 2 (also called a quadratic). For the demand-price data in the file P13_10.xlsx, fit all three of these types of trend lines to the data, and calculate the MAPE for each. Which provides the best fit? (Hint: Note that a polynomial of order 2 is still another of Excels Trend line options.)arrow_forwardAlthough the normal distribution is a reasonable input distribution in many situations, it does have two potential drawbacks: (1) it allows negative values, even though they may be extremely improbable, and (2) it is a symmetric distribution. Many situations are modelled better with a distribution that allows only positive values and is skewed to the right. Two of these that have been used in many real applications are the gamma and lognormal distributions. @RISK enables you to generate observations from each of these distributions. The @RISK function for the gamma distribution is RISKGAMMA, and it takes two arguments, as in =RISKGAMMA(3,10). The first argument, which must be positive, determines the shape. The smaller it is, the more skewed the distribution is to the right; the larger it is, the more symmetric the distribution is. The second argument determines the scale, in the sense that the product of it and the first argument equals the mean of the distribution. (The mean in this example is 30.) Also, the product of the second argument and the square root of the first argument is the standard deviation of the distribution. (In this example, it is 3(10=17.32.) The @RISK function for the lognormal distribution is RISKLOGNORM. It has two arguments, as in =RISKLOGNORM(40,10). These arguments are the mean and standard deviation of the distribution. Rework Example 10.2 for the following demand distributions. Do the simulated outputs have any different qualitative properties with these skewed distributions than with the triangular distribution used in the example? a. Gamma distribution with parameters 2 and 85 b. Gamma distribution with parameters 5 and 35 c. Lognormal distribution with mean 170 and standard deviation 60arrow_forward
- In Problem 12 of the previous section, suppose that the demand for cars is normally distributed with mean 100 and standard deviation 15. Use @RISK to determine the best order quantityin this case, the one with the largest mean profit. Using the statistics and/or graphs from @RISK, discuss whether this order quantity would be considered best by the car dealer. (The point is that a decision maker can use more than just mean profit in making a decision.)arrow_forwardReferring to Example 11.1, if the average bid for each competitor stays the same, but their bids exhibit less variability, does Millers optimal bid increase or decrease? To study this question, assume that each competitors bid, expressed as a multiple of Millers cost to complete the project, follows each of the following distributions. a. Triangular with parameters 1.0, 1.3, and 2.4 b. Triangular with parameters 1.2, 1.3, and 2.2 c. Use @RISKs Define Distributions window to check that the distributions in parts a and b have the same mean as the original triangular distribution in the example, but smaller standard deviations. What is the common mean? Why is it not the same as the most likely value, 1.3?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
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