Practical Management Science
6th Edition
ISBN: 9781337406659
Author: WINSTON, Wayne L.
Publisher: Cengage,
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 7, Problem 66P
Summary Introduction
To show: The
Non-linear programming (NLP):
Non-linear programming (NLP) is used in complex optimization problems where the objectives or constraints or sometimes both are non-linear functions of the decision variables. A model can be termed as non-linear for more than one reason.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionChapter 7 Solutions
Practical Management Science
Ch. 7.3 - Prob. 1PCh. 7.3 - Prob. 2PCh. 7.3 - Pricing Decisions at Madison The Madison Company...Ch. 7.3 - Prob. 4PCh. 7.3 - Prob. 5PCh. 7.3 - Prob. 6PCh. 7.3 - Prob. 7PCh. 7.3 - Prob. 8PCh. 7.3 - Prob. 9PCh. 7.3 - Prob. 10P
Ch. 7.3 - Prob. 11PCh. 7.3 - Prob. 12PCh. 7.3 - Prob. 13PCh. 7.3 - PRICING SUITS AT SULLIVANS Sullivans is a retailer...Ch. 7.3 - Prob. 15PCh. 7.4 - Prob. 16PCh. 7.4 - Prob. 17PCh. 7.4 - Prob. 18PCh. 7.4 - Prob. 19PCh. 7.4 - Prob. 20PCh. 7.4 - Prob. 21PCh. 7.4 - Prob. 22PCh. 7.4 - Prob. 23PCh. 7.5 - Prob. 24PCh. 7.5 - Prob. 25PCh. 7.5 - Prob. 26PCh. 7.5 - Prob. 27PCh. 7.6 - Prob. 28PCh. 7.6 - Prob. 29PCh. 7.6 - Prob. 30PCh. 7.6 - Prob. 31PCh. 7.6 - Prob. 32PCh. 7.6 - Prob. 33PCh. 7.6 - The method for rating teams in Example 7.8 is...Ch. 7.7 - Prob. 35PCh. 7.7 - Prob. 36PCh. 7.7 - Prob. 37PCh. 7.7 - The stocks in Example 7.9 are all positively...Ch. 7.7 - Prob. 39PCh. 7.7 - Prob. 40PCh. 7.7 - Prob. 41PCh. 7.7 - Prob. 42PCh. 7.8 - Given the data in the file Stock Beta.xlsx,...Ch. 7.8 - Prob. 44PCh. 7 - Prob. 45PCh. 7 - Prob. 46PCh. 7 - Another way to derive a demand function is to...Ch. 7 - Prob. 48PCh. 7 - If a monopolist produces q units, she can charge...Ch. 7 - Prob. 50PCh. 7 - Prob. 51PCh. 7 - Prob. 52PCh. 7 - Prob. 53PCh. 7 - Prob. 54PCh. 7 - Prob. 55PCh. 7 - Prob. 56PCh. 7 - A beer company has divided Bloomington into two...Ch. 7 - Prob. 58PCh. 7 - Prob. 59PCh. 7 - Prob. 60PCh. 7 - Prob. 61PCh. 7 - Prob. 62PCh. 7 - Prob. 63PCh. 7 - You have 50,000 to invest in three stocks. Let Ri...Ch. 7 - Prob. 65PCh. 7 - Prob. 66PCh. 7 - Prob. 67PCh. 7 - Prob. 68PCh. 7 - Prob. 69PCh. 7 - Prob. 70PCh. 7 - Based on Grossman and Hart (1983). A salesperson...Ch. 7 - Prob. 73PCh. 7 - Prob. 74PCh. 7 - Prob. 75PCh. 7 - Prob. 76PCh. 7 - Prob. 77PCh. 7 - Prob. 78PCh. 7 - Prob. 79PCh. 7 - Prob. 80PCh. 7 - Prob. 81PCh. 7 - Prob. 82PCh. 7 - Prob. 83PCh. 7 - Prob. 84PCh. 7 - Prob. 85PCh. 7 - Prob. 86PCh. 7 - Prob. 1.1CCh. 7 - Prob. 1.2CCh. 7 - Prob. 1.3CCh. 7 - Prob. 1.4C
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, operations-management and related others by exploring similar questions and additional content below.Similar questions
- Based on Grossman and Hart (1983). A salesperson for Fuller Brush has three options: (1) quit, (2) put forth a low level of effort, or (3) put forth a high level of effort. Suppose for simplicity that each salesperson will sell 0, 5000, or 50,000 worth of brushes. The probability of each sales amount depends on the effort level as described in the file P07_71.xlsx. If a salesperson is paid w dollars, he or she regards this as a benefit of w1/2 units. In addition, low effort costs the salesperson 0 benefit units, whereas high effort costs 50 benefit units. If a salesperson were to quit Fuller and work elsewhere, he or she could earn a benefit of 20 units. Fuller wants all salespeople to put forth a high level of effort. The question is how to minimize the cost of encouraging them to do so. The company cannot observe the level of effort put forth by a salesperson, but it can observe the size of his or her sales. Thus, the wage paid to the salesperson is completely determined by the size of the sale. This means that Fuller must determine w0, the wage paid for sales of 0; w5000, the wage paid for sales of 5000; and w50,000, the wage paid for sales of 50,000. These wages must be set so that the salespeople value the expected benefit from high effort more than quitting and more than low effort. Determine how to minimize the expected cost of ensuring that all salespeople put forth high effort. (This problem is an example of agency theory.)arrow_forwardIn August of the current year, a car dealer is trying to determine how many cars of the next model year to order. Each car ordered in August costs 20,000. The demand for the dealers next year models has the probability distribution shown in the file P10_12.xlsx. Each car sells for 25,000. If demand for next years cars exceeds the number of cars ordered in August, the dealer must reorder at a cost of 22,000 per car. Excess cars can be disposed of at 17,000 per car. Use simulation to determine how many cars to order in August. For your optimal order quantity, find a 95% confidence interval for the expected profit.arrow_forwardSuppose you begin year 1 with 5000. At the beginning of each year, you put half of your money under a mattress and invest the other half in Whitewater stock. During each year, there is a 40% chance that the Whitewater stock will double, and there is a 60% chance that you will lose half of your investment. To illustrate, if the stock doubles during the first year, you will have 3750 under the mattress and 3750 invested in Whitewater during year 2. You want to estimate your annual return over a 30-year period. If you end with F dollars, your annual return is (F/5000)1/30 1. For example, if you end with 100,000, your annual return is 201/30 1 = 0.105, or 10.5%. Run 1000 replications of an appropriate simulation. Based on the results, you can be 95% certain that your annual return will be between which two values?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_forwardA common decision is whether a company should buy equipment and produce a product in house or outsource production to another company. If sales volume is high enough, then by producing in house, the savings on unit costs will cover the fixed cost of the equipment. Suppose a company must make such a decision for a four-year time horizon, given the following data. Use simulation to estimate the probability that producing in house is better than outsourcing. If the company outsources production, it will have to purchase the product from the manufacturer for 25 per unit. This unit cost will remain constant for the next four years. The company will sell the product for 42 per unit. This price will remain constant for the next four years. If the company produces the product in house, it must buy a 500,000 machine that is depreciated on a straight-line basis over four years, and its cost of production will be 9 per unit. This unit cost will remain constant for the next four years. The demand in year 1 has a worst case of 10,000 units, a most likely case of 14,000 units, and a best case of 16,000 units. The average annual growth in demand for years 2-4 has a worst case of 7%, a most likely case of 15%, and a best case of 20%. Whatever this annual growth is, it will be the same in each of the years. The tax rate is 35%. Cash flows are discounted at 8% per year.arrow_forward
- It is surprising (but true) that if 23 people are in the same room, there is about a 50% chance that at least two people will have the same birthday. Suppose you want to estimate the probability that if 30 people are in the same room, at least two of them will have the same birthday. You can proceed as follows. a. Generate random birthdays for 30 different people. Ignoring the possibility of a leap year, each person has a 1/365 chance of having a given birthday (label the days of the year 1 to 365). You can use the RANDBETWEEN function to generate birthdays. b. Once you have generated 30 peoples birthdays, how can you tell whether at least two people have the same birthday? One way is to use Excels RANK function. (You can learn how to use this function in Excels online help.) This function returns the rank of a number relative to a given group of numbers. In the case of a tie, two numbers are given the same rank. For example, if the set of numbers is 4, 3, 2, 5, the RANK function returns 2, 3, 4, 1. (By default, RANK gives 1 to the largest number.) If the set of numbers is 4, 3, 2, 4, the RANK function returns 1, 3, 4, 1. c. After using the RANK function, you should be able to determine whether at least two of the 30 people have the same birthday. What is the (estimated) probability that this occurs?arrow_forwardAn automobile manufacturer is considering whether to introduce a new model called the Racer. The profitability of the Racer depends on the following factors: The fixed cost of developing the Racer is triangularly distributed with parameters 3, 4, and 5, all in billions. Year 1 sales are normally distributed with mean 200,000 and standard deviation 50,000. Year 2 sales are normally distributed with mean equal to actual year 1 sales and standard deviation 50,000. Year 3 sales are normally distributed with mean equal to actual year 2 sales and standard deviation 50,000. The selling price in year 1 is 25,000. The year 2 selling price will be 1.05[year 1 price + 50 (% diff1)] where % diff1 is the number of percentage points by which actual year 1 sales differ from expected year 1 sales. The 1.05 factor accounts for inflation. For example, if the year 1 sales figure is 180,000, which is 10 percentage points below the expected year 1 sales, then the year 2 price will be 1.05[25,000 + 50( 10)] = 25,725. Similarly, the year 3 price will be 1.05[year 2 price + 50(% diff2)] where % diff2 is the percentage by which actual year 2 sales differ from expected year 2 sales. The variable cost in year 1 is triangularly distributed with parameters 10,000, 12,000, and 15,000, and it is assumed to increase by 5% each year. Your goal is to estimate the NPV of the new car during its first three years. Assume that the company is able to produce exactly as many cars as it can sell. Also, assume that cash flows are discounted at 10%. Simulate 1000 trials to estimate the mean and standard deviation of the NPV for the first three years of sales. Also, determine an interval such that you are 95% certain that the NPV of the Racer during its first three years of operation will be within this interval.arrow_forwardIn Problem 11 from the previous section, we stated that the damage amount is normally distributed. Suppose instead that the damage amount is triangularly distributed with parameters 500, 1500, and 7000. That is, the damage in an accident can be as low as 500 or as high as 7000, the most likely value is 1500, and there is definite skewness to the right. (It turns out, as you can verify in @RISK, that the mean of this distribution is 3000, the same as in Problem 11.) Use @RISK to simulate the amount you pay for damage. Run 5000 iterations. Then answer the following questions. In each case, explain how the indicated event would occur. a. What is the probability that you pay a positive amount but less than 750? b. What is the probability that you pay more than 600? c. What is the probability that you pay exactly 1000 (the deductible)?arrow_forward
- W. L. Brown, a direct marketer of womens clothing, must determine how many telephone operators to schedule during each part of the day. W. L. Brown estimates that the number of phone calls received each hour of a typical eight-hour shift can be described by the probability distribution in the file P10_33.xlsx. Each operator can handle 15 calls per hour and costs the company 20 per hour. Each phone call that is not handled is assumed to cost the company 6 in lost profit. Considering the options of employing 6, 8, 10, 12, 14, or 16 operators, use simulation to determine the number of operators that minimizes the expected hourly cost (labor costs plus lost profits).arrow_forwardA European put option allows an investor to sell a share of stock at the exercise price on the exercise data. For example, if the exercise price is 48, and the stock price is 45 on the exercise date, the investor can sell the stock for 48 and then immediately buy it back (that is, cover his position) for 45, making 3 profit. But if the stock price on the exercise date is greater than the exercise price, the option is worthless at that date. So for a put, the investor is hoping that the price of the stock decreases. Using the same parameters as in Example 11.7, find a fair price for a European put option. (Note: As discussed in the text, an actual put option is usually for 100 shares.)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
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Practical Management ScienceOperations ManagementISBN:9781337406659Author:WINSTON, Wayne L.Publisher:Cengage,
Practical Management Science
Operations Management
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:Cengage,