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Quantitative Analysis for Management, Global 13th Edition Solution Manual

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    1 - 1 CHAPTER 1 Introduction to Quantitative Analysis TEACHING SUGGESTIONS Teaching Suggestion 1.1: Importance of Qualitative Factors. Section 1.2 gives students an overview of quantitative analysis. In this section, a number of qualitative factors, including federal legislation and new technology, are discussed. Students can be asked to discuss other qualitative factors that could have an impact on quantitative analysis. Waiting lines and project planning can be used as examples. Teaching Suggestion 1.2: Discussing Other Quantitative Analysis Problems. Section 1.2 covers an application of the quantitative analysis approach. Students can be asked to describe other problems or areas that could benefit from quantitative analysis. Teaching Suggestion 1.3: Discussing Conflicting Viewpoints. Possible problems in the QA approach are presented in this chapter. A discussion of conflicting viewpoints within the organization can help students understand this problem. For example, how many people should staff a registration desk at a university? Students will want more staff to reduce waiting time, while university administrators will want less staff to save money. A discussion of these types of conflicting viewpoints will help students understand some of the problems of using quantitative analysis. Teaching Suggestion 1.4: Difficulty of Getting Input Data. A major problem in quantitative analysis is getting proper input data. Students can be asked to explain how they would get the information they need to determine inventory ordering or carrying costs. Role - playing with students assuming the parts of the analyst who needs inventory costs and the instructor playing the part of a veteran inventory manager can be fun and interesting. Students quickly learn that getting good data can be the most difficult part of using quantitative analysis. Teaching Suggestion 1.5: Dealing with Resistance to Change. Resistance to change is discussed in this chapter. Students can be asked to explain how they would introduce a new system or change within the organization. People resisting new approaches can be a major stumbling block to the successful implementation of quantitative analysis. Students can be asked why some people may be afraid of a new inventory control or forecasting system.

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    1 - 2 SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS 1 - 1. Through an approach that is scientific, logical and rational and hence avoids reliance on guesswork, subjective opinions and intuition . 1 - 2. Quantitative analysis is the scientific approach to managerial decision making. This type of analysis is a logical and rational approach to making decisions. Emotions, guesswork, and whim are not part of the quantitative analysis approach. A number of organizations support the use of the scientific approach: the Institute for Operation Research and Management Science (INFORMS), Decision Sciences Institute, and Academy of Management . 1 - 3 . The three categories of business analytics are descriptive, predictive, and prescriptive. Descriptive analytics provides an indication of how things performed in the past. Predictive analytics uses past data to forecast what will happen in the future. Prescriptive analytics uses optimization and other models to present better ways for a company to operate to reach goals and objectives. 1 - 4. The 7 steps are as follows: ▪ Defining the problem. ▪ Developing a Model. ▪ Acquiring Input Data. ▪ Developing a Solution. ▪ Testing the Solution. ▪ Analyzing the Results. ▪ Implementing the Results . 1 - 5. Although the formal study of quantitative analysis and the refinement of the tools and techniques of the scientific method have occurred only in the recent past, quantitative approaches to decision making have been in existence since the beginning of time. In the early 1900s, Frederick W. Taylor developed the principles of the scientific approach. During World War II, quantitative analysis was intensified and used by the military. Because of the success of these techniques during World War II, interest continued after the war. 1 - 6. Quantitative analysis involves the use of mathematical equations or relationships in analyzing a particular problem. In most cases, the results of quantitative analysis will be one or more numbers that can be used by managers and decision makers in making better decisions. (1) Defining the Problem, (2) Developing a Model, (3) Acquiring Input Data, (4) Developing a Solution, (5) Testing the Solution, (6) Analyzing the Results, and (7) Implementing the Results . 1 - 7. Input data can come from company reports and documents, interviews with employees and other personnel, direct measurement, and sampling procedures. For many problems, a number of different sources are required to obtain data, and in some cases it is necessary to obtain the same data from different sources in order to check the accuracy and consistency of the input data. If the input data are not accurate, the results can be misleading and very costly to the organization. This concept is called “garbage in, garbage out”. 1 - 8. Implementation is the process of taking the solution and incorporating it into the company or organization. This is the final step in the quantitative analysis approach, and if a good job is not done with implementation, all of the effort expended on the previous steps can be wasted .

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    1 - 3 1 - 9. The phrase ‘Garbage in, garbage out’ highlights the importance of acquiring accurate input data. If the input data is inaccurate then no matter how good the model, the results produced will be misleading . 1 - 10. There are a large number of quantitative terms that may not be understood by managers. Examples include PERT, CPM, simulation, the Monte Carlo method, mathematical programming, EOQ, and so on. The student should explain each of the four terms selected in his or her own words. 1 - 11. Answers will vary but may include: (1) lack of commitment by management, (2) resistance to change by management, and (3) lack of commitment by quantitative analysts . 1 - 12. Users need not become involved in technical aspects of the QA technique, but they should have an understanding of what the limitations of the model are, how it works (in a general sense), the jargon involved, and the ability to question the validity and sensitivity of an answer handed to them by an analyst. 1 - 13. Churchman meant that sophisticated mathematical solutions and proofs can be dangerous because people may be afraid to question them. Many people do not want to appear ignorant and question an elaborate mathematical model; yet the entire model, its assumptions and its approach, may be incorrect. 1 - 14. The break - even point is the number of units that must be sold to make zero profits. To compute this, we must know the selling price, the fixed cost, and the variable cost per unit. 1 - 15. f = 350 s = 15 v = 8 a) Total revenue = 20(15) = $300 Total variable cost = 20(8) = $160 b) BEP = f /( s − v ) = 350/(15 − 8) = 50 units Total revenue = 50(15) = $750 1 - 16. f = 150 s = 50 v = 20 BEP = f /( s − v ) = 150/(50 − 20) = 5 units 1 - 17. f = 150 s = 50 v = 15 BEP = f /( s − v ) = 150/(50 − 15) = 4.29 units 1 - 18. f = 400 + 1,000 = 1,400 s = 5 v = 3 BEP = f /( s − v ) = 1400/(5 − 3) = 700 units 1 - 19. BEP = f /( s − v ) 500 = 1400/( s − 3) 500( s − 3) = 1400 s − 3 = 1400/500 s = 2.8 + 3 s = $5.80

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    1 - 4 1 - 20. f = 2400 s = 40 v = 25 BEP = f /( s – v ) = 2400/(40 – 25) = 160 per week Total revenue = 40(160) = $6400 1 - 21. f = 2400 s = 50 v = 25 BEP = f /( s – v ) = 2400/(50 – 25) = 96 per week Total revenue = 50(96) = $4800 1 - 22. f = 2400 s = ? v = 25 BEP = f /( s – v ) 120 = 2400/( s – 25) 120( s – 25) = 2400 s = 45 1 - 23. f = 11000 s = 250 v = 60 BEP = f /( s – v ) = 11000/(250 – 60) = 57.9 1 - 24. a) f = 300 + 75 = 375 s = 20 v = 5 BEP = f /( s – v ) = 375/(20 – 5) = 25 b) f = 200 + 75 = 275 s = 20 v = 5 BEP = f /( s − v ) = 275/(20 − 5) = 18.333 1 - 25. a) Machine 1: f = 600 s = 0.05 v = 0.010 BEP = f /( s – v ) = 600/(0.05 – 0.010) = 15,000 Machine 2: f = 400 s = 0.05 v = 0.015 BEP = f /( s – v ) = 400/(0.05 – 0.015) = 11,428.57 b) Machine 1: Cost = 600 + 0.010(10,000) = $700 Machine 2: Cost = 400 + 0.015(10,000) = $550 c) Machine 1: Cost = 600 + 0.010(30,000) = $900 Machine 2: Cost = 400 + 0.015(30,000) = $850 d) Let X = the number of copies 600 + 0.010X = 400 + 0.015X 600 – 400 = 0.015X – 0.010X 200 = 0.05X X = 40,000 copies

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    1 - 5 1 - 26. a) Proposal A: f = 65,000 s = 18 v = 10 BEP = f /( s – v ) = 65,000/(18 – 10) = 8 , 125 Proposal B: f = 34,000 s = 1 8 v = 1 4 BEP = f /( s − v ) = 34,000 /( 18 − 14 ) = 8,500 b) Proposal A should be chosen.

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    1 - 6 SOLUTION TO FOOD AND BEVERAGES AT SOUTHWESTERN UNIVERSITY FOOTBALL GAMES The total fixed cost per game includes salaries, rental fees, and cost of the workers in the six booths. These are: Salaries = $20,000 Rental fees = 2,400  $2 = $4,800 Booth worker wages = 6  6  5  $7 = $1,260 Total fixed cost per game = $20,000 + $4,800 + $1,260 = $26,060 The cost of this allocated to each food item is shown in the table: Percent Allocated fixed Item revenue cost Soft drink 25% $6,515 Coffee 25% $6,515 Hot dogs 20% $5,212 Hamburgers 20% $5,212 Misc. snacks 10% $2,606 The break - even points for each of these items are found by computing the contribution to profit (profit margin) for each item and dividing this into the allocated fixed cost. These are shown in the next table: Selling Var. Profit Percent Allocated Break even Item price cost margin revenue fixed cost volume Soft drink $1.50 $0.75 $0.75 25% 6515 8686.67 Coffee $2.00 $0.50 $1.50 25% 6515 4343.33 Hot dogs $2.00 $0.80 $1.20 20% 5212 4343.33 Hamburgers $2.50 $1.00 $1.50 20% 5212 3474.67 Misc. snacks $1.00 $0.40 $0.60 10% 2606 4343.33

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    1 - 7 To determine the total sales for each item that is required to break even, multiply the selling price by the break even volume. The results are shown: Selling Break even Dollar volume Item price volume of sales Soft drink $1.50 8686.67 $13,030.00 Coffee $2.00 4343.33 $8,686.67 Hot dogs $2.00 4343.33 $8,686.67 Hamburgers $2.50 3474.67 $8,686.67 Misc. snacks $1.00 4343.33 $4,343.33 Total $43,433.34 Thus, to break even, the total sales must be $43,433.34. If the attendance is 35,000 people, then each person would have to spend $43,433.34/35,000 = $1.24. If the attendance is 60,000, then each person would have to spend $43,433.34/60,000 = $0.72. Both of these are very low values, so we should be confident that this food and beverage operation will at least break even. Note: While this process provides information about break - even points based on the current percent revenues for each product, there is one difficulty. The total revenue using the break - even points will not result in the same percentages (dollar volume of product/total revenue) as originally stated in the problem. A more complex model is available to do this (see p. 308 Jay Heizer and Barry Render, Principles of Operations Management , 9th ed., Upper Saddle River, NJ: Prentice Hall, 2014).

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    2 - 1 CHAPTER 2 Probability Concepts and Applications TEACHING SUGGESTIONS Teaching Suggestion 2.1: Concept of Probabilities Ranging From 0 to 1. People often misuse probabilities by such statements as, “I’m 110% sure we’re going to win the big game.” The two basic rules of probability should be stressed. Teaching Suggestion 2.2: Where Do Probabilities Come From? Students need to understand where probabilities come from. Sometimes they are subjective and based on personal experiences. Other times they are objectively based on logical observations such as the roll of a die. Often, probabilities are derived from historical data — if we can assume the future will be about the same as the past. Teaching Suggestion 2.3: Confusion Over Mutually Exclusive and Collectively Exhaustive Events. This concept is often foggy to even the best of students — even if they just completed a course in statistics. Use practical examples and drills to force the point home. Teaching Suggestion 2.4: Addition of Events That Are Not Mutually Exclusive. The formula for adding events that are not mutually exclusive is P ( A or B ) = P ( A ) + P ( B ) – P ( A and B ). Students must understand why we subtract P ( A and B ). Explain that the intersection has been counted twice. Teaching Suggestion 2.5: Expected Value of a Probability Distribution. A probability is often described by its mean and variance. These important terms should be discussed with such practical examples as heights or weights of students. But students need to be reminded that even if most of the men in class (or the United States) have heights between 5 feet 6 inches and 6 feet 2 inches, there is still some small probability of outliers. Teaching Suggestion 2.6: Bell - Shaped Curve. Stress how important the normal is to a large number of processes in our lives (for example, filling boxes of cereal with 32 ounces of cornflakes). Each normal depends on the mean and standard deviation. Discuss Figures 2.7 and 2.8 to show how these relate to the shape and position of a normal .

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    2 - 2 Teaching Suggestion 2.7: Three Symmetrical Areas Under the Normal Curve. Figure 2.13 is very important, and students should be encouraged to truly comprehend the meanings of ±1, 2, and 3 standard deviation symmetrical areas. They should especially know that managers often speak of 95% and 99% confidence intervals, which roughly refer to ±2 and 3 standard deviation graphs. Clarify that 95% confidence is actually ±1.96 standard deviations, while ±3 standard deviations is actually a 99.7% spread. Teaching Suggestion 2.8: Using the Normal Table to Answer Probability Questions. The IQ example in Figure 2.9 is a particularly good way to treat the subject since everyone can relate to it. Students are typically curious about the chances of reaching certain scores. Go through at least a half - dozen examples until it’s clear that everyone can use Table 2.10. Students get especially confused answering questions such as P ( X  85) since the standard normal table shows only right - hand - side (positive) Z values. The symmetry requires special care.

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    2 - 3 ALTERNATIVE EXAMPLES Alternative Example 2.1: In the past 30 days, Roger’s Rural Roundup has sold either 8, 9, 10, or 11 lottery tickets. It never sold fewer than 8 nor more than 11. Assuming that the past is similar to the future, here are the probabilities: Sales No. Days Probability 8 10 0.333 9 12 0.400 10 6 0.200 11 2 0.067 Total 30 1.000 Alternative Example 2.2: Grades received for a course have a probability based on the professor’s grading pattern. Here are Professor Ernie Forman’s BA205 grades for the past five years. Outcome Probability A 0.25 B 0.30 C 0.35 D 0.03 F 0.02 Withdraw/drop 0.05 1.00 These grades are mutually exclusive and collectively exhaustive. Alternative Example 2.3: P (drawing a 3 from a deck of cards) = 4/52 = 1/13 P (drawing a club on the same draw) = 13/52 = 1/4 These are neither mutually exclusive nor collectively exhaustive.

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    2 - 4 Alternative Example 2.4: In Alternative Example 2.3 we looked at 3s and clubs. Here is the probability for 3 or club: P (3 or club) = P (3) + P (club) – P (3 and club) = 4/52 + 13/52 – 1/52 = 16/52 = 4/13 Alternative Example 2.5: A class contains 30 students. Ten are female (F) and U.S. citizens (U); 12 are male (M) and U.S. citizens; 6 are female and non - U.S. citizens (N); 2 are male and non - U.S. citizens. A name is randomly selected from the class roster and it is female. What is the probability that the student is a U.S. citizen? U.S. Not U.S. Total F 10 6 16 M 12 2 14 Total 22 8 30 P(U | F) = 10/16 = 5/8 Alternative Example 2.6: Your professor tells you that if you score an 85 or better on your midterm exam, there is a 90% chance you’ll get an A for the course. You think you have only a 50% chance of scoring 85 or better. The probability that both your score is 85 or better and you receive an A in the course is P (A and 85) = P (A  85)  P (85) = (0.90)(0.50) = 0.45 = a 45% chance

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    2 - 5 Alternative Example 2.7: An instructor is teaching two sections (classes) of calculus. Each class has 24 students, and on the surface, both classes appear identical. One class, however, consists of students who have all taken calculus in high school. The instructor has no idea which class is which. She knows that the probability of at least half the class getting As on the first exam is only 25% in an average class, but 50% in a class with more math background. A section is selected at random and quizzed. More than half the class received As. Now, what is the revised probability that the class was the advanced one? P (regular class chosen) = 0.5 P (advanced class chosen) = 0.5 P (  1/2 As  regular class) = 0.25 P (  1/2 As  advanced class) = 0.50 P (  1/2 As and regular class) = P (  1/2 As  regular )  P (regular) = (0.25)(0.50) = 0.125 P (  1/2 As and advanced class) = P (  1/2 As  advanced)  P (advanced) = (0.50)(0.5) = 0.25 So P (  1/2 As) = 0.125 + 0.25 = 0.375 ( ) ( ) ( ) advanced and 1/2 As advanced 1/2 As 1/ 2 As 0.25 2 / 3 0.375 P P P   =  = = So there is a 66% chance the class tested was the advanced one.

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    2 - 6 Alternative Example 2.8: Students in a statistics class were asked how many “away” football games they expected to attend in the upcoming season. The number of students responding to each possibility is shown below: Number of games Number of students 5 40 4 30 3 20 2 10 1 0 100 A probability of the results would be: Number of games Probability P ( X ) 5 0.4 = 40/100 4 0.3 = 30/100 3 0.2 = 20/100 2 0.1 = 10/100 1 0.0 = 0/100 1.0 = 100/100 This discrete probability is computed using the relative frequency approach. Probabilities are shown in graph form below.

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    2 - 7 Alternative Example 2.9: Here is how the expected outcome can be computed for the question in Alternative Example 2.8. ( ) ( ) ( ) ( ) 5 1 2 2 1 i i i i E x x P X x P X x P X = = = +  + x 3 P ( X 3 ) + x 4 P ( X 4 ) + x 5 P ( X 5 ) = 5(0.4) + 4(0.3) + 3(0.2) + 2(0.1) + 1(0) = 4.0 Alternative Example 2.10: Here is how variance is computed for the question in Alternative Example 2.8: ( ) ( ) ( ) 5 2 1 variance i i i x E x P x = = −  = (5 – 4) 2 (0.4) + (4 – 4) 2 (0.3) + (3 – 4) 2 (0.2) + (2 – 4) 2 (0.1) + (1 – 4) 2 (0) = (1) 2 (0.4) + (0) 2 (0.3) + ( – 1) 2 (0.2) + ( – 2) 2 (0.1) + ( - 3) 2 (0) = 0.4 + 0.0 + 0.2 + 0.4 + 0.0 = 1.0 The standard deviation is variance 1  = = = 1 Alternative Example 2.11: The length of the rods coming out of our new cutting machine can be said to approximate a normal with a mean of 10 inches and a standard deviation of 0.2 inch. Find the probability that a rod selected randomly will have a length a. of less than 10.0 inches b. between 10.0 and 10.4 inches c. between 10.0 and 10.1 inches d. between 10.1 and 10.4 inches e. between 9.9 and 9.6 inches f. between 9.9 and 10.4 inches g. between 9.886 and 10.406 inches

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    2 - 8 First compute the standard normal , the Z - value: x z   − = Next, find the area under the curve for the given Z - value by using a standard normal table. a. P ( x  10.0) = 0.50000 b. P (10.0  x  10.4) = 0.97725 – 0.50000 = 0.47725 c. P (10.0  x  10.1) = 0.69146 – 0.50000 = 0.19146 d. P (10.1  x  10.4) = 0.97725 – 0.69146 = 0.28579 e. P (9.6  x  9.9) = 0.97725 – 0.69146 = 0.28579 f. P (9.9  x  10.4) = 0.19146 + 0.47725 = 0.66871 g. P (9.886  x  10.406) = 0.47882 + 0.21566 = 0.69448 SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS 2 - 1. There are two basic laws of probability. First, the probability of any event or state of nature occurring must be greater than or equal to zero and less than or equal to 1. Second, the sum of the simple probabilities for all possible outcomes of the activity must equal 1. 2 - 2. Events are mutually exclusive if only one of the events can occur on any one trial. Events are collectively exhaustive if the list of outcomes includes every possible outcome. An example of mutually exclusive events can be seen in flipping a coin. The outcome of any one trial can either be a head or a tail. Thus, the events of getting a head and a tail are mutually exclusive because only one of these events can occur on any one trial. This assumes, of course, that the coin does not land on its edge. The outcome of rolling the die is an example of events that are collectively exhaustive. In rolling a standard die, the outcome can be either 1, 2, 3, 4, 5, or 6. These six outcomes are collectively exhaustive because they include all possible outcomes. Again, it is assumed that the die will not land and stay on one of its edges. 2 - 3. Classical Method of determining probability uses logical method and probability can be identified without performing a series of trials, we can often logically determine what the probabilities of various events should be. For example, the probability of tossing a fair coin once and getting a head is: P (head) = 1/2. Here 1 is the Number of ways of getting a head. And 2 is the Number of possible outcomes (head or tail). Therefore, our probability here is 0.5 = 50% 2 - 4. The addition law for mutually exclusive events is: P ( A or B ) = P ( A ) + P ( B ) The addition law for events that are not mutually exclusive is P ( A or B ) = P ( A ) + P ( B ) – P ( A and B ) In this case we reduce the probability by subtracting the chance of both events occurring together.

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