Quantitative Analysis for Management, Global 13th Edition Solution Manual

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1-1CHAPTER1Introduction to Quantitative AnalysisTEACHING SUGGESTIONSTeaching Suggestion 1.1:Importance of Qualitative Factors.Section 1.2 gives students an overview of quantitative analysis. In this section, a number ofqualitative factors,including federal legislation and new technology, are discussed. Students canbe 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 todescribe 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 conflictingviewpoints within the organization can help students understand this problem. For example, howmany people should staff a registration desk at a university? Students will want more staff toreduce waiting time, while university administrators will want less staff to save money. Adiscussion of these types of conflicting viewpoints will help students understand some of theproblems 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 toexplain how they would get the information they need todetermine inventory ordering orcarrying costs. Role-playing with students assuming the parts of the analyst who needs inventorycosts and the instructor playing the part of a veteran inventory manager can be fun andinteresting. Students quickly learn that getting good data can be the most difficult part of usingquantitative 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 theywouldintroduceanewsystemorchangewithintheorganization.Peopleresistingnewapproaches can be a major stumbling block to the successful implementation of quantitativeanalysis. Students can be asked why some people may be afraid of a new inventory control orforecasting system.

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1-2SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS1-1.Throughan approach that is scientific, logical and rational and hence avoids reliance onguesswork, subjective opinions and intuition.1-2.Quantitative analysis is the scientific approach to managerial decision making. This type ofanalysis is a logical and rational approach to making decisions. Emotions, guesswork, and whimare not part of the quantitative analysis approach. A number of organizations support the use ofthescientificapproach:theInstituteforOperationResearchandManagementScience(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.Predictiveanalytics uses past data to forecast what will happen in the future.Prescriptive analytics usesoptimization and other models to present better ways for a company to operate to reach goals andobjectives.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.Althoughthe formal study of quantitative analysis and the refinement of the tools andtechniques of the scientific method have occurred only in the recent past, quantitative approachesto 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 thesetechniques during World War II, interest continued after the war.1-6.Quantitative analysis involves the useofmathematicalequations or relationshipsinanalyzing a particular problem. In most cases, the results of quantitative analysis will be one ormore 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 aSolution, (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 andother personnel, direct measurement, and sampling procedures. For many problems, a number ofdifferent sources are required to obtain data, and in some cases it is necessary to obtain the samedata from different sources in order to check the accuracy and consistency of the input data. Ifthe 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 companyor organization. This is the final step in the quantitative analysis approach, and if a good job isnot done with implementation, all of the effort expended on the previous steps can be wasted.

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1-31-9.Thephrase ‘Garbage in, garbage out’ highlights the importance of acquiring accurate inputdata. If the input data is inaccurate then no matter how good the model, the results produced willbe misleading.1-10.There are a large number of quantitative terms that may not be understood by managers.ExamplesincludePERT,CPM,simulation,theMonteCarlomethod,mathematicalprogramming, EOQ, and so on. The student should explain each of the four terms selected in hisor her own words.1-11.Answerswillvary 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,butthey shouldhave an understanding of what the limitations of the model are, how it works (in a generalsense), the jargon involved, and the ability to question the validity and sensitivity of an answerhanded to them by an analyst.1-13.Churchman meant that sophisticated mathematical solutions and proofs can be dangerousbecause people may be afraid to question them. Many people do not want to appear ignorant andquestion an elaborate mathematical model; yet the entire model, its assumptions and itsapproach, may be incorrect.1-14.The break-even point is the number of units that must be sold to make zero profits. Tocompute this, we must know the selling price, the fixed cost, and the variable cost per unit.1-15.f=350s=15v=8a)Total revenue=20(15)=$300Total variable cost=20(8)=$160b)BEP=f/(s−v)=350/(15−8)=50 unitsTotal revenue=50(15)=$7501-16.f=150s=50v=20BEP=f/(s−v)=150/(50−20)=5 units1-17.f=150s=50v=15BEP=f/(s−v)=150/(50−15)=4.29 units1-18.f=400+1,000=1,400s=5v=3BEP=f/(s−v)=1400/(5−3)=700 units1-19.BEP=f/(s−v)500=1400/(s−3)500(s−3)=1400s−3=1400/500s=2.8+3s=$5.80

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1-41-20.f=2400s=40v=25BEP=f/(s–v)=2400/(40–25)=160 per weekTotal revenue=40(160)=$64001-21.f=2400s=50v=25BEP=f/(s–v)=2400/(50–25)=96 per weekTotal revenue=50(96)=$48001-22.f=2400s=?v=25BEP=f/(s–v)120=2400/(s–25)120(s–25)=2400s=451-23.f=11000s=250v=60BEP=f/(s–v)=11000/(250–60)=57.91-24.a)f=300 + 75 = 375s=20v=5BEP=f/(s–v)=375/(20–5)=25b)f=200 + 75 = 275s=20v=5BEP=f/(s−v)=275/(20−5)=18.3331-25.a)Machine 1:f=600s=0.05v=0.010BEP=f/(s–v)=600/(0.05–0.010)=15,000Machine 2:f=400s=0.05v=0.015BEP=f/(s–v)=400/(0.05–0.015)=11,428.57b)Machine 1: Cost = 600 + 0.010(10,000) = $700Machine 2:Cost = 400 + 0.015(10,000) = $550c)Machine 1: Cost = 600 + 0.010(30,000) = $900Machine 2: Cost = 400 + 0.015(30,000) = $850d) Let X = the number of copies600 + 0.010X = 400 + 0.015X600–400 = 0.015X–0.010X200 = 0.05XX = 40,000 copies

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1-51-26.a) Proposal A:f=65,000s=18v=10BEP=f/(s–v)=65,000/(18–10)=8,125Proposal B:f=34,000s=18v=14BEP=f/(s−v)=34,000/(18−14)=8,500b)ProposalA should be chosen.

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1-6SOLUTION TO FOOD AND BEVERAGES ATSOUTHWESTERNUNIVERSITY FOOTBALL GAMESThe total fixed cost per game includes salaries, rental fees, and cost of the workers in the sixbooths. These are:Salaries=$20,000Rental fees=2,400$2=$4,800Booth worker wages=665$7=$1,260Total fixed cost per game=$20,000+$4,800+$1,260=$26,060The cost of this allocated to each food item is shown in the table:PercentAllocated fixedItemrevenuecostSoft drink25%$6,515Coffee25%$6,515Hot dogs20%$5,212Hamburgers20%$5,212Misc. snacks10%$2,606The 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 inthe nexttable:SellingVar.ProfitPercentAllocatedBreak evenItempricecostmarginrevenuefixed costvolumeSoft drink$1.50$0.75$0.7525%65158686.67Coffee$2.00$0.50$1.5025%65154343.33Hot dogs$2.00$0.80$1.2020%52124343.33Hamburgers$2.50$1.00$1.5020%52123474.67Misc. snacks$1.00$0.40$0.6010%26064343.33

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1-7To determine the total sales for each item that is required to break even, multiply the selling priceby the break even volume. The results are shown:SellingBreak evenDollar volumeItempricevolumeof salesSoft drink$1.508686.67$13,030.00Coffee$2.004343.33$8,686.67Hot dogs$2.004343.33$8,686.67Hamburgers$2.503474.67$8,686.67Misc. snacks$1.004343.33$4,343.33Total$43,433.34Thus, to break even, the total sales must be $43,433.34. If the attendance is 35,000 people, theneach person would have to spend $43,433.34/35,000=$1.24. If the attendance is 60,000, theneach 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 currentpercent revenues for each product, there is one difficulty. The total revenue using the break-evenpoints will not result in the same percentages (dollar volume of product/total revenue) asoriginally stated in the problem. A more complex model is available to do this (see p. 308 JayHeizer and Barry Render,Principles of Operations Management, 9th ed., Upper Saddle River,NJ: Prentice Hall, 2014).

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2-1CHAPTER2Probability Concepts and ApplicationsTEACHING SUGGESTIONSTeaching 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 thebig 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 andbased on personalexperiences. Other times they are objectively based on logical observationssuch as the roll of a die. Often, probabilities are derived from historical data—ifwe can assumethe 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 instatistics. 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 isP(AorB) =P(A) +P(B)–P(AandB). Students must understand why we subtractP(AandB). Explain that the intersectionhas beencounted twice.Teaching Suggestion 2.5:Expected Value of a Probability Distribution.A probabilityis often described by its mean and variance. These important termsshould be discussed with such practical examples as heights or weights of students. But studentsneed to be reminded that even if most of the men in class (or the United States) have heightsbetween 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 normalis to a large number of processes in our lives (forexample, filling boxes of cereal with 32 ounces of cornflakes). Each normaldependson the mean and standard deviation. Discuss Figures 2.7 and 2.8 to show how these relate to theshape and position of a normal.

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2-2Teaching Suggestion 2.7:Three Symmetrical Areas Under the Normal Curve.Figure 2.13 is very important, and students should be encouraged to truly comprehend themeanings of ±1, 2, and 3 standard deviation symmetrical areas. They should especially know thatmanagers often speak of 95% and 99% confidence intervals, which roughly refer to ±2 and 3standard 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 canrelate to it. Students are typically curious about the chances of reaching certain scores. Gothroughat leasta half-dozen examples until it’s clear that everyone can use Table 2.10. Studentsget especially confused answering questions such asP(X85) since the standard normal tableshows only right-hand-side (positive)Zvalues. The symmetry requires special care.

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2-3ALTERNATIVE EXAMPLESAlternative 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 issimilar to the future, here are the probabilities:SalesNo. DaysProbability8100.3339120.4001060.2001120.067Total301.000Alternative Example 2.2:Grades received for a course have a probability based onthe professor’s grading pattern. Here are Professor Ernie Forman’s BA205 grades forthe past five years.OutcomeProbabilityA0.25B0.30C0.35D0.03F0.02Withdraw/drop0.051.00These grades are mutually exclusive and collectively exhaustive.Alternative Example 2.3:P(drawing a 3 from a deck of cards) = 4/52 = 1/13P(drawing a club on the same draw) = 13/52 = 1/4These are neither mutually exclusive nor collectivelyexhaustive.

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2-4Alternative Example 2.4:In Alternative Example 2.3 we looked at 3s and clubs. Here is theprobability for 3orclub:P(3orclub) =P(3) +P(club)–P(3andclub)= 4/52 + 13/52–1/52= 16/52 = 4/13Alternative 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 andnon-U.S. citizens.A name is randomly selected from the class roster and it is female. What is the probabilitythat the student is a U.S. citizen?U.S.Not U.S.TotalF10616M12214Total22830P(U | F) = 10/16 = 5/8Alternative Example 2.6:Your professor tells you that if you score an 85 or better on yourmidterm exam, there is a 90% chance you’ll get an A for the course. You think you have only a50% chance of scoring 85 or better. The probability thatbothyour score is 85 or betterandyoureceive an A in the course isP(A and 85) =P(A85)P(85) = (0.90)(0.50) = 0.45= a 45% chance

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2-5Alternative Example 2.7:An instructor is teaching two sections (classes) of calculus. Eachclass has 24 students, and on the surface, bothclasses appear identical. One class, however,consists of students who have all taken calculus in high school. The instructor has no idea whichclass is which. She knows that the probability of at least half the class getting As on the firstexam 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) = 0.5P(advanced class chosen) = 0.5P(1/2 Asregular class) = 0.25P(1/2 Asadvanced class) = 0.50P(1/2 As and regular class)=P(1/2 Asregular )P(regular)= (0.25)(0.50) = 0.125P(1/2 As and advanced class)=P(1/2 Asadvanced)P(advanced)= (0.50)(0.5) = 0.25SoP(1/2 As) = 0.125 + 0.25 = 0.375()()()advanced and1/2 Asadvanced1/2 As1/ 2 As0.252 / 30.375PPP===So there is a 66% chance the class tested was the advanced one.

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2-6Alternative Example 2.8:Students in a statistics class were asked howmany “away” footballgames they expected to attend in the upcoming season. The number of students responding toeach possibility is shown below:Number of gamesNumber of students54043032021010100A probabilityof the results would be:Number of gamesProbabilityP(X)50.4 = 40/10040.3 = 30/10030.2 = 20/10020.1 = 10/10010.0 = 0/1001.0 = 100/100Thisdiscreteprobabilityiscomputedusingtherelativefrequencyapproach.Probabilities are shown in graph form below.

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2-7Alternative Example 2.9:Here is how the expected outcome can be computed for the questionin Alternative Example 2.8.()()()()51221iiiiExx PXx PXx PX===++x3P(X3) +x4P(X4) +x5P(X5)= 5(0.4) + 4(0.3) + 3(0.2) + 2(0.1) + 1(0)= 4.0Alternative Example 2.10:Here is how variance is computed for the question in AlternativeExample 2.8:()()()521varianceiiixExP 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= 1.0The standard deviation isvariance1=== 1Alternative Example 2.11:The length of the rods coming out of our new cuttingmachine canbe said to approximate a normalwith a mean of 10 inches and a standard deviationof 0.2 inch. Find the probability that a rod selected randomly will have a lengtha.of less than 10.0 inchesb.between 10.0 and 10.4 inchesc.between 10.0 and 10.1 inchesd.between 10.1 and 10.4 inchese.between 9.9 and 9.6 inchesf.between 9.9 and 10.4 inchesg.between 9.886 and 10.406 inches

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2-8First compute the standard normal, theZ-value:xz−=Next, find the area under the curve for the givenZ-value by using a standard normaltable.a.P(x10.0) = 0.50000b.P(10.0x10.4) = 0.97725–0.50000 = 0.47725c.P(10.0x10.1) = 0.69146–0.50000 = 0.19146d.P(10.1x10.4) = 0.97725–0.69146 = 0.28579e.P(9.6x9.9) = 0.97725–0.69146 = 0.28579f.P(9.9x10.4) = 0.19146 + 0.47725 = 0.66871g.P(9.886x10.406) = 0.47882 + 0.21566 = 0.69448SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS2-1.There are two basic laws of probability. First, the probability of any event or state of natureoccurring must be greater than or equal to zero and less than or equal to 1. Second, the sum ofthe 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. Eventsare collectively exhaustive if the list of outcomes includes every possible outcome. An exampleof mutually exclusive events can be seen in flipping a coin. The outcome of any one trial caneither be a head or a tail. Thus, the events of getting a head and a tail are mutually exclusivebecause only one of these events can occur on any one trial. This assumes, of course, that thecoin does not land on its edge. The outcome of rolling the die is an example of events that arecollectively 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 beidentified without performing a series of trials, we can often logically determine what theprobabilities 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(AorB) =P(A) +P(B)The addition law for events that are not mutually exclusive isP(AorB) =P(A) +P(B)–P(AandB)In this case we reduce the probability by subtracting the chance of both events occurringtogether.

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