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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Document preview page 1

Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 1

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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual

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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 1 preview image1CHAPTER 1INTRODUCTION TO OPERATIONS MANAGEMENTCONCEPTUAL QUESTIONS1.Answer:SloganAemphasizesFit;SloganBemphasizesTiming;SloganCemphasizesPrice;SloganDemphasizesLocation;andSloganEemphasizesPerformance.2.Answer:D.Customer Satisfaction3.Answer:False4.Answer:False5.Answer:Yes6.Answer:C.Fatigue7.Answer:B.How much do we pay the CEO?PROBLEMS AND APPLICATIONS1.Answer:C.Location and Time
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 2 preview image
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 3 preview image22.Answer:B. Fit.Feedback:Fitis a subcomponent of the consumption utility that captures how wellthe product or service matches with the unique characteristics of a given consumer.3.Answer:A.Performance.Feedback:Performance is a subcomponent of the consumption utility that captureshow much an average consumer desires a product or service.4.Answer:D. Timing.Feedback:The "to-go" section is designed for customers to purchase food quicklyand move on their way to their departure gate. The primary focus is on the speed ofservice, which addresses the timing element of the customer utility function.5.Answer:A.Performance.Feedback:The "special edition" coupe has features that give it a higher level ofperformance compared to the standard model. As a result, the"special edition"vehicle clearly emphasizes the performance dimension of the customer utilityfunction.6.Answer:Hotels B, C, and D.Feedback:The only hotel that is Pareto dominated is hotel A-all other are on theefficient frontier. Hotel A is Pareto dominated by hotel B, as B is both cheaper andbetter.7.Answer:Carriers A, B, and C.Feedback:Carrier D is the only one not on the efficient frontier because it isdominated by Carrier A on both measures.
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 4 preview image38.Answer:Dry Cleaner B.Feedback:Dry Cleaner Bis the only one not on the efficient frontier because it isdominated by both Dry Cleaner A and C on both measures.9.Answer:C.Empty bottles.Feedback:The bottle is a part of the soft drink product.10.Answer:B.Bottling machines.Feedback:The bottling machine is a resource used to manufacture the soft drinks.The other items are inputs and are part of the product.11.Answer:C.Feedback:The only item in the list that is a material used in thedoctor's officeoperations is a needle. The rest of the items would be considered resources.12.Answer:B.Feedback:The only item in the list that is used to transform inputs to outputs is theprojector. The rest of the items in the list would beclassified as inputs because theyare materials and supplies used in the operations.13.Answer:D.Feedback:Employee work schedules set a week in advance14.Answer:C.Feedback:Customers incorrectly list information on forms
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 5 preview image415.Answer:A andD.Feedback:Convenience refers to the questions of “when” and “where” the demandwill be fulfilled.16.Answer:B.Feedback:The operational efficiency will affect the price that the firm is able tocharge for its product and service tomaximize its profitability.17.Answer:C.Feedback:Theproductorservicecharacteristicswillaffecthowmucheachconsumerwillliketheoverallproductorservice,whichismeasuredbyconsumption utility.CASEThere is no case for thischapter.
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 6 preview image1CHAPTER2INTRODUCTION TO PROCESSESCONCEPTUAL QUESTIONS1.Answer:C. Number of customers.Feedback:The number of workers, cash registers, and suppliers are unlikely tochange much over the course of a month and do not “flow”through the process ofthe hardware store.2.Answer:D. The number of patients.Feedback:Physicians, beds, and square footage are unlikely to change much over thecourse of a month and do not “flow” through the process of a hospital.3.Answer:The flow rate is 1,000 passengers per day and the flow time is 5 days.4.Answer:The inventory is 15 voters.Feedback:The flow rate is 1,800 / 10 = 180 per hour, or 180 / 60 = 3 per minute.The flow time is 5 minutes.5.Answer:B.Feedback:The flow rate into a process must equal the flow rate out of a process.6.Answer:False.Feedback:Little’s Law applies even if there are fluctuations in inventory, flow rates,and flow times.
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 7 preview image2PROBLEMS AND APPLICATIONS1.Answer:D.Feedback:The number of customers is the appropriate flow unit for processanalysis. Theemployees are resources, and the other two measures are unlikely tochange from week to week.2.Answer:B.Feedback:The number of tax returns completed each week reflects the mainoperation of the accounting firm during tax season. The accountants are resources;the customers with past-due invoices reflect the accounts receivable process andnot the main operation; and the reams of paper received are a result of the firm’spurchasing policies and not necessarily the main operation.3.Answer:A and Dare correctFeedback:The gasoline pumps and employees are resources, not flow units.4.Answer:0.4 callers per minuteFeedback:8 calls divided by 20 minutes = 0.4 calls per minute.5.Answer:4 minutesFeedback:To calculate the flow time of the callers, subtract the callers departuretime from his or her arrival time. 32 total minutes divided by 8 callers = 4 minutes.6.Answer:0.1667 customers per minuteFeedback:Flow rate = 10 customers divided by 60 minutes = 0.1167.7.Answer:8.6 minutes
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 8 preview image3Feedback:To calculate the flow time of the customers, subtract the customersdeparture time from his or her arrival time. 86 total minutes divided by 10customers = 8.6 minutes.8.Answer:4 minutesFeedback:To solve this problem, use Little’s Law. Inventory = Flow rate × Flow time.10 people in line (average inventory) = 2.5 flow rate x flow timeFlow time = 4 minutesThe flow rate is 300 customers divided by 120 minutes = 2.59.Answer:90,000 wafersFeedback:100 per second x 60 seconds per minute x 15 minutes = 90,00010.Answer:360 skiersFeedback:1,800 skiers divided by 60 minutes per hour (flow rate) x 12 minutes(flow time) = 360 skiers11.Answer:8,539 visitorsFeedback:Flow rate = 3,400,000 visitors divided by 365 days = 9,315.07 visitors perdayFlow Rate = 22 hours/ 24 hours per day = .9167 dayInventory = 9,315.07 (flow rate) x 0.9167 (flow time) = 8539.12 visitors per day12.Answer:900,000 patientsFeedback:6 months (flow time) x 150,000 new patients per month = 900,000patients13.Answer:20 chat sessionsFeedback:Flow rate = 240 chats divided by 30 employees = 8Flow time = 5 minutes divided by 60 minutes = 0.833 hour
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 9 preview image4Inventory = Flow Rate x Flow Time, 8 x 0.833= 0.6667 x 30 employees = 20 chats14.Answer:840 unitsFeedback:4,200 units divided by (12 minutes/60 minutes) = 840 units15.Answer:120 skiersFeedback:1,200 beds divided by 10 days = 120 new skiers per day.16.Answer:7.5minutesFeedback:To solve this problem, use Little’s Law. Inventory = Flow rate × Flow time.30 people in line (average inventory) = 240 customers/ 60 minutes (flow rate) xflow time.Flow time = 7.5 minutes17.Answer:8 yearsFeedback:120 associates = 15 new employees x flow time. Flow time = 8CASEAlthough the analysis of the case is relatively simple, the intuition is not always easy tograsp–many students will intuitively believe that the capacity of the faster lift should begreater than the capacity of the slower lift. The main lesson in this case is to get students tounderstand why that intuition is not correct.To begin the case discussion, ask the students their opinion as to who is correct, Mark(unloading capacity should be twice as high on the detachable lift) or Doug (the unloadingcapacity should be the same on the two lifts). Hopefully there are students who supporteach opinion.To resolve the question, begin with the simple process flow diagram:
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 10 preview image5Ask the question “Do all of the skiers that get on the lift at the bottom get off the lift at thetop?” Of course, the answer is “We would hope so!”. So “What does that mean about howthe rate of skiers getting on the lift, Ron, is related to the rate of skiers getting off the lift,Roff?” And the answer there must be that they are equal! If the rate on where faster thanthe rate off, the number of people on the lift would grow and grow and grow. We know thatcan’t happen. Similarly, if the rate off exceeded the rate on, then the number of people onthe lift would shrink and shrink and shrink, leaving the lift eventually with nobody. Whichalso doesn’t happen.So we can add to our process flow diagram:Now it is time to compare the two lifts. We can draw the process flow for each of them,emphasizing that the rate on for each must equal the rate off:LiftSkiersSkiersLiftSkiersSkiersRonRoff=Slow LiftRsRsFast LiftRfRf
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 11 preview image6Now ask students “How can we compare the rates across the two types of lifts?” Theanswer is given in the case–we are told that the rate skiers load onto the slow (fixed grip)lift is the same as the rate they load onto the fast (detachable) lift. That means that Rs= Rf.And that means that the rates that they onload skiers at the top must be the same!Thus, Doug is correct–both lifts have the same capacity to unload skiers at the top eventhough one is faster than the other.And this brings us to Jessica’s question–so what is the difference between the two lifts? Ifyou ask students this question, the likely first response is that skiers spend less time on thefaster lift. And that is correct. But are there other differences? Actually, there are twoadditional differences worth mentioning. The first comes from Little’s Law and the 2ndonerequires a deeper understanding of this process.The first obvious difference is the number of skiers on the lift. According toLittle’s Law, I =R x T. So if the two lifts have the same R, but the faster lift has a smaller T, then the fasterlift must have a smaller I as well:Slow LiftRsRsFast LiftRfRf==Fast LiftTf= seconds on liftIf= # of skiersRRTf< TsIf= R x Tf< R x Ts= Is
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 12 preview image7So fewer people are on the faster lift and they spend less time on the lift but the faster liftand the slower lift bring skiers to the top at the same rate.If students can’t get the next difference between the two lifts, then you can prompt themwith the following question “If the faster lift has fewer skiers than the slower lift, thenwhere are the additional skiers?” Or put another way: “If the ski area attracts a certainnumber of skiers but the faster lift has fewer skiers on it, then where are the other skiers?”The answer is that they are on the slopes! That means that adding a faster lift takes skiersoff the lift but they don’t disappear. Instead, they are on the only other place they can be,the slopes. Which means, somewhat counter-intuitively, that adding a faster lift makes theslopes more crowded (holding the total number of skiers fixed).
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 13 preview image1CHAPTER3PROCESS ANALYSISCONCEPTUAL QUESTIONS1.Answer:B.How long does it take the office to process an application?2.Answer:D.Chefs3.Answer:C.Paying the bill4.Answer:B.Theyare reciprocals of each other.5.Answer:A.Capacity constrained6.Answer:C.The cycle time7.Answer:D.1.008.Answer:B.Equal to9.Answer:A.Machine-paced10.Answer:B.Shorter than the average flow time
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 14 preview image2PROBLEMS AND APPLICATIONS1.(a)Answer:4 customers per hourFeedback:60 minutes/15 minutes = 4 customers per hour;(b)Answer:2 customers per hour(c)Answer:50 percentFeedback:2 demand / 4 capacity = 50%;(d)Answer:30 minutes percustomerFeedback:Cycle time = 1/flow rate, 2/ 60 minutes= 30 minutes per customer2.(a)Answer:72 visits per 9 hour work dayFeedback:12 nurses × 9 hours/1.5 hours per visit = 72 visits;(b)Answer:83 percentFeedback:60 demand/ 72 capacity = 83 utilization;(c)Answer:9 minutes per patientFeedback:(9 hours × 60 minutes) / 60 patients = 9.0 minutes per patient3.(a)
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 15 preview image3(b)Answer:0.333 units per minuteFeedback:1unit / 3 minutes per unit = 0.333 units per minute.(c)Answer:Resource 1Feedback:Resource 1 has the longest processing time.(d)Answer:50%Feedback:Demand = 60 minutes / 6 minutes Resource 1bottleneck = 10 units,Capacity 60 minutes / 3 minutes = 20 unitsUtilization = 10 units / 20 units = 50%(e)Answer:1,208 minutesFeedback:200 units× 6 minutes Resource 1 + 3 minutes Resource 2 + 5 minutesResource 3 = 1,208 minutes.4.(a)(b)Answer:1,200 bottles per hour(c)Answer:Packaging(d)BottlingApplyaLidLabelingPackaging
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 16 preview image4Answer:900 bottles per hour(e)Answer:75%(f)Answer:1,996 + 16 =2,012 seconds5.a)(b)Answer:20 patients per hourFeedback:10 workers × (60 minutes per hour/30 minutes per patient) = 20 patientsper hour.(c)Answer:The dental assistantsFeedback:The dental assistants have the lowest capacity of 12 patients per hour, 3dental assistants × (60 minutes per hour/ 15 minutes per patient) = 12 patients perhour.(d)Answer:12 patientsper hourFeedback:TheFlow Rate = 12 patients per hour, 3 dental assistants × (60 minutesper hour/ 15 minutes per patient) = 12 patients per hour.(e)Answer:50%Feedback:Receptionists capacity = 2 receptionists × (60 minutes per hour/5minutes per patient) = 24 patients per hour.Self-ServeReceptionistDentalAssistantDentist
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 17 preview image5Utilization = 12 patients per hour at the bottleneck/24 patients per hour capacity =50%(f)Answer:100 minutesFeedback:[(5/2) at bottleneck x 20] + 5 Self-Serve + 15 Dental Assistant + 30Dentist = 100 minutes.6.(a)Answer:5 batches perhourFeedback:Since there is only one machine at each process step, the capacity is thereciprocal of the activity time. The capacity of the baking process step is 1 / 12batches per minute * 60 minutes per hour = 5 batches per hour.(b)Answer:CoolingFeedback:Since there is only one machine at each process step, the bottleneck is thestep with the most work to do.(c)Answer:3.33 batches per hourFeedback:The process flow rate is the capacity of the bottleneck, which is thecooling process.The process flow rate is 1 / 18 batches per minute * 60 minutes perhour = 3.33 batches per hour.(d)Answer:83 percentFeedback:The capacity of the mixing process step is 1 / 15 batches per minute * 60minutes per hour = 4 batches per hour. The process flow rate is 3.33 batches perhour. The utilization is the flow rate divided by the capacity of the process step =3.33 batches per hour / 4 batches per hour = 83%.(e)Answer:15 hours to complete 50 batches.Feedback:The time to complete Xunits in a full system is X * Cycle time. The cycletime is 1 / 3.33 hours, so it takes 50 / 3.33 = 15 hours to complete 50 batches.
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 18 preview image67.(a)Answer:Office managersFeedback:The capacity of each resource is the number of employees / activity time.The capacity for the office managers is 2 / 4 = 0.50 loans per hour, which is less thanthe capacities of either of the other two resources.(b)Answer:20 loans per weekFeedback:The flow rate is the capacity of the bottleneck, which is 2 / 4 = 0.50loansper hour * 8 hours per day * 5 days per week = 20 loans per week.(c)Answer:90 percentFeedback:If the demand is 18 loans per week, this is the process flow rate because itis lower than the process capacity. The utilization of the office managers’ resource isFlow rate / Resource capacity = 18 loans per week / 20 loans per week = 90%.(d)Answer:30 hoursFeedback:This is a worker-paced process, so the first loan will take 1 + 7 + 4 = 12hours to complete. The remaining loans have a cycle time of 1 / 0.50 = 2 hours perloan. This means that the remaining loans will take an additional 9 * 2 = 18 hours tocomplete. The total time to complete 10 loans is 12 + 9 * 2 = 30 hours.8.Answer:54 hoursFeedback:Step 3 is the bottleneck step of the process, so the machine-paced processwill be set at a speed of 1/2 hour per unit at each step. The first unit will take 4 * 1/2= 2 hours to complete. The remaining 104 units will have a cycle time of 1/2 hour.The total time required to complete a batch of 105 units is 2 + 104 * 1/2 = 54 hours.9.Answer:128 minutes to serve 20 customersFeedback:The bottleneck is the first step of the process, which means that the cycletime for all customers other than the first one is 5 minutes. The first customer takes
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 19 preview image75 + 15 + 10 + 3 = 33 minutes to complete service. As a result, it takes a total of 33 +19 * 5 = 128 minutes to serve 20 customers.10.(a)Answer:12 customers per hourFeedback:The bottleneck is the Wash process, with a capacity of 12 cars per hour.The flow rate is the minimum of the capacity of the bottleneck and the demand.(b)Answer:60 percentFeedback:Utilization is the flow rate divided by the capacity of the process step. Ifthe demand is 15 cars per hour, the flow rate is 12 cars per hour, because the flowrate is the minimum of the capacity of the bottleneck and the demand. The capacityof the machine that performs the Wax process is 1 / 3 * 60 = 20 cars per hour. Theutilization of this machine is 12 / 20 = 60%.11.(a)Answer:12 per hourFeedback:The bottleneck resource is the license processing station. Each machinecan process 4 customers per hour, and there are three machines; thus, the flow ratethrough this stage (and the process as a whole) is 12 customers per hour. All of theother steps have higher capacities.(b)Answer:15 per hourFeedback:Thecurrent bottleneck is the license processing station. Each machine hasa capacity of 4 customers per hour, so adding one machine would increase thecapacity of this station to 16 customers per hour. The cashiers each have a capacityof 7.5 customers per hour, so the capacity of this resource is only 15 customers perhour. The cashiers will now be the bottleneck, and the process flow rate is 15customers per hour.
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 20 preview image8CASEThe chapter’s ‘Connection side-box’ showcases the Tesla plant. The key goal of this mini-case is that students learn how to take a complex process as shown in the video and thenabstract this information to something simple as a process flow diagram.The process flow diagram can take various forms / there does not exist a uniquely rightanswer. We suggest a diagram such as the following:As far as the calculations are concerned:Cycle time: the process operates for 80 hours per week. It produces 500 cars per week. So,the cycle time is 80 hours / 500 cars = 0.16 hours/car = 9.6 minutes/carThe flow time of the process is the 3-5 days it takes a car to journey through the processInventory can exist between steps and at the beginning as raw materials. However, most ofthe inventory is likely to sit in each of the boxes as work in process inventory. This includesmany cars that are half assembled as well pieces of metal that are waiting to be joinedtogether. The total inventory is I = R*T = 100 cars per day * 4 days = 400 cars.
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 21 preview image1CHAPTER 4PROCESS IMPROVEMENTCONCEPTUAL QUESTIONS1.Answer:A. Costs of direct labor would be 50% lower2.Answer:B. The labor content stays the same3.Answer:A. The idle time would decrease4.Answer:B. No5.Answer:A. True6.Answer:A. Takt time only depends on demand, not capacity. Cycle time doesdepend on capacity.7.Answer:A. True8.Answer:C. The takt time decreases9.Answer:A. The target manpower increases10.Answer:B. The target manpower doubles
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 22 preview image211.Answer:C.Increasing wages for production workers12.Answer:B. False13.Answer:B. False14.Answer:B. False15.Answer:A. True16.Answer:B.LargeFeedback:Low variable costs result in a high unit margin, so every additionalcustomer yields asignificant contribution to profitability.17.Answer:A.SmallFeedback:High variable costs result in a low unit margin, which reduces the impactof each additional customer on profitability.18.Answer:B.FalseFeedback:Efficiency can result in higher revenue as well because the operation canserve more customers.PROBLEMS AND APPLICATIONS1.Answer:$800 per customerFeedback:The Direct Labor cost is (8 hours × $200 per hour)/2 customers = $800per customer.
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 23 preview image32.(a)Answer:$3.60 per unitFeedback:Resource 1 is the bottleneck center at 6 minutes per unit. The DirectLabor cost is (6 minutes per resource × 3 resource centers/60 minutes per hour) ×$12 per hour = $3.60(b)Answer:14 minutes per unitFeedback:Labor content = 6 minutes Resource 1 + 3 minutes Resource 2 + 5minutes Resource 3 = 14 minutes(c)Answer:1 minute per unitFeedback:The idle time at Resource 3 = 6 minutes Resource 1 (bottleneck)–5minutes Resource 3 = 1 minute idle time(d)Answer:77.78percentFeedback:Labor Utilization = 14 demand/18 capacity = 77.78%(e)Answer:3 minutes per unitFeedback:Takt Time = 1/Demand rate = 1/(20 customers/60 minutes) = 3 minutes(f)Answer:4.67minutes per unitFeedback:The target manpower = 14 Labor Content/3 minutes Takt Time = 4.673.(a)Answer:50 minutes per patient(b)Answer:$95.83 per patientFeedback:$1,150 per hour / 12 patients per hour = $95.83per patient4.(a)Answer:60 minutes per customer
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 24 preview image4Feedback:Labor content = 10 minutes Activity 1 + 10 minutes Activity 2 + 10minutes Activity 3 + 25 minutes Activity 4 + 5 minutes Activity 5 = 60 minutes(b)Answer:66.67 percentFeedback:60 minutes labor content / (3 employees x 30 minutes bottleneck) =66.67%(c)Answer:$30 per customerFeedback:$60 per hour / 2 customers per hour = $30 per customer(d)Answer:$26.67 per customerFeedback:$80 per hour / 3 customers per hour = $26.67 per customer(e)Answer:$25 per customerFeedback:$60 per hour /(60/25) customers per hour = $25 per customer5.(a)Answer:395 seconds per watchFeedback:Labor content = 68 seconds Station A + 60 seconds Station B + 70 secondsStation C + 58 seconds Station D + 75 seconds Station E + 64 second Station F = 395seconds(b)Answer:72 seconds per watchFeedback:Takt Time = 1/Demand rate = 1/(50 watches/3600 seconds per hour)=72 seconds(c)Answer:5.49 workersFeedback:395 / 72 = 5.49(d)Answer:0.0watches per hourFeedback:There is noimpact.(e)Answer:D.Movecosmetic inspection(step 14) to station D6.(a)Answer:$1.25 per unitFeedback:5 * 15 / 60 = $1.25 per unit(b)Answer:$25 per hour
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 25 preview image5Feedback:Revenues are $360 per hour and costs are $335 per hour so profits are$25 per hour.(c)Answer:$31 per hourFeedback:Savings = 60 units per hour x $0.10 = $6.00. $6.00 savings + $25 per hour(see part B) = $31.00(d)Answer:$45 per hourFeedback:$20 fixed cost savings + $25 (see part B) = $45 per hour.(e)Answer:$52.27 per hourFeedback:It now takes 55 seconds to make each unit. 3600 seconds per hourdivided 55 seconds per unit = 65.4545 units. Revenues are $392.73-$65.46 parts-$75 labor-$200 fixed coast = 52.CASEThe mini-case for this chapter is the Xootr manufacturing process.The process flowdiagrams are as follows:We compare the two proposals assuming that there are no benefits of specialization.ProposalSpecializedOne worker does it allNumber of workers66Wages per hour6*126*12Processing times6.5., 6.5, 5.5, 5.5, 4, 432BottleneckFirst resourceFirst resource
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 26 preview image6Capacity1/6.5 units per minute6/32 units per minuteCost of direct labor72 / (60/6.5)72 / (6/32)Idle time0, 0, 1, 1, 2.5, 2,50Labor content32 minutes per unit32 minutes per unitLaborutilization=32/(32+7)1Variables that might differ: the question is if it would be possible to cut the processingtimes reflecting specialization gains in the first proposal. Also, the first proposal could beimproved if the line could be better balanced
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 27 preview image1CHAPTER 5PROCESS ANALYSIS WITH MULTIPLE FLOW UNITSCONCEPTUAL QUESTIONS1.C.62.A.23.A.True4.B.False.This really depends on the available capacity.5.B.False.When demand exceeds capacity, the implied utilization is biggerthan 100%.6.A.True7.C.40 / 0.5 = 80 units per day8.B.100%9.B.False.We have to also look at the demand matrix.10.C.The yield decreases.11.D.Minutes or hours of work
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 28 preview image212.A.True13.A.True14.B.False15.D.Scrap is a special case of rework in which flow units have to repeat all resourcesin the process up to the defect.16.A.TruePROBLEMS AND APPLICATIONS1.(a)Answer:50 per week(b)Answer:10 per week(c)Answer:40 per weekFeedback:Weuse the first two steps from Figure SUM1.Step 1: Compute the demand matrix(Group 1)(Group 2)(Administrator)0.2 * 500.8 * 50D =(Senior ac.)0.2 * 500(Junior ac.)00.8 * 50
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 29 preview image3Step2:Computethetotaldemandrates,whichare50perweekfortheadministrator, 10 for the senior accountant, and 40 for the junior accountant.2.(a)Answer:AdministratorFeedback:We continue with Step 3 in Figure SUM1.Step 3: We compute the capacity levels asfollows:Administrator: 1/20 units / minute = 3 units per hour = 120 units per weekSenior accountant: 1/40 units / minute = 1.5 units per hour = 60 units per weekJunior accountant: 1/15 units / minute = 4 units per hour = 160 units per weekStep 4:The levels of implied utilization are:Administrator: 50 units per week / 120 units per week = 0.4166Senior accountant: 10 units per week / 60 units per week = 0.1667Junior accountant: 40 units per week / 160 units per week = 0.25So the administrator is the bottleneck.(b)Answer:The flow rate 10 new cases per week and 40 repeat customers perweek; the capacity for new customers is 24 cases per week and for repeatcustomers is 96 cases per weekFeedback:Step 5: The process is demand constrained and thus the flow rates are thedemand rates, 10 new cases per week and 40 repeat customers per week.The capacity levels are found by dividing the demand rates by the highest impliedutilization. Thus, the capacity of this process with a 20:80 mix would be:New customers: 10 / 0.4166 = 24 cases per weekRepeat customers: 40 / 0.4166 = 96 cases per week3.(a)Answer:The written examFeedback:We follow the 4-step process outlined in Figure SUM2.Step 1: We find the demand matrix by starting with one good unit (a student thatpasses). Given the 40% failure rate, we need 1 / 0.6 = 1.67 students to take the road
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 30 preview image4exam to get one succeed. In the same way, we compute that we need 1.667/ 0.85 =1.96 students to take the written exam, and 1.96 / 0.9 = 2.18 students to show up. Sothe demand matrix is:(ID)(2.18)D =(Written)(1.96)(Road)(1.67)Step 2: The capacity levels are simply computed as the number of workers dividedby the processing times:Capacity(ID) = 4/5 units / minute = 384 units per dayCapacity(Written) = 2/3 units / minute = 320 units per dayCapacity(Road) = 15/20 units / minute = 360 units per dayStep 3: Compute the levels of implied utilizationIU(ID) = 2.18 / 0.8 = 2.72IU(Written) = 1.96 / 0.67 = 2.94IU(Road) = 1.67 / 0.75 = 2.22So the bottleneck is at the written exam.(b)Answer:163 good units per dayFeedback:Step 4: Find the capacity of the process in terms of good units by dividing1 good unit of output bythe highest implied utilization:Capacity(ID) = 1 / 2.94 = 0.34 good units per minute = 163.2 good units per day4.We use the 5-step process outlined in Figure SUM3.(a)Answer:Resource 3Feedback:Step 1: Compute the work-load matrix(Type A)(Type B)(Type C)(Resource 1) (40 * 550 * 560 * 5)(Resource 2) (40 * 450 * 460 * 5)WL =(Resource 3) (40 * 1500)(Resource 4) (050 * 350 * 3)(Resource 5) (40 * 650 * 640 * 4)Step 2: Add up thework-loads for each resource:
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Operations Management, (Mcgraw-hill Education Operations and Decision Sciences) 1st Edition Solution Manual - Page 31 preview image5Resource 1: 750 minutes per dayResource 2: 660 minutes per dayResource 3: 600 minutes per dayResource 4: 300 minutes per dayResource 5: 700 minutes per dayStep 3: Compute the available time per day at each resourceResource 1: 2 * 480 = 960 minutes per dayResource 2: 2 * 480 = 960 minutes per dayResource 3: 1 * 480 = 480 minutes per dayResource 4: 1 * 480 = 480 minutes per dayResource 5: 2 * 480 = 960 minutes per dayStep 4: Compute the implied utilization levels as demand rate in minutes of workdivided by the available time in minutes of work:Resource 1: 750 minutes per day / 960 minutes per day = 0.78Resource 2: 660 minutes per day / 960 minutes per day = 0.69Resource 3: 600 minutes per day / 480 minutes per day = 1.25Resource 4: 300 minutes per day / 480 minutes per day = 0.625Resource 5: 700 minutes per day / 960 minutes per day = 0.73Resource 3 is the bottleneck because it has the highest implied utilization.(b)Answer:The flow rate for A is 32units per day; B is 40 units per day; and C is48 units per day.Feedback:Step 5: The process is capacity constrained. So we have to divide all flowsby 1.25:Flow A = 40 / 1.25 = 32 units per dayFlow B = 50 / 1.25 = 40 units per dayFlow C = 60 /1.25 = 48 units per day5.We use the framework outlined in Figure SUM4.(a)Answer:Step 3Feedback:Step 1: We compute the work-load matrixGoodRework1(0.7 * 50.3 * [5 + 5])2(0.7 * 60.3 * [6 + 6])WL =3(0.7 * 30.3 * [3 + 3])4(0.7 * 40.3 * 4)
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Operations Management
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Operations Management, (Mcgraw-Hill...

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