《運(yùn)營(yíng)管理》課后習(xí)題答案

Solutions_Problems_OM_11e_StevensonChapter 02 - Competitiveness, Strategy, and Productivity3.(1)(2)(3)(4)(5)(6)(7) WeekOutputWorker Cost$12x40Overhead Cost 1.5Material Cost$6Total CostMFP (2) (6)130,0002,8804,3202,7009,9003.03233,6003,3605,0402,82011,2202.99332,2003,3605,0402,76011,1602.89435,4003,8405,7602,88012,4802.84*refer to solved problem #2Multifactor productivity dropped steadily from a high of 3.03 to about 2.84.4.a.Before: 80 5 = 16 carts per worker per hour.After: 84 4 = 21 carts per worker per hour.b.Before: ($10 x 5 = $50) + $40 = $90; hence 80 $90 = .89 carts/$1.After: ($10 x 4 = $40) + $50 = $90; hence 84 $90 = .93 carts/$1.c.Labor productivity increased by 31.25% (21-16)/16).Multifactor productivity increased by 4.5% (.93-.89)/.89).*Machine ProductivityBefore: 80 40 = 2 carts/$1.After: 84 50 = 1.68 carts/$1.Productivity increased by -16% (1.68-2)/2)Chapter 03 - Product and Service Design6.Steps for Making Cash Withdrawal from an ATM1.Insert Card: Magnetic Strip Should be Facing Down2.Watch Screen for Instructions3.Select Transaction Options:1) Deposit2) Withdrawal3) Transfer4) Other4.Enter Information:1) PIN Number2) Select a Transaction and Account3) Enter Amount of Transaction5.Deposit/Withdrawal:1) Depositplace in an envelope (which youll find near or in the ATM) and insert it into the deposit slot2) Withdrawallift the “Withdrawal Door,” being careful to remove all cash6.Remove card and receipt (which serves as the transaction record)8.Technical RequirementsIngredientsHandlingPreparationCustomer RequirementsTasteAppearanceTexture/consistencyChapter 04 - Strategic Capacity Planning for Products and Services2.Actual output = .8 (Effective capacity)Effective capacity = .5 (Design capacity)Actual output = (.5)(.8)(Effective capacity)Actual output = (.4)(Design capacity)Actual output = 8 jobsUtilization = .410. a.Given:10 hrs. or 600 min. of operating time per day.250 days x 600 min. = 150,000 min. per year operating time.Total processing time by machineProductABC148,00064,00032,000248,00048,00036,000330,00036,00024,000460,00060,00030,000Total186,000208,000122,000You would have to buy two “A” machines at a total cost of $80,000, or two “B” machines at a total cost of $60,000, or one “C” machine at $80,000.b.Total cost for each type of machine:A (2): 186,000 min 60 = 3,100 hrs. x $10 = $31,000 + $80,000 = $111,000B (2) : 208,000 60 = 3,466.67 hrs. x $11 = $38,133 + $60,000 = $98,133C(1): 122,000 60 = 2,033.33 hrs. x $12 = $24,400 + $80,000 = $104,400Buy 2 Bsthese have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3adf752b4c4e9h5i6g3.Desired output = 4Operating time = 56 minutesTask# of Following tasksPositional WeightA423B320C218D325E218F429G324H114I05a.First rule: most followers. Second rule: largest positional weight. Assembly Line Balancing Table (CT = 14)Work StationTaskTask TimeTime RemainingFeasible tasks RemainingIF59A,D,GA36B,GG6IID77B, EB25CC41IIIE410HH91IVI59b. First rule: Largest positional weight.Assembly Line Balancing Table (CT = 14)Work StationTaskTask TimeTime RemainingFeasible tasks RemainingIF59A,D,GD72IIG68A, EA35B,EB23IIIC410EE46IVH95II5c.abdcf1. egh1. 4.a.l.2. Minimum Ct = 1.3 minutesTaskFollowing tasksa4b3c3d2e3f2g1h0Work StationEligibleAssignTime RemainingIdle TimeIaA1.1b,c,e, (tie)B0.7C0.4E0.30.3IIdD0.00.0IIIf,gF0.5G0.20.2IVhH0.10.10.63.4.b.1.2. Assign a, b, c, d, and e to station 1: 2.3 minutes no idle timeAssign f, g, and h to station 2: 2.3 minutes3. 4. 7.15438762Chapter 06 - Work Design and Measurement3.ElementPROTNTAFjobST1.90.46.4141.15.4762.851.5051.2801.151.47231.10.83.9131.151.05041.001.161.1601.151.334 Total 4.3328.A = 24 + 10 + 14 = 48 minutes per 4 hours9.a.ElementPROTNTAST11.101.191.3091.151.50521.15.83.9551.151.09831.05.56.5881.15.676b. c.e = .01 minutes Chapter 07- Location Planning and Analysis1.FactorLocal bankSteel millFood warehousePublic school1.Convenience for customersHLMHMH2.Attractiveness of buildingHLMMH3.Nearness to raw materialsLHLM4.Large amounts of powerLHLL5.Pollution controlsLHLL6.Labor cost and availabilityLMLL7.Transportation costsLMHMHM8.Construction costsMHMMHLocation (a)Location (b)4.FactorABCWeightABC1.Business Services9552/918/910/910/92.Community Services7671/97/96/97/93.Real Estate Cost3871/93/98/97/94.Construction Costs5652/910/912/910/95.Cost of Living4781/94/97/98/96.Taxes5551/95/95/94/97.Transportation 6 7 81/9 6/9 7/9 8/9Total3944451.053/955/954/9Each factor has a weight of 1/7.a.Composite Scores 394445777B or C is the best and A is least desirable.b.Business Services and Construction Costs both have a weight of 2/9; the other factors each have a weight of 1/9.5 x + 2 x + 2 x = 1 x = 1/9c.Composite Scores ABC53/955/954/9B is the best followed by C and then A.5.LocationxyA37B82C46D41E 6 4Totals2520=xi=25= 5.0=yi=20= 4.0n5n5Hence, the center of gravity is at (5,4) and therefore the optimal location.Chapter 08 - Management of Quality1.ChecksheetWork TypeFrequencyLube and Oil12Brakes7Tires6Battery4Transmission1Total30127641Lube & OilBrakesTiresBatteryTrans.Paretobreaklunchbreak32102.The run charts seems to show a pattern of errors possibly linked to break times or the end of the shift. Perhaps workers are becoming fatigued. If so, perhaps two 10 minute breaks in the morning and again in the afternoon instead of one 20 minute break could reduce some errors. Also, errors are occurring during the last few minutes before noon and the end of the shift, and those periods should also be given managements attention.Power offPersonLampMissingDidnt turncompletely onNot plugged inOutletdefectiveDefectiveBurned outLooseLamp fails to lightOtherCord4Chapter 9 - Quality Control4.SampleMeanRange179.482.6Mean Chart: A2= 79.96 0.58(1.87)280.142.3= 79.96 1.08380.141.2UCL = 81.04, LCL = 78.88479.601.7Range Chart: UCL = D4= 2.11(1.87) = 3.95580.022.0LCL = D3= 0(1.87) = 0680.381.4Both charts suggest the process is in control: Neither has any points outside the limits.6.n = 200 Control Limits = Thus, UCL is .0234 and LCL becomes 0.Since n = 200, the fraction represented by each data point is half the amount shown. E.g., 1 defective = .005, 2 defectives = .01, etc.Sample 10 is too large. 7.Control limits: UCL is 16.266, LCL becomes 0.All values are within the limits.14. Let USL = Upper Specification Limit, LSL = Lower Specification Limit, = Process mean, s = Process standard deviationFor process H: For process K:Assuming the minimum acceptable is 1.33, since 1.0 < 1.33, the process is not capable.For process T:Since 1.33 = 1.33, the process is capable.Chapter 10 - Aggregate Planning and Master Scheduling7. a.No backlogs are allowedPeriodMar.Apr.MayJun.JulyAug.Sep.TotalForecast50445560504051350OutputRegular40404040404040280Overtime888883851Subcontract2031220019Output - Forecast0440033InventoryBeginning0040003Ending0400030Average022001.51.57Backlog00000000Costs:Regular 3,2003,2003,2003,2003,2003,2003,20022,400Overtime 9609609609609603609606,120Subcontract 28004201,680280002,660Inventory 0202000151570Total4,4404,1804,6005,8404,4403,5754,17531,250b.Level strategyPeriodMar.Apr.MayJun.JulyAug.Sep.TotalForecast50445560504051350OutputRegular40404040404040280Overtime888888856Subcontract222222214Output - Forecast065100101InventoryBeginning0061001Ending0610010Average033.5.50.5.58Backlog000990018Costs:Regular 3,2003,2003,2003,2003,2003,2003,20022,400Overtime 9609609609609609609606,720Subcontract 2802802802802802802801,960Inventory 3035505580Backlog180180360Total4,4404,4704,4754,6254,6204,4454,44531,5208. Period123456TotalForecast160150160180170140960OutputRegular150150150150160160920Overtime1010010101050Subcontract0010100020Output- Forecast01001000InventoryBeginning00101000Ending01010000Average051050020Backlog0000000Costs:Regular 7,5007,5007,5007,5008,0008,00046,000Overtime 75075007507507503,750Subcontract 00800800001,600Inventory 20402080Backlog000000Total8,2508,2708,3409,0709,0508,75051,430Chapter 11 - MRP and ERP1. a.F: 2G: 1H: 1J: 2 x 2 = 4L: 1 x 2 = 2A: 1 x 4 = 4D: 2 x 4 = 8J: 1 x 2 = 2D: 1 x 2 = 2Totals: F = 2; G = 1; H = 1; J = 6; D = 10; L = 2; A = 4StaplerTop AssemblyBase AssemblyCoverSpringSlide AssemblyBaseStrike PadRubber Pad 2SlideSpringb. 4.Master ScheduleDayBeg. Inv.1234567Quantity100150200Table Beg. Inv.1234567Gross requirements100150200Scheduled receiptsProjected on handNet requirements100150200Planned-order receipts100150200Planned-order releases100150200Wood SectionsBeg. Inv.1234567Gross requirements200300400Scheduled receipts100Projected on hand100100Net requirements100300400Planned-order receipts100300400Planned-order releases400400Braces Beg. Inv.1234567Gross requirements300450600Scheduled receiptsProjected on hand60606060Net requirements240450600Planned-order receipts240450600Planned-order releases240450600Legs Beg. Inv.1234567Gross requirements400600800Scheduled receiptsProjected on hand120120120120888871Net requirements280600800Planned-order receipts308660880Planned-order releases96888010.Week1234Material40806070Week1234Labor hr.160320240280Mach. hr.120240180210a. Capacity utilizationWeek1234Labor53.3%106.7%80%93.3%Machine60%120%90%105%b. Capacity utilization exceeds 100% for both labor and machine in week 2, and for machine alone in week 4.Production could be shifted to earlier or later weeks in which capacity is underutilized. Shifting to an earlier week would result in added carrying costs; shifting to later weeks would mean backorder costs.Another option would be to work overtime. Labor cost would increase due to overtime premium, a probable decrease in productivity, and possible increase in accidents.Chapter 12 - Inventory Management2. The following table contains figures on the monthly volume and unit costs for a random sample of 16 items for a list of 2,000 inventory items.DollarItemUnit CostUsageUsageCategoryK34102002,000CK352560015,000AK36361505,400BM101625400CM2020801,600CZ458025016,000AF14203006,000BF953080024,000AF9920601,200CD45105505,500BD4812901,080CD52151101,650CD57401204,800BN0830401,200CP05165008,000BP091030300Ca. See table.b. To allocate control efforts.c. It might be important for some reason other than dollar usage, such as cost of a stockout, usage highly correlated to an A item, etc.3.D = 1,215 bags/yr.S = $10H = $75a.b.Q/2 = 18/2 = 9 bagsc.d. e.Assuming that holding cost per bag increases by $9/bag/yearQ = 17 bags Increase by $1,428.71 $1,350 = $78.714.D = 40/day x 260 days/yr. = 10,400 packagesS = $60 H = $30a.b. c.Yesd. TC200 = 3,000 + 3,120 = $6,120 6,120 6,118.82 (only $1.18 higher than with EOQ, so 200 is acceptable.)7.H = $2/monthS = $55D1 = 100/month (months 16)D2 = 150/month (months 712)a. b.The EOQ model requires this.c.Discount of $10/order is equivalent to S 10 = $45 (revised ordering cost)16 TC74 =$148.32712TC91 =$181.6610.p = 50/ton/dayD= 20 tons/day x 200 days/yr. = 4,000 tons/yr.u = 20 tons/day200 days/yr.S = $100H = $5/ton per yr.a.b.Average is tons approx. 3,098 bagsc.Run length = d.Runs per year = e.Q = 258.2TC = TCorig. = $1,549.00 TCrev. = $ 774.50Savings would be $774.5015.RangePHQD = 4,900 seats/yr.0999$5.00$2.00495H = .4P1,0003,9994.951.98497 NFS = $504,0005,9994.901.96500 NF6,000+4.851.94503 NFCompare TC495 with TC for all lower price breaks:TC495 =495($2) +4,900($50) + $5.00(4,900) = $25,4902495TC1,000 =1,000($1.98) +4,900($50) + $4.95(4,900) = $25,49021,000TC4,000 =4,000($1.96) +4,900($50) + $4.90(4,900) = $27,99124,000TC6,000 =6,000($1.94) +4,900($50) + $4.85(4,900) = $29,62626,0004954975005031,0004,0006,000QuantityTCHence, one would be indifferent between 495 or 1,000 units22.d = 30 gal./dayROP = 170 gal.LT = 4 days, ss = ZsLT = 50 galRisk = 9% Z = 1.34Solving, sLT = 37.313% Z = 1.88, ss=1.88 x 37.31 = 70.14 gal.Chapter 13 - JIT and Lean Operations1.N = ?N =DT(1 + X)D = 80 pieces per hourCT = 75 min. = 1.25 hr.=80(1.25) (1.35) = 3C = 4545X = .354.The smallest daily quantity evenly divisible into all four quantities is 3. Therefore, use three cycles.ProductDaily quantityUnits per cycleA2121/3 = 7B1212/3 = 4C33/3 = 1D1515/3 = 55. a. Cycle 1234A 6655B 3333C 1111D 4455E 2222b. Cycle 1 2A 1111B 6 6C 2 2D 8 8E 4 4c. 4 cycles = lower inventory, more flexibility 2 cycles = fewer changeovers7. Net available time = 480 75 = 405. Takt time = 405/300 units per day = 1.35 minutes.Chapter 15 - Scheduling6.a.FCFS: ABCDSPT: DCBAEDD: CBDACR: ACDBFCFS:Job timeFlow timeDue dateDaysJob(days)(days)(days)tardyA1414200B1024168C7311516D 6 3717 203710644SPT:Job timeFlow timeDue dateDaysJob(days)(days)(days)tardyD66170C713150B1023167A 14 3720 17377924EDD:Job timeFlow timeDue dateDaysJob(days)(days)(days)tardyC77150B1017161D623176A14 3720 178424Critical RatioJobProcessing Time (Days)Due DateCritical Ratio CalculationA1420(20 0) / 14 = 1.43B1016(16 0) /10 = 1.60C715(15 0) / 7 = 2.14D617(17 0) / 6 = 2.83 Job A has the lowest critical ratio, therefore it is scheduled first and completed on day 14. After the completion of Job A, the revised critical ratios are:JobProcessing Time (Days)Due DateCritical Ratio CalculationAB1016(16 14) /10 = 0.20C715(15 14) / 7 = 0.14D617(17 14) / 6 = 0.50 Job C has the lowest critical ratio, therefore it is scheduled next and completed on day 21. After the completion of Job C, the revised critical ratios are:JobProcessing Time (Days)Due DateCritical Ratio CalculationAB1016(16 21) /10 = 0.50CD617(17 21) / 6 = 0.67 Job D has the lowest critical ratio therefore it is scheduled next and completed on day 27. The critical ratio sequence is ACDB and the makespan is 37 days. Critical Ratio sequenceP