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《运营管理》课程习题及答案修订版

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《运营管理》课后习题答案

《运营管理》课后习题答案

Chapter 02 - Competitiveness, Strategy, and Productivity3.(1)(2)(3)(4)(5)(6)(7)Week OutputWorkerCost@$12x40OverheadCost@1.5MaterialCost@$6TotalCostMFP(2) (6)130,0002,8804,3202,7009,900 3.03233,6003,3605,0402,82011,220 2.99332,2003,3605,0402,76011,160 2.89435,4003,8405,7602,88012,480 2.84*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) Deposit—place in an envelope (which you’ll find near or in theATM) and insert it into the deposit slot2) Withdrawal—lift the “Withdrawal Door,” being careful to remove all cash6. Remove card and receipt (which serves as the transaction record)8.Chapter 04 - Strategic Capacity Planning for Products and Services2.%80capacityEffective outputActual Efficiency ==Actual output = .8 (Effective capacity) Effective capacity = .5 (Design capacity) Actual output = (.5)(.8)(Effective capacity) Actual output = (.4)(Design capacity) Actual output = 8 jobs Utilization = .4capacityDesign outputActual =n Utilizatiojobs 204.8capacity Effective output Actual Capacity Design ===10. 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 machineProduc tABC1 48,000 64,000 32,0002 48,000 48,000 36,0003 30,000 36,000 24,0004 60,000 60,000 30,000Total186,00208,00122,000machine181.000,150000,122machine 238.1000,150000,208machine224.1000,150000,186≈==≈==≈==C B A N N NYou 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 =$104,400 Buy 2 Bs —these have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3. Desired output = 4 Operating time = 56 minutesunit per minutes 14hourper units 4hourper minutes 65output Desired time Operating CT ===Tas k # of Following tasksPositional WeightA 4 23 B3203adf752b4c4e9h5i6gC218D325E218F429G324H114I05a. First rule: most followers. Second rule: largest positional weight.Assembly Line Balancing Table (CT = 14)b. First rule: Largest positional weight.Assembly Line Balancing Table (CT = 14)c. %36.805645stations of no. x CT time Total Efficiency ===4.a. l.2. Minimum Ct = 1.3 minutesTask Following tasksa4b3c3d2e3f2g1h03. percent 54.11)3.1(46.CT x N time)(idle percent Idle ==∑=4. 420 min./day 323.1 ( 323)/1.3 min./OT Output rounds to copiers day CT cycle=== b. 1. inutes m 3.224.6N time Total CT ,6.4 time Total ==== 2. Assign a, b, c, d, and e to station 1: 2.3 minutes [no idle time]Assign f, g, and h to station 2: 2.3 minutes3. 420182.6 copiers /2.3OT Output day CT ===4.420 min./dayMaximum Ct is 4.6. Output 91.30 copiers /4.6 min./day cycle==7.Chapter 06 - Work Design and Measurement3. ElementPR OT NT AF job ST1.90.46.4141.15.4762.85 1.5051.2801.151.47231.1.83.9131.151.050 41.01.16 1.1601.151.334Total4.3328. A = 24 + 10 + 14 = 48 minutes per 4 hours.min 125.720.11x70.5ST .min 70.5)95(.6NT 20.24048A =-=====9. a. ElementPR OTNTAST11.10 1.191.309 1.151.50521.15.83.955 1.151.09831.05.56.588 1.15.676b.01.A 00.2z 034.s 83.x ==== 222(.034)67.12~68.01(.83)zs n observations ax ⎛⎫⎛⎫===⎪ ⎪⎝⎭⎝⎭c. e = .01 minutes 47 to round ,24.4601.)034(.2e zs n 22=⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛=Chapter 07- Location Planning and Analysis1.Factor Local bank Steel mill Food warehousePublic school1. Conveniencefor customers H L M –H M –H 2. Attractivenessof building H L M M –H 3. Nearness toraw materials L H L M 4. Large amountsof power L H L L 5. PollutioncontrolsL H L L 6. Labor cost andavailabilityL M L L 7. TransportationcostsL M –H M –H M8. ConstructioncostsMHMM –HLocation (a)Location (b)4.Factor A B C WeightA B C1.Business Services9552/918/910/910/9munityServices7671/97/96/97/93.Real Estate Cost3871/93/98/97/94.ConstructionCosts5652/910/912/910/95.Cost of Living4781/94/97/98/96.Taxes5551/95/95/94/97.Transportation6781/96/97/98/9Total394445 1.053/955/954/9 Each factor has a weight of 1/7.posite Scores394445777B orC 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 = 1/9posite ScoresA B C 53/955/954/9B is the best followed byC and then A.5.Locationx yA37B82C46D41E64Totals2520x=x i=25= 5.0y=y i=20= 4.0 n5n5Hence, the center of gravity is at (5,4) and therefore the optimallocation.Chapter 08 - Management of Quality 1.ChecksheetWork Type FrequencyLube and Oil12Brakes7Tires6Battery4Transmission1Total30Pareto2 .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 management’s attention.break lunch break3 2 1 0• • •• • • ••• • • ••••••• ••• •• • •• • •••4Chapter 9 - Quality Control4. Sample Mean Range1 79.48 2.6 Mean Chart: =X ± A 2R = 79.96 ± 0.58(1.87)2 80.14 2.3 = 79.96 ± 1.083 80.14 1.2UCL = 81.04, LCL = 78.884 79.60 1.7 Range Chart: UCL = D 4R = 2.11(1.87) = 3.95 5 80.02 2.0LCL = D 3R = 0(1.87) = 0680.381.4[Both charts suggest the process is in control: Neither has any points outside the limits.]6. n = 200 Control Limits = np p p )1(2-±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. 857.714110c ==Control limits: 409.8857.7c 3c ±=± UCL is 16.266, LCL becomes 0.All values are within the limits.14. Let USL = Upper Specification Limit, LSL = Lower Specification Limit,0096.)200(1325==p 0138.0096.200)9904(.0096.20096.±=±=X = Process mean, = Process standard deviationFor process H:}{capablenot ,0.193.93.04.1 ,938.min 04.1)32)(.3(1516393.)32)(.3(1.14153<===-=σ-=-=σ-pk C X USL LSL XFor process K:.1}17.1,0.1min{17.1)1)(3(335.3630.1)1)(3(30333===-=σ-=-=σ- C X USL LSL X pk Assuming the minimum acceptable pk C is 1.33, since 1.0 < 1.33, the process is not capable.For process T:33.1}33.1,67.1min{33.1)4.0)(3(5.181.20367.1)4.0)(3(5.165.183===-=σ-=-=σ- C X USL LSL X pk Since 1.33 = 1.33, the process is capable.Chapter 10 - Aggregate Planning and Master Scheduling7. a.No backlogs are allowedPeriod Mar. Apr. May Jun. July Aug. Sep. Total Forecast 50 44 55 60 50 40 51 350Output Regular 40 40 40 40 40 40 40 280 Overtime 8 8 8 8 8 3 8 51 Subcontract 2 0 3 12 2 0 0 19 Output - 0 4 –40 0 3 –3InventoryBeginningEndingAverageBacklogCosts:RegularOvertimeSubcontractInventoryTotalb. Level strategyPeriodForecastOutputRegularOvertimeSubcontractOutput -InventoryBeginningEndingAverageBacklogCosts:RegularOvertimeSubcontractInventoryBacklogTotal8.PeriodForecastOutputRegularOvertimeSubcontractOutput-InventoryBeginningEndingAverageBacklogCosts:RegularOvertimeSubcontractInventoryBacklog Total Chapter 11 - MRP and ERP1. a. F: 2 G : 1 H : 1 J: 2 x 2 = 4 L: 1 x 2 = 2 A: 1 x 4 = 4 D : 2 x 4 = 8 J: 1 x 2 = 2 D : 1 x 2 = 2Totals: F = 2; G = 1; H = 1; J = 6; D = 10; L = 2; A = 44.Master Schedule10.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 inweek 2, and for machine alone in week 4.Production could be shifted to earlier or later weeks in which capacityis underutilized. Shifting to an earlier week would result in addedcarrying costs; shifting to later weeks would mean backorder costs.Another option would be to work overtime. Labor cost would increasedue to overtime premium, a probable decrease in productivity, andpossible increase in accidents.Chapter 12 - Inventory Management2. The following table contains figures on the monthly volume and unitcosts for a random sample of 16 items for a list of 2,000 inventoryitems.K35 25 600 15,000 A K36 36 150 5,400 B M10 16 25 400 C M20 20 801,600C Z45 80 250 16,000 A F14 20 3006,000B F95 30 800 24,000 A F99 20 60 1,200C D45 10 550 5,500 B D48 12 90 1,080 C D52 15 110 1,650 C D57 40 120 4,800 B N08 30 40 1,200 C P05 16 500 8,000 BP091030300Ca. See table.b. To allocate control efforts.c. It might be important for some reason other than dollar usage, suchas cost of a stockout, usage highly correlated to an A item, etc.3. D = 1,215 bags/yr. S = $10 H = $75a. bags HDS Q 187510)215,1(22===b. Q/2 = 18/2 = 9 bagsc.orders ordersbags bags Q D 5.67/ 18 215,1== d .S QD H 2/Q TC +=350,1$675675)10(18215,1)75(218=+=+=e. Assuming that holding cost per bag increases by $9/bag/yearQ ==84)10)(215,1(217 bags71.428,1$71.714714)10(17215,1)84(217=+=+=TC Increase by [$1,428.71 – $1,350] = $78.714. D = 40/day x 260 days/yr. = 10,400 packagesS = $60 H = $30a. oxes b 20496.2033060)400,10(2H DS 2Q 0====b. S QD H 2Q TC +=82.118,6$82.058,3060,3)60(204400,10)30(2204=+=+=c. Yesd. )60(200400,10)30(2200TC 200+=TC 200 = 3,000 + 3,120 = $6,1206,120 – 6,118.82 (only $1.18 higher than with EOQ, so 200 isacceptable.)7.H = $2/monthS = $55D 1 = 100/month (months 1–6)D 2 = 150/month (months 7–12)a. 16.74255)100(2Q :D H DS2Q 010===83.90255)150(2Q :D 02==b. The EOQ model requires this.c. Discount of $10/order is equivalent to S – 10 = $45 (revisedordering cost)1–6TC74 =$148.32180$)45(150100)2(2150TC 145$)45(100100)2(2100TC *140$)45(50100)2(250TC 15010050=+==+==+=7–12TC 91 =$181.66195$)45(150150)2(2150TC *5.167$)45(100150)2(2100TC 185$)45(50150)2(250TC 15010050=+==+==+=10. p = 50/ton/dayu = 20 tons/day 200 days/yr.S = $100H = $5/ton per yr.a. bags] [10,328 tons 40.5162050505100)000,4(2u p p H DS 2Q 0=-=-=b. ]bags 8.196,6 .approx [ tons 84.309)30(504.516)u p (P Q I max ==-=D= 20 tons/day x 200 days/yr. = 4,000 tons/yr.Average is92.154248.309:2I max =tons [approx. 3,098 bags] c. Run length =days 33.10504.516P Q == d. Runs per year = 8] approx .[ 7.754.516000,4QD == e. Q = 258.2TC =S QD H 2I max + TC orig. = $1,549.00 TC rev. = $ 774.50Savings would be$774.5015.RangePHQD = 4,900 seats/yr.0–999$5.00$2.00495H = .4P 1,000–3,9994.951.98497 NFS = $50 4,000–5,9994.901.96500 NF6,000+4.851.94503 NFCompare TC 495 with TC for all lower price breaks:TC 495 =495 ($2)+ 4,900 ($50) + $5.00(4,900) = $25,4902495TC 1,000 =1,000($1.98) +4,900($50) + $4.95(4,900) =$25,49021,000TC 4,000 =4,000($1.96) +4,900($50) + $4.90(4,900) =$27,99124,000TC 6,000 =6,000($1.94) +4,900($50) + $4.85(4,900) =$29,62626,000Hence, one would be indifferent between 495 or 1,000 units22.d = 30 gal./dayROP = 170 gal.QuantityTCLT = 4days,ss = Z d LT = 50 galRisk = 9% Z =1.34Solving, d LT = 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 hour CT = 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.Product Daily quantity Units per cycleA2121/3 = 7B1212/3 = 4C33/3 = 1D1515/3 = 55.a. Cycle 1 2 3 4A 6 6 5 5B 3 3 3 3C 1 1 1 1D 4 4 5 5E 2 2 2 2b. Cycle 1 2A 11 11B 6 6C 2 2D 8 8E 4 4c. 4 cycles = lower inventory, more flexibility2 cycles = fewer changeovers7. Net available time = 480 – 75 = 405. Takt time = 405/300 units per day = 1.35 minutes.Chapter 15 - Scheduling6. a.FCFS:A–B–C–DSPT:D–C–B–AEDD:C–B–D–ACR:A–C–D–BFCFS:Job time FlowtimeDuedate DaysJob(days)(days)(days)tardyA1414200 B1024168 C7311516 D6371720 3710644SPT:Job time FlowtimeDuedate DaysJob(days)(days)(days)tardyD66170C713150B1023167A14372017377924 EDD:and completed on day 14. After the completion of Job A, the revised criticaland completed on day 21. After the completion of Job C, the revised criticalJob D has the lowest critical ratio therefore it is scheduled next and completed on day 27.b.ardi Flow timeAverage flow time Number of jobs Days tardy Average job t ness Number of jobs Flow time Average number of jobs at the center Makespan==∑=FCFS SPT EDD CR26.50 19.75 21.00 24.7511.0 6.00 6.00 9.252.86 2.14 2.27 2.67c. SPT is superior.9.Thus, the sequence is b-a-g-e-f-d-c.。

《运营管理》课程习题及标准答案-修订版(1)

《运营管理》课程习题及标准答案-修订版(1)

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课后习题答案

《运营管理》课后习题答案

Chapter 02 - Competitiveness，Strategy，and Productivity3。

(1）（2）(3) (4）(5）(6）(7)Week Output WorkerCost@$12x40 OverheadCost ＠1。

5MaterialCost@$6TotalCostMFP（2）÷ (6)1 30,000 2，880 4,320 2,700 9,900 3。

032 33，600 3，360 5，040 2,820 11,220 2.993 32，200 3,360 5,040 2，760 11,160 2。

894 35，400 3，840 5,760 2,880 12,480 2。

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 = .89carts/$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。

《运营管理》课后习题标准答案

《运营管理》课后习题标准答案

《运营管理》课后习题答案————————————————————————————————作者：————————————————————————————————日期：2Chapter 02 - Competitiveness, Strategy, and Productivity3. (1) (2) (3) (4) (5) (6) (7)Week Output WorkerCost@$12x40Overhead********MaterialCost@$6TotalCostMFP(2) ÷ (6)1 30,000 2,880 4,320 2,700 9,900 3.032 33,600 3,360 5,040 2,820 11,220 2.993 32,200 3,360 5,040 2,760 11,160 2.894 35,400 3,840 5,760 2,880 12,480 2.84*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) Deposit —place in an envelope (which you’ll find near or in the ATM) andinsert it into the deposit slot2) Withdrawal —lift the “Withdrawal Door,” being careful to remove all cash6. Remove card and receipt (which serves as the transaction record)8.TechnicalRequirements IngredientsHandlingPreparationCustomer RequirementsTaste √√ Appearance√ √√Texture/consistency√√Chapter 04 - Strategic Capacity Planning for Products and Services2. %80capacityEffective outputActual Efficiency ==Actual output = .8 (Effective capacity) Effective capacity = .5 (Design capacity) Actual output = (.5)(.8)(Effective capacity) Actual output = (.4)(Design capacity) Actual output = 8 jobs Utilization = .4capacityDesign outputActual =n Utilizatiojobs 204.8capacity Effective output Actual Capacity Design ===10. 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 machineProductABC 1 48,000 64,000 32,000 2 48,000 48,000 36,000 3 30,000 36,000 24,000 460,000 60,000 30,000 Total 186,000208,000122,000machine181.000,150000,122machine 238.1000,150000,208machine224.1000,150000,186≈==≈==≈==C B A N N NYou 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 = $98,133 C(1): 122,000 ÷ 60 = 2,033.33 hrs. x $12 = $24,400 + $80,000 = $104,400Buy 2 Bs —these have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3.3 adf752 b4 c4 e9 h5 i6 gDesired output = 4Operating time = 56 minutesunit per minutes 14hourper units 4hourper minutes 65output Desired time Operating CT ===Task # of Following tasksPositional WeightA 4 23B 3 20C 2 18D 3 25E 2 18F 4 29G 3 24H 1 14 I5a. First rule: most followers. Second rule: largest positional weight.Assembly Line Balancing Table (CT = 14)Work StationTask Task TimeTime RemainingFeasible tasksRemainingIF 5 9 A,D,G A 3 6 B,G G6 – – II D7 7 B, E B 2 5 C C4 1 – III E 4 10 H H9 1 – IV I59–b. First rule: Largest positional weight.Assembly Line Balancing Table (CT = 14)Work StationTask Task TimeTime RemainingFeasible tasks RemainingIF 5 9 A,D,G D7 2 – II G 6 8 A, E A 3 5 B,E B2 3 – III C 4 10 E E4 6 – IV H 95 I I5–c. %36.805645stations of no. x CT time Total Efficiency ===4. a. l.2. Minimum Ct = 1.3 minutesTask Following tasksa 4b 3c 3d 2e 3f 2g 1habd cfeghWork StationEligible Assign Time RemainingIdle TimeIa A 1.1 b,c,e, (tie)B 0.7C 0.4E 0.3 0.3 II d D 0.0 0.0 IIIf,g F 0.5G 0.2 0.2 IVh H 0.1 0.10.63. percent 54.11)3.1(46.CT x N time)(idle percent Idle ==∑=4. 420 min./day 323.1 ( 323)/1.3 min./OT Output rounds to copiers day CT cycle=== b. 1. inutes m 3.224.6N time Total CT ,6.4 time Total ==== 2. Assign a, b, c, d, and e to station 1: 2.3 minutes [no idle time]Assign f, g, and h to station 2: 2.3 minutes3. 420182.6 copiers /2.3OT Output day CT ===4.420 min./dayMaximum Ct is 4.6. Output 91.30 copiers /4.6 min./day cycle==7. 1 5 4 3 8 762Chapter 06 - Work Design and Measurement3. Element PR OT NT AF job ST1 .90.46.414 1.15 .4762 .85 1.505 1.280 1.15 1.4723 1.10.83.913 1.15 1.05041.00 1.16 1.160 1.15 1.334Total4.3328. A = 24 + 10 + 14 = 48 minutes per 4 hours.min 125.720.11x70.5ST .min 70.5)95(.6NT 20.24048A =-=====9. a. Element PR OT NT A ST1 1.10 1.19 1.309 1.15 1.5052 1.15 .83 .955 1.15 1.09831.05.56.588 1.15 .676b.01.A 00.2z 034.s 83.x ==== 222(.034)67.12~68.01(.83)zs n observations ax ⎛⎫⎛⎫===⎪ ⎪⎝⎭⎝⎭c. e = .01 minutes 47 to round ,24.4601.)034(.2e zs n 22=⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛=Chapter 07- Location Planning and Analysis1. Factor Local bank Steel mill Food warehouse Public school1. Convenience forcustomers H L M–H M–H2. Attractiveness ofbuilding H L M M–H3. Nearness to rawmaterials L H L M4. Large amounts ofpower L H L L5. Pollution controls L H L L6. Labor cost andavailability L M L L7. Transportationcosts L M–H M–H M8. Constructioncosts M H M M–HLocation (a) Location (b)4. Factor A B C Weight A B C1. Business Services 9 5 5 2/9 18/9 10/9 10/92. Community Services 7 6 7 1/9 7/9 6/9 7/93. Real Estate Cost 3 8 7 1/9 3/9 8/9 7/94. Construction Costs 5 6 5 2/9 10/9 12/9 10/95. Cost of Living 4 7 8 1/9 4/9 7/9 8/96. Taxes 5 5 5 1/9 5/9 5/9 4/97. Transportation 6 7 8 1/9 6/9 7/9 8/9Total 39 44 45 1.0 53/9 55/9 54/9 Each factor has a weight of 1/7.a. Composite Scores 39 44 45 7 7 7B orC is the best and A is least desirable.b. Business Services and Construction Costs both have a weight of 2/9; the other factors eachhave a weight of 1/9.5 x + 2 x + 2 x = 1 x = 1/9c. Composite ScoresA B C 53/9 55/9 54/9B is the best followed byC and then A.5.Locationx yA 3 7B 8 2C 4 6D 4 1E 6 4Totals 25 20-x =∑x i= 25 = 5.0 -y =∑y i= 20 = 4.0 n 5 n 5Hence, the center of gravity is at (5,4) and therefore the optimal location.Chapter 08 - Management of Quality1. ChecksheetWork Type FrequencyLube and Oil 12Brakes 7Tires 6Battery 4Transmission 1Total 30Pareto127641 Lube & Oil Brakes Tires Battery Trans.2 .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 management’s attention.4Power Per LamMissDidn’Not OutletDefectBurn LoosLampOtheCordbreak lunch3 2•• •• •• • ••• • ••• •••• ••• •• • •• • •••Chapter 9 - Quality Control4. Sample Mean Range179.48 2.6 Mean Chart: =X ± A 2-R = 79.96 ± 0.58(1.87) 2 80.14 2.3 = 79.96 ± 1.083 80.14 1.2UCL = 81.04, LCL = 78.884 79.60 1.7 Range Chart: UCL = D 4-R = 2.11(1.87) = 3.95 5 80.02 2.0LCL = D 3-R = 0(1.87) = 0680.381.4[Both charts suggest the process is in control: Neither has any points outside the limits.]6. n = 200 Control Limits = np p p )1(2-±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. 857.714110c ==Control limits: 409.8857.7c 3c ±=± UCL is 16.266, LCL becomes 0.All values are within the limits.14. Let USL = Upper Specification Limit, LSL = Lower Specification Limit,X = Process mean, σ = Process standard deviationFor process H:}{capablenot ,0.193.93.04.1 ,938.min 04.1)32)(.3(1516393.)32)(.3(1.14153<===-=σ-=-=σ-pk C X USL LSL X 0096.)200(1325==p 0138.0096.200)9904(.0096.20096.±=±=For process K:.1}17.1,0.1min{17.1)1)(3(335.3630.1)1)(3(30333===-=σ-=-=σ- C X USL LSL X pk Assuming the minimum acceptable pk C is 1.33, since 1.0 < 1.33, the process is not capable.For process T:33.1}33.1,67.1min{33.1)4.0)(3(5.181.20367.1)4.0)(3(5.165.183===-=σ-=-=σ- C X USL LSL X pk Since 1.33 = 1.33, the process is capable.Chapter 10 - Aggregate Planning and Master Scheduling7. a.No backlogs are allowedPeriod Mar. Apr. May Jun. July Aug. Sep. TotalForecast 50 44 55 60 50 40 51 350 Output Regular 40 40 40 40 40 40 40 280 Overtime 8 8 8 8 8 3 8 51 Subcontract 2 0 3 12 2 0 0 19 Output - Forecast 0 4 –4 0 0 3 –3 Inventory Beginning 0 0 4 0 0 0 3 Ending 0 4 0 0 0 3 0 Average 0 2 2 0 0 1.5 1.5 7 Backlog 0 0 0 0 0 0 0 0 Costs: Regular 3,200 3,200 3,200 3,200 3,200 3,200 3,200 22,400 Overtime 960 960 960 960 960 360 960 6,120 Subcontract 280 0 420 1,680280 0 0 2,660 Inventory 0 20 20 0 0 15 15 70 Total4,4404,1804,6005,8404,4403,575 4,17531,250b. Level strategyPeriod Mar. Apr. May Jun. July Aug. Sep. Total Forecast 50 44 55 60 50 40 51 350 OutputRegular 40 40 40 40 40 40 40 280 Overtime 8 8 8 8 8 8 8 56 Subcontract 2 2 2 2 2 2 2 14 Output - Forecast 0 6 –5 –10 0 10 –1InventoryBeginning 0 0 6 1 0 0 1Ending 0 6 1 0 0 1 0Average 0 3 3.5 .5 0 .5 .5 8 Backlog 0 0 0 9 9 0 0 18 Costs:Regular 3,200 3,200 3,200 3,200 3,200 3,200 3,200 22,400 Overtime 960 960 960 960 960 960 960 6,720 Subcontract 280 280 280 280 280 280 280 1,960 Inventory 30 35 5 0 5 5 80 Backlog 180 180 360 Total 4,440 4,470 4,475 4,625 4,620 4,445 4,445 31,520 8.Period 1 2 3 4 5 6 TotalForecast 160 150 160 180 170 140 960OutputRegular 150 150 150 150 160 160 920Overtime 10 10 0 10 10 10 50Subcontract 0 0 10 10 0 0 20Output- Forecast 0 10 0 –10 0 0InventoryBeginning 0 0 10 10 0 0Ending 0 10 10 0 0 0Average 0 5 10 5 0 0 20Backlog 0 0 0 0 0 0 0Costs:Regular 7,500 7,500 7,500 7,500 8,000 8,000 46,000Overtime 750 750 0 750 750 750 3,750Subcontract 0 0 800 800 0 0 1,600Inventory 20 40 20 80Backlog 0 0 0 0 0 0Total 8,250 8,270 8,340 9,070 9,050 8,750 51,430Chapter 11 - MRP and ERP1. a. F: 2 G: 1 H: 1J: 2 x 2 = 4 L: 1 x 2 = 2 A: 1 x 4 = 4D: 2 x 4 = 8 J: 1 x 2 = 2 D: 1 x 2 = 2Totals: F = 2; G = 1; H = 1; J = 6; D = 10; L = 2; A = 4b.4. Master Schedule Day Beg. Inv. 1 2 3 4 5 6 7 Quantity100 150 200 TableBeg. Inv. 1 2 3 4 5 6 7 Gross requirements 100 150 200 Scheduled receipts Projected on hand Net requirements 100 150 200 Planned-order receipts 100 150 200 Planned-order releases 100 150 200Wood Sections Beg. Inv. 1 2 3 4 5 6 7 Gross requirements 200300 400 Scheduled receipts 100 Projected on hand 100100 Net requirements 100 300 400 Planned-order receipts 100 300 400 Planned-order releases400 400Braces Beg. Inv. 1 2 3 4 5 6 7 Gross requirements 300 450 600 Scheduled receipts Projected on hand 60 60 60 60 Net requirements 240 450 600 Planned-order receipts 240 450 600Planned-order releases 240 450 600StaplerTopBaseCoveSpri SlideBase Strik RubberSlidSpriLegs Beg.Inv.1 2 3 4 5 6 7Gross requirements 400 600 800Scheduled receiptsProjected on hand 120 120 120 120 88 88 71 Net requirements 280 600 800Planned-order receipts 308 660 880Planned-order releases 968 88010. Week 1 2 3 4Material 40 80 60 70Week 1 2 3 4Labor hr. 160 320 240 280Mach. hr. 120 240 180 210a. Capacity utilizationWeek 1 2 3 4Labor 53.3% 106.7% 80% 93.3%Machine 60% 120% 90% 105%b. C apacity utilization exceeds 100% for both labor and machine in week 2, and formachine alone in week 4.Production could be shifted to earlier or later weeks in which capacity isunderutilized. 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 toovertime premium, a probable decrease in productivity, and possible increase inaccidents.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 Cost UsageUsageCategoryK34 10 200 2,000 C K35 25 600 15,000 A K36 36 150 5,400 B M10 16 25 400 C M20 20 80 1,600 C Z45 80 250 16,000 A F14 20 300 6,000 B F95 30 800 24,000 A F99 20 60 1,200 C D45 10 550 5,500 B D48 12 90 1,080 C D52 15 110 1,650 C D57 40 120 4,800 B N08 30 40 1,200 C P05 16 500 8,000 BP091030300Ca. See table.b. To allocate control efforts.c. It might be important for some reason other than dollar usage, such as cost of astockout, usage highly correlated to an A item, etc.3. D = 1,215 bags/yr. S = $10 H = $75a. bags HDS Q 187510)215,1(22===b. Q/2 = 18/2 = 9 bagsc.orders ordersbags bags Q D 5.67/ 18 215,1== d . S QD H 2/Q TC +=350,1$675675)10(18215,1)75(218=+=+=e. Assuming that holding cost per bag increases by $9/bag/yearQ ==84)10)(215,1(217 bags71.428,1$71.714714)10(17215,1)84(217=+=+=TCIncrease by [$1,428.71 – $1,350] = $78.714.D = 40/day x 260 days/yr. = 10,400 packagesS = $60 H = $30a. oxes b 20496.2033060)400,10(2H DS 2Q 0====b. S QD H 2Q TC +=82.118,6$82.058,3060,3)60(204400,10)30(2204=+=+=c. Yesd. )60(200400,10)30(2200TC 200+=TC 200 = 3,000 + 3,120 = $6,1206,120 – 6,118.82 (only $1.18 higher than with EOQ, so 200 is acceptable.)7.H = $2/month S = $55D 1 = 100/month (months 1–6)D 2 = 150/month (months 7–12)a. 16.74255)100(2Q :D H DS2Q 010===83.90255)150(2Q :D 02==b. The EOQ model requires this.c. Discount of $10/order is equivalent to S – 10 = $45 (revised ordering cost)1–6 TC74 = $148.32180$)45(150100)2(2150TC 145$)45(100100)2(2100TC *140$)45(50100)2(250TC 15010050=+==+==+=7–12 TC 91 =$181.66195$)45(150150)2(2150TC *5.167$)45(100150)2(2100TC 185$)45(50150)2(250TC 15010050=+==+==+=10. p = 50/ton/day u = 20 tons/day200 days/yr. S = $100 H = $5/ton per yr.a. bags] [10,328 tons 40.5162050505100)000,4(2u p p H DS 2Q 0=-=-=b. ]bags 8.196,6 .approx [ tons 84.309)30(504.516)u p (P Q I max ==-=Average is92.154248.309:2I max =tons [approx. 3,098 bags] c. Run length =days 33.10504.516P Q == d. Runs per year = 8] approx .[ 7.754.516000,4QD == e. Q ' = 258.2TC =S QD H 2I max + TC orig. = $1,549.00 TC rev. = $ 774.50Savings would be $774.50D= 20 tons/day x 20015. RangeP H Q D = 4,900 seats/yr. 0–999 $5.00 $2.00 495 H = .4P 1,000–3,999 4.95 1.98 497 NF S = $50 4,000–5,999 4.90 1.96 500 NF 6,000+4.851.94503 NFCompare TC 495 with TC for all lower price breaks:TC 495 =495 ($2) + 4,900($50) + $5.00(4,900) = $25,490 2 495 TC 1,000 = 1,000 ($1.98) + 4,900($50) + $4.95(4,900) = $25,4902 1,000 TC 4,000 = 4,000 ($1.96) + 4,900($50) + $4.90(4,900) = $27,9912 4,000 TC 6,000 = 6,000 ($1.94) + 4,900($50) + $4.85(4,900) = $29,6262 6,000Hence, one would be indifferent between 495 or 1,000 units 22. d = 30 gal./day ROP = 170 gal. LT = 4 days,ss = Z σd LT = 50 galRisk = 9% Z = 1.34 Solving, σd LT = 37.31 3% 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 hourC T = 75 min. = 1.25 hr. = 80(1.25) (1.35)= 3C = 45 45X = .35• •• •495 497 500 5031,0004,000 6,000QuantityTC4. The smallest daily quantity evenly divisible into all four quantities is 3. Therefore, usethree cycles.Product Daily quantity Units per cycleA 21 21/3 = 7B 12 12/3 = 4C 3 3/3 = 1D 15 15/3 = 55.a. Cycle 1 2 3 4A 6 6 5 5B 3 3 3 3C 1 1 1 1D 4 4 5 5E 2 2 2 2 b. Cycle 1 2A 11 11B 6 6C 2 2D 8 8E 4 4c. 4 cycles = lower inventory, more flexibility2 cycles = fewer changeovers7. Net available time = 480 – 75 = 405. Takt time = 405/300 units per day = 1.35 minutes. Chapter 15 - Scheduling6. a. FCFS: A–B–C–DSPT: D–C–B–AEDD: C–B–D–ACR: A–C–D–BFCFS: Job time Flow time Due date DaysJob (days) (days) (days) tardyA 14 14 20 0B 10 24 16 8C 7 31 15 16D 6 37 17 2037 106 44SPT: Job time Flow time Due date Days Job (days) (days) (days) tardyD 6 6 17 0C 7 13 15 0B 10 23 16 7A 14 37 20 1737 79 24EDD: Job time Flow time Due date DaysJob (days) (days) (days) tardyC 7 7 15 0B 10 17 16 1D 6 23 17 6A 14 37 20 1784 24Critical RatioJob Processing Time(Days) Due Date Critical Ratio CalculationA 14 20 (20 – 0) / 14 = 1.43B 10 16 (16 – 0) /10 = 1.60C 7 15 (15 – 0) / 7 = 2.14D 6 17 (17 – 0) / 6 = 2.83Job 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:Job Processing Time(Days) Due Date Critical Ratio CalculationA –––B 10 16 (16 – 14) /10 = 0.20C 7 15 (15 – 14) / 7 = 0.14D 6 17 (17 – 14) / 6 = 0.50Job 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:Job Processing Time(Days) Due Date Critical Ratio CalculationA –––B 10 16 (16 – 21) /10 = –0.50C –––D 6 17 (17 – 21) / 6 = –0.67Job D has the lowest critical ratio therefore it is scheduled next and completed on day 27. The critical ratio sequence is A –C –D –B and the makespan is 37 days. Critical Ratio sequenceProcessing Time(Days)Flow time Due Date TardinessA 14 14 20 0 C 7 21 15 6 D 6 27 17 10 B1037 16 21 ∑9937b.ardi Flow time Average flow time Number of jobsDays tardy Average job t ness Number of jobs Flow timeAverage number of jobs at the center Makespan==∑=FCFS SPT EDD CR26.50 19.75 21.00 24.75 11.0 6.00 6.00 9.25 2.86 2.142.272.67c. SPT is superior.9.Time (hr.) Sequence of assignment:Order Step 1 Step 2A 1.20 1.40 .80 [C] last (or 7th)B 0.90 1.30 .90 [B] firstC 2.00 0.80 1.20 [A] 2ndD 1.70 1.50 1.30 [G] 3rdE 1.60 1.80 1.60 [E] 4thF 2.20 1.75 1.50 [D] 6th G1.301.401.75[F]5thThus, the sequence is b-a-g-e-f-d-c.。

《运营管理》课程习题和答案解析_修订版

《运营管理》课程习题和答案解析_修订版

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课后习题答案

《运营管理》课后习题答案

Chapter 02 - Competitiveness, S trategy, and Productivity3. (1) (2) (3) (4) (5) (6) (7)Week Output WorkerCost@$12x40Overhead********MaterialCost@$6TotalCostMFP(2) ÷ (6)1 30,000 2,880 4,320 2,700 9,900 3.032 33,600 3,360 5,040 2,820 11,220 2.993 32,200 3,360 5,040 2,760 11,160 2.894 35,400 3,840 5,760 2,880 12,480 2.84*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) Deposit —place in an envelope (which you’ll find near or in the ATM) andinsert it into the deposit slot2) Withdrawal —lift the ―Withdrawal Door,‖ being careful to remove all cash6. Remove card and receipt (which serves as the transaction record)8.Chapter 04 - Strategic Capacity Planning for Products and S ervices2. %80capacityEffectiveoutput Actual Efficiency ==Actual output = .8 (Effective capacity) Effective capacity = .5 (Design capacity) Actual output = (.5)(.8)(Effective capacity) Actual output = (.4)(Design capacity) Actual output = 8 jobs Utilization = .4capacityDesignoutput Actual =n Utilizatiojobs204.8capacityEffectiveoutput Actual Capacity Design ===10. a. Given: 10 hrs. or 600 min. of operating time per day.250 days x 600 min. = 150,000 min. per year operating time.Solutions_Problems_OM_11e_Stevenson3 / 22Total processing time by machineProductABC 1 48,000 64,000 32,000 2 48,000 48,000 36,000 3 30,000 36,000 24,000 460,000 60,000 30,000 Total 186,000208,000122,000machine181.000,150000,122machine238.1000,150000,208machine224.1000,150000,186≈==≈==≈==CBANNNYou 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 = $98,133 C(1): 122,000 ÷ 60 = 2,033.33 hrs. x $12 = $24,400 + $80,000 = $104,400Buy 2 Bs —these have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3.Desired output = 4Operating time = 56 minutesunitper minutes14hourper units 4hourper minutes65outputDesiredtime Operating CT ===Task # of Following tasksPositional WeightA 4 23B 3 20C 2 18D 3 25E 2 18F 4 29G 3 24H 1 14 I5a. First rule: most followers. Second rule: largest positional weight.Assembly Line Balancing Table (CT = 14)Solutions_Problems_OM_11e_Stevenson5 / 22b. First rule: Largest positional weight.Assembly Line Balancing Table (CT = 14)c. %36.805645stationsof no. x CT tim eTotal Efficiency===4. a. l.2. Minimum Ct = 1.3 minutesTask Following tasksa 4b 3c 3d 2e 3f 2g 1h3.percent54.11)3.1(46.CTx N time)(idle percent Idle ==∑=4. 420 m in./d a y 323.1 ( 323)/1.3 m in ./O T O u tp u t r o u n d s to c o p ie r s d a yC Tc y c le===b. 1. inutesm 3.224.6Ntim eTotal CT ,6.4 tim e Total ====2. Assign a, b, c, d, and e to station 1: 2.3 minutes [no idle time]Assign f, g, and h to station 2: 2.3 minutes3. 420 182.6 c o p ie rs /2.3O T O u tp u t d a yC T===4.420 m in./d a y M a x im u m C t is 4.6. O u tp u t 91.30 c o p ie rs /4.6 m in ./d a yc y c le==7.Solutions_Problems_OM_11e_Stevenson7 / 22Chapter 06 - Work Design and Measurement3. Element PR OT NT AF job ST1 .90.46.414 1.15 .4762 .85 1.505 1.280 1.15 1.4723 1.10.83.913 1.15 1.05041.00 1.16 1.160 1.15 1.334Total4.3328. A = 24 + 10 + 14 = 48 minutes per 4 hours.min 125.720.11x 70.5ST .min 70.5)95(.6NT 20.24048A =-=====9. a. Element PR OT NT A ST1 1.10 1.19 1.309 1.15 1.5052 1.15 .83 .955 1.15 1.09831.05.56.588 1.15 .676b.01.A 00.2z 034.s 83.x ==== 222(.034)67.12~68.01(.83)z s n o b s e r v a tio n s a x ⎛⎫⎛⎫=== ⎪ ⎪⎝⎭⎝⎭c. e = .01 minutes47to round ,24.4601.)034(.2e zs n 22=⎪⎭⎫⎝⎛=⎪⎭⎫⎝⎛=Chapter 07- Location Planning and Analysis1. Factor Local bank Steel mill Food warehouse Public school1. Convenience forcustomers H L M–H M–H2. Attractiveness ofbuilding H L M M–H3. Nearness to rawmaterials L H L M4. Large amounts ofpower L H L L5. Pollution controls L H L L6. Labor cost andavailability L M L L7. Transportationcosts L M–H M–H M8. Constructioncosts M H M M–HLocation (a) Location (b)4. Factor A B C Weight A B C1. Business Services 9 5 5 2/9 18/9 10/9 10/92. Community Services 7 6 7 1/9 7/9 6/9 7/93. Real Estate Cost 3 8 7 1/9 3/9 8/9 7/94. Construction Costs 5 6 5 2/9 10/9 12/9 10/95. Cost of Living 4 7 8 1/9 4/9 7/9 8/96. Taxes 5 5 5 1/9 5/9 5/9 4/97. Transportation 6 7 8 1/9 6/9 7/9 8/9Total 39 44 45 1.0 53/9 55/9 54/9 Each factor has a weight of 1/7.a. Composite Scores 39 44 45 7 7 7B orC is the best and A is least desirable.b. Business Services and Construction Costs both have a weight of 2/9; the other factors eachhave a weight of 1/9.5 x + 2 x + 2 x = 1 x = 1/9c. Composite ScoresA B C 53/9 55/9 54/9B is the best followed byC and then A.Solutions_Problems_OM_11e_Stevenson9 / 225. Locationx yA3 7 B 8 2 C4 6 D4 1E 6 4 Totals 25 20-x = ∑x i = 25 = 5.0-y =∑y i = 20= 4.0n5n5Hence, the center of gravity is at (5,4) and therefore the optimal location.Chapter 08 - Management of Quality1.ChecksheetWork Type FrequencyLube and Oil 12 Brakes 7 Tires 6 Battery 4 Transmission1Total30ParetoLube & OilBrakesTiresBatteryTrans.2 .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 b e given management’s attention.4break lunch break3 2 1 0∙ ∙ ∙∙ ∙ ∙ ∙∙∙ ∙ ∙ ∙∙∙∙∙∙∙ ∙∙∙ ∙∙ ∙ ∙∙ ∙ ∙∙∙Solutions_Problems_OM_11e_Stevenson11 / 22Chapter 9 - Quality Control4. Sample Mean Range179.48 2.6 Mean Chart: =X ± A 2-R = 79.96 ± 0.58(1.87) 2 80.14 2.3 = 79.96 ± 1.083 80.14 1.2UCL = 81.04, LCL = 78.884 79.60 1.7 Range Chart: UCL = D 4-R = 2.11(1.87) = 3.95 5 80.02 2.0LCL = D 3-R = 0(1.87) = 0680.381.4[Both charts suggest the process is in control: Neither has any points outside the limits.]6. n = 200 Control Limits = np p p )1(2-±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. 857.714110c == Control limits:409.8857.7c 3c ±=±UCL is 16.266, LCL becomes 0.All values are within the limits.14. Let USL = Upper Specification Limit, LSL = Lower Specification Limit,X= Process mean, σ = Process standard deviationFor process H:}{capablenot ,0.193.93.04.1 ,938.min04.1)32)(.3(1516393.)32)(.3(1.14153<===-=σ-=-=σ-pkCXUSL LSL X0096.)200(1325==p 0138.0096.200)9904(.0096.20096.±=±=For process K:.1}17.1,0.1min{17.1)1)(3(335.3630.1)1)(3(30333===-=σ-=-=σ- CXUSL LSL X pkAssuming the minimum acceptable p kC is 1.33, since 1.0 < 1.33, the process is notcapable.For process T:33.1}33.1,67.1min{33.1)4.0)(3(5.181.20367.1)4.0)(3(5.165.183===-=σ-=-=σ- CXUSL LSL X pkSince 1.33 = 1.33, the process is capable.Chapter 10 - Aggregate Planning and Mas ter Scheduling7. a.No backlogs are allowedPeriodForecast Output Regular Overtime Subcontract Output - Forecast Inventory Beginning Ending Average Backlog Costs: Regular Overtime Subcontract InventoryTotalSolutions_Problems_OM_11e_Stevenson13 / 22b.Level strategyPeriodForecast Output Regular Overtime Subcontract Output - Forecast Inventory Beginning Ending Average Backlog Costs: Regular Overtime Subcontract Inventory BacklogTotal8.Period Forecast Output Regular Overtime Subcontract Output- Forecast Inventory Beginning Ending Average Backlog Costs: Regular Overtime Subcontract Inventory Backlog TotalChapter 11 - MRP and ERP1. a. F: 2G: 1H: 1J: 2 x 2 = 4 L: 1 x 2 = 2 A: 1 x 4 = 4 D: 2 x 4 = 8 J: 1 x 2 = 2 D: 1 x 2 = 2Totals: F = 2; G = 1; H = 1; J = 6; D = 10; L = 2; A = 44. MasterScheduleSolutions_Problems_OM_11e_Stevenson10. Week 1 2 3 4Material 40 80 60 70Week 1 2 3 4Labor hr. 160 320 240 280Mach. hr. 120 240 180 210a. Capacity utilizationWeek 1 2 3 4Labor 53.3% 106.7% 80% 93.3%Machine 60% 120% 90% 105%b. Capacity utilization exceeds 100% for both labor and machine in week 2, and formachine alone in week 4.Production could be shifted to earlier or later weeks in which capacity isunderutilized. 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 toovertime premium, a probable decrease in productivity, and possible increase inaccidents.15 / 22Chapter 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.a. See table.b. To allocate control efforts.c. It might be important for some reason other than dollar usage, such as cost of astockout, usage highly correlated to an A item, etc.3. D = 1,215 bags/yr. S = $10 H = $75a.bagsHDS Q 187510)215,1(22===b. Q/2 = 18/2 = 9 bagsc.ordersordersbags bags QD 5.67/ 18 215,1==d .SQD H 2/Q TC +=Solutions_Problems_OM_11e_Stevenson17 / 22350,1$675675)10(18215,1)75(218=+=+=e. Assuming that holding cost per bag increases by $9/bag/yearQ ==84)10)(215,1(217 bags71.428,1$71.714714)10(17215,1)84(217=+=+=TCIncrease by [$1,428.71 – $1,350] = $78.714.D = 40/day x 260 days/yr. = 10,400 packagesS = $60 H = $30a.ox esb 20496.2033060)400,10(2HDS 2Q====b.SQD H 2Q TC +=82.118,6$82.058,3060,3)60(204400,10)30(2204=+=+=c. Yesd.)60(200400,10)30(2200TC200+=TC 200 = 3,000 + 3,120 = $6,1206,120 – 6,118.82 (only $1.18 higher than with EOQ, so 200 is acceptable.)7.H = $2/monthS = $55D 1 = 100/month (months 1–6) D 2 = 150/month (months 7–12)a.16.74255)100(2Q:D HDS 2Q10===83.90255)150(2Q:D2==b. The EOQ model requires this.c. Discount of $10/order is equivalent to S – 10 = $45 (revised ordering cost)1–6 TC74 = $148.32180$)45(150100)2(2150TC145$)45(100100)2(2100TC*140$)45(50100)2(250TC15010050=+==+==+=7–12 TC 91 =$181.66195$)45(150150)2(2150TC*5.167$)45(100150)2(2100TC185$)45(50150)2(250TC15010050=+==+==+=10. p = 50/ton/day u = 20 tons/day 200 days/yr.S = $100 H = $5/ton per yr.a.bags][10,328 tons 40.5162050505100)000,4(2up p HDS 2Q 0=-=-=b. ]bags 8.196,6 .approx [ tons 84.309)30(504.516)u p (PQ I max ==-=Average is92.154248.309:2I max =tons [approx. 3,098 bags] c. Run length =days33.10504.516PQ ==d. Runs per year = 8]approx .[ 7.754.516000,4QD ==e. Q ' = 258.2TC =SQD H 2I max +TC orig. = $1,549.00 TC rev. = $ 774.50Savings would be $774.50D= 20 tons/day x 200 days/yr. = 4,000 tons/yr.Solutions_Problems_OM_11e_Stevenson19 / 2215.Range PH Q D = 4,900 seats/yr. 0–999 $5.00 $2.00 495 H = .4P 1,000–3,999 4.95 1.98 497 NF S = $50 4,000–5,999 4.90 1.96 500 NF6,000+ 4.85 1.94 503 NFCompare TC 495 with TC for all lower price breaks:TC 495 =495 ($2) + 4,900($50) + $5.00(4,900) = $25,490 2 495 TC 1,000 = 1,000 ($1.98) + 4,900($50) + $4.95(4,900) = $25,4902 1,000 TC 4,000 = 4,000 ($1.96) + 4,900($50) + $4.90(4,900) = $27,9912 4,000 TC 6,000 = 6,000 ($1.94) + 4,900($50) + $4.85(4,900) = $29,6262 6,000Hence, one would be indifferent between 495 or 1,000 units 22. d = 30 gal./day ROP = 170 gal. LT = 4 days,ss = Z σd LT = 50 galRisk = 9% Z = 1.34 Solving, σd LT = 37.31 3% 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 hourC T = 75 min. = 1.25 hr. = 80(1.25) (1.35)= 3C = 45 45X = .35QuantityTC4. The smallest daily quantity evenly divisible into all four quantities is 3. Therefore, usethree cycles.Product Daily quantity Units per cycleA 21 21/3 = 7B 12 12/3 = 4C 3 3/3 = 1D 15 15/3 = 55.a. Cycle 1 2 3 4A 6 6 5 5B 3 3 3 3C 1 1 1 1D 4 4 5 5E 2 2 2 2 b. Cycle 1 2A 11 11B 6 6C 2 2D 8 8E 4 4c. 4 cycles = lower inventory, more flexibility2 cycles = fewer changeovers7. Net available time = 480 – 75 = 405. Takt time = 405/300 units per day = 1.35 minutes. Chapter 15 - Scheduling6. a. FCFS: A–B–C–DSPT: D–C–B–AEDD: C–B–D–ACR: A–C–D–BFCFS: Job time Flow time Due date DaysJob (days) (days) (days) tardyA 14 14 20 0B 10 24 16 8C 7 31 15 16D 6 37 17 2037 106 44Solutions_Problems_OM_11e_Stevenson21 / 22SPT: Job time Flow time Due date Days Job (days) (days) (days) tardy D 6 6 17 0 C 7 13 15 0 B 10 23 16 7 A 14 37 20 17377924EDD:Job D has the lowest critical ratio therefore it is scheduled next and completed on day27.b.a rd i F lo w tim e A v e r a g e flo w tim e N u mb e r o f jo b s D a y s ta r d y A v e r a g e jo b t n e s s N u m b e r o f jo b s F lo w tim e A v e r a g e n u m b e r o f jo b s a t th ec e n te r M a k e s p a n==∑=FCFS SPT EDD CR26.50 19.75 21.00 24.75 11.0 6.00 6.00 9.25 2.862.142.272.67c. SPT is superior.9.Thus, the sequence is b-a-g-e-f-d-c.。

运营管理课程习题及答案修订版(供参考)

运营管理课程习题及答案修订版(供参考)

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课后习题标准答案

《运营管理》课后习题标准答案

《运营管理》课后习题答案————————————————————————————————作者：————————————————————————————————日期：2Chapter 02 - Competitiveness, Strategy, and Productivity3. (1) (2) (3) (4) (5) (6) (7)Week Output WorkerCost@$12x40Overhead********MaterialCost@$6TotalCostMFP(2) ÷ (6)1 30,000 2,880 4,320 2,700 9,900 3.032 33,600 3,360 5,040 2,820 11,220 2.993 32,200 3,360 5,040 2,760 11,160 2.894 35,400 3,840 5,760 2,880 12,480 2.84*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) Deposit —place in an envelope (which you’ll find near or in the ATM) andinsert it into the deposit slot2) Withdrawal —lift the “Withdrawal Door,” being careful to remove all cash6. Remove card and receipt (which serves as the transaction record)8.TechnicalRequirements IngredientsHandlingPreparationCustomer RequirementsTaste √√ Appearance√ √√Texture/consistency√√Chapter 04 - Strategic Capacity Planning for Products and Services2. %80capacityEffective outputActual Efficiency ==Actual output = .8 (Effective capacity) Effective capacity = .5 (Design capacity) Actual output = (.5)(.8)(Effective capacity) Actual output = (.4)(Design capacity) Actual output = 8 jobs Utilization = .4capacityDesign outputActual =n Utilizatiojobs 204.8capacity Effective output Actual Capacity Design ===10. 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 machineProductABC 1 48,000 64,000 32,000 2 48,000 48,000 36,000 3 30,000 36,000 24,000 460,000 60,000 30,000 Total 186,000208,000122,000machine181.000,150000,122machine 238.1000,150000,208machine224.1000,150000,186≈==≈==≈==C B A N N NYou 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 = $98,133 C(1): 122,000 ÷ 60 = 2,033.33 hrs. x $12 = $24,400 + $80,000 = $104,400Buy 2 Bs —these have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3.3 adf752 b4 c4 e9 h5 i6 gDesired output = 4Operating time = 56 minutesunit per minutes 14hourper units 4hourper minutes 65output Desired time Operating CT ===Task # of Following tasksPositional WeightA 4 23B 3 20C 2 18D 3 25E 2 18F 4 29G 3 24H 1 14 I5a. First rule: most followers. Second rule: largest positional weight.Assembly Line Balancing Table (CT = 14)Work StationTask Task TimeTime RemainingFeasible tasksRemainingIF 5 9 A,D,G A 3 6 B,G G6 – – II D7 7 B, E B 2 5 C C4 1 – III E 4 10 H H9 1 – IV I59–b. First rule: Largest positional weight.Assembly Line Balancing Table (CT = 14)Work StationTask Task TimeTime RemainingFeasible tasks RemainingIF 5 9 A,D,G D7 2 – II G 6 8 A, E A 3 5 B,E B2 3 – III C 4 10 E E4 6 – IV H 95 I I5–c. %36.805645stations of no. x CT time Total Efficiency ===4. a. l.2. Minimum Ct = 1.3 minutesTask Following tasksa 4b 3c 3d 2e 3f 2g 1habd cfeghWork StationEligible Assign Time RemainingIdle TimeIa A 1.1 b,c,e, (tie)B 0.7C 0.4E 0.3 0.3 II d D 0.0 0.0 IIIf,g F 0.5G 0.2 0.2 IVh H 0.1 0.10.63. percent 54.11)3.1(46.CT x N time)(idle percent Idle ==∑=4. 420 min./day 323.1 ( 323)/1.3 min./OT Output rounds to copiers day CT cycle=== b. 1. inutes m 3.224.6N time Total CT ,6.4 time Total ==== 2. Assign a, b, c, d, and e to station 1: 2.3 minutes [no idle time]Assign f, g, and h to station 2: 2.3 minutes3. 420182.6 copiers /2.3OT Output day CT ===4.420 min./dayMaximum Ct is 4.6. Output 91.30 copiers /4.6 min./day cycle==7. 1 5 4 3 8 762Chapter 06 - Work Design and Measurement3. Element PR OT NT AF job ST1 .90.46.414 1.15 .4762 .85 1.505 1.280 1.15 1.4723 1.10.83.913 1.15 1.05041.00 1.16 1.160 1.15 1.334Total4.3328. A = 24 + 10 + 14 = 48 minutes per 4 hours.min 125.720.11x70.5ST .min 70.5)95(.6NT 20.24048A =-=====9. a. Element PR OT NT A ST1 1.10 1.19 1.309 1.15 1.5052 1.15 .83 .955 1.15 1.09831.05.56.588 1.15 .676b.01.A 00.2z 034.s 83.x ==== 222(.034)67.12~68.01(.83)zs n observations ax ⎛⎫⎛⎫===⎪ ⎪⎝⎭⎝⎭c. e = .01 minutes 47 to round ,24.4601.)034(.2e zs n 22=⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛=Chapter 07- Location Planning and Analysis1. Factor Local bank Steel mill Food warehouse Public school1. Convenience forcustomers H L M–H M–H2. Attractiveness ofbuilding H L M M–H3. Nearness to rawmaterials L H L M4. Large amounts ofpower L H L L5. Pollution controls L H L L6. Labor cost andavailability L M L L7. Transportationcosts L M–H M–H M8. Constructioncosts M H M M–HLocation (a) Location (b)4. Factor A B C Weight A B C1. Business Services 9 5 5 2/9 18/9 10/9 10/92. Community Services 7 6 7 1/9 7/9 6/9 7/93. Real Estate Cost 3 8 7 1/9 3/9 8/9 7/94. Construction Costs 5 6 5 2/9 10/9 12/9 10/95. Cost of Living 4 7 8 1/9 4/9 7/9 8/96. Taxes 5 5 5 1/9 5/9 5/9 4/97. Transportation 6 7 8 1/9 6/9 7/9 8/9Total 39 44 45 1.0 53/9 55/9 54/9 Each factor has a weight of 1/7.a. Composite Scores 39 44 45 7 7 7B orC is the best and A is least desirable.b. Business Services and Construction Costs both have a weight of 2/9; the other factors eachhave a weight of 1/9.5 x + 2 x + 2 x = 1 x = 1/9c. Composite ScoresA B C 53/9 55/9 54/9B is the best followed byC and then A.5.Locationx yA 3 7B 8 2C 4 6D 4 1E 6 4Totals 25 20-x =∑x i= 25 = 5.0 -y =∑y i= 20 = 4.0 n 5 n 5Hence, the center of gravity is at (5,4) and therefore the optimal location.Chapter 08 - Management of Quality1. ChecksheetWork Type FrequencyLube and Oil 12Brakes 7Tires 6Battery 4Transmission 1Total 30Pareto127641 Lube & Oil Brakes Tires Battery Trans.2 .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 management’s attention.4Power Per LamMissDidn’Not OutletDefectBurn LoosLampOtheCordbreak lunch3 2•• •• •• • ••• • ••• •••• ••• •• • •• • •••Chapter 9 - Quality Control4. Sample Mean Range179.48 2.6 Mean Chart: =X ± A 2-R = 79.96 ± 0.58(1.87) 2 80.14 2.3 = 79.96 ± 1.083 80.14 1.2UCL = 81.04, LCL = 78.884 79.60 1.7 Range Chart: UCL = D 4-R = 2.11(1.87) = 3.95 5 80.02 2.0LCL = D 3-R = 0(1.87) = 0680.381.4[Both charts suggest the process is in control: Neither has any points outside the limits.]6. n = 200 Control Limits = np p p )1(2-±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. 857.714110c ==Control limits: 409.8857.7c 3c ±=± UCL is 16.266, LCL becomes 0.All values are within the limits.14. Let USL = Upper Specification Limit, LSL = Lower Specification Limit,X = Process mean, σ = Process standard deviationFor process H:}{capablenot ,0.193.93.04.1 ,938.min 04.1)32)(.3(1516393.)32)(.3(1.14153<===-=σ-=-=σ-pk C X USL LSL X 0096.)200(1325==p 0138.0096.200)9904(.0096.20096.±=±=For process K:.1}17.1,0.1min{17.1)1)(3(335.3630.1)1)(3(30333===-=σ-=-=σ- C X USL LSL X pk Assuming the minimum acceptable pk C is 1.33, since 1.0 < 1.33, the process is not capable.For process T:33.1}33.1,67.1min{33.1)4.0)(3(5.181.20367.1)4.0)(3(5.165.183===-=σ-=-=σ- C X USL LSL X pk Since 1.33 = 1.33, the process is capable.Chapter 10 - Aggregate Planning and Master Scheduling7. a.No backlogs are allowedPeriod Mar. Apr. May Jun. July Aug. Sep. TotalForecast 50 44 55 60 50 40 51 350 Output Regular 40 40 40 40 40 40 40 280 Overtime 8 8 8 8 8 3 8 51 Subcontract 2 0 3 12 2 0 0 19 Output - Forecast 0 4 –4 0 0 3 –3 Inventory Beginning 0 0 4 0 0 0 3 Ending 0 4 0 0 0 3 0 Average 0 2 2 0 0 1.5 1.5 7 Backlog 0 0 0 0 0 0 0 0 Costs: Regular 3,200 3,200 3,200 3,200 3,200 3,200 3,200 22,400 Overtime 960 960 960 960 960 360 960 6,120 Subcontract 280 0 420 1,680280 0 0 2,660 Inventory 0 20 20 0 0 15 15 70 Total4,4404,1804,6005,8404,4403,575 4,17531,250b. Level strategyPeriod Mar. Apr. May Jun. July Aug. Sep. Total Forecast 50 44 55 60 50 40 51 350 OutputRegular 40 40 40 40 40 40 40 280 Overtime 8 8 8 8 8 8 8 56 Subcontract 2 2 2 2 2 2 2 14 Output - Forecast 0 6 –5 –10 0 10 –1InventoryBeginning 0 0 6 1 0 0 1Ending 0 6 1 0 0 1 0Average 0 3 3.5 .5 0 .5 .5 8 Backlog 0 0 0 9 9 0 0 18 Costs:Regular 3,200 3,200 3,200 3,200 3,200 3,200 3,200 22,400 Overtime 960 960 960 960 960 960 960 6,720 Subcontract 280 280 280 280 280 280 280 1,960 Inventory 30 35 5 0 5 5 80 Backlog 180 180 360 Total 4,440 4,470 4,475 4,625 4,620 4,445 4,445 31,520 8.Period 1 2 3 4 5 6 TotalForecast 160 150 160 180 170 140 960OutputRegular 150 150 150 150 160 160 920Overtime 10 10 0 10 10 10 50Subcontract 0 0 10 10 0 0 20Output- Forecast 0 10 0 –10 0 0InventoryBeginning 0 0 10 10 0 0Ending 0 10 10 0 0 0Average 0 5 10 5 0 0 20Backlog 0 0 0 0 0 0 0Costs:Regular 7,500 7,500 7,500 7,500 8,000 8,000 46,000Overtime 750 750 0 750 750 750 3,750Subcontract 0 0 800 800 0 0 1,600Inventory 20 40 20 80Backlog 0 0 0 0 0 0Total 8,250 8,270 8,340 9,070 9,050 8,750 51,430Chapter 11 - MRP and ERP1. a. F: 2 G: 1 H: 1J: 2 x 2 = 4 L: 1 x 2 = 2 A: 1 x 4 = 4D: 2 x 4 = 8 J: 1 x 2 = 2 D: 1 x 2 = 2Totals: F = 2; G = 1; H = 1; J = 6; D = 10; L = 2; A = 4b.4. Master Schedule Day Beg. Inv. 1 2 3 4 5 6 7 Quantity100 150 200 TableBeg. Inv. 1 2 3 4 5 6 7 Gross requirements 100 150 200 Scheduled receipts Projected on hand Net requirements 100 150 200 Planned-order receipts 100 150 200 Planned-order releases 100 150 200Wood Sections Beg. Inv. 1 2 3 4 5 6 7 Gross requirements 200300 400 Scheduled receipts 100 Projected on hand 100100 Net requirements 100 300 400 Planned-order receipts 100 300 400 Planned-order releases400 400Braces Beg. Inv. 1 2 3 4 5 6 7 Gross requirements 300 450 600 Scheduled receipts Projected on hand 60 60 60 60 Net requirements 240 450 600 Planned-order receipts 240 450 600Planned-order releases 240 450 600StaplerTopBaseCoveSpri SlideBase Strik RubberSlidSpriLegs Beg.Inv.1 2 3 4 5 6 7Gross requirements 400 600 800Scheduled receiptsProjected on hand 120 120 120 120 88 88 71 Net requirements 280 600 800Planned-order receipts 308 660 880Planned-order releases 968 88010. Week 1 2 3 4Material 40 80 60 70Week 1 2 3 4Labor hr. 160 320 240 280Mach. hr. 120 240 180 210a. Capacity utilizationWeek 1 2 3 4Labor 53.3% 106.7% 80% 93.3%Machine 60% 120% 90% 105%b. C apacity utilization exceeds 100% for both labor and machine in week 2, and formachine alone in week 4.Production could be shifted to earlier or later weeks in which capacity isunderutilized. 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 toovertime premium, a probable decrease in productivity, and possible increase inaccidents.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 Cost UsageUsageCategoryK34 10 200 2,000 C K35 25 600 15,000 A K36 36 150 5,400 B M10 16 25 400 C M20 20 80 1,600 C Z45 80 250 16,000 A F14 20 300 6,000 B F95 30 800 24,000 A F99 20 60 1,200 C D45 10 550 5,500 B D48 12 90 1,080 C D52 15 110 1,650 C D57 40 120 4,800 B N08 30 40 1,200 C P05 16 500 8,000 BP091030300Ca. See table.b. To allocate control efforts.c. It might be important for some reason other than dollar usage, such as cost of astockout, usage highly correlated to an A item, etc.3. D = 1,215 bags/yr. S = $10 H = $75a. bags HDS Q 187510)215,1(22===b. Q/2 = 18/2 = 9 bagsc.orders ordersbags bags Q D 5.67/ 18 215,1== d . S QD H 2/Q TC +=350,1$675675)10(18215,1)75(218=+=+=e. Assuming that holding cost per bag increases by $9/bag/yearQ ==84)10)(215,1(217 bags71.428,1$71.714714)10(17215,1)84(217=+=+=TCIncrease by [$1,428.71 – $1,350] = $78.714.D = 40/day x 260 days/yr. = 10,400 packagesS = $60 H = $30a. oxes b 20496.2033060)400,10(2H DS 2Q 0====b. S QD H 2Q TC +=82.118,6$82.058,3060,3)60(204400,10)30(2204=+=+=c. Yesd. )60(200400,10)30(2200TC 200+=TC 200 = 3,000 + 3,120 = $6,1206,120 – 6,118.82 (only $1.18 higher than with EOQ, so 200 is acceptable.)7.H = $2/month S = $55D 1 = 100/month (months 1–6)D 2 = 150/month (months 7–12)a. 16.74255)100(2Q :D H DS2Q 010===83.90255)150(2Q :D 02==b. The EOQ model requires this.c. Discount of $10/order is equivalent to S – 10 = $45 (revised ordering cost)1–6 TC74 = $148.32180$)45(150100)2(2150TC 145$)45(100100)2(2100TC *140$)45(50100)2(250TC 15010050=+==+==+=7–12 TC 91 =$181.66195$)45(150150)2(2150TC *5.167$)45(100150)2(2100TC 185$)45(50150)2(250TC 15010050=+==+==+=10. p = 50/ton/day u = 20 tons/day200 days/yr. S = $100 H = $5/ton per yr.a. bags] [10,328 tons 40.5162050505100)000,4(2u p p H DS 2Q 0=-=-=b. ]bags 8.196,6 .approx [ tons 84.309)30(504.516)u p (P Q I max ==-=Average is92.154248.309:2I max =tons [approx. 3,098 bags] c. Run length =days 33.10504.516P Q == d. Runs per year = 8] approx .[ 7.754.516000,4QD == e. Q ' = 258.2TC =S QD H 2I max + TC orig. = $1,549.00 TC rev. = $ 774.50Savings would be $774.50D= 20 tons/day x 20015. RangeP H Q D = 4,900 seats/yr. 0–999 $5.00 $2.00 495 H = .4P 1,000–3,999 4.95 1.98 497 NF S = $50 4,000–5,999 4.90 1.96 500 NF 6,000+4.851.94503 NFCompare TC 495 with TC for all lower price breaks:TC 495 =495 ($2) + 4,900($50) + $5.00(4,900) = $25,490 2 495 TC 1,000 = 1,000 ($1.98) + 4,900($50) + $4.95(4,900) = $25,4902 1,000 TC 4,000 = 4,000 ($1.96) + 4,900($50) + $4.90(4,900) = $27,9912 4,000 TC 6,000 = 6,000 ($1.94) + 4,900($50) + $4.85(4,900) = $29,6262 6,000Hence, one would be indifferent between 495 or 1,000 units 22. d = 30 gal./day ROP = 170 gal. LT = 4 days,ss = Z σd LT = 50 galRisk = 9% Z = 1.34 Solving, σd LT = 37.31 3% 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 hourC T = 75 min. = 1.25 hr. = 80(1.25) (1.35)= 3C = 45 45X = .35• •• •495 497 500 5031,0004,000 6,000QuantityTC4. The smallest daily quantity evenly divisible into all four quantities is 3. Therefore, usethree cycles.Product Daily quantity Units per cycleA 21 21/3 = 7B 12 12/3 = 4C 3 3/3 = 1D 15 15/3 = 55.a. Cycle 1 2 3 4A 6 6 5 5B 3 3 3 3C 1 1 1 1D 4 4 5 5E 2 2 2 2 b. Cycle 1 2A 11 11B 6 6C 2 2D 8 8E 4 4c. 4 cycles = lower inventory, more flexibility2 cycles = fewer changeovers7. Net available time = 480 – 75 = 405. Takt time = 405/300 units per day = 1.35 minutes. Chapter 15 - Scheduling6. a. FCFS: A–B–C–DSPT: D–C–B–AEDD: C–B–D–ACR: A–C–D–BFCFS: Job time Flow time Due date DaysJob (days) (days) (days) tardyA 14 14 20 0B 10 24 16 8C 7 31 15 16D 6 37 17 2037 106 44SPT: Job time Flow time Due date Days Job (days) (days) (days) tardyD 6 6 17 0C 7 13 15 0B 10 23 16 7A 14 37 20 1737 79 24EDD: Job time Flow time Due date DaysJob (days) (days) (days) tardyC 7 7 15 0B 10 17 16 1D 6 23 17 6A 14 37 20 1784 24Critical RatioJob Processing Time(Days) Due Date Critical Ratio CalculationA 14 20 (20 – 0) / 14 = 1.43B 10 16 (16 – 0) /10 = 1.60C 7 15 (15 – 0) / 7 = 2.14D 6 17 (17 – 0) / 6 = 2.83Job 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:Job Processing Time(Days) Due Date Critical Ratio CalculationA –––B 10 16 (16 – 14) /10 = 0.20C 7 15 (15 – 14) / 7 = 0.14D 6 17 (17 – 14) / 6 = 0.50Job 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:Job Processing Time(Days) Due Date Critical Ratio CalculationA –––B 10 16 (16 – 21) /10 = –0.50C –––D 6 17 (17 – 21) / 6 = –0.67Job D has the lowest critical ratio therefore it is scheduled next and completed on day 27. The critical ratio sequence is A –C –D –B and the makespan is 37 days. Critical Ratio sequenceProcessing Time(Days)Flow time Due Date TardinessA 14 14 20 0 C 7 21 15 6 D 6 27 17 10 B1037 16 21 ∑9937b.ardi Flow time Average flow time Number of jobsDays tardy Average job t ness Number of jobs Flow timeAverage number of jobs at the center Makespan==∑=FCFS SPT EDD CR26.50 19.75 21.00 24.75 11.0 6.00 6.00 9.25 2.86 2.142.272.67c. SPT is superior.9.Time (hr.) Sequence of assignment:Order Step 1 Step 2A 1.20 1.40 .80 [C] last (or 7th)B 0.90 1.30 .90 [B] firstC 2.00 0.80 1.20 [A] 2ndD 1.70 1.50 1.30 [G] 3rdE 1.60 1.80 1.60 [E] 4thF 2.20 1.75 1.50 [D] 6th G1.301.401.75[F]5thThus, the sequence is b-a-g-e-f-d-c.。

《运营管理》课程习题和答案解析_修订版

《运营管理》课程习题和答案解析_修订版

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第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中,处于核心地位的是:（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（ )。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置,加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（)A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（ )A.服务过程B.生产过程 C。

计划过程 D。

管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程 C。

服务过程 D。

生产过程6、制造业企业与服务业企业最主要的一个区别是（ )A.产出的物理性质B.与顾客的接触程度 C。

产出质量的度量 D。

对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发 B。

产品设计 C.生产运营活动 D。

生产系统的选择8、下列哪项不是生产运作管理的目标（ )A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在( ）A。

《运营管理》课程习题和答案解析_修订版

《运营管理》课程习题和答案解析_修订版

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域围2、运营管理中的决策容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课程习题及答案_修订版(1)

《运营管理》课程习题及答案_修订版(1)

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课后习题答案

《运营管理》课后习题答案

Chapter 02 - Competitiveness, Strategy, and Productivity3. (1) (2) (3) (4) (5) (6) (7)Week Output WorkerCost@$12x40OverheadCost @1.5MaterialCost@$6TotalCostMFP(2) ÷ (6)1 30,000 2,880 4,320 2,700 9,900 3.032 33,600 3,360 5,040 2,820 11,220 2.993 32,200 3,360 5,040 2,760 11,160 2.894 35,400 3,840 5,760 2,880 12,480 2.84*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) Deposit —place in an envelope (which you’ll find near or in the ATM) andinsert it into the deposit slot2) Withdrawal —lift the “Withdrawal Door,” being careful to remove all cash6. Remove card and receipt (which serves as the transaction record)8.Chapter 04 - Strategic Capacity Planning for Products and Services2. %80capacityEf f ective outputActual Ef f iciency ==Actual output = .8 (Effective capacity) Effective capacity = .5 (Design capacity) Actual output = (.5)(.8)(Effective capacity) Actual output = (.4)(Design capacity) Actual output = 8 jobs Utilization = .4capacityDesign outputActual =n Utilizatiojobs 204.8capacity Ef f ective output Actual Capacity Design ===10. a. Given: 10 hrs. or 600 min. of operating time per day.250 days x 600 min. = 150,000 min. per year operating time.Solutions_Problems_OM_11e_Stevenson3 / 22Total processing time by machineProductABC 1 48,000 64,000 32,000 2 48,000 48,000 36,000 3 30,000 36,000 24,000 460,000 60,000 30,000 Total 186,000208,000122,000machine181.000,150000,122machine 238.1000,150000,208machine224.1000,150000,186≈==≈==≈==C B A N N NYou 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 = $98,133 C(1): 122,000 ÷ 60 = 2,033.33 hrs. x $12 = $24,400 + $80,000 = $104,400Buy 2 Bs —these have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3.Desired output = 4Operating time = 56 minutesunit per minutes 14hourper units 4hourper minutes 65output Desired time Operating CT ===Task # of Following tasksPositional WeightA 4 23B 3 20C 2 18D 3 25E 2 18F 4 29G 3 24H 1 14 I5a. First rule: most followers. Second rule: largest positional weight.Assembly Line Balancing Table (CT = 14)Solutions_Problems_OM_11e_Stevenson5 / 22b. First rule: Largest positional weight.Assembly Line Balancing Table (CT = 14)c. %36.805645stations of no. x CT time Total Ef f iciency ===4. a. l.2. Minimum Ct = 1.3 minutesTask Following tasksa 4b 3c 3d 2e 3f 2g 1h3. percent 54.11)3.1(46.CT x N time)(idle percent Idle ==∑=4. 420 min./day 323.1 ( 323)/1.3 min./OT Output rounds to copiers day CT cycle=== b. 1. inutes m 3.224.6N time Total CT ,6.4 time Total ==== 2. Assign a, b, c, d, and e to station 1: 2.3 minutes [no idle time]Assign f, g, and h to station 2: 2.3 minutes3. 420182.6 copiers /2.3OT Output day CT ===4.420 min./dayMaximum Ct is 4.6. Output 91.30 copiers /4.6 min./day cycle==7.Solutions_Problems_OM_11e_Stevenson7 / 22Chapter 06 - Work Design and Measurement3.Element PR OT NT AF job ST1 .90.46.414 1.15 .4762 .85 1.505 1.280 1.15 1.4723 1.10.83.913 1.15 1.05041.00 1.16 1.160 1.15 1.334Total4.3328. A = 24 + 10 + 14 = 48 minutes per 4 hours.min 125.720.11x70.5ST .min 70.5)95(.6NT 20.24048A =-=====9. a. Element PR OT NT A ST1 1.10 1.19 1.309 1.15 1.5052 1.15 .83 .955 1.15 1.09831.05.56.588 1.15 .676b.01.A 00.2z 034.s 83.x ==== 222(.034)67.12~68.01(.83)zs n observations ax ⎛⎫⎛⎫===⎪ ⎪⎝⎭⎝⎭ c. e = .01 minutes 47 to round ,24.4601.)034(.2e zs n 22=⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛=Chapter 07- Location Planning and Analysis1. Factor Local bank Steel mill Food warehouse Public school1. Convenience forcustomers H L M–H M–H2. Attractiveness ofbuilding H L M M–H3. Nearness to rawmaterials L H L M4. Large amounts ofpower L H L L5. Pollution controls L H L L6. Labor cost andavailability L M L L7. Transportationcosts L M–H M–H M8. Constructioncosts M H M M–HLocation (a) Location (b)4. Factor A B C Weight A B C1. Business Services 9 5 5 2/9 18/9 10/9 10/92. Community Services 7 6 7 1/9 7/9 6/9 7/93. Real Estate Cost 3 8 7 1/9 3/9 8/9 7/94. Construction Costs 5 6 5 2/9 10/9 12/9 10/95. Cost of Living 4 7 8 1/9 4/9 7/9 8/96. Taxes 5 5 5 1/9 5/9 5/9 4/97. Transportation 6 7 8 1/9 6/9 7/9 8/9Total 39 44 45 1.0 53/9 55/9 54/9 Each factor has a weight of 1/7.a. Composite Scores 39 44 45 7 7 7B orC is the best and A is least desirable.b. Business Services and Construction Costs both have a weight of 2/9; the other factors eachhave a weight of 1/9.5 x + 2 x + 2 x = 1 x = 1/9c. Composite ScoresA B C 53/9 55/9 54/9B is the best followed byC and then A.Solutions_Problems_OM_11e_Stevenson9 / 225. Location x yA3 7 B 8 2 C4 6 D4 1E 6 4 Totals 25 20-x = ∑x i = 25 = 5.0-y =∑y i = 20= 4.0n5n5Hence, the center of gravity is at (5,4) and therefore the optimal location.Chapter 08 - Management of Quality1.ChecksheetWork Type FrequencyLube and Oil 12 Brakes 7 Tires 6 Battery 4 Transmission1Total30ParetoLube & Oil BrakesTiresBatteryTrans.2 .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 b e given management’s attention.4break lunch break3 2 1 0∙ ∙ ∙∙ ∙ ∙ ∙∙∙ ∙ ∙ ∙∙∙∙∙∙∙ ∙∙∙ ∙∙ ∙ ∙∙ ∙ ∙∙∙Solutions_Problems_OM_11e_Stevenson11 / 22Chapter 9 - Quality Control4. Sample Mean Range179.48 2.6 Mean Chart: =X ± A 2-R = 79.96 ± 0.58(1.87) 2 80.14 2.3 = 79.96 ± 1.083 80.14 1.2UCL = 81.04, LCL = 78.884 79.60 1.7 Range Chart: UCL = D 4-R = 2.11(1.87) = 3.95 5 80.02 2.0LCL = D 3-R = 0(1.87) = 0680.381.4[Both charts suggest the process is in control: Neither has any points outside the limits.]6. n = 200 Control Limits = np p p )1(2-±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. 857.714110c == Control limits: 409.8857.7c 3c ±=± 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, σ = Process standard deviationFor process H:}{capablenot ,0.193.93.04.1 ,938.min 04.1)32)(.3(1516393.)32)(.3(1.14153<===-=σ-=-=σ-pk C X USL LSL X 0096.)200(1325==p 0138.0096.200)9904(.0096.20096.±=±=For process K:.1}17.1,0.1min{17.1)1)(3(335.3630.1)1)(3(30333===-=σ-=-=σ- C X USL LSL X pk Assuming the minimum acceptable pk C is 1.33, since 1.0 < 1.33, the process is not capable.For process T:33.1}33.1,67.1min{33.1)4.0)(3(5.181.20367.1)4.0)(3(5.165.183===-=σ-=-=σ- C X USL LSL X pk Since 1.33 = 1.33, the process is capable.Chapter 10 - Aggregate Planning and Master Scheduling7. a. No backlogs are allowedPeriodForecast Output Regular Overtime Subcontract Output - Forecast Inventory Beginning Ending Average Backlog Costs: Regular Overtime Subcontract Inventory TotalSolutions_Problems_OM_11e_Stevenson13 / 22b.Level strategyPeriodForecast Output Regular Overtime Subcontract Output - Forecast Inventory Beginning Ending Average Backlog Costs: Regular Overtime Subcontract Inventory Backlog Total8.Period Forecast Output Regular Overtime Subcontract Output- Forecast Inventory Beginning Ending Average Backlog Costs: Regular Overtime Subcontract Inventory Backlog TotalChapter 11 - MRP and ERP1. a. F: 2G: 1H: 1J: 2 x 2 = 4 L: 1 x 2 = 2 A: 1 x 4 = 4 D: 2 x 4 = 8 J: 1 x 2 = 2 D: 1 x 2 = 2Totals: F = 2; G = 1; H = 1; J = 6; D = 10; L = 2; A = 44. Master ScheduleSolutions_Problems_OM_11e_Stevenson10. Week 1 2 3 4Material 40 80 60 70Week 1 2 3 4Labor hr. 160 320 240 280Mach. hr. 120 240 180 210a. Capacity utilizationWeek 1 2 3 4Labor 53.3% 106.7% 80% 93.3%Machine 60% 120% 90% 105%b. C apacity utilization exceeds 100% for both labor and machine in week 2, and formachine alone in week 4.Production could be shifted to earlier or later weeks in which capacity isunderutilized. 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 toovertime premium, a probable decrease in productivity, and possible increase inaccidents.15 / 22Chapter 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.a. See table.b. To allocate control efforts.c. It might be important for some reason other than dollar usage, such as cost of astockout, usage highly correlated to an A item, etc.3. D = 1,215 bags/yr. S = $10 H = $75a. bags HDS Q 187510)215,1(22===b. Q/2 = 18/2 = 9 bagsc.orders ordersbags bags Q D 5.67/ 18 215,1== d . S QD H 2/Q TC +=Solutions_Problems_OM_11e_Stevenson17 / 22350,1$675675)10(18215,1)75(218=+=+=e. Assuming that holding cost per bag increases by $9/bag/yearQ ==84)10)(215,1(217 bags71.428,1$71.714714)10(17215,1)84(217=+=+=TCIncrease by [$1,428.71 – $1,350] = $78.714.D = 40/day x 260 days/yr. = 10,400 packagesS = $60 H = $30a. oxes b 20496.2033060)400,10(2H DS 2Q 0====b. S QD H 2Q TC +=82.118,6$82.058,3060,3)60(204400,10)30(2204=+=+=c. Yesd. )60(200400,10)30(2200TC 200+=TC 200 = 3,000 + 3,120 = $6,1206,120 – 6,118.82 (only $1.18 higher than with EOQ, so 200 is acceptable.)7.H = $2/month S = $55D 1 = 100/month (months 1–6)D 2 = 150/month (months 7–12)a. 16.74255)100(2Q :D H DS2Q 010===83.90255)150(2Q :D 02==b. The EOQ model requires this.c. Discount of $10/order is equivalent to S – 10 = $45 (revised ordering cost)1–6 TC74 = $148.32180$)45(150100)2(2150TC 145$)45(100100)2(2100TC *140$)45(50100)2(250TC 15010050=+==+==+=7–12 TC 91 =$181.66195$)45(150150)2(2150TC *5.167$)45(100150)2(2100TC 185$)45(50150)2(250TC 15010050=+==+==+=10. p = 50/ton/day u = 20 tons/day200 days/yr.S = $100 H = $5/ton per yr.a. bags] [10,328 tons 40.5162050505100)000,4(2u p p H DS 2Q 0=-=-=b. ]bags 8.196,6 .approx [ tons 84.309)30(504.516)u p (P Q I max ==-=Average is92.154248.309:2I max =tons [approx. 3,098 bags] c. Run length =days 33.10504.516P Q == d. Runs per year = 8] approx.[ 7.754.516000,4QD == e. Q ' = 258.2TC =S QD H 2I max + TC orig. = $1,549.00 TC rev. = $ 774.50Savings would be $774.50D= 20 tons/day x 200 days/yr. = 4,000 tons/yr.Solutions_Problems_OM_11e_Stevenson19 / 2215. RangeP H Q D = 4,900 seats/yr. 0–999 $5.00 $2.00 495 H = .4P 1,000–3,999 4.95 1.98 497 NF S = $50 4,000–5,999 4.90 1.96 500 NF 6,000+ 4.85 1.94503 NFCompare TC 495 with TC for all lower price breaks:TC 495 = 495 ($2) + 4,900($50) + $5.00(4,900) = $25,4902 495 TC 1,000 = 1,000 ($1.98) + 4,900($50) + $4.95(4,900) = $25,4902 1,000 TC 4,000 = 4,000 ($1.96) + 4,900($50) + $4.90(4,900) = $27,9912 4,000 TC 6,000 = 6,000 ($1.94) + 4,900($50) + $4.85(4,900) = $29,6262 6,000Hence, one would be indifferent between 495 or 1,000 units22. d = 30 gal./day ROP = 170 gal. LT = 4 days,ss = Z σd LT = 50 galRisk = 9% Z = 1.34 Solving, σd LT = 37.31 3% 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 hourC T = 75 min. = 1.25 hr. = 80(1.25) (1.35)= 3C = 45 45X = .35QuantityTC4. The smallest daily quantity evenly divisible into all four quantities is 3. Therefore, usethree cycles.Product Daily quantity Units per cycleA 21 21/3 = 7B 12 12/3 = 4C 3 3/3 = 1D 15 15/3 = 55.a. Cycle 1 2 3 4A 6 6 5 5B 3 3 3 3C 1 1 1 1D 4 4 5 5E 2 2 2 2 b. Cycle 1 2A 11 11B 6 6C 2 2D 8 8E 4 4c. 4 cycles = lower inventory, more flexibility2 cycles = fewer changeovers7. Net available time = 480 – 75 = 405. Takt time = 405/300 units per day = 1.35 minutes. Chapter 15 - Scheduling6. a. FCFS: A–B–C–DSPT: D–C–B–AEDD: C–B–D–ACR: A–C–D–BFCFS: Job time Flow time Due date DaysJob (days) (days) (days) tardyA 14 14 20 0B 10 24 16 8C 7 31 15 16D 6 37 17 2037 106 44Solutions_Problems_OM_11e_Stevenson21 / 22SPT: Job time Flow time Due date Days Job (days) (days) (days) tardy D 6 6 17 0 C 7 13 15 0 B 10 23 16 7 A 14 37 20 17377924EDD:Job D has the lowest critical ratio therefore it is scheduled next and completed on day 27.b.ardi Flow time Average flow time Number of jobsDays tardy Average job t ness Number of jobs Flow timeAverage number of jobs at the center Makespan==∑=FCFS SPT EDD CR26.50 19.75 21.00 24.75 11.0 6.00 6.00 9.252.86 2.142.272.67c. SPT is superior.9.Thus, the sequence is b-a-g-e-f-d-c.。

《运营管理》课后习题答案

《运营管理》课后习题答案

Chapter 02 - Competitiveness, Strategy, and Productivity3. (1) (2) (3) (4) (5) (6) (7)Week Output WorkerCost@$12x40OverheadCost @1.5MaterialCost@$6TotalCostMFP(2) ÷ (6)1 30,000 2,880 4,320 2,700 9,900 3.032 33,600 3,360 5,040 2,820 11,220 2.993 32,200 3,360 5,040 2,760 11,160 2.894 35,400 3,840 5,760 2,880 12,480 2.84*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) Deposit —place in an envelope (which you’ll find near or in the ATM) andinsert it into the deposit slot2) Withdrawal —lift the “Withdrawal Door,” being careful to remove all cash6. Remove card and receipt (which serves as the transaction record)8.Chapter 04 - Strategic Capacity Planning for Products and Services2. %80capacityEf f ective outputActual Ef f iciency ==Actual output = .8 (Effective capacity) Effective capacity = .5 (Design capacity) Actual output = (.5)(.8)(Effective capacity) Actual output = (.4)(Design capacity) Actual output = 8 jobs Utilization = .4capacityDesign outputActual =n Utilizatiojobs 204.8capacity Ef f ective output Actual Capacity Design ===10. a. Given: 10 hrs. or 600 min. of operating time per day.250 days x 600 min. = 150,000 min. per year operating time.Solutions_Problems_OM_11e_Stevenson3 / 22Total processing time by machineProductABC 1 48,000 64,000 32,000 2 48,000 48,000 36,000 3 30,000 36,000 24,000 460,000 60,000 30,000 Total 186,000208,000122,000machine181.000,150000,122machine 238.1000,150000,208machine224.1000,150000,186≈==≈==≈==C B A N N NYou 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 = $98,133 C(1): 122,000 ÷ 60 = 2,033.33 hrs. x $12 = $24,400 + $80,000 = $104,400Buy 2 Bs —these have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3.Desired output = 4Operating time = 56 minutesunit per minutes 14hourper units 4hourper minutes 65output Desired time Operating CT ===Task # of Following tasksPositional WeightA 4 23B 3 20C 2 18D 3 25E 2 18F 4 29G 3 24H 1 14 I5a. First rule: most followers. Second rule: largest positional weight.Assembly Line Balancing Table (CT = 14)Solutions_Problems_OM_11e_Stevenson5 / 22b. First rule: Largest positional weight.Assembly Line Balancing Table (CT = 14)c. %36.805645stations of no. x CT time Total Ef f iciency ===4. a. l.2. Minimum Ct = 1.3 minutesTask Following tasksa 4b 3c 3d 2e 3f 2g 1h3. percent 54.11)3.1(46.CT x N time)(idle percent Idle ==∑=4. 420 min./day 323.1 ( 323)/1.3 min./OT Output rounds to copiers day CT cycle=== b. 1. inutes m 3.224.6N time Total CT ,6.4 time Total ==== 2. Assign a, b, c, d, and e to station 1: 2.3 minutes [no idle time]Assign f, g, and h to station 2: 2.3 minutes3. 420182.6 copiers /2.3OT Output day CT ===4.420 min./dayMaximum Ct is 4.6. Output 91.30 copiers /4.6 min./day cycle==7.Solutions_Problems_OM_11e_Stevenson7 / 22Chapter 06 - Work Design and Measurement3.Element PR OT NT AF job ST1 .90.46.414 1.15 .4762 .85 1.505 1.280 1.15 1.4723 1.10.83.913 1.15 1.05041.00 1.16 1.160 1.15 1.334Total4.3328. A = 24 + 10 + 14 = 48 minutes per 4 hours.min 125.720.11x70.5ST .min 70.5)95(.6NT 20.24048A =-=====9. a. Element PR OT NT A ST1 1.10 1.19 1.309 1.15 1.5052 1.15 .83 .955 1.15 1.09831.05.56.588 1.15 .676b.01.A 00.2z 034.s 83.x ==== 222(.034)67.12~68.01(.83)zs n observations ax ⎛⎫⎛⎫===⎪ ⎪⎝⎭⎝⎭ c. e = .01 minutes 47 to round ,24.4601.)034(.2e zs n 22=⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛=Chapter 07- Location Planning and Analysis1. Factor Local bank Steel mill Food warehouse Public school1. Convenience forcustomers H L M–H M–H2. Attractiveness ofbuilding H L M M–H3. Nearness to rawmaterials L H L M4. Large amounts ofpower L H L L5. Pollution controls L H L L6. Labor cost andavailability L M L L7. Transportationcosts L M–H M–H M8. Constructioncosts M H M M–HLocation (a) Location (b)4. Factor A B C Weight A B C1. Business Services 9 5 5 2/9 18/9 10/9 10/92. Community Services 7 6 7 1/9 7/9 6/9 7/93. Real Estate Cost 3 8 7 1/9 3/9 8/9 7/94. Construction Costs 5 6 5 2/9 10/9 12/9 10/95. Cost of Living 4 7 8 1/9 4/9 7/9 8/96. Taxes 5 5 5 1/9 5/9 5/9 4/97. Transportation 6 7 8 1/9 6/9 7/9 8/9Total 39 44 45 1.0 53/9 55/9 54/9 Each factor has a weight of 1/7.a. Composite Scores 39 44 45 7 7 7B orC is the best and A is least desirable.b. Business Services and Construction Costs both have a weight of 2/9; the other factors eachhave a weight of 1/9.5 x + 2 x + 2 x = 1 x = 1/9c. Composite ScoresA B C 53/9 55/9 54/9B is the best followed byC and then A.Solutions_Problems_OM_11e_Stevenson9 / 225. Location x yA3 7 B 8 2 C4 6 D4 1E 6 4 Totals 25 20-x = ∑x i = 25 = 5.0-y =∑y i = 20= 4.0n5n5Hence, the center of gravity is at (5,4) and therefore the optimal location.Chapter 08 - Management of Quality1.ChecksheetWork Type FrequencyLube and Oil 12 Brakes 7 Tires 6 Battery 4 Transmission1Total30ParetoLube & Oil BrakesTiresBatteryTrans.2 .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 b e given management’s attention.4break lunch break3 2 1 0∙ ∙ ∙∙ ∙ ∙ ∙∙∙ ∙ ∙ ∙∙∙∙∙∙∙ ∙∙∙ ∙∙ ∙ ∙∙ ∙ ∙∙∙Solutions_Problems_OM_11e_Stevenson11 / 22Chapter 9 - Quality Control4. Sample Mean Range179.48 2.6 Mean Chart: =X ± A 2-R = 79.96 ± 0.58(1.87) 2 80.14 2.3 = 79.96 ± 1.083 80.14 1.2UCL = 81.04, LCL = 78.884 79.60 1.7 Range Chart: UCL = D 4-R = 2.11(1.87) = 3.95 5 80.02 2.0LCL = D 3-R = 0(1.87) = 0680.381.4[Both charts suggest the process is in control: Neither has any points outside the limits.]6. n = 200 Control Limits = np p p )1(2-±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. 857.714110c == Control limits: 409.8857.7c 3c ±=± 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, σ = Process standard deviationFor process H:}{capablenot ,0.193.93.04.1 ,938.min 04.1)32)(.3(1516393.)32)(.3(1.14153<===-=σ-=-=σ-pk C X USL LSL X 0096.)200(1325==p 0138.0096.200)9904(.0096.20096.±=±=For process K:.1}17.1,0.1min{17.1)1)(3(335.3630.1)1)(3(30333===-=σ-=-=σ- C X USL LSL X pk Assuming the minimum acceptable pk C is 1.33, since 1.0 < 1.33, the process is not capable.For process T:33.1}33.1,67.1min{33.1)4.0)(3(5.181.20367.1)4.0)(3(5.165.183===-=σ-=-=σ- C X USL LSL X pk Since 1.33 = 1.33, the process is capable.Chapter 10 - Aggregate Planning and Master Scheduling7. a. No backlogs are allowedPeriodForecast Output Regular Overtime Subcontract Output - Forecast Inventory Beginning Ending Average Backlog Costs: Regular Overtime Subcontract Inventory TotalSolutions_Problems_OM_11e_Stevenson13 / 22b.Level strategyPeriodForecast Output Regular Overtime Subcontract Output - Forecast Inventory Beginning Ending Average Backlog Costs: Regular Overtime Subcontract Inventory Backlog Total8.Period Forecast Output Regular Overtime Subcontract Output- Forecast Inventory Beginning Ending Average Backlog Costs: Regular Overtime Subcontract Inventory Backlog TotalChapter 11 - MRP and ERP1. a. F: 2G: 1H: 1J: 2 x 2 = 4 L: 1 x 2 = 2 A: 1 x 4 = 4 D: 2 x 4 = 8 J: 1 x 2 = 2 D: 1 x 2 = 2Totals: F = 2; G = 1; H = 1; J = 6; D = 10; L = 2; A = 44. Master ScheduleSolutions_Problems_OM_11e_Stevenson10. Week 1 2 3 4Material 40 80 60 70Week 1 2 3 4Labor hr. 160 320 240 280Mach. hr. 120 240 180 210a. Capacity utilizationWeek 1 2 3 4Labor 53.3% 106.7% 80% 93.3%Machine 60% 120% 90% 105%b. C apacity utilization exceeds 100% for both labor and machine in week 2, and formachine alone in week 4.Production could be shifted to earlier or later weeks in which capacity isunderutilized. 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 toovertime premium, a probable decrease in productivity, and possible increase inaccidents.15 / 22Chapter 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.a. See table.b. To allocate control efforts.c. It might be important for some reason other than dollar usage, such as cost of astockout, usage highly correlated to an A item, etc.3. D = 1,215 bags/yr. S = $10 H = $75a. bags HDS Q 187510)215,1(22===b. Q/2 = 18/2 = 9 bagsc.orders ordersbags bags Q D 5.67/ 18 215,1== d . S QD H 2/Q TC +=Solutions_Problems_OM_11e_Stevenson17 / 22350,1$675675)10(18215,1)75(218=+=+=e. Assuming that holding cost per bag increases by $9/bag/yearQ ==84)10)(215,1(217 bags71.428,1$71.714714)10(17215,1)84(217=+=+=TCIncrease by [$1,428.71 – $1,350] = $78.714.D = 40/day x 260 days/yr. = 10,400 packagesS = $60 H = $30a. oxes b 20496.2033060)400,10(2H DS 2Q 0====b. S QD H 2Q TC +=82.118,6$82.058,3060,3)60(204400,10)30(2204=+=+=c. Yesd. )60(200400,10)30(2200TC 200+=TC 200 = 3,000 + 3,120 = $6,1206,120 – 6,118.82 (only $1.18 higher than with EOQ, so 200 is acceptable.)7.H = $2/month S = $55D 1 = 100/month (months 1–6)D 2 = 150/month (months 7–12)a. 16.74255)100(2Q :D H DS2Q 010===83.90255)150(2Q :D 02==b. The EOQ model requires this.c. Discount of $10/order is equivalent to S – 10 = $45 (revised ordering cost)1–6 TC74 = $148.32180$)45(150100)2(2150TC 145$)45(100100)2(2100TC *140$)45(50100)2(250TC 15010050=+==+==+=7–12 TC 91 =$181.66195$)45(150150)2(2150TC *5.167$)45(100150)2(2100TC 185$)45(50150)2(250TC 15010050=+==+==+=10. p = 50/ton/day u = 20 tons/day200 days/yr.S = $100 H = $5/ton per yr.a. bags] [10,328 tons 40.5162050505100)000,4(2u p p H DS 2Q 0=-=-=b. ]bags 8.196,6 .approx [ tons 84.309)30(504.516)u p (P Q I max ==-=Average is92.154248.309:2I max =tons [approx. 3,098 bags] c. Run length =days 33.10504.516P Q == d. Runs per year = 8] approx.[ 7.754.516000,4QD == e. Q ' = 258.2TC =S QD H 2I max + TC orig. = $1,549.00 TC rev. = $ 774.50Savings would be $774.50D= 20 tons/day x 200 days/yr. = 4,000 tons/yr.Solutions_Problems_OM_11e_Stevenson19 / 2215. RangeP H Q D = 4,900 seats/yr. 0–999 $5.00 $2.00 495 H = .4P 1,000–3,999 4.95 1.98 497 NF S = $50 4,000–5,999 4.90 1.96 500 NF 6,000+ 4.85 1.94503 NFCompare TC 495 with TC for all lower price breaks:TC 495 = 495 ($2) + 4,900($50) + $5.00(4,900) = $25,4902 495 TC 1,000 = 1,000 ($1.98) + 4,900($50) + $4.95(4,900) = $25,4902 1,000 TC 4,000 = 4,000 ($1.96) + 4,900($50) + $4.90(4,900) = $27,9912 4,000 TC 6,000 = 6,000 ($1.94) + 4,900($50) + $4.85(4,900) = $29,6262 6,000Hence, one would be indifferent between 495 or 1,000 units22. d = 30 gal./day ROP = 170 gal. LT = 4 days,ss = Z σd LT = 50 galRisk = 9% Z = 1.34 Solving, σd LT = 37.31 3% 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 hourC T = 75 min. = 1.25 hr. = 80(1.25) (1.35)= 3C = 45 45X = .35QuantityTC4. The smallest daily quantity evenly divisible into all four quantities is 3. Therefore, usethree cycles.Product Daily quantity Units per cycleA 21 21/3 = 7B 12 12/3 = 4C 3 3/3 = 1D 15 15/3 = 55.a. Cycle 1 2 3 4A 6 6 5 5B 3 3 3 3C 1 1 1 1D 4 4 5 5E 2 2 2 2 b. Cycle 1 2A 11 11B 6 6C 2 2D 8 8E 4 4c. 4 cycles = lower inventory, more flexibility2 cycles = fewer changeovers7. Net available time = 480 – 75 = 405. Takt time = 405/300 units per day = 1.35 minutes. Chapter 15 - Scheduling6. a. FCFS: A–B–C–DSPT: D–C–B–AEDD: C–B–D–ACR: A–C–D–BFCFS: Job time Flow time Due date DaysJob (days) (days) (days) tardyA 14 14 20 0B 10 24 16 8C 7 31 15 16D 6 37 17 2037 106 44Solutions_Problems_OM_11e_Stevenson21 / 22SPT: Job time Flow time Due date Days Job (days) (days) (days) tardy D 6 6 17 0 C 7 13 15 0 B 10 23 16 7 A 14 37 20 17377924EDD:Job D has the lowest critical ratio therefore it is scheduled next and completed on day 27.b.ardi Flow time Average flow time Number of jobsDays tardy Average job t ness Number of jobs Flow timeAverage number of jobs at the center Makespan==∑=FCFS SPT EDD CR26.50 19.75 21.00 24.75 11.0 6.00 6.00 9.252.86 2.142.272.67c. SPT is superior.9.Thus, the sequence is b-a-g-e-f-d-c.。

《运营管理》课程习题和答案解析_修订版

《运营管理》课程习题和答案解析_修订版

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课程习题及答案修订版

《运营管理》课程习题及答案修订版

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课程习题及答案_修订版(1)

《运营管理》课程习题及答案_修订版(1)

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

《运营管理》课程习题及答案_修订版(1)

《运营管理》课程习题及答案_修订版(1)

第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

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第1章运营管理概述习题一、单项选择题1、在组织的三大基本职能中，处于核心地位的是：（）A、财务B、营销C、运营D、人力2、产品品种单一、产量大、生产重复程度高的生产类型称为（）。

A、单件生产B、大量生产C、批量生产D、大批量生产3、生产设施按工艺流程布置，加工顺序固定不变，工艺过程的程序化、自动化程度较高的生产类型称为（）A、连续型生产B、间断式生产C、订货式生产D、备货式生产4、有形产品的变换过程通常也称为（）A.服务过程B.生产过程C.计划过程D.管理过程5、无形产品的变换过程有时称为（）A.管理过程B.计划过程C.服务过程D.生产过程6、制造业企业与服务业企业最主要的一个区别是（）A.产出的物理性质B.与顾客的接触程度C.产出质量的度量D.对顾客需求的响应时间7、企业经营活动中的最主要部分是（）A.产品研发B.产品设计C.生产运营活动D.生产系统的选择8、下列哪项不是生产运作管理的目标（）A、质量B、成本C、价格D、柔性9、按照生产要素密集程度和顾客接触程度划分，医院是：（）A、大量资本密集服务B、大量劳动密集服务C、专业资本密集服务D、专业劳动密集服务10、当供不应求时，会出现下列情况：（）A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与服务水平下降二、多项选择题1、服务运营管理的特殊性体现在（）A.设施规模较小B.质量易于度量C.对顾客需求的响应时间短D.产出不可储存E.可服务于有限区域范围内2、运营管理中的决策内容包括（）A.运营战略决策B.运营系统运行决策C.运营组织决策D.运营系统设计决策E.营销决策3、产品结果无论有形还是无形，其共性表现在（）.A.市场畅销B.满足人们某种需要C.投入一定资源D.经过变换实现E..实现价值增值4、企业经营管理的职能有（）.A.财务管理B.技术管理C.运营管理D.营销管理E.人力资源管理5、运营管理的计划职能具体包括以下方面内容（）A.目标B.原因C.人员D.地点E.时间F. 方式三、简答题1、根据生产活动的定义，生产活动有哪些含义？2、从管理的角度来看制造过程和服务过程，二者存在哪些重要异同？3、按照产品品种多少和生产的重复程度划分的生产类型有哪些？特点是什么？4、生产运营系统有哪些的主要特征？试对其进行简单描述。

5、试述资源整合原则的含义。

四、论述题1、试述运营管理的新环境。

2、简述运营管理组织职能的具体内容。

第2章运营战略习题一、单项选择题1、运营战略包括（）A、跟随者B、独具一格C、集中一点D、自制或购买2、产品或服务的开发与设计策略有（）A、低成本和大批量B、多品种和小批量C、自己设计还是请外单位设计D、高质量3、对多品种小批量生产一般采用（）布置。

A、流水线B、按功能C、固定位置D、生产单元4、运营战略属于（）A.公司级战略B.事业部级战略C.职能级战略D.作业层战略5、从市场定位开始入手，而这种市场定位是根据企业对其选择的目标市场的理解以及竞争对手的市场行为而得出来的匹配是（）A.市场驱动的匹配B.运营能力驱动的匹配C.实践领域的匹配D.可持续性二、多项选择题1、一般来说，企业的战略可以划分的层次有（）A.公司发展战略B.公司总体战略C.公司营销战略D.公司经营战略E.职能级战略2、影响运营战略制定的因素有（）A.市场需求及其变化B.技术进步C.企业整体经营目标D.供应市场E.各部门职能战略3、企业根据自身条件可以确定的竞争重点包括（）A.成本B.时间C.空间D.柔性E.质量4、运营战略在整个企业战略体系中形成的基本特征有（）A.从属性B.支撑性C.协调性D.竞争性E.风险性5、运营系统的构成要素中，非结构化要素包括（）A.人力资源及组织设计B.生产能力C.库存策略D.质量管理策略E.生产计划策略三、判断题1、当价格是影响需求的主要因素时，就出现了基于成本的竞争。

2、当质量成为影响需求的主要因素时，降低成本就没有意义了。

3、成本可以无限降低。

4、高质量可能导致低成本和差异化。

5、高效率不能导致产品差异化。

四、简答题1、试述企业三个层次战略涉及的内容。

2、竞争优势一般有哪几类？3、企业制定战略的一般步骤有哪些？4、服务能力的扩大和生产能力的扩大有什么不同之处？5、产品或服务的选择要考虑哪些因素？五、论述题1、试阐述生产运营作战略在企业战略体系中的作用。

2、论述运营战略制定的影响因素。

第3章生产系统设计习题一、单项选择题1、以下哪个条件要求厂址应该接近消费市场（）A.原料易变质B.原料笨重C.产品运输不便D.原料运输不便2、按产品专业化原则建立生产单位，适用于（）A.单件生产B.小批生产C.大批生产D.工程项目3、汽车装配应采用（）A.流水线布置B.定位布置C.对象专业化布置D.以上都不是4、计算机辅助设计是指利用计算机帮助（）A.产品设计B.决策制定C.质量控制D.流程控制5、以下哪项与成组技术紧密相关（）A.数控机床B.机器人C.单元制造D.计算机辅助设计6、下列哪一项是产品标准化的缺点（）A.增加库存品种数B.增加培训费用C.减少设计费用D.减少产品多样性7、哪种加工类型更适合多品种生产（）A.装配线B.单件小批生产C.批量生产D.连续生产8、宜家家具的设计很巧妙，可以通过不同组合变化为多种功能的家具，也可以根据使用者的不同需要及喜好改变摆放方式，这种产品设计的方式是（）A．模块设计 B.稳健设计 C.计算机辅助设计 D.环保设计9、许多制造企业也把工厂建到消费市场附近，以降低运费和损耗，这体现了设施选择的什么原则？（）A．追求总费用最小化 B.接近用户原则C.长远发展原则D.分散与集中原则10、适用于产品标准化程度高的设施布置类型是（）A．工艺布置 B.产品布置 C.定位布置 D.混合布置二、多项选择题1、生产系统的设计原则包括（）A.平行性B.节奏性C.连续性D.经济性 E、比例性2、厂址选择的方法包括（）A.线性规划法B.加权评分法C.本量利分析法D.重心法E.前三项都是3、服务业设施布置的原则包括（）A.过程原则布置B.对象原则布置C.工艺原则布置D.定位布置E.以上都是4、设施选址的方法包括（）等方法。

A．对象专业化法 B.本量利分析法C.重心法D.加权评分法E.工艺专业化法5、厂址选择的主要影响因素有（）A．接近顾客 B.接近原材料C.运输问题D.员工的生活习惯E.基础设施三、判断题1、选择决策只是新企业进行的一次性决策。

（）2、啤酒厂应该建在水源质量有保障的地方。

（）3、按工艺原则布置适于体积庞大、不能移动的加工对象。

（）4、汽车生产流水线是按工艺原则布置的。

（）5、节拍是指零件从开始加工到完成加工所经历的时间。

（）四、简答题1、简述生产系统的构成要素2、厂址选择的影响因素3、简述加权评分法的步骤4、简述产品设计方法5、简述制造设计的基本原则五、计算题1、长江电子公司是生产小型计算器和微型计算机的公司，王某是总裁。

公司现在在L城和R城拥有工厂。

王总评价在中国西部开设第三家工厂的好处。

她希望把D城和O城作为备选城市来进行比较。

小于等于1000个单位的原材料，每个月可以从任意以下三个地点中得到：A、B和C.表3-1给出了现在工厂及未来工厂的月需求，以及从某一供应点购买与运输到指定工厂的总成本。

表1（1）选择D（2）如果选择D城，这些成本是多少？如果选择O城呢？2、真果营养棒制作的原材料包括干果、人工色素、防腐剂和葡萄糖。

这些混合物会被压成薄片，然后刻成各种形状，进行翻卷和包装。

流水线的先后关系和时间要求在表3-4中给出。

为了满足需求，每周40个小时需要生产6000个产品。

在满足先后关系限制的前提下，设计一条可以满足这一生产定额并且包含最少生产中心的流水线。

表2第4章新产品、新服务开发习题一、单项选择题1、从彩色电视机到LED电视机的转变，属于哪种新产品类型（）A.派生产品B.换代产品C.创新产品D.以上都不是2、联邦快递开创了隔夜速递服务，抛弃了传统的地面交通工具而改用了飞机。

属于哪种新服务类型（）A.粉饰型服务B.多样化服务C.创新型服务D.渠道开发型服务3、以下哪项不是产品开发的基本环节（）A.构想方案提出B.价值工程C.产品规划D.工艺设计4、产生新产品构想的直接原因是（）A.顾客的需要B.竞争对手C.政府机关D.科技5、对新产品设计决策限制最严格的部分来自（）A.企业的市场营销目标B.企业条件的限制C.目标市场消费行为D.经营环境的限制6、杜邦公司在其发明了尼龙的生产方法后，又用尼龙制成了尼龙丝袜、尼龙混纺、尼龙帐篷等产品，属于哪种新产品战略（）A.领先型战略B.紧跟型战略C.稳健战略D.以上都不是7、在现有的组织中设立利润中心，属于哪种组织创新（）A.组织过程B.组织结构C.人D.以上都不是8、在新产品开发过程中，充分掌握竞争对手的信息，属于哪种风险控制（）A.对新产品开发活动进行系统管理B.重视新产品上市管理C.增强组织柔性D.合理选择新产品开发项目9、利用科学技术的新发明所生产的、与原有产品完全不同的产品是指（）A.派生产品B.换代产品C.高科技产品D.创新产品10、服务业设计的焦点集中在（）A．设计出好的服务产品 B.保证服务产品的质量C.降低服务产品的成本D.研究如何传递服务的运营流程上二、多项选择题1、新产品开发战略的特征有（）A.全局性B.未来性C.系统性D.竞争性E.相对稳定性2、新产品开发过程包括几个阶段（）A.提出构想方案B.进行可行性研究C.产品设计D.工艺设计E.市场导入3、新产品开发的特殊目标（）A.多元化B.季节性调整C.加速回收投资D.提高产品质量E.提升企业形象4、新产品开发的组织模式有（）A.串行工程B.并行工程C.反向工程D.价值工程E.全部都是5、新产品开发的组织形式包括（）A.新产品委员会B.新产品部C.新产品经理D.研究开发部E.联盟三、判断题1、粉饰型服务是指提供的服务内容没有多大变化，但采用了全新的服务传递方式。

如自动取款机、网上购物等。

（）2、企业进行新产品开发的主要目的是满足现实市场的需求，发掘潜在的市场需求以及创造需求。

（）3、领先型新产品的利润率低于模仿型品牌产品。

（）4、多元化目标是通过开发新产品来分散经营风险，增强市场适应能力。

（）5、顾客期望不等于实际购买，但其期望不易发生变化。

（）四、简答题1、简述新产品、新服务开发的重要性2、简述新服务分类矩阵3、简述新产品开发的发展目标及其类型4、简述新产品的开发过程5、简述产品设计的基本原则五、论述题1、试论述新产品开发战略的类型2、试论述并行工程的基本思想，并比较其与串行工程的区别第5章需求预测习题一、简答题1、影响需求的因素是什么？2、常见的需求预测的方法有哪些？4、测量预测误差有哪些指标？各有什么特点？3、预测误差是指什么？预测误差有哪些来源？5、CPFR是什么？有哪些特点？二、判断题：1、时间序列预测模型是以历史数据分析为基础的对将来的预测。

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