《运营管理》课后习题答案.pptx

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。

第10章 思考与练习参考答案1. 周日早晨，顾客到达面包店的平均速度是每小时16位。

到达分布能够用均值为16的泊松分布表示。

每个店员能够在平均3分钟内接待一名顾客；服务时间基本符合均值为3分钟的指数分布，并用Excel 求解。

（1）到达速度和服务速度是多少？ （2）计算同时接受服务的顾客平均数？（3）假设队中等候的顾客平均数是3.2，计算系统中的顾客平均数（即排队等候的和接受服务的之和）、顾客平均排队等候时间以及花费在系统中的平均时间？（4）求当M=1、2与3时的系统利用率。

答案：（1）问题中已经给出了到达速度 ג =16为顾客/小时；µ=20位顾客/小时。

（2）0.80位顾客/小时。

（3） Ls= Lq+ r=3.2+0.8=4.0位顾客Wq=0.20小时/顾客，或0.20小时×60分钟/小时=12分钟 Ws=0.25小时，或15分钟（4）当M=1时，160.80120ρ==⨯ 当M=2时，160.40220ρ==⨯当M=3时，160.27320ρ==⨯2. Alpha Taxi and Hauling 公司在某机场有7个出租汽车站。

公司发现非周末晚上午夜以后的顾客需求服从泊松分布，均值为每小时6.6人。

服务时间则为均值为每人50分钟的指数分布。

假设每辆出租汽车只有一位顾客，求：（1）系统利用率；（2）等候服务的期望顾客数；（3）顾客等候出租汽车的平均时间，并用Excel 求解。

答案：（1）786.02.176.6=⨯==μλρM （2） 674.1=q L （3）Wq=0.2536小时，即15.22分钟3. 某航空公司计划在一家新开张的商业大厦设售票处。

估计购票或咨询平均每小时48人次，服从泊松分布。

服务时间假定服从负指数分布。

根据以前类似售票机构的经验，服务时间均值平均为2.4分钟。

假设顾客的等待成本为每小时20元，一个售票处的相关成本为每小时8元，问应该设置几个售票柜台使得系统总成本最小？并用Excel 求解。

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。

学习-----好资料Chapter 02 - Competitiveness, Strategy, and Productivity3. (1) (2) (3) (4) (5) (6) (7)Overhead Material Total MFP WorkerCost Week Cost @1.5 Cost@ Output Cost@$6 (6) ?(2)$12x401 9,900 3.03 30,000 2,700 4,320 2,88011,220 3,360 2 5,040 2.99 2,820 33,6003,360 11,160 32,200 5,040 3 2.89 2,76035,40012,4803,8402,8802.8445,760*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 InstructionsSelect Transaction Options: 3.1) DepositWithdrawal 2)Transfer 3)4) Other4. Enter Information:1) PIN Number更多精品文档．学习-----好资料2) 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 A TM) 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 ServicesActual output?Efficiency?80% 2. Effective capacity Actual output = .8 (Effective capacity)Effective capacity = .5 (Design capacity)Actual output = (.5)(.8)(Effective capacity)Actual output = (.4)(Design capacity)Actual output = 8 jobsUtilization = .4Actual output?Utilization Design capacityActual output8??20 jobsDesign Capacity?4Effective .capacitya. Given: 10 hrs. or 600 min. of operating time per day. 10.250 days x 600 min. = 150,000 min. per year operating time.更多精品文档．学习-----好资料Total processing time by machineProduct A B C32,000 48,000 64,000 136,000 48,000 2 48,00024,000 36,000 3 30,00030,000 460,000 60,000122,000Total 186,000 208,000186,000?1.24N??2 machine A150,000208,000?1.?N38?2 machineB150,000122,000?.81?1 machine?N C150,000You would have to buy two “A”machines at a total cost of $80,000, or two “B”machines at a total cost of $60,000, or one “C”machine at $80,000.b.Total cost for each type of machine:A (2): 186,000 min ?60 = 3,100 hrs. x $10 = $31,000 + $80,000 = $111,000B (2) : 208,000 ? 60 = 3,466.67 hrs. x $11 = $38,133 + $60,000 = $98,133C(1): 122,000 ? 60 = 2,033.33 hrs. x $12 = $24,400 + $80,000 = $104,400Buy 2 Bs —these have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3. 2435i更多精品文档．学习-----好资料Desired output = 4Operating time = 56 minutesOperating time56 minutes per hour??14 minutes per CT?unitDesired output4 units per hourTask # of Following tasksPositional Weight23 A 420 3 B18 2 C25 D 318 E 229 4 F24 3 G14 H 15Ia. First rule: most followers. Second rule: largest positional weight.更多精品文档．学习-----好资料First rule: Largest positional weight.b.Total time45 80.36%Efficiency c. CTx no. of stations564.cl.a.dabhgef2. Minimum Ct = 1.3 minutesTask Following tasks4 a3 b3 c2 d3 e2 f1 gh更多精品文档．学习-----好资料6 time).?(idlepercent54 percent???11.Idle 3. ).34(1NxCT420 min./day OTday/to 323)copiers??323.1 (rounds Output? 4.cycle/CT1.3 min.4.6timeTotal ? CT.6,Total time3??2. minutes?41. b.2N Assign a, b, c, d, and e to station 1: 2.3 minutes [no idle time] 2.Assign f, g, and h to station 2: 2.3 minutes420OTday182.6 copiers?/??Output 3. 2.3CT4.420 min./day day/?91.30 copiersMaximum Ct is 4.6. Output?cycle min./4.67. 4 1 5738更多精品文档．学习-----好资料6 2Chapter 06 - Work Design and Measurement3.Element PR OT NT AF STjob.476 .90 .46 1.15 .414 1.85 1.505 1.280 1.15 2 1.4721.10 .83 .913 3 1.15 1.0501.00 1.16 41.160 1.15 1.334Total 4.3328. A = 24 + 10 + 14 = 48 minutes per 4 hours48?.20A?240NT?6(.95)?5.70min.1?7.125min?ST5.70x.1?.209. a. Element PR OT NT A ST1 1.10 1.19 1.309 1.15 1.5052 1.15 .83 .955 1.15 1.098.588 1.05 1.15 .56 .6763b.x?.8322??2(.034)zs??034?.s?observations~68?n?67.12????.01(.83)ax????002.z?01?.A22zs2(.034)????n???46.24, round to 47 e = .01 minutes c. ????e.01????更多精品文档．学习-----好资料Chapter 07- Location Planning and Analysis1. Factor Local bank Steel millFood warehousePublic school1. forConvenience H ML M–H H –customers2. ofAttractiveness ML H –M H building3. rawto Nearness M L H L materials4. ofLarge amounts L HL L powerL H L 5. L Pollution controls6. andcost Labor L M L L availability7. TransportationM M–H M–L H costs8. ConstructionMM–HMHcosts更多精品文档．学习-----好资料Location (a) Location (b)A B C Weight 4. FactorC A B10/918/9 1. 10/9 2/9 5 9 5 Business Services7/9 1/9 7/9 7 6 2. 7 6/9 Community Services7/9 3 8/9 3. 8 1/9 3/9 7 Real Estate Cost2/9 4. 10/9 5 5 12/9 10/96 Construction Costs8/9 4/9 4 7/9 5. 8 7 1/9 Cost of Living4/95 5/9 5/9 1/9 55 6. Taxes7 6 8 1/9 6/9 7. 7/9 8/9 Transportation1.0 444553/9 55/9 54/939TotalEach factor has a weight of 1/7.39 44 45 a. Composite Scores 77 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 each have a weight of 1/9.5 x + 2 x + 2 x = 1 x = 1/9A B C Composite Scores c. 54/955/9 53/9B is the best followed byC and then A.x yLocation 5.A7 32 B 86 C 41D 46 4 ETotals 25 20?y ?x ii??2025 = 4.0= = 5.0= = x = yn5n5Hence, the center of gravity is at (5,4) and therefore the optimal location.Chapter 08 - Management of Quality更多精品文档．好资料学习----- Checksheet 1.FrequencyWork Type12 Lube and Oil7 Brakes6 Tires4 Battery1 Transmission30TotalPareto127641Trans.BatteryBrakesLube & Oil Tires2??3 ??.???2???????????1????????????0break lunch breakThe 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. 更多精品文档．学习-----好资料4PersonCordOtherChapter 9 - Quality Control更多精品文档．学习-----好资料Sample Mean Range4.=? 0.58(1.87) AR = 79.96 ±Mean Chart: X ±2.6 179.48 2 1.08±= 79.96 2.3 80.14 21.2 UCL = 81.04, LCL = 78.883 80.14?1.7 79.60 4 = 2.11(1.87) = 3.95 Range Chart: UCL = DR 4?2.0 5 80.02 = 0(1.87) = 0LCL = DR 3[Both charts suggest the process is in control: Neither has any 1.4680.38points outside the limits.])1?pp(2p?Control Limits = n = 200 6. n)99040096(..25 2.0096??0096p?.?200 )13(2000138.?.0096?Thus, UCL is .0234 and LCL becomes 0.Since n = 200, the fraction represented by each data point ishalf the amount shown. E.g., 1 defective = .005, 2 defectives= .01, etc.Sample 10 is too large.110409.857?8c?3c?7.857?.?7c7. Control limits: 14UCL is 16.266, LCL becomes 0.All values are within the limits.Let USL = Upper Specification Limit, LSL = Lower Specification Limit, 14.X = Process standard deviation= Process mean, ?For process H:1.??LSL1514X93.??)32(3)(.3?USL?X16?15??1.043?(3)(.32)???.0493 min.938,1.?C pk.93?1.0, not capable更多精品文档．学习-----好资料For process K:30??LSL33X0.??1))(13?(333??X36.5USL17.??1)3()(13?0?1.min{1.0,1.17}C?pk C not process is acceptable < is 1.33, since 1.0 1.33, the Assuming the minimum pk capable.For process T:516.X?LSL18.5?671.??)4(3?3)(0. L?X20.1?1833??1.)43?(3)(0.331.33}?1.C?min{1.67,pk Since 1.33 = 1.33, the process is capable.Chapter 10 - Aggregate Planning and Master SchedulingNo backlogs are allowed7. a.TotalJun. May Apr. Period Mar. Aug. July Sep.350 51 50 60 40 55 44 50 Forecast Output280 40 40 Regular 40 40 40 40 4051 3 8 8 8 8 Overtime 8 819 2 3 12 0 0 0 2 Subcontract0 0 4 3 Output - Forecast 3 ––4 0Inventory0 0 Beginning 3 0 4 0 00 3 0 0 Ending 4 0 07 0 0 1.5 1.5 0 2 Average 20 0 0 Backlog 0 0 0 0 0Costs:22,400 Regular 3,200 3,200 3,200 3,200 3,200 3,200 3,200 6,120 960 960 Overtime 960 360 960 960 9602,660 0 280 Subcontract 280 0 420 0 1,68070 0 0 15 Inventory 0 20 15 2031,2504,4405,8403,575 4,175Total4,4404,1804,600Level strategyb.更多精品文档．学习-----好资料Period Mar. Apr. May Jun. July Aug. Sep.Total350 51 40 55 60 Forecast 50 50 44Output280 Regular 40 40 40 40 40 40 4056 8 8 8 8 Overtime 8 8 8 Subcontract Output - Forecast Inventory Beginning Ending 8 Average 18 Backlog Costs: Regular Overtime Subcontract80 Inventory Backlog Total8.Period Forecast Output920 150 160 150 Regular 150 160 15050 0 10 Overtime 10 10 10 1020 10 Subcontract 0 0 0 0 100 0 10 0 –Output- Forecast 10 0Inventory0 Beginning 10 0 0 10 00 0 0 10 10 0 Ending20 0 0 0 10 Average 5 50 0 0 0 Backlog 0 0 0Costs: 46,000 Regular 7,500 7,500 7,500 8,000 8,000 7,5003,750 0 750 Overtime 750 750 750 7501,600 0 800 0 0 800 Subcontract 080 20 Inventory 40 200 Backlog 0 0 0 0 051,4308,3409,0508,2708,250Total9,0708,750Chapter 11 - MRP and ERP1. a. F: 2G: 1H: 1A: L: 1 x 2 = 2 J: 2 x 2 = 4 1 x 4 = 4D: J: 1 x 2 = 2 1 x 2 = 2 D: 2 x 4 = 8Totals: F = 2; G = 1; H = 1; J = 6; D = 10; L = 2; A = 4b.Stapler更多精品文档Base AssemblyTop Assembly学习-----好资料Beg. 4. 1 2 3 4 Day 5 6 7 Master Inv. Schedule 200 Quantity更多精品文档．学习-----好资料4 1 3 2 Week 10.70 40Material 80 602 3 Week 1 4Labor hr. 280 240 320 160Mach. hr. 210240 180120a. Capacity utilizationWeek 1 23 493.3% 106.7% 80% Labor 53.3%105%60%Machine 120% 90%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.Chapter 12 - Inventory Management更多精品文档．-----好资料学习The following table contains figures on the monthly volume and unit costs for a 2. random sample of 16 items for a list of 2,000 inventory items.B 8,000 P05 16 500C30300P0910a. See table.b. To allocate control efforts.c. It might be important for some reason other than dollar usage, such as cost of a stockout, usage highly correlated to an A item, etc.3. D = 1,215 bags/yr.S = $10H = $752DS2(1,215)10 a. bags Q 18H75b. Q/2 = 18/2 = 9 bagsD1,215 bags c.orders5 ??67.Q18 bags/orders DS?/2HTC?Q d.Q181,215(75)?(10)?675??675?$1,350218e. Assuming that holding cost per bag increases by $9/bag/year更多精品文档．学习-----好资料)215)(102(1,?17 bagsQ =842151,17?71TC714.71?$1,428.(84)?(10)?714?172 $1,350] = $78.71 –Increase by [$1,428.71D = 40/day x 260 days/yr. = 10,400 packages4.H = $30 S = $6060400)22DS(10,oxes Q?b??203.96?204 a. 030H DQSTC?H? b.Q2400,20410?82$6,118.,)?3060?3,058.82?30()?(602042Yes c. 40020010,?(30TC60))?(d. 2002002 = 3,000 + 3,120 = $6,120TC 200 6,118.82 (only $1.18 higher than with EOQ, so 200 is acceptable.)6,120 –H = $2/month 7.S = $556) –D = 100/month (months 1112)–D = 150/month (months 7255100)2DS2(16?Q.Q D:??74 a.0102H55)2(150?Q83D:?90.022The EOQ model requires this.b.10 = $45 (revised ordering cost) Discount of $10/order is equivalent to S –c.$148.32–16 TC74 =更多精品文档．学习-----好资料10050*?$140)?(45TC?)(250502100100?TC?(45)$145(2)? 1001002100150?TC?$180(2)?(45)1501502=$181.667–12 TC 9115050185?$?(TC?45)(2)50502150100?.5*TC)?(45)?$167(2 1001002150150?195TC)?(45)?$(21501502p = 50/ton/day 10.u = 20 tons/day D= 20 tons/day x 200 days/yr. = 4,000 tons/yr.200 days/yr.S = $100H = $5/ton per yr.50100,4000)2DSp2( a. bags] [10,328 ?516.40 tonsQ??020?u550Hp?4.Q516b. ? bags]I [approx.6,196.8.u(p?)?(30)?30984 tons max50PI48309.max tons [approx. 3,098 bags] Average is92.:?154224516.QRun length =c. days ??10.3350P000,D4 Runs per year =d. 8] .7.75 ?[approx?4Q516. = 258.2 Q?e.ID max SH?TC =Q2 TC = $1,549.00 orig.774.50= $ TC rev.$774.50Savings would be更多精品文档．-----好资料学习Q P H 15. Range495 $5.00 $2.00 D = 4,900 seats/yr. 0–999497 NF 4.95 1.98 H = .4P 1,000–3,999500 NF 1.96 S = $50 4,000–5,999 4.90503 NF1.946,000+ 4.85with TC for all lower price breaks:Compare TC4954,900495 ($50) + $5.00(4,900) = $25,490($2) + TC = 495495 24,9001,000 ($50) + $4.95(4,900) = $25,490($1.98) + TC = 1,0001,000 24,9004,000 ($50) + $4.90(4,900) = $27,991= ($1.96) + TC4,0004,000 24,9006,000 ($50) + $4.85(4,900) = $29,626= ($1.94) + TC6,0006,0002Hence, one would be indifferent between 495 or 1,000 unitsTC????6,000503 500 1,000 4,000 495 497Quantity= 30 gal./day d 22.= 170 gal. ROPLT = 4 days,= 50 gal?ss = Z LTdZ = 1.34 Risk = 9% = 37.31 ?Solving, LTdss=1.88 x 37.31 = 70.14 gal. Z = 1.88, 3%Chapter 13 - JIT and Lean Operations= ?DT(1 + X) 1. N N = C D = 80 pieces per hour80(1.25) (1.35)= 75 min. = 1.25 hr. T = 3= 45 45 =C= .35X更多精品文档．学习-----好资料4. The smallest daily quantity evenly divisible into all four quantities is 3. Therefore, use three cycles.Product Daily quantityUnits per cycle21/3 = 7 21 A12/3 = 4 12 B3/3 = 1 3 C15/3 = 515D5.a. Cycle 1 2 3 4b. Cycle 1 2A 6 6 5 5 A 11 116 6 3 3 3 B B 32 2 C 1 1 1 1 C8 8 4 D 4 5 5 D2 2 E 4 42 E 2c. 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 Daystardy (days) (days) (days) Job20 14 A 14 016 10 B 24 8C 7 31 15 166 D 17 37 201064437更多精品文档．学习-----好资料SPT: Job time Flow time Due date Days Job (days) (days) (days) tardy0 6 17 D 60 13 15 C 77 16 B 23 1020 A 37 14 17243779Job time Flow time Due date EDD: DaysJob (days) (days) (days) tardyC 7 7 15 01 B 10 16 176 D 17 6 23201437 A172484Critical RatioProcessing Time(Days)Due Date Critical Ratio Calculation Job(20 –14 A 20 0) / 14 = 1.43(16 10 –16 B 0) /10 = 1.60(15 –7 15 0) / 7 = 2.14 C(17 –17D0) / 6 = 2.83 6Job 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:Processing Time(Days)Due DateJob Critical Ratio Calculation–– A –(16 16 10 B –14) /10 = 0.20(15 15 7 C –14) / 7 = 0.14(17 176–14) / 6 = 0.50 DJob 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:Processing Time(Days)Due DateJob Critical Ratio Calculation––– A(16 10 –B 21) /10 = 16 –0.50–C ––(17 –D21) / 6 = 617–0.67更多精品文档．学习-----好资料Job 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.9937b. FCFS SPT EDD CRtime Flow24.75 21.00 26.50 19.75 ?timeAverage flow jobsof Number tardy Days?nesst ardi Average job jobsNumber of 9.25 11.0 6.00 6.00 time ?Flow?the center Average number of jobsat Makespan 2.672.272.86 2.14SPT is superior.c.9.Time (hr.) Sequence of assignment:Order Step 1 Step 2last (or 7th) 1.40 A 1.20 .80 [C]first B 0.90 1.30 .90 [B]2nd 1.20 0.80 [A] C 2.003rd 1.30 [G] 1.70 D 1.504th 1.80 E 1.60 1.60 [E]6th 1.75 1.50 [D] 2.20 F5th1.40G1.301.75[F]Thus, the sequence is b-a-g-e-f-d-c. 更多精品文档．。

Chapter 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 envelop e (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. %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.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)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 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.Chapter 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 30ParetoLube & 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.4break lunch break3 2 1 0• • •• • • ••• • • ••••••• ••• •• • •• • •••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 allowedPeriodForecast Output Regular Overtime Subcontract Output - Forecast Inventory Beginning Ending Average Backlog Costs: Regular Overtime Subcontract Inventory Totalb. Level strategyPeriodForecastOutputRegularOvertimeSubcontractOutput - ForecastInventoryBeginningEndingAverageBacklogCosts:RegularOvertimeSubcontractInventoryBacklogTotal8.PeriodForecastOutputRegularOvertimeSubcontractOutput- ForecastInventoryBeginningEndingAverageBacklogCosts:RegularOvertimeSubcontractInventoryBacklogTotalChapter 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 = 44. Master Schedule10. 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.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 +=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 200 days/yr. = 4,000 tons/yr.15. 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) = $25,490 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 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 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 9.25 2.86 2.142.272.67c. SPT is superior.9.Thus, the sequence is b-a-g-e-f-d-c.。

第1 章运营管理概述习题一、单项选择题1、在组织的三大根本职能中，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、连续式生产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.效劳过程B. 生产过程C.方案过程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. 方案过程C.效劳过程D. 生产过程6、制造业企业与效劳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局部是〔〕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、以下哪项不是生产运作管理的目标〔〕A、质量 B 、本钱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、大量资本密集效劳B、大量劳动密集效劳C、专业资本密集效劳D、专业劳动密集效劳10、当供不应求时，会出现以下情况：〔〕A、供方之间竞争激化B、价格下跌C、出现回扣现象D、质量与效劳水平下降二、多项选择题1、效劳运营管理的特殊性表达在〔〕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. 可效劳于有限区域X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、运营管理的方案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效劳过程，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？试对其进展简单描述。

Chapter 02 - Competitiveness, Strategy, and Productivity3. (1) (2) (3) (4) (5) (6) (7)Week Output WorkerCost@$12x40OverheadCost @MaterialCost@$6TotalCostMFP(2) ÷ (6)1 30,000 2,880 4,320 2,700 9,9002 33,600 3,360 5,040 2,820 11,2203 32,200 3,360 5,040 2,760 11,1604 35,400 3,840 5,760 2,880 12,480*refer to solved problem #2Multifactor productivity dropped steadily from a high of to about .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 % ((21-16)/16).Multifactor productivity increased by % ((./.89).*Machine ProductivityBefore: 80 ÷ 40 = 2 carts/$1.After: 84 ÷ 50 = carts/$1.Productivity increased by -16% (/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. %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, hrs. x $11 = $38,133 + $60,000 = $98,133 C(1): 122,000 ÷ 60 = 2, 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)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 = 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: minutes [no idle time]Assign f, g, and h to station 2: 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. Element PR OT NT AF jobST1 .90 .46 .414 .4762 .853 .83 .913 4Total8. 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 ST12 .83 .9553.56.588.676b.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 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 = -y =∑y i= 20 = 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 30ParetoLube & 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.4break lunch break3 2 1 0• • •• • • ••• • • ••••••• ••• •• • •• • •••Chapter 9 - Quality Control4. Sample Mean Range1Mean Chart: =X ± A 2-R = ± 2 = ±3UCL = , LCL =4 Range Chart: UCL = D 4-R = = 5LCL = D 3-R = 0 = 06[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. ., 1 defective = .005, 2 defectives = .01, etc.Sample 10 is too large.7. 857.714110c ==Control limits: 409.8857.7c 3c ±=± UCL is , 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 , since < , 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 = , 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 Totalb.Level strategyPeriodForecastOutputRegularOvertimeSubcontractOutput - ForecastInventoryBeginningEndingAverageBacklogCosts:RegularOvertimeSubcontractInventoryBacklogTotal8.PeriodForecastOutputRegularOvertimeSubcontractOutput- ForecastInventoryBeginningEndingAverageBacklogCosts:RegularOvertimeSubcontractInventoryBacklogTotalChapter 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.MasterSchedule10. 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 % % 80% %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.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 +=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, – $1,350] = $4.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, (only $ 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 = $180$)45(150100)2(2150TC 145$)45(100100)2(2100TC *140$)45(50100)2(250TC 15010050=+==+==+=7–12 TC 91 =$195$)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 ' =TC =S QD H 2I max + TC orig. = $1, TC rev. = $Savings would be $D= 20 tons/day x 200 days/yr. = 4,000 tons/yr.15. Range PHQ D = 4,900 seats/yr. 0–999 $ $ 495 H = .4P 1,000–3,999 497 NF S = $50 4,000–5,999 500 NF 6,000+503 NFCompare TC 495 with TC for all lower price breaks:TC 495 =495 ($2) + 4,900($50) + $(4,900) = $25,490 2 495 TC 1,000 = 1,000 ($ + 4,900($50) + $(4,900) = $25,4902 1,000 TC 4,000 = 4,000 ($ + 4,900($50) + $(4,900) = $27,9912 4,000 TC 6,000 = 6,000 ($ + 4,900($50) + $(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 gal Risk = 9% Z = Solving, σd LT = 3% Z = , ss= x = gal.Chapter 13 - JIT and Lean Operations1. N = ?N = DT(1 + X)D = 80 pieces per hourC T = 75 min. = hr. = 80 = 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 = 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 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 jobs Days tardy Average job t ness Number of jobs Flow timeAverage number of jobs at the center Makespan==∑=FCFS SPT EDD CRc. SPT is superior.9.Thus, the sequence is b-a-g-e-f-d-c.。

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。

《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.。

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。

Chapter 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 fin d near or in theATM) and insert it into the deposit slot2) Withdrawal —lift the “Withdrawal Door,” being careful to remove all cash 6. 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 machineProductA B C 1 48,000 64,000 32,000 2 48,000 48,000 36,000 3 30,000 36,000 24,000 460,00060,00030,000Total 186,000 208,000 122,000machine181.000,150000,122machine 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 tasks PositionalWeight A 4 23 B 3 20 C 2 18 D 3 25 E 2 18 F 4 29 G 3 24 H 1 14 I5a. First rule: most followers. Second rule: largest positionalweight.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 tasksa 4b 3c 3d2e 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.Chapter 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 1.050 41.00 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. Attractivenessof building H L M M–H3. Nearness to rawmaterials L H L M4. Large amountsof power L H L L5. Pollutioncontrols 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 WeightA 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 8/9 Total 39 44 45 1.0 53/9 55/9 54/9 Each factor has a weight of 1/7.a . Composite Scores39 44 457 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 each have a weight of 1/9.5 x + 2 x + 2 x = 1 x = 1/9c . Composite ScoresA B C53/9 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 optimallocation.Chapter 08 - Management of Quality1. ChecksheetWork Type FrequencyLube and Oil 12Brakes 7Tires 6Battery 4Transmission 1Total 30ParetoLube & 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.4break lunch break3 2 1 0• • •• • • ••• • • ••••••• ••• •• • •• • •••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:0096.)200(1325==p 0138.0096.200)9904(.0096.20096.±=±=}{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 allowedPeriodForecast Output Regular Overtime Subcontract Inventory Beginning Ending Average Backlog Costs:RegularOvertimeSubcontractInventoryTotalb. Level strategyPeriodForecastOutputRegularOvertimeSubcontractInventoryBeginningEndingAverageBacklogCosts:RegularOvertimeSubcontractInventoryBacklogTotal8.PeriodForecastOutputRegularOvertimeSubcontractInventoryBeginningEndingAverageBacklogCosts:RegularOvertimeSubcontractInventoryBacklogTotalChapter 11 - MRP and ERP1. a. F: 2G: 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. 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 week2, 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 increase dueto overtime premium, a probable decrease in productivity, and possibleincrease in accidents.Chapter 12 - Inventory Management2.The following table contains figures on the monthly volume and unit costs for a random sample of 16 items for a list of 2,000 inventory items.a. 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/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 = $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 ==-=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 ' = 258.2D= 20 tons/day x 200 days/yr. = 4,000 tons/yr.TC =S QDH 2I max + TC orig. = $1,549.00 TC rev. = $ 774.50Savings would be$774.5015.RangePHQD = 4,900 seats/yr. 0–999 $5.00 $2.00 495 H = .4P 1,000–3,9994.95 1.98 497 NFS = $50 4,000–5,9994.90 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) = $25,4902 495TC 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.QuantityTCLT = 4 days,ss = Zσd LT = 50 galRisk = 9% Z = 1.34 Solving, σd 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 = 45 45X = .354. The smallest daily quantity evenly divisible into all four quantities is3. Therefore, use three 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 DaysJob (days) (days) (days) tardyD 6 6 17 0C 7 13 15 0B 10 23 16 7A 14 37 20 1737 79 24EDD:completed on day 14. After the completion of Job A, the revised critical ratios are: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 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 2.14 2.27 2.67c. SPT is superior.9.Thus, the sequence is b-a-g-e-f-d-c.。

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.。

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。

一、生产与运作管理(学）的逻辑结构。

二、项目管理中需要注意的二个主要问题是什么？（1）项目组长（经理)的人选(2）授权目前,关于授权的有效的方法是让项目经理根据项目的要求（时间、任务目标）自行确定, 再由主管领导审批的方式，这一方式的优点主要有：①便于对项目组长考核。

②了解项目组长对项目全过程管理的计划能力。

③便于评价项目活动中的各项资源的使用情况等。

三、服务业和制造业在服务时间、服务场所的依赖性和需求稳定性上的差异。

1、尽管服务业的服务能力同制造业的生产能力一样，都受到许多相同性质的因素影响，而且服务设施规模的大小在决定服务能力时也同样起着至关重要的作用.但是，服务业的服务能力和制作业的生产能力之间仍有许多差别,服务能力对服务时间和服务场所的依赖性更大。

2、下面，从时间、场所、需求稳定性三个方面来进行比较:(1）时间制造业:产品可以储存服务业：产品不可以储存,服务必须及时。

（2)场所服务业：离顾客越近越好，不能远离顾客（一个城市的出租车或餐馆，对在另一城市生活的人来说，毫无意义，服务能力必须在服务对象需要的时候出现在他（她）身旁）。

制造业：无所谓，要求不高.（3)需求稳定性服务业：需求不稳性，多变制造业：需求相对稳定,变化速度较服务业慢。

四、企业R&D中的基本决策问题主要包括什么内容？1、R&D领域的选择R＆D领域选择的目的是发现能够最适度发挥企业资本收益、提高企业竞争力的事业领域，是决定如何对新产品、新事业的各种机会进行探索的基本方针。

从企业目前的现有技术和现有市场向新事业领域的探索可以分为三种类型：(1)在现行事业领域，依靠现有的技术开发多种产品,以扩大现有市场；（2）向原有市场推出用新技术开发的新产品；（3)用新技术开发新产品，并加入新市场。

2、R＆D方式的选择选定R&D领域以后，接下来的问题是:采取什么方式进行R&D？这需要根据所选定的开发内容、企业自身的开发能力以及周围可利用资源的情况而定。

《生产与运作管理》教材习题参考答案第七章：P215 练习题1：1）网络图如下图所示：2）最早开始时间和最迟结束时间，如图中标示所示。

工期为：8+4+3=153）关键路线为：1-3-4-7（红色粗线标示）练习2： 解：1)2 关于赶工问题参照教材P202，例7.2。

略第五章：P140答案参见教材：P118，例5.6，略第四章：P95 练习题1：12）车间布置：根据关系分数排序，题目条件或约束，确定好车间7、2、4、1（注意车间1不能位于厂区中间位置，故列于最左下）；车间3与车间2的关系为A ，故安排于其左或右；车间8与车间4关系为A ，可安排于其上或其右；车间3、5、8的安排根据其相邻的所有车间的关系密切程度之和可以计算（略），可知车间5位于车间4上，车间8位于车间4右，车间3位于车间1上。

车间6与车间8的关系为A ，故车间6位于车间8上。

最终结果如图示：第三章：P61练习题1：解：Y=0.5X40+0.35X30+0.20X10=32.5 城市2综合得分高，因此是最佳选择。

练习题3． 解：设采用营口冷库，则如下表所示：其运输总费用=210*10+60*12+70*17+10*20+150*11+160*7=6980 由此可知，采用营口冷库更佳。

练习题5：解：首先，建立地图坐标系 9 8 7 65 4 3 2 11 23 4 5 6 7 8 9 10 11 设处理中心（X ，Y ），则利用重心法公式计算如下：∑∑=ii i Q Q x X /)403025926/()408302254942610(++++⨯+⨯+⨯+⨯+⨯==6∑∑=i i i Q Q y Y /)403025926/()40730625791265(++++⨯+⨯+⨯+⨯+⨯==6处理中心坐标为（6，6），位置如图中红色点标示。

第九章：P302 练习题1：解：这是一个存在折扣的库存模型问题。

1） 无论订购多少，订购单价为10元时，库存维持费为：H=10*10%=1 ，订货费用S=225 年需求量为：D=3600，则：EOQ （10）=12731225360022=⨯⨯=H DS单位2） 当订购量为大于等于500时，库存维持费为：H=9*10%=0.9，订货费用S=225，年需求量为：D=3600，则： EOQ （9）=13429.0225360022=⨯⨯=H DS单位3） 当订购量为大于等于1000时，库存维持费为：H=8*10%=0.8，订货费用S=225，年需求量为：D=3600，则： EOQ （8）=14238.0225360022=⨯⨯=H DS单位以上三种经济订货量单位都大于1000的订量单位，显然都是可行的订货量。

运营管理第九版课后练习题含答案第一章绪论1.1 为什么需要运营管理学科？运营管理学科主要是研究如何有效地组织、管理和改进企业的生产和服务流程，从而提高企业的效率和效益。

运营管理学科的发展, 是由于市场的变化和竞争的加剧对企业提出了更高的要求, 必须精细管理, 持续创新, 增强企业的竞争力。

1.2 运营管理学科的主要研究内容有哪些？运营管理学科的主要研究内容包括：生产计划与控制、库存管理、质量管理、供应链管理、产品设计和开发、过程改进和创新、服务运营管理等。

1.3 运营管理的基本特征是什么？运营管理的基本特征有以下几个方面：•以流程为中心的管理思想•高度重视客户需求•讲究效益与质量的综合平衡•强调创新和改进1.4 运营管理的发展历程有哪些？运营管理的发展历程大致可以分为以下三个阶段：第一阶段：生产管理时期，主要是生产工人实施正常生产，由工头或监工实际管理。

第二阶段：生产管理向生产规划管理转变，以订单计划和统计分析两个为主要手段，提高了生产效率。

第三阶段：生产管理向运营管理转变，注重竞争的实质在于企业的各种资源的管理，因此逐渐扩大管理的范围，并注重企业的战略计划和执行。

1.5 运营管理的目标有哪些？运营管理的主要目标包括：•降低成本并提高效率•增强企业的市场竞争力•提高服务质量•保障社会和环境的责任答案1.为了适应市场的变化和提高企业竞争力，需要建立一门关于企业产品与服务供应链的学科。

2.运营管理的主要研究内容包括：生产计划与控制、库存管理、质量管理、供应链管理、产品设计和开发、过程改进和创新、服务运营管理等。

3.运营管理的基本特征有以流程为中心的管理思想、高度重视客户需求、讲究效益与质量的综合平衡、强调创新和改进。

4.运营管理的发展历程分为生产管理时期、生产规划管理时期及运营管理时期。

5.运营管理的目标包括降低成本并提高效率、增强企业的市场竞争力、提高服务质量、保障社会和环境的责任。