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．。