上海交通大学：《生产计划与控制 Production Planning and Control》教学资源_历年试卷_2011_PPC_Answers

生产计划与控制2010至2011学年第2学期试卷A答案 Answers1 20' Ft=D1+(1-0)F-1 a)Freb=(.20)(23.3)+(.80)25)=24.66 FMarch=(.20(72.3)+(.80)24.745)=34.188 FApr=(.20)(30.3)+(.80)31.88)=33.4104 FMay = (.2015.5)+(.80)31.63)=29.78832 b)FFeb =(.50)23.3)+(.50)25)=24.15 FMarch 48.225 FApr =39.2625 FMa=27.28125 c) Compute MSE for February through April: Month Error (a) Error (b) (g=.20)_ (g=.50) Feb 47.64 48.15 Mar 3.88 17.93 Apr 18.11 23.96 MSE 870.89 1073.31 a =.20 gave a better forecast Answers2 20 Week 12131415161718192021222224 MPS for EP 1 1201127622 56 90210 MPS for EP 2 62689077 2630 54 POR for A 12011276 22 56 90210 POR for F for A 360336228 66168270630 POR for B 240224152 44112180420 POR for G for B 480448304 88224360840 POR for H for B 240224152 44112180420 POR for D 626890 77 2630 54 POR for F for D 1241361801545260108 POR for G for F 372408540462156180324 POR for H for F 248272360308104120216 POR for F 484472408220220330738 POR for G 372408540942604484412224360840 POR for H 248512584460148232396420 Answers3 20' Time Flavor Required Due Date Vanilla 4 days 3 Chocolate 5 days

生产计划与控制 2010 至 2011 学年第 2 学期试卷 A 答案 Answers1 20’ Ft = Dt-1 + (1-)Ft-1 a) FFeb = (.20)(23.3) + (.80)(25) = 24.66 FMarch = (.20)(72.3) + (.80)(24.745) = 34.188 FApr = (.20)(30.3) + (.80)(31.88) = 33.4104 FMay = (.20)(15.5) + (.80)(31.63) = 29.78832 b) FFeb = (.50)(23.3) + (.50)(25) = 24.15 FMarch = 48.225 FApr = 39.2625 FMay = 27.28125 c) Compute MSE for February through April: Month Error (a) Error (b) ( = .20) ( = .50) Feb 47.64 48.15 Mar 3.88 17.93 Apr 18.11 23.96 MSE = 870.89 1073.31  = .20 gave a better forecast Answers2 20’ Week 12 13 14 15 16 17 18 19 20 21 22 22 24 MPS for EP 1 120 112 76 22 56 90 210 MPS for EP 2 62 68 90 77 26 30 54 POR for A 120 112 76 22 56 90 210 POR for F for A 360 336 228 66 168 270 630 POR for B 240 224 152 44 112 180 420 POR for G for B 480 448 304 88 224 360 840 POR for H for B 240 224 152 44 112 180 420 POR for D 62 68 90 77 26 30 54 POR for F for D 124 136 180 154 52 60 108 POR for G for F 372 408 540 462 156 180 324 POR for H for F 248 272 360 308 104 120 216 POR for F 484 472 408 220 220 330 738 POR for G 372 408 540 942 604 484 412 224 360 840 POR for H 248 512 584 460 148 232 396 420 Answers3 20’ Time Flavor Required Due Date Vanilla 4 days 3 Chocolate 5 days 8

Strawberry 2 days 6 Peanut 2 days 12 a)SPT rule minimizes the mean flow time. The optimal sequence IS Strawberry-Peanut-Vanilla-Chocolate or Peanut-Strawberry-Vanilla-Chocolate b)Sequence the jobs using EDD rule. Time Flavor Required Due Date Vanilla 4 days 3* stop. Reject vanilla,then the new sequence is: Time Flavor Required Due Date Strawberry 2 days 6 Chocolate 7 days 8 Peanut 9 days 12 No more tardy jobs. Hence,the number of tardy jobs 1. Answers4 20 Att. Att. Total Month Belts(#)Belts(hrs) Hand(#)Hand(hrs) Case#Case(hrs)Hrs. 2500 5000 1250 3750 240 1440 10190 2 2800 5600 680 2040 380 2280 9920 心 2000 4000 1625 4875 110 660 9535 4 3400 6800 745 2235 75 450 9485 夕 3000 6000 835 2505 126 756 9261 6 1600 3200 375 1125 45 270 4595 b) Cum Cum Min. Working Working Working Demand Workforce Month Days Hours Hours Hours (Ratio) 1 22 154 154 10190 67 20 140 294 20110 69 3 19 133 427 29645 70 4 24 168 595 39130 66 5 21 147 742 48391 66 6 17 119 861 52986 62 Hence the minimum constant work force size is 70.It is probably not advantageous to bring the work force level up to 70 workers.The excess demand can be absorbed by overtime or by employing part-time workers. The firm can be more flexible by keeping the number of full-time employees to a minimum. c)If a plan is to utilize only regular time employees,then excess demand must be absorbed by overtime.Since there are 46 employees with the firm,we obtain (A) (B) (C=46*B) (D) (E=D-C) Working Regular Required Overtime Month Hours Hours Hours Hours 1 154 7084 10,190 3106 2 140 6440 9,920 3408 3 133 6118 9,535 3417 4 168 7728 9,485 1757 5 147 6762 9,261 2499

Strawberry 2 days 6 Peanut 2 days 12 a) SPT rule minimizes the mean flow time. The optimal sequence is Strawberry-Peanut-Vanilla-Chocolate or Peanut-Strawberry-Vanilla-Chocolate b) Sequence the jobs using EDD rule. Time Flavor Required Due Date Vanilla 4 days 3 * Stop. Reject vanilla, then the new sequence is: Time Flavor Required Due Date Strawberry 2 days 6 Chocolate 7 days 8 Peanut 9 days 12 No more tardy jobs. Hence, the number of tardy jobs = 1. Answers4 20’ Att. Att. Total Month Belts(#) Belts(hrs) Hand(#) Hand(hrs) Case# Case(hrs) Hrs. 1 2500 5000 1250 3750 240 1440 10190 2 2800 5600 680 2040 380 2280 9920 3 2000 4000 1625 4875 110 660 9535 4 3400 6800 745 2235 75 450 9485 5 3000 6000 835 2505 126 756 9261 6 1600 3200 375 1125 45 270 4595 b) Cum Cum Min. Working Working Working Demand Workforce Month Days Hours Hours Hours (Ratio) 1 22 154 154 10190 67 2 20 140 294 20110 69 3 19 133 427 29645 70 4 24 168 595 39130 66 5 21 147 742 48391 66 6 17 119 861 52986 62 Hence the minimum constant work force size is 70. It is probably not advantageous to bring the work force level up to 70 workers. The excess demand can be absorbed by overtime or by employing part-time workers. The firm can be more flexible by keeping the number of full-time employees to a minimum. c) If a plan is to utilize only regular time employees, then excess demand must be absorbed by overtime. Since there are 46 employees with the firm, we obtain (A) (B) (C=46*B) (D) (E=D-C) Working Regular Required Overtime Month Hours Hours Hours Hours 1 154 7084 10,190 3106 2 140 6440 9,920 3408 3 133 6118 9,535 3417 4 168 7728 9,485 1757 5 147 6762 9,261 2499

6 119 5474 4,595 Totals: 39,606 14,259 Cost of plan=(39,606)(8.50)+(14,259)14.00) 336.651+ 199,626 =$536.277 d)Here we will determine the size of the work force necessary to meet monthly demands as closely as possible and add additional employees to meet the excess demand. (A) (B) (C) (D=[B/C])(E) (F=D-E) (G=F*C) Demand Working Regular Add'1 Total Hrs. Month Hours Hour Ratio Workers Workers Add.Emply. 1 10,190 154 67 46 21 3234 2 9,920 140 71 46 25 3500 3 9,535 133 72 46 26 3458 4 9,485 168 57 46 11 1848 5 9,261 147 63 46 17 2499 6 4,595 119 39 46 0 Total: 14,539 From part(c),regular employees cost $336,651. Additional employees cost(11)(14,539)=$159,929 Total cost=$336,651+$159,929=$496,580 Answers5 10' The input data for this problem are: 1 P h h' 2500 45000 2.88 2.7200 80 5500 40000 3.24 2.7945 120 1450 26000 3.96 3.7392 60 a)First we compute. (2X80+120+60) =.1373 years Y(2.72(2500)+(2.7945X5500)+(3.7392X1450) b)and c) The order quantities for each item are given by the formula Qj=λT5· Obtain: Q1=343.21 Q2=755.05 Q3=199.06 The respective production times are given by Tj =Qj/Pj. Substituting,one obtains: T1=.007626 T2=.018876

6 119 5474 4,595 Totals: 39,606 14,259 Cost of plan = (39,606)(8.50) + (14,259)(14.00) = 336,651 + 199,626 = $536,277 d) Here we will determine the size of the work force necessary to meet monthly demands as closely as possible and add additional employees to meet the excess demand. (A) (B) (C) (D=[B/C]) (E) (F=D-E) (G=F*C) Demand Working Regular Add'l Total Hrs. Month Hours Hour Ratio Workers Workers Add. Emply. 1 10,190 154 67 46 21 3234 2 9,920 140 71 46 25 3500 3 9,535 133 72 46 26 3458 4 9,485 168 57 46 11 1848 5 9,261 147 63 46 17 2499 6 4,595 119 39 46 0 _______ Total: 14,539 From part (c), regular employees cost $336,651. Additional employees cost (11)(14,539) = $159,929. Total cost = $336,651 + $159,929 = $496,580 Answers5 10’ The input data for this problem are:  P h h K 2500 45000 2.88 2.7200 80 5500 40000 3.24 2.7945 120 1450 26000 3.96 3.7392 60 a) First we compute T *. T * = (2)(80 120  60) (2.72 )(2500 )  (2.7945 )(5500 )  (3.7392 )(1450 ) = .1373 years b) and c) The order quantities for each item are given by the formula Qj = Tj. Obtain: Q1 = 343.21 Q2 = 755.05 Q3 = 199.06 The respective production times are given by Tj = Qj/Pj. Substituting, one obtains: T1 = .007626 T2 = .018876

T3=.007656 It follows that the total up time each cycle is the sum of these three quantities which gives:total up time =.03416.The total idle time each cycle is.1373 -.0342=.1031.The percentage of each cycle which is idle time is thus.1031/.1373=75%. d)Using the formula G(T）=2(K1T+',T2) one obtains,G(T)=$3787.82 annually. Answers6 10' Weekly demand has mean 38 and standard deviation 130 LTD hasμ=38x3=114 ando=530=19.75 a)1-F(R)=h-(500401880 =.004757 p2(400)1976) z=2.59,R=0z+μ=165 b)Must determine optimal Q by iteration E0Q=198=Q0 1-F(R0)= (198)7.52)=.0018838 (400)1976) z=2.90,L(z)=.000542,n(R0)=σL(z)=.0107 Q= 2[K pn(R)] =204 h This value of Q turns out to be optimal.(must iterate once more to obtain R 171). c)Use formula G(Q,R)=h[Q/2 R -u]+K/Q pn(R)/Q Substitute (Q,R)=(500,165)from part (a) =(204,171)from part (b) 0 btain G(500,165)=$2606.75 A cost $641.59 yearly G(204,171) =$1965.16 d)Use the relationship n(R)=(1-B)Q to determine R and the fact that n(R)=o sL(z). Hence,L(z）=M®=L-2=0198=.10026 19.75 From Table B-4,z.90.Since R oz +u we obtain R=(19.75)(.90)+114=132. The imputed shortage cost is: 合=Qh =$4.09. (1-F(R)

T3 = .007656 It follows that the total up time each cycle is the sum of these three quantities which gives: total up time = .03416. The total idle time each cycle is .1373 - .0342 = .1031. The percentage of each cycle which is idle time is thus .1031/.1373 = 75%. d) Using the formula G(T) = (Kj j 1 n  / T  hj '  jT / 2) one obtains, G(T) = $3787.82 annually. Answers6 10’ Weekly demand has mean 38 and standard deviation 130 LTD has  = 38 x 3 = 114 and  = 3 130 = 19.75 a) 1 - F(R) = Qh p  (500)(.40)(18.80) (400)(1976) = .004757 z = 2.59, R = z +  = 165 b) Must determine optimal Q by iteration EOQ = 198 = Q0 1 - F(R0) = (198)(7.52) (400)(1976) = .0018838 z = 2.90, L(z) = .000542, n(R0) = L(z) = .0107 Q = 2K  pn(R) h = 204 This value of Q turns out to be optimal. (must iterate once more to obtain R = 171). c) Use formula G(Q,R) = h[Q/2 + R - ] + K/Q + pn(R)/Q Substitute (Q,R) = (500,165) from part (a) = (204,171) from part (b) Obtain G(500,165) = $2606.75 >  cost = $641.59 yearly G(204,171) = $1965.16 d) Use the relationship n(R) = (1 - )Q to determine R and the fact that n(R) =  sL(z). Hence, L(z) = n(R)   (1  )Q   (.01)(198) 19.75 = .10026 From Table B-4, z  .90. Since R = z +  we obtain R = (19.75)(.90) + 114 = 132. The imputed shortage cost is: ^p = Qh ((1  F(R)) = $4.09