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_2010_PPC_Answers

生产计划与控制2010至2011学年第1学期试卷A答案 Answer 1 a)and b) Forecast Period Actual et (86+75)/2=80.5 3 72 +8.5 (75+72)/2=73.5 4 83 -9.5 etc 77.5 5 132 -54.5 107.5 6 65 42.5 98.5 ) 110 -11.5 c)MAD=25.3 MSE =1014.25 MAPE = =28.0757 Answer 2 入=280 c =2.40 K =45 I=.20 a)Q*= 2K (2)(45)(280) =229 (.2)(2.40) b)T=Q*/=229/280=.8179yrs.(=9.81 months) c)G*=V2Kh=N(2)(45)(280)(.2)2.40)=$109.98 d) 229 3 wks R 9.82m0s

生产计划与控制 2010 至 2011 学年第 1 学期试卷 A 答案 Answer 1 a) and b) Forecast Period Actual et (86 + 75)/2 = 80.5 3 72 +8.5 (75 + 72)/2 = 73.5 4 83 -9.5 etc 77.5 5 132 -54.5 107.5 6 65 42.5 98.5 7 110 -11.5 c) MAD = 25.3 MSE = 1014.25 MAPE = 100 1 n ei Di        = 28.0757 Answer 2  = 280 c = 2.40 K = 45 I = .20 a) Q* = 2K h  (2)(45)(280) (.2)(2.40)  229 b) T = Q*/ = 229/280 = .8179 yrs. (= 9.81 months) c) G* = 2Kh = (2)(45)(280)(.2)(2.40) = $109.98 d) 229 R 3 wks 9.82 mos

r=y=(280)(3/52)=16.15 r=16 units Answer 3 a)c0=.08-.03=.05 cu=.35-.08=.27 Critical ratio 27 =.84375 .05+.27 From the given distribution,we have: f() F(Q)_ 0 .05 .05 5 .10 .15 10 .10 .25 15 .20 .45 20 .25 .70 <----.84375 25 .15 .85 30 .10 .95 35 .05 1.00 Since the critical ratio falls between 20 and 25 the optimal is Q=25 bagels. b)The answers should be close since the given distribution appears to be close to the normal. c)u=xf(x)=(0)(.05)+(5)(.10)+...+(35)(.05)=18 σ2=∑x2f(x)-u2=402.5-(18)2=78.5 (2321032) 0= .36 =8.86 The z value corresponding to a critical ratio of.84375 is 1.01. Hence, Q*=0z+μ=(8.86)(1.01)+18=26.95~27

r =  = (280)(3/52) = 16.15 r = 16 units Answer 3 a) c0 = .08 - .03 = .05 cu = .35 - .08 = .27 Critical ratio = .27 .05  .27 = .84375 From the given distribution, we have: Q f(Q) F(Q) 0 .05 .05 5 .10 .15 10 .10 .25 15 .20 .45 20 .25 .70  - - - - .84375 25 .15 .85 30 .10 .95 35 .05 1.00 Since the critical ratio falls between 20 and 25 the optimal is Q = 25 bagels. b) The answers should be close since the given distribution appears to be close to the normal. c)  = xf(x) = (0)(.05) + (5)(.10) +...+(35)(.05) = 18 2 = x 2f(x) -  2 = 402.5 - (18) 2 = 78.5  = (2)(32)(1032) .36 = 8.86 The z value corresponding to a critical ratio of .84375 is 1.01. Hence, Q* = z +  = (8.86)(1.01) + 18 = 26.95 ~ 27

Answer 4 k 1700/120/46 0.308 thousands of cookies per worker per day Cum Cum *Cum #Units #Units Forecast Forecast *Monthly Monthly *End Working /Workers /Workers Demand Demand #Min Product Product Inv. Month Days (000 (000) (000) (000) Workers (000) (000) (000 26 8.01 8.01 850 850 107 1105 1105 255 2 24 7.39 15.40 1260 2110 138 1020 2125 15 20 6.16 21.56 510 2620 122 850 2975 355 4 16 4.93 26.49 980 3600 136 680 3655 55 Total 680 *Note:These three columns assume a work force of 138 workers in every period a)Minimum constant work force =138 workers b)CH=100 CF=200 CI=0.1 per cookie per month Initial workers =100 Beg.inv.0 workers hired 138 -100 38 End inv.=680 Total cost of this plan is (100)(38)+(.1)(680,000)=$71,800 Answer 5 a) MRP calculations:component A Week 910 11 12 13 14 15 16 17 Gross reqts 200 200 80 80200 400400 400 Net req 200 200 80 80200 400 400 400 Time phased net reg 200 200 80 80200 400400 400 Planned order release 200 200 80 80200 400 400 400 b)MRP calculations:component B Week 910 11121314151617 Gross reqts. 1001004040100200200200 Net reqts. 10010040 40100200200200 Time phased net reqts.100100 4040100200 200200 Planned order release 100 100 4040 100 200200 200 c)MRP calculations:component C Week 78910111213141516

Answer 4 k = 1700/120/46 = 0.308 thousands of cookies per worker per day. Cum Cum *Cum #Units #Units Forecast Forecast *Monthly Monthly *End Working /Workers /Workers Demand Demand #Min Product Product Inv. Month Days (000 (000) (000) (000) Workers (000) (000) (000 1 26 8.01 8.01 850 850 107 1105 1105 255 2 24 7.39 15.40 1260 2110 138 1020 2125 15 3 20 6.16 21.56 510 2620 122 850 2975 355 4 16 4.93 26.49 980 3600 136 680 3655 55 Total 680 *Note: These three columns assume a work force of 138 workers in every period a) Minimum constant work force = 138 workers b) CH = 100 CF = 200 CI = 0.1 per cookie per month Initial # workers = 100 Beg. inv. = 0 # workers hired = 138 - 100 = 38 End inv. = 680 Total cost of this plan is (100)(38) + (.1)(680,000) = $71,800 Answer 5 a) MRP calculations: component A Week 9 10 11 12 13 14 15 16 17 Gross reqts 200 200 80 80 200 400 400 400 Net req 200 200 80 80 200 400 400 400 Time phased net req 200 200 80 80 200 400 400 400 Planned order release 200 200 80 80 200 400 400 400 b) MRP calculations: component B Week 8 9 10 11 12 13 14 15 16 17 Gross reqts. 100 100 40 40 100 200 200 200 Net reqts. 100 100 40 40 100 200 200 200 Time phased net reqts. 100 100 40 40 100 200 200 200 Planned order release 100 100 40 40 100 200 200 200 c) MRP calculations: component C Week 6 7 8 9 10 11 12 13 14 15 16

Planned order release-A 2002008080200400400400 Gross reqts.of c for A 2002008080200400400400 Net reqts.of c for A 2002008080200400400400 Time ph.n.reqts.of C for A 20020080 80200400400400 P.O.release of c for A 2002008080200400400400 Planned order release-B 100100 4040100200200200 Gross reqts of c for B 2002008080200400400400 Net reqts.of c for B 2002008080200400400400 Time ph.n.reqts of c for B 200 200 8080200400400400 P.0.release of C for B 200200 8080200400400400 0 P.O.release of C for A&B 200400280160280600800800400 Answer 6 a)SPT sequence minimizes mean flow time.It is 6-5-1-4-2-7-3. b)First order by EDD which is 1-2-5-3-4-7-6. Processing Completion Job Time Time Due Date 3 4 2 6 9 8 Stop. Place job 2 at the end of the current sequence. Processing Completion Job Time Time Due Date V 0 3 4 5 2 5 11 8 13 12 Stop. Place job 3 at the end of the current sequence

Planned order release-A 200 200 80 80 200 400 400 400 Gross reqts. of C for A 200 200 80 80 200 400 400 400 Net reqts. of C for A 200 200 80 80 200 400 400 400 Time ph.n.reqts. of C for A 200 200 80 80 200 400 400 400 P.O. release of C for A 200 200 80 80 200 400 400 400 Planned order release-B 100 100 40 40 100 200 200 200 Gross reqts of C for B 200 200 80 80 200 400 400 400 Net reqts. of C for B 200 200 80 80 200 400 400 400 Time ph. n. reqts of C for B 200 200 80 80 200 400 400 400 P.O. release of C for B 200 200 80 80 200 400 400 400 0 P.O. release of C for A&B 200 400 280 160 280 600 800 800 400 Answer 6 a) SPT sequence minimizes mean flow time. It is 6-5-1-4-2-7-3. b) First order by EDD which is 1-2-5-3-4-7-6. Processing Completion Job Time Time Due Date 1 3 3 4 2 6 9 8 * Stop. Place job 2 at the end of the current sequence. Processing Completion Job Time Time Due Date 1 3 3 4 5 2 5 11 3 8 13 12 * Stop. Place job 3 at the end of the current sequence

Processing Completion Job Time Time Due Date 1 3 3 4 5 3 夕 11 4 Y 9 15 7 16 21 6 1 17 25 *Stop The optimal sequence is 1-5-4-7-6-2-3 (or 1-5-4-7-6-3-2).Jobs 2 and 3 are tardy. c)The maximum lateness is minimized by EDD which is 1-2-5-3-4-7-6. d)The makespan is the sum of the processing times which is 31

Processing Completion Job Time Time Due Date 1 3 3 4 5 2 5 11 4 4 9 15 7 7 16 21 6 1 17 25 * Stop The optimal sequence is 1-5-4-7-6-2-3 (or 1-5-4-7-6-3-2). Jobs 2 and 3 are tardy. c) The maximum lateness is minimized by EDD which is 1-2-5-3-4-7-6. d) The makespan is the sum of the processing times which is 31