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By: Prof. Y. Peter Chiu

Material Management ~ Mid-Term #1 Solution ~ 4 / 2010. By: Prof. Y. Peter Chiu. r = ( 55, 62, 52, 42, 88, 65 ) C= ( 60, 60, 50, 50, 77, 70 ). #1. Not satisfied ！ No feasible solutions by regular shift in-house production plan.

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By: Prof. Y. Peter Chiu

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  1. Material Management ~ Mid-Term #1Solution ~ 4 / 2010 By: Prof. Y. Peter Chiu

  2. r = ( 55, 62, 52, 42, 88, 65 ) C= ( 60, 60, 50, 50, 77, 70 ) #1 Not satisfied！ No feasible solutions by regular shift in-house production plan. However, one can look into “Overtime” or “Outsourcing” solutions.

  3. #2

  4. # 3 Yes (a) Week 1 2 3 4 5 6 7 Req. Capacities 350 180 150 320 230 210 170 500 500 100 100 450 100 100 (C-R) lot-for-lot Initial Solution 150 320 (50) (220) 220 (110) (70) r’ 350 450 100 100 410 100 100 (C-R’) 150 50 0 0 40 0 0 Costs for Initial Solution : ($105)*7+($0.4)[270+220+180+70]=$1031 (b) One Solution 500 500 0 0 410 100 100 Alternative #1: K=$105 ; h=$0.4 (A) Week 4th lot : by Week 2 2*(0.4)*50=$40 by Week 1 3*(0.4)*50=$60 Saves $5 Total savings $30 & (B) Week 3rd lot : by Week 1 2*(0.4)*100=$80 Saves $25 Costs for this Solution : ($105)*5+($0.4)[150+470+320+180+70]=$1001

  5. # 3 Week 1 2 3 4 5 6 7 Req. Capacities 350 180 150 320 230 210 170 500 500 100 100 450 100 100 r’ 350 450 100 100 410 100 100 (C-R’) 150 50 0 0 40 0 0 (b) Optimal Solution 400 500 0 100 410 100 100 Inventories 50 370 220 0 180 70 0 Alternative #2: the best  Total Savings $45 X (A) Week 3rd lot : by Week 2 $(0.4)(50)=$20 by Week 1 $(0.4)(2)(50)=$40 (B) Week 4th lot : by Week 1 3($0.4)(100)=$120 cost Initial Solution : ($105)*7+($0.4)[270+220+180+70]=$1031 Costs for optimal solution : ($105)*6+($0.4)[50+370+220+180+70]=$986

  6. # 4 (a)

  7. # 4 (b) Starting in Week 1 Order Horizon Total Holding cost 1 2 3 4 0 110 (0.5*220) 290 [0.5*220+2*(0.5)*180] 440 [290+3*(0.5)*100] ∵ K is closer to period 3 ∴ K=320

  8. # 4 Starting in Week 4 (b) Order Horizon Total Holding cost 1 2 3 0 100 (0.5*200) 300 [0.5*200+2*(0.5)*200] K=320 ∵ K is closer to period 3 ∴

  9. # 4 (c) Total Costs = Setup costs + holding costs = 2*($320) + $0.5*(1080) = $ 1180 ( Silver-Meal) = 2*($320) + $0.5*(1180) = $ 1230 ( P.P.B.) Silver/Meal has the lower costs. (4.07% lower)

  10. # 5 Virginia’s Sweaters Inc. has authorized the followingMPS for his exclusive line of cashmere sweaters. She wants to use the MPS record for promising future orders. Current order promises are included. The MPS order quantity is 65 units.Beginning inventory is 24. Complete the Following Projected available & ATP records. Projected available is calculated as Projected available = Beginning inventory + MPS shipment – the greater of the period’s forecast or the customer orders promised for delivery. ATP action bucket =(beginning inventory ) + (MPS shipment)- (Customer orders before next replenishment) ATP future = (MPS shipment) - (Customer orders between current MPS shipment and next scheduled replenishment)

  11. # 6 Virgina. the owner of J.C. Sweaters has received several additional orders to consider. Using the ATP record calculated in Solved Problem #5, calculate which of the new order Jeannette should accept. The new orders are: • (1) 32 units for delivery in period 4. • (2) 30 units for delivery in Period 6. • (3) 36 units for delivery in Period 7. • (4) 64 units for delivery in Period 9.

  12. # 6 • 32 units for delivery in period 4. CAN NOT ACCEPT

  13. # 6 (2) 30 units for delivery in Period 6.ACCEPT the Order

  14. # 6 (3) 36 units for delivery in Period 7.CAN NOT ACCEPT

  15. # 6 (4) 64 units for delivery in Period 9.ACCEPT the Order

  16. The End

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