1 By: Prof. Y. Peter Chiu MRP & JIT ~ HOMEWORK SOLUTION ~

2 What is the MRP Calculations for the slide assemblies ? ( 3 slide assemblies for each valve casing ) Lead Time = 2 weeks Assume On-hand inventory of 270 slide assemblies at the end of week 3 & Scheduled receipts of 78 & 63 at the beginning of week 5 & 7 Solution to MRP Calculations for the slide assemblies Week Gross Requirements Scheduled Receipts On-hand inventory Net Requirement s Time-Phased Net Requirements Planned Order Release (lot for lot)

3 #4 (a) MPS for the computers   

4 #5

5 #6

6 # 9 (b) Week MPS-end item Component B (P.O.R) Component F -Net. Req. Time Phased Net. Req. Ans. →Planned Order Release

7 # 9 (c) Week MPS-end item P.O.R –Comp. B Net Req. –Comp. E Time Phased –Net Req. P.O.R – Comp. E P.O.R –Comp. G Time Phased –Net Req. Net Req. –Comp. G Net Req. –Comp. I Net Req. –Comp. H Time Phased –Net Req. Ans. → P.O.R –Comp. I Ans. → P.O.R –comp. H (d)

8 Month Demand Current Inventory : 4 An ending Inventory should be : 8 h = $ 1 k = $ 40 Month Net. Demand # Starting in Period 1 : C(1) = 40 C(2) = (40+12)/2 = 26 C(3) = [ (4)] /3 = 20 C(4) = [ (4)+3(8)] /4= 21 ∴ (a) Silver-Meal

9 Starting in Period 4 : ∴ Starting in Period 9 : Starting in Period 6 : C(1) = 40 C(2) = [40+15]/2 = 27.5 C(3) = [ (25)] /3 = 35 C(1) = 40 C(2) = [40+20]/2 = 30 C(3) = [ (5)] /3 = 23.3 C(4) = [ (5)+3(10)] /4 = 25 C(1) = 40 C(2) = [40+20]/2 = 30 C(3) = [ (5)] /3 =23.3 C(4) = [ (5)+3(20)] /4 = 32.5 ∴ ∴ ∴ ∴ Using Silver- Meal ; y = [ 18, 0, 0, 23, 0, 50, 0, 0, 35, 0, 0, 20 ] # 14 (a) Silver-Meal

10 C(1) = 40 /2 = 20 C(2) = 52 /14 = 3.71 C(3) = 60 /18 = 3.33 C(4) = 84 /26 = 3.23 C(5) = (84+60) /41 = 3.51 (b) LUC Starting in Period 1 : Starting in Period 7 : Starting in Period 5 : C(1) = 40 /20 = 2 C(2) = (40+5) /25 = 1.80 C(3) = [40+5+2(10)] /35 = 1.86 C(1) = 40 /15 = 2.67 C(2) = 65 /40 = 1.63 C(3) = [65+2(20)] /60 = 1.75 ∴ ∴ ∴ # 14

11 (C) PPB Starting in Period 1 : PeriodHolding cost * (12) = (4) = (8) = 44 Closer to period 4 K = $40 ∴ Starting in Period 9 : C(1) = 40 /10 = 4 C(2) = 60 /30 = 2 C(3) = [60+2(5)] /35 = 2 C(4) = [70+60] /55 = 2.36 ∴ Using LUC ; y = [ 26, 0, 0, 0,40, 0, 25, 0, 35, 0, 0, 20 ] ∴ ∴ (b) LUC # 14

12 Starting in Period 5 : PeriodHolding cost (20) = 65 Closer to period 2 ∴ Starting in Period 7 : Starting in Period 10 : 2323 PeriodHolding cost 5 5+2(12) = 29 PeriodHolding cost (10) = (20) = 85 Closer to period 3 ∴ ∴ ∴ Using PPB ; y = [ 26, 0, 0, 0,40, 0, 35, 0, 0, 45, 0, 0 ] (C) PPB K = $40

13 SM LUC PPB Demand Inv. SM Inv. LUC Inv. PPB Σ = 95 Σ = 104 Σ = 139 Cost of S.M. ($40*5)+($1*95) = $295 Cost of LUC ($40*5)+($1*104) = $304 Cost of PPB ($40*4)+($1*139) = $299 ∴ Silver Meal Method is the least expensive one ! # 14 (d)

14 #17 K=200; h=0.3

15 #17

16 #17

17 #18

18 #18 (cont’d)

19 h = $0.4 K = $180 Starting Inventory Week 6 is 75 Receiving: 30 & 10 in week 8 & 10 # 24 Week Net Demand Demand (a) PPB : Starting in Period 1 : Starting in Period 4 : Holding costs Week Period *0.4= (0.4)(150)= 186 Closer to period 3 ∴ ∴ *0.4= (300)(0.4)= 266 Closer to period 3 K = $180

20 # 24 MRP - Mother Boards Week Net. Req. Time-Phased Net Req. P.O.R. (lot-for-lot) Ans. → ( PPB ) P.O.R For DRAM Ans. → Gross Req. For DRAM Time-Phased Net Req. For DRAM P.O.R , ,

21 #25

22 #25 (cont’d)

23 #28(a) K=$200 ; h=$0.30

24 #28 (b) 17(b) S-M: ($ $754) / ($754) = 40 % 17(c) LUC: ($ $852) / ($852) = 23 % 17(d) PPB: ($ $754) / ($754) = 40 %

25 # CW.3 Consider the example presented previously for the scheduling of the valve casing assembly. Suppose that the production capacity in any week is 50 valve casings. Determine the feasible planned order release for the valve casings. Recall that the time phased net requirements for the valve casings as followed: Week Net Requirements Time-Phased Net Requirements r = Production c= Capacity

26 Week Net Requirements Time-Phased Net Requirements r = Production c= Capacity # CW.3 excess (c-r) = Capacity (62) 5 36 (26) 12 (c-r)’ = [2] Lot-shifting technique (back-shift demand from r j > c j ): (62) 5 36 (26) (c-r)’ = final r ’ = [1] First test for: It is okay!

27 # CW.4 ( continues on #CW.3) Determine the optimal production plan. Suppose that with overtime work on 2 nd shifts, the company could increase the weekly production capacity to 120 valve casings, however, extra cost per week = $105. Where K=$100, is the regular setup cost. The holding cost per valve casing per week is estimated to be $0.65. Determine the optimal production plan. Time-Phased Net Requirements r = Production c= Capacity Week Production c= Capacity (O-T)

28 # CW.4 ( continues on #CW.3) Determine the optimal production plan. Suppose that with overtime work on 2 nd shifts, the company could increase the weekly production capacity to 120 valve casings, however, extra cost per week = $105. Where K=$100, is the regular setup cost. The holding cost per valve casing per week is estimated to be $0.65. Determine the optimal production plan. Time-Phased Net Requirements r = Production c= Capacity final r ’ = (using regular shift) Week Σ= 126 Ending Inventories =

29 Time-Phased Net Requirements r = Production c= Capacity (O-T) excess (c-r) = Capacity Week # CW.4 [1] First, the cost for using regular shift is $100(10) + $0.65 (126) [ lot for lot ] = $1,081.9 [ lot for lot ] r = excess (c-r)’= Capacity final r ’ =

30 # CW.4 Time-Phased Net Requirements r = Week final r ’ = [2] T he cost for using Overtime shift is $205(4) + $0.65(372) = $ Ending Inventories = Less than the cost for using regular shift $1,081.9,  Saved $ 20.10

31 # CW.4 Ending Inventories = [3] To think about the following solution: Time-Phased Net Requirements r = Week Suppose r ’= Σ= [ One OT, 7 regular ] T he cost for using only one Overtime shift on week 4 is $205(1) + $100(7) + $0.65(232) = $ Less than the cost for using regular shift $1,081.9,  Saved $ 26.1 Less than the cost for using all Overtime shift $  Saved $ 6.0 WHY ?

32 # CW.4 Ending Inventories = [4] A Better Solution : Time-Phased Net Requirements r = Week Suppose r ’= Σ= T he cost for using the above solution is $205 (3) + $100 (2) + $0.65(195) = $ Less than the cost for using regular shift $1,081.9,  Saved $ Wow !

33 #33

34 The End The End

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