Solve problem 14 assuming a maximum order size of 20 per month. r Eight Push and
ID: 432730 • Letter: S
Question
Solve problem 14 assuming a maximum order size of 20 per month. r Eight Push and Puill Production Control Systems: MRP and JIT 14. A single inventory item is ordered from an outside supplier. The demand for this item over the next 12 months is 6, 12, 4, 8, is, 25, 20, 5, to po 12. Current inventory of this item is 4, and ending inventory should be us20 holding cost of S1 per period and a setup cost of $40. Determine the order for this item based on 8.Assume a a. Silver-Meal. b. Least unit cost c. Part period balancing.Explanation / Answer
Demand = (6, 12, 4, 8,15, 25, 20, 5, 10, 20, 5, 12)
Subtract starting inventory from period 1 demand and add ending inventory to last period demand
Therefore Net demand = (2, 12, 4, 8, 15, 25, 20, 5, 10, 20, 5, 20)
a) Silver Meal method
Start in period 1:
C(1) = 40
C(2) = (40 + 12)/2 = 26
C(3) = [40 + 12 + (2)(4)]/3 = 20
C(4) = [40 + 12 + (2)(4) + (3)(8)]/4 = 21
Start in period 4:
C(1) = 40
C(2) = (40 + 15)/2 = 27.5
C(3) = [40 + 15 + (2)(25)]/3 = 35
Start in period 6:
C(1) = 40
C(2) = (40 + 20)/2 = 30
C(3) = [40 + 20 + (2)(5)]/3 = 23.3333
C(4) = [40 + 20 + (2)(5) + (3)(10)]/4 = 25
Start in period 9:
C(1) = 40
C(2) = (40 + 20)/2 = 30
C(3) = [40 + 20 + (2)(5)]/3 = 23.3333
C(4) = [40 + 20 + (2)(5) + (3)(20)]/4 = 32.5
Optimal order policy = (2, 12, 4), (8, 15), (25, 20, 5), (10, 20, 5), (20)
Total cost = 20*3+27.5*2+23.33*3+23.33*3+40 = 295
b) Least Unit cost
Start in period 1:
C(1) = 40/2 = 20
C(2) = (40 + 12)/(2 + 12) = 3.71
C(3) = (40 + 12 + 8) /(2 + 12 + 4) = 3.33
C(4) = (40 + 12 + 8 + 24) /(2 + 12 + 4 + 8) = 3.23
C(5) = (40 + 12 + 8 + 24 + 60) /(2 + 12 + 4 + 8 + 15) = 3.51
Start in period 5:
C(1) = 40/15 = 2.67
C(2) = (40 + 25)/(15 + 25) = 1.625
C(3) = (40 + 25 + 40)/(15 + 25 + 20) = 1.75
Start in period 7:
C(1) = 40/20 = 2
C(2) = (40 + 5)/(20 + 5) = 1.8
C(3) = (40 + 5 + 20)/(20 + 5 + 10) = 1.86
Start in period 9:
C(1) = 40/10 = 4
C(2) = (40 + 20)/(10 + 20) = 2
C(3) = (40 + 20 + 10)/(10 + 20 + 5) = 2
C(4) = (40 + 20 + 10 + 60)/(10 + 20 + 5 + 20) = 2.3636
Optimal order policy = (2, 12, 4, 8), (15, 25), (20, 5), (10, 20, 5), (20)
Total cost = 3.23*(2+12+4+8)+1.625*(15+25)+1.8*(20+5)+2*(10+20+5)+40 = 304
c) Part Period Balancing
In this method, we set the order cycle time equal to the number of periods to most closely match total holding cost with ordering cost. We calculate absolute difference between holding and ordering cost in for each period and select lowest value period.
Start in period
Periods
Holding Cost
Abs difference (Holding - Setup)
1
2
12
28
3
20
20
4
44
4 (closest)
5
2
25
15 (closest)
3
60
20
7
2
5
35
3
25
15 (closest)
4
85
45
10
2
5
35
3
45
5 (closest)
Part Period Balancing solution: r = (2, 12, 4, 8), (15, 25), (20, 5, 10), (20, 5, 20)
Total cost = 40*4+44+25+25+45 = $ 299
d) Cost comparison of three methods.
1. Silver Meal method Total cost = $ 295
2. Least Unit Cost method Total cost = $ 304
3. Part period balancing method total cost = $ 299
Silver Meal method yields the least cost.
Start in period
Periods
Holding Cost
Abs difference (Holding - Setup)
1
2
12
28
3
20
20
4
44
4 (closest)
5
2
25
15 (closest)
3
60
20
7
2
5
35
3
25
15 (closest)
4
85
45
10
2
5
35
3
45
5 (closest)
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