1. ALTERNATIVES LOT-SIZING SCHEMES

2. Alternatives Lot-Sizing Schemes
The silver-meal heuristic
Least Unit Cost
Past Period Balancing

3. The Silver-Meal Heuristic
Forward method that requires determining the average cost per period as a function of the number of periods the current order to span.
Minimize the cost per period
Formula : C(j) = (K + hr2 + 2hr3 + … + (j-1)hrj) / j
C(j)  average holding cost and setup cost per period
k  order cost or setup cost
h  holding cost
r  demand

4. Method Start the calculation from period 1 to next period
C(1) = K
C(2) = (K + hr2) / 2
C(3) = (K + hr2 + 2hr3) / 3
Stop the calculation when C(j) > C(j-1)
Set y1 = r1 + r2 + … + rj-1
Start over at period j, repeat step (I) – (III)

5. Example
A machine shop uses the Silver-Meal heuristic to schedule production lot sizes for computer casings. Over the next five weeks the demands for the casing are r = (18, 30, 42, 5, 20). The holding cost is $2 per case per week, and the production setup cost is $80. Find the recommended lot sizing.

6. Step I, II & III r = (18, 30, 42, 5, 20) k = $80 h = $2
Starting in period 1
C(1) = 80
C(2) = [80 + (2)(30)] / 2
= 70
C(3) = [80 + (2)(30) + (2)(2)(42)] / 3 =
Stop the calculation as the C(3) > C(2)
y1 = r1 + r2 =
= 48

7. Step IV = 42 + 5 = 47 Starting in period 3 C(1) = 80
= 45
C(3) = [80 + (2)(5) + (2)(2)(20)] / 3
= Stop
y3 = r3 + r4 =
= 47

8. Since period 5 is the final period, thus no need to start the process again.
Set y5 = r5
= 20
Thus y = (48, 0, 47, 0, 20)

9. Least Unit Cost Similar to Silver-Meal method
Minimize cost per unit of demand
Formula : C(j) = (K + hr2 + 2hr3 + … + (j-1)hrj) / (r1 + r2 + … + rj
C(j)  average holding cost and setup cost per period
k  order cost or setup cost
h  holding cost
r  demand

10. Method Start the calculation from period 1 to next period
C(1) = K / r1
C(2) = (K + hr2) / (r1 + r2)
C(3) = (K + hr2 + 2hr3) / (r1 + r2 + r3 )
Stop the calculation when C(j) > C(j-1)
Set y1 = r1 + r2 + … + rj-1
Start over at period j, repeat step (I) – (III)

11. Step I, II & III r = (18, 30, 42, 5, 20) k = $80 h = $2
Starting in period 1
C(1) = 80 / 18
= 4.44
C(2) = [80 + (2)(30)] / ( )
= 2.92
C(3) = [80 + (2)(30) + (2)(2)(42)] / ( )
= 3.42
Stop the calculation as the C(3) > C(2)
y1 = r1 + r2 =
= 48

12. Step IV = 42 Starting in period 3 C(1) = 80 / 42 = 1.9
= 1.92 Stop
y3 = r3
= 42

13. Step IV = 5 + 20 = 25 Starting in period 4 C(1) = 80 / 5 = 16
= 4.8
y4 = r4 + r5 =
= 25
Thus y = (48, 0, 42, 25, 0)

14. Part Period Balancing
Set the order horizon equal to the number of periods that most closely matches the total holding cost with the setup cost over that period.

15. Example r = (18, 30, 42, 5, 20) Holding cost = $2 per case per week
Setup cost = $80
Starting in period 1
Because 228 exceeds the setup cost of 80, we stop. As 80 is closer to 60 than to 228, the first order horizon is two periods,
y1 = r1 +r2 = = 48
Order horizon
Total holding cost 1 2
2(30) = 60 3
2(30) + 2(2)(42) = 228
closest

16. Order horizon Total holding cost
Starting in period 3
We have exceeded the setup cost of 80, so we stop.
Because 90 is closer to 80 than 10, the order horizon is three periods.
y3 = r3 + r4 + r5 =
= 67
y = (48, 0, 67, 0, 0)
Order horizon
Total holding cost 1 2
2(5) = 10 3
2(5) + 2(2)(20) = 90
closest

17. Comparison of Results Silver – Meal Least Unit Cost
Part Period Balancing
Demand
r = (18, 30, 42, 5, 20)
Solution
y = (48, 0, 47, 0, 20)
y = (48, 0, 42, 25, 0)
y = (48, 0, 67, 0, 0)
Holding inventory
30+5=35
30+20=50
30+5+(2)(20)=75
Holding cost
35(2)=70
50(2)=100
75(2)=150
Setup cost
3(80)=240
2(80)=160
Total Cost
310
340
The Silver Meal and Part Period Balancing heuristics resulted in the same least expensive costs.

18. Exercise 14 – pg 381
A single inventory item is ordered from an outside supplier. The anticipated demand for this item over the next 12 months is 6, 12, 4, 8, 15, 25, 20, 5, 10, 20, 5, 12. Current inventory of this item is 4, and ending inventory should be 8. Assume a holding cost of $1 per period and a setup cost of $40. Determine the order policy for this item based on
Silver-Meal
Least unit cost
Part period balancing
Which lot-sizing method resulted in the lowest cost for the 12 periods?

19. Exercise 14 - pg 381
Demand = (6, 12, 4, 8,15, 25, 20, 5, 10, 20, 5, 12) Starting inventory = 4 Ending inventory = 8 h = 1 K = 40 Net out starting and ending inventories to obtain r = (2, 12, 4, 8,15, 25, 20, 5, 10, 20, 5, 20) a) Silver Meal Start in period 1: C(1) = 40 C(2) = ( )/2 = 26 C(3) = [ (2)(4)]/3 = 20 C(4) = [ (2)(4) + (3)(8)]/4 = 21 stop. y1 = r1 + r2 + r3 = = 18

20. Start in period 4: C(1) = 40 C(2) = (40 + 15)/2 = 27
Start in period 4: C(1) = 40 C(2) = ( )/2 = 27.5 C(3) = [ (2)(25)]/3 = 35 Stop. y4 = r4 + r5 = 8+15 = 23 Start in period 6: C(2) = ( )/2 = 30 C(3) = [ (2)(5)]/3 = C(4) = [ (2)(5) + (3)(10)]/4 = 25 Stop. y6 = r6 + r7 + r8 = = 50 Start in period 9: C(4) = [ (2)(5) + (3)(20)]/4 = 32.5 y9 = r9 + r10 + r11 = =35 y12 = r12 = 20 y= (18, 0, 0, 23, 0, 50, 0, 0, 35, 0, 0, 20)

21. b) Least unit cost Start in period 1: C(1) = 40/2 = 20 C(2) = ( )/(2 + 12) = 3.71 C(3) = ( ) /( ) = 3.33 C(4) = ( ) /( ) = 3.23 C(5) = ( ) /( ) = 3.51 Stop. y1 = r1 + r2 + r3 + r4 = = 26 Start in period 5: C(1) = 40/15 = 2.67 C(2) = ( )/( ) = C(3) = ( )/( ) = 1.75 Stop. y5 = r5 + r6 = = 40 Start in period 7: C(1) = 40/20 = 2 C(2) = (40 + 5)/(20 + 5) = 1.8 C(3) = ( )/( ) = 1.86 Stop. y7 = r7 + r8 = 20+5= 25

22. Start in period 9:
C(1) = 40/10 = 4
C(2) = ( )/( ) = 2
C(3) = ( )/( ) = 2
C(4) = ( )/( ) =
y9 = r9 + r10 + r11 = =35
y12 = r12 = 20
y= (26, 0, 0, 0, 40, 0, 25, 0, 35, 0, 0, 20)

23. c) Part period balancing
h = 1
K = 40
Starting in period 1
 y1 = r1 + r2 + r3 + r4 = = 26
Order horizon
Total holding cost 1 2
1(12) = 12 3
1(12) + 2(1)(4) = 20 4
1(12) + 2(1)(4) + 3(1)(8) = 44
closest

24. Order horizon Total holding cost Order horizon Total holding cost
We start again in period 5
y5 = r5 + r6 = = 40
Start in period 7
 y7 = r7 + r8 + r9= = 35
Order horizon
Total holding cost 1 2
1(25) = 25 3
1(25) + 2(1)(20) = 65
closest
Order horizon
Total holding cost 1 2
1(5) = 5 3
1(5) + 2(1)(10) = 25 4
1(5) + 2(1)(10) + 3(1)(20) = 85
closest

25. Start in period 10
 y10 = r10 + r11 + r12 =
= 45
y = (26, 0, 0, 0, 40, 0, 35, 0, 0, 45, 0, 0)
Order horizon
Total holding cost 1 2
1(5) = 5 3
1(5) + 2(1)(20) = 45
closest

26. Comparison of results Silver – Meal Least Unit Cost
Part Period Balancing
Demand
r = (2, 12, 4, 8,15, 25, 20, 5, 10, 20, 5, 20)
Solution
y= (18, 0, 0, 23, 0, 50, 0, 0, 35, 0, 0, 20)
y = (26, 0, 0, 0, 40, 0, 25, 0, 35, 0, 0, 20)
y = (26, 0, 0, 0, 40, 0, 35, 0, 0, 45, 0, 0)
Holding inventory
12+2(4) (5)+20+2(5)= 95
12+2(4)+3(8)+25+5+
20+2(5)=104
2(10)+5+2(20)= 139
Holding cost
95(1)=95
104(1)=104
139(1)=139
Setup cost
5(40)=200
4(40)=160
Total Cost
295
304
299
The Silver Meal resulted the least expensive cost.

27. Exercise 17 – pg 381
The time-phased net requirements for the base assembly in a table lamp over the next six weeks are
The setup cost for the construction of the base assembly is $200, and the holding cost is $0.30 per assembly per week
Determine the lot sizes using the Silver-Meal heuristic
Determine the lot sizes using the least unit cost heuristic
Determine the lot sizes using part period balancing
Which lot-sizing method resulted in the lowest cost for the 6 periods?
Week 1 2 3 4 5 6
Requirements
335
200
140
440
300

28. Exercise 17 pg 381 y4= r4 + r5 + r6 = 440 + 300 + 200 = 940 Week 1 2 3
h = $0.30
a) Silver Meal
Start in period 1:
C(1) = 200
C(2) = [200 + (200)(0.3)]/2 = 130
C(3) = [(2)(130) + (2)(140)(0.3)]/3 =
C(4) = [(3)(114.67) + (3)(440)(0.3)]/4 = 185 Stop.
  y1= = r1 + r2 + r3 = = 675
Start in period 4:
C(2) = [200 + (300)(0.3)]/2 = 145
C(3) = [(2)(145) + (2)(200)(0.3)]/3 = Stop.
y4= r4 + r5 + r6 = = 940
y = (675, 0, 0, 940, 0, 0)
Week 1 2 3 4 5 6
Requirements
335
200
140
440
300

29. b) Least unit cost
Start in period 1:
C(1) = 200/335 = 0.597
C(2) = [200 + (200)(0.3)]/( ) = 0.486
C(3) = [200 + (200)(0.3) + (140)(2)(0.3)]/( ) = Stop.
y1= r1 + r2 = = 535
Start in period 3:
C(1) = 200/140 = 1.428
C(2) = [200 + (440)(0.3)]/( ) = 0.572
C(3) = [200 + (440)(0.3) + (300)(2)(0.3)]/( ) = Stop.
y3= r3 + r4 = = 580
Start in period 5:
C(1) = 200/300 = 0.67
C(2) = [200 + (200)(0.3)]/( ) = 0.52 Stop.
 y5 = r5 + r6 = = 500
y = (535, 0, 580, 0, 500, 0)

30. c) Part period balancing
Week 1 2 3 4 5 6
Requirements
335
200
140
440
300
K = $200
h = $0.30
Starting in period 1
y1= r1 + r2 + r3 = = 675
Order horizon
Total holding cost 1 2
0.3(200) = 60 3
0.3(200) + 2(0.3)(140) = 144 4
0.3(200) + 2(0.3)(140) + 3(0.3)(440) = 540
closest

31. Order horizon Total holding cost
Starting in period 4
y4= r4 + r5 + r6 = = 940
y = (675, 0, 0, 940, 0, 0)
Order horizon
Total holding cost 1 2
0.3(300) = 90 3
0.3(300) + 2(0.3)(200) = 210
closest

32. Comparison of results Silver – Meal Least Unit Cost
Part Period Balancing
Demand
r = (335, 200, 140, 440, 300, 200)
Solution
y = (675, 0, 0, 940, 0, 0)
y = (535, 0, 580, 0, 500, 0)
Holding inventory
200+2(140) (200) = 1180
= 840
Holding cost
1180(0.3)=354
840(0.3)=252
Setup cost
2(200)=400
3(200)=600
Total Cost
754
852
The Silver Meal and Part Period Balancing heuristics resulted in the same least expensive costs.