ALTERNATIVES LOT-SIZING SCHEMES
Alternatives Lot-Sizing Schemes
 The silver-meal heuristic
 Least Unit Cost
 Past Period Balancing
The Silver-Meal Heuristic
 Forward method that requires determining the average cost per
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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
Method
I.
Start the calculation from period 1 to next period


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C(1) = K
C(2) = (K + hr2) / 2
C(3) = (K + hr2 + 2hr3) / 3
Stop the calculation when C(j) > C(j-1)
III. Set y1 = r1 + r2 + … + rj-1
IV. Start over at period j, repeat step (I) – (III)
II.
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.
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
= 102.67
Stop the calculation as the C(3) > C(2)
•
y1 = r 1 + r 2
= 18 +30
= 48
Step IV
Starting in period 3
•
C(1) = 80
•
C(2) = [80 + (2)(5)] / 2
= 45
•
C(3) = [80 + (2)(5) + (2)(2)(20)] / 3
= 56.67. Stop
•
y3 = r3 + r4
= 42 + 5
= 47
 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)
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
+
Method
I.
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)
III. Set y1 = r1 + r2 + … + rj-1
IV. Start over at period j, repeat step (I) – (III)
II.
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)] / (18 + 30)
= 2.92
•
C(3) = [80 + (2)(30) + (2)(2)(42)] / (18+30+42)
= 3.42
Stop the calculation as the C(3) > C(2)
•
y1 = r1 + r2
= 18 +30
= 48
Step IV
• Starting in period 3
•
•
C(1) = 80 / 42
= 1.9
C(2) = [80 + (2)(5)] / (42 + 5)
= 1.92 Stop
•
y3 = r3
= 42
= 42
Step IV
Starting in period 4
•
C(1) = 80 / 5
= 16
•
C(2) = [80 + (2)(20)] / (5 + 20)
= 4.8
•
y4 = r4 + r5
= 5 + 20
= 25
•
Thus y = (48, 0, 42, 25, 0)
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.
Example
r = (18, 30, 42, 5, 20)
Holding cost = $2 per case per week
Setup cost = $80
Starting in period 1
Order horizon
Total holding cost
1
0
2
2(30) = 60
3
2(30) + 2(2)(42) = 228
closest
 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 = 18 + 30 = 48
Starting in period 3
Order horizon
Total holding cost
1
0
2
2(5) = 10
3
2(5) + 2(2)(20) = 90
• 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
= 42 + 5 + 20
= 67
• y = (48, 0, 67, 0, 0)
closest
Comparison of Results
Silver – Meal
Demand
Solution
Least Unit Cost
Part Period
Balancing
r = (18, 30, 42, 5, 20)
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
3(80)=240
2(80)=160
Total Cost
310
340
310
The Silver Meal and Part Period Balancing heuristics resulted in the same least
expensive costs.
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
a)
b)
c)
d)
Silver-Meal
Least unit cost
Part period balancing
Which lot-sizing method resulted in the lowest cost for the 12
periods?
Exercise 17 – pg 381
• The time-phased net requirements for the base assembly in a table
lamp over the next six weeks are
Week
Requirements
1
2
3
4
5
6
335
200
140
440
300
200
• The setup cost for the construction of the base assembly is $200,
and the holding cost is $0.30 per assembly per week
a)
b)
c)
d)
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?