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 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



C(1) = K
C(2) = (K + hr2) / 2
C(3) = (K + hr2 + 2hr3) / 3
II. 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)
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 = r1 + r2
= 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 )
II. 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)
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
closest
• 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)
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 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) = (40 + 12)/2 = 26
C(3) = [40 + 12 + (2)(4)]/3 = 20
C(4) = [40 + 12 + (2)(4) + (3)(8)]/4 = 21
y1 = r1 + r2 + r3 = 2 + 12 + 4 = 18
stop.
Start in period 4:
C(1) = 40
C(2) = (40 + 15)/2 = 27.5
C(3) = [40 + 15 + (2)(25)]/3 = 35
y4 = r4 + r5 = 8+15 = 23
Stop.
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
y6 = r6 + r7 + r8 = 25+20+5 = 50
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
y9 = r9 + r10 + r11 = 10+20+5=35
y12 = r12 = 20
y= (18, 0, 0, 23, 0, 50, 0, 0, 35, 0, 0, 20)
Stop.
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
y1 = r1 + r2 + r3 + r4 = 2 + 12 + 4 + 8= 26
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
y5 = r5 + r6 = 15+25 = 40
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
y7 = r7 + r8 = 20+5= 25
Stop.
Stop.
Stop.
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
y9 = r9 + r10 + r11 = 10+20+5=35
y12 = r12 = 20
y= (26, 0, 0, 0, 40, 0, 25, 0, 35, 0, 0, 20)
c) Part period balancing
r = (2, 12, 4, 8,15, 25, 20, 5, 10, 20, 5, 20)
h=1
K = 40
Starting in period 1
Order horizon
Total holding cost
1
0
2
1(12) = 12
3
1(12) + 2(1)(4) = 20
4
1(12) + 2(1)(4) + 3(1)(8) = 44
y1 = r1 + r2 + r3 + r4 = 2 + 12 + 4 + 8= 26
closest
•
We start again in period 5
Order horizon
Total holding cost
1
0
2
1(25) = 25
3
1(25) + 2(1)(20) = 65
closest
y5 = r5 + r6 = 15+25 = 40
•
Start in period 7
Order horizon
Total holding cost
1
0
2
1(5) = 5
3
1(5) + 2(1)(10) = 25
4
1(5) + 2(1)(10) + 3(1)(20) = 85
y7 = r7 + r8 + r9= 20+5+10= 35
closest
•
Start in period 10
Order horizon
Total holding cost
1
0
2
1(5) = 5
3
1(5) + 2(1)(20) = 45
y10 = r10 + r11 + r12
= 20+5+20
= 45
y = (26, 0, 0, 0, 40, 0, 35, 0, 0, 45, 0, 0)
closest
Comparison of results
Silver – Meal
Demand
Least Unit Cost
Part Period
Balancing
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)+15+20+2(5)
+20+2(5)= 95
12+2(4)+3(8)+25+5+
20+2(5)=104
12+2(4)+3(8)+25+5+
2(10)+5+2(20)= 139
Holding cost
95(1)=95
104(1)=104
139(1)=139
Setup cost
5(40)=200
5(40)=200
4(40)=160
Total Cost
295
304
299
The Silver Meal resulted the least expensive cost.
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?
Exercise 17 pg 381
Week
Requirements
1
2
3
4
5
6
335
200
140
440
300
200
K = $200
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 = 114.67
C(4) = [(3)(114.67) + (3)(440)(0.3)]/4 = 185 Stop.
y1= = r1 + r2 + r3 = 335 + 200 + 140 = 675
Start in period 4:
C(1) = 200
C(2) = [200 + (300)(0.3)]/2 = 145
C(3) = [(2)(145) + (2)(200)(0.3)]/3 = 136.67 Stop.
y4= r4 + r5 + r6 = 440 + 300 + 200 = 940
y = (675, 0, 0, 940, 0, 0)
b) Least unit cost
Start in period 1:
C(1) = 200/335 = 0.597
C(2) = [200 + (200)(0.3)]/(335 + 200) = 0.486
C(3) = [200 + (200)(0.3) + (140)(2)(0.3)]/(335 + 200 + 140) = 0.51
Stop.
y1= r1 + r2 = 335 + 200 = 535
Start in period 3:
C(1) = 200/140 = 1.428
C(2) = [200 + (440)(0.3)]/(140 + 440) = 0.572
C(3) = [200 + (440)(0.3) + (300)(2)(0.3)]/(140 + 440 + 300) = 0.58
y3= r3 + r4 = 140 + 440 = 580
Start in period 5:
C(1) = 200/300 = 0.67
C(2) = [200 + (200)(0.3)]/(300 + 200) = 0.52
y5 = r5 + r6 = 300 + 200 = 500
y = (535, 0, 580, 0, 500, 0)
Stop.
Stop.
c) Part period balancing
Week
Requirements
1
2
3
4
5
6
335
200
140
440
300
200
K = $200
h = $0.30
• Starting in period 1
Order horizon
Total holding cost
1
0
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
y1= r1 + r2 + r3 = 335 + 200 + 140 = 675
closest
• Starting in period 4
Order horizon
Total holding cost
1
0
2
0.3(300) = 90
3
0.3(300) + 2(0.3)(200) = 210
y4= r4 + r5 + r6 = 440 + 300 + 200 = 940
y = (675, 0, 0, 940, 0, 0)
closest
Comparison of results
Silver – Meal
Demand
Part Period
Balancing
Least Unit Cost
r = (335, 200, 140, 440, 300, 200)
Solution
y = (675, 0, 0, 940,
0, 0)
y = (535, 0, 580, 0,
500, 0)
y = (675, 0, 0, 940,
0, 0)
Holding inventory
200+2(140)+300+
2(200) = 1180
200+440+200 = 840
200+2(140)+300+
2(200) = 1180
1180(0.3)=354
840(0.3)=252
1180(0.3)=354
Setup cost
2(200)=400
3(200)=600
2(200)=400
Total Cost
754
852
754
Holding cost
The Silver Meal and Part Period Balancing heuristics resulted in the same least
expensive costs.