IE652 - Chapter 6
Dynamic Lot Sizing Techniques
1
Dynamic Lot Sizing Methods
Simple Rules
Heuristic Methods
Period order quantity
Fixed period demand
Lot for lot (L4L)
Silver-Meal method (SM)
Least Unit Cost (LUC)
Part Period Balancing (PPB)
Dynamic Programming (Optimum)
Wagner-Whitin
2
A Prototype Example
Suppose for a certain product type you need to produce
weekly demand below:
Week
1
2
Demand 100 75
3
4
5
6
7
175 200 150 100 75
8
100
A = $50 per order
H = $0.5 per unit per week
Assumption:
Lead time is known with certainty (fixed lead time)
3
Period Order Quantity
The average lot size desired is divided by
the average period demand
For weekly demand given above evaluate
POQ for Q = 125 units, 140 units, and 275
units.
4
POQ-Example Solution
Week
1
2
3
4
5
6
7
8
Demand
100
75
175
200
150
100
75
100
•Total demand over 8 periods = 975 units
•Average weekly demand = 975 / 8 = 122 units per week
5
POQ-Example Solution Continued
•Fixed period for order Q is determined as follows:
T=
Q
D
Where, Q = desired order (lot) size
D = average demand over the planning period
T = number of periods (time interval between orders)
6
POQ-Example Solution Continued
For Q = 125
T = fixed period between orders = 125 /122 = 1.02 = 1 week
Week Beginning Inventory Demand Order End Inventory
1
0
100
125
25
2
25
75
125
75
3
75
175
125
25
4
5
6
7
8
25
-50
-75
-50
0
200
150
100
75
100
125
125
125
125
125
-50
-75
-50
0
25
7
POQ-Example Solution Continued
For Q = 140
T = fixed period between orders = 140 /122 = 1.14 = 1 week
Week Beginning Inventory Demand Order End Inventory
1
0
100
140
40
2
40
75
140
105
3
105
175
140
70
4
70
200
140
10
5
6
10
0
150
100
140
140
0
40
7
8
40
105
75
100
140
140
105
145
Total Cost = 8 ($50) + 515 ($0.5) = $657.5
8
POQ-Example Solution Continued
For Q = 275
T = fixed period between orders = 275 /122 = 2.25 = 2 weeks
Week Beginning Inventory Demand Order End Inventory
1
0
100
275
175
2
175
75
--100
3
4
5
6
100
200
0
125
175
200
150
100
275
--275
---
200
0
125
25
7
8
25
225
75
100
275
---
225
125
Total Cost = 4 ($50) + 975 ($0.5) = $687.5
9
Fixed Period Demand
Ordering m periods of demand,
m = selected fixed period
For weekly demand given above evaluate
FPD for T = 2 weeks, 4 weeks, and
8 weeks.
10
FPD-Example Solution
Week
1
2
3
4
5
6
7
8
Demand
100
75
175
200
150
100
75
100
For T = 2 weeks
Q1 = 175 units, Q3 = 375 units, Q5 = 250 units, Q7 = 175 units
For T = 4 weeks
Q1 = 550 units, Q5 = 425 unit
For T = 8 weeks
Q1 = 975 units
11
FPD-Example Solution for T=2
t
1
2
3
4
Beginning Inventory
0
75
0
200
Demand
100
75
175
200
Qt
175
5
6
7
0
100
0
150
100
75
250
8
100
100
375
175
End Inventory
75
0
200
0
100
0
100
0
Total cost = 4 ($50) + 475 ($0.5) = $437.5
12
FPD-Example Solution for T=4
t
1
Beginning Inventory
0
Demand
100
2
3
4
5
450
375
200
0
75
175
200
150
6
7
8
275
175
100
100
75
100
Qt
550
End Inventory
450
425
375
200
0
275
175
100
0
Total cost = 2 ($50) + 1575 ($0.5) = $887.5
13
FPD-Example Solution for T=8
t
1
2
3
4
Beginning Inventory
0
875
800
625
Demand
100
75
175
200
Qt
975
End Inventory
875
800
625
425
5
6
7
425
275
175
150
100
75
275
175
100
8
100
100
0
Total cost = 1 ($50) + 3275 ($0.5) = $1687.5
14
Lot For Lot Rule – L4L
The order quantity is always the demand
for one period
For weekly demand given above evaluate
L4L rule
15
L4L-Example Solution
Week
1
2
Demand
100
75
3
4
5
175 200 150
Lot size per order:
6
7
8
100
75
100
Q1 = 100 units, Q2 = 75 units, Q3 = 175 units, Q4 = 200 units
Q5 = 150 units, Q6 = 100 units, Q7 = 75 units, Q8 = 100 units
16
L4L-Example Solution Continued
t
1
Beginning Inventory
0
Demand
100
Qt
100
End Inventory
0
2
3
4
5
0
0
0
0
75
175
200
150
75
175
200
150
0
0
0
0
6
7
8
0
0
0
100
75
100
100
75
100
0
0
0
Total cost = 8 ($50) + 0 ($0.5) = $400
17
Silver-Meal Method
Heuristic approach to aim at a low-cost
solution that is not necessarily optimal
Aim to achieve the minimum average cost
per period for the m-period span.
The average cost per period includes
ordering and inventory holding costs
18
Silver-Meal Method
The average cost per period is as follows:
K(m) =
1
( A + HD2 + ... +(m- 1 )HDm )
m
Where;
m = number of demand periods to be ordered
in the present time.
A = fixed ordering cost per order
H = inventory holding cost per unit per period
K(m) = average cost per period during m periods
19
Silver-Meal Method
Compute K(m) for m = 1,2,…,m
Stop when, K(m+1) > K(m) , i.e. the period in
which the average cost per period start to
increase.
Order the quantity equals to m periods demand.
Qi = D1 + D2 + … + Dm
Qi is the quantity ordered in period i, and it
covers m periods into the future.
The process repeats at period (m+i) and
continues through the planning horizon.
20
SM-Example
Determine the order quantities for the following lumpy
demands using Silver Meal algorithm
Week
Demand
1
100
2
75
3
175
4
200
A = $50 per order H = $0.5 per unit per week
21
For Q1:
Week
Demand
1
2
3
4
100
75
175
200
m=1, K(1) = 50
m=2, K(2) = 1/2 (50 + 0.5(75)) = 43.75 < K(1)
m=3, K(3) = 1/3 (50 + 0.5(75) + (2)(0.5)(175)) = 87.6 > K(2)
STOP
m=2 is selected for Q1
Q1 = D1 + D2 Q1 = 100 + 75 = 175 units
Next order should arrive in week 3,
So continue for Q3
22
For Q3:
Week
1
2
3
4
Demand
100
75
175
200
m=1, K(1) = 50
m=2, K(2) = 1/2 (50 + 0.5(200)) = 75 > K(1)
STOP
m=1 is selected for Q3
Q3 = D3 Q3 = 175 unit
Next order should arrive in week 4,
So continue for Q4
23
For Q4:
Week
1
2
3
4
Demand
100
75
175
200
Q4 = D4 Q4 = 200 units
24
SM-Example Solution Continued
t
1
2
3
Beginning Inventory
0
75
0
Demand
100
75
175
Qt
175
175
End Inventory
75
0
0
4
0
200
200
0
Total cost = 3 ($50) + 75 ($0.5) = $187.5
25
Least Unit Cost
Similar to SM algorithm except for the total
cost calculation
K ( m) =
A + hD2 + 2hD3 + ... + ( m − 1) hDm
D1 + D2 + D3 + ... + Dm
The stopping rule: K(m+1) > K(m)
Order for m periods into the future : Qi = D1 + D2 + … + Dm
Continue from period (m+i) : Q(m+i)
26
LUC-Example
Determine the order quantities for the following lumpy
demands using Least Unit Cost method
Week
Demand
1
100
2
75
3
175
4
200
A = $50 per order H = $0.5 per unit per week
27
Part Period Balance
Part Period (PP) is defined as the number of the
inventory carrying periods. PP balancing is the
quantity ordered which balance the A and H.
PP(m) = D2 + 2 D3 + ... + (m − 1) Dm
PPF =
A
H
Economic part period factor
The stopping rule: PP(m+1) > PPF
Order for m periods into the future : Qi = D1 + D2 + … + Dm
Continue from period (m+1) : Q(m+1)
28
PPB-Example
Determine the order quantities for the following lumpy
demands using Part Period Balancing method
Week
Demand
1
100
2
75
3
175
4
200
A = $50 per order H = $0.5 per unit per week
29
Wagner-Whitin Algorithm
WW is an optimization procedure based on
dynamic programming to find optimum order
quantity policy Qi with a minimum cost solution.
WW evaluates all possible ways of ordering to
cover demand in each period of the planning
horizon.
Wagner-Whitin replaces EOQ for the case of
lumpy demand.
30
Wagner-Whitin Algorithm
Cost of placing order:
 m

K(t, m) = A + H ∑ (j - t)D j 
 j = t +1

Where;
K(t,m) = total cost of quantity ordered at period t for m periods
A = ordering cost,
H = inventory holding cost per unit per period
Dj = demand at period j
t = 1,2,..,N and m = t,t+1,t+2,…,N
31
Wagner-Whitin Algorithm
For each period minimum cost is defined as:
K*(m) = min t = 1,2,…,m {K*(t-1) + K(t,m)}
K*(0) = 0
K*(N) is defined as the least cost solution.
32
WW-Example Solution
Week
1
2
3
4
Demand
100
75
175
200
A = $50 per order H = $0.5 per unit per week
33
WW-Example Solution
Week
1
2
3
4
Demand
100
75
175
200
K*(0) = 0
For m=1
K*(1) = K*(0) + K(1,1) = 0 + A = 50
For m=2
K*(2) = min
K*(2) = 87.5
K*(0) + K(1,2) = 0 + (A+HD2) = 50+0.5(75) = 87.5
K*(1) + K(2,2) = 50 + A = 50 + 50 = 100
34
WW-Example Solution
Week
1
2
3
4
Demand
100
75
175
200
For m=3
K*(0) + K(1,3)= 0+[50+0.5(75)+2(0.5)(175)]=262.5
K*(3) = min
K*(1) + K(2,3) = 50 +[50+0.5(175)]= 187.5
K*(2) + K(3,3) = 87.5 + 50 = 137.5
K*(3) = 137.5
K*(0) + K(1,m)
K*(1) + K(2,m)
should not be
considered for m>3
35
WW-Example Solution
Week
1
2
3
4
Demand
100
75
175
200
For m=4
K*(2) + K(3,4) = 87.5 +[50+0.5(200)]= 237.5
K*(4) = min
K*(3) + K(4,4) = 137.5 + 50 = 187.5
K*(4) = 187.5
K*(2) + K(3,m)
should not be
considered for m>4
36
WW-Example Solution
Week
1
2
3
4
Demand
100
75
175
200
For m=4
K*(2) + K(3,4) = 87.5 +[50+0.5(200)]= 237.5
K*(4) = min
K*(3) + K(4,4) = 137.5 + 50 = 187.5
K*(4) = 187.5
Q4 = D4 = 200
Continue with K*(3)
37
WW-Example Solution
Week
1
2
3
4
Demand
100
75
175
200
For m=3
K*(0) + K(1,3)= 0+[50+0.5(75)+2(0.5)(175)]=262.5
K*(3) = min
K*(1) + K(2,3) = 50 +[50+0.5(175)]= 187.5
K*(2) + K(3,3) = 87.5 + 50 = 137.5
K*(3) = 137.5
Q3 = D3 = 175
Continue with K*(2)
38
WW-Example Solution
Week
1
2
3
4
Demand
100
75
175
200
For m=2
K*(2) = min
K*(0) + K(1,2) = 0 + (A+HD2) = 50+0.5(75) = 87.5
K*(1) + K(2,2) = 50 + A = 50 + 50 = 100
K*(2) = 87.5
Q1 = D1 + D2 = 175
39
WW-Example Solution
t
1
2
3
4
Beginning Inventory
0
75
0
0
Demand
100
75
175
200
Qt
175
175
200
End Inventory
75
0
0
0
Total cost = 3 ($50) + 75 ($0.5) = $187.5
40