IE652 - Chapter 6 Dynamic Lot Sizing Techniques
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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.
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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
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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)
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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.
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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
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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
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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
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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
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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
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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
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For Q4: Week
1
2
3
4
Demand
100
75
175
200
Q4 = D4 Q4 = 200 units
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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
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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
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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
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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.
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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.
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WW-Example Solution Week
1
2
3
4
Demand
100
75
175
200
A = $50 per order H = $0.5 per unit per week
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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
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