## IE652 - Chapter 6. Dynamic Lot Sizing Methods. Dynamic Lot Sizing Techniques. Simple Rules Period order quantity Fixed period demand Lot for lot (L4L)

IE652 - Chapter 6 Dynamic Lot Sizing Techniques 1 Dynamic Lot Sizing Methods  Simple Rules     Heuristic Methods     Period order quan...
Author: Kerry Charles
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. 

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)

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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

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

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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

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

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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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