1. Dynamic Lot Size Models
A Production Engineering Special
presented by the ENM Faculty
Fine print: This is section 7.2 and Appendix 7-A in the text. You will want to read it!

2. Why would demands fluctuate?
Material Requirements Planning (MRP) methodology (planned order releases)
optimal production lot sizes
produce to meet contract orders by specified dates
seasonal demands
replacement parts
trends in demands
shifts in market place (competitor, advertising, discounts, etc.
change in sales force

3. The Problem A known set of time-varying demands
Setup or order costs independent of order or lot size (Q)
Holding costs proportional to the number of time periods item is maintained in inventory
What order quantities or production lot sizes will minimize order + holding cost over the planning horizon?

4. Methods for dealing with “lumpy” demands
(D1, D2, …,Dn)
A set of lumpy demands
Production Smoothing
Use EOQ (assumes constant demand)
Simple rules
fixed period demand
period order quantity
lot for lot reordering
Heuristic rules
Silver-Meal method
least unit cost
part period balancing (PPB)
Wagner-Whitin algorithm
Transportation Problem

5. Some Assumptions Demands, Dj, are known for periods j =1, …n
Demand Dj must be satisfied in period j and available at the start of the period
Replenishments arrive at the beginning of a period
No quantity discounts
Unit costs do not change over planning horizon
no shortages permitted
lead-times are known and constant
entire order quantity arrives at the same time
items are independent of one another
carrying costs applies only to inventory carried over from one time period to another

6. Using EOQ Setup cost is $132 Holding costs are $.60 per item per week
Wk1
Wk2
Wk3
Wk4
Wk5
Wk6
Wk7
Wk8
Wk9
Wk10 42 32 12 26
112 45 14 76 38
Setup cost is $132
Holding costs are $.60 per item per week

7. Using EOQ Total Setup cost is $132 x 4 = $528
Wk1
Wk2
Wk3
Wk4
Wk5
Wk6
Wk7
Wk8
Wk9
Wk10 42 32 12 26
112 45 14 76 38 Q
139
End Inv 97 55 23 11
124
106 92 16
117
Sum =
653
Total Setup cost is $132 x 4 = $528
Total Holding costs are $.60 x ( ) = $321.60
Total cost = $849.60

8. Fixed Period Demand (D1, D2, …,Dn)
Rule: Order m months worth of demands.
Example: m = 3
1st order quantity: D1 + D2 + D3
2nd order quantity: D4 + D5 + D6

9. Fixed Period Demand Q End Inv
Wk1
Wk2
Wk3
Wk4
Wk5
Wk6
Wk7
Wk8
Wk9
Wk10 42 32 12 26
112 45 14 76 38 84 44
138 59
114 Q
End Inv
M = 2; cost = 5 x x .60 = $790.80
Wk1
Wk2
Wk3
Wk4
Wk5
Wk6
Wk7
Wk8
Wk9
Wk10 42 32 12 26
112 45 14 76 38
154
285 70
173
128
114 Q
End Inv
M = 5; cost = 2 x x .60 = $683.40

10. Period Order Quantity (POQ)
(D1, D2, …,Dn)
1. Establish an average lot size - L.
2. Determine the average demand - Davg
3. Set m = L / Davg
4. Order in period j, Dj+1 , Dj+2 , …,Dj+m
Example: week
demands
Davg = 120 / 6 = 20 If L = 40 then m = 40 / 20 = 2
order: D1 + D2 , D3+ D4 , D5 + D6

11. Period Order Quantity (POQ)
(D1, D2, …,Dn)
1 L = 150 (perhaps breakpoint for quantity discounting)
2. Davg = 43.9  44
3. Set m = L / Davg = 150 / 44 = 3.4  3
Wk1
Wk2
Wk3
Wk4
Wk5
Wk6
Wk7
Wk8
Wk9
Wk10 42 32 12 26
112 45 14 76 38
116
150
135 74
138 90 Q
End Inv
cost =4 x x .60 = $804.00

12. Lot for Lot (L4L) (D1, D2, …,Dn) Set order quantity equal to Dj
That is, order for each period , the expected demands.
Wk1
Wk2
Wk3
Wk4
Wk5
Wk6
Wk7
Wk8
Wk9
Wk10 42 32 12 26
112 45 14 76 38 Q
End Inv
cost = 10 x 132 = $1,320
Note that holding costs are zero!

13. Silver-Meal Heuristic
Try to minimize average cost per period
costs includes ordering (set-up) (K) and holding cost (h).
Assume K and h are constant over the planning horizon.
assume holding costs occur at the end of the period
assume quantity needed in a period is used at the beginning of the period

14. Silver-Meal Heuristic
(D1, D2, …,Dn)
Continue until
cost / period
starts increasing
Order quantity is then D1 + D2 + … + Dm m

15. Silver-Meal Heuristic
Example: K = $90 ; h = 1.20 per unit per month
Month
Demands
Order Period Ordering cost Holding Cost Avg cost / period
one month
two months (30) = / 2 = 63
three months (30)
+2.4(23)= / 3 = 60.4
Order quantity = = 73
M=3
Four months (19)
= / 4 = 62.4

16. Repeat for next order Example: K = $90 ; h = 1.20 per unit per month
Demands
Order Period Ordering cost Holding Cost Avg cost / period
one month
two months (32) = / 2 = 64.2
three months (32)
+2.4(28)= / 3 = 65.2
Order quantity = = 51
M=2

17. Least Unit Cost Heuristic
Compute average cost per unit demanded rather
average cost per period.

18. Example: K = $90 ; h = 1.20 per unit per month
Demands
Order Period Ordering cost Holding Cost Avg cost / unit
one month /20 = 4.50
two months (30) = / 50 = 2.52
three months (30)
+2.4(23)= / 73 = 2.48
Order quantity = = 73
M=3
Four months (19)
= / 92 = 2.71

19. Part Period Balancing (PPB)
Attempts to minimize the sum of the variable cost for all lots.
Definition:
Part period = one unit held in inventory for one period
PPm = part period for m periods
PP1 = 0
PP2 = D2
PP3 = D2 + 2D3
PPm = D2 + 2D3 + … + (m-1) Dm

20. Continued Part Period Balancing (PPB)
Inventory holding cost = h (PPm)
Find m so that K h(PPm) or PPm K / h
Order quantity = Q = D1 + D2 + … + Dm

21. Example problem continued
Example: K = $90 ; h = 1.20 per unit per month
Month
Demands
K / h = 90 / 1.2 = 75
PP1 = 0
PP2 = 30
PP3 = (30) +2(23)= 76 stop!
Q1 = = 73
Starting month 4:
PP1 = 0
PP2 = (32)
PP3 = (28) = 88 stop
Q2 = = 79

22. An Old Favorite PPm  K/h = 132 / .6 = 220 pp1 = 0 pp2 = 45 pp1 = 0
Wk1
Wk2
Wk3
Wk4
Wk5
Wk6
Wk7
Wk8
Wk9
Wk10 42 32 12 26
112 45 14 76 38
PPm  K/h = 132 / .6 = 220
pp1 = 0
pp2 = 42
Pp3 = 42 + (2) 32 = 106
pp4 = pp3 + 3(12) = 142
pp5 = pp4 + 4(26) = 246
Q1 = = 154
pp1 = 0
pp2 = 45
pp3 = 45 + (2) 14 = 73
pp4 = (76) = 301
Q6 = = 247
Q10 = 38
Wk1
Wk2
Wk3
Wk4
Wk5
Wk6
Wk7
Wk8
Wk9
Wk10 42 32 12 26
112 45 14 76 38
154
247 70
135 90
cost =3 x x .60 = $724.20

23. The Wagner-Whitin Algorithm
Our Feature Presentation
The Wagner-Whitin Algorithm
The Whole Thing!
The Big Enchilada

24. The General Problem Qt = production or order quantity in period t
It = inventory at the end of period t
Ct(Qt) = cost of production in period t
ht(It) = holding cost from period t to t+1

25. The Linear Problem Qt = production or order quantity in period t
It = inventory at the end of period t
Kt = fixed cost of production in period t
Ct = cost of production in period t
ht = holding cost per unit carried from period t to t+1

26. A Simpler Problem?

27. It-1 Qt = 0 Property 1: A replenishment only takes place when
the inventory level is zero. Therefore
Qk = 0, or Dk or Dk + Dk+1 or … or Dk + Dk+1 + … + Dn
It-1 Qt = 0
Property 2: There is an upper limit to how far before a
period j we would include its requirements, Dj in a
replenishment quantity. That is, the carrying costs
become so high that it is less expensive to have a second
replenishment occur.

28. A Wagner-Whitin Example
The Maka Parte Company makes parts for General Motors
automobiles. One part they make is a vulcanized tri-solenoid
distributor. A primary component used in the manufacture of
this distributor is a silicon computer chip. This chip is purchased
from a vendor - Outspeak Corp. Ordering costs are $70 and holding
costs are $ .5 per item per month. Demands for the next six months
based upon an exponential smoothing model with seasonal effects are:
Nov Dec Jan Feb Mar Apr

29. A Wagner-Whitin Example
n = 6 (Apr)
Q6 = 110; f = $70
Order Cost = $70
Holding cost = .5
n = 5 (Mar/Apr)
Q5 = 86, 196
f5(86) = =140
f5(196) = (110) = 125
Nov Dec Jan Feb Mar Apr
n = 4 (Feb/Mar/Apr)
Q4 = 78, 164, 274
f4(78) = = 195
f4(164) = (86) + 70 = 183
f4(274) = (86) (110) = 223

30. A Wagner-Whitin Example
Order Cost = $70
Holding cost = .5
Nov Dec Jan Feb Mar Apr
n = 3 (Jan/Feb/Mar/Apr)
Q3 = 94, 172, 258, 368
f3(94) = = 253
f3(172) = (78) = 234
f3(258) = (78) (86) + 70 = 265
f3(368) = (78) (86) + 1.5(110) = 360

31. A Wagner-Whitin Example
Order Cost = $70
Holding cost = .5
Nov Dec Jan Feb Mar Apr
n = 2 (Dec/Jan/Feb/Mar/Apr)
Q2 = 80, 174, 252, 338, 448
f2(80) = = 304
f2(174) = (94) = 300
f2(252) = (94) (78) = 320
f2(338) = (94) (78) + 1.5(86) + 70 = 394
f2(448) = (94) (78) + 1.5(86) + 2(110) = 544

32. A Wagner-Whitin Example
Order Cost = $70 Holding cost = .5
n = 1 (Nov/Dec/Jan/Feb/Mar/Apr)
Q1 = 120, 200, 294, 372, 458, 568
f1(120) = = 370
f1(200) = (80) = 344
f1(294) = (80) (94) = 387
f1(372) = (80) (94) + 1.5(78) = 446
f1(458) = (80) (94) + 1.5(78) + 2(86)+70= 563
f1(568) = (80) (94)
+ 1.5(78) + 2(86) + 2.5(110) = 768
Nov Dec Jan Feb Mar Apr
Q1 = 200; Q3 = 172; Q5 = 196 ; Cost = $344

33. Capacity Constraints

34. Why isn’t Wagner-Whitin used more frequently?
relatively complex
needs a well-defined ending point
all information out to end point needed even to compute initial quantity
within MRP systems using rolling schedules, the solution will keep changing
the assumption that replenishments can be made only at discrete intervals
computational requirements

35. No Fixed Cost

36. The Transportation Problem
xtj = number of units produced in month t
to satisfy month j demands

37. The Example
h = $2 per unit per month

38. What can we conclude from all of this?
Most heuristics outperform EOQ
the Silver-Meal heuristic incurs an average cost penalty relative to Wagner-Whitin of less than 1 percent.
Significant costs penalties using Silver-Meal will incur if
demand pattern drops rapidly over several periods
when there are a large number of periods having no demand

39. Can we have some really neat homework problems? Huh?
Text: Chapter 7: problems 13, 14, 17, 18, 19, 22

40. Safety Stock
When demand or lead-time is random (or both), then safety stock
may be established as a “hedge” against uncertain demands.
For the deterministic case: R = D L
For the stochastic case: R = LTDavg + s where
LTD = a random variable, the lead-time demand,
LTDavg = average lead-time demand and s is the safety stock.

41. Safety Stock based on Fill Rate
Shortage probability
Pr{LTD > R) = p
LTD R
LTDavg s
Fill rate criterion: set s = z STD
where STD = standard deviation of the lead-time demand distribution
then
R = LTDavg + z STD

42. But I need to know when
demands are lumpy, don’t I?
Compute the variability coefficient,
v = variance of demand per period
square of average demand per period
If V < .25 , use EOQ with Davg
else use a DLS method