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!
1
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
2
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?
3
Methods for dealing with “lumpy” demands
A set of lumpy demands
(D1, D2, …,Dn)
• 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
4
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
5
Using EOQ
Wk1
Wk2
Wk3
Wk4
Wk5
Wk6
Wk7
Wk8
Wk9
Wk10
42
42
32
12
26
112
45
14
76
38
• Setup cost is $132
• Holding costs are $.60 per item per week
Total demand  439
D  43.9, K  132 / h  .6
(2)(132)(43.9)
Q
 139
.6
6
Using EOQ
Q
End
Inv
Wk1
Wk2
Wk3
Wk4
Wk5
Wk6
Wk7
Wk8
Wk9
Wk10
42
139
42
0
32
0
12
0
26
139
112
0
45
139
14
0
76
0
38
139
97
55
23
11
124
12
106
92
16
117
Sum =
653
• Total Setup cost is $132 x 4 = $528
• Total Holding costs are $.60 x (653-117) = $321.60
• Total cost = $849.60
7
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
8
Fixed Period Demand
Q
End Inv
Wk1
Wk2
Wk3
Wk4
Wk5
Wk6
Wk7
Wk8
Wk9
Wk10
42
42
32
12
26
112
45
14
76
38
84
44
138
59
114
42
12
112
14
38
• M = 2; cost = 5 x 132 + 218 x .60 = $790.80
Q
End Inv
Wk1
Wk2
Wk3
Wk4
Wk5
Wk6
Wk7
Wk8
Wk9
Wk10
42
42
32
12
26
112
45
14
76
38
128
114
38
0
154
112
285
70
38
26
0
173
• M = 5; cost = 2 x 132 + 699 x .60 = $683.40
9
Period Order Quantity (POQ)
(D1, D2, …,Dn)
1.
2.
3.
4.
Establish an average lot size - L.
Determine the average demand - Davg
Set m = L / Davg
Order in period j, Dj+1 , Dj+2 , …,Dj+m
Example:
demands
1
19
Davg = 120 / 6 = 20
2
14
week
3
21
4
25
5
18
6
23
If L = 40 then m = 40 / 20 = 2
order: D1 + D2 , D3+ D4 , D5 + D6
10
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
Q
End Inv
Wk1
Wk2
Wk3
Wk4
Wk5
Wk6
Wk7
Wk8
Wk9
Wk10
42
42
32
12
26
112
45
14
76
38
116
74
150
32
0
138
135
112
0
90
38
14
0
0
cost =4 x 132 + 460 x .60 = $804.00
11
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
Q
End Inv
42
42
0
42
42
0
32
32
0
12
12
0
26
26
0
112
112
0
45
45
0
14
14
0
76
76
0
38
38
0
cost = 10 x 132 = $1,320
Note that holding costs are zero!
12
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
13
Silver-Meal Heuristic (D1, D2, …,Dn) Continue until
Order quantity is then D1 + D2 + … + Dm
cost / period
starts increasing
Order period
Ordering cost
Holding cost
Cost / period
One period
K
0
K/1
Two periods
K
hD2
(K+hD2) / 2
Three periods
K
hD2 + 2hD3
(K+ hD2 +
2hD3) / 3
Four periods
K
hD2 + 2hD3 +
3hD4
(K+ hD2 +
2hD3 + 3hD4)
/4
14
m
Silver-Meal Heuristic
Example: K = $90 ; h = 1.20 per unit per month
Month
Demands
1
20
2
30
3
23
4
19
5
32
6
28
Order Period Ordering cost Holding Cost Avg cost / period
one month
90
two months
90
1.2(30) = 36
126 / 2 = 63
three months
90
1.2(30)
+2.4(23)=91.2
181.2 / 3 = 60.4
Four months
90
90
91.2 + 3.6 (19)
= 159.6
M=3
249.6 / 4 = 62.4
Order quantity = 20 + 30 + 23 = 73
15
Repeat for next order
Example: K = $90 ; h = 1.20 per unit per month
Month
Demands
1
20
2
30
3
23
4
19
5
32
6
28
Order Period Ordering cost Holding Cost Avg cost / period
one month
90
90
two months
90
1.2(32) = 38.4
three months
90
1.2(32)
+2.4(28)=105.6
128.4 / 2 = 64.2
M=2
195.6 / 3 = 65.2
Order quantity = 19 + 32 = 51
16
Least Unit Cost Heuristic
Compute average cost per unit demanded rather
average cost per period.
Order period
Ordering cost
Holding cost
Cost / unit
One period
K
0
K / D1
Two periods
K
hD2
(K+hD2) / (D1
+ D2)
Three periods
K
hD2 + 2hD3
Four periods
K
hD2 + 2hD3 +
3hD4
(K+ hD2 +
2hD3) / (D1 +
D2 + D3)
(K+ hD2 +
2hD3 + 3hD4) /
(D1+D2+D3+D)
17
Example: K = $90 ; h = 1.20 per unit per month
Month
Demands
Order Period
1
20
2
30
Ordering cost
3
23
4
19
Holding Cost
5
32
6
28
Avg cost / unit
one month
90
two months
90
1.2(30) = 36
126/ 50 = 2.52
three months
90
1.2(30)
+2.4(23)=91.2
181.2 / 73 = 2.48
Four months
90
90/20 = 4.50
91.2 + 3.6 (19)
= 159.6
M=3
249.6 / 92 = 2.71
Order quantity = 20 + 30 + 23 = 73
18
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
19
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
20
Example problem continued
Example: K = $90 ; h = 1.20 per unit per month
Month
Demands
1
20
2
30
3
23
4
19
5
32
6
28
K / h = 90 / 1.2 = 75
PP1 = 0
PP2 = 30
PP3 = (30) +2(23)= 76 stop!
Q1 = 20 + 30 + 23 = 73
Starting month 4:
PP1 = 0
PP2 = (32)
PP3 =32 + 2 (28) = 88 stop
Q2 = 19 + 32 + 28 = 79
21
An Old Favorite
Wk1 Wk2 Wk3 Wk4 Wk5 Wk6 Wk7 Wk8 Wk9 Wk10
42
42
32
12
26
112
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 = 42 + 42 + 32 + 12 + 26 = 154
45
14
76
38
pp1 = 0
pp2 = 45
pp3 = 45 + (2) 14 = 73
pp4 = 73 + 3(76) = 301
Q6 = 112 + 45 + 14 + 76 = 247
Q10 = 38
Wk1 Wk2 Wk3 Wk4 Wk5 Wk6 Wk7 Wk8 Wk9 Wk10
42
154
112
42
32
12
26
70
38
26
0
112
247
135
45
14
76
90
76
0
38
38
0
22
cost =3 x 132 + 328 x .60 = $724.20
Our Feature Presentation
The Wagner-Whitin Algorithm
The Whole Thing!
The Big Enchilada
23
The General Problem
n
Min
 C (Q )  h  I 
t 1
t
t
t
t
subject to :
I t  I t 1  Qt  Dt

 t  1, 2,..., n
Qt , I t  0,1, 2,3,...
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
24
The Linear Problem
n
Min
K
t 1
t
 ct Qt  ht I t
subject to :
I t  I t 1  Qt  Dt

 t  1, 2,..., n
Qt , I t  0,1, 2,3,...
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
25
A Simpler Problem?
n
Min
K
t 1
t
 cQt  h I t 
subject to :
I t  I t 1  Qt  Dt

 t  1, 2,..., n
Qt , I t  0,1, 2,3,...
n
Min
K
t 1
 h It 
t
n
since
 cQ
t 1
t
n
n
t 1
t 1
 c Qt  c Dt
26
n
Min
K
t 1
t
 h It
subject to :
I t  I t 1  Qt  Dt

 t  1, 2,..., n
Qt , I t  0,1, 2,3,...
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.
27
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
120
Dec
80
Jan
94
Feb
78
Mar
86
Apr
110
28
A Wagner-Whitin Example
n = 6 (Apr)
Q6 = 110;
f = $70
n = 5 (Mar/Apr)
Q5 = 86, 196
f5(86) = 70 + 70 =140
f5(196) = 70 + .5 (110) = 125
Order Cost = $70
Holding cost = .5
Nov Dec Jan Feb Mar Apr
120 80 94 78 86 110
n = 4 (Feb/Mar/Apr)
Q4 = 78, 164, 274
f4(78) = 70 + 125 = 195
f4(164) = 70 + .5 (86) + 70 = 183
f4(274) = 70 + .5(86) + 1.00 (110) = 223
29
A Wagner-Whitin Example
Order Cost = $70
Holding cost = .5
Nov Dec Jan Feb Mar Apr
120 80 94 78 86 110
n = 3 (Jan/Feb/Mar/Apr)
Q3 = 94, 172, 258, 368
f3(94) = 70 + 183 = 253
f3(172) = 70 + .5 (78) + 125 = 234
f3(258) = 70 + .5(78) + 1.00 (86) + 70 = 265
f3(368) = 70 + .5(78) + 1.00 (86) + 1.5(110) = 360
30
A Wagner-Whitin Example
Order Cost = $70
Holding cost = .5
Nov Dec Jan Feb Mar Apr
120 80 94 78 86 110
n = 2 (Dec/Jan/Feb/Mar/Apr)
Q2 = 80, 174, 252, 338, 448
f2(80) = 70 + 234 = 304
f2(174) = 70 + .5 (94) + 183 = 300
f2(252) = 70 + .5(94) + 1.00 (78) + 125 = 320
f2(338) = 70 + .5(94) + 1.00 (78) + 1.5(86) + 70 = 394
f2(448) = 70 + .5(94) + 1.00 (78) + 1.5(86) + 2(110) = 544
31
A Wagner-Whitin Example
Order Cost = $70 Holding cost = .5
Nov Dec Jan Feb Mar Apr
n = 1 (Nov/Dec/Jan/Feb/Mar/Apr)
120 80 94 78 86 110
Q1 = 120, 200, 294, 372, 458, 568
f1(120) = 70 + 300 = 370
f1(200) = 70 + .5 (80) + 234 = 344
f1(294) = 70 + .5(80) + 1.00 (94) + 183 = 387
f1(372) = 70 + .5(80) + 1.00 (94) + 1.5(78) + 125 = 446
f1(458) = 70 + .5(80) + 1.00 (94) + 1.5(78) + 2(86)+70= 563
f1(568) = 70 + .5(80) + 1.00 (94)
+ 1.5(78) + 2(86) + 2.5(110) = 768
Q1 = 200;
Q3 = 172;
Q5 = 196 ; Cost = $344
32
Capacity Constraints
n
Min
 C (Q )  h  I 
t 1
t
t
t
t
subject to :
Qt  Pt


I t  I t 1  Qt  Dt
 t  1, 2,..., n
Qt , I t  0,1, 2,3,...
33
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
34
No Fixed Cost
n
Min
c Q  h I
t 1
t
t
t
t
subject to :
Qt  Pt

 t  1, 2,..., n
I t  I t 1  Qt  Dt 
35
The Transportation Problem
n
n
Min z   ctj xtj
t 1 j t
where ctj  ct  ht 1  ht  2  ...  h j 1
subject to:
n
x
j t
tj
j
x
t 1
ij
 Pt
t  1, 2,..., n
 Dj
j  1, 2,..., n
xtj = number of units produced in month t
to satisfy month j demands
36
The Example
Month
Shift
March
R
O
R
O
R
O
R
O
R
O
R
O
April
May
June
July
August
Demands
March April
10
12
M
M
M
M
M
M
M
M
M
M
16
12
14
11
13
M
M
M
M
M
M
M
M
24
May
June
July
14
16
13
15
12
14
M
M
M
M
M
M
16
16
18
15
17
14
16
13
15
M
M
M
M
24
18
20
17
19
16
18
15
17
14
16
M
M
16
August Dum supply prod
cost
20
0
16
10
22
0
6
12
19
0
15
11
21
0
5
13
18
0
17
12
20
0
6
14
17
0
16
13
19
0
6
15
16
0
18
14
18
0
8
16
15
0
12
15
17
0
5
17
24
10
130
h = $2 per unit per month
37
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
38
Can we have some
really neat
homework
problems? Huh?
Text: Chapter 7: problems 13, 14, 17, 18, 19, 22
39
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.
40
Safety Stock based on Fill Rate
Shortage probability
Pr{LTD > R) = p
R
LTDavg
LTD
s
Fill rate criterion: set s = z STD
where STD = standard deviation of the lead-time demand distribution
then
R = LTDavg + z STD
41
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
n
V
n Dt2
t 1

1
F
I
DJ
G

H K
2
n
t
If V < .25 , use EOQ with Davg
else use a DLS method
t 1
42