• Lot for lot (also l-f-l or L4L) policy is the most basic approach for planned order releases. However it is most of the time impractical and uneconomical.
• Fixed batch size policy is more reasonable if the setup times are considerably long.
• EOQ policy is fine when demand is almost constant.
However when demand pattern is dynamic it becomes unlikely to have close-to-optimal solutions with the EOQ approach. As an example, consider the following:
– Recalling the net requirements of subassembly D in the previous example: 0 in week 1, 10 in w2, 480 in w3, 440 in w4, 800 in w5,
400 in w6, and 0 in weeks 7 and 8. Let the setup and holding costs be: A=$400, h=$0.5.

– Average demand for the planning horizon is:
• D = (0+10+480+440+800+400+0+0)/8 ~= 266 items per week.
– So EOQ value becomes:
• EOQ = sqrt(2*A*D/h) ~= 652 and the new schedule of order releases will become:
Week
Subassembly D requirements (1 week lead time)
1
Gross requirements
Projected inventory balance 110
2
110
3
120
642
4
480
162
5
440
374
10 480 440 Net requirements
Planned receipts (EOQ)
Planned order release (EOQ) 652
652
652
652
652
6
800
226
800
652
652
7
400
478
400
652
478
8

• Heuristics for dynamic lot sizing:
– Silver-Meal heuristic (S-M).
– Least unit cost heuristic (LUC).
– Part period balancing (PPB).
• Optimal lot sizing: The Wagner-Whitin (W-W) algorithm:
– Dynamic programming approach.
– Assume no capacity restrictions.
– The algorithm owes its efficiency to a peculiarity which is known as the Wagner-Whitin property.
– Wagner-Whitin property says that there is an optimal solution for the dynamic lot sizing problem in which for each period k either the beginning inventory or the production quantity is zero. The property states that at least one optimal solution will satisfy this condition: (More formally I k-1
*Q k
= 0 for all k = 1,2, … n .)

– We will search through all solutions which satisfy this condition, because it restricts the solution set to enable efficient computation.
– W-W property guarantees that the optimal solution which we are going to find for the restricted problem is also optimal for the unrestricted problem. We can restate the property alternatively as:
– Wagner-Whitin property (identical restatement): An optimal lotsizing policy has the property that each value of Q is exactly the sum of a set of future demands.
• W-W algorithm:
– We start from the final period, and by definition f n+ 1
= 0.
Iteratively we compute: f k
= min j>k
( c kj
+ f j
) for k = 1 , … n. “c kj
” is defined to be the cost of producing in period k sufficient to fulfill the accumulated demand in periods k through j -1.
– When f
1 is determined the algorithm is terminated. f
1 optimal value of the dynamic lot-sizing problem. The yields the corresponding solution of Q
1
, Q
2
, … Q n production schedule.
gives us the optimal

• Fixed setup cost, A is $50 and holding cost, h is $0.5. For convenience variable unit production cost is ignored (assumed to be unchanging throughout the planning horizon.
Period, k 1
Predicted demand, D 100
Setup cost, A
Holding cost, h
50
0.5
W-W example
2
100
50
0.5
3
50
50
0.5
4
50
50
0.5
5
210
50
0.5
1 c_k,j values k
1
4
5
2
3 j 2
50
3
100
50
4
150
75
50
5
225
125
75
50
6
645
440
285
155
50

Wagner-Whitin solution
• f
6
= 0.
• f
5
= min ( c
56
+ f
6
) = 50 + 0 = 50.
• f
4
= min ( c
45
+ f
5
, c
46
+ f
6
) = min ( 50 + 50 , 155 + 0 ) = 100.
• f
3
= min (
0) = 125.
c
34
+ f
4
, c
35
+ f
5
, c
36
+ f
6
) = min ( 50 + 100 , 75 + 50 , 285 +
• f
2
= min ( c
23
+ f
3
, c
24
+ f
4
, c
25
+ f
100 , 125 + 50 , 440 + 0 ) = 175.
5
, c
26
+ f
6
) = min ( 50 + 125 , 75 +
• f
1
= min ( c
12
+ f
2
, c
13
+ f
3
, c
14
+ f
4
, c
15
+ f
5
, c
16
+ f
6
100 + 125 , 150 + 100 , 225 + 50 , 645 + 0 ) = 225.
) = min ( 50 + 175 ,
• Optimal solution has a cost of $225.
• So W-W algorithm gives alternative optimal solutions of this dynamic lot sizing problem (2 examples with cost $225 listed below):
– Q = (100, 100, 100, 0, 210)
– Q = (200, 0, 100, 0, 210)

• Silver-Meal heuristic (consider the same example):
– K ( m ) = (A + hD
2
+ 2hD
3
+ 3hD
4
+ … + ( m -1)hD m
)/m
– And we stop accumulating demand when K ( m +1) > K ( m ).
– So,
• m =1 K (1) = 50.
• m =2 K (2) = (50 + (0.5)100) / 2 = 50, continue.
• m =3 K (3) = (50 + (0.5)100 + (2)(0.5)50) / 3 = 50, continue.
• m =4 K (4) = (50 + (0.5)100 + (2)(0.5)50 + (3)(0.5)50) / 4 = 56.25 so stop!
Q
1
= 100+100+50 = 250.
• Continue from the 4 th month, m =1 K (1) = 50.
• m =2 K (2) = (50 + (0.5) 210) / 2 = 72.5 so stop! Q
4
= 50.
• Continue from the 5 th month, m =1 K (1) = 50, but we are already at the final month, so we conclude that Q
5
= 210.
• Hence S-M solution is: Q = (250, 0, 0, 50, 210) and it costs us 50 + 50
+ 50 + (0.5)100 + (2)(0.5)50 = $250. This solution is not optimal!

• Least Unit Cost heuristic
– K’ ( m ) = (A + hD
2
+ 2hD
3
+ 3hD
4
+ … + ( m -1)hD m
)/(D
1
+ ... +D m
)
– And we stop accumulating demand when K’ ( m +1) > K’ ( m ).
– So,
• m =1 K’ (1) = 50/100 = 0.5.
• m =2 K’ (2) = (50 + (0.5)100) / (100+100) = 0.5, continue.
• m =3 K’ (3) = (50 + (0.5)100 + (2)(0.5)50) / (100+100+50) = 0.6 so stop!
Q
1
= 100 + 100 = 200.
• Continue from the 3 rd month, m =1 K’ (1) = 50/50 = 1.
• m =2 K’ (2) = (50 + (0.5)50) / (50+50) = 0.75, continue.
• m =3 K’ (3) = (50 + (0.5)50 + (2)(0.5)210) / (50+50+210) = 0.92 so stop!
Q
3
= 50 + 50 + 210 = 310.
• We stop since all demand is met. So LUC solution is: Q = (200, 0,
310, 0, 0) and it costs us 50 + 50 + (0.5)100 + (0.5)50 + (2)(0.5)210
= $385. This solution is not optimal!

• Part Period Balancing
– PP m
= 0 + D
2
+ 2D
3
+ 3D
4
+ … + ( m -1)D m
– We stop accumulating when PP m exceeds A/h. We should choose the m value which gives the closest PP m value to A/h.
– So, A/h = 50 / 0.5 = 100 and:
– PP
1
= 0.
– PP
2
= 0 + 100 = 100, continue.
– PP
3
= 0 + 100 + (2)(50) = 200, stop!
Q
1
= 100 + 100 = 200.
– Continue from the 3 rd month, PP
1
= 0.
– PP
2
= 0 + 50 = 50, continue.
– PP
3
= 0 + 50 + (2)(210) = 470, stop!
Q
3
= 50 + 50 = 100.
– Since month 5 is the ending period we conclude that Q
5
= 210.
– So the PPB solution is: Q = (200, 0, 100, 0, 210) and the cost of this solution is 50 + 50 + 50 + (0.5)(100) + (0.5)(50) = $225.
This solution is optimal (compare with W-W solution).