Lot sizing in MRP systems • 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:

Subassembly D requirements (1 week lead time) 1 2 3 4 5 Week 120 480 440 Gross requirements 110 110 642 162 374 Projected inventory balance 10 480 440 Net requirements 652 652 Planned receipts (EOQ) 652 652 652 Planned order release (EOQ)

6 800 226 800 652 652

7 400 478 400 652

8 478

Alternative lot sizing techniques • 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 Ik-1*Qk = 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 fn+1 = 0. Iteratively we compute: fk = minj>k(ckj + fj) for k = 1, … n. “ckj” is defined to be the cost of producing in period k sufficient to fulfill the accumulated demand in periods k through j-1. – When f1 is determined the algorithm is terminated. f1 yields the optimal value of the dynamic lot-sizing problem. The corresponding solution of Q1, Q2, … Qn gives us the optimal production schedule.

Wagner-Whitin example • 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 Predicted demand, D Setup cost, A Holding cost, h c_k,j values k 1 2 3 4 5

j

W-W example 1 2 100 100 50 50 0.5 0.5

1

3 50 50 0.5

4 50 50 0.5

5 210 50 0.5

2

3

4

5

6

50

100 50

150 75 50

225 125 75 50

645 440 285 155 50

Wagner-Whitin solution • • • • • • • •

f6 = 0. f5 = min ( c56 + f6 ) = 50 + 0 = 50. f4 = min ( c45 + f5 , c46 + f6 ) = min ( 50 + 50 , 155 + 0 ) = 100. f3 = min ( c34 + f4 , c35 + f5 , c36 + f6 ) = min ( 50 + 100 , 75 + 50 , 285 + 0) = 125. f2 = min ( c23 + f3 , c24 + f4 , c25 + f5 , c26 + f6 ) = min ( 50 + 125 , 75 + 100 , 125 + 50 , 440 + 0 ) = 175. f1 = min (c12 + f2 , c13 + f3 , c14 + f4 , c15 + f5 , c16 + f6 ) = min ( 50 + 175 , 100 + 125 , 150 + 100 , 225 + 50 , 645 + 0 ) = 225. 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)

Lot-sizing Heuristics • Silver-Meal heuristic (consider the same example): – K(m) = (A + hD2 + 2hD3 + 3hD4 + … + (m-1)hDm)/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! Q1 = 100+100+50 = 250. Continue from the 4th month, m=1 K(1) = 50. m=2 K(2) = (50 + (0.5) 210) / 2 = 72.5 so stop! Q4 = 50. Continue from the 5th month, m=1 K(1) = 50, but we are already at the final month, so we conclude that Q5 = 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 + hD2 + 2hD3 + 3hD4 + … + (m-1)hDm)/(D1+ ... +Dm) – 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! Q1 = 100 + 100 = 200. • Continue from the 3rd 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! Q3 = 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 – PPm = 0 + D2 + 2D3 + 3D4 + … + (m-1)Dm – We stop accumulating when PPm exceeds A/h. We should choose the m value which gives the closest PPm value to A/h. – So, A/h = 50 / 0.5 = 100 and: – PP1 = 0. – PP2 = 0 + 100 = 100, continue. – PP3 = 0 + 100 + (2)(50) = 200, stop! Q1 = 100 + 100 = 200. – Continue from the 3rd month, PP1 = 0. – PP2 = 0 + 50 = 50, continue. – PP3 = 0 + 50 + (2)(210) = 470, stop! Q3 = 50 + 50 = 100. – Since month 5 is the ending period we conclude that Q5 = 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).