Supply Chains: Planning with Dynamic Demand

Transcription

1 Department of Industrial Engineering Supply Chains: Planning with Dynamic Demand Jayant Rajgopal, Ph.D., P.E. Department of Industrial Engineering University of Pittsburgh Pittsburgh, PA 15261

2 PRODUCTION / INVENTORY CONTROL OF ITEMS WITH INDEPENDENT, DYNAMIC DEMAND If the demand rate is not reasonably steady over time then the EOQ model would not be appropriate. Suppose we have demand forecasts for the next n time periods as D 1, D 2,, D n and these are very different Example: (n=6) Given A=$50, h=$0.30/unit/unit time t D

3 DYNAMIC DEMAND MODELS Simple (simplistic?) approaches: 1. Lot-for-lot: no holding cost; total cost 6*50=$ Approximate EOQ based on average demand: ( )/6=40, so EOQ (2*40*50/0.3) 0.5 =115. Produce in batches of around 115 at a time maybe ( ) in period 1 and ( ) in period Production order quantity (POQ): a common approach in practice where we convert the approximate EOQ into a time supply and produces for this many periods each time: EOQ/(average demand) = 115/40 3. So produce for three periods at a time. None of these work too well in practice Q: When is dynamic demand sufficiently dynamic to preclude use of the EOQ model? A: When the coefficient of variation (std. dev. mean) is more than 20-25%. (For our example c.v. = 26/40=65%)

4 PRODUCTION / INVENTORY CONTROL WITH DYNAMIC DEMAND Important Property: Each production run covers the demand across some integer number of future periods. As a consequence, if there is any inventory at the end of a period, there can never be any production in the next period, i.e., (I t-1 )*(P t ) = 0 If I t =0, then t is called a regeneration point If P t >0, then t is called a replenishment point

5 PRODUCTION / INVENTORY CONTROL WITH DYNAMIC DEMAND EXAMPLE: Suppose we have 5 periods in our planning horizon. Some possible production patterns are P 1 P 2 P 3 P 4 P 5 D 1 +D 2 +D D 4 +D 5 0 D 1 D 2 +D 3 0 D 4 D 5 D 1 +D 2 0 D 3 +D 4 0 D 5 D 1 D 2 D 3 D 4 D 5 D 1 +D 2 +D 3 +D 4 +D Corresponding end-of-period inventory levels are: I 1 I 2 I 3 I 4 I 5 D 2 +D 3 D 3 0 D D D 2 0 D D 2 +D 3 +D 4 +D 5 D 3 +D 4 +D 5 D 4 +D 5 D 5 0

6 Consider a production run. Let s Some replenishment period (in which the production run took place; thus s-1 was a regeneration point) t Period until which demand is covered by the run, i.e. t is the next regeneration point HC(s,t) = Holding Cost incurred during the run due to the production (in period s) of an amount equal to D s +D s D t = h s (D s+1 +D s D t ) + h s+1 (D s+2 +D s D t ) + h s+3 (D s+3 +D s D t ) h t-1 (D t ) + h t (0) Then the total variable cost for the production run is equal to A s +HC(s,t). Note that A j is the setup cost in period j and h j is the holding cost in period j - if these are the same in all periods we can replace all A j with A and h j with h.

7 Heuristics: The general approach is: 1. Start with s=j=1 2. Increase j successively to s+1, s+2, etc., i.e., evaluate production runs of D s, (D s +D s+1 ), (D s +D s+1 +D s+2 ), etc., until we stop for some value of j; say j=t. Thus the production run covers demand for periods s through t. Different heuristics use different criteria to decide when to stop 3. Reset s=j=t+1 and repeat Step 2. Continue the process until the end of the planning horizon is reached We will study three different heuristics that work well in practice

8 Part-Period Balancing Heuristic Department of Industrial Engineering End the run at j for which HC(s,j) is as close as possible to A s. Since HC(s,j) is monotone increasing in j, we find the first j for which HC(s,j) exceeds A s. Then t is equal to either j or j-1, depending on whether HC(s,j) or HC(s,j-1) is closer to A s. Note that this method tries to choose a run for which setup and holding costs are balanced as far as possible.

9 Silver-Meal Heuristic End the run at j for which the total variable cost per unit time = TVCUT(s,j) first reaches a minimum, i.e., t is the first value of j for which TVCUT(s,j+1)>TVCUT(s,j). Here we define TVCUT(s,j) = [A s + HC(s,j)] [j-s+1] TVC for run No of periods covered by run

10 Least Unit Cost Heuristic End the run at j for which the total variable cost per unit of demand = TVCUD(s,j) first reaches a minimum, i.e., t is the first value of j for which TVCUD(s,j+1)>TVCUD(s,j). Here we define TVCUD(s,j) = [A s + HC(s,j)] [D s +D s D j ] TVC for run Demand covered by run

11 DYNAMIC LOT SIZING - AN EXAMPLE Given: A=$50, h=i*c= 0.2*18= 3.6 $/unit/year = 3.6/12= $0.30/unit/month. t D Part Period Balancing Heuristic s=1 j=1, P s =10; HC(1,1) = 0.3*0 = 0 < 50 j=2, P s =50; HC(1,2) = 0.3*(40+0) = 12 < 50 j=3, P s =130; HC(1,3) = 0.3*( ) = 60 > 50! Since 60 is closer (than 12 is) to 50, we have t=3 P1=130, P2=P3=0 Reset: s=4 j=4, P s =60; HC(4,4) = 0.3*0 = 0 < 50 j=5, P s =90; HC(4,5) = 0.3*(30+0) = 9 < 50 j=6, P s =110; HC(4,6) = 0.3*( ) = 21 < 50 Would like to go on, but we've reached the end of the planning horizon STOP, with t=6 P 4 =110, P 5 =P 6 =0 Variable Cost = ( ) = $181 (setups) (inventory)------

12 DYNAMIC LOT SIZING EXAMPLE (cont d) Silver-Meal Heuristic s=1 j=1, P s =10; TVCUT(1,1) = [50+0.3*0] 1 = 50 j=2, P s =50; TVCUT(1,2) j=3, P s =130; TVCUT(1,3) Thus t=2 and P1=60, P2=0 Reset s=3 j=3, P s =80; TVCUT(3,3) = [50+0.3*0] 1 = 50 j=4, P s =140; TVCUT(3,4) j=5, P s =170; TVCUT(3,5) j=6, P s =190; TVCUT(3,6) = [50+0.3*(40+0)] 2 = 31 (<50 - continue) = [50+0.3*( )] 3 = (> 31 - reset!) = [50+0.3*(60+0)] 2 = 34 (<50 - continue) = [50+0.3*( )] 3 = (<34 - continue) = 0+0.3*( )] 4 = 26 (< continue) Would go on, but reached end of planning horizon STOP with t=6 and P 3 =190, P 4 =P 5 =P 6 =0. Variable Cost = 2* *( ) = $166 (setups) (inventory)------

13 DYNAMIC LOT SIZING EXAMPLE (cont d) The Least Unit Cost Heuristic s=1 j=1, P s =10; TVCUD(1,1) = [50+0.3*0] 10 = 5 j=2, P s =50; TVCUD(1,2) = [50+0.3*(40+0)] 50 = 1.24 (<5 - continue) j=3, P s =130; TVCUD(1,3) = [50+0.3*( )] 130 = 0.85 (< continue) j=4,p s =190; TVCUD(1,4) = [50+0.3*( )] 190 = (> reset) Thus t=3 and P 1 =130, P 2 =P 3 =0 Reset s=4 j=4, P s =60; TVCUD(4,4) = [50+0.3*0] 60 = j=5, P s =90; TVCUD(4,5) = [50+0.3*(30+0)] 90 = (< continue) j=6, P s =110; TVCUD(4,6) = [50+0.3*( )] 110 = (< continue) Would go on, but reached end of planning horizon STOP with t=6 and P 4 =110, P 5 =P 6 =0. Variable Cost = 2* *( ) = $181 (setups) (inventory)------

14 An Optimal Approach using Dynamic Programming: The Wagner-Whitin Algorithm P 1 P 2 P n Define I 0 I 1 1 I 2 2 I n-1 I n n D 1 D 2 D n F(t) = minimum cost of satisfying demands in periods 1 through t. (So of course, we seek F(n) ) s t* = the best period in which to start the production run that satisfies demand for period t. The general recursive equation is { + (, } F ( t) = Mins t F ( s 1) + A HC s t) * F (1) = A, s1 = 1 F (0) = 0

15 Wagner-Whitin Algorithm t=1 F(1) = 50, and =1 t=2 F(2) = Min {[F(0)+ A+ HC(1,2)], [F(1)+ A+HC(2,2)} = Min {[ )], [ ]} = Min {62, 100}=62; s 2* =1 t=3 F(3) = Min { [F(0)+ A+ HC(1,3)], [F(1)+ A+HC(2,3)], [F(2)+ A+HC(3,3)] } = Min { [ )], [ ], [ ] } = Min {110, 124, 112} = 110 s 3* =1

16 t=4 t=5 F(4) = Min { [F(0)+ A+HC(1,4)], [F(1)+ A+HC(2,4)], [F(2)+ A+HC(3,4)], [F(3)+ A+HC(4,4)] } = Min { [ ], [ ], [ ], [ ] } = Min {164, 160, 130, 160} = 130 s 4* =3 F(5) = Min { [F(0)+ A+HC(1,5)], [F(1)+ A+HC(2,5))], [F(2)+ A+HC(3,5))], [F(3)+ A+HC(4,5)], [F(4)+ A+HC(5,5) } = Min { [ * * *90+0.3*30)], [ * *90+0.3*30], [ *90+0.3*30], [ *30], [ ]} = Min {200, 187, 148, 169, 180} = 148 s 5* =3

17 t=6 F(6) = Min { [F(0)+ A+HC(1,6)], [F(1)+ A+HC(2,6)], [F(2)+ A+HC(3,6)], [F(3)+ A+HC(4,6)] [F(4)+ A+HC(5,6)], [F(5)+ A+HC(6,6)] } =Min { [ * * * *50+0.3*20)], [ * * *50+0.3*20], [ * *50+0.3*20], [ *50+0.3*20], [ *20], [ ]} = Min {230, 211, 166, 181, 186, 198} = 166 s 6* =3 So the optimal plan has production in periods s 6* (=3) and s 2* (=1) so that P 3 =D 3 +D 4 +D 5 +D 6 =190 and P 1 =D 1 +D 2 =50.

18 P 3 =D 3 +D 4 +D 5 +D 6 =190 and P 1 =D 1 +D 2 =50. The DP algorithm implicitly evaluates all the paths shown in the graph below, where each node represents a possible regeneration point

19 INCORPORATING LEAD TIMES Consider an independent demand item with dynamic demand for which we have planned orders using one of the methods we just saw. In this context, A would be the order processing cost for each order. Suppose there is a lead time L for these orders to be delivered from the source say a factory (to a distribution center), or perhaps a distribution center (to a store). We need to ensure that orders to the source are timed correctly for the correct amounts in order to ensure timely delivery

20 An example: DISTRIBUTION REQUIREMENTS PLANNING (DRP) EXAMPLE: Consider a warehouse that supplies several stores every week with some item. Demand for the item is highly dynamic because of differences in what each store orders ever week and because not all stores order every week. The total demand at the warehouse over the next 12 weeks was considered and orders were planned using the Silver-Meal heuristic. Suppose that the lead time to obtain a shipment of items from the supplier in China is 2 weeks

21 DRP Example ITEM RX-2100 Lead Time 2 weeks Safety Stock 20 Current Inventory 25 Order lot sizes based on Silver-Meal heuristic WEEK Requirements Scheduled Receipts Projected Ending 25 Balance Planned Receipts Planned Order Release