Deterministic Variable Demand Inventory. Chen Zhou Assoc. Prof For Prof. Reveliotis Sept 11, 2008

Transcription

1 Deterministic Variable Demand Inventory Chen Zhou Assoc. Prof For Prof. Reveliotis Sept 11, 2008

2 Demand and replenishment Demand can be Nearly deterministic Variable Replenishment is in discrete intervals Per week, per day, per month Decisions How much to order in which period Objective min cost of Ordering Carrying

3 Example Demand for a industrial pump in the following 10 weeks are given in the table The fixed ordering cost K = $132/order. The carrying cost is h = $0.6/week/unit.

4 Extreme solutions Everyweek vs. One order Costs in extreme cases Order each week: 10 * K = 1320 One order: Find inventories from the end TC = h * sum inventories = 1,406.2 Week Demand Inventory Find inventories from the beginning h*( *2 + 12* * * 9) = Can we do better?

5 EOQ λ = average(d)/week = 43.9 / week EOQ = , use 139 Week Demand EOQ release Inventory TC = 132 * * sum inventories = % savings over order every week Can we do better with some adjustment? To be fair, I 10 = 0. savings of 70.2, TC = See inventory chart >

6 Beginning inventory chart Can we do better?

7 Wagner-Whitin property In optimum policy, either I t - 1 = 0, or y t = 0, or I t - 1 * y t = 0 In the above example Week Demand WW Adj EOQ Inventory TC = 780, > 40% savings over order everyweek Can we do better?

8 Max No. of potential solutions w. WW property n period I 0 = I n = 0

9 Greedy heuristics Silver-Meal algorithm When we increase the number of periods covered by a setup The setup cost shared by each period goes down Carrying cost goes up There exists a saddle point for the cost/period We find a release and start over again Cost/period K periods

10 Procedure Let C(i) be the cost/period if the current release produces for i periods. Assume I 0 = 0 C(1) = K C(2) = (K + h r 2 ) / 2 C(3) = (K + h r h r 3 ) / 3, or (2*C(2) + 2 h r 3 ) / 3

11 Example From 1 st period C(1) = K = 132 C(2) = (K + h r 2 ) / 2 = ( * 42) / 2 = 76.2 < 132 C(3) = ( (42+2*32)) / 3 = 65.2 < 76.2 C(4) = 54.3 < 65.2 C(5) = > 54.3, stop y 1 = = 128 (save as adjusted EOQ) Start from 5 th period C(1) = K = 132 C(2) = 99.6 C(3) = 84.4 C(4) = 69.6 C(5) = y 5 = 197 Start from period 9 y 5 = 114 Total cost = 650, or 57% savings over order everyweek

12 Is Silver-Meal algorithm optimum? Can not be proved but can be disapprove by counter evidence The solution is myopic or greedy Greedy algorithm is often simple and sometimes provides very good solutions For heuristic, one can generate similar measures Min cost per part Restart when setup cost = carrying cost

13 If the assumptions holds, how to find optimum? Wagner-Whitin algorithm A form of dynamic programming The idea is to perform minimum computation and at the same time evaluate all potential solutions or without omitting any possible good solutions. Based on Wagner-Whitin property, what is the total number of possible solutions?

14 Wagner-Whitin Algorithm Given: Find Demands in periods 1, 2, n Fixed setup cost K $/setup Fixed holding cost h $/unit/period I 0 = I n = 0 (can be relaxed) The optimum policy n-period problem Similar to dynamic programming

15 Algorithm Let C(j) = minimum cost for j period problem Total cost of 1 period problem Total cost of 2-period problem

16 Algorithm Total cost of 3 period problem 4 period problem

17 Rational We have effectively evaluated all potentially good solutions Each time you eliminate a branch from min operation, you removed a lot of unnecessary calculations. We only perform partial solutions

18 Tabular implementation of the algorithm Cols: partial planning horizon, 1, 2, j. Rows: the period number in which the last order is placed i \ j 1 2 j n 1 2 j n Partial solutions Newly added Latest setup

19 Tabular implementation of the algorithm Rows: we do not have to continue in rows once minimum is reached Upon completion, trace the order releases from the back. Find the minimum cost in jth period. The corresponding row number, say j, is the period with last release. Omit the columns j, j+1, n. Repeat with n - m problem until all orders are found.

20 E.g. K = 100/setup, h = $2/unit/period

21 Single period problem C(1) K = 100/setup, h = $2/unit/period

22 2-period problem K = 100/setup, h = $2/unit/period C(2)

23 3-Period problem C(3) K = 100/setup, h = $2/unit/period

24 4-Period problem C(4) K = 100/setup, h = $2/unit/period

25 5-Period problem K = 100/setup, h = $2/unit/period No need to compute C(5)

26 Continue K = 100/setup, h = $2/unit/period C(5) Note: cost increases faster in upper rows C(6) C(7) C(8)

27 Back trace releases Find min from last col Locate its row, the last setup From period 8, min = 580, last release should be in 8 or 6 Let s follow 6, y 6 = = 75, forget last 3 periods In period 5, the minimum is 340 with last release at 4, y 4 = 70 y 1 = 60 Totol cost = 580. You can try the other one. Find release

28 Back trace releases Find min from last col Find min from last col Locate its row Think it as a 5-period problem From period 8, min = 580, last release should be in 8 or 6 Let s follow 6, y 6 = = 75, forget last 3 periods In period 5, the minimum is 340 with last release at 4, y 4 = 70 y 1 = 60 Totol cost = 580. You can try the other one.

29 Final result From period 8, min = 580, last release should be in 8 or 6 Let s follow 6, y 6 = = 75, forget last 3 periods In period 5, the minimum is 340 with last release at 4, y 4 = 70 y 1 = 60 Totol cost = 580. You can try the other one.

30 Example problem: K = 132, h = 0.6

31 Policy comparisons for the example

32 Some discussions What is the total number of possible order policies for an n period problem? In WW, we evaluate up to (n+1)n/2 partial evaluations We did not miss any possible good solutions Optimality: For n period problems satisfy assumptions: optimum In reality: rolling horizon problem

33 Summary Deterministic and variable demand Ordering vs. carrying (no shortage) WW property WW algorithm Complexity Optimality

34 Questions