# Touches the demand curve before an order is placed 2

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touches the demand curve before an order is placed (2) the optimal cost of ordering in period j to cover demand in period t is independent of how demand was covered before period j the inventory costs are generated from disjoint curvy-triangular areas at periods where inventory = 0 (3) in each period, the only necessary ordering choices are partial sums of future demands (that will be carried toward future periods) where (x)=1 if x>0; 0 otherwise ( ) D T t T time Cum’l items j Every lot-size decision corresponds to an headway decision, and we only need to find the optimal timing of orders!

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LSA: Chapter 3 Brief tutorial: Dynamic Programming (shortest path) Simple example: Driving from Los Angeles to Boston, going through two cities, each from one of the two clusters There is a travel cost between any two possible cities stage LA Solution method: Simplistic enumeration: Evaluate cost for O( m*n) possible paths Dynamic programming (shortest path) Label each city with a “cost-to-go” value Evaluate label cost for O( m+n ) nodes state Denver Salt Lake City Phoenix Chicago Indianapolis Nashville Atlanta Boston Cluster 1 m cities Cluster 2 n cities LSA: Chapter 3 Brief tutorial: Dynamic Programming (shortest path) A simple deterministic dynamic program System starts from initial state s 0 to arrive at final state s n after n sequential stages There is a cost c k ( i , j ) for system to transition from state i to state j at stage k (under one or more control actions a ) . There is a cost-to-go value associated with each stage- state point V k ( i ) , for all i , k . The optimality condition (Bellman’s Equation) n stage i k 1 + k j s 0 State = condition of the system s n { } 1 ( ) min ( , ) ( ) , k k k j V i c i j V j + = + i n n k = , 0 ,..., 1 , 0
Common Characteristics of Dynamic Programming Problems 1. The problem can be divided into stages with a decision required at each stage. Stage: the layer of cities; Decision: where to go next? 2. Each stage has a number of states associated with it. State: the node reached. 3. The decision at one stage transforms one state into a state in the next stage according to some dynamics rule. 4. There may or may not be random disturbances that affect the transition between stages. 5. Given the current state, the optimal decision for each of the remaining states does not depend on the previous states or decisions – memory-less. While looking forward, it was not necessary to know how you have got to a node, only that you did. 6. There are costs associated with each state at each stage 7. There exists a recursive relationship that identifies the optimal decision for stage j, given that stage j+1 has already been solved. Such decisions are often identified as policies. 8. The final stage must be solvable by itself. The big skill in dynamic programming, and the art involved, is to take a problem and determine stages and states so that all of the above hold. If you can, then the recursive relationship makes finding the values relatively easy. Stages and states, however, can be defined in multiple ways. The following are some examples.

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• Fall '16

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