HW8_solution

# HW8_solution - n-I Â E x-D n-W Â d n-K Â 1 d n 6 = d n-1 Î±...

This preview shows page 1. Sign up to view the full content.

solution of homework 8 Problem 1 Let V n ( x ) denote the minimal cost from period n to the end of the planning horizon given that the inventory level at the beginning of period n is x. Let us use the notation as the following: W: wholesale price of purchasing R: Retail price S: Salvage sale price α : Discount factor K: setup cost if order Then we have the following dynamic optimality formulation: V n ( x ) = min y x { K · 1 { y>x } + W ( y - x ) - R · E [min { x,D n } ] + α · E [ V n +1 (( x - D n ) + + ( y - x ))] } The boundary condition is: V N ( x ) = - R · E [min { x,D N } ] - S · E [( x - D N ) + ] Problem 2 Let V n ( x,d ) denote the maximum revenue from period n to the end of the planning horizon given that the initial inventory level at the beginning of period n is x and the amount of order in period n-1 is d.( n = 2 , 3 ,..., 13). Let V 1 ( x ) denote the maximum revenue from period 1 to the end of the planning horizon given that the initial inventory level at the beginning of period 1 is x. Then we have the following dynamic optimality formulation: V n ( x,d n - 1 ) = max d n 0 { R · E [min { x,D
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n } ]-I Â· E [( x-D n ) + ]-W Â· d n-K Â· 1 { d n 6 = d n-1 } + Î± Â· E [ V n +1 (( x-D n ) + + d n ,d n )] } The boundary condition is: V 13 ( x,d 12 ) = W Â· x We want to calculate V 1 ( x ), then: V 1 ( x ) = max d 1 â‰¥ { R Â· E [min { x,D 1 } ]-I Â· E [( x-D 1 ) + ]-W Â· d 1 + Î± Â· E [ V 2 (( x-D 1 ) + + d 1 ,d 1 )] } Based on the above dynamic optimality formulation, we can calculate d 1 ,d 2 ,...,d 11 and d 12 is always 0. Problem 3 Let V n ( i ) denote the minimum expected total discounted cost from period n to period N given that the state at the beginning of period n is i. Then the dynamic optimality formulation is: V n ( i ) = min { C R + Î± Â· K X j =1 P 1 j V n +1 ( j ) ,C ( i ) + Î± Â· K X j = i P ij V n +1 ( j ) } i = 1 , 2 ,...,K-1 V n ( K ) = C R + Î± Â· K X j =1 P 1 j V n +1 ( j ) The boundary condition is: V N +1 ( i ) = 0 1...
View Full Document

## This note was uploaded on 03/16/2012 for the course IEOR 466 taught by Professor Richard during the Spring '12 term at Columbia.

Ask a homework question - tutors are online