# hw3_sol.pdf - IEOR-E4707 Spring 2019 Homework 3 Solutions...

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IEOR-E4707, Spring 2019: Homework 3 Solutions March 9, 2019 Problem 1. Solution: Denote by V ( x ) the shortest distance from the node x to D , d ( x, y ) the assigned distance between two adjacent nodes x and y , and n ( x ) the node succeeding x on the optimal path starting from x . Clearly, V ( D ) = 0. Applying dynamic programming, we have V ( I ) = d ( I, D ) + V ( D ) = 3 , n ( I ) = D, V ( J ) = d ( J, D ) + V ( D ) = 6 , n ( J ) = D, V ( E ) = min { d ( E, I ) + V ( I ) , d ( E, J ) + V ( J ) } = 7 , n ( E ) = I, V ( F ) = min { d ( F, I ) + V ( I ) , d ( F, J ) + V ( J ) } = 8 , n ( F ) = I, V ( G ) = d ( G, J ) + V ( J ) = 8 , n ( G ) = J, V ( H ) = min { d ( H, I ) + V ( I ) , d ( H, J ) + V ( J ) } = 8 , n ( H ) = I, V ( A ) = min { d ( A, E ) + V ( E ) , d ( A, F ) + V ( F ) } = 10 , n ( A ) = E, V ( B ) = min { d ( B, F ) + V ( F ) , d ( B, G ) + V ( G ) } = 12 , n ( B ) = F, V ( C ) = min { d ( C, G ) + V ( G ) , d ( C, H ) + V ( H ) } = 11 , n ( C ) = H, V ( O ) = min { d ( O, A ) + V ( A ) , d ( O, B ) + V ( B ) , d ( O, C ) + V ( C ) } = 13 , n ( O ) = A. Therefore the shortest distance from O to D is V ( O ) = 13. Searching backward, we obtain the shortest path: O A E I D. Problem 2. Write down the HJB equation for the following deterministic optimal control problem. Maximize J ( u ) = - x T s.t ˙ x t = u t x t x 0 = 1 u t [ - 1 , 1] .