HW4 Solution

# HW4 Solution - (100%, 80%, 64%) of the current state. At...

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MIE375H1F - Financial Engineering - HW4 Solutions Nick Yeung November 10, 2010 Question 5.3 The formulation for the Two Period Budget is: Decision Variables Let x i be 1 if we invest in project i and 0 otherwise. Let y be the excess cash after Year 1 Objective Function max 7 X i =1 NPV i · x i max 150 x 1 + 200 x 2 + 100 x 3 + 100 x 4 + 120 x 5 + 150 x 6 + 240 x 7 Constraints In Year 1: 90 x 1 + 80 x 2 + 50 x 3 + 20 x 4 + 40 x 5 + 80 x 6 + 80 x 7 + y 250 In Year 2: 58 x 1 + 80 x 2 + 100 x 3 + 64 x 4 + 50 x 5 + 20 x 6 + 100 x 7 250 + 1 . 1 y x i = { 0 , 1 } , x = 1 ... 7 y 0 Solving the above Integer Program by using Branch and Bound or other IP methods, we get the solution to invest in projects 1,4,5,7 (or 4,5,6,7) for a total NPV of \$610,000. 1

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Question 5.5 Trinomial Lattice: 16 Nodes, Trinomial Tree: 40 Nodes Question 5.9 5.9a States are the barrels of oil remaining. From each node, you can move to next year’s states depending on the three decisions (no pump, pump, enhanced pump) in which the states would be
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Unformatted text preview: (100%, 80%, 64%) of the current state. At each state, we would consider: max { no pump, pump, enhanced pump } At each node, you would consider which of the three options yields the largest PV (including the future decisions) 2 At the top node (100) of Year 2, you would consider: max { no pump, pump, enhanced pump } = max { , (100 , 000 * . 2) * 10-50 , 000 , (100 , 000 * . 36)-120 , 000 } = max { , 150 , 000 , 240 , 000 } = 240 , 000 (enhanced pumping) Continue with all the nodes using the Dynamic Programming Algorithm 5.9b The maximum present value is \$366,740. The strategy is to have Enhanced Pumping for the ﬁrst two years and normal pumping for the last year. 3...
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## This note was uploaded on 04/20/2011 for the course MIE 375 taught by Professor R.kwon during the Spring '11 term at University of Toronto- Toronto.

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HW4 Solution - (100%, 80%, 64%) of the current state. At...

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