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S4 - C&O 350 Linear Optimization Fall 2009 Solutions to...

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C&O 350 Linear Optimization – Fall 2009 – Solutions to Assignment 4 Question # Max. marks Part marks 1 8 2 6 2 8 6 2 3 8 4 4 4 8 2 2 2 2 5 8 4 4 6 (Bonus) (5) Total 40 + (5) 1. 8 marks = 2+6 Solution: (a) We start with: z + x 1 - x 2 + x 3 + x 4 + x 5 - x 6 = 0 x 1 + x 4 + 6 x 6 = 9 3 x 1 + x 2 - 4 x 3 + 2 x 6 = 2 x 1 + 2 x 3 + x 5 + 2 x 6 = 6 The tableau with basis { 1 , 2 , 3 } is: z - 9 4 x 4 + 5 2 x 5 - 29 x 6 = - 77 2 x 1 + x 4 + 6 x 6 = 9 x 2 - 5 x 4 + 2 x 5 - 24 x 6 = - 31 x 3 - 1 2 x 4 + 1 2 x 5 - 2 x 6 = - 3 2 This does not give us a feasible solution. The tableau with basis { 1 , 4 , 5 } is: z - 2 3 x 2 - 7 3 x 3 - 25 3 x 6 = - 43 3 - 1 3 x 2 + 4 3 x 3 + x 4 + 16 3 x 6 = 25 3 x 1 + 1 3 x 2 - 4 3 x 3 + 2 3 x 6 = 2 3 - 1 3 x 2 + 10 3 x 3 + x 5 + 4 3 x 6 = 16 3 This gives us the feasible solution x * = [ 2 3 , 0 , 0 , 25 3 , 16 3 , 0] T . We will use this tableau to start the simplex method in (b). (b) There are many possible solutions to this. We’ll present one that follows Dantzig’s rule (see page 78 of coursenotes). Starting with the tableau that corresponds to the basis { 1 , 4 , 5 } , we can choose between x 2 , x 3 or x 6 as our entering variable. Since max { c 2 , c 3 , c 6 } = max 2 3 , 7 3 , 25 3 = c 6 , so Dantzig’s rule chooses x 6 to enter. Next, we compute t = min 25 3 / 16 3 , 2 3 / 2 3 , 16 3 / 4 3 = min 25 16 , 1 , 4 = 1, so x 1 leaves. z + 25 2 x 1 + 7 2 x 2 - 19 x 3 = - 6 - 8 x 1 - 3 x 2 + 12 x 3 + x 4 = 3 3 2 x 1 + 1 2 x 2 - 2 x 3 + x 6 = 1 - 2 x 1 - x 2 + 6 x 3 + x 5 = 4 1
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Now the only varible that has positive reduced cost is x 3 , so x 3 enters our basis. We compute min 3 12 , - , 4 6 = 1 4 , so x 4 leaves. The next tableau is z - 1 6 x 1 - 5 4 x 2 + 19 12 x 4 = - 5 4 - 2 3 x 1 - 1 4 x 2 + x 3 + 1 12 x 4 = 1 4 1 6 x 1 + 1 6 x 4 + x 6 = 3 2 2 x 1 + 1 2 x 2 - 1 2 x 4
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