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Rec7sol_2010

# Rec7sol_2010 - Fall 2010 Optimization I(ORIE 3300/5300...

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Fall 2010 Optimization I (ORIE 3300/5300) Recitation 7 Solution Problem 1 After adding slack variables x 4 , x 5 , x 6 , the standard equality form of this problem is: z - 3 x 2 + x 3 = 0 - x 2 + x 3 + x 4 = 0 x 1 + 2 x 2 - x 3 + x 5 = 4 - 5 x 1 + 2 x 2 + x 3 + x 6 = 6 So, c = ( 0 3 - 1 0 0 0 ) T A = 0 - 1 1 1 0 0 1 2 - 1 0 1 0 - 5 2 1 0 0 1 b = 0 4 6 We are given that the basis is B = [4 , 2 , 3] and we would like to generate all of the entries in the final tableau. In order to do this, we will use Theorem 12.2 and find the unique tableau for the above system corresponding to basis B. This will take the form: z - ( c T N - y T A N ) x N = y T b x B + A - 1 B A N x N = A - 1 B b where the vector y solves the equation A T B y = c B . ? First, c N = ( 0 0 0 ) T , A N = 0 0 0 1 1 0 - 5 0 1 , and x N = x 1 x 5 x 6 . We also have that y = ( A T B ) - 1 c B = 1 0 0 - 1 2 2 1 - 1 1 - 1 0 3 - 1 . So, y = 0 1 . 25 0 . 25 And, z - ( c T N - y T A N ) x N = z + 1 . 25 x 5 + 0 . 25 x 6 Next, y T b = 6 . 5 So, the row corresponding to the objective function in our tableau is: z + 1 . 25 x 5 + 0 . 25 x 6 = 6 . 5 1

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Fall 2010 Optimization I (ORIE 3300/5300) ? Now, we will solve for the rows in the tableau corresponding to the constraints. We have x B = x 4 x 2 x 3 , A - 1 B = 1 0 . 75 - 0 . 25 0 0 . 25 0 . 25 0 - 0 . 50 0 . 50 , and A N = 0 0 0 1 1 0 - 5 0 1 . So, A - 1 B A N = 2 0 . 75 - 0 . 25 - 1 0 . 25 0 . 25 - 3 - 0 . 50 0 . 50 , and A - 1 B b = 1 . 5 2 . 5 1 We have x B + A - 1 B A N x N = A - 1 B b
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Rec7sol_2010 - Fall 2010 Optimization I(ORIE 3300/5300...

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