HW8_Solutions

# HW8_Solutions - ORIE 3300/5300 ASSIGNMENT 8 SOLUTION Fall...

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Unformatted text preview: ORIE 3300/5300 ASSIGNMENT 8 SOLUTION Fall 2008 Problem 1 We are starting with A = 2 1 1 1 0 3 1 2 0 1 , b = 3 5 , c = 3 2 3 0 0 T . Iteration 1 For the basis B = [4 , 5], we have: c B = , A B = 1 0 0 1 , A- 1 B = 1 0 0 1 , ˆ x B = 3 5 . Step 1: We first compute the vector y : y = ( A T B )- 1 c B = ( A- 1 B ) T c B = 1 0 0 1 = . Step 2: To choose an entering index k , we note j c j A T j y 1 3 > 2 2 > 3 3 > so we can choose k = 1 as the entering index according to the least index rule. Step 3: We next compute the vector d : d = A- 1 B A k = 1 0 0 1 2 3 = 2 3 . Step 4: To choose a leaving index by the ratio test, we note t = min { 3 2 , 5 3 } = 1 . 5 , ( i = 4 , 5) so we can choose r = B (1) = 4 as the leaving index. Step 5: We update the basis and the current values of the basic variables: ˆ x 1 ← t = 1 . 5 ˆ x 5 ← ˆ x 5- td 5 = 5- 3 · 1 . 5 = 0 . 5 B ← B ∪ { k } \ { r } = [1 , 5] . 1 Iteration 2: For the basis B = [1 , 5], we have: c B = 3 , A B = 2 0 3 1 , A- 1 B = 1 / 2 0- 3 / 2 1 , ˆ x B = 1 . 5 . 5 . Step 1: We first compute the vector y : y = ( A T B )- 1 c B = ( A- 1 B ) T c B = 1 / 2- 3 / 2 1 3 = 3 / 2 ....
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## This homework help was uploaded on 02/20/2009 for the course ORIE 3300 taught by Professor Todd during the Fall '08 term at Cornell.

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HW8_Solutions - ORIE 3300/5300 ASSIGNMENT 8 SOLUTION Fall...

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