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Preliminära svar till tentamen TNE052 2007-05-29

Preliminära svar till tentamen TNE052 2007-05-29

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(iii) (2p)Interval x =(x 1 ,x 2 ,x 3 )0 ≤ v ≤ 2 x =(0, 0, 0)2 ≤ v ≤ 7 3x =(0, 0, 1)73≤ v ≤ 3 x =(0, 1, 1)3 ≤ v x =(1, 1, 1)Duala funktionen blir⎧⎪⎨h(v) =⎪⎩7v om 0 ≤ v ≤ 22v + 10 om 2 ≤ v ≤ 7 3−v + 17 om 7 3 ≤ v ≤ 3−2v + 20 om 3 ≤ vOptimallösningen i det duala problemet är v ∗ = 7 3 ,ochh(v∗ )= 443 .4

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