gropt - max 0 x 1 +0 x 2-5 x 3 +0 x 4-10 3 x 5 s.t. x 1 , x...

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Suppose we are interested in an l.p. problem of the form: max c 1 x 1 + c 2 x 2 + c 3 x 3 + c 4 x 4 + c 5 x 5 s.t. 2 x 1 +2 x 2 + x 3 = 80 4 x 1 +2 x 2 + x 4 = 120 3 x 1 +6 x 2 + x 5 = 210 x 1 0 , x 2 0 x 3 0 , x 4 0 x 5 0 It is easy to check that the following two propoesd solutions are feasible: x 0 = (0 , 0 , 80 , 120 , 210) , ˆ x = (10 , 30 , 0 , 20 , 0). Whether one or the other is optimal will depend on the objective function coeFcients c 1 , ··· , c 5 . Note that if c 1 0 , c 2 0, and c 3 = c 4 = c 5 = 0, then x 0 is optimal. For example take c 1 = 0 , c 2 = 0 , c 3 = 0 , c 4 = 0 , c 5 = 0 On the other hand, if c 1 = c 2 = c 4 = 0 and c 3 0 , c 5 0, then ˆ x is optimal. For example take c 1 = 0 , c 2 = 0 , c 3 = - 5 , c 4 = 0 , c 5 = - 10 3 . How do we know this is true? 1
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Because ˆ x = (10 , 30 , 0 , 20 , 0) is feasible for max c 1 x 1 + c 2 x 2 + c 3 x 3 + c 4 x 4 + c 5 x 5 s.t. 2 x 1 +2 x 2 + x 3 = 80 4 x 1 +2 x 2 + x 4 = 120 3 x 1 +6 x 2 + x 5 = 210 x 1 0 , x 2 0 x 3 0 , x 4 0 x 5 0 and clearly optimal for
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Unformatted text preview: max 0 x 1 +0 x 2-5 x 3 +0 x 4-10 3 x 5 s.t. x 1 , x 2 x 3 , x 4 x 5 Note that the magic numbers 5 and 10 3 were reported to us by AMPL in its solution of Grandmas sasusage problem. We are about to see how AMPL got them, as well as the optimal production levels for Type1 sausage (10 batches) and Type2 sausage(30 batches), which happen to be the same as out x 1 and x 2 . The magic comes from linear algebra. 2...
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gropt - max 0 x 1 +0 x 2-5 x 3 +0 x 4-10 3 x 5 s.t. x 1 , x...

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