2phase - Example BigM and 2 Phase 1 Formulating the problem...

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Unformatted text preview: Example BigM and 2 Phase 1 Formulating the problem min z = x 1 + x 2 + x 3 s.t. x 1 + x 2 + x 3 3 2 x 1 + x 2 = 3 2 x 1 + x 2 + 3 x 3 5 x 1 ,x 2 We need to reformulate the above problem so that there is a feasible solution with x 1 , x 2 , and x 3 equal to zero. First, note that x 3 can be negative. We will change this first by replacing x 3 with x 3 = x + 3- x- 3 (where both x + 3 and x- 3 are 0). This gives us: min z = x 1 + x 2 + x + 3- x- 3 s.t. x 1 + x 2 + x + 3- x- 3 3 2 x 1 + x 2 = 3 2 x 1 + x 2 + 3 x + 3- 3 x- 3 5 x 1 , x 2 , x + 3 , x- 3 Next, lets get rid of the constraint (and turn it into an = constraint). This is done by adding a surplus variable, s 3 . min z = x 1 + x 2 + x + 3- x- 3 s.t. x 1 + x 2 + x + 3- x- 3 3 2 x 1 + x 2 = 3 2 x 1 + x 2 + 3 x + 3- 3 x- 3- s 3 = 5 x 1 , x 2 , x + 3 , x- 3 , s 3 Now we have 2 equalites (neither of which can be satisfied when x 1 , x 2 , and x 3 are equal to zero. We must add artifical variables s 2 and s 3 to make the origin feasible. This gives us: min z = x 1 + x 2 + x + 3- x- 3 s.t. x 1 + x 2 + x + 3- x- 3 3 2 x 1 + x 2 + s 2 = 3 2 x 1 + x 2 + 3 x + 3- 3 x- 3- s 3 + s 3 = 5 x 1 , x 2 , x + 3 , x- 3 , s 3 3 , s 2 , s 3 1 Note that now there are solutions feasible to the problem with...
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2phase - Example BigM and 2 Phase 1 Formulating the problem...

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