2phase

# 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 ˆ s 2 and ˆ 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 + ˆ s 2 = 3 2 x 1 + x 2 + 3 x + 3- 3 x- 3- s 3 + ˆ s 3 = 5 x 1 , x 2 , x + 3 , x- 3 , s 3 3 , ˆ s 2 , ˆ s 3 ≥ 1 Note that now there are solutions feasible to the problem with...
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## This note was uploaded on 03/12/2009 for the course MSCI 331 taught by Professor Ostrowski during the Winter '09 term at Waterloo.

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

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