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Various violations that can occur in linear programming problems and provides remedies for them. These violations include non-standard formulations, minimization problems, negative right-hand sides, and constraints with the equality or inequality signs. Examples are provided to illustrate these concepts.
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What do you do if your problem formulation does not have the Standard Form?
This is an important issue because the simplex procedure we described relies very much on the standard form, eg
x (^) j ≥ 0 , j = 1,..., n
max x Z = c (^) j x (^) j j = 1
n
1. Multiply the coefficients of the objective function by -1 and maximize the new objective function. 2. Change the simplex algorithm a bit. Remark : Australia is a free country so in principle you can use either of these two approaches. In his notes, Moshe prefers the second approach. I prefer the first.
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x
1 2 3 1 2 3 1 2 3 1 2 3
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max ' x
Z = − x (^) 1 + x (^) 2 + 2 x 3
x x x x x x x x x x x x
1 2 3 1 2 3 1 2 3 1 2 3
2 10 2 4 2 40 2 3 30 0
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After setting the simplex tableau and conducting two pivot operations we obtain the above final tableau. The optimal solution is thus: x=(10/3,0,70/3,30,0,0) The optimal value of the modified objective function is: Z’ = 130/ Thus the optimal value of the original objective function is equal to Z = - Z’ = - 130/3.
BV Eq. # (^) x 1 x 2 x 3 x 4 x 5 x 6 RHS x 4 1 0 4 0 1 1/2^0 x 3 2 0 7/3^1 0 1/3^ 1/3^ 70/ x 1 3 1 1/3^0 0 - 1/6 1/3^ 10/ Z' Z' 0 10/3 0 0 5/6 1/3 130/
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Modification in the Simplex Algorithm for opt=min:
1. Optimality Test :
2. Greedy Rule :
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(Continued)
x
1 2 3 1 2 3 1 2 3 1 2 3
(^13) Step 4 (Variable out):
The (usual) Ratio Test identifies x 5 as the leaving basic variable. Step 5 (Pivoting): As usual New tableau: BV Eq. # (^) x 1 x 2 x 3 x 4 x (^) 5 x 6 RHS x 4 1 0 4 0 1 1/2^0 x 3 2 - 1 2 1 0 1/2^0 x 6 3 3 1 0 0 - 1/2 1 10 Z Z (^1) - 3 0 0 - 1 0 - 40
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BV Eq. # (^) x 1 x 2 x 3 x 4 x (^) 5 x 6 RHS x 4 1 0 4 0 1 1/2^0 x 3 2 - 1 2 1 0 1/2^0 x 6 3 3 1 0 0 - 1/2 1 10 Z Z (^1) - 3 0 0 - 1 0 - 40 BV Eq. # (^) x 1 x 2 x 3 x 4 x (^) 5 x 6 RHS x 4 1 0 4 0 1 1/2^0 x 3 2 0 7/3^1 0 1/3^ 1/3^ 70/ x 1 3 1 1/3^0 0 - 1/6 1/3^ 10/ Z Z (^0) - 10/3 0 0 - 5/6 - 1/3 - 130/ All the reduced costs are nonpositive , so we stop. Optimal solution: x=(10/3,0,70/3,30,0,0) Optimal value of the objective function: Z* = -130/3.
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Clearly, if x’j > x”j then x (^) j>0, whereas if x’j<x”j then x (^) j<0. And if x’j=x”j then x (^) j=0. Thus, x (^) j is indeed unrestricted in sign ( URS )
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x
1 2 1 2 1 1 2
(^19) Violation #
This is handled by multiplying the respective constraint by -1 and taking care of the inequality sign if necessary (changing <= to >= and >= to <=).
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Observe that in fixing this violation we created another one!
