# hw10s - ENGRI 1101 Engineering Applications of OR Fall 08...

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ENGRI 1101 Engineering Applications of OR Fall ’08 Homework 10 Homework 10 Solutions 1. The linear program in this problem is: max 12 - 2 x 4 - x 2 + x 3 s.t. x 1 = 6 - x 4 - x 2 x 5 = 10 - x 4 - x 2 - 2 x 3 x 6 = 2 - x 4 - x 2 x 7 = 4 - x 2 x 1 , . . . , x 7 0 (a) The problem is already in dictionary form, and the solution corresponding to the initial dictionary is x 2 = x 3 = x 4 = 0 , x 1 = 6 , x 5 = 10 , x 6 = 2 , x 7 = 4 , which is feasible, since all variables have nonnegative values. Objective function value correspond- ing to this solution is 12. As the next step, we increase x 3 , which is the only variable with a positive coefficient in the objective function in the current dictionary. The only bound on the amount of increase of x 3 comes from the second constraint: to keep x 5 0, we must have x 3 5. Exchanging the roles of x 5 and x 3 in the dictionary, we get: x 3 = 5 - 1 2 x 4 - 1 2 x 2 - 1 2 x 5 , and substituting the above expression for x 3 into the objective and remaining constraints (actually, x 3 does not appear in any other constraints, so our task in this example is simplified), we get the updated dictionary: max 17 - 5 2 x 4 - 3 2 x 2 - 1 2 x 5 s.t. x 1 = 6 - x 4 - x 2 x 3 = 5 - 1 2 x 4 - 1 2 x 2 - 1 2 x 5 x 6 = 2 - x 4 - x 2 x 7 = 4 - x 2 x 1 , . . . , x 7 0 .

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