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# GE330practice2 - Solution(a The standard form is as follows...

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Problem 1. Is each of the following linear programs in the standard form? If it is not, convert it to the standard form. (a) min 3 x 1 - 7 x 2 + 4 x 3 s.t. x 1 + x 2 + x 3 3 x 1 + 2 x 2 + 3 x 3 5 x 1 , x 2 , x 3 0 Solution: min 3 x 1 - 7 x 2 + 4 x 3 s.t. x 1 + x 2 + x 3 - s 1 = 3 x 1 + 2 x 2 + 3 x 3 + s 2 = 5 x 1 , x 2 , x 3 , s 1 , s 2 0 (b) min 2 x 1 - 4 x 2 + 5 x 3 s.t. x 1 + x 2 - x 3 = 4 x 1 + 2 x 2 - 3 x 3 = 5 x 1 - x 2 = 1 x 1 0 , x 2 0 Solution: min 2 x 1 + 4 y 1 + 5 y 2 - 5 y 3 s.t. x 1 - y 1 - y 2 + y 3 = 4 x 1 - 2 y 1 - 3 y 2 + 3 y 3 = 5 x 1 + y 1 = 1 x 1 , y 1 , y 2 , y 3 0 (c) min 3 x 1 + 6 x 2 + 4 x 3 s.t. x 1 - x 2 = 2 x 3 x 1 + x 2 + x 3 - 5 = 0 x i 0 , i = 1 , 2 , 3 Solution: min 3 x 1 + 6 x 2 + 4 x 3 s.t. x 1 - x 2 - 2 x 3 = 0 x 1 + x 2 + x 3 = 5 x i 0 , i = 1 , 2 , 3 (d) min 5 x 1 + 2 x 2 + 2 x 3 s.t. x 1 - x 2 + x 3 = 4 x 1 + x 2 - x 3 = 1 x i 0 , i = 1 , 2 , 3 Solution: This is already in the standard form. 1

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Problem 2. Consider the following LP: max z = x 1 + 3 x 2 s.t. x 1 + 2 x 2 4 3 x 1 + 2 x 2 8 x 1 , x 2 0 (a) Convert the problem to the standard form. (b) Determine all the basic solutions of the problem, and classify them as feasible and infeasible. (c) Use direct substitution in the objective function to determine the optimum basic feasible solution. (d) Verify graphically that the solution obtained in (c) is the optimum LP solution. (e) Show how the infeasible basic solutions are represented on the graphical solution space.
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Unformatted text preview: Solution: (a) . The standard form is as follows max z = x 1 + 3 x 2 s.t. x 1 + 2 x 2 + s 1 = 4 3 x 1 + 2 x 2 + s 2 = 8 x 1 ,x 2 ,s 1 ,s 2 ≥ (b) . The basic solutions is listed in the following table Basic variables Basic solution Feasible? Objective Value ( x 1 ,x 2 ) (2,1) Yes 5 ( x 1 ,s 1 ) (8/3,4/3) Yes 8/3 ( x 1 ,s 2 ) (4,-4) No-( x 2 ,s 1 ) (4,-4) No-( x 2 ,s 2 ) (2,4) Yes 6 ( s 1 ,s 2 ) (4,8) Yes (c) . x 2 = 2, s 2 = 4, x 1 = s 1 = 0 is the optimal solution. (d) . The graph is as follows: ABCD is the feasible region, and the optimal solution is point B . 1 2 3 4 5 x 1 x 2 1 2 3 4 s 1 = 0 s 2 = 0 A B C D E F (e) . The two infeasible basic solutions are E and F . E : x 1 = 4 ,s 2 =-4, F : x 2 = 4 ,s 1 =-4. 2...
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GE330practice2 - Solution(a The standard form is as follows...

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