

*New code written for removing artificial column from Iteration table (on 17Sep17).
It may be possible, some working problem may not work properly. If you find any such problem then mail me immediately with the problem, so i will try my best to improve the software as soon as possible.

Solve the Linear programming problem using
BigM method

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Find


 max Z = 3x1 + 5x2 + 4x3
subject to 2x1 + 3x2 <= 8 2x2 + 5x3 <= 10 3x1 + 2x2 + 4x3 <= 15 and x1,x2,x3 >= 0
 min Z = x1 + x2
subject to 2x1 + 4x2 >= 4 x1 + 7x2 >= 7 and x1,x2 >= 0
 max Z = 5x1 + 10x2 + 8x3
subject to 3x1 + 5x2 + 2x3 <= 60 4x1 + 4x2 + 4x3 <= 72 2x1 + 4x2 + 5x3 <= 100 and x1,x2,x3 >= 0
 max Z = 4x1 + 3x2
subject to 2x1 + x2 <= 1000 x1 + x2 <= 800 x1 <= 400 x2 <= 700 and x1,x2 >= 0
 min Z = 600x1 + 500x2
subject to 2x1 + x2 >= 80 x1 + 2x2 >= 60 and x1,x2 >= 0
 min Z = 5x1 + 3x2
subject to 2x1 + 4x2 <= 12 2x1 + 2x2 = 10 5x1 + 2x2 >= 10 and x1,x2 >= 0
 max Z = x1 + 2x2 + 3x3  x4
subject to x1 + 2x2 + 3x3 = 15 2x1 + x2 + 5x3 = 20 x1 + 2x2 + x3 + x4 = 10 and x1,x2,x3,x4 >= 0
 max Z = 3x1 + 9x2
subject to x1 + 4x2 <= 8 x1 + 2x2 <= 4 and x1,x2 >= 0
 max Z = 3x1 + 2x2 + x3
subject to 2x1 + 5x2 + x3 = 12 3x1 + 4x2 = 11 and x2,x3 >= 0 and x1 unrestricted in sign
 max Z = 3x1 + 3x2 + 2x3 + x4
subject to 2x1 + 2x2 + 5x3 + x4 = 12 3x1 + 3x2 + 4x3 = 11 and x1,x2,x3,x4 >= 0
 max Z = 6x1 + 4x2
subject to 2x1 + 3x2 <= 30 3x1 + 2x2 <= 24 x1 + x2 >= 3 and x1,x2 >= 0
 max Z = 3x1 + 5x2
subject to x1  2x2 <= 6 x1 <= 10 x2 >= 1 and x1,x2 >= 0
 max Z = 6x1 + 4x2
subject to x1 + x2 <= 5 x2 >= 8 and x1,x2 >= 0
 max Z = 6x1 + 4x2
subject to x1  x2 >= 5 x2 >= 8 and x1,x2 >= 0





Solution 


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Solve the following LP problem
using BigM method





