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16.05.18 - Dual simplex method new implementation(1. infeasible solution and 2. fails to get solution)
Solve the Linear programming problem using
Graphical method calculator
Type your linear programming problem
OR
Total Variables :
Total Constraints :
Click On Generate
Mode :
Fraction
Decimal
Solve after converting Min function to Max function
Zj-Cj (display in steps)
Alternate Solution (if exists)
Artificial Column Remove
Tooltip for calculation steps
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Find
1. Simplex (BigM) method
2. TwoPhase method
3. Dual simplex method
4. Integer simplex method
5. Graphical method
6. Primal to Dual
7. Branch and Bound method
8. 0-1 Integer programming problem
max z = -2x1 - x2
subject to
-3x1 - x2 <= -3
-4x1 - 3x2 <= -6
-x1 - 2x2 <= -3
and x1,x2 >= 0
max z = 15x1 + 10x2
subject to
4x1 + 6x2 <= 360
3x1 <= 180
5x2 <= 200
and x1,x2 >= 0
max z = 2x1 + x2
subject to
x1 + 2x2 <= 10
x1 + x2 <= 6
x1 - x2 <= 2
x1 - 2x2 <= 1
and x1,x2 >= 0
max z = -x1 + 2x2
subject to
x1 - x2 <= -1
-0.5x1 + x2 <= 2
and x1,x2 >= 0
max z = 40x1 + 80x2
subject to
2x1 + 3x2 <= 48
x1 <= 15
x2 <= 10
and x1,x2 >= 0
max z = 60x1 + 40x2
subject to
x1 <= 25
x2 <= 35
2x1 + x2 <= 60
and x1,x2 >= 0
min z = 3x1 + 2x2
subject to
5x1 + x2 >= 10
x1 + x2 >= 6
x1 + 4x2 >= 12
and x1,x2 >= 0
min z = 600x1 + 400x2
subject to
3x1 + 3x2 >= 40
3x1 + x2 >= 40
2x1 + 5x2 >= 44
and x1,x2 >= 0
min z = 4x1 + 3x2
subject to
200x1 + 100x2 >= 4000
x1 + 2x2 >= 50
40x1 + 40x2 >= 1400
and x1,x2 >= 0
min z = -x1 + 2x2
subject to
-x1 + 3x2 <= 10
x1 + x2 <= 6
x1 - x2 <= 2
and x1,x2 >= 0
max z = -15x1 - 10x2
subject to
-3x1 - 5x2 <= -5
-5x1 - 2x2 <= -3
and x1,x2 >= 0
max z = 600x1 + 500x2
subject to
2x1 + x2 >= 80
x1 + 2x2 >= 60
and x1,x2 >= 0
max z = 3x1 + 2x2
subject to
x1 - x2 >= 1
x1 + x2 >= 3
and x1,x2 >= 0
max z = 5x1 + 4x2
subject to
x1 - 2x2 <= 1
x1 + 2x2 >= 3
and x1,x2 >= 0
max z = -4x1 + 3x2
subject to
x1 - x2 <= 0
x1 <= 4
and x1,x2 >= 0
max z = 3x1 + 4x2
subject to
x1 - x2 = -1
-x1 + x2 <= 0
and x1,x2 >= 0
max z = 6x1 - 4x2
subject to
2x1 + 4x2 <= 4
4x1 + 8x2 >= 16
and x1,x2 >= 0
max z = x1 + 1/2x2
subject to
3x1 + 2x2 <= 12
5x1 = 10
x1 + x2 >= 8
-x1 + x2 >= 4
and x1,x2 >= 0
max z = 3x1 + 2x2
subject to
-2x1 + 3x2 <= 9
3x1 - 2x2 <= -20
and x1,x2 >= 0
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Graphical Method
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