simplex method adapting to other forms. until now, we have dealt with the standard form of the...
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Simplex Method
Adapting to Other Forms
Until now, we have dealt with the standard form of the Simplex method
What if the model has a non-standard form? Equality Constraints x1 + x2 = 8
Greater than Constraints x1 + x2 ≥ 8
Minimizing How do we get the initial BF solution?
Original Form
Maximize Z = 3x1 + 5x2
Subject to:x1 ≤ 4
2x2 ≤ 12
3x1 + 2x2 = 18
x1 ≥ 0, x2 ≥ 0
Augmented Form
Maximize Z = 3x1 + 5x2
Subject to:Z - 3x1 - 5x2 = 0
x1 + x3 = 4
2x2 + x4 = 12
3x1 + 2x2 = 18
x1 , x2 , x3 , x4 ≥ 0
Original Form
Maximize Z = 3x1 + 5x2
Subject to:x1 ≤ 4
2x2 ≤ 12
3x1 + 2x2 = 18
x1 ≥ 0, x2 ≥ 0
Artificial Form (Big M Method)
Maximize Z = 3x1 + 5x2
Subject to:Z - 3x1 - 5x2 + Mx5 = 0x1 + x3 = 4
2x2 + x4 = 12
3x1 + 2x2 + x5 = 18
x1 , x2 , x3 , x4 , x5 ≥ 0
Basic Variable
Coefficient of:Right Side
Z x1 x2 x3 x4 x5
Z 1 -3 -5 0 0 M 0x3 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 3 2 0 0 1 18
Z -3x1 -5x2 + Mx5 = 0x1 + x3 = 4
x2 + x4 = 123x1 2x2 + x5 = 18
Basic Variable
Coefficient of:Right Side
Z x1 x2 x3 x4 x5
Z 1 -3M-3 -2M-5 0 0 0 -18Mx3 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 3 2 0 0 1 18
Select Initial Pointnonbasic variables: x1 and x2 (origin)Initial BF solution: (x1, x2, x3, x4, x5) = (0,0,4,12,18M)
To get to initial point, remove x5 coefficient (M) from Z - 3x1 - 5x2 + Mx5 = 0
(-M) ( 3x1 + 2x2 + x5 = 18) (-3M-3)x1 + (-2M-5)x2 = -18M
Basic Variable
Coefficient of:Right Side
Z x1 x2 x3 x4 x5
Z 1 -3M-3 -2M-5 0 0 0 -18Mx3 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 3 2 0 0 1 18
Optimality Test: Are all coefficients in row (0) ≥ 0?
If yes, then STOP – optimal solutionIf no, then continue algorithm
Basic Variable
Coefficient of:Right Side
Z x1 x2 x3 x4 x5
Z 1 -3M-3 -2M-5 0 0 0 -18Mx3 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 3 2 0 0 1 18
Select Entering Basic VariableChoose variable with negative coefficient having largest absolute value
Select Leaving Basic Variable1. Select coefficient in pivot column > 0
2. Divide Right Side value by this coefficient3. Select row with smallest ratio
4/1 = 4
18/3 = 6
Basic Variable
Coefficient of:Right Side
Z x1 x2 x3 x4 x5
Z 1 -3M-3 -2M-5 0 0 0 -18Mx3 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 3 2 0 0 1 18Z 1 0 -2M-5 3M+3 0 0
-6M+12
x1 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 0 2 -3 0 1 6
Basic Variable
Coefficient of:Right Side
Z x1 x2 x3 x4 x5
Z 1 0 -2M-5 3M+3 0 0-
6M+12
x1 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 0 2 -3 0 1 6
BF Solution: (x1, x2, x3, x4, x5) = (4,0,0,12,6)Optimality Test: Are all coefficients in row (0) ≥ 0?
If yes, then STOP – optimal solutionIf no, then continue algorithm
Basic Variable
Coefficient of:Right Side
Z x1 x2 x3 x4 x5
Z 1 0 -2M-5 3M+3 0 0-
6M+12
x1 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 0 2 -3 0 1 6
Select Entering Basic VariableChoose variable with negative coefficient having largest absolute value
Select Leaving Basic Variable1. Select coefficient in pivot column > 0
2. Divide Right Side value by this coefficient3. Select row with smallest ratio
12/2 = 6
6/2 = 3
Basic Variable
Coefficient of:Right Side
Z x1 x2 x3 x4 x5
Z 1 0 -2M-5 3M+3 0 0-
6M+12
x1 0 1 0 1 0 0 4x4 0 0 2 0 1 0 12x5 0 0 2 -3 0 1 6Z 1 0 0 -9/2 0 M+5/2 27 x1 0 1 0 1 0 0 4 x4 0 0 0 3 1 -1 6 x2 0 0 1 -3/2 0 1/2 3
Basic Variable
Coefficient of:Right Side
Z x1 x2 x3 x4 x5
Z 1 0 0 -9/2 0 M+5/2 27 x1 0 1 0 1 0 0 4 x4 0 0 0 3 1 -1 6 x2 0 0 1 -3/2 0 1/2 3
BF Solution: (x1, x2, x3, x4, x5) = (4,3,0,6,0)Optimality Test: Are all coefficients in row (0) ≥ 0?
If yes, then STOP – optimal solutionIf no, then continue algorithm
Basic Variable
Coefficient of:Right Side
Z x1 x2 x3 x4 x5
Z 1 0 0 -9/2 0 M+5/2 27 x1 0 1 0 1 0 0 4 x4 0 0 0 3 1 -1 6 x2 0 0 1 -3/2 0 1/2 3
Select Entering Basic VariableChoose variable with negative coefficient having largest absolute value
Select Leaving Basic Variable1. Select coefficient in pivot column > 0
2. Divide Right Side value by this coefficient3. Select row with smallest ratio
4/1 = 4
6/3 = 2
Basic Variable
Coefficient of:Right Side
Z x1 x2 x3 x4 x5
Z 1 0 0 -9/2 0 M+5/2 27 x1 0 1 0 1 0 0 4 x4 0 0 0 3 1 -1 6 x2 0 0 1 -3/2 0 1/2 3Z 1 0 0 0 3/2 M+1 36 x1 0 1 0 0 -1/3 1/3 2 x3 0 0 0 1 1/3 -1/3 2 x2 0 0 1 0 1/2 0 6
Basic Variable
Coefficient of:Right Side
Z x1 x2 x3 x4 x5
Z 1 0 0 0 3/2 M+1 36 x1 0 1 0 0 -1/3 1/3 2 x3 0 0 0 1 1/3 -1/3 2 x2 0 0 1 0 1/2 0 6
BF Solution: (x1, x2, x3, x4, x5) = (2,6,2,0,0)Optimality Test: Are all coefficients in row (0) ≥ 0?
If yes, then STOP – optimal solutionIf no, then continue algorithm
Minimize Z = 3x1 + 5x2
Multiply by -1
Maximize -Z = -3x1 - 5x2
x1 - x2 ≤ -1
Multiply by -1
-x1 + x2 ≥ 1
x1 - x2 ≥ 1
x1 - x2 - x5 ≥ 1
Change Inequality
x1 - x2 - x5 ≤ 1
x1 - x2 - x5 + x6 ≤ -1Big M
AugmentedForm
Original Form
Minimize Z = 4x1 + 5x2
Subject to:3x1 + x2 ≤ 27
5x1 + 5x2 = 60
6x1 + 4x2 ≥ 60
x1 ≥ 0, x2 ≥ 0
Adaption Form
Minimize Z = 4x1 + 5x2
Maximize -Z = -4x1 - 5x2
Subject to:-Z + 4x1 + 5x2 + Mx4 + Mx6 = 03x1 + x2 + x3 = 27
5x1 + 5x2 + x4 = 60
6x1 + 4x2 - x5 + x6 = 60