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Simplex Method Adapting to Other Forms. Introduction to Operations Research. Until now, we have dealt with the standard form of the Simplex method What if the model has a non-standard form? Equality Constraints x 1 + x 2 = 8 Greater than Constraints x 1 + x 2 ≥ 8 Minimizing
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Simplex Method Adapting to Other Forms Introduction to Operations Research
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? Simplex Method
Original Form Maximize Z = 3x1 + 5x2 Subject to: x1 ≤ 4 2x2 ≤ 12 3x1 + 2x2 = 18 x1 ≥ 0, x2 ≥ 0 Equality Constraint 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 Adapting Equality Constraint Artificial Form (Big M Method) Maximize Z = 3x1 + 5x2 Subject to: • Z - 3x1 - 5x2 + Mx5 = 0 • x1 + x3= 4 • 2x2 + x4 = 12 • 3x1 + 2x2+ x5 = 18 • x1 , x2 , x3 , x4 , x5 ≥ 0
Iterations of the Simplex METHOD 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 Select Initial Point nonbasic variables: x1 and x2 (origin) Initial BF solution: (x1, x2, x3, x4, x5) = (0,0,4,12,18M)
Iterations of the Simplex METHOD Optimality Test: Are all coefficients in row (0) ≥ 0? If yes, then STOP – optimal solution If no, then continue algorithm
Iterations of the Simplex METHOD 4/1 = 4 18/3 = 6 Select Entering Basic Variable Choose variable with negative coefficient having largest absolute value Select Leaving Basic Variable 1. Select coefficient in pivot column > 0 2. Divide Right Side value by this coefficient 3. Select row with smallest ratio
Iterations of the Simplex METHOD 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 solution If no, then continue algorithm
Iterations of the Simplex METHOD 12/2 = 6 6/2 = 3 Select Entering Basic Variable Choose variable with negative coefficient having largest absolute value Select Leaving Basic Variable 1. Select coefficient in pivot column > 0 2. Divide Right Side value by this coefficient 3. Select row with smallest ratio
Iterations of the Simplex METHOD 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 solution If no, then continue algorithm
Iterations of the Simplex METHOD 4/1 = 4 6/3 = 2 Select Entering Basic Variable Choose variable with negative coefficient having largest absolute value Select Leaving Basic Variable 1. Select coefficient in pivot column > 0 2. Divide Right Side value by this coefficient 3. Select row with smallest ratio
Iterations of the Simplex METHOD 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 solution If no, then continue algorithm
Minimize Z = 3x1 + 5x2 Multiply by -1 Maximize -Z = -3x1 - 5x2 Adapting for Minimization
x1 - x2 ≤ -1 Multiply by -1 -x1 + x2 ≥ 1 Negative Right-Hand Side
x1 - x2 ≥ 1 x1 - x2 - x5≥ 1 Change Inequality x1 - x2 - x5≤ 1 x1 - x2 - x5 + x6≤ -1 Adapting for a ≥ Constraint Augmented Form Big M
Original Form Minimize Z = 4x1 + 5x2 Subject to: 3x1 + x2 ≤ 27 5x1 + 5x2 = 60 6x1 + 4x2 ≥ 60 x1 ≥ 0, x2 ≥ 0 ExampLe of Adapting form Adaption Form Minimize Z = 4x1 + 5x2 Maximize -Z = -4x1 - 5x2 Subject to: • -Z + 4x1 + 5x2 + Mx4 + Mx6 = 0 • 3x1+ x2+ x3= 27 • 5x1 + 5x2 + x4 = 60 • 6x1 + 4x2- x5 + x6 = 60