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LECTURE 14 MINIMIZATION TWO PHASE METHOD BY DR. ARSHAD ZAHEER Simplex Method

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Two Phase Method

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Illustration Minimize f=4x 1 + x 2 (Objective Function) Subject to: (Constraints) 3x 1 + x 2 = 3 4x 1 + 3x 2 ≥ 6 x 1 + 2x 2 ≤ 4 x 1, x 2 ≥ 0(Non-Negativity Constraints)

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Inequalities Constraints in Equation Form Let S 1 and S 2 be the surplus and slack variables for second and third constraints, respectively. Minimizef =4x 1 + x 2 (Objective Function) Subject to:(Constraints) 3x 1 + x 2 = 3………….(1) 4x 1 + 3x 2 – S 1 = 6………….(2) x 1 + 2x 2 + S 2 = 4………….(3) x 1, x 2, S 1, S 2≥ 0

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Initial solution Arbitrary values =# of Variables -- # of Equations 4-2 = 2 Let x1= 0, x2 = 0 Putting above values in objective function (f =5x 1 + 4x 2 ) and equation 1-3, f = 0 0 = 3 (Contradiction) S 1 = -6 (Violation) S 2 = 4 This is against the basic rules of Mathematics as the 0 cannot be equal to 3. This is against the non negativity constraint that X must be non zero. This situation cannot be called as initial feasible solution because it is not satisfying the condition. In order to make it feasible we need to add Artificial variables In equations having contradictions and violation

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Artificial Variables Let A 1 and A 2 be the artificial variables for first and second equation respectively. Minimize: f =4x 1 + x 2 (Objective Function) Subject to:(Constraints) 3x 1 + x 2 + A 1 = 3………….(4) 4x 1 + 3x 2 – S 1 + A 2 = 6………….(5) x 1 + 2x 2 + S 2 = 4………….(6) x 1, x 2, S 1, S 2, A 1 + A 2 ≥ 0

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Solution of Artificial Variable Questions which involve artificial variables can not solve straight away. We have to use following two methods to solve it 1. Penalty Method or M-Method (Big M Method) 2. Two Phase Method In this lecture we will solve this problem by Two Phase Method

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Solution by Two Phase Method Phase I: In Phase I we introduce artificial objective function Capital “F” as the sum of artificial variables introduced in the constraints. We substitute the values of artificial variables from the constraints and get artificial objective function. We have Two artificial variables so the artificial objective function would be as under; F = A1 + A2

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Artificial Objective Function A1 Use Equation no 4 to find the value of A1 3x1+ x2 + A1= 3 A1 = 3 - 3x1- x2 A2 Use Equation no 5 to find the value of A2 4x 1 + 3x 2 – S 1 + A 2 = 6 A 2 = 6 - 4x 1 - 3x 2 + S 1

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F = A 1 + A 2 Putting the values of A1 and A2 F = ( 3 – 3X1- X2) + (6 – 4X1- 3X2 + S1) =9 - 7x 1 -4 x 2 + S 1 Minimize f = 4x1+ x2 F=9 - 7x 1 -4 x 2 + S 1 Subject to: 3x1+ x2 + A1 = 3………….(4) 4x1+ 3x2 – S1 + A2= 6………….(5) x1+ 2x2 + S2 = 4………….(6) x 1, x 2, S 1, S 2, A 1 + A 2 ≥0

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Initial Feasible Solution Arbitrary values =# of Variables -- # of Equation 6-3 = 3 Let x1= 0, x2 = 0, S1 = 0 Putting above values in objective functions and equation 4-6, F = 9 f = 0 A1 = 3 A2 = 6 S2 = 4 This solution is called initial feasible solution because it satisfies the all Non negativity constraint and also do not have any contradiction or violation

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Initial Tableau BasicsX1X1 X2X2 S1S1 S2S2 A1A2RHSRatio A /3=1(Min) A /4=1.5 S /1=4 f F ↑ In artificial function of minimization we select the maximum positive as the entering variable which is X1 For leaving variable the rule is same as maximization, the variable with minimum ratio;A1

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Optimal solution for Artificial objective Function The solution of artificial objective function is said to be optimal when artificial objective functions coefficients become non-positive or zero

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Calculation [11/ /3 0 1] New Pivot Row = 1 XOld Pivot Row Pivot No. New Pivot Row = 1 X[ ] 3

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Calculation New Row = Old Row – Pivot Column Coefficient x New Pivot Row New A2 Row = [ ] - (4)[11/3001/301] = [05/3-10-4/312] New S2 Row =[ ] - (1) [11/3001/301] = [05/301-1/303] New f Row = [ ] - (-4) [11/3001/301] = [01/3004/304] New F Row = [ ] - (7) [11/3001/301] = [05/3-10-7/302]

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Tableau I BasicsX1X2X2 S1S1 S2S2 A1A2RHSRatio X111/ ÷1/3=3 A205/30-4/312 2 ÷5/3=1.2 (Min) S205/301-1/3033/5 ÷ 3=1.8 f01/3004/304 F05/30-7/302 ↑ Still there is one positive coefficient so we need to make further tableau

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Calculation [0 1-3/5 0 -4/5 3/56/5] New Pivot Row = 1 XOld Pivot Row Pivot No. New Pivot Row = 1 X[05/3-10-4/312] 5/3

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Calculation New Row = Old Row – Pivot Column Coefficient x New Pivot Row New X1 Row = [11/3001/301] -(1/3)[01-3/50-4/53/56/5] = [101/503/5-1/53/5] New S2 Row = [05/301-1/303] -(5/3)[01-3/5 0-4/53/56/5] = [ ]

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Calculation New f Row = [01/ /3 04] -(1/3)[01-3/5 0-4/53/56/5] = [001/5 08/5-1/518/5] New F Row [05/3-10-7/302] -(5/3)[01-3/50-4/53/56/5] = [ ]

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Tableau II BasicsX1X2X2 S1S1 S2S2 A1A2RHS X1101/503/5-1/53/5 X201-3/50-4/53/56/5 S f001/508/5-1/518/5 F Now there is no positive value in the artificial function so this is the end of Phase I

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Phase II In phase two rewrite the previous tableau by dropping the artificial objective function and artificial variables BasicsX1X2X2 S1S1 S2S2 RHSRatio X1101/503/53/5÷1/5=3 X201-3/506/5 S /1=1(Min) f001/5018/5 ↑

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Calculation [ ] New Pivot Row = 1 XOld Pivot Row Pivot No. New Pivot Row = 1 X[00111] 1

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Calculation New Row = Old Row – Pivot Column Coefficient x New Pivot Row New X1 Row = [101/5 03/5] -(1/5) [ ] = [100-1/5 2/5] New X2 Row = [01-3/5 06/5] -(-3/5)[ ] = [01 03/59/5]

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Calculation New f Row = [00 1/5 0 18/5] -(1/5) [ ] = [ /5 17/5]

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Tableau BasicsX1X2X2 S1S2S2 RHS X1100-1/52/5 X20003/59/5 S f000-1/517/5 Now there is no positive value in the objective function so this is the optimal point

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Optimal Solution X1 = 2/5 X2 = 9/5 f = 17/5 Cross checking of maximization point put values of X1 and X2 from above solution into original objective function f=4x 1 + x 2 =4 (2/5) + (9/5) =17/5

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