Simplex Method MINIMIZATION BY DR ARSHAD ZAHEER Penalty

Simplex Method MINIMIZATION BY DR. ARSHAD ZAHEER

Penalty Method Big M Method

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

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

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

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

Solution of Artificial Variable Questions in which we introduce 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 Penalty Method

Solution by Penalty Method (Big M Method) Let M > >0 (M is very large positive No. ) f =4 x 1+ x 2 + MA 1 + MA 2 ……. <7> we put penalty to artificial variables �A 1 Use Equation no 4 to find the value of A 1 3 x 1+ x 2 + A 1= 3 A 1 = 3 - 3 x 1 - x 2

�A 2 Use Equation no 5 to find the value of A 2 4 x 1+ 3 x 2 – S 1 + A 2= 6 - 4 x 1 - 3 x 2 + S 1 Putting these values of A 1 and A 2 in eq. <7> we get f = 4 x 1+ x 2 + M(3 - 3 x 1 - x 2)+ M(6 - 4 x 1 - 3 x 2 + S 1) =4 x 1+ x 2 + 3 M– 3 Mx 1 - Mx 2)+ 6 M– 4 Mx 1 - 3 Mx 2 + MS 1 =(4 – 7 M)x 1+ (1 – 4 M)X 2 + MS 1 + 9 M = 9 M +(– 7 M + 4)x 1+ (– 4 M+1)X 2 + MS 1

Problem set f = (– 7 M + 4)x 1+ (– 4 M+1)X 2 + MS 1+ 9 M Subject to: 3 x 1+ x 2+ A 1 = 3 …………. 4 x 1+ 3 x 2 – S 1 + A 2= 6 …………. x 1+ 2 x 2+ S 24 …………. x 1, x 2, S 1 , S 2 , A 1 + A 2 ≥ 0 (4) (5) (6)

Initial Feasible Solution Arbitrary values =# of Variables -- # of Equation 6 3=3 This solution is called initial feasible Let x 1= 0, x 2 = 0, S 1 = 0 solution because it satisfies the Non Putting above values in objective functions negativity constraint and also do not and equation 4 -6, have any contradiction or violation f = 9 M A 1 = 3 A 2 = 6 S 2 = 4

Initial Tableau We Thisselect tableau the entering is not satisfied and leaving because variable, our criteria the entering for minimization variable will is be non with positivity the highest of all positive the coefficients objective of function objectivecoefficient function so and we leaving will iterate it variable further will be with minimum ratio

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

Calculation New Row = Old Row – Pivot Column Coefficient x New Pivot Row New A 2 Row = [4 -1 0 0 0 0 1/3 -4/3 1 0 1 6] 1] 2] New S 2 Row =[1 0 0 0 1 0 1/3 -1/3 0 0 0 4] 1] 3] -M 0 0 0 3 - (4)[1 1/3 = [0 5/3 2 - (1) [1 1/3 = [0 5/3 New f Row = [7 M-4 4 M-1 - (7 M-4) [1 1/3 [0 5 M+1/3 0 0 1/3 0 -7 M+4/3 0 9 M] 1] 2 M+4]

Tableau 1 Basis X 1 X 2 S 1 S 2 A 2 RHS Ratio This tableau does not. A 1 fulfill condition X 1 1 1÷ 1/3=3 1/3 0 0 1/3 0 1 of. A 2 optimality because our criteria 2÷ 5/3= for 0 5/3 -1 non 0 -4/3 1 minimization is positivity of 2 all 1. 2(min) the S 2 coefficients of 0 objective function so 3÷ 5/3= we 0 5/3 1 -1/3 0 3 1. 8 will iterate it further f 0 5 M+1/3 -M 0 -7 M+4/3 0 2 M+4

Calculation New Pivot Row = New Pivot 1 x [0 Row = 5/3 = = 3 5 x [0 [0 1 X Old Pivot Row Pivot No. 5/3 -1 0 -4/3 1 2] 1 -3/5 0 -4/5 3/5 6/5]

Calculation New Row = Old Row – Pivot Column Coefficient x New Pivot Row New X 1 Row = [1 0 -3/5 1/5 0 0 0 1/3 -4/5 3/5 0 3/5 -1/5 1] 6/5] 3/5] New S 2 Row =[0 0 -3/5 1 1 0 1 -1/3 -4/5 1 0 3/5 -1 3] 6/5] 1] 0 0 -7 M+4/3 -4/5 0 3/5 2 M+4] 6/5] 1/3 - (1/3) [0 1 = [1 0 5/3 -(5/3) [0 1 = [0 0 New f Row = [0 5 M+1/3 -M -(5 M+1/3) [0 1 -3/5 0 0 1/5 0 -15 M+24/15 -5 M-1/5 18

Tableau 2 Basis X 1 X 2 S 1 S 2 A 1 A 2 RHS Ratio 3/5÷ 1/5 because 3/5 =3 This tableau 1 0 1/5 is 0 not 3/5 satisfied -1/5 our for 0 minimization is non X 2 criteria 0 1 -3/5 -4/5 3/5 6/5 positivity of all the coefficients 1/1=1 of S 2 0 0 1 1 1 -1 1 objective function so we will iterate(min) it f 0 0 1/5 0 -15 M+24 -5 M-1/5 18/5 further /15

Calculation New Pivot Row = = 1 1 1 X Old Pivot Row Pivot No. X [0 0 1 1 1 -1 1]

Calculation New Row = Old Row – Pivot Column Coefficient x New Pivot Row New X 1 Row = [1 1/5 1 0 0 1 -1/5 3/5 1 2/5 -1 0 3/5] 1] 2/5] New X 2 Row =[0 -3/5 1 0 0 1 3/5 -4/5 1 -1/5 3/5 -1 0 6/5] 1] 9/5] 0 -(1/5) [0 0 = [1 0 1 -(3/5) [0 0 = [0 1 New f Row = [0 -(1/5) [0 [0 0 1 5 1 0 0 -15 M+24 -5 M-1 18 15 5 5 1 -1/5 1 -1 -15 M+21/15 1] -M 17/5]

Tableau 3 This tableau is SATISFIED because our Basis X 1 X 2 for S 1 S 2 A 1 A 2 RHS Ratio criteria minimization is non X 1 1 0 0 -1/5 2/5 0 2/5 positivity of all the coefficients of X 2 0 1 0 3/5 -1/5 0 9/5 objective function which is attained S 1 0 0 1 1 1 -1 1 f 0 0 0 -1/5 -15 M+21 /15 -M 17/5

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

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