Question #185367

Minimize Z = 0.2x1+ O.lx2 + 0.3x3

Subject to constraint

0. Sx1+ 0.2x2 + 0.7x3 = 0.420

0. 3x1+ 0.2x2 + O.Sx3 ;:::: 0.280

Xv Xz, X3 ;:::: 0.


(i) W1ite second initial basic solution of the p1imal problem using M-Method.


(ii) Solve the dual from optimal Primal table calculated in (i).


1
Expert's answer
2021-04-28T03:40:40-0400

(i) Minimize  Z=0.2x1+0.1x2+0.3x3Minimize\ \ Z=0.2x_1+0.1x_2+0.3x_3\\

Subject to constraint

0.5x1+0.2x2+0.7x3=0.4200.3x1+0.2x2+0.5x30.280x1,x2,x300. 5x_1+ 0.2x_2 + 0.7x_3 = 0.420\\ 0. 3x_1+ 0.2x_2 + 0.5x_3 \geq 0.280\\x_1,x_2,x_3\geq 0


The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate


1. As the constraint-1 is of type '=' we should add artificial variable A1


2. As the constraint-2 is of type '≥' we should subtract surplus variable S1 and add artificial variable A2


After introducing surplus, artificial variablesMin Z=0.2x1+0.1x2+0.3x3+0S1+MA1+MA2subject to0.5x1+0.2x2+0.7x3+A1=0.420.3x1+0.2x2+0.5x3S1+A2=0.28and  x1,x2,x3,S1,A1,A20\text{After introducing surplus, artificial variables}\\ Min\ Z = 0.2 x_1 + 0.1 x_2 + 0.3 x_3 + 0 S_1 + M A_1 + M A_2\\ subject \ to\\ 0.5 x_1 + 0.2 x_2 + 0.7 x_3 + A_1 = 0.42\\ 0.3 x_1 + 0.2 x_2 + 0.5 x_3 - S_1 + A_2 = 0.28\\ and\ \ x_1,x_2,x_3,S_1,A_1,A_2≥0





Positive maximum ZjCjZ_j-C_j  is 1.2M-0.3 and its column index is 3. So, the entering variable is x3.


Minimum ratio is 0.56 and its row index is 2. So, the leaving basis variable is A2.


∴ The pivot element is 0.5.


Entering =x3, Departing =A2, Key Element =0.5






Positive maximum Zj-Cj is 1.4M-0.6 and its column index is 4. So, the entering variable is S1.


Minimum ratio is 0.02 and its row index is 1. So, the leaving basis variable is A1.


∴ The pivot element is 1.4.


Entering =S1, Departing =A1, Key Element =1.4





Positive maximum Zj-Cj is 0.0143 and its column index is 1. So, the entering variable is x1.


Minimum ratio is 0.35 and its row index is 1. So, the leaving basis variable is S1.


∴ The pivot element is 0.0571.


Entering =x1, Departing =S1, Key Element =0.0571





Since all Zj-Cj≤0


Hence, optimal solution is arrived with value of variables as :

x1=0.35,x2=0,x3=0.35


Min Z=0.175



(ii) Dual is :

Max Z=0.42y1+0.28y2subject to 0.5y1+0.3y20.20.2y1+0.2y20.10.7y1+0.5y20.3and y20; y1    unrestricted in signSince y1 is unrestricted in sign,introduce the nonnegative variables y1,y1′′so that y1=y1y1′′; y1,y1′′0.Max\ Z = 0.42 y _1 + 0.28 y_2\\ subject\ to\ \\ 0.5 y _1 + 0.3 y_2 ≤ 0.2\\ 0.2 y_1 + 0.2 y_2 ≤ 0.1\\ 0.7 y_1 + 0.5 y_2 ≤ 0.3\\\\ and\ y _2≥0;\ y _1\ \ \ \text{ unrestricted in sign}\\ Since \ y _1 \ is \ unrestricted\ in\ sign, introduce\ the\ non-negative\ variables\ y_1′,y_1′′\\ so \ that\ y_1=y_1′-y_1′′;\ y_1′,y_1′′≥0.


The standard form of the LP problem becomes

Max Z=0.42y10.42y1′′+0.28y2subject to0.5y10.5y1′′+0.3y20.20.2y10.2y1′′+0.2y20.10.7y10.7y1′′+0.5y20.3and  y1,y1′′,y20;Max\ Z = 0.42 y _1′ - 0.42 y_1′′ + 0.28 y_2\\ subject\ to\\ 0.5 y_1′ - 0.5 y_1′′ + 0.3 y_2 ≤ 0.2\\ 0.2 y_1′ - 0.2 y_1′′ + 0.2 y_2 ≤ 0.1\\ 0.7 y_1′ - 0.7 y_1′′ + 0.5 y_2 ≤ 0.3\\ and \ \ y_1′,y_1′′,y_2≥0;


The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate


1. As the constraint-1 is of type '≤' we should add slack variable S1


2. As the constraint-2 is of type '≤' we should add slack variable S2


3. As the constraint-3 is of type '≤' we should add slack variable S3



After introducing slack variables

Max Z=0.42y10.42y1′′+0.28y2+0S1+0S2+0S3subject to0.5y10.5y1′′+0.3y2+S1=0.20.2y10.2y1′′+0.2y2+S2=0.10.7y10.7y1′′+0.5y2+S3=0.3and  y1,y1′′,y2,S1,S2,S30Max \ Z = 0.42 y_1′ - 0.42 y_1′′ + 0.28 y_2 + 0 S_1 + 0 S_2 + 0 S_3\\ subject\ to\\ 0.5 y_1′ - 0.5 y_1′′ + 0.3 y_2 + S_1 = 0.2\\ 0.2 y_1′ - 0.2 y_1′′ + 0.2 y_2 + S_2 = 0.1\\ 0.7 y_1′ - 0.7 y_1′′ + 0.5 y_2 + S_3 = 0.3\\ and\ \ y_1′,y_1′′,y_2,S_1,S_2,S_3≥0




Negative minimum Zj-Cj is -0.42 and its column index is 1. So, the entering variable is y1′.


Minimum ratio is 0.4 and its row index is 1. So, the leaving basis variable is S1.


∴ The pivot element is 0.5.


Entering =y1′, Departing =S1, Key Element =0.5





Negative minimum Zj-Cj is -0.028 and its column index is 3. So, the entering variable is y2.


 Minimum ratio is 0.25 and its row index is 2. So, the leaving basis variable is S2.


∴ The pivot element is 0.08.


Entering = y2, Departing =S2, Key Element =0.08




Since all ZjCj0Z_j-C_j≥0


Hence, optimal solution is arrived with value of variables as :

y1=0.25,y1′′=0,y2=0.25y_1′=0.25,y_1′′=0,y_2=0.25


Max Z=0.175



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