Answer to Question #237073 in Operations Research for opr

Question #237073

Solve the following linear programming problem using the simplex method:

š‘€š‘Žš‘„š‘–š‘šš‘–š‘§š‘’ š‘§ = 14š‘„ + 15š‘¦

š‘ š‘¢š‘š‘—š‘’š‘š‘” š‘”š‘œ

13š‘„ + 15š‘¦ ≤ 80

āˆ’12š‘„ āˆ’ 17š‘¦ ≄ āˆ’120

š‘„ ≄ 0, š‘¦ ≄ 0


1
Expert's answer
2021-09-28T02:21:40-0400

Replace "x" by "x_1" and "y" by "x_2."

Problem is: "\\operatorname{Max} Z=14 x_{1}+15 x_{2}"

subject to:

"\\begin{array}{rllll} \n& 13 & x_{1} & + & 15 & x_{2} & \\leq 80 \\\\\n- & 12 & x_{1} & - & 17 & x_{2} & \\leq-120\n\\end{array}"

Here, "b_{2}=-120<0"

so multiply this constraint byĀ "-1" to make "b_{2}>0 ."

"12 x_{1}+17 \\quad x_{2} \\geq 120\n \n \\ and \\ x_{1}, x_{2} \\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 slack variableĀ "S_1"


2. As constraint-2 is of type '≄' we should subtract surplus variableĀ "S_2" Ā and add artificial variableĀ "A_1."


After introducing slack, surplus,artificial variables:


"\\operatorname{Max} Z=14 x_{1}+15 x_{2}+0 S_{1}+0 S_{2}-M A_{1}"

subject to:

"13 x_{1}+15 x_{2}+S_{1} \\quad=80 \\\\ 12 x_{1}+17 x_{2} \\quad-S_{2}+A_{1}=120 \\\\ and \\ x_{1}, x_{2}, S_{1}, S_{2}, A_{1} \\geq 0"




Negative minimumĀ Zj-CjĀ isĀ -17M-15Ā and its column index isĀ 2. So,Ā the entering variable isĀ x2.


Minimum ratio isĀ 5.3333Ā and its row index isĀ 1. So,Ā the leaving basis variable isĀ S1.


∓ The pivot element is 15.


EnteringĀ =x2, DepartingĀ =S1, Key ElementĀ =15






Since allĀ Zj-Cj≄0


Hence, the optimal solution arrives with value of variables as :

x1=0,x2=5.3333


MaxĀ Z=80


But this solution is not feasible

because the final solution violates theĀ 2ndĀ constraintĀ Ā 12Ā x1Ā +Ā 17Ā x2 ≄ 120.


and the artificial variableĀ A1Ā appears in the basis with a positive valueĀ of 29.3333


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