Question

In: Operations Management

Find the optimum solution to the following LP using the Simplex Algorithm. Use Big M method....

Find the optimum solution to the following LP using the Simplex Algorithm. Use Big M method.

??? ?=5?1+2?2+?3 ??
?1 + ?2 ≤ 6
?1 − ?2 + 2?3 ≤ 2

3?1 + 2?2 = 12

?1,?2,?3 ≥0

Solutions

Expert Solution

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Solution:

Problem is

Max Z = 5 x1 + 2 x2 + x3
subject to
x1 + x2 6
x1 - x2 + 2 x3 2
3 x1 + 2 x2 = 12
and x1,x2,x3≥0;



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

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 artificial variable A1

After introducing slack,artificial variables

Max Z = 5 x1 + 2 x2 + x3 + 0 S1 + 0 S2 - M A1
subject to
x1 + x2 + S1 = 6
x1 - x2 + 2 x3 + S2 = 2
3 x1 + 2 x2 + A1 = 12
and x1,x2,x3,S1,S2,A1≥0


Iteration-1 Cj 5 2 1 0 0 -M
B CB XB x1 x2 x3 S1 S2 A1 MinRatio
XBx1
S1 0 6 1 1 0 1 0 0 61=6
S2 0 2 (1) -1 2 0 1 0 21=2
A1 -M 12 3 2 0 0 0 1 123=4
Z=-12M Zj -3M -2M 0 0 0 -M
Zj-Cj -3M-5 -2M-2 -1 0 0 0



Negative minimum Zj-Cj is -3M-5 and its column index is 1. So, the entering variable is x1.

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

The pivot element is 1.

Entering =x1, Departing =S2, Key Element =1






Iteration-2 Cj 5 2 1 0 0 -M
B CB XB x1 x2 x3 S1 S2 A1 MinRatio
XBx2
S1 0 4 0 2 -2 1 -1 0 42=2
x1 5 2 1 -1 2 0 1 0 ---
A1 -M 6 0 (5) -6 0 -3 1 65=1.2
Z=-6M+10 Zj 5 -5M-5 6M+10 0 3M+5 -M
Zj-Cj 0 -5M-7 6M+9 0 3M+5 0



Negative minimum Zj-Cj is -5M-7 and its column index is 2. So, the entering variable is x2.

Minimum ratio is 1.2 and its row index is 3. So, the leaving basis variable is A1.

The pivot element is 5.

Entering =x2, Departing =A1, Key Element =5






Iteration-3 Cj 5 2 1 0 0
B CB XB x1 x2 x3 S1 S2 MinRatio
S1 0 1.6 0 0 0.4 1 0.2
x1 5 3.2 1 0 0.8 0 0.4
x2 2 1.2 0 1 -1.2 0 -0.6
Z=18.4 Zj 5 2 1.6 0 0.8
Zj-Cj 0 0 0.6 0 0.8



Since all Zj-Cj≥0

Hence, optimal solution is arrived with value of variables as :
x1=3.2,x2=1.2,x3=0
Max Z=18.4

PLEASE RATE THE ANSWER, I have spent 2 hours to solve and format !!

PLEASE RATE THE ANSWER, I have spent 2 hours to solve and format !!

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