In: Operations Management
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
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 !!
PLEASE RATE THE ANSWER, I have spent 2 hours to solve and format !!
Solution:
Problem is
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subject to | |||||||||||||||||||||||||||||||||
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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
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subject to | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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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 !!
PLEASE RATE THE ANSWER, I have spent 2 hours to solve and format !!