Work through the simplex method (in algebraic form)
step by step to solve the following problem.
Maximize Z = 4?1 + 3?2 + 6?3,
Subject to
3?1 + ?2 + 3?3 ≤ 30
2?1 + 2?2 + 3?3 ≤ 40
and
?1 ≥ 0, ?2≥ 0, ?3 ≥ 0.
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
After introducing slack variables
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subject to | ||||||||||||||||||||||||||||||||||
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and x1,x2,x3,S1,S2≥0 |
Iteration-1 | Cj | 4 | 3 | 6 | 0 | 0 | ||
B | CB | XB | x1 | x2 | x3 | S1 | S2 | MinRatio XB/x3 |
S1 | 0 | 30 | 3 | 1 | 2 | 1 | 0 | 30/2=15 |
S2 | 0 | 40 | 2 | 2 | (3) | 0 | 1 | 40/3=13.3333→ |
Z=0 | Zj | 0 | 0 | 0 | 0 | 0 | ||
Zj-Cj | -4 | -3 | -6↑ | 0 | 0 |
Negative minimum Zj-Cj is -6
and its column index is 3. So, the entering variable is
x3.
Minimum ratio is 13.3333 and its row index is 2. So, the leaving
basis variable is S2.
∴ The pivot element is 3.
Entering =x3, Departing =S2, Key Element =3
R2(old) = | 40 | 2 | 2 | 3 | 0 | 1 |
R2(new)=R2(old)÷3 | 13.3333 | 0.6667 | 0.6667 | 1 | 0 | 0.3333 |
R1(new)=R1(old) - 2R2(new)
R1(old) = | 30 | 3 | 1 | 2 | 1 | 0 |
R2(new) = | 13.3333 | 0.6667 | 0.6667 | 1 | 0 | 0.3333 |
2×R2(new) = | 26.6667 | 1.3333 | 1.3333 | 2 | 0 | 0.6667 |
R1(new)=R1(old) - 2R2(new) | 3.3333 | 1.6667 | -0.3333 | 0 | 1 | -0.6667 |
Iteration-2 | Cj | 4 | 3 | 6 | 0 | 0 | ||
B | CB | XB | x1 | x2 | x3 | S1 | S2 | MinRatio |
S1 | 0 | 3.3333 | 1.6667 | -0.3333 | 0 | 1 | -0.6667 | |
x3 | 6 | 13.3333 | 0.6667 | 0.6667 | 1 | 0 | 0.3333 | |
Z=80 | Zj | 4 | 4 | 6 | 0 | 2 | ||
Zj-Cj | 0 | 1 | 0 | 0 | 2 |
Since all Zj-Cj≥0
Hence, optimal solution is arrived with value of variables as
:
x1=0,x2=0,x3=13.3333
Max Z=80
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