MAXIMIZE C= 15x+11y+4z
Subject To,
8x+10y+17z ≤ 62
15x+18y+11z ≤ 124
x, y, z ≥ 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.
Re-writing the problem introducing the slack variables we get-
MAXIMISE C= 15x+11y+4z+0S1+0S2
Subject To,
8x+10y+17z+ S1 = 62
15x+18y+11z+ S2 = 124
x, y, z, S1, S2 ≥ 0
BV |
Basic Table |
|||||||
Zj |
15 |
11 |
4 |
0 |
0 |
Solution |
Replacement Ratio |
|
CB |
x |
y |
z |
s1 |
s2 |
|||
s1 |
0 |
8 |
10 |
17 |
1 |
0 |
62 |
(62/8) = 7.75 |
s2 |
0 |
15 |
18 |
11 |
0 |
1 |
124 |
(124/15) = 8.267 |
Cj |
0 |
0 |
0 |
0 |
0 |
0 |
||
Cj-Zj |
-15 |
-11 |
-4 |
0 |
0 |
Maximization requires All Cj-Zj ≥ 0.
Since All Cj-Zj in Basic Table is less than or equal to zero, we will form the Iteration- 1 follows.
In Iteration-1:
x1 is entering variable and
s1 is departing variable
Key Element 8
BV |
Iteration 1 |
|||||||
Zj |
15 |
11 |
4 |
0 |
0 |
Solution |
Replacement Ratio |
|
CB |
x |
y |
z |
s1 |
s2 |
|||
x |
15 |
1 |
1.25 |
2.125 |
0.125 |
0 |
7.75 |
|
s2 |
0 |
0 |
-0.75 |
-20.875 |
-1.875 |
1 |
7.75 |
|
Cj |
15 |
18.75 |
31.875 |
1.875 |
0 |
116.25 |
||
Cj-Zj |
0 |
7.75 |
27.875 |
1.875 |
0 |
Since all Zj-Cj ≥ 0
Hence, optimal solution is arrived with value of variables
as:
x =
7.75, y
= 0,
z =
0
Max C = 15x + 11y + 4z =
(15x7.75) + (11x0)+ (4x0) =
116.25
You can also see the below images of the table
Basic Table:
Iteration 1:
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