Suppose following simplex table: 2 3 1 -1 0 0 -100 x1 x2 x3 x4 x5 x6 x7 b x5 0 7/2 0 0 3/2 1 1/2 -3/2 14 x3 1 5/2 0 1 1/2 0 1/2 1/2 13 x2 3 0 1 0 1 0 0 1 17 Z - C We can say, that: Select one: The solution is not optimal. The solution is optimal. The solution can be the initial solution.
Suppose following simplex table: 2 3 1 -1 0 0 -100 x1 x2 x3 x4 x5 x6 x7 b x5 0 7/2 0 0 3/2 1 1/2 -3/2 14 x3 1 5/2 0 1 1/2 0 1/2 1/2 13 x2 3 0 1 0 1 0 0 1 17 Z - C We can say, that: Select one: The solution is not optimal. The solution is optimal. The solution can be the initial solution.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
100%
Suppose following simplex table:
2 |
3 |
1 |
-1 |
0 |
0 |
-100 |
||||
x1 |
x2 |
x3 |
x4 |
x5 |
x6 |
x7 |
b |
|||
x5 |
0 |
7/2 |
0 |
0 |
3/2 |
1 |
1/2 |
-3/2 |
14 |
|
x3 |
1 |
5/2 |
0 |
1 |
1/2 |
0 |
1/2 |
1/2 |
13 |
|
x2 |
3 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
17 |
|
Z - C |
We can say, that:
Select one:
The solution is not optimal.
The solution is optimal.
The solution can be the initial solution.
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