written 6.3 years ago by
teamques10
★ 64k
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• modified 4.0 years ago |
Using simplex method solve the following L.P.P
Max. $Z = 4x_1 +3x_2 + 6x_3$, subject to $2x_1 + 3x_2 +2x_3 ≤ 440 ; 4x_1 +3x_3≤470 ; 2x_1 + 5x_2 ≤ 430$
$x_1 , x_2 , x_3 ≥ 0 $
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written 6.1 years ago by
teamques10
★ 64k
|
• modified 6.0 years ago |
We express the given problem in standard form ,taking slack variables
$ s_1, s_2, s_3$
$z – 4x_1 – 3x_2 – 6x_3 + 0s_1 +0s_2+ 0s_3 = 0\\ 2x_1 + 3x_2 + 2x_3 +s_1 +0s_2+0s_3 = 440 \\ 4x_1 +0x_2 +3x_3 +0s_1 +s_2+0s_3 = 470 \\ 2x_1 + 5x_2+0x_3+0s_1 +0s_2+s_3 = 430$
Now we form the simplex table as
| Iteration no. | Basic variables | $\frac{coefficients of}{x_1,x_2,x_3,s_1,s_2,s_3}$ | R.H.S solution | Ratio |
|---|---|---|---|---|
| 0 | z | -4, -3, -6, 0, 0, 0 | 0 | |
| $s_2$ leaves | $s_1$ | 2, 3, 2, 1, 0, 0 | 440 | 440/2=220 |
| $x_3$ enters | $s_2$ | 4, 0, 3*, 0, 1, 0 | 470 | 470/3=156.67 |
| $s_3$ | 2, 5, 0, 0, 0, 1 | 430 | 430/0 = - | |
| 1 | z | 4, -3, 0, 0, 2, 0 | 940 | |
| $s_1$ leaves | $s_1$ | -2/3, 3*, 0, 1, - 2/3, 0 | 380/3 | 380/9=42.2 |
| $x_2$ enters | $x_3$ | 4/3, 0, 1, 0, 1/3, 0 | 470/3 | |
| $s_3$ | 2, 5, 0, 0, 0, 1 | 430 | 430/5=86 | |
| 2 | z | 10/3, 0, 0, 1, 4/3, 0 | 3200/3 | |
| $x_2$ | -2/9, 1, 0, 1/3, -2/9, 0 | 380/9 | ||
| $x_3$ | 4/3, 0, 1, 0, 1/3, 0 | 470/3 | ||
| $s_3$ | 28/9, 0, 0, -5/3, 10/9, 1 | 1970/9 |
Since all the elements in the row of z becomes positive ,we get solution from R.H.S.
$x_2=380/9 , x_3 = 470/3 , x_1=0 , Z_{Max} = 3200/3 $
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