please Solve This!!
Sice optimal table is
Z | x1 | x2 | x3 | x4 | rhs | |
1 | 2 | 1 | 0 | 0 | 10 | |
0 | 3 | 2 | 1 | 0 | 3 | |
0 | 4 | 3 | 0 | 1 | 5 |
Optimal solution is z=10 for x1=x2=0, x3=3,x4=5
To find the second best basic feasible solution i.e.bfs
One pivot away from the optimal solution we choose x2 as the entering variable bcoz it gives the smallest decrease in objective value z and corresponding leaving variable is x3 according to the ratio test
After two iteration new table is determined by applying
z | x1 | x2 | x3 | x4 | rhs | |
1 | 1/2 | 0 | -1/2 | 0 | 17/2 | |
0 | 3/2 | 1 | 1/2 | 0 | 3/2 | |
0 | -1/2 | 0 | -3/2 | 1 | 1/2 |
So second best bfs is x2=3/2 ; x4=1/2 ;z=17/2
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