Mixing
Proving certificate of optimality
$begingroup$
suppose we have the following linear program max{$c^Tx+z: Ax=b, x geq0$} and that B is an optimal basis with $bar{x}$ being the basic feasible solution corresponding to B. How would I be able to prove that $y=A_B^{-T}c_B$ is a certificate of optimality for $bar{x}$?
I know that I have to show
1) $A^Ty geq c$
2) $c^Tx=y^Tb$
I have been able to show 2) but am struggling on showing 1)
linear-algebra convex-optimization linear-programming
asked Jan 28 at 17:11
SkrrrrrttttSkrrrrrtttt
387110
$endgroup$
add a comment |
$begingroup$
suppose we have the following linear program max{$c^Tx+z: Ax=b, x geq0$} and that B is an optimal basis with $bar{x}$ being the basic feasible solution corresponding to B. How would I be able to prove that $y=A_B^{-T}c_B$ is a certificate of optimality for $bar{x}$?
I know that I have to show
1) $A^Ty geq c$
2) $c^Tx=y^Tb$
I have been able to show 2) but am struggling on showing 1)
linear-algebra convex-optimization linear-programming
asked Jan 28 at 17:11
SkrrrrrttttSkrrrrrtttt
387110
$endgroup$
add a comment |
$begingroup$
suppose we have the following linear program max{$c^Tx+z: Ax=b, x geq0$} and that B is an optimal basis with $bar{x}$ being the basic feasible solution corresponding to B. How would I be able to prove that $y=A_B^{-T}c_B$ is a certificate of optimality for $bar{x}$?
I know that I have to show
1) $A^Ty geq c$
2) $c^Tx=y^Tb$
I have been able to show 2) but am struggling on showing 1)
linear-algebra convex-optimization linear-programming
asked Jan 28 at 17:11
SkrrrrrttttSkrrrrrtttt
387110
$endgroup$
suppose we have the following linear program max{$c^Tx+z: Ax=b, x geq0$} and that B is an optimal basis with $bar{x}$ being the basic feasible solution corresponding to B. How would I be able to prove that $y=A_B^{-T}c_B$ is a certificate of optimality for $bar{x}$?
I know that I have to show
1) $A^Ty geq c$
2) $c^Tx=y^Tb$
I have been able to show 2) but am struggling on showing 1)
linear-algebra convex-optimization linear-programming
linear-algebra convex-optimization linear-programming
asked Jan 28 at 17:11
SkrrrrrttttSkrrrrrtttt
387110
asked Jan 28 at 17:11
SkrrrrrttttSkrrrrrtttt
387110
asked Jan 28 at 17:11
SkrrrrrttttSkrrrrrtttt
387110
asked Jan 28 at 17:11
SkrrrrrttttSkrrrrrtttt
387110
asked Jan 28 at 17:11
SkrrrrrttttSkrrrrrtttt
387110
387110
add a comment |
add a comment |
1 Answer
1
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$begingroup$
From $y=A_B^{-T}c_B$ you multiply with $A_B^T$ to get $A_B^Ty = c_B$.
The tableau is optimal, so $c_B^T A_B^{-1} A_N - c_N^T geq 0$, so $y^T A_N geq c_N^T$.
Combining the two you get $A^Ty geq c$.
answered Jan 29 at 1:47
LinAlgLinAlg
10.1k1521
$endgroup$
$begingroup$
why does the inequality follow from the fact that the basis is optimal?
$endgroup$
– Skrrrrrtttt
Jan 29 at 1:48
$begingroup$
@Skrrrrrtttt look at the first row of the simplex tableau
$endgroup$
– LinAlg
Jan 29 at 1:55
$begingroup$
sorry can you explain it a little more depth please
$endgroup$
– Skrrrrrtttt
Jan 29 at 1:58
1
$begingroup$
@Skrrrrrtttt just perform one iteration of revised simplex, you should find the formula in the step where you check for optimality.
$endgroup$
– LinAlg
Jan 29 at 2:00
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
From $y=A_B^{-T}c_B$ you multiply with $A_B^T$ to get $A_B^Ty = c_B$.
The tableau is optimal, so $c_B^T A_B^{-1} A_N - c_N^T geq 0$, so $y^T A_N geq c_N^T$.
Combining the two you get $A^Ty geq c$.
answered Jan 29 at 1:47
LinAlgLinAlg
10.1k1521
$endgroup$
$begingroup$
why does the inequality follow from the fact that the basis is optimal?
$endgroup$
– Skrrrrrtttt
Jan 29 at 1:48
$begingroup$
@Skrrrrrtttt look at the first row of the simplex tableau
$endgroup$
– LinAlg
Jan 29 at 1:55
$begingroup$
sorry can you explain it a little more depth please
$endgroup$
– Skrrrrrtttt
Jan 29 at 1:58
1
$begingroup$
@Skrrrrrtttt just perform one iteration of revised simplex, you should find the formula in the step where you check for optimality.
$endgroup$
– LinAlg
Jan 29 at 2:00
add a comment |
$begingroup$
From $y=A_B^{-T}c_B$ you multiply with $A_B^T$ to get $A_B^Ty = c_B$.
The tableau is optimal, so $c_B^T A_B^{-1} A_N - c_N^T geq 0$, so $y^T A_N geq c_N^T$.
Combining the two you get $A^Ty geq c$.
answered Jan 29 at 1:47
LinAlgLinAlg
10.1k1521
$endgroup$
$begingroup$
why does the inequality follow from the fact that the basis is optimal?
$endgroup$
– Skrrrrrtttt
Jan 29 at 1:48
$begingroup$
@Skrrrrrtttt look at the first row of the simplex tableau
$endgroup$
– LinAlg
Jan 29 at 1:55
$begingroup$
sorry can you explain it a little more depth please
$endgroup$
– Skrrrrrtttt
Jan 29 at 1:58
1
$begingroup$
@Skrrrrrtttt just perform one iteration of revised simplex, you should find the formula in the step where you check for optimality.
$endgroup$
– LinAlg
Jan 29 at 2:00
add a comment |
$begingroup$
From $y=A_B^{-T}c_B$ you multiply with $A_B^T$ to get $A_B^Ty = c_B$.
The tableau is optimal, so $c_B^T A_B^{-1} A_N - c_N^T geq 0$, so $y^T A_N geq c_N^T$.
Combining the two you get $A^Ty geq c$.
answered Jan 29 at 1:47
LinAlgLinAlg
10.1k1521
$endgroup$
From $y=A_B^{-T}c_B$ you multiply with $A_B^T$ to get $A_B^Ty = c_B$.
The tableau is optimal, so $c_B^T A_B^{-1} A_N - c_N^T geq 0$, so $y^T A_N geq c_N^T$.
Combining the two you get $A^Ty geq c$.
answered Jan 29 at 1:47
LinAlgLinAlg
10.1k1521
answered Jan 29 at 1:47
LinAlgLinAlg
10.1k1521
answered Jan 29 at 1:47
LinAlgLinAlg
10.1k1521
answered Jan 29 at 1:47
LinAlgLinAlg
10.1k1521
10.1k1521
$begingroup$
why does the inequality follow from the fact that the basis is optimal?
$endgroup$
– Skrrrrrtttt
Jan 29 at 1:48
$begingroup$
@Skrrrrrtttt look at the first row of the simplex tableau
$endgroup$
– LinAlg
Jan 29 at 1:55
$begingroup$
sorry can you explain it a little more depth please
$endgroup$
– Skrrrrrtttt
Jan 29 at 1:58
1
$begingroup$
@Skrrrrrtttt just perform one iteration of revised simplex, you should find the formula in the step where you check for optimality.
$endgroup$
– LinAlg
Jan 29 at 2:00
add a comment |
$begingroup$
why does the inequality follow from the fact that the basis is optimal?
$endgroup$
– Skrrrrrtttt
Jan 29 at 1:48
$begingroup$
@Skrrrrrtttt look at the first row of the simplex tableau
$endgroup$
– LinAlg
Jan 29 at 1:55
$begingroup$
sorry can you explain it a little more depth please
$endgroup$
– Skrrrrrtttt
Jan 29 at 1:58
1
$begingroup$
@Skrrrrrtttt just perform one iteration of revised simplex, you should find the formula in the step where you check for optimality.
$endgroup$
– LinAlg
Jan 29 at 2:00
$begingroup$
why does the inequality follow from the fact that the basis is optimal?
$endgroup$
– Skrrrrrtttt
Jan 29 at 1:48
$begingroup$
why does the inequality follow from the fact that the basis is optimal?
$endgroup$
– Skrrrrrtttt
Jan 29 at 1:48
$begingroup$
@Skrrrrrtttt look at the first row of the simplex tableau
$endgroup$
– LinAlg
Jan 29 at 1:55
$begingroup$
@Skrrrrrtttt look at the first row of the simplex tableau
$endgroup$
– LinAlg
Jan 29 at 1:55
$begingroup$
sorry can you explain it a little more depth please
$endgroup$
– Skrrrrrtttt
Jan 29 at 1:58
$begingroup$
sorry can you explain it a little more depth please
$endgroup$
– Skrrrrrtttt
Jan 29 at 1:58
1
1
$begingroup$
@Skrrrrrtttt just perform one iteration of revised simplex, you should find the formula in the step where you check for optimality.
$endgroup$
– LinAlg
Jan 29 at 2:00
$begingroup$
@Skrrrrrtttt just perform one iteration of revised simplex, you should find the formula in the step where you check for optimality.
$endgroup$
– LinAlg
Jan 29 at 2:00
add a comment |
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