Mixing
Computing initial feasible basis in the simplex algorithm
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My textbook introduces the following method to compute initial feasible basis in the simplex algorithm:
What is the implication for the original LP if the auxiliary LP's objective function can't attain the value $0$, or if it is infeasible?
linear-programming
asked Feb 2 at 14:37
ensbanaensbana
252214
$endgroup$
add a comment |
$begingroup$
My textbook introduces the following method to compute initial feasible basis in the simplex algorithm:
What is the implication for the original LP if the auxiliary LP's objective function can't attain the value $0$, or if it is infeasible?
linear-programming
asked Feb 2 at 14:37
ensbanaensbana
252214
$endgroup$
add a comment |
$begingroup$
My textbook introduces the following method to compute initial feasible basis in the simplex algorithm:
What is the implication for the original LP if the auxiliary LP's objective function can't attain the value $0$, or if it is infeasible?
linear-programming
asked Feb 2 at 14:37
ensbanaensbana
252214
$endgroup$
My textbook introduces the following method to compute initial feasible basis in the simplex algorithm:
What is the implication for the original LP if the auxiliary LP's objective function can't attain the value $0$, or if it is infeasible?
linear-programming
linear-programming
asked Feb 2 at 14:37
ensbanaensbana
252214
asked Feb 2 at 14:37
ensbanaensbana
252214
asked Feb 2 at 14:37
ensbanaensbana
252214
asked Feb 2 at 14:37
ensbanaensbana
252214
asked Feb 2 at 14:37
ensbanaensbana
252214
252214
add a comment |
add a comment |
1 Answer
1
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If the auxiliary problem can’t attain the value zero, then the original problem is infeasible.
The auxiliary problem can’t be infeasible. For a general LP with constraints $Ax=b$ with $xgeq0$, the auxiliary problem has constraints $Ax+s=b$, $xgeq0$, $s$ free**. There is always the feasible solution $x=0$ and $s=b$.
**If you require $sgeq0$, then you can write the slack variables $s$ as $s^+-s^-$ for $s^+,s^-geq0$.
answered Feb 2 at 15:57
David M.David M.
2,232421
$endgroup$
$begingroup$
If the auxiliary problem can't attain the value zero, is it possible that the original problem is not infeasible, but unbounded?
$endgroup$
– ensbana
Feb 2 at 22:44
$begingroup$
I'm working with this auxiliary program: maximize $z = -x_6 - x_7$, subject to: $-2x_1 + 2x_2 + 2x_3 - x_4 + x_6 = 2$, $2x_1 - 2x_2 + x_3 + x_5 + x_7 =2$, $x_1,x_2,x_3,x_4,x_5,x_6,x_7 geq 0$. Let $x_6, x_7$ be the basic variables, we obtain $z = -4 + 3x_3 - x_4 + x_5$, so there's no possibility of further improving $z$. I also used an online solver to solve the original LP, and it says the program is unbounded.
$endgroup$
– ensbana
Feb 2 at 22:47
$begingroup$
No, if the original problem is unbounded, the auxiliary problem will be feasible. I’m not really sure what your second comment is describing. Add it to the question with some details (e.g. what’s the original LP, etc.)
$endgroup$
– David M.
Feb 3 at 1:36
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If the auxiliary problem can’t attain the value zero, then the original problem is infeasible.
The auxiliary problem can’t be infeasible. For a general LP with constraints $Ax=b$ with $xgeq0$, the auxiliary problem has constraints $Ax+s=b$, $xgeq0$, $s$ free**. There is always the feasible solution $x=0$ and $s=b$.
**If you require $sgeq0$, then you can write the slack variables $s$ as $s^+-s^-$ for $s^+,s^-geq0$.
answered Feb 2 at 15:57
David M.David M.
2,232421
$endgroup$
$begingroup$
If the auxiliary problem can't attain the value zero, is it possible that the original problem is not infeasible, but unbounded?
$endgroup$
– ensbana
Feb 2 at 22:44
$begingroup$
I'm working with this auxiliary program: maximize $z = -x_6 - x_7$, subject to: $-2x_1 + 2x_2 + 2x_3 - x_4 + x_6 = 2$, $2x_1 - 2x_2 + x_3 + x_5 + x_7 =2$, $x_1,x_2,x_3,x_4,x_5,x_6,x_7 geq 0$. Let $x_6, x_7$ be the basic variables, we obtain $z = -4 + 3x_3 - x_4 + x_5$, so there's no possibility of further improving $z$. I also used an online solver to solve the original LP, and it says the program is unbounded.
$endgroup$
– ensbana
Feb 2 at 22:47
$begingroup$
No, if the original problem is unbounded, the auxiliary problem will be feasible. I’m not really sure what your second comment is describing. Add it to the question with some details (e.g. what’s the original LP, etc.)
$endgroup$
– David M.
Feb 3 at 1:36
add a comment |
$begingroup$
If the auxiliary problem can’t attain the value zero, then the original problem is infeasible.
The auxiliary problem can’t be infeasible. For a general LP with constraints $Ax=b$ with $xgeq0$, the auxiliary problem has constraints $Ax+s=b$, $xgeq0$, $s$ free**. There is always the feasible solution $x=0$ and $s=b$.
**If you require $sgeq0$, then you can write the slack variables $s$ as $s^+-s^-$ for $s^+,s^-geq0$.
answered Feb 2 at 15:57
David M.David M.
2,232421
$endgroup$
$begingroup$
If the auxiliary problem can't attain the value zero, is it possible that the original problem is not infeasible, but unbounded?
$endgroup$
– ensbana
Feb 2 at 22:44
$begingroup$
I'm working with this auxiliary program: maximize $z = -x_6 - x_7$, subject to: $-2x_1 + 2x_2 + 2x_3 - x_4 + x_6 = 2$, $2x_1 - 2x_2 + x_3 + x_5 + x_7 =2$, $x_1,x_2,x_3,x_4,x_5,x_6,x_7 geq 0$. Let $x_6, x_7$ be the basic variables, we obtain $z = -4 + 3x_3 - x_4 + x_5$, so there's no possibility of further improving $z$. I also used an online solver to solve the original LP, and it says the program is unbounded.
$endgroup$
– ensbana
Feb 2 at 22:47
$begingroup$
No, if the original problem is unbounded, the auxiliary problem will be feasible. I’m not really sure what your second comment is describing. Add it to the question with some details (e.g. what’s the original LP, etc.)
$endgroup$
– David M.
Feb 3 at 1:36
add a comment |
$begingroup$
If the auxiliary problem can’t attain the value zero, then the original problem is infeasible.
The auxiliary problem can’t be infeasible. For a general LP with constraints $Ax=b$ with $xgeq0$, the auxiliary problem has constraints $Ax+s=b$, $xgeq0$, $s$ free**. There is always the feasible solution $x=0$ and $s=b$.
**If you require $sgeq0$, then you can write the slack variables $s$ as $s^+-s^-$ for $s^+,s^-geq0$.
answered Feb 2 at 15:57
David M.David M.
2,232421
$endgroup$
If the auxiliary problem can’t attain the value zero, then the original problem is infeasible.
The auxiliary problem can’t be infeasible. For a general LP with constraints $Ax=b$ with $xgeq0$, the auxiliary problem has constraints $Ax+s=b$, $xgeq0$, $s$ free**. There is always the feasible solution $x=0$ and $s=b$.
**If you require $sgeq0$, then you can write the slack variables $s$ as $s^+-s^-$ for $s^+,s^-geq0$.
answered Feb 2 at 15:57
David M.David M.
2,232421
answered Feb 2 at 15:57
David M.David M.
2,232421
answered Feb 2 at 15:57
David M.David M.
2,232421
answered Feb 2 at 15:57
David M.David M.
2,232421
2,232421
$begingroup$
If the auxiliary problem can't attain the value zero, is it possible that the original problem is not infeasible, but unbounded?
$endgroup$
– ensbana
Feb 2 at 22:44
$begingroup$
I'm working with this auxiliary program: maximize $z = -x_6 - x_7$, subject to: $-2x_1 + 2x_2 + 2x_3 - x_4 + x_6 = 2$, $2x_1 - 2x_2 + x_3 + x_5 + x_7 =2$, $x_1,x_2,x_3,x_4,x_5,x_6,x_7 geq 0$. Let $x_6, x_7$ be the basic variables, we obtain $z = -4 + 3x_3 - x_4 + x_5$, so there's no possibility of further improving $z$. I also used an online solver to solve the original LP, and it says the program is unbounded.
$endgroup$
– ensbana
Feb 2 at 22:47
$begingroup$
No, if the original problem is unbounded, the auxiliary problem will be feasible. I’m not really sure what your second comment is describing. Add it to the question with some details (e.g. what’s the original LP, etc.)
$endgroup$
– David M.
Feb 3 at 1:36
add a comment |
$begingroup$
If the auxiliary problem can't attain the value zero, is it possible that the original problem is not infeasible, but unbounded?
$endgroup$
– ensbana
Feb 2 at 22:44
$begingroup$
I'm working with this auxiliary program: maximize $z = -x_6 - x_7$, subject to: $-2x_1 + 2x_2 + 2x_3 - x_4 + x_6 = 2$, $2x_1 - 2x_2 + x_3 + x_5 + x_7 =2$, $x_1,x_2,x_3,x_4,x_5,x_6,x_7 geq 0$. Let $x_6, x_7$ be the basic variables, we obtain $z = -4 + 3x_3 - x_4 + x_5$, so there's no possibility of further improving $z$. I also used an online solver to solve the original LP, and it says the program is unbounded.
$endgroup$
– ensbana
Feb 2 at 22:47
$begingroup$
No, if the original problem is unbounded, the auxiliary problem will be feasible. I’m not really sure what your second comment is describing. Add it to the question with some details (e.g. what’s the original LP, etc.)
$endgroup$
– David M.
Feb 3 at 1:36
$begingroup$
If the auxiliary problem can't attain the value zero, is it possible that the original problem is not infeasible, but unbounded?
$endgroup$
– ensbana
Feb 2 at 22:44
$begingroup$
If the auxiliary problem can't attain the value zero, is it possible that the original problem is not infeasible, but unbounded?
$endgroup$
– ensbana
Feb 2 at 22:44
$begingroup$
I'm working with this auxiliary program: maximize $z = -x_6 - x_7$, subject to: $-2x_1 + 2x_2 + 2x_3 - x_4 + x_6 = 2$, $2x_1 - 2x_2 + x_3 + x_5 + x_7 =2$, $x_1,x_2,x_3,x_4,x_5,x_6,x_7 geq 0$. Let $x_6, x_7$ be the basic variables, we obtain $z = -4 + 3x_3 - x_4 + x_5$, so there's no possibility of further improving $z$. I also used an online solver to solve the original LP, and it says the program is unbounded.
$endgroup$
– ensbana
Feb 2 at 22:47
$begingroup$
I'm working with this auxiliary program: maximize $z = -x_6 - x_7$, subject to: $-2x_1 + 2x_2 + 2x_3 - x_4 + x_6 = 2$, $2x_1 - 2x_2 + x_3 + x_5 + x_7 =2$, $x_1,x_2,x_3,x_4,x_5,x_6,x_7 geq 0$. Let $x_6, x_7$ be the basic variables, we obtain $z = -4 + 3x_3 - x_4 + x_5$, so there's no possibility of further improving $z$. I also used an online solver to solve the original LP, and it says the program is unbounded.
$endgroup$
– ensbana
Feb 2 at 22:47
$begingroup$
No, if the original problem is unbounded, the auxiliary problem will be feasible. I’m not really sure what your second comment is describing. Add it to the question with some details (e.g. what’s the original LP, etc.)
$endgroup$
– David M.
Feb 3 at 1:36
$begingroup$
No, if the original problem is unbounded, the auxiliary problem will be feasible. I’m not really sure what your second comment is describing. Add it to the question with some details (e.g. what’s the original LP, etc.)
$endgroup$
– David M.
Feb 3 at 1:36
add a comment |
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