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How to use the barrier method for equality (or positive) constraints?
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How to use the barrier method for equality (or positive) constraints?
https://optimization.mccormick.northwestern.edu/index.php/Interior-point_method_for_LP#Barrier_Method
My initial reasoning was that since the example there starts with constraints $f_i(x) leq 0$ and then takes
$$phi(x)=-sum_{i=1}^m log(-f_i(x))$$
Then if one'd start with $f_i(x)=0$, then reasonably one doesn't need to flip the sign so could one then simply do
$$phi(x)=-sum_{i=1}^m log(f_i(x))$$
for constraints of the form $f_i(x)=0$ or $f_i(x) geq 0$.
However, without being entirely familiar with the barrier method, is this enough?
optimization
asked Jan 30 at 21:09
mavaviljmavavilj
2,84911138
$endgroup$
add a comment |
$begingroup$
How to use the barrier method for equality (or positive) constraints?
https://optimization.mccormick.northwestern.edu/index.php/Interior-point_method_for_LP#Barrier_Method
My initial reasoning was that since the example there starts with constraints $f_i(x) leq 0$ and then takes
$$phi(x)=-sum_{i=1}^m log(-f_i(x))$$
Then if one'd start with $f_i(x)=0$, then reasonably one doesn't need to flip the sign so could one then simply do
$$phi(x)=-sum_{i=1}^m log(f_i(x))$$
for constraints of the form $f_i(x)=0$ or $f_i(x) geq 0$.
However, without being entirely familiar with the barrier method, is this enough?
optimization
asked Jan 30 at 21:09
mavaviljmavavilj
2,84911138
$endgroup$
add a comment |
$begingroup$
How to use the barrier method for equality (or positive) constraints?
https://optimization.mccormick.northwestern.edu/index.php/Interior-point_method_for_LP#Barrier_Method
My initial reasoning was that since the example there starts with constraints $f_i(x) leq 0$ and then takes
$$phi(x)=-sum_{i=1}^m log(-f_i(x))$$
Then if one'd start with $f_i(x)=0$, then reasonably one doesn't need to flip the sign so could one then simply do
$$phi(x)=-sum_{i=1}^m log(f_i(x))$$
for constraints of the form $f_i(x)=0$ or $f_i(x) geq 0$.
However, without being entirely familiar with the barrier method, is this enough?
optimization
asked Jan 30 at 21:09
mavaviljmavavilj
2,84911138
$endgroup$
How to use the barrier method for equality (or positive) constraints?
https://optimization.mccormick.northwestern.edu/index.php/Interior-point_method_for_LP#Barrier_Method
My initial reasoning was that since the example there starts with constraints $f_i(x) leq 0$ and then takes
$$phi(x)=-sum_{i=1}^m log(-f_i(x))$$
Then if one'd start with $f_i(x)=0$, then reasonably one doesn't need to flip the sign so could one then simply do
$$phi(x)=-sum_{i=1}^m log(f_i(x))$$
for constraints of the form $f_i(x)=0$ or $f_i(x) geq 0$.
However, without being entirely familiar with the barrier method, is this enough?
optimization
optimization
asked Jan 30 at 21:09
mavaviljmavavilj
2,84911138
asked Jan 30 at 21:09
mavaviljmavavilj
2,84911138
asked Jan 30 at 21:09
mavaviljmavavilj
2,84911138
asked Jan 30 at 21:09
mavaviljmavavilj
2,84911138
asked Jan 30 at 21:09
mavaviljmavavilj
2,84911138
2,84911138
add a comment |
add a comment |
1 Answer
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$begingroup$
A constraint $f(x) geq 0$ is equivalent to $-f(x) leq 0$. Of course $f$ should be concave if you expect globally optimal solutions.
An equality constraint does not fit in the barrier method, since the feasible set for such a constraint has a nonempty interior.
The page you link to also has a description of the primal-dual method. That method can cope with equality constraints (in fact, all inequality constraints are first reformulated as equality constraints).
answered Jan 30 at 21:31
LinAlgLinAlg
10.1k1521
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$begingroup$
Would I need to convert the equality into inequality then? And how should I do that?
$endgroup$
– mavavilj
Jan 30 at 21:54
$begingroup$
@mavavilj no, if you transform it into two inequalities, the feasible set is still not full dimensional
$endgroup$
– LinAlg
Jan 30 at 21:54
$begingroup$
Good, but this doesn't answer my question. Since I'm not switching $-f(x)$ to $f(x)$, but $g(-f(x))$ to $g(f(x))$.
$endgroup$
– mavavilj
Feb 25 at 16:56
$begingroup$
@mavavilj the barrier term is $-sum_i log(f_i(x))$ for constraints of the type $f_i(x)geq 0$ (this is obvious from the first line in my answer). However, you need to use a different method for equality constraints, because the logarithmic function forces a strict inequality which you cannot have for an equality constraint.
$endgroup$
– LinAlg
Feb 25 at 17:12
add a comment |
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1 Answer
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1 Answer
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$begingroup$
A constraint $f(x) geq 0$ is equivalent to $-f(x) leq 0$. Of course $f$ should be concave if you expect globally optimal solutions.
An equality constraint does not fit in the barrier method, since the feasible set for such a constraint has a nonempty interior.
The page you link to also has a description of the primal-dual method. That method can cope with equality constraints (in fact, all inequality constraints are first reformulated as equality constraints).
answered Jan 30 at 21:31
LinAlgLinAlg
10.1k1521
$endgroup$
$begingroup$
Would I need to convert the equality into inequality then? And how should I do that?
$endgroup$
– mavavilj
Jan 30 at 21:54
$begingroup$
@mavavilj no, if you transform it into two inequalities, the feasible set is still not full dimensional
$endgroup$
– LinAlg
Jan 30 at 21:54
$begingroup$
Good, but this doesn't answer my question. Since I'm not switching $-f(x)$ to $f(x)$, but $g(-f(x))$ to $g(f(x))$.
$endgroup$
– mavavilj
Feb 25 at 16:56
$begingroup$
@mavavilj the barrier term is $-sum_i log(f_i(x))$ for constraints of the type $f_i(x)geq 0$ (this is obvious from the first line in my answer). However, you need to use a different method for equality constraints, because the logarithmic function forces a strict inequality which you cannot have for an equality constraint.
$endgroup$
– LinAlg
Feb 25 at 17:12
add a comment |
$begingroup$
A constraint $f(x) geq 0$ is equivalent to $-f(x) leq 0$. Of course $f$ should be concave if you expect globally optimal solutions.
An equality constraint does not fit in the barrier method, since the feasible set for such a constraint has a nonempty interior.
The page you link to also has a description of the primal-dual method. That method can cope with equality constraints (in fact, all inequality constraints are first reformulated as equality constraints).
answered Jan 30 at 21:31
LinAlgLinAlg
10.1k1521
$endgroup$
$begingroup$
Would I need to convert the equality into inequality then? And how should I do that?
$endgroup$
– mavavilj
Jan 30 at 21:54
$begingroup$
@mavavilj no, if you transform it into two inequalities, the feasible set is still not full dimensional
$endgroup$
– LinAlg
Jan 30 at 21:54
$begingroup$
Good, but this doesn't answer my question. Since I'm not switching $-f(x)$ to $f(x)$, but $g(-f(x))$ to $g(f(x))$.
$endgroup$
– mavavilj
Feb 25 at 16:56
$begingroup$
@mavavilj the barrier term is $-sum_i log(f_i(x))$ for constraints of the type $f_i(x)geq 0$ (this is obvious from the first line in my answer). However, you need to use a different method for equality constraints, because the logarithmic function forces a strict inequality which you cannot have for an equality constraint.
$endgroup$
– LinAlg
Feb 25 at 17:12
add a comment |
$begingroup$
A constraint $f(x) geq 0$ is equivalent to $-f(x) leq 0$. Of course $f$ should be concave if you expect globally optimal solutions.
An equality constraint does not fit in the barrier method, since the feasible set for such a constraint has a nonempty interior.
The page you link to also has a description of the primal-dual method. That method can cope with equality constraints (in fact, all inequality constraints are first reformulated as equality constraints).
answered Jan 30 at 21:31
LinAlgLinAlg
10.1k1521
$endgroup$
A constraint $f(x) geq 0$ is equivalent to $-f(x) leq 0$. Of course $f$ should be concave if you expect globally optimal solutions.
An equality constraint does not fit in the barrier method, since the feasible set for such a constraint has a nonempty interior.
The page you link to also has a description of the primal-dual method. That method can cope with equality constraints (in fact, all inequality constraints are first reformulated as equality constraints).
answered Jan 30 at 21:31
LinAlgLinAlg
10.1k1521
answered Jan 30 at 21:31
LinAlgLinAlg
10.1k1521
answered Jan 30 at 21:31
LinAlgLinAlg
10.1k1521
answered Jan 30 at 21:31
LinAlgLinAlg
10.1k1521
10.1k1521
$begingroup$
Would I need to convert the equality into inequality then? And how should I do that?
$endgroup$
– mavavilj
Jan 30 at 21:54
$begingroup$
@mavavilj no, if you transform it into two inequalities, the feasible set is still not full dimensional
$endgroup$
– LinAlg
Jan 30 at 21:54
$begingroup$
Good, but this doesn't answer my question. Since I'm not switching $-f(x)$ to $f(x)$, but $g(-f(x))$ to $g(f(x))$.
$endgroup$
– mavavilj
Feb 25 at 16:56
$begingroup$
@mavavilj the barrier term is $-sum_i log(f_i(x))$ for constraints of the type $f_i(x)geq 0$ (this is obvious from the first line in my answer). However, you need to use a different method for equality constraints, because the logarithmic function forces a strict inequality which you cannot have for an equality constraint.
$endgroup$
– LinAlg
Feb 25 at 17:12
add a comment |
$begingroup$
Would I need to convert the equality into inequality then? And how should I do that?
$endgroup$
– mavavilj
Jan 30 at 21:54
$begingroup$
@mavavilj no, if you transform it into two inequalities, the feasible set is still not full dimensional
$endgroup$
– LinAlg
Jan 30 at 21:54
$begingroup$
Good, but this doesn't answer my question. Since I'm not switching $-f(x)$ to $f(x)$, but $g(-f(x))$ to $g(f(x))$.
$endgroup$
– mavavilj
Feb 25 at 16:56
$begingroup$
@mavavilj the barrier term is $-sum_i log(f_i(x))$ for constraints of the type $f_i(x)geq 0$ (this is obvious from the first line in my answer). However, you need to use a different method for equality constraints, because the logarithmic function forces a strict inequality which you cannot have for an equality constraint.
$endgroup$
– LinAlg
Feb 25 at 17:12
$begingroup$
Would I need to convert the equality into inequality then? And how should I do that?
$endgroup$
– mavavilj
Jan 30 at 21:54
$begingroup$
Would I need to convert the equality into inequality then? And how should I do that?
$endgroup$
– mavavilj
Jan 30 at 21:54
$begingroup$
@mavavilj no, if you transform it into two inequalities, the feasible set is still not full dimensional
$endgroup$
– LinAlg
Jan 30 at 21:54
$begingroup$
@mavavilj no, if you transform it into two inequalities, the feasible set is still not full dimensional
$endgroup$
– LinAlg
Jan 30 at 21:54
$begingroup$
Good, but this doesn't answer my question. Since I'm not switching $-f(x)$ to $f(x)$, but $g(-f(x))$ to $g(f(x))$.
$endgroup$
– mavavilj
Feb 25 at 16:56
$begingroup$
Good, but this doesn't answer my question. Since I'm not switching $-f(x)$ to $f(x)$, but $g(-f(x))$ to $g(f(x))$.
$endgroup$
– mavavilj
Feb 25 at 16:56
$begingroup$
@mavavilj the barrier term is $-sum_i log(f_i(x))$ for constraints of the type $f_i(x)geq 0$ (this is obvious from the first line in my answer). However, you need to use a different method for equality constraints, because the logarithmic function forces a strict inequality which you cannot have for an equality constraint.
$endgroup$
– LinAlg
Feb 25 at 17:12
$begingroup$
@mavavilj the barrier term is $-sum_i log(f_i(x))$ for constraints of the type $f_i(x)geq 0$ (this is obvious from the first line in my answer). However, you need to use a different method for equality constraints, because the logarithmic function forces a strict inequality which you cannot have for an equality constraint.
$endgroup$
– LinAlg
Feb 25 at 17:12
add a comment |
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