Measuring infeasibility in convex optimization, relations with dual problem
$begingroup$
A question regarding convex optimization and (maybe) duality.
I have a problem in the form
begin{align}
x^* = mathrm{arg} min_x f(x) quad text{s.t.} quad A x leq b, C x = d,
end{align}
where $f$ is either quadratic or, if it makes things easier, linear. Let's also assume that $P = { x mid A x leq b }$ is bounded and not empty.
I would like to design an auxiliary optimization problem that:
- if the original problem is feasible returns the solution $x^*$ of the original problem,
- if the original problem is infeasible returns the minimum perturbation (according to some norm) $e^*$ that I should add to $d$ to make the problem feasible (remember $P neq emptyset$).
A naive approach could be, for example, something like
begin{align}
min_{x,e} f(x) + M | e | quad text{s.t.} quad A x leq b, C x = d + e,
end{align}
where $M$ is "big enough". This however is very unpractical since the high value of $M$ would lead to numeric problems.
Do you have any better idea?
Do you think duality can help in some way here?
If I solve the dual of the original problem, and I detect unboundedness (we know that, under these assumptions, the primal infeasible implies the dual unbounded) can I elaborate the result to get something related to $e^*$?
For example, could an extreme ray of the dual feasible set help?
Thank you very much!
convex-optimization linear-programming duality-theorems quadratic-programming
asked Nov 28 '18 at 18:55
Tobia MarcucciTobia Marcucci
354112
$endgroup$
add a comment |
$begingroup$
A question regarding convex optimization and (maybe) duality.
I have a problem in the form
begin{align}
x^* = mathrm{arg} min_x f(x) quad text{s.t.} quad A x leq b, C x = d,
end{align}
where $f$ is either quadratic or, if it makes things easier, linear. Let's also assume that $P = { x mid A x leq b }$ is bounded and not empty.
I would like to design an auxiliary optimization problem that:
- if the original problem is feasible returns the solution $x^*$ of the original problem,
- if the original problem is infeasible returns the minimum perturbation (according to some norm) $e^*$ that I should add to $d$ to make the problem feasible (remember $P neq emptyset$).
A naive approach could be, for example, something like
begin{align}
min_{x,e} f(x) + M | e | quad text{s.t.} quad A x leq b, C x = d + e,
end{align}
where $M$ is "big enough". This however is very unpractical since the high value of $M$ would lead to numeric problems.
Do you have any better idea?
Do you think duality can help in some way here?
If I solve the dual of the original problem, and I detect unboundedness (we know that, under these assumptions, the primal infeasible implies the dual unbounded) can I elaborate the result to get something related to $e^*$?
For example, could an extreme ray of the dual feasible set help?
Thank you very much!
convex-optimization linear-programming duality-theorems quadratic-programming
asked Nov 28 '18 at 18:55
Tobia MarcucciTobia Marcucci
354112
$endgroup$
$begingroup$
If the primal is infeasible, the dual may also be infeasible. Why not solve 2 problems?
$endgroup$
– LinAlg
Nov 28 '18 at 19:03
$begingroup$
You could add a binary variable to deal with infeasibility, but I doubt it is faster.
$endgroup$
– LinAlg
Nov 28 '18 at 19:17
$begingroup$
In the quadratic case, the dual is always feasible, but in general you are right: the dual can also be infeasible. I guess solving two problems is always an option. I wanted to understand if I'm trowing away some useful information from the solution of the dual of the original problem.
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:20
$begingroup$
This would be actually a lower level QP/LP solver of a branch and bound code for mixed integer programming (reason why I have the solution of the dual almost for free). Yes, the binary would work, but in practice it's equivalent to solve two problems...
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:23
add a comment |
$begingroup$
A question regarding convex optimization and (maybe) duality.
I have a problem in the form
begin{align}
x^* = mathrm{arg} min_x f(x) quad text{s.t.} quad A x leq b, C x = d,
end{align}
where $f$ is either quadratic or, if it makes things easier, linear. Let's also assume that $P = { x mid A x leq b }$ is bounded and not empty.
I would like to design an auxiliary optimization problem that:
- if the original problem is feasible returns the solution $x^*$ of the original problem,
- if the original problem is infeasible returns the minimum perturbation (according to some norm) $e^*$ that I should add to $d$ to make the problem feasible (remember $P neq emptyset$).
A naive approach could be, for example, something like
begin{align}
min_{x,e} f(x) + M | e | quad text{s.t.} quad A x leq b, C x = d + e,
end{align}
where $M$ is "big enough". This however is very unpractical since the high value of $M$ would lead to numeric problems.
Do you have any better idea?
Do you think duality can help in some way here?
If I solve the dual of the original problem, and I detect unboundedness (we know that, under these assumptions, the primal infeasible implies the dual unbounded) can I elaborate the result to get something related to $e^*$?
For example, could an extreme ray of the dual feasible set help?
Thank you very much!
convex-optimization linear-programming duality-theorems quadratic-programming
asked Nov 28 '18 at 18:55
Tobia MarcucciTobia Marcucci
354112
$endgroup$
A question regarding convex optimization and (maybe) duality.
I have a problem in the form
begin{align}
x^* = mathrm{arg} min_x f(x) quad text{s.t.} quad A x leq b, C x = d,
end{align}
where $f$ is either quadratic or, if it makes things easier, linear. Let's also assume that $P = { x mid A x leq b }$ is bounded and not empty.
I would like to design an auxiliary optimization problem that:
- if the original problem is feasible returns the solution $x^*$ of the original problem,
- if the original problem is infeasible returns the minimum perturbation (according to some norm) $e^*$ that I should add to $d$ to make the problem feasible (remember $P neq emptyset$).
A naive approach could be, for example, something like
begin{align}
min_{x,e} f(x) + M | e | quad text{s.t.} quad A x leq b, C x = d + e,
end{align}
where $M$ is "big enough". This however is very unpractical since the high value of $M$ would lead to numeric problems.
Do you have any better idea?
Do you think duality can help in some way here?
If I solve the dual of the original problem, and I detect unboundedness (we know that, under these assumptions, the primal infeasible implies the dual unbounded) can I elaborate the result to get something related to $e^*$?
For example, could an extreme ray of the dual feasible set help?
Thank you very much!
convex-optimization linear-programming duality-theorems quadratic-programming
convex-optimization linear-programming duality-theorems quadratic-programming
asked Nov 28 '18 at 18:55
Tobia MarcucciTobia Marcucci
354112
asked Nov 28 '18 at 18:55
Tobia MarcucciTobia Marcucci
354112
asked Nov 28 '18 at 18:55
Tobia MarcucciTobia Marcucci
354112
asked Nov 28 '18 at 18:55
Tobia MarcucciTobia Marcucci
354112
asked Nov 28 '18 at 18:55
Tobia MarcucciTobia Marcucci
354112
354112
$begingroup$
If the primal is infeasible, the dual may also be infeasible. Why not solve 2 problems?
$endgroup$
– LinAlg
Nov 28 '18 at 19:03
$begingroup$
You could add a binary variable to deal with infeasibility, but I doubt it is faster.
$endgroup$
– LinAlg
Nov 28 '18 at 19:17
$begingroup$
In the quadratic case, the dual is always feasible, but in general you are right: the dual can also be infeasible. I guess solving two problems is always an option. I wanted to understand if I'm trowing away some useful information from the solution of the dual of the original problem.
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:20
$begingroup$
This would be actually a lower level QP/LP solver of a branch and bound code for mixed integer programming (reason why I have the solution of the dual almost for free). Yes, the binary would work, but in practice it's equivalent to solve two problems...
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:23
add a comment |
$begingroup$
If the primal is infeasible, the dual may also be infeasible. Why not solve 2 problems?
$endgroup$
– LinAlg
Nov 28 '18 at 19:03
$begingroup$
You could add a binary variable to deal with infeasibility, but I doubt it is faster.
$endgroup$
– LinAlg
Nov 28 '18 at 19:17
$begingroup$
In the quadratic case, the dual is always feasible, but in general you are right: the dual can also be infeasible. I guess solving two problems is always an option. I wanted to understand if I'm trowing away some useful information from the solution of the dual of the original problem.
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:20
$begingroup$
This would be actually a lower level QP/LP solver of a branch and bound code for mixed integer programming (reason why I have the solution of the dual almost for free). Yes, the binary would work, but in practice it's equivalent to solve two problems...
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:23
$begingroup$
If the primal is infeasible, the dual may also be infeasible. Why not solve 2 problems?
$endgroup$
– LinAlg
Nov 28 '18 at 19:03
$begingroup$
If the primal is infeasible, the dual may also be infeasible. Why not solve 2 problems?
$endgroup$
– LinAlg
Nov 28 '18 at 19:03
$begingroup$
You could add a binary variable to deal with infeasibility, but I doubt it is faster.
$endgroup$
– LinAlg
Nov 28 '18 at 19:17
$begingroup$
You could add a binary variable to deal with infeasibility, but I doubt it is faster.
$endgroup$
– LinAlg
Nov 28 '18 at 19:17
$begingroup$
In the quadratic case, the dual is always feasible, but in general you are right: the dual can also be infeasible. I guess solving two problems is always an option. I wanted to understand if I'm trowing away some useful information from the solution of the dual of the original problem.
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:20
$begingroup$
In the quadratic case, the dual is always feasible, but in general you are right: the dual can also be infeasible. I guess solving two problems is always an option. I wanted to understand if I'm trowing away some useful information from the solution of the dual of the original problem.
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:20
$begingroup$
This would be actually a lower level QP/LP solver of a branch and bound code for mixed integer programming (reason why I have the solution of the dual almost for free). Yes, the binary would work, but in practice it's equivalent to solve two problems...
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:23
$begingroup$
This would be actually a lower level QP/LP solver of a branch and bound code for mixed integer programming (reason why I have the solution of the dual almost for free). Yes, the binary would work, but in practice it's equivalent to solve two problems...
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Well, if the original problem is infeasible, you can just solve for
begin{align}
min_{x,e} |e|: & \
Ax&le b, \
e &= d-Cx.
end{align}
answered Nov 28 '18 at 19:11
HansHans
4,98021032
$endgroup$
$begingroup$
Unfortunately I do not know a priori if the problem is infeasible. Your approach would then boil down to just solve two problems, right?
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:25
$begingroup$
You will have to first check the feasibility of the original problem anyway. Besides, this new problem is always feasible.
$endgroup$
– Hans
Nov 28 '18 at 19:36
$begingroup$
I do not agree that I have to first check the feasibility of the original problem anyway: the example I described with the big M in the objective function is a single problem, where I do not need that check first. No?
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:44
$begingroup$
Your requirement is to solve for the original problem if the domain is feasible. Your objective function with the big M violates that requirement. The bigger the M is, the further away your solution deviates from the true solution, when the original problem is feasible. My formulation, on the other hand, fulfills exactly your two requirements.
$endgroup$
– Hans
Nov 28 '18 at 19:56
1
$begingroup$
It's clear that solving two problems as you are suggesting would work: my question is in fact how to avoid that...
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 20:33
|
show 3 more comments
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$begingroup$
Well, if the original problem is infeasible, you can just solve for
begin{align}
min_{x,e} |e|: & \
Ax&le b, \
e &= d-Cx.
end{align}
answered Nov 28 '18 at 19:11
HansHans
4,98021032
$endgroup$
$begingroup$
Unfortunately I do not know a priori if the problem is infeasible. Your approach would then boil down to just solve two problems, right?
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:25
$begingroup$
You will have to first check the feasibility of the original problem anyway. Besides, this new problem is always feasible.
$endgroup$
– Hans
Nov 28 '18 at 19:36
$begingroup$
I do not agree that I have to first check the feasibility of the original problem anyway: the example I described with the big M in the objective function is a single problem, where I do not need that check first. No?
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:44
$begingroup$
Your requirement is to solve for the original problem if the domain is feasible. Your objective function with the big M violates that requirement. The bigger the M is, the further away your solution deviates from the true solution, when the original problem is feasible. My formulation, on the other hand, fulfills exactly your two requirements.
$endgroup$
– Hans
Nov 28 '18 at 19:56
1
$begingroup$
It's clear that solving two problems as you are suggesting would work: my question is in fact how to avoid that...
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 20:33
|
show 3 more comments
$begingroup$
Well, if the original problem is infeasible, you can just solve for
begin{align}
min_{x,e} |e|: & \
Ax&le b, \
e &= d-Cx.
end{align}
answered Nov 28 '18 at 19:11
HansHans
4,98021032
$endgroup$
$begingroup$
Unfortunately I do not know a priori if the problem is infeasible. Your approach would then boil down to just solve two problems, right?
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:25
$begingroup$
You will have to first check the feasibility of the original problem anyway. Besides, this new problem is always feasible.
$endgroup$
– Hans
Nov 28 '18 at 19:36
$begingroup$
I do not agree that I have to first check the feasibility of the original problem anyway: the example I described with the big M in the objective function is a single problem, where I do not need that check first. No?
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:44
$begingroup$
Your requirement is to solve for the original problem if the domain is feasible. Your objective function with the big M violates that requirement. The bigger the M is, the further away your solution deviates from the true solution, when the original problem is feasible. My formulation, on the other hand, fulfills exactly your two requirements.
$endgroup$
– Hans
Nov 28 '18 at 19:56
1
$begingroup$
It's clear that solving two problems as you are suggesting would work: my question is in fact how to avoid that...
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 20:33
|
show 3 more comments
$begingroup$
Well, if the original problem is infeasible, you can just solve for
begin{align}
min_{x,e} |e|: & \
Ax&le b, \
e &= d-Cx.
end{align}
answered Nov 28 '18 at 19:11
HansHans
4,98021032
$endgroup$
Well, if the original problem is infeasible, you can just solve for
begin{align}
min_{x,e} |e|: & \
Ax&le b, \
e &= d-Cx.
end{align}
answered Nov 28 '18 at 19:11
HansHans
4,98021032
answered Nov 28 '18 at 19:11
HansHans
4,98021032
answered Nov 28 '18 at 19:11
HansHans
4,98021032
answered Nov 28 '18 at 19:11
HansHans
4,98021032
4,98021032
$begingroup$
Unfortunately I do not know a priori if the problem is infeasible. Your approach would then boil down to just solve two problems, right?
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:25
$begingroup$
You will have to first check the feasibility of the original problem anyway. Besides, this new problem is always feasible.
$endgroup$
– Hans
Nov 28 '18 at 19:36
$begingroup$
I do not agree that I have to first check the feasibility of the original problem anyway: the example I described with the big M in the objective function is a single problem, where I do not need that check first. No?
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:44
$begingroup$
Your requirement is to solve for the original problem if the domain is feasible. Your objective function with the big M violates that requirement. The bigger the M is, the further away your solution deviates from the true solution, when the original problem is feasible. My formulation, on the other hand, fulfills exactly your two requirements.
$endgroup$
– Hans
Nov 28 '18 at 19:56
1
$begingroup$
It's clear that solving two problems as you are suggesting would work: my question is in fact how to avoid that...
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 20:33
|
show 3 more comments
$begingroup$
Unfortunately I do not know a priori if the problem is infeasible. Your approach would then boil down to just solve two problems, right?
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:25
$begingroup$
You will have to first check the feasibility of the original problem anyway. Besides, this new problem is always feasible.
$endgroup$
– Hans
Nov 28 '18 at 19:36
$begingroup$
I do not agree that I have to first check the feasibility of the original problem anyway: the example I described with the big M in the objective function is a single problem, where I do not need that check first. No?
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:44
$begingroup$
Your requirement is to solve for the original problem if the domain is feasible. Your objective function with the big M violates that requirement. The bigger the M is, the further away your solution deviates from the true solution, when the original problem is feasible. My formulation, on the other hand, fulfills exactly your two requirements.
$endgroup$
– Hans
Nov 28 '18 at 19:56
1
$begingroup$
It's clear that solving two problems as you are suggesting would work: my question is in fact how to avoid that...
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 20:33
$begingroup$
Unfortunately I do not know a priori if the problem is infeasible. Your approach would then boil down to just solve two problems, right?
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:25
$begingroup$
Unfortunately I do not know a priori if the problem is infeasible. Your approach would then boil down to just solve two problems, right?
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:25
$begingroup$
You will have to first check the feasibility of the original problem anyway. Besides, this new problem is always feasible.
$endgroup$
– Hans
Nov 28 '18 at 19:36
$begingroup$
You will have to first check the feasibility of the original problem anyway. Besides, this new problem is always feasible.
$endgroup$
– Hans
Nov 28 '18 at 19:36
$begingroup$
I do not agree that I have to first check the feasibility of the original problem anyway: the example I described with the big M in the objective function is a single problem, where I do not need that check first. No?
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:44
$begingroup$
I do not agree that I have to first check the feasibility of the original problem anyway: the example I described with the big M in the objective function is a single problem, where I do not need that check first. No?
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:44
$begingroup$
Your requirement is to solve for the original problem if the domain is feasible. Your objective function with the big M violates that requirement. The bigger the M is, the further away your solution deviates from the true solution, when the original problem is feasible. My formulation, on the other hand, fulfills exactly your two requirements.
$endgroup$
– Hans
Nov 28 '18 at 19:56
$begingroup$
Your requirement is to solve for the original problem if the domain is feasible. Your objective function with the big M violates that requirement. The bigger the M is, the further away your solution deviates from the true solution, when the original problem is feasible. My formulation, on the other hand, fulfills exactly your two requirements.
$endgroup$
– Hans
Nov 28 '18 at 19:56
1
1
$begingroup$
It's clear that solving two problems as you are suggesting would work: my question is in fact how to avoid that...
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 20:33
$begingroup$
It's clear that solving two problems as you are suggesting would work: my question is in fact how to avoid that...
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 20:33
|
show 3 more comments
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$begingroup$
If the primal is infeasible, the dual may also be infeasible. Why not solve 2 problems?
$endgroup$
– LinAlg
Nov 28 '18 at 19:03
$begingroup$
You could add a binary variable to deal with infeasibility, but I doubt it is faster.
$endgroup$
– LinAlg
Nov 28 '18 at 19:17
$begingroup$
In the quadratic case, the dual is always feasible, but in general you are right: the dual can also be infeasible. I guess solving two problems is always an option. I wanted to understand if I'm trowing away some useful information from the solution of the dual of the original problem.
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:20
$begingroup$
This would be actually a lower level QP/LP solver of a branch and bound code for mixed integer programming (reason why I have the solution of the dual almost for free). Yes, the binary would work, but in practice it's equivalent to solve two problems...
$endgroup$
– Tobia Marcucci
Nov 28 '18 at 19:23