Find at least one solution to matrix inequality











up vote
1
down vote

favorite












I have the following problem posed:
find at least one vector $mathbf{x}$ such that
$$
Amathbf{x} + mathbf{b} geqslant mathbf{0}
$$

for a given matrix $A$ and vector $mathbf{b}$. Nothing is known about dimensions of the matrix. You may assume the set of solutions for this problem is non-empty and that the rank of the matrix is complete, i.e. $mbox{rk}(A) = mbox{min}(m, n)$. The inequality is understood as applied component-wise.



Do you know any algorithm capable of solving such problem? I checked textbooks on linear programming, but found nothing specifically dealing with this problem. Thanks for any comments!










share|cite|improve this question






















  • You could always just solve $A mathbf x = - mathbf b$, which evidently gives you an $mathbf x$ that satisfies the wanted inequality.
    – MisterRiemann
    Nov 21 at 17:40












  • @MisterRiemann what if $A$ is non-square?
    – MajinSaha
    Nov 21 at 17:41















up vote
1
down vote

favorite












I have the following problem posed:
find at least one vector $mathbf{x}$ such that
$$
Amathbf{x} + mathbf{b} geqslant mathbf{0}
$$

for a given matrix $A$ and vector $mathbf{b}$. Nothing is known about dimensions of the matrix. You may assume the set of solutions for this problem is non-empty and that the rank of the matrix is complete, i.e. $mbox{rk}(A) = mbox{min}(m, n)$. The inequality is understood as applied component-wise.



Do you know any algorithm capable of solving such problem? I checked textbooks on linear programming, but found nothing specifically dealing with this problem. Thanks for any comments!










share|cite|improve this question






















  • You could always just solve $A mathbf x = - mathbf b$, which evidently gives you an $mathbf x$ that satisfies the wanted inequality.
    – MisterRiemann
    Nov 21 at 17:40












  • @MisterRiemann what if $A$ is non-square?
    – MajinSaha
    Nov 21 at 17:41













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have the following problem posed:
find at least one vector $mathbf{x}$ such that
$$
Amathbf{x} + mathbf{b} geqslant mathbf{0}
$$

for a given matrix $A$ and vector $mathbf{b}$. Nothing is known about dimensions of the matrix. You may assume the set of solutions for this problem is non-empty and that the rank of the matrix is complete, i.e. $mbox{rk}(A) = mbox{min}(m, n)$. The inequality is understood as applied component-wise.



Do you know any algorithm capable of solving such problem? I checked textbooks on linear programming, but found nothing specifically dealing with this problem. Thanks for any comments!










share|cite|improve this question













I have the following problem posed:
find at least one vector $mathbf{x}$ such that
$$
Amathbf{x} + mathbf{b} geqslant mathbf{0}
$$

for a given matrix $A$ and vector $mathbf{b}$. Nothing is known about dimensions of the matrix. You may assume the set of solutions for this problem is non-empty and that the rank of the matrix is complete, i.e. $mbox{rk}(A) = mbox{min}(m, n)$. The inequality is understood as applied component-wise.



Do you know any algorithm capable of solving such problem? I checked textbooks on linear programming, but found nothing specifically dealing with this problem. Thanks for any comments!







linear-programming matrix-equations geometric-inequalities






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 21 at 17:16









MajinSaha

385




385












  • You could always just solve $A mathbf x = - mathbf b$, which evidently gives you an $mathbf x$ that satisfies the wanted inequality.
    – MisterRiemann
    Nov 21 at 17:40












  • @MisterRiemann what if $A$ is non-square?
    – MajinSaha
    Nov 21 at 17:41


















  • You could always just solve $A mathbf x = - mathbf b$, which evidently gives you an $mathbf x$ that satisfies the wanted inequality.
    – MisterRiemann
    Nov 21 at 17:40












  • @MisterRiemann what if $A$ is non-square?
    – MajinSaha
    Nov 21 at 17:41
















You could always just solve $A mathbf x = - mathbf b$, which evidently gives you an $mathbf x$ that satisfies the wanted inequality.
– MisterRiemann
Nov 21 at 17:40






You could always just solve $A mathbf x = - mathbf b$, which evidently gives you an $mathbf x$ that satisfies the wanted inequality.
– MisterRiemann
Nov 21 at 17:40














@MisterRiemann what if $A$ is non-square?
– MajinSaha
Nov 21 at 17:41




@MisterRiemann what if $A$ is non-square?
– MajinSaha
Nov 21 at 17:41










2 Answers
2






active

oldest

votes

















up vote
1
down vote













Consider the linear optimization problem
$$min_{x} { 0^Tx : Ax geq b}.$$
This can be solved in many ways. Let me name 3:




  1. An infeasible interior point method. This solves
    $$min_{sgeq 0, x} { 0^Tx : Ax - s = b}.$$


  2. A feasible interior point method. Consider the extended linear optimization problem:
    $$min_{x,ygeq 0} { e^Ty : Ax+b+y geq 0}.$$
    A feasible starting point for this problem is $x=0$, $y=max{0,-b}$.


  3. With the two phase simplex method in which you only need the first phase. This is essentially solving
    $$min_{x^- geq 0,x^+geq 0, y geq 0} { c^Ty : A(x^+-x^-) + By geq b},$$
    where $B$ is a diagonal matrix with $B_{ii}=1$ if $b_igeq 0$, $-1$ otherwise, and $c_i=1$ if $b_ileq 0$, $0$ otherwise.







share|cite|improve this answer





















  • Thanks. I just found the mention of two-phase approach before reading your answer. I'll take a look at all three.
    – MajinSaha
    Nov 21 at 23:28


















up vote
0
down vote













Linear programming is your method of choice.



A good way to visualize what is going on is to consider the rows of $A$ as vectors $A_i$. So you have a set of inequalities $ A_i x + b_i geq 0$. Each of these inequalities defines a halfspace in $n$-dimensional space, with a boundary which is a hyperplane with normal vector $A_i$ and distance from the origin $b_i/ |A_i|$. Your matrix inequality equation is then the intersection of all of these halfspaces. If the intersection is not empty, the intersection space is a polyhedron.



Finding an $x$ is exactly the question whether this intersection is empty or not. In linear programming, this is called "interior point method" (you find it on wikipedia and elsewhere) and it is a well-known procedure for finding a feasible point to start any optimization procedure (which you do not have).






share|cite|improve this answer























  • Yes, I read about IP approach before. Currently, I'm only aware of a method to initialize a feasible starting point for IP for a linear program $mbox{min} (mathbf{c}^T mathbf{x})$, but such method requires the information on $mathbf{c}$. It minimizes some simple quadratic and involves lagrange multipliers to present the starting point for the IP opimization. My question though is a bit general, and I don't know how to use that for my case.
    – MajinSaha
    Nov 21 at 19:01












  • This is not called "interior point method". An interior point method is an optimization method, not a method to find a feasible point. On the contrary: they need a starting point within the interior.
    – LinAlg
    Nov 21 at 21:56










  • I am aware of that. I probably didn't explain clearly what I meant. Anyways, thanks for commenting.
    – MajinSaha
    Nov 21 at 23:26










  • Well, as I explained, you are looking for an interior point in a polyhedron (if it exists). So you may choose a point at the boundary of such polyhedron which is the intersection of at most $n$ halfplanes which are not linearly dependent, so such a point obeys the halfplane constraints with equality. If the number $m$ of constraints is larger than $n$, then the remaining constraints must be satisfied with inequality. Only then (if it fails) may there be a problem. So, use an active set method or use slack variables as Phase I of this IP problem to check feasibility.
    – Andreas
    2 days ago











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














 

draft saved


draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008047%2ffind-at-least-one-solution-to-matrix-inequality%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













Consider the linear optimization problem
$$min_{x} { 0^Tx : Ax geq b}.$$
This can be solved in many ways. Let me name 3:




  1. An infeasible interior point method. This solves
    $$min_{sgeq 0, x} { 0^Tx : Ax - s = b}.$$


  2. A feasible interior point method. Consider the extended linear optimization problem:
    $$min_{x,ygeq 0} { e^Ty : Ax+b+y geq 0}.$$
    A feasible starting point for this problem is $x=0$, $y=max{0,-b}$.


  3. With the two phase simplex method in which you only need the first phase. This is essentially solving
    $$min_{x^- geq 0,x^+geq 0, y geq 0} { c^Ty : A(x^+-x^-) + By geq b},$$
    where $B$ is a diagonal matrix with $B_{ii}=1$ if $b_igeq 0$, $-1$ otherwise, and $c_i=1$ if $b_ileq 0$, $0$ otherwise.







share|cite|improve this answer





















  • Thanks. I just found the mention of two-phase approach before reading your answer. I'll take a look at all three.
    – MajinSaha
    Nov 21 at 23:28















up vote
1
down vote













Consider the linear optimization problem
$$min_{x} { 0^Tx : Ax geq b}.$$
This can be solved in many ways. Let me name 3:




  1. An infeasible interior point method. This solves
    $$min_{sgeq 0, x} { 0^Tx : Ax - s = b}.$$


  2. A feasible interior point method. Consider the extended linear optimization problem:
    $$min_{x,ygeq 0} { e^Ty : Ax+b+y geq 0}.$$
    A feasible starting point for this problem is $x=0$, $y=max{0,-b}$.


  3. With the two phase simplex method in which you only need the first phase. This is essentially solving
    $$min_{x^- geq 0,x^+geq 0, y geq 0} { c^Ty : A(x^+-x^-) + By geq b},$$
    where $B$ is a diagonal matrix with $B_{ii}=1$ if $b_igeq 0$, $-1$ otherwise, and $c_i=1$ if $b_ileq 0$, $0$ otherwise.







share|cite|improve this answer





















  • Thanks. I just found the mention of two-phase approach before reading your answer. I'll take a look at all three.
    – MajinSaha
    Nov 21 at 23:28













up vote
1
down vote










up vote
1
down vote









Consider the linear optimization problem
$$min_{x} { 0^Tx : Ax geq b}.$$
This can be solved in many ways. Let me name 3:




  1. An infeasible interior point method. This solves
    $$min_{sgeq 0, x} { 0^Tx : Ax - s = b}.$$


  2. A feasible interior point method. Consider the extended linear optimization problem:
    $$min_{x,ygeq 0} { e^Ty : Ax+b+y geq 0}.$$
    A feasible starting point for this problem is $x=0$, $y=max{0,-b}$.


  3. With the two phase simplex method in which you only need the first phase. This is essentially solving
    $$min_{x^- geq 0,x^+geq 0, y geq 0} { c^Ty : A(x^+-x^-) + By geq b},$$
    where $B$ is a diagonal matrix with $B_{ii}=1$ if $b_igeq 0$, $-1$ otherwise, and $c_i=1$ if $b_ileq 0$, $0$ otherwise.







share|cite|improve this answer












Consider the linear optimization problem
$$min_{x} { 0^Tx : Ax geq b}.$$
This can be solved in many ways. Let me name 3:




  1. An infeasible interior point method. This solves
    $$min_{sgeq 0, x} { 0^Tx : Ax - s = b}.$$


  2. A feasible interior point method. Consider the extended linear optimization problem:
    $$min_{x,ygeq 0} { e^Ty : Ax+b+y geq 0}.$$
    A feasible starting point for this problem is $x=0$, $y=max{0,-b}$.


  3. With the two phase simplex method in which you only need the first phase. This is essentially solving
    $$min_{x^- geq 0,x^+geq 0, y geq 0} { c^Ty : A(x^+-x^-) + By geq b},$$
    where $B$ is a diagonal matrix with $B_{ii}=1$ if $b_igeq 0$, $-1$ otherwise, and $c_i=1$ if $b_ileq 0$, $0$ otherwise.








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 21 at 21:54









LinAlg

7,6491520




7,6491520












  • Thanks. I just found the mention of two-phase approach before reading your answer. I'll take a look at all three.
    – MajinSaha
    Nov 21 at 23:28


















  • Thanks. I just found the mention of two-phase approach before reading your answer. I'll take a look at all three.
    – MajinSaha
    Nov 21 at 23:28
















Thanks. I just found the mention of two-phase approach before reading your answer. I'll take a look at all three.
– MajinSaha
Nov 21 at 23:28




Thanks. I just found the mention of two-phase approach before reading your answer. I'll take a look at all three.
– MajinSaha
Nov 21 at 23:28










up vote
0
down vote













Linear programming is your method of choice.



A good way to visualize what is going on is to consider the rows of $A$ as vectors $A_i$. So you have a set of inequalities $ A_i x + b_i geq 0$. Each of these inequalities defines a halfspace in $n$-dimensional space, with a boundary which is a hyperplane with normal vector $A_i$ and distance from the origin $b_i/ |A_i|$. Your matrix inequality equation is then the intersection of all of these halfspaces. If the intersection is not empty, the intersection space is a polyhedron.



Finding an $x$ is exactly the question whether this intersection is empty or not. In linear programming, this is called "interior point method" (you find it on wikipedia and elsewhere) and it is a well-known procedure for finding a feasible point to start any optimization procedure (which you do not have).






share|cite|improve this answer























  • Yes, I read about IP approach before. Currently, I'm only aware of a method to initialize a feasible starting point for IP for a linear program $mbox{min} (mathbf{c}^T mathbf{x})$, but such method requires the information on $mathbf{c}$. It minimizes some simple quadratic and involves lagrange multipliers to present the starting point for the IP opimization. My question though is a bit general, and I don't know how to use that for my case.
    – MajinSaha
    Nov 21 at 19:01












  • This is not called "interior point method". An interior point method is an optimization method, not a method to find a feasible point. On the contrary: they need a starting point within the interior.
    – LinAlg
    Nov 21 at 21:56










  • I am aware of that. I probably didn't explain clearly what I meant. Anyways, thanks for commenting.
    – MajinSaha
    Nov 21 at 23:26










  • Well, as I explained, you are looking for an interior point in a polyhedron (if it exists). So you may choose a point at the boundary of such polyhedron which is the intersection of at most $n$ halfplanes which are not linearly dependent, so such a point obeys the halfplane constraints with equality. If the number $m$ of constraints is larger than $n$, then the remaining constraints must be satisfied with inequality. Only then (if it fails) may there be a problem. So, use an active set method or use slack variables as Phase I of this IP problem to check feasibility.
    – Andreas
    2 days ago















up vote
0
down vote













Linear programming is your method of choice.



A good way to visualize what is going on is to consider the rows of $A$ as vectors $A_i$. So you have a set of inequalities $ A_i x + b_i geq 0$. Each of these inequalities defines a halfspace in $n$-dimensional space, with a boundary which is a hyperplane with normal vector $A_i$ and distance from the origin $b_i/ |A_i|$. Your matrix inequality equation is then the intersection of all of these halfspaces. If the intersection is not empty, the intersection space is a polyhedron.



Finding an $x$ is exactly the question whether this intersection is empty or not. In linear programming, this is called "interior point method" (you find it on wikipedia and elsewhere) and it is a well-known procedure for finding a feasible point to start any optimization procedure (which you do not have).






share|cite|improve this answer























  • Yes, I read about IP approach before. Currently, I'm only aware of a method to initialize a feasible starting point for IP for a linear program $mbox{min} (mathbf{c}^T mathbf{x})$, but such method requires the information on $mathbf{c}$. It minimizes some simple quadratic and involves lagrange multipliers to present the starting point for the IP opimization. My question though is a bit general, and I don't know how to use that for my case.
    – MajinSaha
    Nov 21 at 19:01












  • This is not called "interior point method". An interior point method is an optimization method, not a method to find a feasible point. On the contrary: they need a starting point within the interior.
    – LinAlg
    Nov 21 at 21:56










  • I am aware of that. I probably didn't explain clearly what I meant. Anyways, thanks for commenting.
    – MajinSaha
    Nov 21 at 23:26










  • Well, as I explained, you are looking for an interior point in a polyhedron (if it exists). So you may choose a point at the boundary of such polyhedron which is the intersection of at most $n$ halfplanes which are not linearly dependent, so such a point obeys the halfplane constraints with equality. If the number $m$ of constraints is larger than $n$, then the remaining constraints must be satisfied with inequality. Only then (if it fails) may there be a problem. So, use an active set method or use slack variables as Phase I of this IP problem to check feasibility.
    – Andreas
    2 days ago













up vote
0
down vote










up vote
0
down vote









Linear programming is your method of choice.



A good way to visualize what is going on is to consider the rows of $A$ as vectors $A_i$. So you have a set of inequalities $ A_i x + b_i geq 0$. Each of these inequalities defines a halfspace in $n$-dimensional space, with a boundary which is a hyperplane with normal vector $A_i$ and distance from the origin $b_i/ |A_i|$. Your matrix inequality equation is then the intersection of all of these halfspaces. If the intersection is not empty, the intersection space is a polyhedron.



Finding an $x$ is exactly the question whether this intersection is empty or not. In linear programming, this is called "interior point method" (you find it on wikipedia and elsewhere) and it is a well-known procedure for finding a feasible point to start any optimization procedure (which you do not have).






share|cite|improve this answer














Linear programming is your method of choice.



A good way to visualize what is going on is to consider the rows of $A$ as vectors $A_i$. So you have a set of inequalities $ A_i x + b_i geq 0$. Each of these inequalities defines a halfspace in $n$-dimensional space, with a boundary which is a hyperplane with normal vector $A_i$ and distance from the origin $b_i/ |A_i|$. Your matrix inequality equation is then the intersection of all of these halfspaces. If the intersection is not empty, the intersection space is a polyhedron.



Finding an $x$ is exactly the question whether this intersection is empty or not. In linear programming, this is called "interior point method" (you find it on wikipedia and elsewhere) and it is a well-known procedure for finding a feasible point to start any optimization procedure (which you do not have).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 21 at 21:58









LinAlg

7,6491520




7,6491520










answered Nov 21 at 17:44









Andreas

7,5441037




7,5441037












  • Yes, I read about IP approach before. Currently, I'm only aware of a method to initialize a feasible starting point for IP for a linear program $mbox{min} (mathbf{c}^T mathbf{x})$, but such method requires the information on $mathbf{c}$. It minimizes some simple quadratic and involves lagrange multipliers to present the starting point for the IP opimization. My question though is a bit general, and I don't know how to use that for my case.
    – MajinSaha
    Nov 21 at 19:01












  • This is not called "interior point method". An interior point method is an optimization method, not a method to find a feasible point. On the contrary: they need a starting point within the interior.
    – LinAlg
    Nov 21 at 21:56










  • I am aware of that. I probably didn't explain clearly what I meant. Anyways, thanks for commenting.
    – MajinSaha
    Nov 21 at 23:26










  • Well, as I explained, you are looking for an interior point in a polyhedron (if it exists). So you may choose a point at the boundary of such polyhedron which is the intersection of at most $n$ halfplanes which are not linearly dependent, so such a point obeys the halfplane constraints with equality. If the number $m$ of constraints is larger than $n$, then the remaining constraints must be satisfied with inequality. Only then (if it fails) may there be a problem. So, use an active set method or use slack variables as Phase I of this IP problem to check feasibility.
    – Andreas
    2 days ago


















  • Yes, I read about IP approach before. Currently, I'm only aware of a method to initialize a feasible starting point for IP for a linear program $mbox{min} (mathbf{c}^T mathbf{x})$, but such method requires the information on $mathbf{c}$. It minimizes some simple quadratic and involves lagrange multipliers to present the starting point for the IP opimization. My question though is a bit general, and I don't know how to use that for my case.
    – MajinSaha
    Nov 21 at 19:01












  • This is not called "interior point method". An interior point method is an optimization method, not a method to find a feasible point. On the contrary: they need a starting point within the interior.
    – LinAlg
    Nov 21 at 21:56










  • I am aware of that. I probably didn't explain clearly what I meant. Anyways, thanks for commenting.
    – MajinSaha
    Nov 21 at 23:26










  • Well, as I explained, you are looking for an interior point in a polyhedron (if it exists). So you may choose a point at the boundary of such polyhedron which is the intersection of at most $n$ halfplanes which are not linearly dependent, so such a point obeys the halfplane constraints with equality. If the number $m$ of constraints is larger than $n$, then the remaining constraints must be satisfied with inequality. Only then (if it fails) may there be a problem. So, use an active set method or use slack variables as Phase I of this IP problem to check feasibility.
    – Andreas
    2 days ago
















Yes, I read about IP approach before. Currently, I'm only aware of a method to initialize a feasible starting point for IP for a linear program $mbox{min} (mathbf{c}^T mathbf{x})$, but such method requires the information on $mathbf{c}$. It minimizes some simple quadratic and involves lagrange multipliers to present the starting point for the IP opimization. My question though is a bit general, and I don't know how to use that for my case.
– MajinSaha
Nov 21 at 19:01






Yes, I read about IP approach before. Currently, I'm only aware of a method to initialize a feasible starting point for IP for a linear program $mbox{min} (mathbf{c}^T mathbf{x})$, but such method requires the information on $mathbf{c}$. It minimizes some simple quadratic and involves lagrange multipliers to present the starting point for the IP opimization. My question though is a bit general, and I don't know how to use that for my case.
– MajinSaha
Nov 21 at 19:01














This is not called "interior point method". An interior point method is an optimization method, not a method to find a feasible point. On the contrary: they need a starting point within the interior.
– LinAlg
Nov 21 at 21:56




This is not called "interior point method". An interior point method is an optimization method, not a method to find a feasible point. On the contrary: they need a starting point within the interior.
– LinAlg
Nov 21 at 21:56












I am aware of that. I probably didn't explain clearly what I meant. Anyways, thanks for commenting.
– MajinSaha
Nov 21 at 23:26




I am aware of that. I probably didn't explain clearly what I meant. Anyways, thanks for commenting.
– MajinSaha
Nov 21 at 23:26












Well, as I explained, you are looking for an interior point in a polyhedron (if it exists). So you may choose a point at the boundary of such polyhedron which is the intersection of at most $n$ halfplanes which are not linearly dependent, so such a point obeys the halfplane constraints with equality. If the number $m$ of constraints is larger than $n$, then the remaining constraints must be satisfied with inequality. Only then (if it fails) may there be a problem. So, use an active set method or use slack variables as Phase I of this IP problem to check feasibility.
– Andreas
2 days ago




Well, as I explained, you are looking for an interior point in a polyhedron (if it exists). So you may choose a point at the boundary of such polyhedron which is the intersection of at most $n$ halfplanes which are not linearly dependent, so such a point obeys the halfplane constraints with equality. If the number $m$ of constraints is larger than $n$, then the remaining constraints must be satisfied with inequality. Only then (if it fails) may there be a problem. So, use an active set method or use slack variables as Phase I of this IP problem to check feasibility.
– Andreas
2 days ago


















 

draft saved


draft discarded



















































 


draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008047%2ffind-at-least-one-solution-to-matrix-inequality%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Berounka

Sphinx de Gizeh

Different font size/position of beamer's navigation symbols template's content depending on regular/plain...