If the solution set of linear programming problem is unbounded, can you find that out in finite steps?











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Let $(P) maxleft{c^T cdot x mid A cdot x leq b, x geq
0right}$
be an arbitrary linear programming problem and $M$ its
solution set. Is it possible to find out if $M$ is unbounded (in
finite steps)?




I think the answer is yes because we have linear program here and that's why it's possible to tell in polynomial time if it has feasible solution or infeasible. So there must be an algorithm to do this in polynomial time e.g. in finite steps.



But I don't know any algorithm to do this. So far I've read about the simplex method in a book and if I understood correctly, if you run simplex method on the LP above, you will see in the table that the problem is unbounded because there will appear one or more variables where you can pivot these to $infty$.



But this is only thought is it correct like that? Maybe it's possible with another way too?










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  • 1




    Yes, it is possible. You may look here. NB(!): the standard form there is minimization and equality $Ax=b$.
    – A.Γ.
    Nov 10 at 11:14















up vote
1
down vote

favorite













Let $(P) maxleft{c^T cdot x mid A cdot x leq b, x geq
0right}$
be an arbitrary linear programming problem and $M$ its
solution set. Is it possible to find out if $M$ is unbounded (in
finite steps)?




I think the answer is yes because we have linear program here and that's why it's possible to tell in polynomial time if it has feasible solution or infeasible. So there must be an algorithm to do this in polynomial time e.g. in finite steps.



But I don't know any algorithm to do this. So far I've read about the simplex method in a book and if I understood correctly, if you run simplex method on the LP above, you will see in the table that the problem is unbounded because there will appear one or more variables where you can pivot these to $infty$.



But this is only thought is it correct like that? Maybe it's possible with another way too?










share|cite|improve this question


















  • 1




    Yes, it is possible. You may look here. NB(!): the standard form there is minimization and equality $Ax=b$.
    – A.Γ.
    Nov 10 at 11:14













up vote
1
down vote

favorite









up vote
1
down vote

favorite












Let $(P) maxleft{c^T cdot x mid A cdot x leq b, x geq
0right}$
be an arbitrary linear programming problem and $M$ its
solution set. Is it possible to find out if $M$ is unbounded (in
finite steps)?




I think the answer is yes because we have linear program here and that's why it's possible to tell in polynomial time if it has feasible solution or infeasible. So there must be an algorithm to do this in polynomial time e.g. in finite steps.



But I don't know any algorithm to do this. So far I've read about the simplex method in a book and if I understood correctly, if you run simplex method on the LP above, you will see in the table that the problem is unbounded because there will appear one or more variables where you can pivot these to $infty$.



But this is only thought is it correct like that? Maybe it's possible with another way too?










share|cite|improve this question














Let $(P) maxleft{c^T cdot x mid A cdot x leq b, x geq
0right}$
be an arbitrary linear programming problem and $M$ its
solution set. Is it possible to find out if $M$ is unbounded (in
finite steps)?




I think the answer is yes because we have linear program here and that's why it's possible to tell in polynomial time if it has feasible solution or infeasible. So there must be an algorithm to do this in polynomial time e.g. in finite steps.



But I don't know any algorithm to do this. So far I've read about the simplex method in a book and if I understood correctly, if you run simplex method on the LP above, you will see in the table that the problem is unbounded because there will appear one or more variables where you can pivot these to $infty$.



But this is only thought is it correct like that? Maybe it's possible with another way too?







optimization linear-programming






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asked Nov 10 at 10:57









tenepolis

4011316




4011316








  • 1




    Yes, it is possible. You may look here. NB(!): the standard form there is minimization and equality $Ax=b$.
    – A.Γ.
    Nov 10 at 11:14














  • 1




    Yes, it is possible. You may look here. NB(!): the standard form there is minimization and equality $Ax=b$.
    – A.Γ.
    Nov 10 at 11:14








1




1




Yes, it is possible. You may look here. NB(!): the standard form there is minimization and equality $Ax=b$.
– A.Γ.
Nov 10 at 11:14




Yes, it is possible. You may look here. NB(!): the standard form there is minimization and equality $Ax=b$.
– A.Γ.
Nov 10 at 11:14










1 Answer
1






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oldest

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up vote
4
down vote



accepted










$$max c^Tx$$



subject to



$$Axle b$$



$$x ge 0$$



Using the simplex algorithm, we can determine if there is a solution, $x^*$.



Now consider



$$max e^Tx$$



subject to



$$Axle b$$
$$c^Tx=c^Tx^*$$
$$x ge 0$$



where $e$ is the all-one vector.



We check if the solution is unbounded, if it is unbounded, then $M$ is unbounded.






share|cite|improve this answer























  • Oh hi you again :p That $e$ is the same as from my previous question, right (the "all ones vector")?
    – tenepolis
    Nov 10 at 14:01






  • 1




    ah yes, it's my habit to use that notation.
    – Siong Thye Goh
    Nov 10 at 14:06











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










$$max c^Tx$$



subject to



$$Axle b$$



$$x ge 0$$



Using the simplex algorithm, we can determine if there is a solution, $x^*$.



Now consider



$$max e^Tx$$



subject to



$$Axle b$$
$$c^Tx=c^Tx^*$$
$$x ge 0$$



where $e$ is the all-one vector.



We check if the solution is unbounded, if it is unbounded, then $M$ is unbounded.






share|cite|improve this answer























  • Oh hi you again :p That $e$ is the same as from my previous question, right (the "all ones vector")?
    – tenepolis
    Nov 10 at 14:01






  • 1




    ah yes, it's my habit to use that notation.
    – Siong Thye Goh
    Nov 10 at 14:06















up vote
4
down vote



accepted










$$max c^Tx$$



subject to



$$Axle b$$



$$x ge 0$$



Using the simplex algorithm, we can determine if there is a solution, $x^*$.



Now consider



$$max e^Tx$$



subject to



$$Axle b$$
$$c^Tx=c^Tx^*$$
$$x ge 0$$



where $e$ is the all-one vector.



We check if the solution is unbounded, if it is unbounded, then $M$ is unbounded.






share|cite|improve this answer























  • Oh hi you again :p That $e$ is the same as from my previous question, right (the "all ones vector")?
    – tenepolis
    Nov 10 at 14:01






  • 1




    ah yes, it's my habit to use that notation.
    – Siong Thye Goh
    Nov 10 at 14:06













up vote
4
down vote



accepted







up vote
4
down vote



accepted






$$max c^Tx$$



subject to



$$Axle b$$



$$x ge 0$$



Using the simplex algorithm, we can determine if there is a solution, $x^*$.



Now consider



$$max e^Tx$$



subject to



$$Axle b$$
$$c^Tx=c^Tx^*$$
$$x ge 0$$



where $e$ is the all-one vector.



We check if the solution is unbounded, if it is unbounded, then $M$ is unbounded.






share|cite|improve this answer














$$max c^Tx$$



subject to



$$Axle b$$



$$x ge 0$$



Using the simplex algorithm, we can determine if there is a solution, $x^*$.



Now consider



$$max e^Tx$$



subject to



$$Axle b$$
$$c^Tx=c^Tx^*$$
$$x ge 0$$



where $e$ is the all-one vector.



We check if the solution is unbounded, if it is unbounded, then $M$ is unbounded.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 10 at 14:07

























answered Nov 10 at 12:58









Siong Thye Goh

93.3k1462114




93.3k1462114












  • Oh hi you again :p That $e$ is the same as from my previous question, right (the "all ones vector")?
    – tenepolis
    Nov 10 at 14:01






  • 1




    ah yes, it's my habit to use that notation.
    – Siong Thye Goh
    Nov 10 at 14:06


















  • Oh hi you again :p That $e$ is the same as from my previous question, right (the "all ones vector")?
    – tenepolis
    Nov 10 at 14:01






  • 1




    ah yes, it's my habit to use that notation.
    – Siong Thye Goh
    Nov 10 at 14:06
















Oh hi you again :p That $e$ is the same as from my previous question, right (the "all ones vector")?
– tenepolis
Nov 10 at 14:01




Oh hi you again :p That $e$ is the same as from my previous question, right (the "all ones vector")?
– tenepolis
Nov 10 at 14:01




1




1




ah yes, it's my habit to use that notation.
– Siong Thye Goh
Nov 10 at 14:06




ah yes, it's my habit to use that notation.
– Siong Thye Goh
Nov 10 at 14:06


















 

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