Is the canonical form of LP programs unambiguous?

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Is the canonical form of LP programs unambiguous?



Because oddly I've seen some notes describe the canonical form as replacing $leq$ constraints with $=$. While some other notes said that canonical forms "only have $leq$ constraints".



So what's the canonical form of LP programs?







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  • 1




    I'm not sure that the terminology is completely standard, but I thought that a canonical form LP has the form: minimize $c^T x$ subject to $Ax=b, x geq 0$.
    – littleO
    Aug 17 at 6:47










  • @littleO Perhaps one can also intepret that the $=$ requirement is just a stricter version of "only have $leq$ constraints". Perhaps the difference it makes is not that big in practical problems.
    – mavavilj
    Aug 17 at 6:50















up vote
0
down vote

favorite












Is the canonical form of LP programs unambiguous?



Because oddly I've seen some notes describe the canonical form as replacing $leq$ constraints with $=$. While some other notes said that canonical forms "only have $leq$ constraints".



So what's the canonical form of LP programs?







share|cite|improve this question
















  • 1




    I'm not sure that the terminology is completely standard, but I thought that a canonical form LP has the form: minimize $c^T x$ subject to $Ax=b, x geq 0$.
    – littleO
    Aug 17 at 6:47










  • @littleO Perhaps one can also intepret that the $=$ requirement is just a stricter version of "only have $leq$ constraints". Perhaps the difference it makes is not that big in practical problems.
    – mavavilj
    Aug 17 at 6:50













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Is the canonical form of LP programs unambiguous?



Because oddly I've seen some notes describe the canonical form as replacing $leq$ constraints with $=$. While some other notes said that canonical forms "only have $leq$ constraints".



So what's the canonical form of LP programs?







share|cite|improve this question












Is the canonical form of LP programs unambiguous?



Because oddly I've seen some notes describe the canonical form as replacing $leq$ constraints with $=$. While some other notes said that canonical forms "only have $leq$ constraints".



So what's the canonical form of LP programs?









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 17 at 6:42









mavavilj

2,474731




2,474731







  • 1




    I'm not sure that the terminology is completely standard, but I thought that a canonical form LP has the form: minimize $c^T x$ subject to $Ax=b, x geq 0$.
    – littleO
    Aug 17 at 6:47










  • @littleO Perhaps one can also intepret that the $=$ requirement is just a stricter version of "only have $leq$ constraints". Perhaps the difference it makes is not that big in practical problems.
    – mavavilj
    Aug 17 at 6:50













  • 1




    I'm not sure that the terminology is completely standard, but I thought that a canonical form LP has the form: minimize $c^T x$ subject to $Ax=b, x geq 0$.
    – littleO
    Aug 17 at 6:47










  • @littleO Perhaps one can also intepret that the $=$ requirement is just a stricter version of "only have $leq$ constraints". Perhaps the difference it makes is not that big in practical problems.
    – mavavilj
    Aug 17 at 6:50








1




1




I'm not sure that the terminology is completely standard, but I thought that a canonical form LP has the form: minimize $c^T x$ subject to $Ax=b, x geq 0$.
– littleO
Aug 17 at 6:47




I'm not sure that the terminology is completely standard, but I thought that a canonical form LP has the form: minimize $c^T x$ subject to $Ax=b, x geq 0$.
– littleO
Aug 17 at 6:47












@littleO Perhaps one can also intepret that the $=$ requirement is just a stricter version of "only have $leq$ constraints". Perhaps the difference it makes is not that big in practical problems.
– mavavilj
Aug 17 at 6:50





@littleO Perhaps one can also intepret that the $=$ requirement is just a stricter version of "only have $leq$ constraints". Perhaps the difference it makes is not that big in practical problems.
– mavavilj
Aug 17 at 6:50
















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