Is the canonical form of LP programs unambiguous?
Clash Royale CLAN TAG#URR8PPP
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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?
linear-programming
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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?
linear-programming
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 $Ax=b, x geq 0$.
â 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.
â mavavilj
Aug 17 at 6:50
add a comment |Â
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?
linear-programming
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?
linear-programming
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 $Ax=b, x geq 0$.
â 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.
â mavavilj
Aug 17 at 6:50
add a comment |Â
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 $Ax=b, x geq 0$.
â 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.
â 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 $Ax=b, x geq 0$.
â 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 $Ax=b, x geq 0$.
â 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.
â 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.
â mavavilj
Aug 17 at 6:50
add a comment |Â
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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 $Ax=b, x geq 0$.
â 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.
â mavavilj
Aug 17 at 6:50