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What exactly is a basis matrix in LPP?
The Next CEO of Stack OverflowSimplex method : Improving a basic feasible solution by BazaraaLP: nonbasic solution made into basic solution, help me with this terminologyHow to justify that a basic feasible solution to a Linear Program corresponds to an extreme point of the feasible region?linear programming problem given initial solutionSimplex method state after first phaseBasic and non basic variables in linear programmingWhat is underlying meaning of basis in simplex method?When using the simplex method , how do we know that the number of basic variables will be exactly equal to n+1?Recover linear programming basisReduced Costs and Nonbasic Indices in Linear Programming
$begingroup$
I'm studying simplex for solving a LPP. $A$ is a $mtimes n$ matix. $$Ax=b$$Then the book says:
Let $x$ be a basic feasible solution to the standard form problem.
Let $$B(1), B(2),cdots B(m)$$ be the indices of the basic variables and let $$B=[A_B(1)cdots A_B(m)]$$ be the corresponding basis matrix.
I can't understand what's this $B$ a basis for. Thanks!
linear-programming
edited Mar 18 at 11:54
MarianD
1,7391617
asked Mar 18 at 11:33
Ankit KumarAnkit Kumar
1,542221
$endgroup$
add a comment |
$begingroup$
I'm studying simplex for solving a LPP. $A$ is a $mtimes n$ matix. $$Ax=b$$Then the book says:
Let $x$ be a basic feasible solution to the standard form problem.
Let $$B(1), B(2),cdots B(m)$$ be the indices of the basic variables and let $$B=[A_B(1)cdots A_B(m)]$$ be the corresponding basis matrix.
I can't understand what's this $B$ a basis for. Thanks!
linear-programming
edited Mar 18 at 11:54
MarianD
1,7391617
asked Mar 18 at 11:33
Ankit KumarAnkit Kumar
1,542221
$endgroup$
add a comment |
$begingroup$
I'm studying simplex for solving a LPP. $A$ is a $mtimes n$ matix. $$Ax=b$$Then the book says:
Let $x$ be a basic feasible solution to the standard form problem.
Let $$B(1), B(2),cdots B(m)$$ be the indices of the basic variables and let $$B=[A_B(1)cdots A_B(m)]$$ be the corresponding basis matrix.
I can't understand what's this $B$ a basis for. Thanks!
linear-programming
edited Mar 18 at 11:54
MarianD
1,7391617
asked Mar 18 at 11:33
Ankit KumarAnkit Kumar
1,542221
$endgroup$
I'm studying simplex for solving a LPP. $A$ is a $mtimes n$ matix. $$Ax=b$$Then the book says:
Let $x$ be a basic feasible solution to the standard form problem.
Let $$B(1), B(2),cdots B(m)$$ be the indices of the basic variables and let $$B=[A_B(1)cdots A_B(m)]$$ be the corresponding basis matrix.
I can't understand what's this $B$ a basis for. Thanks!
linear-programming
linear-programming
edited Mar 18 at 11:54
MarianD
1,7391617
asked Mar 18 at 11:33
Ankit KumarAnkit Kumar
1,542221
edited Mar 18 at 11:54
MarianD
1,7391617
asked Mar 18 at 11:33
Ankit KumarAnkit Kumar
1,542221
edited Mar 18 at 11:54
MarianD
1,7391617
edited Mar 18 at 11:54
MarianD
1,7391617
edited Mar 18 at 11:54
MarianD
1,7391617
1,7391617
asked Mar 18 at 11:33
Ankit KumarAnkit Kumar
1,542221
asked Mar 18 at 11:33
Ankit KumarAnkit Kumar
1,542221
asked Mar 18 at 11:33
Ankit KumarAnkit Kumar
1,542221
1,542221
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that the columns of $B$ are independent and the columns of $B$ indeed span $mathbbR^m$.
We can always solve $$Bx_B=b$$ for any given $b$.
answered Mar 18 at 11:39
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
Ya, true. But since we are in $mathbbR^n$, why do we care about $mathbbR^m$?
$endgroup$
– Ankit Kumar
Mar 18 at 11:56
$begingroup$
the BFS of the standard form of LP has this property that $n-m$ entries are $0$ for sure, those are the non-basic variables. Those potentially non-zero positions are obtained by solving the linear system above.
$endgroup$
– Siong Thye Goh
Mar 18 at 12:00
$begingroup$
Okay. I got it, thanks!
$endgroup$
– Ankit Kumar
Mar 18 at 12:01
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that the columns of $B$ are independent and the columns of $B$ indeed span $mathbbR^m$.
We can always solve $$Bx_B=b$$ for any given $b$.
answered Mar 18 at 11:39
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
Ya, true. But since we are in $mathbbR^n$, why do we care about $mathbbR^m$?
$endgroup$
– Ankit Kumar
Mar 18 at 11:56
$begingroup$
the BFS of the standard form of LP has this property that $n-m$ entries are $0$ for sure, those are the non-basic variables. Those potentially non-zero positions are obtained by solving the linear system above.
$endgroup$
– Siong Thye Goh
Mar 18 at 12:00
$begingroup$
Okay. I got it, thanks!
$endgroup$
– Ankit Kumar
Mar 18 at 12:01
add a comment |
$begingroup$
Note that the columns of $B$ are independent and the columns of $B$ indeed span $mathbbR^m$.
We can always solve $$Bx_B=b$$ for any given $b$.
answered Mar 18 at 11:39
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
Ya, true. But since we are in $mathbbR^n$, why do we care about $mathbbR^m$?
$endgroup$
– Ankit Kumar
Mar 18 at 11:56
$begingroup$
the BFS of the standard form of LP has this property that $n-m$ entries are $0$ for sure, those are the non-basic variables. Those potentially non-zero positions are obtained by solving the linear system above.
$endgroup$
– Siong Thye Goh
Mar 18 at 12:00
$begingroup$
Okay. I got it, thanks!
$endgroup$
– Ankit Kumar
Mar 18 at 12:01
add a comment |
$begingroup$
Note that the columns of $B$ are independent and the columns of $B$ indeed span $mathbbR^m$.
We can always solve $$Bx_B=b$$ for any given $b$.
answered Mar 18 at 11:39
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
Note that the columns of $B$ are independent and the columns of $B$ indeed span $mathbbR^m$.
We can always solve $$Bx_B=b$$ for any given $b$.
answered Mar 18 at 11:39
Siong Thye GohSiong Thye Goh
103k1468119
edited Mar 18 at 11:55
answered Mar 18 at 11:39
Siong Thye GohSiong Thye Goh
103k1468119
answered Mar 18 at 11:39
Siong Thye GohSiong Thye Goh
103k1468119
answered Mar 18 at 11:39
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
$begingroup$
Ya, true. But since we are in $mathbbR^n$, why do we care about $mathbbR^m$?
$endgroup$
– Ankit Kumar
Mar 18 at 11:56
$begingroup$
the BFS of the standard form of LP has this property that $n-m$ entries are $0$ for sure, those are the non-basic variables. Those potentially non-zero positions are obtained by solving the linear system above.
$endgroup$
– Siong Thye Goh
Mar 18 at 12:00
$begingroup$
Okay. I got it, thanks!
$endgroup$
– Ankit Kumar
Mar 18 at 12:01
add a comment |
$begingroup$
Ya, true. But since we are in $mathbbR^n$, why do we care about $mathbbR^m$?
$endgroup$
– Ankit Kumar
Mar 18 at 11:56
$begingroup$
the BFS of the standard form of LP has this property that $n-m$ entries are $0$ for sure, those are the non-basic variables. Those potentially non-zero positions are obtained by solving the linear system above.
$endgroup$
– Siong Thye Goh
Mar 18 at 12:00
$begingroup$
Okay. I got it, thanks!
$endgroup$
– Ankit Kumar
Mar 18 at 12:01
$begingroup$
Ya, true. But since we are in $mathbbR^n$, why do we care about $mathbbR^m$?
$endgroup$
– Ankit Kumar
Mar 18 at 11:56
$begingroup$
Ya, true. But since we are in $mathbbR^n$, why do we care about $mathbbR^m$?
$endgroup$
– Ankit Kumar
Mar 18 at 11:56
$begingroup$
the BFS of the standard form of LP has this property that $n-m$ entries are $0$ for sure, those are the non-basic variables. Those potentially non-zero positions are obtained by solving the linear system above.
$endgroup$
– Siong Thye Goh
Mar 18 at 12:00
$begingroup$
the BFS of the standard form of LP has this property that $n-m$ entries are $0$ for sure, those are the non-basic variables. Those potentially non-zero positions are obtained by solving the linear system above.
$endgroup$
– Siong Thye Goh
Mar 18 at 12:00
$begingroup$
Okay. I got it, thanks!
$endgroup$
– Ankit Kumar
Mar 18 at 12:01
$begingroup$
Okay. I got it, thanks!
$endgroup$
– Ankit Kumar
Mar 18 at 12:01
add a comment |
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