What is the dual for the following LP
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0
down vote
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Minimize $c^Tx$
subject to, $b_l leq Ax leq b_u \ lleq x leq u$
where $A$ is an $m times n$ matrix.
My approach was to separate each inequality into two.
So,
$b_l leq Ax\
Axleq b_u\
lleq x\
xleq u$
And the dual objective would then be,
Maximize $b_l^Ty+b_u^Ty+l^Ty+u^Ty$. I was not able to figure out the constraints
linear-programming
asked Nov 21 at 11:27
Who cares
14513
add a comment |
up vote
0
down vote
favorite
Minimize $c^Tx$
subject to, $b_l leq Ax leq b_u \ lleq x leq u$
where $A$ is an $m times n$ matrix.
My approach was to separate each inequality into two.
So,
$b_l leq Ax\
Axleq b_u\
lleq x\
xleq u$
And the dual objective would then be,
Maximize $b_l^Ty+b_u^Ty+l^Ty+u^Ty$. I was not able to figure out the constraints
linear-programming
asked Nov 21 at 11:27
Who cares
14513
Rewrite as $min c^Tx$ subject to $Bxle d$, $x$ free.
– A.Γ.
Nov 21 at 12:13
@A.Γ. I can make $x$ free via $x=x_1-x_2$,$x_1,x_2 geq 0$ but how do I remove the lower bounds?
– Who cares
Nov 21 at 12:23
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Minimize $c^Tx$
subject to, $b_l leq Ax leq b_u \ lleq x leq u$
where $A$ is an $m times n$ matrix.
My approach was to separate each inequality into two.
So,
$b_l leq Ax\
Axleq b_u\
lleq x\
xleq u$
And the dual objective would then be,
Maximize $b_l^Ty+b_u^Ty+l^Ty+u^Ty$. I was not able to figure out the constraints
linear-programming
asked Nov 21 at 11:27
Who cares
14513
Minimize $c^Tx$
subject to, $b_l leq Ax leq b_u \ lleq x leq u$
where $A$ is an $m times n$ matrix.
My approach was to separate each inequality into two.
So,
$b_l leq Ax\
Axleq b_u\
lleq x\
xleq u$
And the dual objective would then be,
Maximize $b_l^Ty+b_u^Ty+l^Ty+u^Ty$. I was not able to figure out the constraints
linear-programming
linear-programming
asked Nov 21 at 11:27
Who cares
14513
asked Nov 21 at 11:27
Who cares
14513
asked Nov 21 at 11:27
Who cares
14513
asked Nov 21 at 11:27
Who cares
14513
asked Nov 21 at 11:27
Who cares
14513
14513
Rewrite as $min c^Tx$ subject to $Bxle d$, $x$ free.
– A.Γ.
Nov 21 at 12:13
@A.Γ. I can make $x$ free via $x=x_1-x_2$,$x_1,x_2 geq 0$ but how do I remove the lower bounds?
– Who cares
Nov 21 at 12:23
add a comment |
Rewrite as $min c^Tx$ subject to $Bxle d$, $x$ free.
– A.Γ.
Nov 21 at 12:13
@A.Γ. I can make $x$ free via $x=x_1-x_2$,$x_1,x_2 geq 0$ but how do I remove the lower bounds?
– Who cares
Nov 21 at 12:23
Rewrite as $min c^Tx$ subject to $Bxle d$, $x$ free.
– A.Γ.
Nov 21 at 12:13
Rewrite as $min c^Tx$ subject to $Bxle d$, $x$ free.
– A.Γ.
Nov 21 at 12:13
@A.Γ. I can make $x$ free via $x=x_1-x_2$,$x_1,x_2 geq 0$ but how do I remove the lower bounds?
– Who cares
Nov 21 at 12:23
@A.Γ. I can make $x$ free via $x=x_1-x_2$,$x_1,x_2 geq 0$ but how do I remove the lower bounds?
– Who cares
Nov 21 at 12:23
add a comment |
1 Answer
1
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2
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accepted
Let $I$ be the identity matrix. Define
$$
B=begin{bmatrix}A\-A\I\-Iend{bmatrix},qquad d=begin{bmatrix}b_l\-b_u\l\-uend{bmatrix}
$$
then all the constraints can be written as $Bxge d$.
Calculate the dual problem to
$$
min c^Txquadtext{subject to }Bxge d.
$$
There are different ways to do it in optimization courses that covers LP, but let's do it via Lagrange duality. The Lagrange function is
$$
L(x,y)=c^Tx+y^T(d-Bx)=d^Ty+(c-B^Ty)^Tx,quad yge 0.
$$
The dual problem is
$$
max_{yge 0}min_x L(x,y).
$$
The inner minimum is easily calculated as
$$
min_x L(x,y)=
begin{cases}
d^Ty & text{ if }c-B^Ty=0,\
-infty & text{ otherwise}.
end{cases}
$$
Now for maximization the part with $-infty$ is not relevant, so we have the dual problem
$$
max d^Tyquadtext{subject to }B^Ty=c, yge 0.
$$
answered Nov 21 at 16:51
A.Γ.
21.7k22455
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $I$ be the identity matrix. Define
$$
B=begin{bmatrix}A\-A\I\-Iend{bmatrix},qquad d=begin{bmatrix}b_l\-b_u\l\-uend{bmatrix}
$$
then all the constraints can be written as $Bxge d$.
Calculate the dual problem to
$$
min c^Txquadtext{subject to }Bxge d.
$$
There are different ways to do it in optimization courses that covers LP, but let's do it via Lagrange duality. The Lagrange function is
$$
L(x,y)=c^Tx+y^T(d-Bx)=d^Ty+(c-B^Ty)^Tx,quad yge 0.
$$
The dual problem is
$$
max_{yge 0}min_x L(x,y).
$$
The inner minimum is easily calculated as
$$
min_x L(x,y)=
begin{cases}
d^Ty & text{ if }c-B^Ty=0,\
-infty & text{ otherwise}.
end{cases}
$$
Now for maximization the part with $-infty$ is not relevant, so we have the dual problem
$$
max d^Tyquadtext{subject to }B^Ty=c, yge 0.
$$
answered Nov 21 at 16:51
A.Γ.
21.7k22455
add a comment |
up vote
2
down vote
accepted
Let $I$ be the identity matrix. Define
$$
B=begin{bmatrix}A\-A\I\-Iend{bmatrix},qquad d=begin{bmatrix}b_l\-b_u\l\-uend{bmatrix}
$$
then all the constraints can be written as $Bxge d$.
Calculate the dual problem to
$$
min c^Txquadtext{subject to }Bxge d.
$$
There are different ways to do it in optimization courses that covers LP, but let's do it via Lagrange duality. The Lagrange function is
$$
L(x,y)=c^Tx+y^T(d-Bx)=d^Ty+(c-B^Ty)^Tx,quad yge 0.
$$
The dual problem is
$$
max_{yge 0}min_x L(x,y).
$$
The inner minimum is easily calculated as
$$
min_x L(x,y)=
begin{cases}
d^Ty & text{ if }c-B^Ty=0,\
-infty & text{ otherwise}.
end{cases}
$$
Now for maximization the part with $-infty$ is not relevant, so we have the dual problem
$$
max d^Tyquadtext{subject to }B^Ty=c, yge 0.
$$
answered Nov 21 at 16:51
A.Γ.
21.7k22455
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $I$ be the identity matrix. Define
$$
B=begin{bmatrix}A\-A\I\-Iend{bmatrix},qquad d=begin{bmatrix}b_l\-b_u\l\-uend{bmatrix}
$$
then all the constraints can be written as $Bxge d$.
Calculate the dual problem to
$$
min c^Txquadtext{subject to }Bxge d.
$$
There are different ways to do it in optimization courses that covers LP, but let's do it via Lagrange duality. The Lagrange function is
$$
L(x,y)=c^Tx+y^T(d-Bx)=d^Ty+(c-B^Ty)^Tx,quad yge 0.
$$
The dual problem is
$$
max_{yge 0}min_x L(x,y).
$$
The inner minimum is easily calculated as
$$
min_x L(x,y)=
begin{cases}
d^Ty & text{ if }c-B^Ty=0,\
-infty & text{ otherwise}.
end{cases}
$$
Now for maximization the part with $-infty$ is not relevant, so we have the dual problem
$$
max d^Tyquadtext{subject to }B^Ty=c, yge 0.
$$
answered Nov 21 at 16:51
A.Γ.
21.7k22455
Let $I$ be the identity matrix. Define
$$
B=begin{bmatrix}A\-A\I\-Iend{bmatrix},qquad d=begin{bmatrix}b_l\-b_u\l\-uend{bmatrix}
$$
then all the constraints can be written as $Bxge d$.
Calculate the dual problem to
$$
min c^Txquadtext{subject to }Bxge d.
$$
There are different ways to do it in optimization courses that covers LP, but let's do it via Lagrange duality. The Lagrange function is
$$
L(x,y)=c^Tx+y^T(d-Bx)=d^Ty+(c-B^Ty)^Tx,quad yge 0.
$$
The dual problem is
$$
max_{yge 0}min_x L(x,y).
$$
The inner minimum is easily calculated as
$$
min_x L(x,y)=
begin{cases}
d^Ty & text{ if }c-B^Ty=0,\
-infty & text{ otherwise}.
end{cases}
$$
Now for maximization the part with $-infty$ is not relevant, so we have the dual problem
$$
max d^Tyquadtext{subject to }B^Ty=c, yge 0.
$$
answered Nov 21 at 16:51
A.Γ.
21.7k22455
answered Nov 21 at 16:51
A.Γ.
21.7k22455
answered Nov 21 at 16:51
A.Γ.
21.7k22455
answered Nov 21 at 16:51
A.Γ.
21.7k22455
21.7k22455
add a comment |
add a comment |
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Rewrite as $min c^Tx$ subject to $Bxle d$, $x$ free.
– A.Γ.
Nov 21 at 12:13
@A.Γ. I can make $x$ free via $x=x_1-x_2$,$x_1,x_2 geq 0$ but how do I remove the lower bounds?
– Who cares
Nov 21 at 12:23