multiplication of nonzero scalar in a constraint of the primal
Suppose we have primal and its dual in standard form, that is
begin{align*}
(P) max z = cx \
st ; ; Ax = b \
; ; ; x geq 0 \
end{align*}
begin{align*}
(D) min z = by \
st ; ; yA geq c \
; ; ; y ; ; ; free \
end{align*}
Where $A$ is an $m $ by $n$ matrix an $x$ is an n vector and $y$ is an m vector.
Suppose we multiply one of the constraints of the primal by some number $alpha > 0$. Does this affect the solution of the dual?
Thoughts:
Since a constraint is of the form $a_{ij} cdot x $, take one of the $i's$, say we multiply
$$ a_{i1}x_1 + a_{i2} x_2 + ... + a_{in} x_n $$
by $alpha $
Once we set up our tableau, once we divide this row by $alpha$, then in the LFH, we would have
$$ frac{ b_i}{alpha} $$
the ith component of the vector $b$. Doesnt it change the solution in the primal tableau? Since solutions are the same for primal and dual???
linear-programming
asked Nov 2 at 7:10
Neymar
374113
add a comment |
Suppose we have primal and its dual in standard form, that is
begin{align*}
(P) max z = cx \
st ; ; Ax = b \
; ; ; x geq 0 \
end{align*}
begin{align*}
(D) min z = by \
st ; ; yA geq c \
; ; ; y ; ; ; free \
end{align*}
Where $A$ is an $m $ by $n$ matrix an $x$ is an n vector and $y$ is an m vector.
Suppose we multiply one of the constraints of the primal by some number $alpha > 0$. Does this affect the solution of the dual?
Thoughts:
Since a constraint is of the form $a_{ij} cdot x $, take one of the $i's$, say we multiply
$$ a_{i1}x_1 + a_{i2} x_2 + ... + a_{in} x_n $$
by $alpha $
Once we set up our tableau, once we divide this row by $alpha$, then in the LFH, we would have
$$ frac{ b_i}{alpha} $$
the ith component of the vector $b$. Doesnt it change the solution in the primal tableau? Since solutions are the same for primal and dual???
linear-programming
asked Nov 2 at 7:10
Neymar
374113
add a comment |
Suppose we have primal and its dual in standard form, that is
begin{align*}
(P) max z = cx \
st ; ; Ax = b \
; ; ; x geq 0 \
end{align*}
begin{align*}
(D) min z = by \
st ; ; yA geq c \
; ; ; y ; ; ; free \
end{align*}
Where $A$ is an $m $ by $n$ matrix an $x$ is an n vector and $y$ is an m vector.
Suppose we multiply one of the constraints of the primal by some number $alpha > 0$. Does this affect the solution of the dual?
Thoughts:
Since a constraint is of the form $a_{ij} cdot x $, take one of the $i's$, say we multiply
$$ a_{i1}x_1 + a_{i2} x_2 + ... + a_{in} x_n $$
by $alpha $
Once we set up our tableau, once we divide this row by $alpha$, then in the LFH, we would have
$$ frac{ b_i}{alpha} $$
the ith component of the vector $b$. Doesnt it change the solution in the primal tableau? Since solutions are the same for primal and dual???
linear-programming
asked Nov 2 at 7:10
Neymar
374113
Suppose we have primal and its dual in standard form, that is
begin{align*}
(P) max z = cx \
st ; ; Ax = b \
; ; ; x geq 0 \
end{align*}
begin{align*}
(D) min z = by \
st ; ; yA geq c \
; ; ; y ; ; ; free \
end{align*}
Where $A$ is an $m $ by $n$ matrix an $x$ is an n vector and $y$ is an m vector.
Suppose we multiply one of the constraints of the primal by some number $alpha > 0$. Does this affect the solution of the dual?
Thoughts:
Since a constraint is of the form $a_{ij} cdot x $, take one of the $i's$, say we multiply
$$ a_{i1}x_1 + a_{i2} x_2 + ... + a_{in} x_n $$
by $alpha $
Once we set up our tableau, once we divide this row by $alpha$, then in the LFH, we would have
$$ frac{ b_i}{alpha} $$
the ith component of the vector $b$. Doesnt it change the solution in the primal tableau? Since solutions are the same for primal and dual???
linear-programming
linear-programming
asked Nov 2 at 7:10
Neymar
374113
asked Nov 2 at 7:10
Neymar
374113
asked Nov 2 at 7:10
Neymar
374113
asked Nov 2 at 7:10
Neymar
374113
asked Nov 2 at 7:10
Neymar
374113
374113
add a comment |
add a comment |
1 Answer
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Multiplication by a non-zero scalar is equivalent to multiplication of an elementary matrix, $E$.
begin{align*}
(P') max z = c^Tx \
st ; ; (EA)x = (Eb) \
; ; ; x geq 0 \
end{align*}
The dual is
begin{align*}
(D') min z = (Eb)^Ty \
st ; ; y^T(EA) geq c \
; ; ; y ; ; ; free \
end{align*}
Suppose $w$ is the original dual solution, then $y=E^{-T}w.$
For the operation of multiplication by a scalar, we have $E^T=E$.
Hence $y=E^{-1}w$. That is if we multiply $alpha$ to the $i$-th constraint, now for the dual solution, we would divide $w_i$ by $alpha$ and we can keep the rest to be the same.
answered Nov 2 at 7:33
Siong Thye Goh
99.1k1464117
Can you help me with this related problem: math.stackexchange.com/questions/2989440/…
– Neymar
Nov 8 at 15:11
at first glance, it seems that doing that operation might cause the primal problem to become infeasible.
– Siong Thye Goh
Nov 8 at 15:24
But isnt it doing the same operations as the previous case but this time to the dual?
– Neymar
Nov 8 at 18:06
the duality is in inequality form, multiplying an inequality and adding it to another one might change the feasible set. Also, if you multiply by a negative number, the inequality get flipped.
– Siong Thye Goh
Nov 8 at 18:09
You are right :/
– Neymar
Nov 8 at 18:23
|
show 4 more comments
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Multiplication by a non-zero scalar is equivalent to multiplication of an elementary matrix, $E$.
begin{align*}
(P') max z = c^Tx \
st ; ; (EA)x = (Eb) \
; ; ; x geq 0 \
end{align*}
The dual is
begin{align*}
(D') min z = (Eb)^Ty \
st ; ; y^T(EA) geq c \
; ; ; y ; ; ; free \
end{align*}
Suppose $w$ is the original dual solution, then $y=E^{-T}w.$
For the operation of multiplication by a scalar, we have $E^T=E$.
Hence $y=E^{-1}w$. That is if we multiply $alpha$ to the $i$-th constraint, now for the dual solution, we would divide $w_i$ by $alpha$ and we can keep the rest to be the same.
answered Nov 2 at 7:33
Siong Thye Goh
99.1k1464117
Can you help me with this related problem: math.stackexchange.com/questions/2989440/…
– Neymar
Nov 8 at 15:11
at first glance, it seems that doing that operation might cause the primal problem to become infeasible.
– Siong Thye Goh
Nov 8 at 15:24
But isnt it doing the same operations as the previous case but this time to the dual?
– Neymar
Nov 8 at 18:06
the duality is in inequality form, multiplying an inequality and adding it to another one might change the feasible set. Also, if you multiply by a negative number, the inequality get flipped.
– Siong Thye Goh
Nov 8 at 18:09
You are right :/
– Neymar
Nov 8 at 18:23
|
show 4 more comments
Multiplication by a non-zero scalar is equivalent to multiplication of an elementary matrix, $E$.
begin{align*}
(P') max z = c^Tx \
st ; ; (EA)x = (Eb) \
; ; ; x geq 0 \
end{align*}
The dual is
begin{align*}
(D') min z = (Eb)^Ty \
st ; ; y^T(EA) geq c \
; ; ; y ; ; ; free \
end{align*}
Suppose $w$ is the original dual solution, then $y=E^{-T}w.$
For the operation of multiplication by a scalar, we have $E^T=E$.
Hence $y=E^{-1}w$. That is if we multiply $alpha$ to the $i$-th constraint, now for the dual solution, we would divide $w_i$ by $alpha$ and we can keep the rest to be the same.
answered Nov 2 at 7:33
Siong Thye Goh
99.1k1464117
Can you help me with this related problem: math.stackexchange.com/questions/2989440/…
– Neymar
Nov 8 at 15:11
at first glance, it seems that doing that operation might cause the primal problem to become infeasible.
– Siong Thye Goh
Nov 8 at 15:24
But isnt it doing the same operations as the previous case but this time to the dual?
– Neymar
Nov 8 at 18:06
the duality is in inequality form, multiplying an inequality and adding it to another one might change the feasible set. Also, if you multiply by a negative number, the inequality get flipped.
– Siong Thye Goh
Nov 8 at 18:09
You are right :/
– Neymar
Nov 8 at 18:23
|
show 4 more comments
Multiplication by a non-zero scalar is equivalent to multiplication of an elementary matrix, $E$.
begin{align*}
(P') max z = c^Tx \
st ; ; (EA)x = (Eb) \
; ; ; x geq 0 \
end{align*}
The dual is
begin{align*}
(D') min z = (Eb)^Ty \
st ; ; y^T(EA) geq c \
; ; ; y ; ; ; free \
end{align*}
Suppose $w$ is the original dual solution, then $y=E^{-T}w.$
For the operation of multiplication by a scalar, we have $E^T=E$.
Hence $y=E^{-1}w$. That is if we multiply $alpha$ to the $i$-th constraint, now for the dual solution, we would divide $w_i$ by $alpha$ and we can keep the rest to be the same.
answered Nov 2 at 7:33
Siong Thye Goh
99.1k1464117
Multiplication by a non-zero scalar is equivalent to multiplication of an elementary matrix, $E$.
begin{align*}
(P') max z = c^Tx \
st ; ; (EA)x = (Eb) \
; ; ; x geq 0 \
end{align*}
The dual is
begin{align*}
(D') min z = (Eb)^Ty \
st ; ; y^T(EA) geq c \
; ; ; y ; ; ; free \
end{align*}
Suppose $w$ is the original dual solution, then $y=E^{-T}w.$
For the operation of multiplication by a scalar, we have $E^T=E$.
Hence $y=E^{-1}w$. That is if we multiply $alpha$ to the $i$-th constraint, now for the dual solution, we would divide $w_i$ by $alpha$ and we can keep the rest to be the same.
answered Nov 2 at 7:33
Siong Thye Goh
99.1k1464117
answered Nov 2 at 7:33
Siong Thye Goh
99.1k1464117
answered Nov 2 at 7:33
Siong Thye Goh
99.1k1464117
answered Nov 2 at 7:33
Siong Thye Goh
99.1k1464117
99.1k1464117
Can you help me with this related problem: math.stackexchange.com/questions/2989440/…
– Neymar
Nov 8 at 15:11
at first glance, it seems that doing that operation might cause the primal problem to become infeasible.
– Siong Thye Goh
Nov 8 at 15:24
But isnt it doing the same operations as the previous case but this time to the dual?
– Neymar
Nov 8 at 18:06
the duality is in inequality form, multiplying an inequality and adding it to another one might change the feasible set. Also, if you multiply by a negative number, the inequality get flipped.
– Siong Thye Goh
Nov 8 at 18:09
You are right :/
– Neymar
Nov 8 at 18:23
|
show 4 more comments
Can you help me with this related problem: math.stackexchange.com/questions/2989440/…
– Neymar
Nov 8 at 15:11
at first glance, it seems that doing that operation might cause the primal problem to become infeasible.
– Siong Thye Goh
Nov 8 at 15:24
But isnt it doing the same operations as the previous case but this time to the dual?
– Neymar
Nov 8 at 18:06
the duality is in inequality form, multiplying an inequality and adding it to another one might change the feasible set. Also, if you multiply by a negative number, the inequality get flipped.
– Siong Thye Goh
Nov 8 at 18:09
You are right :/
– Neymar
Nov 8 at 18:23
Can you help me with this related problem: math.stackexchange.com/questions/2989440/…
– Neymar
Nov 8 at 15:11
Can you help me with this related problem: math.stackexchange.com/questions/2989440/…
– Neymar
Nov 8 at 15:11
at first glance, it seems that doing that operation might cause the primal problem to become infeasible.
– Siong Thye Goh
Nov 8 at 15:24
at first glance, it seems that doing that operation might cause the primal problem to become infeasible.
– Siong Thye Goh
Nov 8 at 15:24
But isnt it doing the same operations as the previous case but this time to the dual?
– Neymar
Nov 8 at 18:06
But isnt it doing the same operations as the previous case but this time to the dual?
– Neymar
Nov 8 at 18:06
the duality is in inequality form, multiplying an inequality and adding it to another one might change the feasible set. Also, if you multiply by a negative number, the inequality get flipped.
– Siong Thye Goh
Nov 8 at 18:09
the duality is in inequality form, multiplying an inequality and adding it to another one might change the feasible set. Also, if you multiply by a negative number, the inequality get flipped.
– Siong Thye Goh
Nov 8 at 18:09
You are right :/
– Neymar
Nov 8 at 18:23
You are right :/
– Neymar
Nov 8 at 18:23
|
show 4 more comments
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