Question to the solution of “Indicator Variable if x is in specific range”
$begingroup$
This question is to query the solution provided by Erwin Kalvelagen to the post
Indicator Variable if x is in specific range
and
conditional constraint: if $x in [a,b]=> z=1$
(Sorry for not adding comments under the original posts since I don't have enough reputations to do so ...)
In Erwin Kalvelagen's solution, we can observe that $z = 0$ does guarantee that either $x<a$ or $x>b$. However, it does not guarantee that $z = 1$ leads to $x in [a, b]$.
For example, considering when $x<a$, $z=1$, $delta = 1$, the first inequality is $$x leq a - 0.001 + M + M $$ which is correct.
The second inequality :
$$
x geq b + 0.001 - 0 - M
$$
which is also correct (since $b - M$ is a very big negative number)
However, now the $x$ does NOT belong to the interval $[a, b]$, hence $z = 1$ is NOT an indicator to its belongingness.
Could someone help to address it?
Or let me know where I am wrong?
Thank you very much!
optimization linear-programming mixed-integer-programming
asked Dec 6 '18 at 23:00
Yu WangYu Wang
84
$endgroup$
add a comment |
$begingroup$
This question is to query the solution provided by Erwin Kalvelagen to the post
Indicator Variable if x is in specific range
and
conditional constraint: if $x in [a,b]=> z=1$
(Sorry for not adding comments under the original posts since I don't have enough reputations to do so ...)
In Erwin Kalvelagen's solution, we can observe that $z = 0$ does guarantee that either $x<a$ or $x>b$. However, it does not guarantee that $z = 1$ leads to $x in [a, b]$.
For example, considering when $x<a$, $z=1$, $delta = 1$, the first inequality is $$x leq a - 0.001 + M + M $$ which is correct.
The second inequality :
$$
x geq b + 0.001 - 0 - M
$$
which is also correct (since $b - M$ is a very big negative number)
However, now the $x$ does NOT belong to the interval $[a, b]$, hence $z = 1$ is NOT an indicator to its belongingness.
Could someone help to address it?
Or let me know where I am wrong?
Thank you very much!
optimization linear-programming mixed-integer-programming
asked Dec 6 '18 at 23:00
Yu WangYu Wang
84
$endgroup$
add a comment |
$begingroup$
This question is to query the solution provided by Erwin Kalvelagen to the post
Indicator Variable if x is in specific range
and
conditional constraint: if $x in [a,b]=> z=1$
(Sorry for not adding comments under the original posts since I don't have enough reputations to do so ...)
In Erwin Kalvelagen's solution, we can observe that $z = 0$ does guarantee that either $x<a$ or $x>b$. However, it does not guarantee that $z = 1$ leads to $x in [a, b]$.
For example, considering when $x<a$, $z=1$, $delta = 1$, the first inequality is $$x leq a - 0.001 + M + M $$ which is correct.
The second inequality :
$$
x geq b + 0.001 - 0 - M
$$
which is also correct (since $b - M$ is a very big negative number)
However, now the $x$ does NOT belong to the interval $[a, b]$, hence $z = 1$ is NOT an indicator to its belongingness.
Could someone help to address it?
Or let me know where I am wrong?
Thank you very much!
optimization linear-programming mixed-integer-programming
asked Dec 6 '18 at 23:00
Yu WangYu Wang
84
$endgroup$
This question is to query the solution provided by Erwin Kalvelagen to the post
Indicator Variable if x is in specific range
and
conditional constraint: if $x in [a,b]=> z=1$
(Sorry for not adding comments under the original posts since I don't have enough reputations to do so ...)
In Erwin Kalvelagen's solution, we can observe that $z = 0$ does guarantee that either $x<a$ or $x>b$. However, it does not guarantee that $z = 1$ leads to $x in [a, b]$.
For example, considering when $x<a$, $z=1$, $delta = 1$, the first inequality is $$x leq a - 0.001 + M + M $$ which is correct.
The second inequality :
$$
x geq b + 0.001 - 0 - M
$$
which is also correct (since $b - M$ is a very big negative number)
However, now the $x$ does NOT belong to the interval $[a, b]$, hence $z = 1$ is NOT an indicator to its belongingness.
Could someone help to address it?
Or let me know where I am wrong?
Thank you very much!
optimization linear-programming mixed-integer-programming
optimization linear-programming mixed-integer-programming
asked Dec 6 '18 at 23:00
Yu WangYu Wang
84
asked Dec 6 '18 at 23:00
Yu WangYu Wang
84
asked Dec 6 '18 at 23:00
Yu WangYu Wang
84
asked Dec 6 '18 at 23:00
Yu WangYu Wang
84
asked Dec 6 '18 at 23:00
Yu WangYu Wang
84
84
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You seem to additionally want $z=1 Rightarrow x in [a,b]$. This can be linearized as:
$$begin{align}
& x ge a - M(1-z)\
& x le b + M(1-z)
end{align}
$$
answered Dec 6 '18 at 23:24
LinAlgLinAlg
10k1521
$endgroup$
$begingroup$
Thanks for your answer. So basically, if we want to have double-binding between $z = 1$ and $x in [a, b]$. Then we need to have all 4 inequalities: $$x<a + M delta + M z$$ $$x> b - M (1 - delta) - M z$$ $$x > a - M (1 - z)$$ $$x < b + M (1- z)$$. Right?
$endgroup$
– Yu Wang
Dec 7 '18 at 0:13
$begingroup$
@YuWang that's right. Note that you need non-strict inequalities in linear optimization.
$endgroup$
– LinAlg
Dec 7 '18 at 0:26
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
$begingroup$
You seem to additionally want $z=1 Rightarrow x in [a,b]$. This can be linearized as:
$$begin{align}
& x ge a - M(1-z)\
& x le b + M(1-z)
end{align}
$$
answered Dec 6 '18 at 23:24
LinAlgLinAlg
10k1521
$endgroup$
$begingroup$
Thanks for your answer. So basically, if we want to have double-binding between $z = 1$ and $x in [a, b]$. Then we need to have all 4 inequalities: $$x<a + M delta + M z$$ $$x> b - M (1 - delta) - M z$$ $$x > a - M (1 - z)$$ $$x < b + M (1- z)$$. Right?
$endgroup$
– Yu Wang
Dec 7 '18 at 0:13
$begingroup$
@YuWang that's right. Note that you need non-strict inequalities in linear optimization.
$endgroup$
– LinAlg
Dec 7 '18 at 0:26
add a comment |
$begingroup$
You seem to additionally want $z=1 Rightarrow x in [a,b]$. This can be linearized as:
$$begin{align}
& x ge a - M(1-z)\
& x le b + M(1-z)
end{align}
$$
answered Dec 6 '18 at 23:24
LinAlgLinAlg
10k1521
$endgroup$
$begingroup$
Thanks for your answer. So basically, if we want to have double-binding between $z = 1$ and $x in [a, b]$. Then we need to have all 4 inequalities: $$x<a + M delta + M z$$ $$x> b - M (1 - delta) - M z$$ $$x > a - M (1 - z)$$ $$x < b + M (1- z)$$. Right?
$endgroup$
– Yu Wang
Dec 7 '18 at 0:13
$begingroup$
@YuWang that's right. Note that you need non-strict inequalities in linear optimization.
$endgroup$
– LinAlg
Dec 7 '18 at 0:26
add a comment |
$begingroup$
You seem to additionally want $z=1 Rightarrow x in [a,b]$. This can be linearized as:
$$begin{align}
& x ge a - M(1-z)\
& x le b + M(1-z)
end{align}
$$
answered Dec 6 '18 at 23:24
LinAlgLinAlg
10k1521
$endgroup$
You seem to additionally want $z=1 Rightarrow x in [a,b]$. This can be linearized as:
$$begin{align}
& x ge a - M(1-z)\
& x le b + M(1-z)
end{align}
$$
answered Dec 6 '18 at 23:24
LinAlgLinAlg
10k1521
answered Dec 6 '18 at 23:24
LinAlgLinAlg
10k1521
answered Dec 6 '18 at 23:24
LinAlgLinAlg
10k1521
answered Dec 6 '18 at 23:24
LinAlgLinAlg
10k1521
10k1521
$begingroup$
Thanks for your answer. So basically, if we want to have double-binding between $z = 1$ and $x in [a, b]$. Then we need to have all 4 inequalities: $$x<a + M delta + M z$$ $$x> b - M (1 - delta) - M z$$ $$x > a - M (1 - z)$$ $$x < b + M (1- z)$$. Right?
$endgroup$
– Yu Wang
Dec 7 '18 at 0:13
$begingroup$
@YuWang that's right. Note that you need non-strict inequalities in linear optimization.
$endgroup$
– LinAlg
Dec 7 '18 at 0:26
add a comment |
$begingroup$
Thanks for your answer. So basically, if we want to have double-binding between $z = 1$ and $x in [a, b]$. Then we need to have all 4 inequalities: $$x<a + M delta + M z$$ $$x> b - M (1 - delta) - M z$$ $$x > a - M (1 - z)$$ $$x < b + M (1- z)$$. Right?
$endgroup$
– Yu Wang
Dec 7 '18 at 0:13
$begingroup$
@YuWang that's right. Note that you need non-strict inequalities in linear optimization.
$endgroup$
– LinAlg
Dec 7 '18 at 0:26
$begingroup$
Thanks for your answer. So basically, if we want to have double-binding between $z = 1$ and $x in [a, b]$. Then we need to have all 4 inequalities: $$x<a + M delta + M z$$ $$x> b - M (1 - delta) - M z$$ $$x > a - M (1 - z)$$ $$x < b + M (1- z)$$. Right?
$endgroup$
– Yu Wang
Dec 7 '18 at 0:13
$begingroup$
Thanks for your answer. So basically, if we want to have double-binding between $z = 1$ and $x in [a, b]$. Then we need to have all 4 inequalities: $$x<a + M delta + M z$$ $$x> b - M (1 - delta) - M z$$ $$x > a - M (1 - z)$$ $$x < b + M (1- z)$$. Right?
$endgroup$
– Yu Wang
Dec 7 '18 at 0:13
$begingroup$
@YuWang that's right. Note that you need non-strict inequalities in linear optimization.
$endgroup$
– LinAlg
Dec 7 '18 at 0:26
$begingroup$
@YuWang that's right. Note that you need non-strict inequalities in linear optimization.
$endgroup$
– LinAlg
Dec 7 '18 at 0:26
add a comment |
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