integer programming linearization
0
$begingroup$
I have two variables. $g$ is a binary variable and $s$ is a continuous variable. Goal is to linearize this $gs geq0$ a.k.a if $s geq 0, g = 1:: text{or}:: sleq0, g = 0$. How can I linearize this ?
mixed-integer-programming
edited Dec 12 '18 at 6:24
Yadati Kiran
1,7911619
asked Dec 12 '18 at 5:39
Jinghan YangJinghan Yang
1
$endgroup$
add a comment |
0
$begingroup$
I have two variables. $g$ is a binary variable and $s$ is a continuous variable. Goal is to linearize this $gs geq0$ a.k.a if $s geq 0, g = 1:: text{or}:: sleq0, g = 0$. How can I linearize this ?
mixed-integer-programming
edited Dec 12 '18 at 6:24
Yadati Kiran
1,7911619
asked Dec 12 '18 at 5:39
Jinghan YangJinghan Yang
1
$endgroup$
$begingroup$
When $sge 0$, $gsge 0$ is satisfied by both $g=1$ and $g=0$. Is that what you want, or do you want the following: $s>0 implies g=1$; $s<0 implies g=0$; $s=0$ implies nothing about $g$ (which could be either 0 or 1)?
$endgroup$
– prubin
Dec 13 '18 at 16:38
$begingroup$
Sorry about the confusion. $s > 0 Rightarrow g = 1; s leq 0 Rightarrow g = 0 $, a.k.a I want get $ g = 0$ if $s = 0$. Thank you for your reply.
$endgroup$
– Jinghan Yang
Dec 14 '18 at 22:01
$begingroup$
You cannot get exactly what you want. In order to get $g=0$ when $s=0$, you would have to settle for $sge epsilon implies g=1$ and $0 < s < epsilon$ infeasible, for some $epsilon > 0$. $epsilon$ can be small but not too small (i.e., not small enough that it is within rounding tolerance of 0 from the solver's perspective). Is that acceptable?
$endgroup$
– prubin
Dec 16 '18 at 16:01
add a comment |
0
0
0
$begingroup$
I have two variables. $g$ is a binary variable and $s$ is a continuous variable. Goal is to linearize this $gs geq0$ a.k.a if $s geq 0, g = 1:: text{or}:: sleq0, g = 0$. How can I linearize this ?
mixed-integer-programming
edited Dec 12 '18 at 6:24
Yadati Kiran
1,7911619
asked Dec 12 '18 at 5:39
Jinghan YangJinghan Yang
1
$endgroup$
I have two variables. $g$ is a binary variable and $s$ is a continuous variable. Goal is to linearize this $gs geq0$ a.k.a if $s geq 0, g = 1:: text{or}:: sleq0, g = 0$. How can I linearize this ?
mixed-integer-programming
mixed-integer-programming
edited Dec 12 '18 at 6:24
Yadati Kiran
1,7911619
asked Dec 12 '18 at 5:39
Jinghan YangJinghan Yang
1
edited Dec 12 '18 at 6:24
Yadati Kiran
1,7911619
asked Dec 12 '18 at 5:39
Jinghan YangJinghan Yang
1
edited Dec 12 '18 at 6:24
Yadati Kiran
1,7911619
edited Dec 12 '18 at 6:24
Yadati Kiran
1,7911619
edited Dec 12 '18 at 6:24
Yadati Kiran
1,7911619
1,7911619
asked Dec 12 '18 at 5:39
Jinghan YangJinghan Yang
1
asked Dec 12 '18 at 5:39
Jinghan YangJinghan Yang
1
asked Dec 12 '18 at 5:39
Jinghan YangJinghan Yang
1
1
$begingroup$
When $sge 0$, $gsge 0$ is satisfied by both $g=1$ and $g=0$. Is that what you want, or do you want the following: $s>0 implies g=1$; $s<0 implies g=0$; $s=0$ implies nothing about $g$ (which could be either 0 or 1)?
$endgroup$
– prubin
Dec 13 '18 at 16:38
$begingroup$
Sorry about the confusion. $s > 0 Rightarrow g = 1; s leq 0 Rightarrow g = 0 $, a.k.a I want get $ g = 0$ if $s = 0$. Thank you for your reply.
$endgroup$
– Jinghan Yang
Dec 14 '18 at 22:01
$begingroup$
You cannot get exactly what you want. In order to get $g=0$ when $s=0$, you would have to settle for $sge epsilon implies g=1$ and $0 < s < epsilon$ infeasible, for some $epsilon > 0$. $epsilon$ can be small but not too small (i.e., not small enough that it is within rounding tolerance of 0 from the solver's perspective). Is that acceptable?
$endgroup$
– prubin
Dec 16 '18 at 16:01
add a comment |
$begingroup$
When $sge 0$, $gsge 0$ is satisfied by both $g=1$ and $g=0$. Is that what you want, or do you want the following: $s>0 implies g=1$; $s<0 implies g=0$; $s=0$ implies nothing about $g$ (which could be either 0 or 1)?
$endgroup$
– prubin
Dec 13 '18 at 16:38
$begingroup$
Sorry about the confusion. $s > 0 Rightarrow g = 1; s leq 0 Rightarrow g = 0 $, a.k.a I want get $ g = 0$ if $s = 0$. Thank you for your reply.
$endgroup$
– Jinghan Yang
Dec 14 '18 at 22:01
$begingroup$
You cannot get exactly what you want. In order to get $g=0$ when $s=0$, you would have to settle for $sge epsilon implies g=1$ and $0 < s < epsilon$ infeasible, for some $epsilon > 0$. $epsilon$ can be small but not too small (i.e., not small enough that it is within rounding tolerance of 0 from the solver's perspective). Is that acceptable?
$endgroup$
– prubin
Dec 16 '18 at 16:01
$begingroup$
When $sge 0$, $gsge 0$ is satisfied by both $g=1$ and $g=0$. Is that what you want, or do you want the following: $s>0 implies g=1$; $s<0 implies g=0$; $s=0$ implies nothing about $g$ (which could be either 0 or 1)?
$endgroup$
– prubin
Dec 13 '18 at 16:38
$begingroup$
When $sge 0$, $gsge 0$ is satisfied by both $g=1$ and $g=0$. Is that what you want, or do you want the following: $s>0 implies g=1$; $s<0 implies g=0$; $s=0$ implies nothing about $g$ (which could be either 0 or 1)?
$endgroup$
– prubin
Dec 13 '18 at 16:38
$begingroup$
Sorry about the confusion. $s > 0 Rightarrow g = 1; s leq 0 Rightarrow g = 0 $, a.k.a I want get $ g = 0$ if $s = 0$. Thank you for your reply.
$endgroup$
– Jinghan Yang
Dec 14 '18 at 22:01
$begingroup$
Sorry about the confusion. $s > 0 Rightarrow g = 1; s leq 0 Rightarrow g = 0 $, a.k.a I want get $ g = 0$ if $s = 0$. Thank you for your reply.
$endgroup$
– Jinghan Yang
Dec 14 '18 at 22:01
$begingroup$
You cannot get exactly what you want. In order to get $g=0$ when $s=0$, you would have to settle for $sge epsilon implies g=1$ and $0 < s < epsilon$ infeasible, for some $epsilon > 0$. $epsilon$ can be small but not too small (i.e., not small enough that it is within rounding tolerance of 0 from the solver's perspective). Is that acceptable?
$endgroup$
– prubin
Dec 16 '18 at 16:01
$begingroup$
You cannot get exactly what you want. In order to get $g=0$ when $s=0$, you would have to settle for $sge epsilon implies g=1$ and $0 < s < epsilon$ infeasible, for some $epsilon > 0$. $epsilon$ can be small but not too small (i.e., not small enough that it is within rounding tolerance of 0 from the solver's perspective). Is that acceptable?
$endgroup$
– prubin
Dec 16 '18 at 16:01
add a comment |
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$begingroup$
When $sge 0$, $gsge 0$ is satisfied by both $g=1$ and $g=0$. Is that what you want, or do you want the following: $s>0 implies g=1$; $s<0 implies g=0$; $s=0$ implies nothing about $g$ (which could be either 0 or 1)?
$endgroup$
– prubin
Dec 13 '18 at 16:38
$begingroup$
Sorry about the confusion. $s > 0 Rightarrow g = 1; s leq 0 Rightarrow g = 0 $, a.k.a I want get $ g = 0$ if $s = 0$. Thank you for your reply.
$endgroup$
– Jinghan Yang
Dec 14 '18 at 22:01
$begingroup$
You cannot get exactly what you want. In order to get $g=0$ when $s=0$, you would have to settle for $sge epsilon implies g=1$ and $0 < s < epsilon$ infeasible, for some $epsilon > 0$. $epsilon$ can be small but not too small (i.e., not small enough that it is within rounding tolerance of 0 from the solver's perspective). Is that acceptable?
$endgroup$
– prubin
Dec 16 '18 at 16:01