Enforce constraints in nth visited node
$begingroup$
I have a problem similar to the tsp problem where :
$x_{i,j} in left{0,1right}$, is 1 if I visit node $j$ immediately after node $i$.
Now suppose that I need to enforce constraints for the n-th visited node, how would I do that in an efficient way, given the fact that I do not know a-priori which node I will visit first, second etc?
To elaborate, this would be part of my program:
$min sum_i sum_j c_{i,j} y_{i,j}$
st. $ ~ y_{i,j} le M ~ x_{i,j} ~~~ forall ~ i,j $
$ ~ y_{i,j} ge P_n~ x_{i,j} ~~~~ $ if $j$ is the n-th visited node
$ y in mathbb{R}_+ $
mixed-integer-programming
asked Dec 19 '18 at 12:22
nikferrarinikferrari
1
$endgroup$
add a comment |
$begingroup$
I have a problem similar to the tsp problem where :
$x_{i,j} in left{0,1right}$, is 1 if I visit node $j$ immediately after node $i$.
Now suppose that I need to enforce constraints for the n-th visited node, how would I do that in an efficient way, given the fact that I do not know a-priori which node I will visit first, second etc?
To elaborate, this would be part of my program:
$min sum_i sum_j c_{i,j} y_{i,j}$
st. $ ~ y_{i,j} le M ~ x_{i,j} ~~~ forall ~ i,j $
$ ~ y_{i,j} ge P_n~ x_{i,j} ~~~~ $ if $j$ is the n-th visited node
$ y in mathbb{R}_+ $
mixed-integer-programming
asked Dec 19 '18 at 12:22
nikferrarinikferrari
1
$endgroup$
add a comment |
$begingroup$
I have a problem similar to the tsp problem where :
$x_{i,j} in left{0,1right}$, is 1 if I visit node $j$ immediately after node $i$.
Now suppose that I need to enforce constraints for the n-th visited node, how would I do that in an efficient way, given the fact that I do not know a-priori which node I will visit first, second etc?
To elaborate, this would be part of my program:
$min sum_i sum_j c_{i,j} y_{i,j}$
st. $ ~ y_{i,j} le M ~ x_{i,j} ~~~ forall ~ i,j $
$ ~ y_{i,j} ge P_n~ x_{i,j} ~~~~ $ if $j$ is the n-th visited node
$ y in mathbb{R}_+ $
mixed-integer-programming
asked Dec 19 '18 at 12:22
nikferrarinikferrari
1
$endgroup$
I have a problem similar to the tsp problem where :
$x_{i,j} in left{0,1right}$, is 1 if I visit node $j$ immediately after node $i$.
Now suppose that I need to enforce constraints for the n-th visited node, how would I do that in an efficient way, given the fact that I do not know a-priori which node I will visit first, second etc?
To elaborate, this would be part of my program:
$min sum_i sum_j c_{i,j} y_{i,j}$
st. $ ~ y_{i,j} le M ~ x_{i,j} ~~~ forall ~ i,j $
$ ~ y_{i,j} ge P_n~ x_{i,j} ~~~~ $ if $j$ is the n-th visited node
$ y in mathbb{R}_+ $
mixed-integer-programming
mixed-integer-programming
asked Dec 19 '18 at 12:22
nikferrarinikferrari
1
asked Dec 19 '18 at 12:22
nikferrarinikferrari
1
asked Dec 19 '18 at 12:22
nikferrarinikferrari
1
asked Dec 19 '18 at 12:22
nikferrarinikferrari
1
asked Dec 19 '18 at 12:22
nikferrarinikferrari
1
1
add a comment |
add a comment |
1 Answer
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The following may be more cumbersome than necessary, but I think it will work. (I do recommend checking it carefully, though.)
I'll assume that the starting point for your trip is a (possibly artificial) node with index 0, since you said you do not know which node is visited first. Start by adding the Miller-Tucker-Zemlin constraints. That adds (continuous) variables $u_i$ with domain $[1,dots,m]$ (where $m$ is the number of nodes). Fix $u_0=0$ and constraints$$u_i - u_j + 1le m(1-x_{ij})$$for all $i,j$ (including $i=0$ but excluding $j=0$ if this is a closed tour). The result is that $u_i$ will equal $k$ if and only if node $i$ is the $k$-th stop.
Now add binary variables $z_i,(i=1,dots,m)$ and add the constraint$$sum_{i=1}^m z_i=1.$$Variable $z_i$ will be 1 if and only if node $i$ is the $n$-th node visited. To enforce this, add$$u_j ge n z_j,forall j=1,dots,m$$(which ensures that $z_j=1$ implies that node $j$ is visited no earlier than the $n$-th stop) and$$u_ile n-1 + (m-n+1)(2-x_{ij}-z_j), forall i,jin{0,dots,m},jneq 0.$$This last constraint ensures that if $z_j=1$ and $x_{ij}=1$, $u_i le n-1$. Combine those and you have that $z_j=1$ implies $j$ is at least stop $n$ and its predecessor was at most stop $n-1$, so node $j$ must be stop $n$.
Now you just have to frame your conditional constraints using $z$ as the indicator of which stop is stop $n$.
answered Dec 20 '18 at 17:36
prubinprubin
1,565125
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$begingroup$
The following may be more cumbersome than necessary, but I think it will work. (I do recommend checking it carefully, though.)
I'll assume that the starting point for your trip is a (possibly artificial) node with index 0, since you said you do not know which node is visited first. Start by adding the Miller-Tucker-Zemlin constraints. That adds (continuous) variables $u_i$ with domain $[1,dots,m]$ (where $m$ is the number of nodes). Fix $u_0=0$ and constraints$$u_i - u_j + 1le m(1-x_{ij})$$for all $i,j$ (including $i=0$ but excluding $j=0$ if this is a closed tour). The result is that $u_i$ will equal $k$ if and only if node $i$ is the $k$-th stop.
Now add binary variables $z_i,(i=1,dots,m)$ and add the constraint$$sum_{i=1}^m z_i=1.$$Variable $z_i$ will be 1 if and only if node $i$ is the $n$-th node visited. To enforce this, add$$u_j ge n z_j,forall j=1,dots,m$$(which ensures that $z_j=1$ implies that node $j$ is visited no earlier than the $n$-th stop) and$$u_ile n-1 + (m-n+1)(2-x_{ij}-z_j), forall i,jin{0,dots,m},jneq 0.$$This last constraint ensures that if $z_j=1$ and $x_{ij}=1$, $u_i le n-1$. Combine those and you have that $z_j=1$ implies $j$ is at least stop $n$ and its predecessor was at most stop $n-1$, so node $j$ must be stop $n$.
Now you just have to frame your conditional constraints using $z$ as the indicator of which stop is stop $n$.
answered Dec 20 '18 at 17:36
prubinprubin
1,565125
$endgroup$
add a comment |
$begingroup$
The following may be more cumbersome than necessary, but I think it will work. (I do recommend checking it carefully, though.)
I'll assume that the starting point for your trip is a (possibly artificial) node with index 0, since you said you do not know which node is visited first. Start by adding the Miller-Tucker-Zemlin constraints. That adds (continuous) variables $u_i$ with domain $[1,dots,m]$ (where $m$ is the number of nodes). Fix $u_0=0$ and constraints$$u_i - u_j + 1le m(1-x_{ij})$$for all $i,j$ (including $i=0$ but excluding $j=0$ if this is a closed tour). The result is that $u_i$ will equal $k$ if and only if node $i$ is the $k$-th stop.
Now add binary variables $z_i,(i=1,dots,m)$ and add the constraint$$sum_{i=1}^m z_i=1.$$Variable $z_i$ will be 1 if and only if node $i$ is the $n$-th node visited. To enforce this, add$$u_j ge n z_j,forall j=1,dots,m$$(which ensures that $z_j=1$ implies that node $j$ is visited no earlier than the $n$-th stop) and$$u_ile n-1 + (m-n+1)(2-x_{ij}-z_j), forall i,jin{0,dots,m},jneq 0.$$This last constraint ensures that if $z_j=1$ and $x_{ij}=1$, $u_i le n-1$. Combine those and you have that $z_j=1$ implies $j$ is at least stop $n$ and its predecessor was at most stop $n-1$, so node $j$ must be stop $n$.
Now you just have to frame your conditional constraints using $z$ as the indicator of which stop is stop $n$.
answered Dec 20 '18 at 17:36
prubinprubin
1,565125
$endgroup$
add a comment |
$begingroup$
The following may be more cumbersome than necessary, but I think it will work. (I do recommend checking it carefully, though.)
I'll assume that the starting point for your trip is a (possibly artificial) node with index 0, since you said you do not know which node is visited first. Start by adding the Miller-Tucker-Zemlin constraints. That adds (continuous) variables $u_i$ with domain $[1,dots,m]$ (where $m$ is the number of nodes). Fix $u_0=0$ and constraints$$u_i - u_j + 1le m(1-x_{ij})$$for all $i,j$ (including $i=0$ but excluding $j=0$ if this is a closed tour). The result is that $u_i$ will equal $k$ if and only if node $i$ is the $k$-th stop.
Now add binary variables $z_i,(i=1,dots,m)$ and add the constraint$$sum_{i=1}^m z_i=1.$$Variable $z_i$ will be 1 if and only if node $i$ is the $n$-th node visited. To enforce this, add$$u_j ge n z_j,forall j=1,dots,m$$(which ensures that $z_j=1$ implies that node $j$ is visited no earlier than the $n$-th stop) and$$u_ile n-1 + (m-n+1)(2-x_{ij}-z_j), forall i,jin{0,dots,m},jneq 0.$$This last constraint ensures that if $z_j=1$ and $x_{ij}=1$, $u_i le n-1$. Combine those and you have that $z_j=1$ implies $j$ is at least stop $n$ and its predecessor was at most stop $n-1$, so node $j$ must be stop $n$.
Now you just have to frame your conditional constraints using $z$ as the indicator of which stop is stop $n$.
answered Dec 20 '18 at 17:36
prubinprubin
1,565125
$endgroup$
The following may be more cumbersome than necessary, but I think it will work. (I do recommend checking it carefully, though.)
I'll assume that the starting point for your trip is a (possibly artificial) node with index 0, since you said you do not know which node is visited first. Start by adding the Miller-Tucker-Zemlin constraints. That adds (continuous) variables $u_i$ with domain $[1,dots,m]$ (where $m$ is the number of nodes). Fix $u_0=0$ and constraints$$u_i - u_j + 1le m(1-x_{ij})$$for all $i,j$ (including $i=0$ but excluding $j=0$ if this is a closed tour). The result is that $u_i$ will equal $k$ if and only if node $i$ is the $k$-th stop.
Now add binary variables $z_i,(i=1,dots,m)$ and add the constraint$$sum_{i=1}^m z_i=1.$$Variable $z_i$ will be 1 if and only if node $i$ is the $n$-th node visited. To enforce this, add$$u_j ge n z_j,forall j=1,dots,m$$(which ensures that $z_j=1$ implies that node $j$ is visited no earlier than the $n$-th stop) and$$u_ile n-1 + (m-n+1)(2-x_{ij}-z_j), forall i,jin{0,dots,m},jneq 0.$$This last constraint ensures that if $z_j=1$ and $x_{ij}=1$, $u_i le n-1$. Combine those and you have that $z_j=1$ implies $j$ is at least stop $n$ and its predecessor was at most stop $n-1$, so node $j$ must be stop $n$.
Now you just have to frame your conditional constraints using $z$ as the indicator of which stop is stop $n$.
answered Dec 20 '18 at 17:36
prubinprubin
1,565125
answered Dec 20 '18 at 17:36
prubinprubin
1,565125
answered Dec 20 '18 at 17:36
prubinprubin
1,565125
answered Dec 20 '18 at 17:36
prubinprubin
1,565125
1,565125
add a comment |
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