linearization of objective functionHow to linearize the product of a binary and a non-negative continuous...
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linearization of objective function
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linearization of objective function
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$begingroup$
$src_{h,s}$, $dst_{h,s}$, $ch_{h,s}$ are constants.
$a_{h,s}$, $x_{i,j,s}$ are binary variables.
$wt_{h,s}$ are continuous variables.
$mini.$
$$
sum_{h in H} sum_{s in S} (src_{h,s} + ch_{h,s} + dst_{h,s} + wt_{h,s}) times a_{h,s}
$$
$s.t.$
$$
forall i in H hspace{0.3cm} forall j in H hspace{0.3cm} forall s in S hspace{0.3cm}
$$
$$
wt_{j,s} geq ((src_{i,s} + ch_{i,s}+wt_{i,s}) - src_{j,s}) times x_{i,j,s}
$$
$$
forall i in H hspace{0.3cm} forall j in H hspace{0.3cm}
x_{ij} + x_{ji} leq 1
$$
$$
forall i in H hspace{0.3cm} forall j in H hspace{0.3cm} forall s in S hspace{0.3cm}
$$
$$
x_{ij} + x_{ji} geq a_{i,s} + a_{j,s} + 1
$$
$$
forall h in H sum_{s in S} b_{h,s} leq 1
$$
I want to use a LP solver on this problem but there are continuous variable $wt_{h,s}$ and Boolean variable $a_{h,s}$ together in objective function, how to separate them.
I have found a link for linearization in constraints, (https://www.leandro-coelho.com/linearization-product-variables/) but how to linearize in objective function.
Also in first constraint there are two continuous variable $wt_{j,s}$ and $wt_{i,s}$, is it possible to linearize it.
linear-programming optimization linearization
edited 6 hours ago
Simon
4471 silver badge12 bronze badges
asked 8 hours ago
anoop yadavanoop yadav
1084 bronze badges
$endgroup$
add a comment |
$begingroup$
$src_{h,s}$, $dst_{h,s}$, $ch_{h,s}$ are constants.
$a_{h,s}$, $x_{i,j,s}$ are binary variables.
$wt_{h,s}$ are continuous variables.
$mini.$
$$
sum_{h in H} sum_{s in S} (src_{h,s} + ch_{h,s} + dst_{h,s} + wt_{h,s}) times a_{h,s}
$$
$s.t.$
$$
forall i in H hspace{0.3cm} forall j in H hspace{0.3cm} forall s in S hspace{0.3cm}
$$
$$
wt_{j,s} geq ((src_{i,s} + ch_{i,s}+wt_{i,s}) - src_{j,s}) times x_{i,j,s}
$$
$$
forall i in H hspace{0.3cm} forall j in H hspace{0.3cm}
x_{ij} + x_{ji} leq 1
$$
$$
forall i in H hspace{0.3cm} forall j in H hspace{0.3cm} forall s in S hspace{0.3cm}
$$
$$
x_{ij} + x_{ji} geq a_{i,s} + a_{j,s} + 1
$$
$$
forall h in H sum_{s in S} b_{h,s} leq 1
$$
I want to use a LP solver on this problem but there are continuous variable $wt_{h,s}$ and Boolean variable $a_{h,s}$ together in objective function, how to separate them.
I have found a link for linearization in constraints, (https://www.leandro-coelho.com/linearization-product-variables/) but how to linearize in objective function.
Also in first constraint there are two continuous variable $wt_{j,s}$ and $wt_{i,s}$, is it possible to linearize it.
linear-programming optimization linearization
edited 6 hours ago
Simon
4471 silver badge12 bronze badges
asked 8 hours ago
anoop yadavanoop yadav
1084 bronze badges
$endgroup$
1
$begingroup$
Linearize the objective function the same way you would a constraint. Having two continuous variables in the first constraint doesn't add any complications because one of these variable appears "by itself", i.e., not multiplied by another variable, and therefore that variable already appears linearly.
$endgroup$
– Mark L. Stone
8 hours ago
1
$begingroup$
Maybe take a look at this question: How to linearize the product of a binary and a non-negative continuous variable?
$endgroup$
– EhsanK
6 hours ago
$begingroup$
Is this $$ sum_{h in H} sum_{s in S} (src_{h,s} + ch_{h,s} + dst_{h,s} ) times a_{h,s} + ( wt_{h,s} - (1 - a_{h,s}) times infty ) $$ correct linearization of objective function, but what about the bounds.
$endgroup$
– anoop yadav
6 hours ago
add a comment |
$begingroup$
$src_{h,s}$, $dst_{h,s}$, $ch_{h,s}$ are constants.
$a_{h,s}$, $x_{i,j,s}$ are binary variables.
$wt_{h,s}$ are continuous variables.
$mini.$
$$
sum_{h in H} sum_{s in S} (src_{h,s} + ch_{h,s} + dst_{h,s} + wt_{h,s}) times a_{h,s}
$$
$s.t.$
$$
forall i in H hspace{0.3cm} forall j in H hspace{0.3cm} forall s in S hspace{0.3cm}
$$
$$
wt_{j,s} geq ((src_{i,s} + ch_{i,s}+wt_{i,s}) - src_{j,s}) times x_{i,j,s}
$$
$$
forall i in H hspace{0.3cm} forall j in H hspace{0.3cm}
x_{ij} + x_{ji} leq 1
$$
$$
forall i in H hspace{0.3cm} forall j in H hspace{0.3cm} forall s in S hspace{0.3cm}
$$
$$
x_{ij} + x_{ji} geq a_{i,s} + a_{j,s} + 1
$$
$$
forall h in H sum_{s in S} b_{h,s} leq 1
$$
I want to use a LP solver on this problem but there are continuous variable $wt_{h,s}$ and Boolean variable $a_{h,s}$ together in objective function, how to separate them.
I have found a link for linearization in constraints, (https://www.leandro-coelho.com/linearization-product-variables/) but how to linearize in objective function.
Also in first constraint there are two continuous variable $wt_{j,s}$ and $wt_{i,s}$, is it possible to linearize it.
linear-programming optimization linearization
edited 6 hours ago
Simon
4471 silver badge12 bronze badges
asked 8 hours ago
anoop yadavanoop yadav
1084 bronze badges
$endgroup$
$src_{h,s}$, $dst_{h,s}$, $ch_{h,s}$ are constants.
$a_{h,s}$, $x_{i,j,s}$ are binary variables.
$wt_{h,s}$ are continuous variables.
$mini.$
$$
sum_{h in H} sum_{s in S} (src_{h,s} + ch_{h,s} + dst_{h,s} + wt_{h,s}) times a_{h,s}
$$
$s.t.$
$$
forall i in H hspace{0.3cm} forall j in H hspace{0.3cm} forall s in S hspace{0.3cm}
$$
$$
wt_{j,s} geq ((src_{i,s} + ch_{i,s}+wt_{i,s}) - src_{j,s}) times x_{i,j,s}
$$
$$
forall i in H hspace{0.3cm} forall j in H hspace{0.3cm}
x_{ij} + x_{ji} leq 1
$$
$$
forall i in H hspace{0.3cm} forall j in H hspace{0.3cm} forall s in S hspace{0.3cm}
$$
$$
x_{ij} + x_{ji} geq a_{i,s} + a_{j,s} + 1
$$
$$
forall h in H sum_{s in S} b_{h,s} leq 1
$$
I want to use a LP solver on this problem but there are continuous variable $wt_{h,s}$ and Boolean variable $a_{h,s}$ together in objective function, how to separate them.
I have found a link for linearization in constraints, (https://www.leandro-coelho.com/linearization-product-variables/) but how to linearize in objective function.
Also in first constraint there are two continuous variable $wt_{j,s}$ and $wt_{i,s}$, is it possible to linearize it.
linear-programming optimization linearization
linear-programming optimization linearization
edited 6 hours ago
Simon
4471 silver badge12 bronze badges
asked 8 hours ago
anoop yadavanoop yadav
1084 bronze badges
edited 6 hours ago
Simon
4471 silver badge12 bronze badges
asked 8 hours ago
anoop yadavanoop yadav
1084 bronze badges
edited 6 hours ago
Simon
4471 silver badge12 bronze badges
edited 6 hours ago
Simon
4471 silver badge12 bronze badges
edited 6 hours ago
Simon
4471 silver badge12 bronze badges
4471 silver badge12 bronze badges
asked 8 hours ago
anoop yadavanoop yadav
1084 bronze badges
asked 8 hours ago
anoop yadavanoop yadav
1084 bronze badges
asked 8 hours ago
anoop yadavanoop yadav
1084 bronze badges
1084 bronze badges
1
$begingroup$
Linearize the objective function the same way you would a constraint. Having two continuous variables in the first constraint doesn't add any complications because one of these variable appears "by itself", i.e., not multiplied by another variable, and therefore that variable already appears linearly.
$endgroup$
– Mark L. Stone
8 hours ago
1
$begingroup$
Maybe take a look at this question: How to linearize the product of a binary and a non-negative continuous variable?
$endgroup$
– EhsanK
6 hours ago
$begingroup$
Is this $$ sum_{h in H} sum_{s in S} (src_{h,s} + ch_{h,s} + dst_{h,s} ) times a_{h,s} + ( wt_{h,s} - (1 - a_{h,s}) times infty ) $$ correct linearization of objective function, but what about the bounds.
$endgroup$
– anoop yadav
6 hours ago
add a comment |
1
$begingroup$
Linearize the objective function the same way you would a constraint. Having two continuous variables in the first constraint doesn't add any complications because one of these variable appears "by itself", i.e., not multiplied by another variable, and therefore that variable already appears linearly.
$endgroup$
– Mark L. Stone
8 hours ago
1
$begingroup$
Maybe take a look at this question: How to linearize the product of a binary and a non-negative continuous variable?
$endgroup$
– EhsanK
6 hours ago
$begingroup$
Is this $$ sum_{h in H} sum_{s in S} (src_{h,s} + ch_{h,s} + dst_{h,s} ) times a_{h,s} + ( wt_{h,s} - (1 - a_{h,s}) times infty ) $$ correct linearization of objective function, but what about the bounds.
$endgroup$
– anoop yadav
6 hours ago
1
1
$begingroup$
Linearize the objective function the same way you would a constraint. Having two continuous variables in the first constraint doesn't add any complications because one of these variable appears "by itself", i.e., not multiplied by another variable, and therefore that variable already appears linearly.
$endgroup$
– Mark L. Stone
8 hours ago
$begingroup$
Linearize the objective function the same way you would a constraint. Having two continuous variables in the first constraint doesn't add any complications because one of these variable appears "by itself", i.e., not multiplied by another variable, and therefore that variable already appears linearly.
$endgroup$
– Mark L. Stone
8 hours ago
1
1
$begingroup$
Maybe take a look at this question: How to linearize the product of a binary and a non-negative continuous variable?
$endgroup$
– EhsanK
6 hours ago
$begingroup$
Maybe take a look at this question: How to linearize the product of a binary and a non-negative continuous variable?
$endgroup$
– EhsanK
6 hours ago
$begingroup$
Is this $$ sum_{h in H} sum_{s in S} (src_{h,s} + ch_{h,s} + dst_{h,s} ) times a_{h,s} + ( wt_{h,s} - (1 - a_{h,s}) times infty ) $$ correct linearization of objective function, but what about the bounds.
$endgroup$
– anoop yadav
6 hours ago
$begingroup$
Is this $$ sum_{h in H} sum_{s in S} (src_{h,s} + ch_{h,s} + dst_{h,s} ) times a_{h,s} + ( wt_{h,s} - (1 - a_{h,s}) times infty ) $$ correct linearization of objective function, but what about the bounds.
$endgroup$
– anoop yadav
6 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Piecewise linearization methods have been widely applied to convert a nonlinear programming problem into a linear programming problem or a mixed-integer convex programming problem for obtaining an approximated global optimal solution. In the transformation process, extra binary variables, continuous variables, and constraints are introduced to reformulate the original problem. These extra variables and constraints mainly determine the solution efficiency of the converted problem.[source]
answered 7 hours ago
Oguz ToragayOguz Toragay
1,5691 silver badge20 bronze badges
$endgroup$
add a comment |
$begingroup$
Add some additional continuous variables $s_{h,s}$ to your model and use those variables in the objective, instead of the products.
Add the following constraints for each $s_{h,s}$:
This constraint ensures that $s_{h,s}$ is at most equal to the sum:
$s_{h,s} leq src_{h,s}+ch_{h,s}+dst_{h,s}+wt_{h,s}$
This constraint ensures that $s_{h,s}$ will be at least the sum when $a_{h,s}=1$:
$s_{h,s} geq src_{h,s}+ch_{h,s}+dst_{h,s}+wt_{h,s} - M times (1 - a_{h,s}) $
This constraints ensures that $s_{h,s}=0$ when $a_{h,s}=0$:
$s_{h,s} leq M times a_{h,s} $
Some notes about this:
- I assumed that your constants and variables are all nonnegative.
- You should pick small values for the constant $M$ to make it all work (e.g. $src_{h,s}+ch_{h,s}+dst_{h,s}+UB(wt_{h,s})$). Picking much larger values leads to lower performance and might even introduce numerical problems.
- If your solver of choice supports indicator constraints, you could also formulate it using those.
answered 4 hours ago
SimonSimon
4471 silver badge12 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Piecewise linearization methods have been widely applied to convert a nonlinear programming problem into a linear programming problem or a mixed-integer convex programming problem for obtaining an approximated global optimal solution. In the transformation process, extra binary variables, continuous variables, and constraints are introduced to reformulate the original problem. These extra variables and constraints mainly determine the solution efficiency of the converted problem.[source]
answered 7 hours ago
Oguz ToragayOguz Toragay
1,5691 silver badge20 bronze badges
$endgroup$
add a comment |
$begingroup$
Piecewise linearization methods have been widely applied to convert a nonlinear programming problem into a linear programming problem or a mixed-integer convex programming problem for obtaining an approximated global optimal solution. In the transformation process, extra binary variables, continuous variables, and constraints are introduced to reformulate the original problem. These extra variables and constraints mainly determine the solution efficiency of the converted problem.[source]
answered 7 hours ago
Oguz ToragayOguz Toragay
1,5691 silver badge20 bronze badges
$endgroup$
add a comment |
$begingroup$
Piecewise linearization methods have been widely applied to convert a nonlinear programming problem into a linear programming problem or a mixed-integer convex programming problem for obtaining an approximated global optimal solution. In the transformation process, extra binary variables, continuous variables, and constraints are introduced to reformulate the original problem. These extra variables and constraints mainly determine the solution efficiency of the converted problem.[source]
answered 7 hours ago
Oguz ToragayOguz Toragay
1,5691 silver badge20 bronze badges
$endgroup$
Piecewise linearization methods have been widely applied to convert a nonlinear programming problem into a linear programming problem or a mixed-integer convex programming problem for obtaining an approximated global optimal solution. In the transformation process, extra binary variables, continuous variables, and constraints are introduced to reformulate the original problem. These extra variables and constraints mainly determine the solution efficiency of the converted problem.[source]
answered 7 hours ago
Oguz ToragayOguz Toragay
1,5691 silver badge20 bronze badges
answered 7 hours ago
Oguz ToragayOguz Toragay
1,5691 silver badge20 bronze badges
answered 7 hours ago
Oguz ToragayOguz Toragay
1,5691 silver badge20 bronze badges
answered 7 hours ago
Oguz ToragayOguz Toragay
1,5691 silver badge20 bronze badges
1,5691 silver badge20 bronze badges
add a comment |
add a comment |
$begingroup$
Add some additional continuous variables $s_{h,s}$ to your model and use those variables in the objective, instead of the products.
Add the following constraints for each $s_{h,s}$:
This constraint ensures that $s_{h,s}$ is at most equal to the sum:
$s_{h,s} leq src_{h,s}+ch_{h,s}+dst_{h,s}+wt_{h,s}$
This constraint ensures that $s_{h,s}$ will be at least the sum when $a_{h,s}=1$:
$s_{h,s} geq src_{h,s}+ch_{h,s}+dst_{h,s}+wt_{h,s} - M times (1 - a_{h,s}) $
This constraints ensures that $s_{h,s}=0$ when $a_{h,s}=0$:
$s_{h,s} leq M times a_{h,s} $
Some notes about this:
- I assumed that your constants and variables are all nonnegative.
- You should pick small values for the constant $M$ to make it all work (e.g. $src_{h,s}+ch_{h,s}+dst_{h,s}+UB(wt_{h,s})$). Picking much larger values leads to lower performance and might even introduce numerical problems.
- If your solver of choice supports indicator constraints, you could also formulate it using those.
answered 4 hours ago
SimonSimon
4471 silver badge12 bronze badges
$endgroup$
add a comment |
$begingroup$
Add some additional continuous variables $s_{h,s}$ to your model and use those variables in the objective, instead of the products.
Add the following constraints for each $s_{h,s}$:
This constraint ensures that $s_{h,s}$ is at most equal to the sum:
$s_{h,s} leq src_{h,s}+ch_{h,s}+dst_{h,s}+wt_{h,s}$
This constraint ensures that $s_{h,s}$ will be at least the sum when $a_{h,s}=1$:
$s_{h,s} geq src_{h,s}+ch_{h,s}+dst_{h,s}+wt_{h,s} - M times (1 - a_{h,s}) $
This constraints ensures that $s_{h,s}=0$ when $a_{h,s}=0$:
$s_{h,s} leq M times a_{h,s} $
Some notes about this:
- I assumed that your constants and variables are all nonnegative.
- You should pick small values for the constant $M$ to make it all work (e.g. $src_{h,s}+ch_{h,s}+dst_{h,s}+UB(wt_{h,s})$). Picking much larger values leads to lower performance and might even introduce numerical problems.
- If your solver of choice supports indicator constraints, you could also formulate it using those.
answered 4 hours ago
SimonSimon
4471 silver badge12 bronze badges
$endgroup$
add a comment |
$begingroup$
Add some additional continuous variables $s_{h,s}$ to your model and use those variables in the objective, instead of the products.
Add the following constraints for each $s_{h,s}$:
This constraint ensures that $s_{h,s}$ is at most equal to the sum:
$s_{h,s} leq src_{h,s}+ch_{h,s}+dst_{h,s}+wt_{h,s}$
This constraint ensures that $s_{h,s}$ will be at least the sum when $a_{h,s}=1$:
$s_{h,s} geq src_{h,s}+ch_{h,s}+dst_{h,s}+wt_{h,s} - M times (1 - a_{h,s}) $
This constraints ensures that $s_{h,s}=0$ when $a_{h,s}=0$:
$s_{h,s} leq M times a_{h,s} $
Some notes about this:
- I assumed that your constants and variables are all nonnegative.
- You should pick small values for the constant $M$ to make it all work (e.g. $src_{h,s}+ch_{h,s}+dst_{h,s}+UB(wt_{h,s})$). Picking much larger values leads to lower performance and might even introduce numerical problems.
- If your solver of choice supports indicator constraints, you could also formulate it using those.
answered 4 hours ago
SimonSimon
4471 silver badge12 bronze badges
$endgroup$
Add some additional continuous variables $s_{h,s}$ to your model and use those variables in the objective, instead of the products.
Add the following constraints for each $s_{h,s}$:
This constraint ensures that $s_{h,s}$ is at most equal to the sum:
$s_{h,s} leq src_{h,s}+ch_{h,s}+dst_{h,s}+wt_{h,s}$
This constraint ensures that $s_{h,s}$ will be at least the sum when $a_{h,s}=1$:
$s_{h,s} geq src_{h,s}+ch_{h,s}+dst_{h,s}+wt_{h,s} - M times (1 - a_{h,s}) $
This constraints ensures that $s_{h,s}=0$ when $a_{h,s}=0$:
$s_{h,s} leq M times a_{h,s} $
Some notes about this:
- I assumed that your constants and variables are all nonnegative.
- You should pick small values for the constant $M$ to make it all work (e.g. $src_{h,s}+ch_{h,s}+dst_{h,s}+UB(wt_{h,s})$). Picking much larger values leads to lower performance and might even introduce numerical problems.
- If your solver of choice supports indicator constraints, you could also formulate it using those.
answered 4 hours ago
SimonSimon
4471 silver badge12 bronze badges
answered 4 hours ago
SimonSimon
4471 silver badge12 bronze badges
answered 4 hours ago
SimonSimon
4471 silver badge12 bronze badges
answered 4 hours ago
SimonSimon
4471 silver badge12 bronze badges
4471 silver badge12 bronze badges
add a comment |
add a comment |
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1
$begingroup$
Linearize the objective function the same way you would a constraint. Having two continuous variables in the first constraint doesn't add any complications because one of these variable appears "by itself", i.e., not multiplied by another variable, and therefore that variable already appears linearly.
$endgroup$
– Mark L. Stone
8 hours ago
1
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Maybe take a look at this question: How to linearize the product of a binary and a non-negative continuous variable?
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– EhsanK
6 hours ago
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Is this $$ sum_{h in H} sum_{s in S} (src_{h,s} + ch_{h,s} + dst_{h,s} ) times a_{h,s} + ( wt_{h,s} - (1 - a_{h,s}) times infty ) $$ correct linearization of objective function, but what about the bounds.
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– anoop yadav
6 hours ago