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Decision Variable Value from a Set (Gurobi)

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4

$begingroup$

Is there a way to set a decision variable to take values from a set?

Example:

decision variable $x in 0,50,100$

So this variable can only take one of these three values and not more.

I have found a documentation online which unfortunately does not state such a thing:

Gurobi Documentation

For this example, I could probably set two different decision variables, with lower bound 0 and 50 or 100, respectively. I am not satisfied with this method, though.

This is how it would probably look like in Java:

//GRBVar addVar(double lb, double ub, double obj, char type, String name)
GRBVar x = model.addVar(50.0,50.0,null,GRB.SEMIINT,"x");
GRBVar y = model.addVar(100.0,100.0,null,GRB.SEMIINT,"y");

Note:

A semi-continous variable has the property that it takes a value of 0, or a value between the specified lower and upper bounds. A semi-integer variable adds the additional restriction that the variable also take an integral value (GRB.SEMIINT).

integer-programming gurobi

share|improve this question

edited 3 hours ago

LarrySnyder610♦

6,5921515 silver badges6767 bronze badges

asked 9 hours ago

GeorgiosGeorgios

3331313 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I removed the tag or-tools because this question is not about OR Tools, Google's OR solver. If I was wrong to do so, please add it back, along with clarification in the question. Thanks!
  $endgroup$
  – LarrySnyder610♦
  3 hours ago
  

  
  
  
  

add a comment
 | 

4

$begingroup$

Is there a way to set a decision variable to take values from a set?

Example:

decision variable $x in 0,50,100$

So this variable can only take one of these three values and not more.

I have found a documentation online which unfortunately does not state such a thing:

Gurobi Documentation

For this example, I could probably set two different decision variables, with lower bound 0 and 50 or 100, respectively. I am not satisfied with this method, though.

This is how it would probably look like in Java:

//GRBVar addVar(double lb, double ub, double obj, char type, String name)
GRBVar x = model.addVar(50.0,50.0,null,GRB.SEMIINT,"x");
GRBVar y = model.addVar(100.0,100.0,null,GRB.SEMIINT,"y");

Note:

A semi-continous variable has the property that it takes a value of 0, or a value between the specified lower and upper bounds. A semi-integer variable adds the additional restriction that the variable also take an integral value (GRB.SEMIINT).

integer-programming gurobi

share|improve this question

edited 3 hours ago

LarrySnyder610♦

6,5921515 silver badges6767 bronze badges

asked 9 hours ago

GeorgiosGeorgios

3331313 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I removed the tag or-tools because this question is not about OR Tools, Google's OR solver. If I was wrong to do so, please add it back, along with clarification in the question. Thanks!
  $endgroup$
  – LarrySnyder610♦
  3 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Is there a way to set a decision variable to take values from a set?

Example:

decision variable $x in 0,50,100$

So this variable can only take one of these three values and not more.

I have found a documentation online which unfortunately does not state such a thing:

Gurobi Documentation

For this example, I could probably set two different decision variables, with lower bound 0 and 50 or 100, respectively. I am not satisfied with this method, though.

This is how it would probably look like in Java:

//GRBVar addVar(double lb, double ub, double obj, char type, String name)
GRBVar x = model.addVar(50.0,50.0,null,GRB.SEMIINT,"x");
GRBVar y = model.addVar(100.0,100.0,null,GRB.SEMIINT,"y");

Note:

A semi-continous variable has the property that it takes a value of 0, or a value between the specified lower and upper bounds. A semi-integer variable adds the additional restriction that the variable also take an integral value (GRB.SEMIINT).

integer-programming gurobi

share|improve this question

edited 3 hours ago

LarrySnyder610♦

6,5921515 silver badges6767 bronze badges

asked 9 hours ago

GeorgiosGeorgios

3331313 bronze badges

$endgroup$

//GRBVar addVar(double lb, double ub, double obj, char type, String name)
GRBVar x = model.addVar(50.0,50.0,null,GRB.SEMIINT,"x");
GRBVar y = model.addVar(100.0,100.0,null,GRB.SEMIINT,"y");

share|improve this question

edited 3 hours ago

LarrySnyder610♦

6,5921515 silver badges6767 bronze badges

asked 9 hours ago

GeorgiosGeorgios

3331313 bronze badges

share|improve this question

edited 3 hours ago

LarrySnyder610♦

6,5921515 silver badges6767 bronze badges

asked 9 hours ago

GeorgiosGeorgios

3331313 bronze badges

edited 3 hours ago

LarrySnyder610♦

6,5921515 silver badges6767 bronze badges

edited 3 hours ago

LarrySnyder610♦

6,5921515 silver badges6767 bronze badges

asked 9 hours ago

GeorgiosGeorgios

3331313 bronze badges

asked 9 hours ago

GeorgiosGeorgios

3331313 bronze badges

2 Answers
2

active

oldest

votes

4

$begingroup$

I don't think directly, but one way to model that would be to:

• define a parameter as the vector of the possible values: $p = [0,50,100]$




• define three binary decision variables that each correspond to whether a value in the set is selected ($y_i in 0,1, i = 1,2,3$)




• enforce constraints:

$sum_iin 1,2,3 y_i = 1$ (select exactly 1 of the values to use)

$x=sum_iin 1,2,3 p_iy_i $ (apply the chosen value to the variable $x$)

share|improve this answer

answered 8 hours ago

E. TuckerE. Tucker

1,02222 silver badges1919 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Quite similar to my suggestion. Nevertheless, thanks for a clear answer.
  $endgroup$
  – Georgios
  8 hours ago
  

  
  
  
  

add a comment
 | 

2

$begingroup$

In the case of the example that you provided,(and generally, if you have a pattern in the set of available values for $x$), I am assuming that you have the mentioned pattern in the values of the set, you can define your variable $x$ as the integer and then multiply the value of $x$ by a constant. To explain what I mean:

$x in 0, 50, 100$ can be equivalently replaced by $x' in 0, 1, 2$ and then $x = 50 times x'$ in all constraints and objective function.

But if the general form of decision variable's set is not like the example that you mentioned, this approach won't work. Comment on this solution if my assumption is not correct.

share|improve this answer

answered 8 hours ago

Oguz ToragayOguz Toragay

3,58711 gold badge33 silver badges3030 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Brilliant answer if the values are the same as in my example. What if I had values such as $x in 0,2,5,11$ that are not succeeding one another?
  $endgroup$
  – Georgios
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Georgios One option is to create three binary DVs: x_gt_0, x_gt_2, x_gt_5, constrained s.t. x_gt_5 <= x_gt_2 <= x_gt_0, and constrain x = 2*x_gt_0 + (5-2)*x_gt_2 + (11-5)*x_gt_5.
  $endgroup$
  – Geoffrey Brent
  5 hours ago
  

  
  
  
  

add a comment
 | 

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4

$begingroup$

I don't think directly, but one way to model that would be to:

• define a parameter as the vector of the possible values: $p = [0,50,100]$




• define three binary decision variables that each correspond to whether a value in the set is selected ($y_i in 0,1, i = 1,2,3$)




• enforce constraints:

$sum_iin 1,2,3 y_i = 1$ (select exactly 1 of the values to use)

$x=sum_iin 1,2,3 p_iy_i $ (apply the chosen value to the variable $x$)

share|improve this answer

answered 8 hours ago

E. TuckerE. Tucker

1,02222 silver badges1919 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Quite similar to my suggestion. Nevertheless, thanks for a clear answer.
  $endgroup$
  – Georgios
  8 hours ago
  

  
  
  
  

add a comment
 | 

4

$begingroup$

I don't think directly, but one way to model that would be to:

• define a parameter as the vector of the possible values: $p = [0,50,100]$




• define three binary decision variables that each correspond to whether a value in the set is selected ($y_i in 0,1, i = 1,2,3$)




• enforce constraints:

$sum_iin 1,2,3 y_i = 1$ (select exactly 1 of the values to use)

$x=sum_iin 1,2,3 p_iy_i $ (apply the chosen value to the variable $x$)

share|improve this answer

answered 8 hours ago

E. TuckerE. Tucker

1,02222 silver badges1919 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Quite similar to my suggestion. Nevertheless, thanks for a clear answer.
  $endgroup$
  – Georgios
  8 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

I don't think directly, but one way to model that would be to:

• define a parameter as the vector of the possible values: $p = [0,50,100]$




• define three binary decision variables that each correspond to whether a value in the set is selected ($y_i in 0,1, i = 1,2,3$)




• enforce constraints:

$sum_iin 1,2,3 y_i = 1$ (select exactly 1 of the values to use)

$x=sum_iin 1,2,3 p_iy_i $ (apply the chosen value to the variable $x$)

share|improve this answer

answered 8 hours ago

E. TuckerE. Tucker

1,02222 silver badges1919 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

E. TuckerE. Tucker

1,02222 silver badges1919 bronze badges

answered 8 hours ago

E. TuckerE. Tucker

1,02222 silver badges1919 bronze badges

answered 8 hours ago

E. TuckerE. Tucker

1,02222 silver badges1919 bronze badges

2

$begingroup$

In the case of the example that you provided,(and generally, if you have a pattern in the set of available values for $x$), I am assuming that you have the mentioned pattern in the values of the set, you can define your variable $x$ as the integer and then multiply the value of $x$ by a constant. To explain what I mean:

$x in 0, 50, 100$ can be equivalently replaced by $x' in 0, 1, 2$ and then $x = 50 times x'$ in all constraints and objective function.

But if the general form of decision variable's set is not like the example that you mentioned, this approach won't work. Comment on this solution if my assumption is not correct.

share|improve this answer

answered 8 hours ago

Oguz ToragayOguz Toragay

3,58711 gold badge33 silver badges3030 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Brilliant answer if the values are the same as in my example. What if I had values such as $x in 0,2,5,11$ that are not succeeding one another?
  $endgroup$
  – Georgios
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Georgios One option is to create three binary DVs: x_gt_0, x_gt_2, x_gt_5, constrained s.t. x_gt_5 <= x_gt_2 <= x_gt_0, and constrain x = 2*x_gt_0 + (5-2)*x_gt_2 + (11-5)*x_gt_5.
  $endgroup$
  – Geoffrey Brent
  5 hours ago
  

  
  
  
  

add a comment
 | 

2

$begingroup$

In the case of the example that you provided,(and generally, if you have a pattern in the set of available values for $x$), I am assuming that you have the mentioned pattern in the values of the set, you can define your variable $x$ as the integer and then multiply the value of $x$ by a constant. To explain what I mean:

$x in 0, 50, 100$ can be equivalently replaced by $x' in 0, 1, 2$ and then $x = 50 times x'$ in all constraints and objective function.

But if the general form of decision variable's set is not like the example that you mentioned, this approach won't work. Comment on this solution if my assumption is not correct.

share|improve this answer

answered 8 hours ago

Oguz ToragayOguz Toragay

3,58711 gold badge33 silver badges3030 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Brilliant answer if the values are the same as in my example. What if I had values such as $x in 0,2,5,11$ that are not succeeding one another?
  $endgroup$
  – Georgios
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Georgios One option is to create three binary DVs: x_gt_0, x_gt_2, x_gt_5, constrained s.t. x_gt_5 <= x_gt_2 <= x_gt_0, and constrain x = 2*x_gt_0 + (5-2)*x_gt_2 + (11-5)*x_gt_5.
  $endgroup$
  – Geoffrey Brent
  5 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

In the case of the example that you provided,(and generally, if you have a pattern in the set of available values for $x$), I am assuming that you have the mentioned pattern in the values of the set, you can define your variable $x$ as the integer and then multiply the value of $x$ by a constant. To explain what I mean:

$x in 0, 50, 100$ can be equivalently replaced by $x' in 0, 1, 2$ and then $x = 50 times x'$ in all constraints and objective function.

But if the general form of decision variable's set is not like the example that you mentioned, this approach won't work. Comment on this solution if my assumption is not correct.

share|improve this answer

answered 8 hours ago

Oguz ToragayOguz Toragay

3,58711 gold badge33 silver badges3030 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

Oguz ToragayOguz Toragay

3,58711 gold badge33 silver badges3030 bronze badges

answered 8 hours ago

Oguz ToragayOguz Toragay

3,58711 gold badge33 silver badges3030 bronze badges

answered 8 hours ago

Oguz ToragayOguz Toragay

3,58711 gold badge33 silver badges3030 bronze badges