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Modeling the Round (Nearest Integer) function

When to use indicator constraints versus big-M approaches in solving (mixed-)integer programsModeling floor function exactlyTightness of an LP relaxation without using objective functionIs deciding the presence of mixed-integer points in the relative interior of a polyhedron in NP?How to linearize min function as a constraint?Expressing a chain of boolean ORs using ILPRepresenting an indicator function: binary variables and “indicator constraints”Decoding a Deep Neural Network as an Analytical Expression for Optimization PurposeValid Inequalities and Strong InequalitiesIn the context of LASSO regression, how to introduce a constraint for max number of selected betas?

3

$begingroup$

Modeling various non-differentiable functions is quite common knowledge including $abs$, $min$ and $max$ functions. How would one go about modeling the nearest integer function , say in an inequality constraint:

$lfloor{x}rceil leq C$

mixed-integer-programming modeling

share|improve this question

asked 8 hours ago

Josh AllenJosh Allen

6133 bronze badges

New contributor

Josh Allen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have a specific rounding rule in mind for .5 cases? The Wikipedia page on 'Nearest integer function' (en.wikipedia.org/wiki/Nearest_integer_function) says "On most computer implementations, the selected rule is to round half-integers to the nearest even integer"
  $endgroup$
  – Dipayan Banerjee
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DipayanBanerjee Yes, let's assume the nearest even integer rule for this example.
  $endgroup$
  – Josh Allen
  8 hours ago
  
  
  

  
  
  
  

add a comment
 | 

3

$begingroup$

Modeling various non-differentiable functions is quite common knowledge including $abs$, $min$ and $max$ functions. How would one go about modeling the nearest integer function , say in an inequality constraint:

$lfloor{x}rceil leq C$

mixed-integer-programming modeling

share|improve this question

asked 8 hours ago

Josh AllenJosh Allen

6133 bronze badges

New contributor

Josh Allen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have a specific rounding rule in mind for .5 cases? The Wikipedia page on 'Nearest integer function' (en.wikipedia.org/wiki/Nearest_integer_function) says "On most computer implementations, the selected rule is to round half-integers to the nearest even integer"
  $endgroup$
  – Dipayan Banerjee
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DipayanBanerjee Yes, let's assume the nearest even integer rule for this example.
  $endgroup$
  – Josh Allen
  8 hours ago
  
  
  

  
  
  
  

add a comment
 | 

$begingroup$

Modeling various non-differentiable functions is quite common knowledge including $abs$, $min$ and $max$ functions. How would one go about modeling the nearest integer function , say in an inequality constraint:

$lfloor{x}rceil leq C$

mixed-integer-programming modeling

share|improve this question

asked 8 hours ago

Josh AllenJosh Allen

6133 bronze badges

New contributor

Josh Allen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

asked 8 hours ago

Josh AllenJosh Allen

6133 bronze badges

New contributor

Josh Allen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked 8 hours ago

Josh AllenJosh Allen

6133 bronze badges

New contributor

Josh Allen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

Josh AllenJosh Allen

6133 bronze badges

New contributor

Josh Allen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

Josh AllenJosh Allen

6133 bronze badges

4 Answers
4

active

oldest

votes

3

$begingroup$

This is a hack of Robert Schwarz's answer, to accommodate the "round .5 to even" rule. We introduce integer variable $y$ and binary variable $z$, along with the constraints $$2y+z le x le 2y + z + 1$$ and $$x + z - 0.5 le C.$$ If $x$ is noninteger, the first constraint finds the nearest integers on either side. If the floor of $x$ is even (meaning a fraction of one half would round down), $z=0$ and we require that $x - 0.5 le C$. If the floor of $x$ is odd (meaning a fraction of one half would round up), $z=1$ and we require that $x + 0.5 le C$.

There is an ambiguity when $x$ is integer. For instance, if $x = 3$, both $(y,z)=(1,1)$ and $(y,z)=(1,0)$ satisfy the constraint. Fortunately, the solver will bail us out: if it "wants" to have $x=3$ be feasible (and if $C=3$), it will choose $(y,z)=(1,0)$, so that $x=3=C$ satisfies the second constraint.

share|improve this answer

answered 5 hours ago

prubinprubin

5,22199 silver badges3434 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

Assuming finite bounds on $x$, this could be modeled with disjunctions on the many cases to which $x$ can be rounded.

For example, if $x in [ 0, 2 ]$, we would have:
$$
begin{cases}
x le 0.5, & 0 le C \
0.5 le x le 1.5, & 1 le C \
1.5 le x , & 2 le C
end{cases}
$$

The disjunction itself could be modeled with auxiliary binary variables and big-$M$ constraints, for example.

Note that rounding in this context has some ambiguity, as a value of $0.5$ may be rounded either to $0$ or $1$, because strict inequalities can usually not be enforced by MIP solvers.

share|improve this answer

answered 8 hours ago

Robert SchwarzRobert Schwarz

1,07533 silver badges1414 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

As an alternative solution, I propose to add an auxiliary integral variable $y in mathbb{Z}$ that should play the role of the rounded $x$.

For your example of the inequality, I would add:
$$
begin{array}{rll}
y &le& C \
x - 0.5&le& y
end{array}
$$

share|improve this answer

answered 8 hours ago

Robert SchwarzRobert Schwarz

1,07533 silver badges1414 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is close, but trips over the "round-to-even" rule. We can assume that $C$ is integer. Suppose that $C=3$. $x=3.5$, $y=3$ satisfies both inequalities, but $leftlceil xrightrfloor =4$.
  $endgroup$
  – prubin
  6 hours ago
  

  
  
  
  

add a comment
 | 

0

$begingroup$

Assuming that the value $C + 0.5$ is valid, you could model this by simply adding the constraint

$$x leq C + 0.5$$

Otherwise, you define a constant say $epsilon = 10^{-7}$ and do

$$x leq C + 0.5 - epsilon$$

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Claudio ContardoClaudio Contardo

82622 silver badges99 bronze badges

$endgroup$

add a comment
 | 

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3

$begingroup$

This is a hack of Robert Schwarz's answer, to accommodate the "round .5 to even" rule. We introduce integer variable $y$ and binary variable $z$, along with the constraints $$2y+z le x le 2y + z + 1$$ and $$x + z - 0.5 le C.$$ If $x$ is noninteger, the first constraint finds the nearest integers on either side. If the floor of $x$ is even (meaning a fraction of one half would round down), $z=0$ and we require that $x - 0.5 le C$. If the floor of $x$ is odd (meaning a fraction of one half would round up), $z=1$ and we require that $x + 0.5 le C$.

There is an ambiguity when $x$ is integer. For instance, if $x = 3$, both $(y,z)=(1,1)$ and $(y,z)=(1,0)$ satisfy the constraint. Fortunately, the solver will bail us out: if it "wants" to have $x=3$ be feasible (and if $C=3$), it will choose $(y,z)=(1,0)$, so that $x=3=C$ satisfies the second constraint.

share|improve this answer

answered 5 hours ago

prubinprubin

5,22199 silver badges3434 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

This is a hack of Robert Schwarz's answer, to accommodate the "round .5 to even" rule. We introduce integer variable $y$ and binary variable $z$, along with the constraints $$2y+z le x le 2y + z + 1$$ and $$x + z - 0.5 le C.$$ If $x$ is noninteger, the first constraint finds the nearest integers on either side. If the floor of $x$ is even (meaning a fraction of one half would round down), $z=0$ and we require that $x - 0.5 le C$. If the floor of $x$ is odd (meaning a fraction of one half would round up), $z=1$ and we require that $x + 0.5 le C$.

There is an ambiguity when $x$ is integer. For instance, if $x = 3$, both $(y,z)=(1,1)$ and $(y,z)=(1,0)$ satisfy the constraint. Fortunately, the solver will bail us out: if it "wants" to have $x=3$ be feasible (and if $C=3$), it will choose $(y,z)=(1,0)$, so that $x=3=C$ satisfies the second constraint.

share|improve this answer

answered 5 hours ago

prubinprubin

5,22199 silver badges3434 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

This is a hack of Robert Schwarz's answer, to accommodate the "round .5 to even" rule. We introduce integer variable $y$ and binary variable $z$, along with the constraints $$2y+z le x le 2y + z + 1$$ and $$x + z - 0.5 le C.$$ If $x$ is noninteger, the first constraint finds the nearest integers on either side. If the floor of $x$ is even (meaning a fraction of one half would round down), $z=0$ and we require that $x - 0.5 le C$. If the floor of $x$ is odd (meaning a fraction of one half would round up), $z=1$ and we require that $x + 0.5 le C$.

There is an ambiguity when $x$ is integer. For instance, if $x = 3$, both $(y,z)=(1,1)$ and $(y,z)=(1,0)$ satisfy the constraint. Fortunately, the solver will bail us out: if it "wants" to have $x=3$ be feasible (and if $C=3$), it will choose $(y,z)=(1,0)$, so that $x=3=C$ satisfies the second constraint.

share|improve this answer

answered 5 hours ago

prubinprubin

5,22199 silver badges3434 bronze badges

$endgroup$

share|improve this answer

answered 5 hours ago

prubinprubin

5,22199 silver badges3434 bronze badges

answered 5 hours ago

prubinprubin

5,22199 silver badges3434 bronze badges

answered 5 hours ago

prubinprubin

5,22199 silver badges3434 bronze badges

2

$begingroup$

Assuming finite bounds on $x$, this could be modeled with disjunctions on the many cases to which $x$ can be rounded.

For example, if $x in [ 0, 2 ]$, we would have:
$$
begin{cases}
x le 0.5, & 0 le C \
0.5 le x le 1.5, & 1 le C \
1.5 le x , & 2 le C
end{cases}
$$

The disjunction itself could be modeled with auxiliary binary variables and big-$M$ constraints, for example.

Note that rounding in this context has some ambiguity, as a value of $0.5$ may be rounded either to $0$ or $1$, because strict inequalities can usually not be enforced by MIP solvers.

share|improve this answer

answered 8 hours ago

Robert SchwarzRobert Schwarz

1,07533 silver badges1414 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

Assuming finite bounds on $x$, this could be modeled with disjunctions on the many cases to which $x$ can be rounded.

For example, if $x in [ 0, 2 ]$, we would have:
$$
begin{cases}
x le 0.5, & 0 le C \
0.5 le x le 1.5, & 1 le C \
1.5 le x , & 2 le C
end{cases}
$$

The disjunction itself could be modeled with auxiliary binary variables and big-$M$ constraints, for example.

Note that rounding in this context has some ambiguity, as a value of $0.5$ may be rounded either to $0$ or $1$, because strict inequalities can usually not be enforced by MIP solvers.

share|improve this answer

answered 8 hours ago

Robert SchwarzRobert Schwarz

1,07533 silver badges1414 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Assuming finite bounds on $x$, this could be modeled with disjunctions on the many cases to which $x$ can be rounded.

For example, if $x in [ 0, 2 ]$, we would have:
$$
begin{cases}
x le 0.5, & 0 le C \
0.5 le x le 1.5, & 1 le C \
1.5 le x , & 2 le C
end{cases}
$$

The disjunction itself could be modeled with auxiliary binary variables and big-$M$ constraints, for example.

Note that rounding in this context has some ambiguity, as a value of $0.5$ may be rounded either to $0$ or $1$, because strict inequalities can usually not be enforced by MIP solvers.

share|improve this answer

answered 8 hours ago

Robert SchwarzRobert Schwarz

1,07533 silver badges1414 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

Robert SchwarzRobert Schwarz

1,07533 silver badges1414 bronze badges

answered 8 hours ago

Robert SchwarzRobert Schwarz

1,07533 silver badges1414 bronze badges

answered 8 hours ago

Robert SchwarzRobert Schwarz

1,07533 silver badges1414 bronze badges

1

$begingroup$

As an alternative solution, I propose to add an auxiliary integral variable $y in mathbb{Z}$ that should play the role of the rounded $x$.

For your example of the inequality, I would add:
$$
begin{array}{rll}
y &le& C \
x - 0.5&le& y
end{array}
$$

share|improve this answer

answered 8 hours ago

Robert SchwarzRobert Schwarz

1,07533 silver badges1414 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is close, but trips over the "round-to-even" rule. We can assume that $C$ is integer. Suppose that $C=3$. $x=3.5$, $y=3$ satisfies both inequalities, but $leftlceil xrightrfloor =4$.
  $endgroup$
  – prubin
  6 hours ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

As an alternative solution, I propose to add an auxiliary integral variable $y in mathbb{Z}$ that should play the role of the rounded $x$.

For your example of the inequality, I would add:
$$
begin{array}{rll}
y &le& C \
x - 0.5&le& y
end{array}
$$

share|improve this answer

answered 8 hours ago

Robert SchwarzRobert Schwarz

1,07533 silver badges1414 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is close, but trips over the "round-to-even" rule. We can assume that $C$ is integer. Suppose that $C=3$. $x=3.5$, $y=3$ satisfies both inequalities, but $leftlceil xrightrfloor =4$.
  $endgroup$
  – prubin
  6 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

As an alternative solution, I propose to add an auxiliary integral variable $y in mathbb{Z}$ that should play the role of the rounded $x$.

For your example of the inequality, I would add:
$$
begin{array}{rll}
y &le& C \
x - 0.5&le& y
end{array}
$$

share|improve this answer

answered 8 hours ago

Robert SchwarzRobert Schwarz

1,07533 silver badges1414 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

Robert SchwarzRobert Schwarz

1,07533 silver badges1414 bronze badges

answered 8 hours ago

Robert SchwarzRobert Schwarz

1,07533 silver badges1414 bronze badges

answered 8 hours ago

Robert SchwarzRobert Schwarz

1,07533 silver badges1414 bronze badges

0

$begingroup$

Assuming that the value $C + 0.5$ is valid, you could model this by simply adding the constraint

$$x leq C + 0.5$$

Otherwise, you define a constant say $epsilon = 10^{-7}$ and do

$$x leq C + 0.5 - epsilon$$

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Claudio ContardoClaudio Contardo

82622 silver badges99 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

Assuming that the value $C + 0.5$ is valid, you could model this by simply adding the constraint

$$x leq C + 0.5$$

Otherwise, you define a constant say $epsilon = 10^{-7}$ and do

$$x leq C + 0.5 - epsilon$$

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Claudio ContardoClaudio Contardo

82622 silver badges99 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Assuming that the value $C + 0.5$ is valid, you could model this by simply adding the constraint

$$x leq C + 0.5$$

Otherwise, you define a constant say $epsilon = 10^{-7}$ and do

$$x leq C + 0.5 - epsilon$$

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Claudio ContardoClaudio Contardo

82622 silver badges99 bronze badges

$endgroup$

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Claudio ContardoClaudio Contardo

82622 silver badges99 bronze badges

answered 6 hours ago

Claudio ContardoClaudio Contardo

82622 silver badges99 bronze badges

answered 6 hours ago

Claudio ContardoClaudio Contardo

82622 silver badges99 bronze badges