I am wondering the syntax for logical constraints in Google OR-Tools. I have a nurse scheduling project I have done in CPLEX that I am translating over to Google OR-Tools. I have come across the documentation for channeling constraints in Google OR-Tools, but I am confused. Can you help me understand how I would implement this CPLEX logical constraint in Google OR-Tools? I have an attempt, but it is not working as intended :(
Context:
working_assignment_vars_long[r,h,i] is a binary decision variable that denotes whether nurse i in role r is working at the 15-minute interval h (i.e. 1:15PM).
lunch_break_assignment_vars_long[r,h,i] is a binary decision variable that denotes whether nurse i in role r is on break at the 15-minute interval h (i.e. 1:15PM).
simple_break_assignment_vars_long[r,h,i] is a binary decision variable that denotes whether nurse i in role r is on break at 15-minute interval h.
Thus, this constraint in CPLEX is saying that if a given nurse is working 31 or less 15-minute intervals, then they should have 0 lunch breaks and 1 simple break.
CPLEX logical constraint:
for r, num_role in role_dict.items():
for i in range(0,num_role):
model.add_if_then(
model.sum(model.working_assignment_vars_long[r,h,i] for h in range(0,144))
<= 31,
(model.sum(model.lunch_break_assignment_vars_long[r,h,i] for h in range(0,144)) ==
0) + (model.sum(model.simple_break_assignment_vars_long[r,h,i] for h in range(0,144)) ==
1) >= 2,
'less_than_8_hr_0_lunch_break_1_simple_break'
)
I have attempted this in OR-Tools with the following code:
for r, num_role in role_dict.items():
for i in range(0,num_role):
b=model.NewBoolVar(‘b’)
model.Add(sum(working.assignment_vars_long[r,h,i] for h in range(0,144)) <= 31).OnlyEnforceIf(b)
model.Add(sum(lunch_break_assignment_vars_long[r,h,i] for h in range(0,144)) == 0).OnlyEnforceIf(b)
model.Add(sum(simple_break_assignment_vars_long[r,h,i] for h in range(0,144)) == 1).OnlyEnforceIf(b)
This is not working as intended though :( as I see nurses working less than 32 15-minute intervals with both no lunch breaks and no simple breaks. Any insight/help is greatly appreciated. I have been stuck on this problem for so long now :(
In the official documentation it is documented how to do an If-Then-Else expression.
You aren't constraining b.Not():
model.Add(sum(working.assignment_vars_long[r,h,i] for h in range(0,144)) > 31).OnlyEnforceIf(b.Not())
Related
Hello fellows, i am learning Julia and integer programing but i am stuck at one point
How to model "then" in julia-jump for integer programing leanring.
Stuck here here
#Define the variables of the model
#variable(mo, x[1:N,1:S], Bin)
#variable(mo, a[1:S]>=0)
#Assignment constraint
#constraint(mo, [i=1:N], sum(x[i,j] for j=1:S) == 1)
##constraint (mo, PLEASE HELP )
In cases like this you usually need to use Big-M constraints
So this will be:
a_ij >= s_i^2 - M*(1-x_ij)
where M is a "big enough" number. This means that if x_ij == 0 the inequality will always be true (and hence kind of turned-off). On the other hand when x_ij == 1 the M-part will be zeroed and the equation will hold.
In JuMP terms the code will look like this:
const M = 10_000
#constraint(mo, [i=1:N, j=1:S], a[i, j] >= s[i]^2 - M*(1 - x[i, j]))
However, if s[i] is an external parameter rather than model variable you could simply use x[i,j] <= a[j]/s[i]^2 proposed by #DanGetz. However when s[i] is #variable you really want to avoid dividing or multiplying variables by each other. So this big M approach is more general across use cases.
I am trying to formulate a linear program that will assign different number of employees to start in different days. Each group of employees starting on a day will get two days off during the week. However, the schedule is unknown. For example, employees starting Monday can be off any two days in the week. Since the number that will start on day (i) is unknown and whether they will have a day off or not is unknown, I will have the product of two decision variables - one is an integer xi (employees starting on day i) and a binary variable yij (whether the employees starting on day i have a day off on day j).
I am done with formulation and here it is:
Decision variables 1: xi (employees starting on day i)
Decision variables 2: yij (1 if employees starting on day i are working on day j, or 0 if employees starting on day i are off on day j)
Objective function:
Minimize total employees-- sum (i in 1..7) xi
Subject to:
xi*yij >= Requiredj (the number of available workers on day j have to satisfy the demand on day j)
I am trying to code this on CPLEX but i dont know how to make xi*yij linear and write the code....can anyone please help me?
Thank you.
In How to with OPL How to multiply a decision variable by a boolean decision variable in CPLEX ?
// suppose we want b * x <= 7
dvar int x in 2..10;
dvar boolean b;
dvar int bx;
maximize x;
subject to
{
// Linearization
bx<=7;
2*b<=bx;
bx<=10*b;
bx<=x-2*(1-b);
bx>=x-10*(1-b);
// if we use CP we could write directly
// b*x<=7
// or rely on logical constraints within CPLEX
// (b==1) => (bx==x);
// (b==0) => (bx==0);
}
I have 4 non negative real variable that are A, B, C and X. Based on the current problem that I have, I notice that the variable X must belong to the interval of [B,C] and the relation will be a bunch of if-else conditions like this:
If A < B:
x = B
elseif A > C:
x = C
elseif B<=A<=C:
x = A
As you can see, it quite difficult to reformulate as a Mixed Integer Programming problem with corresponding decision variable (d1, d2 and d3). I have try reading some instructions regarding if-then formulation using big M method at this site:
https://www.math.cuhk.edu.hk/course_builder/1415/math3220/L2%20(without%20solution).pdf but it seem that this problem is more challenging than their tutorial.
Could you kindly provide me with a formulation for this situation ?
Thank you very much !
For Colombia's Observatorio Fiscal[1], I am coding a simple tax minimization problem, using CLP(R) (in SWI-Prolog). I want to use minimize/1 to find the least solution first. It is instead listing the bigger solution first. Here is the code:
:- use_module(library(clpr)).
deduction(_,3). % Anyone can take the standard deduction.
deduction(Who,D) :- itemizedDeduction(Who,D). % Or they can itemize.
income(joe,10). % Joe makes $10 a year.
itemizedDeduction(joe,4). % He can deduct more if he itemizes.
taxableIncome(Who,TI) :-
deduction(Who,D),
income(Who,I),
TI is I - D,
minimize(TI).
Here is what an interactive session looks like:
?- taxableIncome(joe,N).
N = 7 ;
N = 6 ;
false.
If I switch the word "minimize" to "maximize" it behaves identically. If I include no minimize or maximize clause, it doesn't look for a third solution, but otherwise it behaves the same:
?- taxableIncome(joe,N).
N = 7 ;
N = 6.
[1] The Observatorio Fiscal is a new organization that aims to model the Colombian economy, in order to anticipate the effects of changes in the law, similar to what the Congressional Budget Office or the Tax Policy Center do in the United States.
First, let's add the following definition to the program:
:- op(950,fy, *).
*_.
Using (*)/1, we can generalize away individual goals in the program.
For example, let us generalize away the minimize/1 goal by placing * in front:
taxableIncome(Who,TI) :-
deduction(Who,D),
income(Who,I),
TI #= I - D,
* minimize(TI).
We now get:
?- taxableIncome(X, Y).
X = joe,
Y = 7 ;
X = joe,
Y = 6.
This shows that CLP(R) in fact has nothing to do with this issue! These answers show that everything is already instantiated at the time minimize/1 is called, so there is nothing left to minimize.
To truly benefit from minimize/1, you must express the task in the form of CLP(R)—or better: CLP(Q)— constraints, then apply minimize/1 on a constrained expression.
Note also that in SWI-Prolog, both CLP(R) and CLP(Q) have elementary mistakes, and you cannot trust their results.
Per Mat's response, I rewrote the program expressing the constraints using CLP. The tricky bit was that I had to first collect all (both) possible values for deduction, then convert those values into a CLP domain. I couldn't get that conversion to work in CLP(R), but I could in CLP(FD):
:- use_module(library(clpfd)).
deduction(_,3). % Anyone can take the same standard deduction.
deduction(Who,D) :- % Or they can itemize.
itemizedDeduction(Who,D).
income(joe,10).
itemizedDeduction(joe,4).
listToDomain([Elt],Elt).
listToDomain([Elt|MoreElts],Elt \/ MoreDom) :-
MoreElts \= []
, listToDomain(MoreElts,MoreDom).
taxableIncome(Who,TI) :-
income(Who,I)
, findall(D,deduction(Who,D),DList)
, listToDomain(DList,DDomain)
% Next are the CLP constraints.
, DD in DDomain
, TI #= I-DD
, labeling([min(TI)],[TI]).
I want to model that departures from a node can only take place in a "every n hours" manner. I've started to model this using two variables - starttime[i,j,k] shows when vehicle k departured i with j as destination, x[i,j,k] is a binary variable having value 1 if vehicle k drove from i to j, and 0 otherwise. The model is:
maximize maxdrive: sum{i in V, j in V, k in K} traveltime[i,j]*x[i,j,k];
subject to TimeConstraint {k in K}:
sum{i in V, j in V} (traveltime[i,j]+servicetime[i])*x [i,j,k] <= 1440;
subject to StartTime{i in V,j in V, k in K}:
starttime[i,j,k] + traveltime[i,j] - 9000 * (1 - x[i,j,k]) <= starttime[j,i,k];
subject to yvar{i in V, j in V}:
sum{k in K} x[i,j,k] <= maxVisits[i,j];
subject to Constraint1{i in V, j in V, k in K, g in V, h in K}:
starttime[i,j,k] + TimeInterval[i]*x[i,j,k] <= starttime[i,g,h];
The constraint in question is "Constraint1" where i is the origin node, j the destination node, and k is the vehicle. The index g is used to show that the later departure can be to any destination node. TimeInterval corresponds to the interval intended, i.e. if TimeInterval at i is 2 hours, the starttime of the next vehicle to departure from i must not be less than 2 hours from previous departure. The origins corresponds to specific products (only available from said origin node) whereas I want the vehicles to not be bounded to a specific origin node - they should be able to jump between nodes to utilize backhauling etc. In other words, I want to conduct this constraint without having restraints on the vehicles themselves but rather the origin nodes.
The objective function to "maximize the traveltime" may seem strange, but the objective function is rather obsolete really. If the constraints are met, the solution is adequate. To maximize traveltime is merely an attempt to "force" the x variables to become 1.
The question is: how can I do this? With this formulation, all x[i,j,k] variables dissappears from the answer (without this constraint, some of the binary variables x becomes 1 and the other 0. The solution meets the maxVisits requirement. With the constraint all x variables becomes 0 and all starttimes becomes 0 as well. MINTO (The solver) doesn't state that the problem is infeasible either.
Also, how to separate the vehicles so the program recognizes that it is a comparison between all departures? I would rather to not include time dimensions, at it would give so much more variables.
EDIT: After trying a new model using a non-linear solver I've seen some strange results. Specifically, I'm using the limit 1440 (minutes) as an upper bound as to for how long a vehicle can operate each day. Using this model below the solution is 0 for every variable, but the starttime for all combinations of i,j,k is 720 (half of 1440). Does anyone have any clue in regards of what causing this solution? How did this constraint remove the link between starttime being higher than 0 requiring that x must be 1.
subject to StartTimeSelf{i in V, j in V, k in K, g in K, h in V}:
starttime[i,j,k]*x[i,j,k] + TimeInterval[i]*x[i,j,k] + y[i,k] <= starttime[i,h,g]*x[i,j,k];