Add constraint
回答済みI have the following code:
GRBLinExpr lhs = 0.0;
for (int ti = 0; ti < alltask_CountVars.Count; ti++)
{
int taskProVal = GetTaskPropertyVal();
for (int diffidx = 0; diffidx < 3; diffidx++)
{
lhs.AddTerm(taskProVal, alltask_CountVars[ti][diffidx]); //
}
}
model.AddConstr(lhs == goal , "constraint1");
Here "alltask_CountVars" is GRB variable.
I have "goal" as a array of value and the constraint is if lhs is equals to any value in the "goal" array.
How do I set this constraint?
thank you.
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Gurobi Staff
You would have to introduce an additional binary variable for every entry of your \(\texttt{goal}\) array. The binary variables would be used to check whether the equality holds as described in the stackexchange post In an integer program, how I can force a binary vairable to equal 1 if some condition holds. Then you can use the addGenConstrOr method to check whether at least one of the equality constraints is fulfilled.
Best regards,
Jaromił0 -
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Thank you for answering my question.
I still have questions about introducing an additional binary variable for every entry of my goal array. I forget to say my goal array is some random integer value(e.g.,[1,41,5,37]) the post you provide assume the input variable x is binary or continuous. This might be an issue?
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I forget to say my goal array is some random integer value(e.g.,[1,41,5,37]) the post you provide assume the input variable x is binary or continuous. This might be an issue?
Integer values should work the same as continuous but without the "numerical inaccuracy" part. So, since all your \(\texttt{goal}\) values are integer, you could use \(\delta=0.5\).
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Hello Jaromil,
According to the link, how do I code these two formulation in Gurobi
Mz1 >= x-b+ sigma (1)
M(1-z1)> b-x-sigma(2)
As far as I understand, M and sigma is some constant value, z1 is a GRB binary VAR, X and b is GRB VAR.
Could you give me an example?
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mip.R is a good example to start with.
The two inequalities would be constructed via
model$A <- matrix(c(1,-1,-M,-1,1,M), nrow=2, ncol=3, byrow=T)
model$rhs <- c(0.5,M+0.5)
model$sense <- c('<', '<')The above constructs
\[\begin{align}
x - b - Mz_1 &\leq 0.5\\
-x + b + Mz_1 &\leq M + 0.5
\end{align}\]Best regards,
Jaromił0 -
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Thank you for giving me the example in R!
I have tried to code the relationship without matrix( because my original code is in C#)
x−b−Mz1≤0.5
−x+b+Mz1≤M+0.5
I follow your suggest and introduce an additional binary variable for every entry of my goal array. Then I encode these two inequalities by using "AddGenConstrIndicator" method.
GRBLinExpr fixedLhs = 0.0;
for (int ti = 0; ti < alltask_CountVars.Count; ti++)
{
int taskProVal = GetTaskPropertyVal();
for (int diffidx = 0; diffidx < 3; diffidx++)
{
fixedLhs.AddTerm(taskProVal, alltask_CountVars[ti][diffidx]); //
}
}for (int i = 0; i < goals.Count; i++)
{
//if x>b−sigma, then forces z1=1 , where x is lhs, b is goal[i], z1 is binary var
model.AddGenConstrIndicator(z1[i], 1, fixedLhs, GRB.GREATER_EQUAL, goals[i] -0.5, "");
//If x<b−sigma , then (2) forces z1=0
model.AddGenConstrIndicator(z1[i], 0, fixedLhs, GRB.LESS_EQUAL, goals[i] - 0.5, "");
//skip z2 here
}
//in the end check min_z1+min_z2 -1 >=1
model.AddConstr(min_z1+min_z2 -1, GRB.GREATER_EQUAL, 1.0, "min_z1+min_z2 -1 >= 1");The optimization result seems fine, except it only gives solutions that satisfied one goal. (I also set pool search mode = 2. ) I expect the optimizer will give me all solutions that satisfy the constraint " if lhs is equals to any value in the "goal" array."
Thank you in advance!
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It seems like you implemented only constraints (1) and (2) from In an integer program, how I can force a binary vairable to equal 1 if some condition holds. Note that you also need constraints (3) and (4). Alternatively, you could use the formulation posted in the last comment of the stackexchange thread.
I expect the optimizer will give me all solutions that satisfy the constraint " if lhs is equals to any value in the "goal" array."
You have exactly one \(\texttt{fixedLhs}\) object. So as long as there are not 2 same \(\texttt{goal}\) values, it is not possible that more that one \(\texttt{goal}\) entry equals \(\texttt{fixedLhs}\) at the same time.
Best regards,
Jaromił0 -
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