Originally posted by: felycite28

Hello everyone ,

I have small question .

I am computing a summation

po=sum( a in t)p[a] and tying to minimum value greater than po in a list ; My po=65 it is good so far , after thta I have to find 70 from the list

I did :

int A=...;
range a=1..A;

range k=1..K;

AR=[20 30 40 50 60 70]

po=sum( a in t)p[a]

minValue=(min(k in a : AR[k] >= po));

and it did not work ? Can you give me an idea to correct it or ay other suggestion to find 70 ?

Thanks in advance

Hi,

 int A=6;
range k=1..A;

int AR[k]=[20 ,30, 40, 50, 60, 70];

int po=65;

int MinValue=AR[min(k in k : AR[k]>=po)k];

execute
{
writeln(MinValue);
}

gives

70

regards

Originally posted by: felycite28

Thanks Alex ,

In my case , po is a decision variable , this time is 65, next time it might be 80, I should have told you before . That swhhy when I try the way you showed I get this error:
Variable de décision (ou expression) "po" non autorisée. 

I could not know how can I handle it ?

Thanks

Is not it possible to compare with a decision variable and the elements of a list?

Hi,

Logical constraints could help.

Then you could write

int A=6;
range K=1..A;

int AR[K]=[20 ,30, 40, 50, 60, 70];

dvar int po;
dvar boolean ARofkismorethanpo[K];
dvar int indexMin in K;
dvar int MinValue;

subject to
{
po==65;

forall(k in K) ARofkismorethanpo[k]==(AR[k]>=po);

indexMin==A+1-max(k in K) ARofkismorethanpo[k]*(A+1-k);

forall(i in K) (i==indexMin) => (MinValue==AR[i]);

}

int MinValue2=AR[min(k in K : AR[k]>=po)k];
execute
{
writeln(MinValue);
writeln(MinValue2);
}

regards

https://www.linkedin.com/pulse/how-opl-alex-fleischer/

Originally posted by: felycite28

Than you so much Alex, Million thanks ,

Now, I need to ask my real question. I found 70 as you taught, I want to reach another matrix which is data again ,  related this 70.Basically,

I did something:

int A=6;

t=3;
range K=1..A;

int AR[K]=[20 ,30, 40, 50, 60, 70];

M[k][t]=[[1 2 3 ]// this line corresponds 20

            [0.2 0.3 0.4]// this line corresponds 30 so on

           [.................]

          [ 2   5  7]] // this line corresponds 70

So I should be able to choose this last line which is related to 70? I will use this line as data for the constraint.

How can I do that ? I tried to use M[indexMin][t] but did not work .

Thank you so much in advance

Hi,

in the how to with OPL link you have

https://www.ibm.com/developerworks/community/forums/html/threadTopic?id=2be2ec22-db4b-4a2c-b164-615b9f735dc9&ps=25

which could help

regards

Originally posted by: felycite28

Yes , it helped but it still got en error , I have to use indexMin again for the other operation . Imagine a; data. b is also decision variable. Y[t] is the latest line which will correspond the latest line , corresponds 70  in the previous post. It will multiply with 2,5,7 in order.

min (sum z in t) Y[z]*a*b

int A=6;

z=3;
range K=1..A;

int AR[K]=[20 ,30, 40, 50, 60, 70];

M[k][z]=[[1 2 3 ]// this line corresponds 20

            [0.2 0.3 0.4]// this line corresponds 30 so on

           [.................]

          [ 2   5  7]] // this line corresponds 70

I did:

forall (z in t, i in K)(i==indexMin)=>(Y[z]==M[i][z]);

It did not turn result, and said

CPLEX Error  5002: Q in objective is not positive semi-definite.  since I described the Y[z] as dvar, it gives this result but It is not that much decision variable . Is there any other way to do it ? It must be , I guess.

Thank you so much

Hi,

can you post your model so that other users could help ?

regards