Decision Optimization

Originally posted by: ClaraD

Hi,

I am having trouble creating scenarios in a stochastic problem.
Let's say I have two objects and each of them can have two values. What I want is to create four scenarios where I have these values combined. 

I had thought about having a variable  like this:
int w[Objects][Weights] = [[7,8], [6,7]]; //Objects' weights

In the first scenario I take the values 7 and 6.
In the second scenario I take the values 7 and 7.
In the third scenario I take the values 8 and 6.
In the fourth scenario I take the values 8 and 7.

but I don't know how to select these scenarios. In the constraints I had to manually create them, but this is an easy example and I would like to increase the number of objects and its values.

Here is what I am trying:

...
range scenarios = 1..(NbObjects*NbWeights); //in this case it's 2*2
dvar boolean x[Objects];//if positive then it goes into the knapsack
dvar float+ z[scenarios];
dvar float+ y[scenarios];
int b = 14; //knapsack capacity
...
subject to {
ct1: //7-6 13
     z[1] >= (w[1][1]*x[1]+w[2][1]*x[2])-b; //13-14 -1
     z[1] >= -(w[1][1]*x[1]+w[2][1]*x[2])+b; //-13+14 1
     y[1] == (((w[1][1]*x[1]+w[2][1]*x[2])-b) + z[1])/2; //I only want the positive ones

     ct2: //7-7 14
     z[2] >= (w[1][1]*x[1]+w[2][2]*x[2])-b;
     z[2] >= -(w[1][1]*x[1]+w[2][2]*x[2])+b;
     y[2] == (((w[1][1]*x[1]+w[2][2]*x[2])-b) + z[2])/2;
     ct3: ...
     ct4: ....

     My question is:
     Is this the right way to store the different objects and values? Is there a way to put the constraints in a loop?

 Thank you so much for your help.

     PS: I have been reading in this forum for a while to find for possible solutions and I see that the site should be theoretically down, but I cannot find the equivalent of this forum in the new IBM site.

Hi,

could

Stochastic optimization

in

https://www.linkedin.com/pulse/making-decision-optimization-simple-alex-fleischer/

help ?

regards

Originally posted by: ClaraD

Thank you Alex,

I had already looked at this example that you provided. I've combined this example with another one you wrote about tuples:https://www.ibm.com/developerworks/community/forums/html/topic?id=9748cb9c-fa58-471a-8f9b-fba546df609b&ps=25

I am now trying to adapt my code to see if by saving the data in tuples I can find a solution. I am confident that this is the good way to go, thanks for pointing me in this direction.

Regards,

Clara

PS: Just in case it's useful to someone I'm gonna leave the code about the stochastic buses with tuples here:

int nbKids=300;
{int} nbKidsScenarii={nbKids+i*10 | i in -10..2};
float proba[nbKidsScenarii]=[ 1, 1, 2, 2, 2 ,3 ,3 ,4,  5 ,10 ,50 ,10, 7];
assert sum(s in nbKidsScenarii) proba[s]==100; // total probability is 100
tuple bus{
key int nbSeats;
float cost;}

// This is a tuple set
{bus} buses={<40,500>,<30,400>};
// asserts help make sure data is fine
assert forall(b in buses) b.nbSeats>0;
assert forall(b in buses) b.cost>0;

float costIncreaseIfLastMinute=1.1;

// number of buses booked in advance
// decision variable array
dvar int+ nbBus[buses];
// number of buses booked at the last minute which is far more expensive
// we call those recourse decision
dvar int+ nbBusonTop[buses][nbKidsScenarii] ;

minimize
   sum(b in buses) b.cost*nbBus[b]
 +
1/100*costIncreaseIfLastMinute*
sum(a in nbKidsScenarii,b in buses) proba[a]*b.cost*(nbBusonTop[b][a]);
 
subject to
{
 forall(a in nbKidsScenarii)
   (sum(b in buses) b.nbSeats*(nbBus[b]+nbBusonTop[b][a]))>=a;
}

Results:

nbBus = [6
         0];
nbBusonTop = [[0 0 0 0 0 0 0 0 1 0 0 1 2]
             [0 0 0 0 0 1 1 1 0 2 2 1 0]];