Dynamic array type in Xpress Python interface

Hello,

the distinction between array and dynamic array in Mosel disappears when one uses Python, because the array of variables or constraints is created on the spot, based on a set of indices. If your question is about how to create an array with a sparse index set, I would suggest using dictionaries like in the following example:

import xpress as xp
 
S = set(['NY', 'AK', 'WA', 'ME'])
 
cost = {'NY': 4, 'AK': 3, 'WA': 7, 'ME': 11}
 
x = {i: xp.var() for i in S}
 
con1 = xp.Sum(x) <= 1
obj = xp.Sum(cost[i] * x[i] for i in S)
 
p = xp.problem()
 
p.addVariable(x)
p.addConstraint(con1)
p.setObjective(obj)

import xpress as xp
 
S1 = set(['a', 'b', 'c', 'd'])
S2 = set(['y', 'z', 't', 'w'])
 
A = [2,2,4,5]
 
x = {(i,j): xp.var(ub=A[i]) for i in range(len(A)) for j in S}
 
x2 = xp.vars(S1, S2, ub=1)

Hello Pietro,

Thank you for your reply and please have a look of the example I gave in this post which I thought will be helpful to clarify my question.

In my case, I am iterating the optimization and dynamically adding a set of variable X into the objective function as well as adding a set of constraints Constr1. The question is the dimension of variable X and constriant Constr1 is unknown.

In Xpress IVE, the code snippet looks as follows:

declaration
prob:mmproblem
I=1..NI # Set of index for X 
iter:integer
 
X:dynamic array (range,range) of mpvar
sol_X:dynamic array(range,range) of real
 
Constr1:dynamic array(set of mpproblem,range,range) of linctr
Obj:array(set of mpproblem) of linctr
end-declaration
 
[#############]
# Define Optimization
...
[#############]
iter:=0
repeat
    [#############]
    # Solve Optimization
    with prob do 
        minimize( Obj(prob) )
    end-do
    # Get Output...
    if iter>=1 then
        forall(i in I) create( sol_X(i,iter) )
        forall(i in I, j in 1..iter) sol_X(i,j):=getsol( X(i,j) )
    end-if
    [#############]
    iter+=1
    forall(i in I) create( X(i,iter) ) # Create a new set of variable
 
    with prob do
        Obj(prob)+=sum(i in I)( X(i,iter) ) # Add the new set of variable into Obj 
 
         if iter>=2 then
             # Create and define a new set of constraint based on the X which is just solved
             forall(i in I) create( Constr1(prob,i,iter-1) )  
             forall(i in I, j in 1..iter-1) Constr1(prob,i,j):= X(i,j)<=sol_X(i,j)
         end-if
    end-do
until condition

Hello Yijun,

I translated the script you provided to Python, I hope this helps clarify. As you can see in Python you can amend an item to a dictionary at any time as (i,j) in this point is just a key to a dictionary, but we can use it as our index set.

import xpress as xp
 
X = {}
sol_X = {}
 
# Define optimization problem here
 
iter = 0
condition = True
objective = 0  # Initially, the objective function is null
 
while condition:
 
    prob.solve()  # Optimization sense has been specified in the
                  # definition of prob
 
    if iter >= 1:
        for i in I:
            for j in range(1, iter + 1):
                sol_X[i,j] = prob.getSolution(X[i, j])
 
    iter += 1
 
    # Create new variables
    for i in I:
        X[i,iter] = xp.var()
        prob.addVariable(X[i, iter])
 
    # Update objective function
    objective += xp.Sum(X[i,iter] for i in I)
    prob.setObjective(objective)
 
    if iter >= 2:
        prob.addConstraint(X[i, j] <= sol_X[i, j] for i in I for j in range(1, iter))
 
    condition = ...  # update exit condition here

Using dictionaries for both X and sol_X, while ensuring both have the same set of keys (which you can get with X.keys() and sol_X.keys()).

I hope this helps. Regards,

Pietro

Many thanks Pietro, very helpful indeed!

Best wishes,

Yijun