# Construct a problem from scratch with variables of various # types. Adds indicator constraints and shows how to retrieve # such data once it has been added to the problem using the # API functions. # # (C) Fair Isaac Corp., 1983-2024 from __future__ import print_function import xpress as xp N = 40 S = range(N) m = xp.problem("test restriction") m.controls.miprelstop = 0 # # All variables used in this example # v1 = m.addVariable(lb=0, ub=10, threshold=5, vartype=xp.continuous) v2 = m.addVariable(lb=1, ub=7, threshold=5, vartype=xp.continuous) v3 = m.addVariable(lb=5, ub=10, threshold=7, vartype=xp.semicontinuous) v4 = m.addVariable(lb=1, ub=7, threshold=3, vartype=xp.semiinteger) vb = m.addVariable(vartype=xp.integer, lb=0, ub=1) y = [m.addVariable(name="y{0}".format(i), lb=0, ub=2*N) for i in S] cc = xp.constraint(body=v1 - v2, lb=2, ub=15) cc0 = xp.constraint(body=v1 + v2, lb=2, ub=15) # Adds both y, a vector (list) of variables, and v1 and v2, two scalar # variables. m.addConstraint(cc) # Indices of variables can be retrieved both using their name and # their Python object. print("index of y[0] from name: ", m.getIndexFromName(xp.names_column, "y0")) print("index of y[0]: ", m.getIndex(y[0])) # Indicator constraints consist of a tuple with a condition on a # binary variable and a constraint). ind1 = (vb == 1, v1 + v2 >= 6) ind2 = (vb == 1, v1 + v3 >= 7) # Adds the first indicator constraint m.addIndicator(ind1) # Adds another indicator constraint and the second one defined above m.addIndicator((vb == 1, v1 + v3 <= 10), ind2) print("get index: var v1 -->", m.getIndex(v1), "; con cc -->", m.getIndex(cc)) ii_inds = [] ii_comps = [] m.getindicators(ii_inds, ii_comps, 1, 3) print("getind: ", ii_inds, ii_comps) # objective overwritten at each setObjective() m.setObjective(xp.Sum([i*y[i] for i in S])) m.optimize() # Retrieve a solution: first declare an empty string, then call the # getmipsol() function to fill it up. mipsol = [] m.getmipsol(mipsol) s1 = m.getSolution(v1, v2, y[10:30]) # get a subset of the solutions s2 = m.getSolution(S) # can get it with indices as well print("v1: ", m.getSolution(v1), ", v2: ", m.getSolution(v2), "; sol vector: ", m.getSolution(), "; obj: ", m.getObjVal(), sep="") # default separator between strings is " " # Adds yet another constraint to the problem and saves it, then # removes an SOS and saves another version m.addConstraint((1,25 * v1 - 2,5*v2 + 4,3) * (3,1 * v2 - 2 * v1 - 5,2) + 72,5 * v1**2 + 73 * v2**2 <= 1950) m.write("restriction", "lp") m.optimize()