New to the python library PULP and I'm finding the documentation somewhat unhelpful, as it does not include examples using lists of variables. I've tried to create an absolutely minimalist example below to illustrate my confusion.
import pulp
IDENTIFIERS = ['A','B','C','D','E']
PRICES = dict( zip( IDENTIFIERS, [100.0, 99.0, 100.5, 101.5, 200.0 ] ) )
n = len( IDENTIFIERS )
x = pulp.LpVariable.dicts( "x", indexs = IDENTIFIERS, lowBound=0, upBound=1, cat='Integer', indexStart=[] )
prob = pulp.LpProblem( "Minimalist example", pulp.LpMaximize )
prob += pulp.lpSum( [ x[i]*PRICES[i] for i in IDENTIFIERS ] ), " Objective is sum of prices of selected items "
prob += pulp.lpSum( [ x[i] for i in IDENTIFIERS ] )==2, " Constraint is that we choose two items "
prob.solve()
for ident in IDENTIFIERS:
if x[ident]==1:
print ident + " is in the basket "
The output is:
A is in the basket
B is in the basket
C is in the basket
D is in the basket
E is in the basket
The optimizer is not recognizing the constraint that we only add two values.
Zero-one linear programming (or binary integer programming) involves problems in which the variables are restricted to be either 0 or 1. Any bounded integer variable can be expressed as a combination of binary variables.
PuLP is a python library which can be used to solve linear programming problems. Linear Programming is used to solve optimization problems and has uses in various industries such as Manufacturing, Transportation, Food Diets etc.
Integer programming models are often classified as being either mixed-integer programming models, pure-integer programming models, or zero-one integer programming models .
Bases: pulp.pulp.LpProblem. Contains the subproblem generated by converting a fixed constraint. into an elastic constraint.
I'll leave this here in case anyone else is just as silly, but actually the above example works fine. I had merely failed to examine the results correctly. Instead:
def printProb( prob ):
for v in prob.variables():
print v.name, "=", v.varValue
print "Status:", pulp.LpStatus[ prob.status ]
reveals that the solution is correct.
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