CODEWITHSUNDEEP
linear-programmingpulp
CSSguy
CSSguy

Reputation: 5

PuLP yields not-optimal solution for a simple constraint problem

PuLP solver doesnt yield the optimal result, while it was possible. Is it ill-defined problem or is there a problem in the library or constraints?

Normally it was giving optimal for most of the constraints possibility, but its not for this specific case.

from pulp import *
prob = LpProblem("minimization", LpMinimize)

#introducing vars
a=LpVariable("a", 0, None)
b=LpVariable("b", 0, None)
c=LpVariable("c", 0, None)
d=LpVariable("d", 0, None)
e=LpVariable("e", 0, None)

#introducing main obj and constraints
#basically addition of all vars should be minimized to 1.
prob+=a+b+c+d+e>=1,"main objective"
prob+=a<=0.3,"a const"
prob+=b<=0.3,"b const"
prob+=c<=0.3,"c const"
prob+=d<=0.3,"d const"   #can change to 0
prob+=e==0.1,"e const"

"""
for this const. 
the res is:
a=0.3
b=0.3
c=0.3
d=0.3
e=0.1
a+b+c+d+e=1.3

if you change prob+=d<=0.3 to prob+=d<=0
then:
a=0.3
b=0.3
c=0.3
d=0.0
e=0.1
a+b+c+d+e=1

"""


#solves the problem
status = prob.solve()
#checks if optimal or infeasible
print("status:",LpStatus[prob.status])

#shows all the vars
for variable in prob.variables():
    print("{} = {}".format(variable.name, variable.varValue))

for this const.

the res is:

a=0.3

b=0.3

c=0.3

d=0.3

e=0.1

a+b+c+d+e=1.3

It should be 1, as it is LpMinimize.

if you change prob+=d<=0.3 to prob+=d<=0

then:

a=0.3

b=0.3

c=0.3

d=0.0

e=0.1

a+b+c+d+e=1

Upvotes: 0

Views: 839

Answers (1)

Erwin Kalvelagen
Erwin Kalvelagen

Reputation: 16724

This is a constraint and not an objective:

prob+=a+b+c+d+e>=1,"main objective"

Essentially you don't have an objective so the solver just looks for a feasible solution.

Try

prob+=a+b+c+d+e,"main objective"
prob+=a+b+c+d+e>=1,"constraint"

Now you are both minimizing a+b+c+d+e as well as having a constraint on this.

Conclusion: I think PuLP is correct here.

Upvotes: 3

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