Error during integer programming with pulp: 'LpVariable' has no len()

Question

Trying to solve the Blending and Mixing problem to exercise optimization with python (using gurobi and pulp).

Sadly I ran into the following error message:

python Blending_problem.py 
Traceback (most recent call last):
  File "Blending_problem.py", line 24, in <module>
    LP += calcium_content   == (n_Limestone*0.38    +n_Corn*0.001   +n_Soy*0.002)   /Total_weight   #kg calcium
  File "/home/bruno/.local/lib/python2.7/site-packages/pulp/pulp.py", line 800, in __div__
    if len(other):
TypeError: object of type 'LpVariable' has no len()

What could be wrong? This is the code:

import pulp
from gurobipy import *
LP = pulp.LpProblem('LP',pulp.LpMinimize)  

Cost=pulp.LpVariable("Cost",lowBound=0,cat=pulp.LpContinuous)

Total_weight=pulp.LpVariable("Total_weight",cat=pulp.LpInteger)

#relative amounts of nutrients
calcium_content=pulp.LpVariable("calcium_content",cat=pulp.LpContinuous,lowBound=0.008,upBound=0.012)
protein_content=pulp.LpVariable("protein_content",cat=pulp.LpContinuous,lowBound=0.22)
fiber_content=pulp.LpVariable("fiber_content",cat=pulp.LpContinuous,upBound=0.05)

#ingredient units
n_Limestone=pulp.LpVariable("n_Limestone",cat=pulp.LpInteger,lowBound=0)
n_Corn=pulp.LpVariable("n_Corn",cat=pulp.LpInteger,lowBound=0)
n_Soy=pulp.LpVariable("n_Soy",cat=pulp.LpInteger,lowBound=0)

#obj
LP += n_Limestone*10 +n_Corn*30.5 +n_Soy*90

LP += Total_weight == n_Limestone+n_Corn+n_Soy

LP += calcium_content   == (n_Limestone*0.38    +n_Corn*0.001   +n_Soy*0.002)   /Total_weight   #kg calcium
LP += protein_content   == (n_Limestone*0       +n_Corn*0.09    +n_Soy*0.5  )   /Total_weight   #kg protein
LP += fiber_content     == (n_Limestone*0       +n_Corn*0.02    +n_Soy*0.08 )   /Total_weight   #kg calcium


status = LP.solve(pulp.solvers.GUROBI(mip=True, msg=True, timeLimit=None,epgap=None))

print( 'LP status: ' + pulp.LpStatus[status] + '')

print(str(n_Limestone.value())+"kg Lime, "+str(n_Corn.value())+"kg Corn, "+str(n_Soy.value())+"kg Soy")

cheesus · Accepted Answer · 2019-07-28 22:16:20Z

As stated by kabdulla, the problem is not solvable like it was posted since division on Lp variables is not possible.

I was able to get the correct results by removing the Integer constraints of the n_.. values, removing the total_weight variable and including the constraint 1 = n_Limestone+n_Corn+n_Soy. The resulting script looks as follows:

import pulp
from gurobipy import *
LP = pulp.LpProblem('LP',pulp.LpMinimize)  

Cost=pulp.LpVariable("Cost",lowBound=0,cat=pulp.LpContinuous)


#relative amounts of nutrients
calcium_content=pulp.LpVariable("calcium_content",cat=pulp.LpContinuous,lowBound=0.008,upBound=0.012)
protein_content=pulp.LpVariable("protein_content",cat=pulp.LpContinuous,lowBound=0.22)
fiber_content=pulp.LpVariable("fiber_content",cat=pulp.LpContinuous,upBound=0.05)

#ingredient units
n_Limestone=pulp.LpVariable("n_Limestone",cat=pulp.LpContinuous,lowBound=0)
n_Corn=pulp.LpVariable("n_Corn",cat=pulp.LpContinuous,lowBound=0)
n_Soy=pulp.LpVariable("n_Soy",cat=pulp.LpContinuous,lowBound=0)
    #obj
LP += n_Limestone*10 +n_Corn*30.5 +n_Soy*90

LP += calcium_content   == (n_Limestone*0.38    +n_Corn*0.001   +n_Soy*0.002)       #kg calcium
LP += protein_content   == (n_Limestone*0       +n_Corn*0.09    +n_Soy*0.5  )       #kg protein
LP += fiber_content     == (n_Limestone*0       +n_Corn*0.02    +n_Soy*0.08 )       #kg calcium
LP += n_Limestone + n_Corn + n_Soy              == 1 


status = LP.solve(pulp.solvers.GUROBI(mip=True, msg=True, timeLimit=None,epgap=None))
print( 'LP status: ' + pulp.LpStatus[status] + '')

print(str(n_Limestone.value())+"kg Lime, "+str(n_Corn.value())+"kg Corn, "+str(n_Soy.value())+"kg Soy")

kabdulla · Accepted Answer · 2019-07-28 20:05:23Z

0

That's not a particularly helpful error message, but at least part of your problem is that the problem as implemented is not linear - you are dividing one problem variable by another in the 'content' constraints. You'll need to reformulate to avoid this division/multiplication of two variables.

answered Jul 28, 2019 at 20:05

kabdulla

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1 Comment

cheesus Over a year ago

Yes its not helpful at all. I appreciate your input, it got me on the track to solving the problem :)

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