So I'm trying to make sense of a scenario in my class exercise which is to find the max and min value of a function. I have two vectors, w and v, of weights which are to sum to 1. The vectors are w = [0.6, 0.2, 0.2]^T v = [0.8, -0.2, 0.4]^T
These vectors form a linear combination of weights M = Aw + Bv, and A and B must sum to 1. The function we are then optimizing is r = [0.1, 0.2, 0.1] • M
The constraints are as follows: 0 ≤ (0.6A + 0.8B) <= 1 , 0 ≤ (0.2A - 0.2B) <= 1 , 0 ≤ (0.2A + 0.4B) <= 1
The answer we should get are A = B = .5 for the minimum value of r which is 0.1. For the maximum we should get A = 2, B = -1 with r = 0.16. But the values I'm getting for the max are A = 3.5714286, B = -1.4285714, and for the min I'm getting A = B = 0.
Below is the code.
import pulp as p
from pulp import *
problem = LpProblem('Car Factory', LpMaximize)
A = LpVariable('Amound of w', cat=LpContinuous)
B = LpVariable('Amount of v', cat=LpContinuous)
#Objective Function
problem += (0.1)*(0.6*A + 0.8*B) + (0.2)*(0.2*A - 0.2*B) + (0.1)*(0.2*A + 0.4*B) , 'Objective Function'
#Constraints
problem += (0.6*A + 0.8*B) <= 1 , 'A'
problem += (0.6*A + 0.8*B) >= 0 , 'AL'
problem += (0.2*A - 0.2*B) <= 1, 'B'
problem += (0.2*A - 0.2*B) >= 0, 'BL'
problem += (0.2*A + 0.4*B) <= 1, 'C'
problem += (0.2*A + 0.4*B) >= 0, 'CL'
problem.solve()
print("Amount of w: ", A.varValue)
print("Amount of v: ", B.varValue)
print("total: ", value(problem.objective))
I'm sure it has to do with the set up which I'm just not seeing. And also is there a more efficient way to put this together?
I think you are missing a constraint, which would explain your deviation from the expected result. Where is your constraint that:
A + B == 1
Also, you are importing pulp twice, which may cause some confusion in the namespace of your code. Do one or the other, not both.
On expressing the problem more efficiently...? Nahh. You could treat your two column vectors as arrays of length 3 and do the math in your objective a bit differently, but it probably isn't worth it and your variables are just scalars, so I'd write it as you did. Now if the vectors were much larger, or if the variables were vectors, sure, I'd do something else.
pulp doesn't naturally handle vectors (like numpy arrays) to my knowledge. If you are going to be doing a lot of optimization in vector-matrix format and you are comfortable with the linear algebra, you might look at cvxpy which handles them naturally. If you're in a class that uses pulp, it's just fine to learn the basics.