pulp solution
After some research, I don't think your objective function is linear. I recreated the problem in the Python pulp library but pulp doesn't like that we're dividing by a float and 'LpAffineExpression'. This answer suggests that linear programming "doesn't understand divisions" but that comment is in context of adding constraints, not the objective function. That comment pointed me to "Mixed Integer Linear Fractional Programming (MILFP)" and on Wikipedia.
Here's how you could do it in pulp if it actually worked (maybe someone can figure out why):
import pulp
data = [(481.79, 5), (412.04, 4), (365.54, 3)] #, (375.88, 3), (379.75, 3), (632.92, 5), (127.89, 1), (835.71, 6), (200.21, 1)]
x = pulp.LpVariable.dicts('x', range(len(data)), lowBound=0, upBound=7, cat=pulp.LpInteger)
numerator = dict((i,tup[0]) for i,tup in enumerate(data))
denom_int = dict((i,tup[1]) for i,tup in enumerate(data))
problem = pulp.LpProblem('Mixed Integer Linear Programming', sense=pulp.LpMinimize)
# objective function (doesn't work)
# TypeError: unsupported operand type(s) for /: 'float' and 'LpAffineExpression'
problem += sum([numerator[i] / (denom_int[i] + x[i]) for i in range(len(data))])
problem.solve()
for v in problem.variables():
print(v.name, "=", v.varValue)
brute solution with scipy.optimize
You can use brute
and ranges of slice
s for each x
in your function. If you have 3 x
s in your function, you'll also have 3 slice
s in your ranges tuple. The key to all of this is to add the step size of 1
to the slice(start, stop,
step
)
so slice(#, #, 1)
.
from scipy.optimize import brute
import itertools
def f(x):
return (481.79/(5+x[0]))+(412.04/(4+x[1]))+(365.54/(3+x[2]))
ranges = (slice(0, 9, 1),) * 3
result = brute(f, ranges, disp=True, finish=None)
print(result)
itertools solution
Or you can use itertools to generate all combinations:
combinations = list(itertools.product(*[[0,1,2,3,4,5,6,7,8]]*3))
values = []
for combination in combinations:
values.append((combination, f(combination)))
best = [c for c,v in values if v == min([v for c,v in values])]
print(best)
Note: this is a scaled-down version of your original function for example purposes.