Let's say I have the following (always binary) options:

import numpy as np
a=np.array([1, 1, 0, 0, 1, 1, 1])
b=np.array([1, 1, 0, 0, 1, 0, 1])
c=np.array([1, 0, 0, 1, 0, 0, 0])
d=np.array([1, 0, 1, 1, 0, 0, 0])

And I want to find the optimal combination of the above that get's me to at least, with minimal above:

req = np.array([50,50,20,20,100,40,10])

For example:

final = X1*a   X2*b   X3*c   X4*d

1. Does this map to a known operational research problem? Or does it fall under mathematical programming?
2. Is this NP-hard, or exactly solveable in a reasonable amount of time (I've assumed it's combinatorally impossible to solve exactly)
3. Are there know solutions to this?

Note: The actual length of arrays are longer - think ~50, and the number of options are ~20 My current research has led me to some combination of the assignment problem and knapsack, but not too sure.

CodePudding user response：

It's a covering problem, easily solvable using an integer program solver (I used OR-Tools below). If the X variables can be fractional, substitute NumVar for IntVar. If the X variables are 0--1, substitute BoolVar.

import numpy as np

a = np.array([1, 1, 0, 0, 1, 1, 1])
b = np.array([1, 1, 0, 0, 1, 0, 1])
c = np.array([1, 0, 0, 1, 0, 0, 0])
d = np.array([1, 0, 1, 1, 0, 0, 0])
opt = [a, b, c, d]
req = np.array([50, 50, 20, 20, 100, 40, 10])


from ortools.linear_solver import pywraplp

solver = pywraplp.Solver.CreateSolver("SCIP")
x = [solver.IntVar(0, solver.infinity(), "x{}".format(i)) for i in range(len(opt))]
extra = [solver.NumVar(0, solver.infinity(), "y{}".format(j)) for j in range(len(req))]
for j, (req_j, extra_j) in enumerate(zip(req, extra)):
    solver.Add(extra_j == sum(opt_i[j] * x_i for (opt_i, x_i) in zip(opt, x)) - req_j)
solver.Minimize(sum(extra))
status = solver.Solve()
if status == pywraplp.Solver.OPTIMAL:
    print("Solution:")
    print("Objective value =", solver.Objective().Value())
    for i, x_i in enumerate(x):
        print("x{} = {}".format(i, x[i].solution_value()))
else:
    print("The problem does not have an optimal solution.")

Output:

Solution:
Objective value = 210.0
x0 = 40.0
x1 = 60.0
x2 = -0.0
x3 = 20.0