您可以使用线性编程程序包，例如 PulP . (请注意，这还要求您安装LP库，例如 GLPK ).

You can use a linear programming package like PuLP. (note this also requires you to install an LP library like GLPK).

在这里，您将使用它来解决您给出的示例:

Here's how you would use it to solve the example you gave:

import pulp

prob = pulp.LpProblem("example", pulp.LpMinimize)

# Variable represent number of times device i is used
n1 = pulp.LpVariable("n1", 0, 5, cat="Integer")
n2 = pulp.LpVariable("n2", 0, 5, cat="Integer")
n3 = pulp.LpVariable("n3", 0, 5, cat="Integer")
n4 = pulp.LpVariable("n4", 0, 5, cat="Integer")
n5 = pulp.LpVariable("n5", 0, 5, cat="Integer")

# Device params
Device1=[8,8,4,4,200]
Device2=[16,0,16,0,250]
Device3=[8,0,4,4,300]
Device4=[16,8,4,4,300]
Device5=[8,8,2,2,150]

# The objective function that we want to minimize: the total cost
prob += n1 * Device1[-1] + n2 * Device2[-1] + n3 * Device3[-1] + n4 * Device4[-1] + n5 * Device5[-1]

# Constraint that we use no more than 5 devices
prob += n1 + n2 + n3 + n4 + n5 <= 5

Target = [24, 12, 16, 8]

# Constraint that the total I/O for all devices exceeds the target
for i in range(4):
    prob += n1 * Device1[i] + n2 * Device2[i] + n3 * Device3[i] + n4 * Device4[i] + n5 * Device5[i] >= Target[i]

# Actually solve the problem, this calls GLPK so you need it installed
pulp.GLPK().solve(prob)

# Print out the results
for v in prob.variables():
    print v.name, "=", v.varValue

运行它的速度非常快，我得到n1 = 2和n2 = 1，其他均为0.

Running this is extremely fast, and I get that n1 = 2 and n2 = 1 and the others are 0.