Question

尝试使用scipy.optimize.linprog优化成本函数，其中成本系数是变量的函数;例如

成本= c1 * x1 + c2 * x2（x1，x2是变量）

例如

如果x1 = 1，则c1 = 0.5

如果x1 = 2，则c1 = 1.25

等

感谢您的帮助

*只是为了澄清*

我们正在寻找变量的最低成本;喜; i = 1,2,3，... xi是正整数。

然而，每个xi的成本系数是xi的值的函数。成本是x1 * f1（x1）+ x2 * f2（x2）+ ... + c0

fi - 是一个＆＃34;率＆＃34;表;例如 - f1（0）= 0; f1（1）= 2.00; f1（2）= 3.00等。

xi受到约束，并且它们不能为负数且不能超过qi =＆gt;

0＆lt; = xi＆lt; = qi

为xi

的每个可能值计算fi（）值

我希望它澄清了模型。

Answer 1

这是一些原型代码，向您展示如何解决您的问题（关于配方和性能;前者在代码中可见）。

该实现使用cvxpy进行建模（仅凸面编程），并基于混合整数方法。

代码

    import numpy as np
    from cvxpy import *

    """
    x0 == 0 -> f(x) = 0
    x0 == 1 -> f(x) = 1
    ...
    x1 == 0 -> f(x) = 1
    x1 == 1 -> f(x) = 4
    ...
    """
    rate_table = np.array([[0, 1, 3, 5], [1, 4, 5, 6], [1.3, 1.7, 2.25, 3.0]])
    bounds_x = (0, 3)  # inclusive; bounds are needed for linearization!

    # Vars
    # ----
    n_vars = len(rate_table)
    n_values_per_var = [len(x) for x in rate_table]

    I = Bool(n_vars, n_values_per_var[0])  # simplified assumption: rate-table sizes equal
    X = Int(n_vars)
    X_ = Variable(n_vars, n_values_per_var[0])  # X_ = mul_elemwise(I*X) broadcasted

    # Constraints
    # -----------
    constraints = []

    # X is bounded
    constraints.append(X >= bounds_x[0])
    constraints.append(X <= bounds_x[1])

    # only one value in rate-table active (often formulated with SOS-type-1 constraints)
    for i in range(n_vars):
        constraints.append(sum_entries(I[i, :]) <= 1)

    # linearization of product of BIN * INT (INT needs to be bounded!)
    # based on Erwin's answer here:
    # https://www.or-exchange.org/questions/10775/how-to-linearize-product-of-binary-integer-and-integer-variables
    for i in range(n_values_per_var[0]):
        constraints.append(bounds_x[0] * I[:, i] <= X_[:, i])
        constraints.append(X_[:, i] <= bounds_x[1] * I[:, i])
        constraints.append(X - bounds_x[1]*(1-I[:, i]) <= X_[:, i])
        constraints.append(X_[:, i] <= X - bounds_x[0]*(1-I[:, i]))

    # Fix chosings -> if table-entry x used -> integer needs to be x
    # assumptions:
    # - table defined for each int
    help_vec = np.arange(n_values_per_var[0])
    constraints.append(I * help_vec == X)

    # ONLY FOR DEBUGGING -> make simple max each X solution infeasible
    constraints.append(sum_entries(mul_elemwise([1, 3, 2], square(X))) <= 15)

    # Objective
    # ---------
    objective = Maximize(sum_entries(mul_elemwise(rate_table, X_)))

    # Problem & Solve
    # ---------------
    problem = Problem(objective, constraints)
    problem.solve()  # choose other solver if needed, e.g. commercial ones like Gurobi, Cplex
    print('Max-objective: ', problem.value)
    print('X:\n' + str(X.value))

输出

('Max-objective: ', 20.70000000000001)
X:
[[ 3.]
 [ 1.]
 [ 1.]]

观

转换目标max: x0*f(x0) + x1*f(x1) + ...
- into：x0*f(x0==0) + x0*f(x0==1) + ... + x1*f(x1==0) + x1*f(x1==1)+ ...
引入二元变量来制定：
- f(x0==0) as I[0,0]*table[0,0]
- f(x1==2) as I[1,2]*table[0,2]
添加约束以限制上述I仅为每个变量x_i设置一个非零条目（只有一个扩展的目标组件将处于活动状态）
线性化产品x0*f(x0==0) == x0*I[0,0]*table(0,0)（整数*二进制*常量）
修复表查找：使用索引为x（x0）的表条目应该导致x0 == x
- 假设表格中没有空白，可以将其表述为I * help_vec == X) help_vec == vector(lower_bound, ..., upper_bound)

cvxpy 会自动（通过构造）证明我们的公式凸，这是大多数求解器所需要的（通常不容易识别）。

只是为了好玩：更大问题和商业解决方案

生成的输入：

def gen_random_growing_table(size):
    return np.cumsum(np.random.randint(1, 10, size))
SIZE = 100
VARS = 100
rate_table = np.array([gen_random_growing_table(SIZE) for v in range(VARS)])
bounds_x = (0, SIZE-1)  # inclusive; bounds are needed for linearization!
...
...
constraints.append(sum_entries(square(X)) <= 150)

输出：

Explored 19484 nodes (182729 simplex iterations) in 129.83 seconds
Thread count was 4 (of 4 available processors)

Optimal solution found (tolerance 1.00e-04)
Warning: max constraint violation (1.5231e-05) exceeds tolerance
Best objective -1.594000000000e+03, best bound -1.594000000000e+03, gap 0.0%
('Max-objective: ', 1594.0000000000005)

使用scipy.optimize.linprog进行线性编程 - 变量系数

1 个答案:

代码

输出

观

只是为了好玩：更大问题和商业解决方案