scipy最小化SLSQP-'LSQ子问题中的奇异矩阵C' [英] scipy minimize SLSQP - 'Singular matrix C in LSQ subproblem'

问题描述

我正在尝试使用SciPy解决一个非常基本的优化问题.这个问题是受约束的，并且范围是可变的，我很确定它是线性的.

I'm trying to solve a pretty basic optimization problem using SciPy. The problem is constrained and with variable bounds and I'm pretty sure it's linear.

当我运行以下代码时，执行失败，并显示错误消息"LSQ子问题中的奇异矩阵C".有谁知道可能是什么问题?预先感谢.

When I run the following code the execution fails with the error message 'Singular matrix C in LSQ subproblem'. Does anyone know what the problem might be? Thanks in advance.

我将在此处添加简短的代码说明. 我在代码的开头定义了一个需求"向量.该向量描述了在一段时间内建立索引的特定产品的需求.我想弄清楚的是如何下订单，以便在某些约束下满足这一需求.这些约束是；

I'll add a short description of what the code should do here. I define a 'demand' vector at the beginning of the code. This vector describes the demand of a certain product indexed over some period of time. What I want to figure out is how to place a set of orders so as to fill this demand under some constraints. These constraints are;

• 如果在特定时间点有需求(需求指数)，我们必须有库存商品
• 在下订单后的4个时间单位"之前，我们无法下订单
• 我们不能以最近4个时间单位下订单

这是我的代码；

from scipy.optimize import minimize
import numpy as np

demand = np.array([5, 10, 10, 7, 3, 7, 1, 0, 0, 0, 8])
orders = np.array([0.] * len(demand))

def objective(orders):
  return np.sum(orders)

def items_in_stock(orders):
  stock = 0
  for i in range(len(orders)):
    stock += orders[i]
    stock -= demand[i]
    if stock < 0.:
      return -1.
  return 0.

def four_weeks_order_distance(orders):
  for i in range(len(orders)):
    if orders[i] != 0.:
      num_orders = (orders[i+1:i+5] != 0.).any()
      if num_orders:
        return -1.
  return 0.

def four_weeks_from_end(orders):
  if orders[-4:].any():
    return -1.
  else:
    return 0.

con1 = {'type': 'eq', 'fun': items_in_stock}
con2 = {'type': 'eq', 'fun': four_weeks_order_distance}
con3 = {'type': 'eq', 'fun': four_weeks_from_end}
cons = [con1, con2, con3]

b = [(0, 100)]
bnds = b * len(orders)

x0 = orders
x0[0] = 10.

minimize(objective, x0, method='SLSQP', bounds=bnds, constraints=cons)

推荐答案

尽管我不是运筹学研究员，但我相信这是因为您实施的约束不是连续的.我做了很少的更改，以使约束现在本质上是连续的.

Though I am not an Operational Researcher, I believe it is because of the fact that the constraints you implemented are not continuous. I made little changes so that the constraints are now continuous in nature.

from scipy.optimize import minimize
import numpy as np

demand = np.array([5, 10, 10, 7, 3, 7, 1, 0, 0, 0, 8])
orders = np.array([0.] * len(demand))

def objective(orders):
    return np.sum(orders)


def items_in_stock(orders):
    """In-equality Constraint: Idea is to keep the balance of stock and demand.
    Cumulated stock should be greater than demand. Also, demand should never cross the stock.
    """
    stock = 0
    stock_penalty = 0
    for i in range(len(orders)):
        stock += orders[i]
        stock -= demand[i]
        if stock < 0:
            stock_penalty -= abs(stock)
    return stock_penalty


def four_weeks_order_distance(orders):
    """Equality Constraint: An order can't be placed until four weeks after any other order.
    """
    violation_count = 0
    for i in range(len(orders) - 6):
        if orders[i] != 0.:
            num_orders = orders[i + 1: i + 5].sum()
            violation_count -= num_orders
    return violation_count


def four_weeks_from_end(orders):
    """Equality Constraint: No orders in the last 4 weeks
    """
    return orders[-4:].sum()


con1 = {'type': 'ineq', 'fun': items_in_stock} # Forces value to be greater than zero. 
con2 = {'type': 'eq', 'fun': four_weeks_order_distance} # Forces value to be zero. 
con3 = {'type': 'eq', 'fun': four_weeks_from_end} # Forces value to be zero. 
cons = [con1, con2, con3]

b = [(0, 100)]
bnds = b * len(orders)

x0 = orders
x0[0] = 10.

res = minimize(objective, x0, method='SLSQP', bounds=bnds, constraints=cons,
               options={'eps': 1})

结果

  status: 0
 success: True
    njev: 22
    nfev: 370
     fun: 51.000002688311334
       x: array([  5.10000027e+01,   1.81989405e-15,  -6.66999371e-16,
         1.70908182e-18,   2.03187432e-16,   1.19349893e-16,
         1.25059614e-16,   4.55582386e-17,   6.60988392e-18,
         3.37907550e-17,  -5.72760251e-18])
 message: 'Optimization terminated successfully.'
     jac: array([ 1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  0.])
     nit: 23

[ round(l, 2) for l in res.x ]
[51.0, 0.0, -0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, -0.0]

因此，该解决方案建议在第一周下达所有订单.

So, the solution suggests to make all the orders in the first week.

• 它避免了缺货的情况
• 单一购买(订单)会在下订单后的四周内遵守无订单.
• 最近4周没有购买

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