scipy minimize SLSQP - 'Singular matrix C in LSQ subproblem'

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南笙 2021-01-07 04:02

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.

When I

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  • 2021-01-07 04:35

    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})
    
    

    Results

      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]
    

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

    • It avoids the out of stock situation
    • Single purchase(order) respects the no order in the next four week after an order.
    • No last 4 week purchase
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