Question

我制定了MILP来建模PuLP中的多周期多客户调度问题。目的是安排一周内要生产的多批次产品，以便通过最大程度地减少人工成本来生产所有批次的产品。每天都有规律的轮班和加班（由于劳动力成本的变化，使用起来更昂贵）。

我使用了两个都被创建为“整数”变量的变量。但是，我的问题解决方案中仍然会得到十进制和负值。这怎么可能？请在下面查看我的代码。

'''
Multi-Period Scheduling Problem
'''
from pulp import *
import itertools

创建参数和数据：

'''SETS'''

#list of batch_id's waiting for production
batch_ids = ['b1', 'b2', 'b3', 'b4', 'b5', 'b6']

#list of the days available in a week
days = ['mon', 'tue', 'wed', 'thu', 'fri']

#list of shifts in a day
shifts= ['rt', 'ot']

'''PARAMETERS'''

#batch quantity needed
batch_size= {    
            'b1': 45,
            'b2': 60,
            'b3': 120,
            'b4': 80,
            'b5': 230,
            'b6': 40
        }

#regular time and over time values
time_dict= {'rt': 450, 'ot': 300}

#dict to store minutes available per day per shift
mins_per_day= {d: {s: time_dict[s] for s in shifts} for d in days}

#cycle time in minutes per unit per batch
cycle_time = {    
            'b1': 45,
            'b2': 30,
            'b3': 10,
            'b4': 80,
            'b5': 13,
            'b6': 35
        }

'''minimum needed units per batch per day; as of now this is a fixed   qty= 10, but we can create a separate
dict to store different qty per batch if needed and then use it in the    dict comprehension here'''
min_qty= {d:{b: 5 for b in batch_ids} for d in days}

pay_by_shift= {'ot': 0.8625, 'rt': 0.575}

#cost per minute per shift per day:
pay_dict= {d: {s: pay_by_shift[s] for s in shifts} for d in days}

#indexes for batch and day combinations
shifts_ind = [(d,s) for d in days for s in shifts]

#indexes for the make variable:
make_ind= [(d,b) for d in days for b in  batch_ids]

定义变量：

'''VARIABLES'''

#number of units of a batch scheduled for production per day per shift
make= LpVariable.dicts("Make Per Day",(days, batch_ids, shifts),0, None,     cat= 'Integer')

#binary variable to decide wether OT is scheduled on a given day or not
y= LpVariable.dicts("Use Shift",(days, shifts), 0, 1, cat= 'Integer')

'''model formulation'''

#create model object with a minimize objective
prob= LpProblem("FlexLine Problem",LpMinimize)

定义目标函数

#objective function
prob += lpSum([y[d][s] for d in days for s in shifts]) + \
        lpSum([make[d][b][s]* cycle_time[b]* pay_dict[d][s] for d in    days for b in batch_ids for s in shifts]), "Total labor cost per week"

定义约束：

'''CONSTRAINTS'''

#demand constraint
for b in batch_ids:
    prob += lpSum([make[d][b][s] for d in days for s in shifts]) ==     batch_size[b] 

#time constraint
for (d,s) in shifts_ind:
prob += lpSum([make[d][b][s] * cycle_time[b] for b in batch_ids]) <=    mins_per_day[d][s] * y[d][s]

#minimum per day constraint
for (d,b) in make_ind:
    prob += lpSum([make[d][b][s] for s in shifts]) >= min_qty[d][b]

#linking constraint
for (d,s) in shifts_ind:
    prob += lpSum([make[d][b][s] for b in batch_ids]) <= 100000 * y[d][s]

求解模型：

prob.solve()
Out: -1

print ("Status:", LpStatus[prob.status])
Out: Status: Infeasible

打印变量值：

for v in prob.variables():
print (v.name, "=", v.varValue)

Out:Make_Per_Day_fri_b1_ot = 0.0
Make_Per_Day_fri_b1_rt = 5.0
Make_Per_Day_fri_b2_ot = 0.0
Make_Per_Day_fri_b2_rt = -5.5
Make_Per_Day_fri_b3_ot = 0.0
Make_Per_Day_fri_b3_rt = 5.0
Make_Per_Day_fri_b4_ot = 1.5625
Make_Per_Day_fri_b4_rt = 3.4375
Make_Per_Day_fri_b5_ot = 0.0
Make_Per_Day_fri_b5_rt = 5.0    
Make_Per_Day_fri_b6_ot = 5.0
Make_Per_Day_fri_b6_rt = 0.0
Make_Per_Day_mon_b1_ot = 0.0
Make_Per_Day_mon_b1_rt = 5.0
Make_Per_Day_mon_b2_ot = 0.0
Make_Per_Day_mon_b2_rt = -5.5
Make_Per_Day_mon_b3_ot = 0.0
Make_Per_Day_mon_b3_rt = 5.0
Make_Per_Day_mon_b4_ot = 2.9375
Make_Per_Day_mon_b4_rt = 2.0625
Make_Per_Day_mon_b5_ot = 5.0
Make_Per_Day_mon_b5_rt = 0.0
Make_Per_Day_mon_b6_ot = 0.0
Make_Per_Day_mon_b6_rt = 5.0
Make_Per_Day_thu_b1_ot = 25.0
Make_Per_Day_thu_b1_rt = 0.0
Make_Per_Day_thu_b2_ot = 61.0
Make_Per_Day_thu_b2_rt = 0.0
Make_Per_Day_thu_b3_ot = 131.5
Make_Per_Day_thu_b3_rt = 0.0
Make_Per_Day_thu_b4_ot = 60.0
Make_Per_Day_thu_b4_rt = 0.0
Make_Per_Day_thu_b5_ot = 210.0
Make_Per_Day_thu_b5_rt = 0.0
Make_Per_Day_thu_b6_ot = 16.142857
Make_Per_Day_thu_b6_rt = 12.857143
Make_Per_Day_tue_b1_ot = 5.0
Make_Per_Day_tue_b1_rt = 0.0
Make_Per_Day_tue_b2_ot = 0.83333333
Make_Per_Day_tue_b2_rt = 4.1666667
Make_Per_Day_tue_b3_ot = 5.0
Make_Per_Day_tue_b3_rt = 0.0
Make_Per_Day_tue_b4_ot = 0.0
Make_Per_Day_tue_b4_rt = 5.0
Make_Per_Day_tue_b5_ot = 0.0
Make_Per_Day_tue_b5_rt = 5.0
Make_Per_Day_tue_b6_ot = 0.0
Make_Per_Day_tue_b6_rt = -4.0
Make_Per_Day_wed_b1_ot = 5.0
Make_Per_Day_wed_b1_rt = 0.0
Make_Per_Day_wed_b2_ot = 5.0
Make_Per_Day_wed_b2_rt = 0.0
Make_Per_Day_wed_b3_ot = -26.5
Make_Per_Day_wed_b3_rt = 0.0
Make_Per_Day_wed_b4_ot = 1.5625
Make_Per_Day_wed_b4_rt = 3.4375
Make_Per_Day_wed_b5_ot = 5.0
Make_Per_Day_wed_b5_rt = 0.0
Make_Per_Day_wed_b6_ot = 0.0
Make_Per_Day_wed_b6_rt = 5.0
Use_Shift_fri_ot = 1.0
Use_Shift_fri_rt = 1.0
Use_Shift_mon_ot = 1.0
Use_Shift_mon_rt = 1.0
Use_Shift_thu_ot = 41.216667
Use_Shift_thu_rt = 1.0
Use_Shift_tue_ot = 1.0
Use_Shift_tue_rt = 1.0
Use_Shift_wed_ot = 1.0
Use_Shift_wed_rt = 1.0

重申一下，我看到负值，小数以及二进制变量显示的值都超过1。

Answer 1

您的解决方案状态不可行或不理想（似乎不可行）。因此，变量值没有任何意义。您可以减小批量大小，并尝试找到可行的解决方案和工作模型。同样使用“ writeLP”，您可以用数字检查模型。

Python PuLP中的混合整数线性程序不可行解决方案

1 个答案: