带有Python纸浆性能问题求解器的MILP模型速度非常慢_Python_Linear Programming_Pulp_Mixed Integer Programming

带有Python纸浆性能问题求解器的MILP模型速度非常慢

python

带有Python纸浆性能问题求解器的MILP模型速度非常慢,python,linear-programming,pulp,mixed-integer-programming,Python,Linear Programming,Pulp,Mixed Integer Programming,我最近开始学习Python中的线性规划，我用纸浆创建了我的第一个优化程序我正在处理一个生产过程的调度问题。目标是通过为一天中的每一小时创建理想的生产计划，并为一年中的所有天创建此计划，从而最大限度地降低每天的生产成本我遇到的问题是，算法的执行需要很长时间（几个小时），而且经常会卡住。而且，我觉得随着时间的推移，它会变得越来越慢我希望得到关于如何提高代码性能的建议我解决问题的方法： def set_supplyoptions(): cols = ['option', 'min_capaci

我最近开始学习Python中的线性规划，我用纸浆创建了我的第一个优化程序

我正在处理一个生产过程的调度问题。目标是通过为一天中的每一小时创建理想的生产计划，并为一年中的所有天创建此计划，从而最大限度地降低每天的生产成本

我遇到的问题是，算法的执行需要很长时间（几个小时），而且经常会卡住。而且，我觉得随着时间的推移，它会变得越来越慢

我希望得到关于如何提高代码性能的建议

我解决问题的方法：

def set_supplyoptions():
cols = ['option', 'min_capacity', 'max_capacity']
options_list = [{'option':'o', 'min_capacity': 0, 'max_capacity':146},
                {'option':'l30', 'min_capacity': 0, 'max_capacity':30},
                {'option':'l50', 'min_capacity': 31, 'max_capacity':50},
                {'option':'l90', 'min_capacity': 51, 'max_capacity':90},
                {'option':'l150', 'min_capacity': 91, 'max_capacity':150},
                {'option':'l230', 'min_capacity': 151, 'max_capacity':230},
                {'option':'a15', 'min_capacity': 0, 'max_capacity':15},
                {'option':'a30', 'min_capacity': 0, 'max_capacity':30},
                {'option':'a45', 'min_capacity': 0, 'max_capacity':45},
                {'option':'a60_below53', 'min_capacity': 0, 'max_capacity':52},
                {'option':'a_flow_below53', 'min_capacity': 0, 'max_capacity':52},
                {'option':'a_flow_above53', 'min_capacity': 0, 'max_capacity':8},
                {'option':'l_total', 'min_capacity': 0, 'max_capacity':230},
                {'option':'a_total', 'min_capacity': 0, 'max_capacity':60}]

options = pd.DataFrame(options_list, columns=cols)  
options = options.set_index('option')

定义我的最小化问题，为一整天创建一个最优的时间表
循环我的代码以在一年中的所有365天内运行此优化
在每个循环结束时，我获取当天的优化生产计划并将其附加到一个数据框中，因此最后我有一个数据框，其中包含一年中所有日期的生产计划

我处理的是3种生产资产（“a”、“l”和“o”），每种资产都有几种生产模式。我将每个资产模式组合定义为一个“选项”，总共有14个选项。每个选项可以每小时变化一次，并且有一个整数值（生产量）和一个二进制值（开/关），导致大约14 x 2 x 24=672个变量。该问题包含大约1250个约束

我的代码有200多行，所以我有点犹豫是否要在这里分享所有内容，但我将在下面分享最重要的部分

定义供应选项：

def set_supplyoptions():
cols = ['option', 'min_capacity', 'max_capacity']
options_list = [{'option':'o', 'min_capacity': 0, 'max_capacity':146},
                {'option':'l30', 'min_capacity': 0, 'max_capacity':30},
                {'option':'l50', 'min_capacity': 31, 'max_capacity':50},
                {'option':'l90', 'min_capacity': 51, 'max_capacity':90},
                {'option':'l150', 'min_capacity': 91, 'max_capacity':150},
                {'option':'l230', 'min_capacity': 151, 'max_capacity':230},
                {'option':'a15', 'min_capacity': 0, 'max_capacity':15},
                {'option':'a30', 'min_capacity': 0, 'max_capacity':30},
                {'option':'a45', 'min_capacity': 0, 'max_capacity':45},
                {'option':'a60_below53', 'min_capacity': 0, 'max_capacity':52},
                {'option':'a_flow_below53', 'min_capacity': 0, 'max_capacity':52},
                {'option':'a_flow_above53', 'min_capacity': 0, 'max_capacity':8},
                {'option':'l_total', 'min_capacity': 0, 'max_capacity':230},
                {'option':'a_total', 'min_capacity': 0, 'max_capacity':60}]

options = pd.DataFrame(options_list, columns=cols)  
options = options.set_index('option')

变量：

# production per supply option per hour
production = pulp.LpVariable.dicts("production",
                                     ((hour, option) for hour in hours for option in options.index),
                                     lowBound=0,
                                     cat='Integer')

# status of supply options per hour (in use or not in use)
options_status = pulp.LpVariable.dicts("options_status",
                                     ((hour, option) for hour in hours for option in options.index),
                                     cat='Binary')

# use status of A tranches on day (used or not used)
a_status_15 = pulp.LpVariable('a_stat_15', cat='Binary')
a_status_30 = pulp.LpVariable('a_stat_30', cat='Binary')
a_status_45 = pulp.LpVariable('a_stat_45', cat='Binary')
a_status_60 = pulp.LpVariable('a_stat_60', cat='Binary')

opt_model = pulp.LpProblem("Optimizer", pulp.LpMinimize)

opt_model += pulp.lpSum(
    #O variable
    [0.2*demand.loc[hour, 'price']*production[hour, 'o'] + 
    #O fixed
    3*demand.loc[hour, 'price'] +
    #L30
    ( 12 )*options_status[hour, 'l30']*demand.loc[hour, 'price'] + 
    #L50
    ( 2*options_status[hour, 'l50']+0.13*production[hour, 'l50'] )*demand.loc[hour, 'price'] +
    #L90
    ( 15*options_status[hour, 'l90']+0.13*production[hour, 'l90'] )*demand.loc[hour, 'price'] +
    #L150
    ( 8*options_status[hour, 'l150']+0.2*production[hour, 'l150'] )*demand.loc[hour, 'price'] +
    #L230
    ( -3*options_status[hour, 'l230']+0.3*production[hour, 'l230'] )*demand.loc[hour, 'price'] +
    #L fixed
    ( 7*(1-options_status[hour, 'ltotal'])*demand.loc[hour, 'price'] ) +
    #A base unit price
    (10*production[hour, 'a_flow_below53']+8.88*production[hour, 'a_flow_above53'])*(c/20) for hour in hours]
    #A daily fee
    + (a_status_15 * 1000 + a_status_30 * 2000 + a_status_45 * 3 + a_status_60 * 4000)*(c/20))

# Sum of production == Demand
for hour in hours:
    opt_model += production[(hour, 'o')] + production[(hour, 'l_total')] + production[(hour, 'a_total')] == int(demand.loc[hour, 'demand'])

# Production <=  Max capacity AND => Min capacity
for hour in hours:
    for option in options.index:
        opt_model += production[(hour, option)] >= options_status[(hour, option)]
        opt_model += production[(hour, option)] >= options.loc[option, 'min_capacity'] * options_status[(hour, option)]
        opt_model += production[(hour, option)] <= options.loc[option, 'max_capacity'] * options_status[(hour, option)]

# Constraints L
for hour in hours:
    opt_model += production[(hour, 'l30')] + production[(hour, 'l0')] + production[(hour, 'l90')] + production[(hour, 'l150')] + production[(hour, 'l230')] == production[(hour, 'l_total')]
    opt_model += options_status[(hour, 'l30')] + options_status[(hour, 'l50')] + options_status[(hour, 'l90')] + options_status[(hour, 'l150')] + options_status[(hour, 'l230')] <= 1

# Constraints A
M = 999
for hour in hours:
    # total flow = sum of tranches
    opt_model += production[(hour, 'a_flow_below53')] == production[(hour, 'a_15')] + production[(hour, 'ap_30')] + production[(hour, 'a_45')] + production[(hour, 'ap_60_below53')] 
    opt_model += production[(hour, 'a_total')] == production[(hour, 'a_flow_below53')] + production[(hour, 'a_flow_above53')]

    # only 1 tranche can be active at the time
    opt_model += production[(hour, 'a_15')] <= M * a_status_15
    opt_model += production[(hour, 'a_30')] <= M * a_status_30 
    opt_model += production[(hour, 'a_45')] <= M * a_status_45
    opt_model += production[(hour, 'a_60_below53')] + production[(hour, 'a_flow_above53')] <= M * a_status_60

    # a_60_above53 can only be active if a_60_below53 == 52
    opt_model += production[(hour, 'a_flow_above53')] <= M * options_status[(hour, 'a_flow_above53')]
    opt_model += (52 - production[(hour, 'a_flow_below53')] ) <= M * (1-options_status[(hour, 'a_flow_above53')])

# only 1 'mode' can be used per day (corresponding to daily fee)
opt_model += a_status_15 + a_status_30 + a_status_45 + a_status_60 <= 1

目标函数：

# production per supply option per hour
production = pulp.LpVariable.dicts("production",
                                     ((hour, option) for hour in hours for option in options.index),
                                     lowBound=0,
                                     cat='Integer')

# status of supply options per hour (in use or not in use)
options_status = pulp.LpVariable.dicts("options_status",
                                     ((hour, option) for hour in hours for option in options.index),
                                     cat='Binary')

# use status of A tranches on day (used or not used)
a_status_15 = pulp.LpVariable('a_stat_15', cat='Binary')
a_status_30 = pulp.LpVariable('a_stat_30', cat='Binary')
a_status_45 = pulp.LpVariable('a_stat_45', cat='Binary')
a_status_60 = pulp.LpVariable('a_stat_60', cat='Binary')

opt_model = pulp.LpProblem("Optimizer", pulp.LpMinimize)

opt_model += pulp.lpSum(
    #O variable
    [0.2*demand.loc[hour, 'price']*production[hour, 'o'] + 
    #O fixed
    3*demand.loc[hour, 'price'] +
    #L30
    ( 12 )*options_status[hour, 'l30']*demand.loc[hour, 'price'] + 
    #L50
    ( 2*options_status[hour, 'l50']+0.13*production[hour, 'l50'] )*demand.loc[hour, 'price'] +
    #L90
    ( 15*options_status[hour, 'l90']+0.13*production[hour, 'l90'] )*demand.loc[hour, 'price'] +
    #L150
    ( 8*options_status[hour, 'l150']+0.2*production[hour, 'l150'] )*demand.loc[hour, 'price'] +
    #L230
    ( -3*options_status[hour, 'l230']+0.3*production[hour, 'l230'] )*demand.loc[hour, 'price'] +
    #L fixed
    ( 7*(1-options_status[hour, 'ltotal'])*demand.loc[hour, 'price'] ) +
    #A base unit price
    (10*production[hour, 'a_flow_below53']+8.88*production[hour, 'a_flow_above53'])*(c/20) for hour in hours]
    #A daily fee
    + (a_status_15 * 1000 + a_status_30 * 2000 + a_status_45 * 3 + a_status_60 * 4000)*(c/20))

# Sum of production == Demand
for hour in hours:
    opt_model += production[(hour, 'o')] + production[(hour, 'l_total')] + production[(hour, 'a_total')] == int(demand.loc[hour, 'demand'])

# Production <=  Max capacity AND => Min capacity
for hour in hours:
    for option in options.index:
        opt_model += production[(hour, option)] >= options_status[(hour, option)]
        opt_model += production[(hour, option)] >= options.loc[option, 'min_capacity'] * options_status[(hour, option)]
        opt_model += production[(hour, option)] <= options.loc[option, 'max_capacity'] * options_status[(hour, option)]

# Constraints L
for hour in hours:
    opt_model += production[(hour, 'l30')] + production[(hour, 'l0')] + production[(hour, 'l90')] + production[(hour, 'l150')] + production[(hour, 'l230')] == production[(hour, 'l_total')]
    opt_model += options_status[(hour, 'l30')] + options_status[(hour, 'l50')] + options_status[(hour, 'l90')] + options_status[(hour, 'l150')] + options_status[(hour, 'l230')] <= 1

# Constraints A
M = 999
for hour in hours:
    # total flow = sum of tranches
    opt_model += production[(hour, 'a_flow_below53')] == production[(hour, 'a_15')] + production[(hour, 'ap_30')] + production[(hour, 'a_45')] + production[(hour, 'ap_60_below53')] 
    opt_model += production[(hour, 'a_total')] == production[(hour, 'a_flow_below53')] + production[(hour, 'a_flow_above53')]

    # only 1 tranche can be active at the time
    opt_model += production[(hour, 'a_15')] <= M * a_status_15
    opt_model += production[(hour, 'a_30')] <= M * a_status_30 
    opt_model += production[(hour, 'a_45')] <= M * a_status_45
    opt_model += production[(hour, 'a_60_below53')] + production[(hour, 'a_flow_above53')] <= M * a_status_60

    # a_60_above53 can only be active if a_60_below53 == 52
    opt_model += production[(hour, 'a_flow_above53')] <= M * options_status[(hour, 'a_flow_above53')]
    opt_model += (52 - production[(hour, 'a_flow_below53')] ) <= M * (1-options_status[(hour, 'a_flow_above53')])

# only 1 'mode' can be used per day (corresponding to daily fee)
opt_model += a_status_15 + a_status_30 + a_status_45 + a_status_60 <= 1

约束条件：

# production per supply option per hour
production = pulp.LpVariable.dicts("production",
                                     ((hour, option) for hour in hours for option in options.index),
                                     lowBound=0,
                                     cat='Integer')

# status of supply options per hour (in use or not in use)
options_status = pulp.LpVariable.dicts("options_status",
                                     ((hour, option) for hour in hours for option in options.index),
                                     cat='Binary')

# use status of A tranches on day (used or not used)
a_status_15 = pulp.LpVariable('a_stat_15', cat='Binary')
a_status_30 = pulp.LpVariable('a_stat_30', cat='Binary')
a_status_45 = pulp.LpVariable('a_stat_45', cat='Binary')
a_status_60 = pulp.LpVariable('a_stat_60', cat='Binary')

opt_model = pulp.LpProblem("Optimizer", pulp.LpMinimize)

opt_model += pulp.lpSum(
    #O variable
    [0.2*demand.loc[hour, 'price']*production[hour, 'o'] + 
    #O fixed
    3*demand.loc[hour, 'price'] +
    #L30
    ( 12 )*options_status[hour, 'l30']*demand.loc[hour, 'price'] + 
    #L50
    ( 2*options_status[hour, 'l50']+0.13*production[hour, 'l50'] )*demand.loc[hour, 'price'] +
    #L90
    ( 15*options_status[hour, 'l90']+0.13*production[hour, 'l90'] )*demand.loc[hour, 'price'] +
    #L150
    ( 8*options_status[hour, 'l150']+0.2*production[hour, 'l150'] )*demand.loc[hour, 'price'] +
    #L230
    ( -3*options_status[hour, 'l230']+0.3*production[hour, 'l230'] )*demand.loc[hour, 'price'] +
    #L fixed
    ( 7*(1-options_status[hour, 'ltotal'])*demand.loc[hour, 'price'] ) +
    #A base unit price
    (10*production[hour, 'a_flow_below53']+8.88*production[hour, 'a_flow_above53'])*(c/20) for hour in hours]
    #A daily fee
    + (a_status_15 * 1000 + a_status_30 * 2000 + a_status_45 * 3 + a_status_60 * 4000)*(c/20))

# Sum of production == Demand
for hour in hours:
    opt_model += production[(hour, 'o')] + production[(hour, 'l_total')] + production[(hour, 'a_total')] == int(demand.loc[hour, 'demand'])

# Production <=  Max capacity AND => Min capacity
for hour in hours:
    for option in options.index:
        opt_model += production[(hour, option)] >= options_status[(hour, option)]
        opt_model += production[(hour, option)] >= options.loc[option, 'min_capacity'] * options_status[(hour, option)]
        opt_model += production[(hour, option)] <= options.loc[option, 'max_capacity'] * options_status[(hour, option)]

# Constraints L
for hour in hours:
    opt_model += production[(hour, 'l30')] + production[(hour, 'l0')] + production[(hour, 'l90')] + production[(hour, 'l150')] + production[(hour, 'l230')] == production[(hour, 'l_total')]
    opt_model += options_status[(hour, 'l30')] + options_status[(hour, 'l50')] + options_status[(hour, 'l90')] + options_status[(hour, 'l150')] + options_status[(hour, 'l230')] <= 1

# Constraints A
M = 999
for hour in hours:
    # total flow = sum of tranches
    opt_model += production[(hour, 'a_flow_below53')] == production[(hour, 'a_15')] + production[(hour, 'ap_30')] + production[(hour, 'a_45')] + production[(hour, 'ap_60_below53')] 
    opt_model += production[(hour, 'a_total')] == production[(hour, 'a_flow_below53')] + production[(hour, 'a_flow_above53')]

    # only 1 tranche can be active at the time
    opt_model += production[(hour, 'a_15')] <= M * a_status_15
    opt_model += production[(hour, 'a_30')] <= M * a_status_30 
    opt_model += production[(hour, 'a_45')] <= M * a_status_45
    opt_model += production[(hour, 'a_60_below53')] + production[(hour, 'a_flow_above53')] <= M * a_status_60

    # a_60_above53 can only be active if a_60_below53 == 52
    opt_model += production[(hour, 'a_flow_above53')] <= M * options_status[(hour, 'a_flow_above53')]
    opt_model += (52 - production[(hour, 'a_flow_below53')] ) <= M * (1-options_status[(hour, 'a_flow_above53')])

# only 1 'mode' can be used per day (corresponding to daily fee)
opt_model += a_status_15 + a_status_30 + a_status_45 + a_status_60 <= 1

#生产总额==需求
以小时为单位的小时：
opt_model+=生产[（小时，'o'）]+生产[（小时，'l_总计'）]+生产[（小时，'a_总计'）]==int（demand.loc[小时，'demand']））
#最小生产能力
以小时为单位的小时：
对于options.index中的选项：
opt_模型+=生产[（小时，选项）]>=选项_状态[（小时，选项）]
opt_model+=生产[（小时，选项）]>=options.loc[选项，“最小容量”]*选项_状态[（小时，选项）]
opt_模型+=生产[（小时，选项）]LP问题是NP难问题。Python对于您的问题来说可能太慢了。您可以考虑以下建议中的一个或多个（按您的附加工作顺序）
减少约束的数量，它可以更快地生成解决方案。消除不可信的解决方案
通过组合输入或约束来重新建模您的问题
使用其他可能更快的软件或库。您可以考虑CPLEX或GLPK。
 
LP问题的求解速度非常快，除非你有大量的变量。正是整数和二进制变量减慢了速度
最有可能是你的生产
变量杀死了你。我建议将这些变量视为线性/连续变量（“线性松弛”），以快速获得解决方案（然后可以向下/向上取整以获得可行的解决方案）
另一个选项可能是增加允许的最佳性差距，或设置超时限制。默认的最优性差距将非常小，这意味着MILP的分支和边界解算器将继续运行。求解者通常会相对较快地获得好的解，然后花费大量时间试图证明解的最优性。-您可以为解算器提供超时以获得近似解：
prob.solve（palp.solvers.palp\u CBC\u CMD（maxSeconds=120））
LP问题的求解速度非常快，除非变量数量非常多。正是整数和二进制变量减慢了速度。在我看来，是你的生产
变量害死了你。我建议将这些变量视为线性/连续变量（“线性松弛”），以快速获得解决方案（然后可以向下/向上取整以获得可行的解决方案）。另一种选择可能是增加允许的最优性差距。默认值将非常低，这意味着MILP的分支和绑定解算器将继续运行。谢谢你，Kabdulla，谢谢你的帮助！你的建议真的解决了我的问题性能问题：）很高兴听到有帮助的评论。我将粘贴它作为答案，以便您可以接受：）LP问题在P中。