Python數學規劃案例一

問題、模型、數據、算法、結果，統一地表述，是習慣也是效率。

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數學規劃模型表述習慣

采用五個部分：Set, Data, Variable, Objective, Constraints；每個Notation，采用一個主字符，上標表示含義，下標來自Set；已知的大寫，未知的小寫，下標用小寫；妥善選擇Notation中的每一個字符；對Notation、Objective、Constratins進行分組。模型即代碼、即注釋、即文章。

數學規劃模型編碼習慣

Ptyhon程序設計代碼與模型表述嚴格一致 Set, Data, Variable, Objective, Constraints盡量相應“翻譯”；將模型、數據預處理、模型結果后處理，分離開來，保持模型的干凈、純粹。代碼即模型、即文檔、即文章。

一個例子

預算約束下營養配餐。取自：

...\IBM\ILOG\CPLEX_Studio129\python\examples\mp\modeling\diet.py

模型

采用五個部分：Set, Data, Variable, Objective, Constraints建模。

代碼

根據diet.py改寫的代碼如下。

# -*- coding: utf-8 -*-

from collections import namedtuple

from docplex.mp.model import Model

from docplex.util.environment import get_environment

# ----------------------------------------------------------------------------

# Initialize the problem data

# ----------------------------------------------------------------------------

FOODS = [

("Roasted Chicken", 0.84, 0, 10),

("Spaghetti W/ Sauce", 0.78, 0, 10),

("Tomato,Red,Ripe,Raw", 0.27, 0, 10),

("Apple,Raw,W/Skin", .24, 0, 10),

("Grapes", 0.32, 0, 10),

("Chocolate Chip Cookies", 0.03, 0, 10),

("Lowfat Milk", 0.23, 0, 10),

("Raisin Brn", 0.34, 0, 10),

("Hotdog", 0.31, 0, 10)

]

NUTRIENTS = [

("Calories", 2000, 2500),

("Calcium", 800, 1600),

("Iron", 10, 30),

("Vit_A", 5000, 50000),

("Dietary_Fiber", 25, 100),

("Carbohydrates", 0, 300),

("Protein", 50, 100)

]

FOOD_NUTRIENTS = [

("Roasted Chicken", 277.4, 21.9, 1.8, 77.4, 0, 0, 42.2),

("Spaghetti W/ Sauce", 358.2, 80.2, 2.3, 3055.2, 11.6, 58.3, 8.2),

("Tomato,Red,Ripe,Raw", 25.8, 6.2, 0.6, 766.3, 1.4, 5.7, 1),

("Apple,Raw,W/Skin", 81.4, 9.7, 0.2, 73.1, 3.7, 21, 0.3),

("Grapes", 15.1, 3.4, 0.1, 24, 0.2, 4.1, 0.2),

("Chocolate Chip Cookies", 78.1, 6.2, 0.4, 101.8, 0, 9.3, 0.9),

("Lowfat Milk", 121.2, 296.7, 0.1, 500.2, 0, 11.7, 8.1),

("Raisin Brn", 115.1, 12.9, 16.8, 1250.2, 4, 27.9, 4),

("Hotdog", 242.1, 23.5, 2.3, 0, 0, 18, 10.4)

]

Food = namedtuple("Food", ["name", "unit_cost", "qmin", "qmax"])

Nutrient = namedtuple("Nutrient", ["name", "qmin", "qmax"])

# ----------------------------------------------------------------------------

# Build the model

# ----------------------------------------------------------------------------

def build_diet_model(**kwargs):

# Create tuples with named fields for foods and nutrients

F = [f[0] for f in FOODS]

C = {f[0]:f[1] for f in FOODS}

Fmin = {f[0]:f[2] for f in FOODS}

Fmax = {f[0]:f[3] for f in FOODS}

N = [n[0] for n in NUTRIENTS]

Nmin = {n[0]:n[1] for n in NUTRIENTS}

Nmax = {n[0]:n[2] for n in NUTRIENTS}

D = {(F[f],N[n]): FOOD_NUTRIENTS[f][n+1] for f in range(len(F)) for n in range(len(N))}

# Model

mdl = Model(name='diet', **kwargs)

# Decision variables, limited to be >= Food.qmin and <= Food.qmax

x = mdl.continuous_var_dict(F, lb=Fmin, ub=Fmax, name=F) # 2,4

# Limit range of nutrients, and mark them as KPIs

for n in N:

y = mdl.sum(x[f] * D[f, n] for f in F)

mdl.add_range(Nmin[n], y, Nmax[n]) # 3

mdl.add_kpi(y, publish_name="Total %s" % n)

for f in F:

mdl.add_kpi(x[f], publish_name="Food %s" % f)

# Minimize cost

mdl.minimize(mdl.sum(x[f] * C[f] for f in F)) # 1

return mdl

# ----------------------------------------------------------------------------

# Solve the model and display the result

# ----------------------------------------------------------------------------

if __name__ == '__main__':

mdl = build_diet_model()

mdl.print_information()

mdl.export_as_lp()

if mdl.solve():

mdl.float_precision = 3

print("* model solved as function:")

mdl.print_solution()

mdl.report_kpis()

# Save the CPLEX solution as "solution.json" program output

with get_environment().get_output_stream("diet2.json") as fp:

mdl.solution.export(fp, "json")

else:

print("* model has no solution")

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