#!/usr/bin/python
# encoding: utf-8

"""
@author:
@contact: zhangcdnuli@163.com
@file: 最優解.py
@time: 2020/7/14 14:12
"""



"""
做推廣方案其實是運籌學中的求最優解問題，運用運籌學的知識可以制定出最佳的投放方案。
如果這個月老闆給你1千萬的推廣預算，讓你給分配到各個管道和品類，要確認分配比例，可以讓這1千萬的效果最大化。

已知條件，該企業有4個主要管道，分別是電商站、某音、某度和線下。
電商站不能少於1百萬的投入，投入產出比（ROI）是1.5
某音不能高於3百萬的投入，投入產出比是2
某度不能少於2百萬的投入，投入產出比是0.7
線下不能低於1百萬的投入，投入產出比是0.1

共有5個主要品類，分別為服裝、家電、百貨、美妝和餐飲具
服裝不能少於100萬的投入，ROI是1.2
家電不能高於300萬的投入，ROI是2
百貨不能少於100萬的投入，ROI是1.4
美妝不能高於350萬的投入，ROI是1.5
餐飲不能少於50萬的投入，ROI是0.8

請把1千萬的推廣預算合理地分配到各個管道和品類
要明確到某個品類要在某個管道中投放多少預算。
"""

from scipy.optimize import linprog

"""
管道投入分佈
max z = 1.5X_1 + 2X_2 + 0.7X_3 + 0.1X_4
X_1 + X_2 + X_3 + X_4 = 1000
X_1 >= 100
X_2 <= 300
X_3 >= 200
X_4 >= 100
"""
c1 = [-1.5,-2,-0.7,-0.1]
A_eq1 = [[1,1,1,1]]
b_eq1 = [1000]
x1 = (100,1000)
x2 = (0,300)
x3 = (200,1000)
x4 = (100,1000)
res1 = linprog(c1,A_eq=A_eq1,b_eq=b_eq1,bounds=(x1,x2,x3,x4))
# 獲取全部x的值
x = res1.get('x')
# 管道投產最大值
fun1 = abs(round(res1.get('fun')))
# 電商站投入份額
x_1 = round(x[0])
# 某音投入份額
x_2 = round(x[1])
# 某度投入份額
x_3 = round(x[2])
# 線下投入份額
x_4 = round(x[3])

print('品類投產最大值:',fun1)
print('電商站投入份額:',x_1)
print('某音投入份額:',x_2)
print('某度投入份額:',x_3)
print('線下投入份額:',x_4)
print('===================================================================')

"""
品類投入分佈
max z = 1.2Y_1 + 2Y_2 + 1.4Y_3 + 1.5Y_4 + 0.8Y_5
Y_1 + Y_2 + Y_3 + Y_4 + Y_5 = 1000
Y_1 >= 100
Y_2 <= 300
Y_3 >= 100
Y_4 <= 350
Y_5 >= 50
"""

c2 = [-1.2,-2,-1.4,-1.5,-0.8]
A_eq2 = [[1,1,1,1,1]]
b_eq2 = [1000]
y1 = (100,1000)
y2 = (0,300)
y3 = (100,1000)
y4 = (0,350)
y5 = (50,1000)
res2 = linprog(c2,A_eq=A_eq2,b_eq=b_eq2,bounds=(y1,y2,y3,y4,y5))
# 獲取全部y的值
y = res2.get('x')
# 品類投產最大值
fun2 = abs(round(res2.get('fun')))
# 服裝投入份額
y_1 = round(y[0])
# 家電投入份額
y_2 = round(y[1])
# 百貨投入份額
y_3 = round(y[2])
# 美妝投入份額
y_4 = round(y[3])
# 餐飲投入份額
y_5 = round(y[4])

print('品類投產最大值:',fun2)
print('服裝投入份額:',y_1)
print('家電投入份額:',y_2)
print('百貨投入份額:',y_3)
print('美妝投入份額:',y_4)
print('餐飲投入份額:',y_5)
print('===================================================================')

# 獲取最小值
def least(num1,num2):
    lea = 0
    if num1 <= num2 :
        lea = num1
    else:
        lea = num2
    return lea

"""
合併管道和品類的投入份額
獲取每個品類在各個管道投入的份額
Max z = 1.8a1 + 2a2 + 0.84a3 + 0.12a4
        + 3b1 + 4b2 + 1.4b3 + 0.2b4
        + 2.1c1 + 2.8c2 + 0.98c3 + 0.14c4
        + 2.25d1 + 3d2 + 1.05d3 + 0.15d4
        + 1.2e1 + 1.6e2 + 0.56e3 + 0.08e4
a1 + a2 + a3 + a4 = 100
b1 + b2 + b3 + b4 = 300
c1 + c2 + c3 + c4 = 200
d1 + d2 + d3 + d4 = 350
e1 + e2 + e3 + e4 = 50
a1 + b1 + c1 + d1 + e1 = 400
a2 + b2 + c2 + d2 + e2 = 300
a3 + b3 + c3 + d3 + e3 = 200
a4 + b4 + c4 + d4 + e4 = 100
"""

c = [-1.8,-2,-0.84,-0.12,-3,-4,-1.4,-0.2,-2.1,-2.8,-0.98,-0.14,-2.25,-3,-1.05,-0.15,-1.2,-1.6,-0.56,-0.08]
cc = [1.8,2,0.84,0.12,3,4,1.4,0.2,2.1,2.8,0.98,0.14,2.25,3,1.05,0.15,1.2,1.6,0.56,0.08]
A = [
    [1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1],
    [1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0],
    [0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0],
    [0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0],
    [0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1]
     ]
b = [y_1,y_2,y_3,y_4,y_5,x_1,x_2,x_3,x_4]
a1 = (0,least(x_1,y_1))
a2 = (0,least(x_2,y_1))
a3 = (0,least(x_3,y_1))
a4 = (0,least(x_4,y_1))
b1 = (0,least(x_1,y_2))
b2 = (0,least(x_2,y_2))
b3 = (0,least(x_3,y_2))
b4 = (0,least(x_4,y_2))
c1 = (0,least(x_1,y_3))
c2 = (0,least(x_2,y_3))
c3 = (0,least(x_3,y_3))
c4 = (0,least(x_4,y_3))
d1 = (0,least(x_1,y_4))
d2 = (0,least(x_2,y_4))
d3 = (0,least(x_3,y_4))
d4 = (0,least(x_4,y_4))
e1 = (0,least(x_1,y_5))
e2 = (0,least(x_2,y_5))
e3 = (0,least(x_3,y_5))
e4 = (0,least(x_4,y_5))
res = linprog(c, A_eq=A, b_eq=b, bounds=(a1,a2,a3,a4,b1,b2,b3,b4,c1,c2,c3,c4,d1,d2,d3,d4,e1,e2,e3,e4))
z = res.get('x')
# 個品類在各個管道投入的份額的投產最大值
fun = abs(round(res.get('fun')))
# 服裝電商站投入份額
a_1 = round(z[0])
# 服裝某音投入份額
a_2 = round(z[1])
# 服裝某度投入份額
a_3 = round(z[2])
# 服裝線下投入份額
a_4 = round(z[3])
# 家電電商站投入份額
b_1 = round(z[4])
# 家電某音投入份額
b_2 = round(z[5])
# 家電某度投入份額
b_3 = round(z[6])
# 家電線下投入份額
b_4 = round(z[7])
# 百貨電商站投入份額
c_1 = round(z[8])
# 百貨某音投入份額
c_2 = round(z[9])
# 百貨某度投入份額
c_3 = round(z[10])
# 百貨線下投入份額
c_4 = round(z[11])
# 美妝電商站投入份額
d_1 = round(z[12])
# 美妝某音投入份額
d_2 = round(z[13])
# 美妝某度投入份額
d_3 = round(z[14])
# 美妝線下投入份額
d_4 = round(z[15])
# 餐飲電商站投入份額
e_1 = round(z[16])
# 餐飲某音投入份額
e_2 = round(z[17])
# 餐飲某度投入份額
e_3 = round(z[18])
# 餐飲線下投入份額
e_4 = round(z[19])

print('每個品類在各個管道投入的份額的投產最大值為:',fun)
print('服裝投入份額管道分佈')
print('電商站投入份額:',a_1,'某音投入份額:',a_2,'某度投入份額:',a_3,'線下投入份額:',a_4)
print('家電投入份額管道分佈')
print('電商站投入份額:',b_1,'某音投入份額:',b_2,'某度投入份額:',b_3,'線下投入份額:',b_4)
print('百貨投入份額管道分佈')
print('電商站投入份額:',c_1,'某音投入份額:',c_2,'某度投入份額:',c_3,'線下投入份額:',c_4)
print('美妝投入份額管道分佈')
print('電商站投入份額:',d_1,'某音投入份額:',d_2,'某度投入份額:',d_3,'線下投入份額:',d_4)
print('餐飲投入份額管道分佈')
print('電商站投入份額:',e_1,'某音投入份額:',e_2,'某度投入份額:',e_3,'線下投入份額:',e_4)