Linear Programming (Simplex LP) PuLP?

问题

In Python only, and using data from a Pandas dataframe, how can I use PuLP to solve linear programming problems the same way I can in Excel? How much budget should be allocated to each Channel under the New Budget column so we maximize the total number of estimated successes? I'm really looking for a concrete example using data from a dataframe and not really high-level advice.

Problem Data Setup

    Channel  30-day Cost  Trials  Success  Cost Min  Cost Max  New Budget
0  Channel1      1765.21    9865      812    882.61   2647.82           0
1  Channel2      2700.00   15000      900   1350.00   4050.00           0
2  Channel3      2160.00   12000      333   1080.00   3240.00           0

This is a Maximization problem.

The objective function is:

objective_function = sum((df['New Budget']/(df['30-day Cost']/df['Trials']))*(df['Success']/df['Trials']))

The constraints are:

The sum of df['New Budget'] must equal 5000
The New Budget for a given channel can go no lower than the Cost Min
The New Budget for a given channel can go no higher than the Cost Max

Any ideas how to translate this pandas dataframe solver linear problem using PuLP or any other solver approach? The end-result would be what you see in the image below.

回答1:

In general you create a dictionary of variables (x in this case) and a model variable (mod in this case). To create the objective you use sum over the variables times some scalars, adding that result to mod. You construct constraints by again computing linear combinations of variables, using >=, <=, or ==, and adding that constraint to mod. Finally you use mod.solve() to get the solutions.

import pulp

# Create variables and model
x = pulp.LpVariable.dicts("x", df.index, lowBound=0)
mod = pulp.LpProblem("Budget", pulp.LpMaximize)

# Objective function
objvals = {idx: (1.0/(df['30-day Cost'][idx]/df['Trials'][idx]))*(df['Success'][idx]/float(df['Trials'][idx])) for idx in df.index}
mod += sum([x[idx]*objvals[idx] for idx in df.index])

# Lower and upper bounds:
for idx in df.index:
    mod += x[idx] >= df['Cost Min'][idx]
    mod += x[idx] <= df['Cost Max'][idx]

# Budget sum
mod += sum([x[idx] for idx in df.index]) == 5000.0

# Solve model
mod.solve()

# Output solution
for idx in df.index:
    print idx, x[idx].value()
# 0 2570.0
# 1 1350.0
# 2 1080.0

print 'Objective', pulp.value(mod.objective)
# Objective 1798.70495012

Data:

import numpy as np
import pandas as pd
idx = [0, 1, 2]
d = {'channel': pd.Series(['Channel1', 'Channel2', 'Channel3'], index=idx),
     '30-day Cost': pd.Series([1765.21, 2700., 2160.], index=idx),
     'Trials': pd.Series([9865, 1500, 1200], index=idx),
     'Success': pd.Series([812, 900, 333], index=idx),
     'Cost Min': pd.Series([882.61, 1350.00, 1080.00], index=idx),
     'Cost Max': pd.Series([2647.82, 4050.00, 3240.00], index=idx)}
df = pd.DataFrame(d)
df
#    30-day Cost  Cost Max  Cost Min  Success  Trials   channel
# 0      1765.21   2647.82    882.61      812    9865  Channel1
# 1      2700.00   4050.00   1350.00      900    1500  Channel2
# 2      2160.00   3240.00   1080.00      333    1200  Channel3

来源：https://stackoverflow.com/questions/33160262/linear-programming-simplex-lp-pulp

标签

python

pandas

dataframe

linear-programming

pulp