# Optimizing price, maximizing revenue

### Problem statement

Price and quantity sold are the two determinants of business revenue/profit. At higher price the revenue is expected to be high. But this is not the case all the time. We know from our everyday experience, as price of something goes up, people have less tendency to buy it.

The reverse is also true, that is, as price is down, sales goes up (think what happens in a block buster sales event in a nearby shopping mall). But sales going up doesn’t always mean that the revenue will also go up, because of the trade-off created by price drop.

Setting a right price of products/services is one of the most important decisions a business can make. Under-pricing and over-pricing both can hurt a company’s bottom line. So where is the sweet spot, the right price, that maximizes revenue and profit?

With a simple example let’s examine how to optimization price to maximize revenue/profit.

### Import libraries

import numpy as np import pandas as pd from pandas import DataFrame import matplotlib.pyplot as plt import seaborn as sns import statsmodels.api as sm from statsmodels.compat import lzip from statsmodels.formula.api import ols 

### Data

Data in this analysis came from here [Susan Li has a nice blog post about price elasticity of demand (i.e. sensitivity of demand to change in price) explained using the same dataset]. This is a time series, quarterly data of beef sales (i.e. quantity sold) and its corresponding price.

This data will be used first to find price elasticity of demand and then to use this information to find at what price the profit is maximized.

# Load data data_source = https://raw.githubusercontent.com/susanli2016/Machine-Learning-with... beef = pd.read_csv('data_source')
# View first few rows beef.tail(5)
# demand curve estimation sns.lmplot( x="Price", y="Quantity", data=beef, fit_reg=True, size=4)
# fit OLS model model = ols("Quantity ~ Price", data=beef).fit() # print model summary  print(model.summary())

Price = [320, 330,340,350, 360, 370, 380, 390] # a range of different prices to find the optimum one Cost = 80 # a fixed cost in this case Revenue = [] for i in Price:         quantity_demanded=30.05-0.0465*i         Revenue.append((i-cost)*quantity_demanded) # profit function  #create data frame of price and revenue  profit = pd.DataFrame({"Price": Price, "Revenue": Revenue}) #plot revenue against price plt.plot(profit["Price"], profit["Revenue"])

# price at which the revenue is maximum profit[profit['Revenue'] == profit['Revenue'].max()]
[Note: This is also posted in my personal blog here]