Poisson Distribution- Business application

Introduction

When we start learning statistics, we come across various types of Probability distributions. We learn about them in theory and also solve some theoretical examples. There are many articles/videos that provide examples and basic explanation about Poisson Distribution. What might be more interesting is to use these distributions in real life to solve business problems. Seems exciting right?

In this article I will try to explain the business application of Poisson Distribution by creating a virtual business scenario using Python.

Poisson Distribution

The only parameter of the Poisson distribution is lambda, the event rate, (or rate parameter). This represents the expected number of events in an interval. If we have a rate in events / time, we can get to the expected events by multiplying the time.

We find the probability of a number of events, x in an interval using the Poisson Probability Density Function:

Different business applications of Poisson Distribution

1. Check for adequate customer service staffing

2. Use the Poisson formula to evaluate whether it is financially viable to keep a store open 24 hours a day.

3. Review and evaluate business insurance coverage.

Let’s work through an example.

Business problem definition

We will use Poisson distribution to find if its financially viable to keep a supermarket open 24 hours a day.

Virtual business scenario

Let’s consider a supermarket that is open 24*7. The store employees 2 employees for every shift.

Shift timings include : 8 am — 4 pm, 4 pm-12 pm, 12 pm- 8 am

For each shift 2 employees work at the supermarket with a monthly salary of 12000 INR.

In every shift, the supermarket spends certain amount in electricity and employee salary per day. Let’s first calculate the stores electricity consumption per shift.

Supermarket electricity consumption

Consider the store has-

1) 10 Fans : An average size ceiling fan has rated power of 70 watts and for 8 hours of operation they consumes 0.56 kWh of power.

2) 14 Tube-lights of 40 watts each : 40 watts of power over 8 hours amounts to 8 x 40 = 320 watt-hours of energy, or 320/1000 = .32 kilowatt-hour (kWh) of energy.

3) 2 commercial refrigerator 300 watts each: 300 watts of power over 8 hours amounts to 8 x 300 = 2400 watt-hours of energy, or 2400/1000 = 2.4 kilowatt-hour (kWh) of energy.

4) Billing computer and other small electric appliances: 30 watts of power over 8 hours amounts to 8 x 30 = 240 watt-hours of energy, or 240/1000 = 0.24 kilowatt-hour (kWh) of energy.

Energy consumption calculation

Energy Consumption Charges/unit

Electricity cost as per appliances:

1) Fans = 0.56*10*4.67 = 26.152 INR
2) Tube-lights = 14*0.32*4.67 = 20.9216 INR
3) Commercial refrigerators = 2.4*2*4.67 = 22.41 INR
4) Billing computer and other small electric appliances = 0.24 * 4.67 = 1.12 INR

Overall Energy cost/shift= 70.602 INR

Employee cost for 8 hour midnight shift:

Monthly salary = 12000 INR

Per day salary = 400

For 2 employees= 800 INR

Overall cost of store for a shift = 800 + 70.602 = 870.6 INR

So for the supermarket to be in profit they need to cover the minimum cost of 870.6 INR in every shift and also earn reasonable profit from the sales.

Each shift will have different average sales as the number of customers varies as per the time.The supermarket will be definitely kept opened for the 1st 2 shifts, the main question arises that is it viable to keep the store open at night time as well. As generally the sales will be less during the night shift.

Consider the average sales of the store at night is 8000 INR. (Just an assumption).

The profit margin for the grocery stores in India ranges from 2% to 20%.

Let’s assume that the supermarket makes a 11% profit for every sale.

This percentage may vary with different items but let’s consider an average of 11%.

So, considering the sales of 8000 INR, 11% will be 880 INR.

According to this scenario the store makes an overall profit of 880–870 = 10 INR for the night shift. (this is based on assumption that average for night shift is 8000 INR)

This will be valid if the store makes minimum sales of 8000 INR every night. This seems a very low profit for a supermarket. We can find out the probability of maximum/minimum profit that the supermarket can make in the night shift using Poisson distribution.

Using Poisson distribution let’s find out what is the minimum sales the store can make which will be the deciding factor to answer if its viable to keep the supermarket open during night shift.

Here the lambda (λ) = 8000 , but as a general rule while using Poisson distribution we observe that as lambda(λ) increases the distribution starts approximating towards Normal distribution. As a rule of thumb we can use that when λ >20 the distribution can be approximated towards normal with μ= 8000 and variance(σ^2) =8000

Normal approximation to Poisson distribution

Code for this article can be found here.

Using the above approximation we will find out the probability of the minimum sales required in order to cover the overall cost of the supermarket. The supermarket requires a minimum profit of 870 INR to cover its cost for the midnight shift. So, with the profit margin of 11% on sales, the supermarket needs to sale items worth 7909 INR.

Sales worth 7909 INR is the minimum requirement for the supermarket to at least cover the shift costs. We will find the probability of this.

Calculating probability that the store makes minimum 7909 INR

Using normal distribution it is difficult to calculate the exact probability that the supermarket makes 7909 INR, so we need to convert the distribution to Standard Normal Distribution with a μ = 0 and σ = 1

We can do this using Z score:

Here, we need to apply continuity correction as we are approximating Poisson to normal distribution.

Continuity correction rules

We need to find P(X>=7909) -> P(X > 7908.5) , according to continuity correction rules.

So, Z score for :

P( X > 7908.5)

= P( Z > ((7908.5–8000)/89.44))

= P( Z > -1.02)

= P( Z < 1.02)

= 0.8461(using Z score table)

There is 84.61% chance that the supermarket will be able to cover its cost.

Conclusion

So it is viable for the supermarket to be open at night as well as it has high probability that it can cover its overall cost. Similarly we can calculate different profit margin wise probability for the supermarket. This is one business application of Poisson distribution, I have also solved the same for another business problem which can be found here .