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Statistics helping to Optimize Purchase & Stock






The chain management department of a furniture retail chain works on improving the company’s purchase and stock logistics. In particular, they want to improve on a few items of relatively high value and large physical volume and that are sold less frequently than most of the other items.

On one hand, they wish to minimize the economical value tied to the items and on the other hand, they wish to have them available when on demand.

In order to get started with this new analytical approach, they select to run it for a selected item which is a special couch with a leather cover.

In this initial process, they decide to base the analysis on non seasonal sales.

Question 1:

What are the following probabilities, knowing out historical data that our average sale per week of Model 1 is 2.5:

  1. The probability of selling maximum 4 sofas during one week?
  2. The probability of selling maximum 8 sofas during two weeks?

Answer to Question 1:

Let be the number of sold Model 1 Sofas per week. We will assume that the number of sold items per week follows the Poisson distribution, \(X \sim \operatorname{Pois} \left({\lambda}=2.5\right)\) whereas the following conditions are met:
– the number of sold Model 1 Sofas occurs in a fixed interval of time
– there are two possible outcomes (sold/ not sold)
– the average (lambda) is known

Question 1.a can then be answered using these approaches :
In order to find the probability of Model 1 selling maximum 4 units per week, P(X \(\geq\) 4), we can look it up in Poisson distribution table, or we can use the formula \[\begin{equation*}\sum _{i=0} ^{4}\frac {\lambda ^x}{x!} e^{-\lambda}\end{equation*}\]
Therefore:

\(\begin{equation*}\frac {2.5^0}{0!} e^{2.5}\end{equation*}\) + \(\begin{equation*}\frac {2.5^1}{1!} e^{2.5}\end{equation*}\)\(\begin{equation*}\frac {2.5^4}{4!} e^{2.5}\end{equation*}\) = 0.891178

In R, we can use the code:

ppois(4, lambda=2.5)
## [1] 0.891178

## <ScaleContinuous>
##  Range:  
##  Limits:    0 --    1
# PLOTS
# Plot 1: Histogram
week1 <- ppois(4, lambda=2.5)
hist(week1)

h <- hist(week1, 
         prob = TRUE, # flipside of "freq = FALSE"
         ylim = c(0, 0.3),
         xlim = c(0, 10),
         col = "#E5E5E5",
         border = 0,
         main = "The probability of selling max. 4 Model 1 in one week ")

The result is that the probability for selling maximum 4 Model 1 in one week is 0.891178 = 89,12%

We will assume that the number of sold items per week follows the Poisson distribution whereas the following conditions are complied with: – the number of sold items occurs in a fixed interval of time – 2 possible outcomes (sold/ not sold) – the average (lambda) is known

Sales of this sofa can occur 0, 1, 2, … times in during a week. The average sale an interval is designated the lambda symbol: ….. Lambda is the event rate, or the rate parameter. The probability of observing k events in an interval is given by the Poisson equation:

-.,-.,-.,-.,-.,-.,-., meaning that our equation is

So, to answer question 1a – what is the probability of selling maximum 4 sofas one week, our formula becomes: -.,-.,-.,-.,-.,-.,-.,