Poisson Distribution for Drill Stock Management

In summary, there is a 5% chance that the store will not have enough drills to meet demand before the next delivery.
  • #1
jsmith613
614
0

Homework Statement


A hardwhere store sells on average 8 drills per week.
The store receives ONE delivery of drills at the same time each week.
Find the no. of drills that need to be in stock after a delivery for there to be at most a 5% chance of the store NOT having sufficent drills to meet demand before the next delivery.


Homework Equations


The Attempt at a Solution



Ok so we know that Y≈Po(8)
where Y = no. expected to sell
lets say n = demand no.

thus we want
p(n > Y) < 0.05
so
P(Y<n) < 0.05

BUT THE ANSWER SAYS
P(Y>n) < 0.05

How is this possible
thanks
 
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  • #2
Aren't your Y and n thesame thing. The number expected to sell surely equals the demand??

I think your n should be something like "drills in store".
 
  • #3
micromass said:
Aren't your Y and n thesame thing. The number expected to sell surely equals the demand??

I think your n should be something like "drills in store".

how can no. expected = demand

the Poisson value = E(X) but demand could be greater than or smaller than this, surely
 
  • #4
ahh
Y is a distribution which means it could take on OTHER values but not def.
so Y = demand
n = no. in stock
with Y-P(8)

So
P(Y>n) <= 0.05

I see
ok thanks
 
  • #5
jsmith613 said:

Homework Statement


A hardwhere store sells on average 8 drills per week.
The store receives ONE delivery of drills at the same time each week.
Find the no. of drills that need to be in stock after a delivery for there to be at most a 5% chance of the store NOT having sufficent drills to meet demand before the next delivery.


Homework Equations


The Attempt at a Solution



Ok so we know that Y≈Po(8)
where Y = no. expected to sell
lets say n = demand no.

thus we want
p(n > Y) < 0.05
so
P(Y<n) < 0.05

BUT THE ANSWER SAYS
P(Y>n) < 0.05

How is this possible
thanks

You need to distinguish between sales and demand. If demand = X~Po(8) and we start the week with stock N, then sales S = min(X,N); that is, sales = demand if demand <= N, and sales = N otherwise. You need to choose N so that P(X > N) <= 0.05; that is, there is a chance of no more than 5% that demand exceeds on-hand stock. (By the way, that is <= 0.05, not < 0.05, because that is what 'no more than' means.)

RGV
 
  • #6
cheers :)
 

1. What is the Poisson Distribution and how is it related to drill stock management?

The Poisson Distribution is a probability distribution that is used to model the likelihood of a certain number of events occurring within a specified time or space. It is often used in drill stock management to predict the number of drill bits that will break or wear out during a drilling operation.

2. How do you calculate the parameters of the Poisson Distribution for drill stock management?

The parameters of the Poisson Distribution for drill stock management can be calculated using the formula λ = np, where λ is the expected number of events (in this case, the expected number of drill bit failures), n is the number of time intervals, and p is the probability of an event occurring in each time interval.

3. What is the significance of the mean and variance in the Poisson Distribution for drill stock management?

The mean and variance in the Poisson Distribution for drill stock management represent the expected number of drill bit failures and the amount of variability in this number, respectively. These values are important for predicting and managing the inventory of drill bits needed for a drilling operation.

4. How can the Poisson Distribution be used to optimize drill stock management?

The Poisson Distribution can be used to optimize drill stock management by providing a probability distribution of the expected number of drill bit failures. This information can be used to determine the appropriate number of drill bits to have on hand, reducing the risk of running out of bits during a drilling operation.

5. Are there any limitations to using the Poisson Distribution for drill stock management?

While the Poisson Distribution is a useful tool for drill stock management, it does have some limitations. For example, it assumes that the probability of an event occurring is constant over time, which may not always be the case in real-world scenarios. Additionally, it may not be a good fit for situations where the number of events is very low or very high.

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