Poisson Process - Number of cars that a petrol station can service

[Usagi]

Question:

A single-pump petrol station is running low on petrol. The total volume of petrol remaining for sale is 100 litres.

Suppose cars arrive to the station according to a Poisson process with rate $$\lambda$$, and that each car fills independently of all other cars and of the arrival process, an amount of petrol that is distributed as a uniform random variable over $$(0, 50)$$ - assume for example that all car tanks have a capacity of 50 litres and drivers decide "at random" when to refill. We assume that service is instantaneous so that there are no queues at the station.

(a) On average, how many cars will the petrol station fully service (sell the full amount requested) before it runs out of petrol (and before any refilling occurs)?

(b) How much time will it take on average before the station runs out of petrol (and before any refilling occurs)?

Attempt:

I'm not exactly sure where to start with this question part (a). Let $$U$$ be uniformly distributed over $$(0,50)$$, then each time a car arrives at the petrol station, the total volume of petrol decreases by $$U$$. So define $$U_1$$ to be the amount of petrol that the first arrival (an "arrival" here being when a car arrives at the petrol station and refills) and $$U_2$$ be that of the second arrival, and so on. Then each $$U_i$$ is identically and independently distributed as $$U$$. So by the $$N$$-th arrival, the station will have $$100-\sum_{i=1}^N U_i$$ litres of petrol remaining. We stop once $$100-\sum_{i=1}^N U_i=0$$ and we basically need to find $$E[N]$$?----------That's all I've got so far, if someone can provide a solution, that would be good.

 

[Fermat1]

Question:

A single-pump petrol station is running low on petrol. The total volume of petrol remaining for sale is 100 litres.

Suppose cars arrive to the station according to a Poisson process with rate $$\lambda$$, and that each car fills independently of all other cars and of the arrival process, an amount of petrol that is distributed as a uniform random variable over $$(0, 50)$$ - assume for example that all car tanks have a capacity of 50 litres and drivers decide "at random" when to refill. We assume that service is instantaneous so that there are no queues at the station.

(a) On average, how many cars will the petrol station fully service (sell the full amount requested) before it runs out of petrol (and before any refilling occurs)?

(b) How much time will it take on average before the station runs out of petrol (and before any refilling occurs)?

Attempt:

I'm not exactly sure where to start with this question part (a). Let $$U$$ be uniformly distributed over $$(0,50)$$, then each time a car arrives at the petrol station, the total volume of petrol decreases by $$U$$. So define $$U_1$$ to be the amount of petrol that the first arrival (an "arrival" here being when a car arrives at the petrol station and refills) and $$U_2$$ be that of the second arrival, and so on. Then each $$U_i$$ is identically and independently distributed as $$U$$. So by the $$N$$-th arrival, the station will have $$100-\sum_{i=1}^N U_i$$ litres of petrol remaining. We stop once $$100-\sum_{i=1}^N U_i=0$$ and we basically need to find $$E[N]$$?----------That's all I've got so far, if someone can provide a solution, that would be good.

Let $Y$ be the amount of petrol put in by a random driver so Y has density function $f(y)=\frac{1}{50}$ It is easy to verify $E(Y)=25$ so, on average four cars can be served before the petrol runs out.

Let$X$ be the number of cars arriving in some time interval B, so $E(X)={\lambda}$ . Then the average time before running out is $\frac{4}{\lambda}B$

 

[Usagi]

Thanks, how about part (a)?

 

[Fermat1]

Thanks, how about part (a)?

I've done both parts but I reckon you missed some information. The rate ${\lambda}$ must be qualified with a time period. E.g ${\lambda}$ cars per hour.

 

[Usagi]

I can confirm I didn't miss anything. I've copied the q exactly as it is. Also i think part a) is harder than it seems. This is my working so far:

So, define $U_k$ as the random variable that denotes the amount of petrol that car $k$ fills, $k = 1, 2, 3, \cdots$. Thus, $U_k, k = 1, 2, 3, \cdots$ are independently and identically distributed as a uniform random variable over $(0,50)$.

Let $N$ be the random variable that denotes the number of cars that the petrol station can *fully* service.

Now I don't quite get the question. Say we have the following scenario:

Car 1 comes with 10L remaining in its tank, so it will fill up 40L, hence the amount of petrol left in the station is now 100-40 = 60L.

Car 2 comes with 10L remaining in its tank, so it will fill up 40L, hence the amount of petrol left in the station is now 60-40 = 20L.

Car 3 comes with 20L remaining in its tank, so it will fill up 30L, but the petrol station only has 20L left, so does this mean Car 3 just leaves the petrol station filling 0L? My gut feeling is that this cannot happen because each car can only fill an amount BETWEEN 0 and 50, ie, (0,50) [note that the end points are not included].

Hence in this scenario, the petrol station runs "out" of petrol at N=3 because it does not have enough to FULLY service Car 3, even though it still has 20L left in the pump. Thus, the petrol station can only service N=2 cars.

Is this interpretation correct? If so, how do I find $E[N]$?----------working on further... Not sure if right

Define $G_N = \sum_{k=1}^N U_k$, then $E[N] = \sum_{n=1}^{\infty} n P(N=n) = \sum_{n=1}^{\infty} n P(G_n \le 100)$

This equivalence comes from the fact that the event $\{N=n\}$ will only happen if $100 - G_n \ge 0$.

However, two questions remain, what is the distribution of $G_n$? (How do I derive the distribution of the sum of $n$ iid uniform random variables and second, how do I compute the infinite sum?