I'm stuck at the following question: Customers with items to repair arrive at a repair facility according to a Poisson process with rate λ. The repair time of an item has a uniform distribution on [a, b]. There are ample repair facilities so that each defective item immediately enters repair. The exact repair time can be determined upon arrival of the item. If the repair time of an item takes longer than τ time units with τ a given number between a and b, then the customer gets a loaner for the defective item until the item returns from repair. A sufficiently large supply of loaners is available. What is the average number of loaners which are out?
I know that the customers who need a loaner arrive according to a Poisson process with rate $\lambda \cdot \frac{b-\tau}{b-a}$, and that the expected loan-time is $\frac{1}{2}(b + \tau)$. I just can't figure out how to calculate the expected number of loaners that are out, because you also have to account for loaners that are being brought back.
I came up with the answer.
The probability that a customer needs a loaner is $\mathbb{P}[X\geq \tau] = \frac{b-\tau}{b-a}]$ where $X$ denotes the repair time of an item. So, the customers that need a loaner arrive according to a Poisson process with rate $\lambda \cdot \frac{b-\tau}{b-a}$. The average repair time for an item that promises the customer a loaner is $\frac{\tau+b}{2}$. The average number of loaners given out at a certain moment is equal to the average number of items, that take longer than $\tau$ to repair, that are being repaired at a certain moment. This tells us that the average number of loaners given out is $ \lambda \cdot \frac{b-\tau}{b-a} \cdot \frac{\tau+b}{2} = \lambda \cdot \frac{b^2 - \tau^2}{2(b-a)} $