Suppose a computer has $s$ processors with identical independent exponential processing times with rate $\mu$. Instructions are processed on a first-come first-serve basis as soon as a processor becomes free. Instructions arrive with independent exponentially distributed inter-arrival times with rate $\lambda$.
Suppose that an instruction finds itself first in line with all processors busy. Denote $W$ the waiting time of the instruction and $T_P$ its processing time once accepted by a processor. Let $T = W + T_P$ be the total time elapsed between the arrival of the instruction and the completion of its processing.
I need to find the distribution of $W$ and calculate $\mathbb{E}[T]$.
My Thoughts
I've found several documents that state that the expected value of the waiting time of a customer in the queue is $$W_q = \left[ \frac{1}{(s-1)!}\left(\frac{\lambda}{\mu}\right)^s\frac{\lambda \mu}{(s\mu-\lambda)^2}\right]p_0,$$ where $p_0$ is the probability of $0$ customers in the queueing system. Now how can this expected value help me to find the distribution of $W$?
This is much easier since you know the exact state on arrival of this particular instruction. Since it's first in line and all the processors are full, the waiting time $W$ is the time until the first of the $s$ instructions being processed is done. Since all is exponential/memoryless, the wait time till the first finishes is exponential with rate $\mu s.$ (This is from the superposition principle for Poisson processes, or more simply from the fact that the minimum of $s$ independent exponentials with rate $\mu$ is an exponential with rate $\mu s$.)
Thus we have $E(T) = \frac{1}{\mu s}+ \frac{1}{\mu}$