Suppose that you have an M/M/1/FIFO queue where jobs arrive with two poisson arrival processes P1, P2. The service rate of the queue is $\mu$. The arrival rate of P1 and P2 is $\lambda_1$ and $\lambda_2$ respectively.
What is the mean delay for the jobs of each process?
If I let:
$\lambda \equiv \lambda_1 + \lambda_2$
be the total arrival rate and
$D \equiv \frac{1}{\mu-\lambda}$
be the mean delay for any job in the queue
then I have the intuition that the mean delay of packets from the first process should be
$D_1 = \frac{1}{\mu\frac{\lambda_1}{\lambda}-\lambda_1} = \frac{\lambda}{\lambda_1} D$
because the packets of the first flow should be serviced $\frac{\lambda_1}{\lambda}$ % of the time. However, I think this is wrong because if I take the mean delay between the proceses:
$D' \equiv \frac{\lambda_1}{\lambda} D_1 + \frac{\lambda_2}{\lambda} D_2 = 2D$
which can't be the case so I'm doing something wrong. So what is it/ where can I find an analysis?
Edit: I'm interested in the mean delay of a job of P1(P2) when both processes act on the queue.
Answer:
Suppose that there are two Poisson processes operating independently, with arrival rates $\lambda_1$ and $\lambda_2$ respectively. $N_{1}(t)$ and $N_{2}(t)$ are the respective cumulative numbers of arrivals through time t. Then the combined or pooled process has a cumulative number of arrivals equal to $N(t) = N_{1}(t)) + N_{2}(t)$. A fundamental property of independent Poisson processes is that their pooled process is also a Poisson process with arrival-rate parameter equal to the sum of the individual arrival rates. Thus, N(t) has a Poisson distribution with mean $(\lambda_1 + \lambda_2)t$.
The mean delay of the pooled process = $D = \frac{1}{\mu - \lambda} = \frac{1}{\mu - \lambda_1 - \lambda_2}$. Get $\mu$ from this in terms of D and respective $\lambda$
That will be $\mu = \frac{1}{D} +\lambda$
For the independent process if it were to act independently with the server would have a mean delay such as this
$$D_1 = \frac{1}{\mu - \lambda_1}$$ $$D_2 = \frac{1}{\mu - \lambda_2}$$
Substitute the value of $\mu$ in this and you would get
$$D_1 = \frac{D}{(1+D\lambda_2)}$$
$$D_2 = \frac{D}{(1+D\lambda_1)}$$
Now equate in the result $D = D_1 +D_2$
You get $D = \frac{1}{\sqrt{\lambda_1 \lambda_2}}$
Substituting the value of D in the equation of $D_1$ and $ D_2$
The mean delay of the two processes are
$$D_1 = \frac{1}{\mu-\lambda_1} = \frac{1}{\sqrt{\lambda_1 \lambda_2}+\lambda_2}$$ $$D_2 = \frac{1}{\mu-\lambda_2} = \frac{1}{\sqrt{\lambda_1 \lambda_2}+\lambda_1}$$
That is going to solve your worry of $D = D_1 + D_2$