One vs multiple servers - problem Consider the following problem:
We have a simple queueing system with $\lambda%$ - probabilistic intensity of queries per some predefined time interval.
Now, we can arrange the system as a single high-end server ($M/M/1$, which can handle the queries with the intensity of $2\mu$) or as two low-end servers ($M/M/2$, each server working with intensity of $\mu$).
So, the question is - which variant is better in terms of overall performance?
I suspect that it's the first one, but, unfortunately, my knowledge of queuing / probability theory isn't enough.
Thank you.

You need to specify what you mean by "overall performance", but for most measures the two server system will have better performance.  Intuitively, a "complicated" customer, one that has a long service time will shut down the M/M/1 queue but only criple the M/M/2 queue.  
If we let the utiliztion be $$\rho=\frac{\lambda}{2\mu}$$  then some of the usual performance measures are
$L_q$ the average length of the queue, $W_q$ the average waiting time, and $\pi_0$ the probability that the queue is empty.  For the M/M/1 queue these measures are
$$L_q=\frac{\rho^2}{1-\rho}$$
$$W_q=\frac{\rho^2}{\lambda(1-\rho)}$$
$$\pi_0=1-\rho$$
and for the M/M/2 queue
$$L_q=\frac{2\rho^3}{1-\rho^2}$$
$$W_q=\frac{2\rho^3}{\lambda(1-\rho^2)}$$
$$\pi_0=\frac{1-\rho}{1+\rho}$$
So, the system is empty more often in the M/M/1 queue, but the expected wait time and the expected queue length are less for the M/M/2 (as $\frac{2\rho}{1+\rho}<1$). 
