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.
 A: 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 $\frac{2\rho}{1+\rho}<1$). 
A: If "overall performance" is the expected time a client/customer/query spend in the M/M system, then the single server system outperforms the second one.
The reasoning is simple: the M/M/1 system functions in "full" intensity even with a single query at the system; the M/M/2 system needs two queries present to reach the highest service intensity. 
So, queries arriving at an empty system spend less time on the M/M/1. [Queries arriving at a system with at least one query present spend them same time on average]
A: To answer the question we will compare a few parameters of the systems:


*

*$\rho$ - the servers utilization: In general the utilization for
an $M/M/c$ system is 
$$
\rho=\frac{\lambda}{c\mu}
$$
For the $M/M/1$ system:
$$
\rho=\frac{\lambda}{\mu}
$$


for the $M/M/2$ system we have a rate of $2\lambda$ so 
$$
\rho=\frac{2\lambda}{2\cdot\mu}=\frac{\lambda}{\mu}
$$
so the utilization is the same.


*$P_{0}$ - the probability that there are no costumers in the queue
(all servers are available): for the $M/M/1$ system:
$$
P_{0}=(1-\rho)\rho^{0}=1-\rho
$$
for the $M/M/2$ system: 


$$
P_{0}=\frac{1}{\sum_{n=0}^{c-1}\frac{(c\rho)^{n}}{n!}+(c\rho)^{c}\cdot\frac{1}{c!}\cdot\frac{1}{1-\rho}}
$$
for $c=2$
$$
P_{0}=\frac{1}{\sum_{n=0}^{1}\frac{(c\rho)^{n}}{n!}+(2\rho)^{2}\cdot\frac{1}{2}\cdot\frac{1}{1-\rho}}
$$
$$
P_{0}=\frac{1}{1+2\rho+2\rho^{2}\cdot\frac{1}{1-\rho}}
$$
$$
=\frac{1}{1+2\rho+\frac{2\rho^{2}}{1-\rho}}
$$
$$
=\frac{1}{\frac{1-\rho+2\rho(1-\rho)+2\rho^{2}}{1-\rho}}
$$
$$
=\frac{1-\rho}{1-\rho+2\rho(1-\rho)+2\rho^{2}}
$$
$$
=\frac{1-\rho}{1-\rho+2\rho-2\rho^{2}+2\rho^{2}}
$$
$$
=\frac{1-\rho}{1+\rho}
$$
we have concluded that both systems have the same utilization $\rho$
thus, since $1+\rho>1$ (unless $\lambda=0$ and in that case the
entire discussion is degenerated) 
$$
P_{0,M/M/2}=\frac{1-\rho}{1+\rho}<1-\rho=P_{0,M/M/1}
$$
thus the system is more empty in the $M/M/1$ system!


*$L$ - the average number of customers in the system: In the $M/M/1$
system:


$$
L_{M/M/1}=\frac{\rho}{1-\rho}
$$
In the $M/M/2$ system
$$
L=2\rho+\frac{(2\rho)^{3}}{2\cdot2\cdot(1-\rho)^{2}}\cdot\frac{1-\rho}{1+\rho}
$$
$$
=2\rho+\frac{2\rho^{3}}{1-\rho}\cdot\frac{1}{1+\rho}
$$
$$
=2\rho+\frac{2\rho^{3}}{1-\rho^{2}}
$$
$$
=\frac{2\rho}{1-\rho^{2}}=\frac{2}{1+\rho}\cdot\frac{\rho}{1-\rho}
$$
thus 
$$
L_{M/M/2}=\frac{2}{1+\rho}\cdot L_{M/M/1}
$$
the question of 
$$
L_{M/M/2}>L_{M/M/1}
$$
comes down to does $\lambda>\mu$ .


*$w$ - the average time a costumer is in the system: 
$$
w_{M/M/1}=\frac{1}{\mu(1-\rho)}=\frac{\rho}{\lambda(1-\rho)}
$$
$$
w_{M/M/2}=\frac{L_{M/M/2}}{2\lambda}=\frac{\rho}{\lambda(1-\rho)(1+\rho)}=\frac{1}{1+\rho}W_{M/M/1}
$$


thus, on average, a customer waits more in an $M/M/2$ system.


*$W_{Q}$ - the average time a customer is in a queue
$$
W_{Q,M/M/1}=w_{M/M/1}-\frac{1}{\mu}
$$


and 
$$
W_{Q,M/M/2}=w_{M/M/2}-\frac{1}{\mu}
$$
and 
$$
W_{Q,M/M/2}>W_{Q,M/M/2}
$$
iff 
$$
w_{M/M/2}>w_{M/M/1}
$$


*$L_{Q}$ - the average number of customers in queue
$$
L_{Q,M/M/1}=\frac{\rho^{2}}{1-\rho}
$$


and 
$$
L_{Q,M/M/2}=2\lambda W_{Q,M/M/2}=2\lambda(w-\frac{1}{\mu})
$$
$$
=2\lambda w-2\frac{\lambda}{\mu}
$$
$$
=2\lambda(\frac{\rho}{\lambda(1-\rho)(1+\rho)})-2\rho
$$
$$
=\frac{2\rho}{(1-\rho)(1+\rho)}-2\rho
$$
$$
=\frac{2\rho-2\rho(1-\rho^{2})}{(1-\rho)(1+\rho)}
$$
$$
=\frac{2\rho-2\rho+2\rho^{3}}{(1-\rho)(1+\rho)}
$$
$$
=\frac{2\rho^{3}}{(1-\rho)(1+\rho)}
$$
thus $L_{Q,M/M/2}>L_{Q,M/M/1}$ iff 
$$
\frac{L_{Q,M/M/2}}{L_{Q,M/M/1}}>1
$$
iff 
$$
\frac{2\rho^{3}}{(1-\rho)(1+\rho)}\cdot\frac{1-\rho}{\rho^{2}}>1
$$
iff
$$
\frac{2\rho}{1+\rho}>1
$$
iff 
$$
2\rho>1+\rho
$$
iff
$$
\rho>1
$$
iff
$$
\lambda>\mu
$$
To conclude - the utilization of both systems are the same, the system
is more empty in the $M/M/1$ system and the average number of customers
in the system, the average time a costumer is in the system, the average
time a customer is in a queue and the average number of customers
in queue are greater in the $M/M/2$ system iff $\lambda>\mu$.
