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 $\lambda>\mu$.