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 $\lambda>\mu$.