Residence time in multi server system
$Q$ represents the average of a discrete random variable.  Sometimes there are no customers waiting in line, sometimes there's one, and sometimes there are many.  So there is some probability distribution for $Q$.  The difference between a two-server queue and a single-server queue that's twice as fast, is the expected service time when the queue is empty except for the one customer receiving service.  That's the transition from state 1 to state 0 in the Markov chain:



Intuitively, when a customer reaches the head of the queue the service in the two-server system will take twice as long as with the single server that's twice as fast.  The expected service time with the two-servers is $S$, while the expected service time with the single fast server is $S/2$.  What's less intuitive is that customers will, on average, spend somewhat less time waiting to get to the head of the queue with the two-server system.

To see this, consider the case of a customer that arrives when there is one customer currently being serviced.  In the two-server case the arriving customer immediately gets the other server.  The total expected response time in this case is $S$, but that's all service time, no waiting time.  In the fast single-server case the arriving customer waits, on average $S/2$ time for the preceding customer to finish and then gets, on average, $S/2$ amount of service.  The total expected response time in this case is also $S$, but it's half service time and half waiting time.  Let's work out the exact difference:

Fast Single-Server

This is an M/M/1 queue with arrival rate $X$ and completion rate $S/2$.  The steady state distribution, $\pi_i$ for the Markov chain in the fast single-server case is
$$\pi_i = (1-U)U^i$$
where $U$, the server utilization, is $XS/2$.

Then the expected number of customers in the queue is
$$ Q_f = \sum_{i=0}^\infty \pi_i i    =
       \sum_{i=0}^\infty (1-U)U^i i = 
  (1-U)U\sum_{i=0}^\infty U^{i-1} i      =
  \frac{(1-U)U}{(1-U)^2} = \frac{U}{1-U}.$$

Then by Little's law
$$ R_f = \frac{Q_f}{X} = \frac{U}{X(1-U)} = \frac{XS/2}{X(1-U)} = \frac{S/2}{1-U}, $$
but we want to represent $R$ as service time + waiting time, so
$$ R_f = \frac{S}{2} - \frac{S}{2} + \frac{S/2}{1-U} = \frac{S}{2} + \frac{-S/2 + SU/2 + S/2}{1-U} = \frac{S}{2} + \frac{US/2}{1-U} = \frac{S}{2} + \frac{S}{2}Q_f. $$

So we have that the waiting time is
$$ W_f = \frac{S}{2}Q_f = \frac{SU}{2(1-U)}.$$

Two Slow Servers

This is an M/M/2 queue.  We will again let $U = S X/2$ (this is Little's law for the utilization of each of the two servers, the reason the result is the same as in the fast-server case is that now $X$ is halved rather than $S$ being halved.)  The steady state distribution is:
$$ \pi_0 = \frac{1-U}{1+U} $$
and
$$ \pi_i = 2 U^i \pi_0, \;\;\; i \geq 1.$$

Now
$$ Q_s = \sum_{i=0}^\infty \pi_i i = \frac{2U}{1-U^2}.$$
(Derivation left as an exercise.)

Then by Little's law we get
$$ R_s = \frac{Q_s}{X} = \frac{S}{1-U^2} = \frac{2}{1+U}R_f, $$
which we want to represent as service time ($S$) plus waiting time.

$$ R_s = S - S + \frac{S}{1-U^2} = S + \frac{SU^2}{1-U^2} = S + \frac{S}{2} U Q_s, $$
and the waiting time is
$$ W_s = \frac{S}{2} U Q_s = \frac{SU^2}{1-U^2} = \frac{2U}{1+U} W_f.$$

Summary

The two slow servers always have a service time twice that of the fast server.  The two slower servers have a total response time that is $2/(1+U)$ times the fast servers' total response time.  ($0\leq U&lt;1$, so the response time is always worse.)  But the two slow servers have a waiting time that is $2U/(1+U)$ times the fast server's waiting time, and so there's less waiting time with two slow servers.