Assume then that we have established some acceptable average number of processes in the system, call it n. In a steady-state, there are always exactly n processes in the system--equivalently, as soon as one process completes, another one arrives to replace it. Since we are looking at average behavior, we can assume that these n processes are all ``average'' processes and thus also indistinguishable. We can then turn our attention to some more careful analysis/dimensioning of our system. As we have seen, each process alternates between CPU bursts and I/O bursts; since our n processes are ``eternal,'' each will be in an infinite cycle between CPU and I/O bursts. Assuming some knowledge about our CPU and I/O servers and about the characteristic CPU and I/O bursts of our ``average'' processes, we can then analyze the CPU vs. I/O behavior of our system and draw conclusions as to CPU and I/O device use.
For illustration, suppose that we have n=4, 1 CPU server, and 2 I/O servers;
further assume that the service rate of the CPU is 
processes per unit
time, while the service rate of each of the two I/O servers is 
processes per unit time. (These and are unrelated to those
used in the previous section.) Since we have 4 processes, our system can
be in one of 5 states: (a) all 4 processes are in CPU burst; (b) 3 processes
are in CPU burst, 1 is in I/O burst; (c) 2 processes are in CPU burst, 2 in I/O
burst; (d) 1 process is in CPU burst, 3 processes are in I/O burst; and (e)
all 4 processes are in I/O burst. (We need no more states, because the
processes are indistinguishable and because the two I/O devices are
identical.) Naturally, since we only have one CPU server and two I/O servers,
processes will often be waiting for a server of one or the other kind.
As before, we can observe such a system on a time scale small enough that,
between any two observations, at most one event (CPU burst completion or I/O
burst completion) can have taken place. However, the completion rates are
doubled when both I/O processors are active; hence the state diagram looks as
follows.
As before, we can assign a probability to each state; let us denote these
probabilities by 
, 
, 
, 
, and 
. (Naturally, the system
must be in one of these 5 states, so that we have 
.) \
We can write one equation for each state, as before, by assuming steady-state
conditions and writing that the net flow into a state equals the net flow
out of that state. As a result, we get 5 simple linear equations in 5 unknowns;
the algebra gets messy, but the principle remains simple. In our system,
letting 
and 
, we get
From these data, we can compute a number of useful quantities. For instance,
the probability of having all three devices busy is just 
;
the average number of processes waiting for CPU service is 
;
the average number of processes waiting for I/O service is 
;
the percentage of time spent waiting for a server (which is also the percentage
of processes waiting at any given time) is
; and so forth. Such quantities indicate
whether the hardware is being well used; where potential bottlenecks are; and
what the impact of additional resources would be (more or faster servers).
More complex systems, with several unequal servers, can be modelled with the same technique.