- A is the arrival time distribution
- B is the service time distribution
- c is the number of servers
- K is the capacity of the queue
- m is the number of customers
- Z is the queue discipline

If the capacity and the number of customers are unbounded and the queue is FCFS this is shortened to A/B/c. Common distribution abbreviations are:

- G for a completely general distribution
- GI for a general distribution where interarrival or service times are independent
- D for deterministic, or constant
- M for memoryless, which leads to Poisson arrivals and exponential service
- E for Erlang
- H for hyperexponential

rq_{0} = mq_{1} and

(r + m)q_{i} = rq_{i-1} + mq_{i+1} for i > 0

We can rewrite the second equation as

rq_{i }- mq_{i+1} = rq_{i-1} - mq_{i}

which are all zero by the first equation, so

q_{i} = (r/m) q_{i-1} = (r/m)^{i}q_{0}

The sum of these probabilities is a geometric series that must sum to 1,
so q_{0} = 1 - r/m, assuming r < m. If the arrival rate is greater than
or equal to the service rate, there is no stationary distribution and the queue
will grow without bound.

We can now evaluate the following (assuming r < m):

- The server utilization is the proportion of time the server is busy.
Assuming that the request at the head of the queue is the one being served
and that it is not removed from the queue until service is completed then
the server is busy exactly when the queue is nonempty, so the server
utilization is r/m = 1 - q
_{0}. - We can use the fact that the queue length is a geometric random variable with parameter r/m to compute the average number of requests in the system as r/(m-r).
- Similarly the variance of the number of requests in the system is rm/(m-r)
^{2} - Little's formula (which we don't prove) states that the average number of requests in the system is the arrival rate, r, times the average response time. Solving for the average response time gives 1/(m-r)

rq_{0} = mq_{1}

(r + im)q_{i} = rq_{i-1} + (i+1)mq_{i+1}
for 0 < i < c and

(r + cm)q_{i} = rq_{i-1} + cmq_{i+1}
for i >= c

Solving these in the same way as above gives

q_{i} = q_{0} (r/m)^{i} /i! for i < c and

q_{i} = q_{0} (r/m)^{i} /(c! c^{i-c})
for i >= c

Solving for q_{0} gives

1/q_{0} = cm(r/m)^{c}/(c! (cm-r) ) + sum from 0 to c-1 of
(r/m)^{i}/i! for r < cm

From this we can derive the following:

- The average number of requests in the system is
r/m + (r/m)

^{c}* rmcq_{0}/(cm - r)^{2}c! - If M is the number of busy servers, then P(M = i) = q
_{i}for i < c and P(M = c) = cmq_{c}/(cm - r) = cmr^{c}q_{0}/(cm - r)c!m^{c}, so E[M] = r/m - The probability that a new request has to wait is P(M = c), given by the above formula, known as Erlang's C formula
- Little's formula can again be used to calculate the average response time from the average number of requests in the system

**References**:

Operating Systems, Second edition

H. M. Deitel

Addison-Wesley, 1990

Probability and Statistics with Reliability, Queuing, and Computer Science
Applications

Kishor S. Trivedi

Prentice-Hall, 1982.