5.3.2 General Model
We consider a region B served by two response units, unit 1
and unit 2. Each unit, while not busy at a service request, is located
at a fixed home location or facility. The system operates in the
following way:
 If X is any subset of B, we assume that
service requests arrive in X according to a Poisson process with
parameter (X), with arrivals in any two
disjoint subsets being independent.
 The region is partitioned into two primary response areas,
area 1 comprising the set A and area 2 comprising the set B 
A. Requests for service arrive from area 1 with rate (A) _{1}, and from area 2 with rate (B  A)
_{2}. Set =
_{1} + _{2}.
 If an area n service request arrives when both
units are available, it is served by unit n, n = 1, 2.
 If an area n service request arrives when exactly
one unit is available, it is served by the available unit, n = 1,
2.
 If an area n service request arrives when neither
unit is available, the request is lost. As discussed above, this
assumption usually implies that the request is handled by some backup
service system.
 The system is operating in the steady state.
 The service times for all service requests are identically
distributed with a finite average 1 / ,
independent of the history or the state of the system at arrival, the
location of the request, and the identity of the serving unit.
Each server can be in one of two states: 0, corresponding to
available or free; and 1, corresponding to busy
servicing a request. There are thus 2^{2} system states, as
follows:
00:  both units are available 
01:  unit 1 is busy and unit 2 is available 
10:  unit 1 is free and unit 2 is busy 
ll:  both units are busy 
(Note that the status of unit n is given by the nth
digit from the right in the binarystate depiction.) As we will see in
the next section, the binarystate classification extends in a natural
way to N servers, in which case the system has
2^{N} states.
The statetransition diagram for this fourstate process is shown in
Figure 5. 1. While the steadystate probabilities do not depend on the
exact form of the service time distribution [CHAI 731, we will assume
negative exponential service times, in order to write the equilibrium
equations of balance in accordance with the procedures of Chapter 4:
 Conservation of flow about state 00
P_{00} = (P_{01} +
P_{10})
 Conservation of flow about state 01
P_{01}( + ) = P_{00}_{1} + P_{11}
 Conservation of flow about state 11
P_{11}2 = P_{01} + P_{10}
 Sum of probabilities
P_{00} + P_{01}
P_{10} + P_{11} = 1
Substituting (5.8) into (5.6) yields
Substituting (5.10) into (5.7) yields
By symmetry we must have
Finally, substitution into (5.9) yields (after some manipulation)
We notice that P_{00} is the same as P{S_{0}} in an
M / M / 2 system with no waiting space. This is just what we
expect: the twoserver system that we are analyzing is simply an M /
M / 2 system with distinguishable servers with no waiting space.
Defining = /2,
_{1} = _{1}/,
_{2} = _{2}/,
we have for the state probabilities
We define the workload of unit n to be the steadystate
probability that unit n is servicing a request, and we denote it
by _{n} Clearly,
The difference in workload , which is a measure of the extent to which one unit
works harder than the other, is thus
Another quantity of interest is
f_{nj} = fraction of answered service requests that take
unit n to response area j
This "interarea dispatch frequency" allows us to compute mean travel
times. Define
T_{n}(C) average time for
unit n to travel to a service request in region C
(C) = average systemwide travel
time, assuming that unit 1's primary response area is the region
C
Then, for instance, T_{n}(B) is the average travel
time if every serviced request (in B) is answered from location
n. Since region A is unit 1's primary response area, the
average systemwide travel time to serviced requests can be written
How do we obtain the f_{nj}'s? Invoking an argument of
Chapter 2, consider a long time interval T. In the steady state,
the average total number of requests serviced is (1 
P_{11})T; the
average number of requests that take unit 1 to region 2, for instance,
is equal to _{2}T
multiplied by the fraction of time the system is in state
P_{10} Thus, the fraction of all serviced requests
arriving in T that take unit 1 to area 2 is
In a similar manner, we find that
Substituting (5.15) and (5.16) into (5.14), we obtain for the mean
travel time,
This formula can be used to calculate mean travel time for any
proposed primary response areas A and B  A.
