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:

1. 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.

2. 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.

3. If an area n service request arrives when both units are available, it is served by unit n, n = 1, 2.

4. If an area n service request arrives when exactly one unit is available, it is served by the available unit, n = 1, 2.

5. 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.

6. The system is operating in the steady state.

7. 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 2² system states, as follows:

(Note that the status of unit n is given by the nth digit from the right in the binary-state depiction.) As we will see in the next section, the binary-state classification extends in a natural way to N servers, in which case the system has 2^N states.

The state-transition diagram for this four-state process is shown in Figure 5. 1. While the steady-state 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:

1. Conservation of flow about state 00
   P₀₀ = (P₀₁ + P₁₀)

2. Conservation of flow about state 01
   P₀₁( + ) = P₀₀₁ + P₁₁

3. Conservation of flow about state 11
   P₁₁2 = P₀₁ + P₁₀

4. Sum of probabilities
   P₀₀ + P₀₁ P₁₀ + P₁₁ = 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 P₀₀ is the same as P{S₀} in an M / M / 2 system with no waiting space. This is just what we expect: the two-server 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 steady-state 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 system-wide 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 system-wide 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 - P₁₁)T; the average number of requests that take unit 1 to region 2, for instance, is equal to ₂T multiplied by the fraction of time the system is in state P₁₀ 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.