service times will be general, we will be using the discrete phase type distribution to represent them based on the fact any discrete general distributions, with finite support, can be represented by a Markov chain (Alfa 2004).
1.2. The Geo/Geo/1 case
Consider the simple Geo/Geo/1 system where the probability of arrival is a and of service completion is b. Let Xn(a, b), n = 0,1,2, ∙∙∙ be the number in the system at time n for a specified pair (a, b), with
For a stable system with
and
where
Next we consider the cases of a = f(b). We require that the following condition applies all the time in this section:
[1.1]
We consider two main cases: one when f(∙) is a decreasing function and the second one when it is an increasing function.
1.2.1. Arrival probability as a function of service completion probability
When we consider the case where a = f (b), we have
and
where
Next, we study the behavior of this system as a function of b.
1.2.1.1. Consider x0(b)
By differentiating x0(b) with respect to b, we have
PROPOSITION 1.1.– By studying the derivative of x0(b) w.r.t. b, we observe
1) if f’(b) is negative, then an increase in b leads to an increase in the probability that the system is empty;
2) if f’(b) is positive, then the probability that the system is empty could be increasing or decreasing, depending on whether respectively. Hence, an increase in the arrival rate as a function of the service rate does not necessarily mean a decrease in the emptiness probability.
We obtain additional information by considering the rate of increase in the probability of empty system as a rate of increase in the service completion probability, i.e.
PROPOSITION 1.2.– If f(b) is convex and f’(b) < 0, then x0(b) is definitely concave. if f(b) is decreasing fast, then x0(b) is increasing slowly, i.e. f(b) decreasing faster than increases. If, however, f(b) is convex and
1.2.1.2. Consider E(X(b))
Next let us consider E(X(b)). By differentiating E(X(b)) w.r.t. b, wehave
Define
For a stable system, we know that b–f(b) and 1–f(b) are always positive, hence both A(b) and B(b) are always positive if
We thus have that the behavior of E(X(b)) depends entirely on f’(b), A(b) and B(b) and whether
PROPOSITION 1.3.– For a stable system,
1) if then E(X(b)) will be decreasing as b increases;
2) if then E(X(b)) will be increasing as b increases.
The first part of proposition 1.3 is straightforward in that an increase in the service rate reduces the arrival, thereby reducing the traffic intensity and decreasing the expected number in the system. However, an increase to the service rate could also lead to an increase in the mean number in the system as observed in the second part of the proposition, which can be explained by the rate at which the arrival probability increases compared to the service completion probability (b), especially if that rate is high.
1.2.1.3. Special cases
CASE 1: Consider the case where f(b) is a decreasing function of b, say for example a = f(b) = 1 – b2, with the strict condition that f(b) = 1 – b2 < b. For this, we have
In this case, as b increases we see that E(X(b)) decreases as well. This is because the arrival probability decreases at a faster rate than service completion increases, i.e. f′(b) = – 2b.
REMARK 1.1.– For the case where a = f(b) = 1 – b2, the arrival probability is a decreasing concave function of the service completion probability and the mean number in the system is decreasing in a convex form.
CASE 2: Next consider the case where f(b) is an increasing function of b, say for example a = f(b) = b2, with the strict condition that f(b) = b2 < b. In this case, we have
In this case, as b increases we see that E(X(b)) increases because the arrival probability increases faster than service completion increases, i.e. f′(b) = 2 b.
REMARK 1.2.– For the case where a = f(b) = b2, with arrival probability increasing