It is the traditional machine interference problem, where the broken machines has to wait and the single repairman fixes the failed machine in FIFO order. Assume the the operating times are xponentially distributed with parameter and the repair rate is . All random variables are supposed to be independent of each other.

Let denote the number of customers in the system at time , which is a birth-death process with birth rates

and with death rate

Thus for the distribution we have

where

and

Since the state space is finite the steady-state distribution always exists but then more repairmen is needed.

For numerical calculation other forms are preferred that is why we introduce some notations.

Let denote the Poisson distribution with parameter and let denote its cumulative distribution function, that is

First we show that

where

By elementary calculations we have

Hence a very important consequence is

The main performance measures can be obtained as follows

1. Utilization of the server and the throughput of the system

For the utilization of the server we have

By using the cumulative distribution function this cab be written as

For the throughput of the system we obtain

2. Mean number of customers in the system can be calculated as

In other form

3. Mean queue length, mean number of customers waiting can be derived as

4. Mean number of customers in the source can be calculated as

5. Mean busy period of the server

Since

thus

In computer science and reliability theory application we often need the following measure.

6. Utilization of a given source ( machine, terminal )

The utilization of the th source is defined by

Then

Hence the overall utilization of the sources is

Thus the utilization of any source is

This can be obtained in the following way as well,

since the source are homogeneous we have

7. Mean waiting time

By using the result of Tomkó we have

Thus

and

which the Little's law for the mean waiting time. Hence

The mean response can be obtained as

It is easy to prove that

which is the Little's law for the mean response time. Clearly we have

8. Further relations

and thus

It should be noted that the utilization of the server plays a key role in the calculation of all the main performance measures.

Distribution at the arrival instants

In the following we find the steady-state distribution of the system at arrival instants and in contrary to the infinity-source model is not he same as the distribution at a random point. To show this use the Bayes's theorem, that is

irrespective to the number of servers. It should be noted that this relation shows a very important result, namely that at arrivals the distribution of the system containing n sources is not he same as its distribution at random points, but equals to the random point distribution of a system with n-1 sources.

Distribution at the departure instants

We are interested in the distribution of the number of customers a departing customer leaves behind in the system. This calculations are independent of the number of servers. By applying the Bayes's theorem we have

in the case when there is customer left in the syste

if the system becomes empty.

Recursive Relations

Similarly to the previous arguments it is easy to see that the density function of the response time can be obtained as

Hence the mean value is

Similarly, for the waiting time we have

thus its mean is

which is clear.

We want to verify the correctness of the formula

As we have shown earlier the utilization can be expressed by the Erlang's loss formula, hence

Using the well-known recursive relation we have

Since

thus

After substitution we have

Therefore

Finally

which is a recursion for the mean number of customers in the system.

Now we able to prove our relation regarding the mean response time. Keeping in mind the recursive relation for we get

which was proved earlier.

Now let us show how we can verify directly. It can easily be seen that

that is there is a recursion for the utilization as well. It is also very important because by using this recursion all the main performance measures can be obtained. Thus if are given we can use the recursion for and finally substitute it into the corresponding formula. Thus

Since

we proceed

which shows the correctness of the formula.

In the following let us show to compute ecursively. As we have seen

we have to know how can be expressed in term of .

It can be shown very easily, namely

The initial values are

Now the iteration proceeds as

that is we use a double iteration. The main advantage is that only the mean values are needed. This method is referred to as mean value analysis.

In the previous section we have derived a recursion for and thus we may expect that there is direct recursive relation for the other mean values as well since they depends on the utilization. As a next step we find a recursion for the mean number of customers in the source. It is quite easy since

By using this relation for the utilization of the source can be expressed as

For the mean number of customers in the system we have

Since

thus after substitution we get

Finally find the recursion for the mean response time. Starting with

using that

substituting into the recursion for we obtain

Obviously the missing initial values are

Distribution Function of the Response Time and Waiting Time

This subsection is devoted to one of the major problems in finite-source queueing systems. To find the distribution function of the response and waiting time is not easy. As it is expected the theorem of total probability should be used. Let us determine the density function and then the distribution function. As we did many times in earlier chapters the law of total probability should be applied for the conditional density functions and the distribution at the arrival instants. So we can write

Similarly for the waiting time

To get the distribution function we have to calculate the integral

Using the substitution , , .

Hence

Similarly for the waiting time we have

Now let us determine the distribution function by the help of the conditional distribution functions. Clearly we have to know the distribution function of the Erlang distributions, thus we can proceed as

Meantime we have used that

and thus

can be written as

During the calculations we could see that the derivative of is , which can be used to find the density function, that is

Generating Function of the Customers in the System

Using the definition the generating function can be calculated as

This could be derived in the following way. Let denote by the number of customers in the source. As we have proved earlier its distribution can be obtained as the distribution of an Erlang loss system with traffic intensity .Since the generating function of this system has been obtained we can use this fact. Thus

To verify the formula let us compute the mean number of customers in the system. By the property of the generating function we have

thus

Laplace-transorm of the Response Time and Waiting Time

Solution 1.

By the law of the total Laplace-transforms we have

since the conditional response time is Erlang distributed with parameters . Substituting we get

Solution 2.

Let us calculate by the help of the density function. Since the denominator is a constant we have to determine the Laplace-transform of the numerator, that is

By using the binomial theorem we get

Since

thus

Solution 3.

The Laplace-transform of the numerator be can obtained as

Substituting we get

and thus

Substituting again thus

therefore

That is all 3 solutions gives the same result. Thus, in principle the higher moments of the response time can be evaluated.

Since , thus

Example 12.1. Consider machines with mean lifetime of hours. Let their mean repair time be 4 hours. Find the performance measures.

Solution:

per hour, per hour, , ,

Figure 12.1. Values of

Example 12.2. Change the mean lifetime to hours in the previous Example. Find the performance measures.

Solution:

, , , , which shows that a single repairman is not enough. We should increase the number of repairmen.

Figure 12.2. Values of

All these measures demonstrate what we have expected because is greater than 1.To decide how many repairmen is needed there are different criterias as we shall see in Section 12.4.

To avoid this congestion we must ensure the condition where is the number of repairmen.