In the next section several simple systems are investigated with the aim that understanding their performance more complicated ones could be analyzed.

Example 4.2. Let us consider a component having two states (0 if it operating, 1 if it failed ) and let us suppose that at time it is operating. Find the probability that at time it is failed assuming that the operating times are exponentially distributed random variables with parameter and they are independent of the the repair times that are exponentially distributed random variables with parameter .

Solution:

To formulate the problem in mathematical terms let introduce the following notations.

Let

furthermore its distribution is denoted by

Then by the help of the theorem of total probability and the memoryless property of the exponential distribution we get

It is easy to see that after substitution and rearranging the terms we have

Thus

Hence the calculations have reduced to

which is a first-order inhomogeneous linear differential equation with constant coefficients. Its solution can be obtained in different ways. In the next part we show how to get the solution by applying the Laplace-transform method.

In the mean time we shall use the following properties of the Laplace-transform

and the Laplace-transform of a constant c is .

Taking the Laplace-transform of both side and keeping in mind these properties the transformed differential equation can be written as

By using the method of partial fractions we get

resulting

thus

Therefore

Knowing that

inverting the terms we obtain the solution, namely

If the initial condition is , then the solution is

Steady-state (stationary) distribution

Let . Then taking the limits in the corresponding equations it is easy to see that the solution is

Notice that we have the same distribution and the the initial condition has no effect on the limiting distribution.

Example 4.3. Find by the help of the steady-state balance equations.

Solution:

Since in steady-state the functions do not depend on the time their derivatives are zero. Hence from the corresponding differential equation and the normalizing condition we have

Example 4.4. Find by the help of the expectations of the operating and repair times.

Solution:

Let , denote the operating times, repair times of the component, respectively. Let us assume that all these times are independent of each other.

As the time goes the states of the component alternate, and create so called cycle which are independent of each other. It can be proved that the stationary distribution that the component is operating is the ratio of the mean operating time and the mean cycle time.

In the case of exponentially distributed times

In the reliability theory the distribution of the time to the first system failure plays a very important role. Obviously, this distribution should depend on the initial condition of the system. The aim of the next Example to illustrate this topic.

Example 4.5. Let us consider a system consisting of two components having exponentially distributed operating times with parameter . The failed component is maintained by a single repairman and the repair times are supposed to be exponentially distributed random variables with parameter . If both components are failed the system is said to be failed and the whole operation stops. Assuming that at the beginning all components are operating and they are independent of each other find the mean time to the first system failure.

Solution:

As in the previous Example let denote the number of failed components, . Since if the process enters to state the system stops it is an absorbing state. It is not difficult to see that the transition rates between the states can be illustrated as follows

Figure 4.1. Transition rates in Example

It can easily be verified that for the distribution of the system the following differential equations with initial condition can be written

It is enough to solve and since

Taking the Laplace-transform at both sides and using the initial condition we have

Thus

Distribution of the time to the first system failure

Let denote the time to the first system failure.

It is easy to see

Thus

since

Therefore

Specially, if , which means that there is no repair, this formula simplifies to

as we could see in the case of a parallel system.

Of course, the density function of the time to the first system failure can be obtained this way. Let us see what to do get it.

Determination of density function

To get let us notice that . Using the properties of the Laplace-transform this can be transformed to

Alternatively, at balance equation

taking the Laplace-transform we have

where

Thus

Therefore

Example 4.6. Modify the initial condition and let us assume that the system operation start with 1 operating component. Find the mean time to the first system failure.

Solution:

Let us notice that we have the same system of differential equations just the initial condition has changed. That is

Similarly to the previous solution we have

Thus

Hence the mean is

In particular, in the case of non-maintained system, that in when it reduces to which shows that our calculation is correct.

Let denote the mean time to the first system failure with initial state . On the basic of the previous calculations we have

It is easy to see that , since

which was expected.

Example 4.7. Find the steady-state probability that components are operating in a system containing of independent components.

Solution:

The key to this problem is the binomial distribution with parameters since in steady-state the probability that a given component is operating is . Since we have components the probability in question is

Mean operation time of a parallel system

Let denote the steady-state availability coefficient of a system, that is the steady-state probability that the system is operating. As we have seen a parallel system is operating if there exists operating component. In other words it is failed if all the components are failed. In the case of a system containing of independent components having exponentially distributed operating time, repair times with parameters , respectively, this probability is given by

Thus

Let denote the mean sojourn time of the system in failed state and let denote the mean operating time of the system. Then

Therefore

In the case of a parallel system it has the form of

The term is the mean time while all the components are failed. Since this time is the minimum of the repair times it is exponentially distributed with parameter because the repair times are exponentially distributed with parameter .

Example 4.8. Let us consider a system with two components and two repairmen. Let denote the number of failed components. Assuming independent exponentially distributed operating and repair times derive the corresponding set of differential equations.

Solution:

Similarly as we have done earlier it is easy to see that the transition rates can be written as it is illustrated and hence the differential equations can easily be derived in the usual way, that is we have

Figure 4.2. components, 2 repairmen

Example 4.9. Let as consider the previous Example with a single repairman. Find the steady-state distribution, the mean operating time of the system, and the mean busy period of the repairman.

Solution:

Of course we have to modify the repair rates, as it is illustrated, but after that the steady-state balance equations can easily be derived in the usual way as follows

Figure 4.3. 2 components, 1 repairman

It is easy to verify that the solution is

For the mean operating time we have the general formula, namely

Since we have only a single repairman and the repair time is exponentially distributed thus . The availability coefficient .

To answer the third question let us introduce the following notations. Let be the mean idle time and be the mean busy period of the server. Since

thus

This time , and in the case of components it is , since the idle time is the minimum of the operating times of the components.

Example 4.10. Compare the mean operation time and of the systems with and repairmen.

Solution:

In the case of two repairmen

As we have calculated earlier

As we have shown in Example 4.5 the mean time to the first system failure starting with two operating components is

It is easy to see that

In the case of a single repairman

Thus

Surprisingly there is no difference in the two cases. Of course the steady-state distributions are different, but nevertheless the mean value is the same.

So far we have investigated systems with homogeneous components. In the next section we are dealing with systems with heterogeneous elements resulting more complicated set of differential equations and formulas.

Example 4.11. Let us consider a system with heterogeneous components and with 2 repairman. The th component has exponentially distributed operating times and repair times with parameter and , respectively, i=1,2.

Assuming that the involved random variables are independent of each other find the transient distribution of the system starting with 2 operating components. Furthermore, in steady-state compute the mean operating time of this parallel system.

What is the mean operating time without repair ?

Solution:

To describe the behavior of the system we need more sophisticated notations since we have to keep in mind the heterogeneity of the components. Thus let us denote by the state when both components are operating, by when component with index is failed, by when component with index is failed, and finally by when both components are failed. The transition rates are illustrated in the following Figure, showing a more complicated situation. By the help of these rates the corresponding differential equations can be written in the traditional way.

Figure 4.4. Heterogeneous case with 2 repairmen

Transient distribution

After elementary calculations we can verify that the solution is

Further performance measures are

Steady-state distribution

Let . As we have seen earlier

Thus it is easy to see that

Therefore the mean operation time of the system is

Example 4.12. Let as consider the previous system with a single repairman.

Find the steady-state performance measures under different service disciplines.

Assuming that at the beginning both components are operating find the mean time to the first system failure in the case of a parallel system.

Fist-In First-Out (FIFO) discipline

As usual, first we have to introduce the states of the system keeping in mind the order of arrivals of the failed components. Thus

The transition rates are illustrated in the following Figure. The set of steady-state balance equations can be written as usual.

Figure 4.5. FIFO discipline

The equations and the normalizing condition are

After these by elementary but lengthly calculations one can verify that the solution is

Hence the distribution of the number of failed components can be computed as

It is easy to see that the main performance measures are

Processor Sharing (PS) discipline

Under this discipline the order of arrivals is not significant and that is why if both components are failed it is denoted by . However, in this state the repair intensity is halved.

The transition rates are illustrated in the following Figure

Figure 4.6. Processor Sharing discipline

The steady-state balance equations and the normalizing condition can be written as

The solution is much simpler, namely

The main performance measures are

Preemptive Priority discipline

Under this discipline component with index has preemptive priority over component with index . This means if component fails when component is under repair the service process stops and the service of component starts immediately. In other words, service of component can be carried out only when component is operating. It is easy to see that the states remain the same but the transition rates are different as it is illustrated in the following Figure.

Figure 4.7. Preemptive Priority discipline

For the steady-state balance equations we have

The distribution of the number of failed components can be obtained as

It is not too difficult to verify that the solution is of the form

For the performance measures we get

Knowing the distribution in all 3 disciplines it is quite simple to get the distribution for the homogeneous case. Namely, we obtain

as we have seen in the earlier Examples.

Mean time to the first system failure

To get the distribution of the time to the first system failure one can easily see that the service discipline has no effect on it if we have only two components. Omitting the tiresome Laplace-transform method the mean time can be obtained relative simple by probabilistic reasoning. To do so we need the following notation.

Let denote the mean time to the first system failure starting from state , . By the theorem of total expectation and the properties of the exponential distribution the following equations can be written

After elementary calculations we obtain

Solution:

In particular, if , that is when there is no repair this formula reduces to the result of Example 2.3, that is

By applying the theorem of total moments the second moment of these times could be calculated and thus the variance of could be obtained.

In particular, if , that is when there is no repair this formula could reduce to the result of Example 2.3.