Exercise 9.5. Solve the following first-order inhomogeneous linear differential equation

Solution:

The homogeneous part is

A particular solution of the inhomogeneous part can be obtained by appying the method of variation of parameters ( or variation of constant ), that is

Thus a particular solution is

Hence for the general solution we have

Taking into account the initial condition we obtain thus .

Therefore

If the initial condition is then the solution is

Taking the limit as for the steady-state distribution we get

Exercise 9.6. Find the probability that at time , components are operating provided that at the beginning components were operational and were failed.

Solution:

For the steady-state distribution we have

Cold reserve

Exercise 9.7. Let us consider a system containing a main component having an exponentially distributed operating time with parameter As soon as it fails a reserve unit start operation with the same probabilistic manner. There are 2 repairmen and the service times are supposed to be exponentially distributed random variables with parameter Assuming that the involved random variables are independent find the main stead-state performance measures of the system.

Solution:

It is easy to see that the transition rates are the following

Figure 9.1. Cold reserve

In stationary case let us introduce the usual notations, that is denote by the probability that components are failed.

The solution can be obtained up to a multiplicative constant

which must satisfy the normalizing condition, that is

The mean operating time of the system is

Warm reserve

Figure 9.2. Warm reserve

Exercise 9.8. In this case both components can operate simultaneously but the reserve's failure rate is (). Find the usual performance measures.

Solution:

In this case the balance equations are

The solution is

where

Finally

Exercise 9.9. Let us consider a component which in case of failure needs a detection time before repairing. This time is supposed to be an exponentially distributed random variable with parameter . Find the steady-state distribution of the system.

Figure 9.3. Detection time

Solution:

Similarly to the previous parts it easy to see that we can obtain the steady-state balance equations as

Thus

Using the normalizing condition we have

Exercise 9.10. Assume that we have a two-component parallel-redundant system with a single repair facility. The operating times for both components are supposed to be exponentially distributed random variables with parameter and the repair times are also exponentially distributed with parameter .When both components have failed, the system is considered to have failed and no recovery is possible, in other words it is a parallel system with repairs. Let us denote by , the number of failed components and let the system start from state 0, that is both components are operating. Find the mean time to the first system failure supposing that the involved random variables are independent of each other.

Figure 9.4. Parallel-redundant system

Solution:

The transient probability distribution of the system can be obtained from the following balance equations

For the differential equations we have

with the above initial conditions. This case, however, we need the time dependent solution, that is we have to solve the system of differential equations. Using Laplace-transform we obtain

After simple calculations we get

Since , it is clear that thus we have

By inversion can be obtained, that is at time t the system is failed since there is no operating component.

Let be denote the time to the first system failure.

Then means that the operating time of the system is less than . Hence the reliability function of the system is

Using the technique of Laplace-transform we get

The denominator can be written in the form and thus we can use the method of partial ratios. So

Therefore

Hence

The mean time to the first system failure can be computed in the following way

It should be noted that can be determined without the density function, since its Laplace-transform is known.

Thus

In the case when the component are not repaired, that is when we get a parallel system treated earlier. After substitution we have

and then

This can be interpreted as follows. The system failure time is the sum of the time of the first failure of the components and the residual operating time of the second component. The first failure is exponentially distributed with parameter and the remaining time is also exponentially distributed with parameter , furthermore they are independent of each other.

Exercise 9.11. Let us modify the previous system in the following way. The repairs are carried out by 2 repairmen and assume that the repair starts when both components are failed. Find the steady-state characteristics of the system.

Figure 9.5. Exercise

Solution: Let us introduce the following notations

It is easy to see that the steady-state balance equations are

For the solution we obtain

Using the normalizing condition we have

The availability A of the system is

Furthermore, for the mean operating time of the system we get