Mathematical Methods of Calculating the Reliability of Standby Systems with Renewal

The mathematical methods of calculating the reliability of renewable systems based on the use of Markov's random processes (RYABININ, L., 1976). The simplest case of calculating the reliability of a renewable system without redundant units has been given. The availability and the reliability of a lightly loaded standby redundant system have been considered. Explicit results are obtained in the case of two main units and two standby redundant units.


The Reliability of a Non-Redundant System with Renewal
Complete and partial failures repeatedly occur in any system during the time of its operation. Each time a failure is remedied the capability of the system is restored and the system continues to serve its purpose, the time of operation of a system between two failures forms only an insignificant part of its service. Life is determined by the total duration of operation until the system is entirely worn out (when it is acknowledged as fully unsuitable for further operation).
Maintenance is eliminating the failure by repair; the repairs of the failed units are carried out by a group of servicing elements.
As in the case of redundancy, renewal is a means of increasing the reliability of systems intended for prolonged use. Today, many systems rank precisely among the category of renewable systems and it is therefore necessary to know the methods of calculating the reliability of a system with renewal.
Let us consider a renewable non-redundant system intended for prolonged functioning and which can be either in an operative or inoperative state at each specific moment of time t of its period of operation, t 0 , depending on the tasks imposed on it. The loss of capability, i.e., failure of the system, may occur in the operative state of the system (when it is functioning according to its purpose) and in the inoperative state (when the system may remain idle at a given moment). In the latter case failure may be eliminated before need arises of starting it again, or everything left as it is (because repairs require more time than may be permitted by the circumstances).
The problem is to determine the probability B(t) of finding the system in failure-free operation at the moment t and the probability R(T) of unfailing operation of this system during the time T beginning from the moment t . The first probability describes the reliability of the system from the point of view of its preparedness for immediate action at any moment of time t ,and the second probability describes the reliability of the system from the standpoint of its unfailing functioning in the course of the required interval of time T.
We solve this problem with the two essential assumptions: (i) The life time of the system is a random variable obeying the exponential distribution with rate λ = constant.
(ii) The repair time of the system is a random variable obeying exponential distribution with rate μ = constant.
Denote by E the state of the system at any moment; E equal to E 0 when the system is operative and equal to E 1 when the system is inoperative. Let P 0 (t) , P 1 (t) be the probabilities that the system is in the operative and    Vol. 14, No. 4;2020 entirely determined by the intensity of failure λ. In the absence of renewal (μ=0), formula (1.6) is transformed into the expression.
The efficiency of renewal can be estimated from the quantitative aspect by the coefficient K e (t) which is equal to the ratio of the probabilities determined from formulas (1.6) and (1.8): or if t is large enough, then = (1.10) If t=0,then = 1 1 + As could "be expected this efficiency increases with a growth in the intensity of renewal μ , the intensity of failures λ , and time t .

The Availability of Standby Redundant System with Renewal
Let us consider a renewable reserved system consisting of N operating and m standby single-type units. This system functions as follows. The task is posed with the system performed by a group of N units. When one of the units in this group fails it is instantaneously replaced by a redundant one and the faulty unit goes for renewal. The servicing system is composed of L units. If all the servicing units are engaged, the faulty unit queues for renewal. The renewed unit is kept as a standby. In the general case, redundant units may also prove faulty. In this case they are also delivered to the servicing system for renewal. This system is known as the lightly loaded standby system with renewal.
Let us now describe the mathematical aspects of this system. The system fails when m+1 units have failed. Let us take it for granted that the system is in state E K when the number of the failed units is K. Obviously, the system is in good order in the states E 0 , E 1 , E m . The state E m+1 is the state of failure of the system; the states, E m+2 , E S (S>N+m) are regarded as impossible.
Let the life time of a main unit be distributed according to the exponential law with parameter λ , and that of a standby unit with parameter λ , where the coefficient may range from 0 to 1. Also, let the time needed to renew one unit by one servicing unit has an exponential distribution with parameter μ. If the system is in a state E K when there are N operating units, and k failed units out of the m standby units, the intensity of failures among the operating and standby elements will be Let us denote the probability of finding the system in the state E k at any moment t by P K (t) (t). We shall naturally be interested in the probabilities of all possible states P 0 (t), P 1 (t), P m=1 (t) in order to determine the system availability. According to the condition of the problem, the probabilities P m+2 (t) =…=P s (t) =0.
We derive the difference differential equations governing the system. By connecting the probabilities at time t + ∆ with those at time t and taking limits as Solving the system (2.4) consecutively for k=1.2…, m+1, we find that From the normalized conditio ∑ p = 1, we get The availability as the probability of finding not more than m units faulty at any moment are equal to Explicit results for the case of two main units (N=2) and two standby units (m=2), in which, case λ=0.1×1/h μ=1×1/h and =0.1 are given below.
Let us consider some particular cases for the solution of this problem assuming the different natures of the standby (lightly loaded, loaded and nonleaded) and also the different kinds of repair (unlimited and limited).
With the lightly loaded standby for r=0.5 from formula (2.1) we have n o = 3, n 1 =2.5 and n 2 =2 In the case of an unlimited renewal, the system availability can be found from formula (2.7): For the limited renewal (L=1), the coefficients are everywhere equal to unity when μ is in the system of equations (2.3). It can be shown that in this case when r=1.
where S is the total number of single-type units functioning in the system. When S > N + m some of the units are inoperative (being renewed or queueing for maintenance). Then The table shows that the results of calculating the availability numbers of this system correctly describe the relationships between the values of B (∞) for various cases of reserving and renewal

The Reliability of a Lightly Loaded Standby Renewable System
Let us take a system that satisfied the conditions indicated in the previous section and find the probability R(T) of unfailing operation of the system within the times interval [0,T]. The system passed into the state failure with the failure of the (m+l) -th unit and the operating state with the renewal of the N-th. unit.
The failure of the (m+1)-th. unit may occur for the first time before the moment T+∆T in the following two mutually exclusive cases: (a) either the failure occurs before the moment T; this event has a probability p m+1 (T), (b) or the m-th failure occurs before this moment, and during the time ∆T one more, (m+1)-th, failure occurs; this event has probability p m (T)n m λ ∆T.

P m+1 (T+∆T) = P m+1 (T)+n m λ ∆TP m (T) (3.1)
When ∆T  0 we get in the limit In order to solve this equation it is necessary know the value of P m (t) which is the probability of finding m units faulty at the moment T not necessarily for the first time, But this value will provide the solution for the system of differential equations (2.3) if m is taken instead of m+1.
Having determined the probability P m+1 (T) this probability of unfailing operation of the system with a repairable standby can be evaluated from the formula.

R(T)=1-P m+1 (T) (3.3)
Applying the Laplace transformation to the system of equations (2.3) and taking account of equation ( where * ( ) is Laplace transform of the probability P K (t).
Let us solve this system with respect to * + 1( ) for the initial conditions.
(3.7) and ∆ (k) (z) is the determinant of system (3.4) in which the column of the coefficients of * ( ) is replaced by the column of free terms for the accepted initial conditions (3.5); (3.8) Writing the determinant (3.7) as the product of factors: where the are the roots of the equation and writing the determinant ∆ (K) (z) in the form of a polynomial. where: B 0 =1.
In order to obtain all the roots of Z i , let us solve equation (3.10 ) of degree (m+1) . The coefficients B k , B k-1 ,...B 1 can be found from the expression.
It can be shown that the coefficient Bk is equal to the product. which when inserted into the first term of expression (3.13), will give because the free term of the algebraic equation (3.10) is present in the numerator and the denominator of (3.16).
Taking account of equality (3.16) , a formula can be found from (3.13) for the probability of unfailing operation of the system with a repairable standby , which is true for the initial conditions (3.5): The mean life time of the system in concern is equal to

Special Case of Two Main Units and Two Standby Redundant Units
Let us now consider several quantitative examples for a system composed of two operating (N=2) and two standby (m=2)

Lightly Loaded Standby
a-Only two operating units are in good order at the initial moment, and two standby units can be renewed in the process of functioning of the system. The standby units will be in relieved operating conditions with a coefficient r =0.05. In this case we have in conformity with formula (2.2) ‫؛‬ n o = 3, n 1 = 2.5, n 2 = 2. The determinant (3.7). will be equal to Solving the cubic equation ∆( ) + 3.75 + 2.98 + 0.015 = 0 We obtain the following values for the roots: Z 0 = -2.610845, Z 1 = 0.005005, Z 2 = -1.13415 The determinant (3.14) will be equal to It can be seen that B k = B 2 = 0.075 and B k-1 =B 1 = 1.55. and Now we use formula (3.17) and find Moduli of the roots Z i are substituted into formula (3.17) because the signs have been taken into account during its derivation.

From formula (3.22)
Moduli of the roots Z i are substituted into formula (3.17) because the signs have been taken into account during its derivation.
We obtain from formula (3.22) M = 200h c-Let us consider the case (a) with the initial conditions P 0 (0) = 1, P i (0) =0, i.e., when all units are operable at the initial moment T=0. We find from formula (3.20) that

Conclusion
The mathematical methods of calculating the reliability of renewable systems based on the use of Markov's random processes. The availability, the reliability and the mean life time of non-redundant and different standby redundant systems with renewal are obtained.
Then we conclusion that renewal together with standby is an effective means of increasing the reliability of a system. In the above cases the mean Life time of the system is equal to: i.e., the mean life time appreciably increases with standby and renewal.