Cost Benefit Analysis of a Two Unit System Model with the Concept with Service Facility-FCFS

INTRODUCTION

In literature, the stochastic behavior of cold standby system has been widely discussed by many researchers including, Osaki and Nakagawa [1971] discussed a two-unit standby redundant system with standby failure. Nakagawa and Osaki [1975] analyzed stochastic behavior of a two-unit priority standby redundant system with repair. Subramanian et al. [1976] explored reliability of a repairable system with standby failure. Gopalan and Nagarwalla [1985] evaluated cost benefit analysis of one server two unit cold standby system with repair and age replacement. Gupta and Goel [1989] studied profit analysis of two-unit priority standby system with administrative delay in repair. Reliability and availability analysis of a system with standby and common cause failures have been explained by Dhillon [1992]. Lam [1997] studied a maintenance model for two–unit redundant system. Malik [2009] discussed reliability modelling and cost-benefit analysis of a system – A case study. Dhankhar and Malik [2011] studied cost- benefit analysis of system reliability models with server failure during inspection and repair. Bhardwaj and Kaur [2014] analyzed reliability and profit of a redundant system with possible renewal of standby subject to inspection. Recently, Grewal et al. [2017] obtained economic analysis of a system having duplicate cold standby unit with priority to repair of original unit. Rohila and Kumar [2018] analyzed cost benefit of industry having duplicate cold standby unit with different failure rate. It is assumed that the operative unit may fail directly from normal mode and on the contrary, the cold standby unit may be out of order owing to remain unused for a longer period of time or due to any other reason. So far, the cold standby systems with the possibility of server failure have been debated much by many scholars, but the standby unit failure is also of high significance during endeavour. The failure of standby unit highly affects the reliability and availability of the system. Although, a two unit redundant system with standby failure has been much discussed, the concept of standby failure needs more emphasis due to its significant benefaction during study. S0: This state is a regenerative state in which the system is in operative mode with one unit working and another unit is kept as cold standby. There are two possible ways of transition to another state as follows: (i) S0 S1: the operative unit may get failed with rate ‗λ‘ and this failed unit goes for under repair. The standby unit takes its place with probability ‗a‘ which is the probability that cold standby unit is ready for use. (ii) S0S3: the operative unit may get failed with rate ‗λ‘ and the standby unit takes its place but standby unit also found out of order with probability ‗b‘ that standby unit in inoperable situation. S1: This is also a regenerative state in which the system is in operative mode with one unit and another failed unit is under repair. There are three possible ways of transition to another state as follows: (i) S1 S4: the operative unit may get failed with rate ‗λ‘ (ii) S1 S0: the failed unit gets repaired with distribution g(t). (iii) S1 S2: It may be possible that server gets failed /tired with rate ‗μ‘ while repairing the unit. S2: S2 is a regenerative state in which the system is in operative mode with one unit operating and another failed unit is waiting for repair due to server is for taking refreshment/treatment. There are two possible ways of transition to another state are as follows: (i) S2 S1: the server repaired the unit after taking refreshment/treatment. (ii) S2 S7: the operating unit may gets failed with rate ‗λ‘ during the server is busy in taking refreshment/treatment. S3: This is also a regenerative and failed state in which the operative unit failed and waiting for repair, same time standby unit is also out of order due to unused for a long period of time. Server is busy for repairing cold standby unit with condition that server is not allowed to take refreshment when busy with repairing of cold standby unit. At last server repaired (i) S3  S1: the unit gets repaired by the distribution g(t). S4: This is a non-regenerative failed state in which the failed unit under repair continuously from previous state while recent failed unit is waiting for repair due to server can repair one unit at a time. The server may feel the need of refreshment to improve his efficiency with rate ‗μ‘ or server repaired the failed unit. So there are two possible ways of transition state as follows: (i) S4  S1: the unit is repaired that was under continuous repair from the previous state with pdf ‗g(t)‘. (ii) S4  S5: During repairing of the failed unit, server may go for refreshment before completion of his job with rate ‗μ‘. S5: It is non-regenerative failed state in which one failed unit is waiting for repair; another failed unit is also waiting for repair continuously from previous state because server is busy to getting refreshment for increase his efficiency. There is only one possible transition that after taking refreshment, server resumes his work i.e. (i) S5  S6: server got the refreshment with pdf ‗f(t)‘. S6: It is a non-regenerative failed state, in which one failed unit is under repair continuously from previous state while another failed unit is waiting for repair continuously from previous state. There are two possible ways of transition to another state as follows: (i) S6  S5: Again the server may get tired during his repair work of the unit and goes for refreshment to increase his efficiency. (ii) S6  S1: the unit is repaired by the server and the unit assumed a new. S7: This is a non-regenerative failed state in which one failed unit is waiting for repair and another failed unit is waiting for repair continuously from previous state. The server is busy with having refreshment/treatment continuously from previous state. There is only one possible way to transit as follows:

i) S7  S6: The server gets his refreshment or treatment with pdf ‗f(t)‘.

NOTATIONS

E : Set of regenerative states {S0, S1, S2, S3}.

continuously from previous state. Sut / SUT: The server is busy with taking refreshment/continuously busy with taking refreshment from previous state. FWRFwr/ : The failed unit is waiting for repair / waiting for repair continuously from previous state because server is busy with taking refreshment. / µ: Constant failure rate of unit / rate by which server feels tiredness. a / b: Probability that cold standby unit is operable / not operable. Csur : The cold standby unit is under repair. f(t) / F(t): pdf / cdf of refreshment rate by which the server recovers his freshness. g(t)/G(t): pdf / cdf of repair rate of the failed unit. qij(t)/Qij(t) : pdf / cdf of direct transition time from a state Si to a state Sj without visiting any other state. qi,jk(t)/Qi,jk(t) : pdf / cdf of first passage time from a state Si to a state Sj visiting state Sk once in (0,t]. qi,jkrs(t)/ Qi,jkrs(t) : pdf / cdf of first passage time from a state Si to a state Sj visiting state Sk , Sr and Ss once in (0,t]. qi,j;k(r,s)(t)/ Qi,j;k(r,s)(t) : pdf / cdf of first passage time from a regenerative state Si to a regenerative state Sj or to a failed state Sj visiting state Sk , Sr and Ss once or more than one time in (0,t]. Mi(t) : Probability that the system is up initially in state ESi is up at time ‗t‘ without visiting to any other state. )(tWi : Probability that the server is busy in state Si up to time ‗t‘ without making any transition to any other regenerative state or before returning to the same state via one or more states. ijm : Contribution to mean sojourn time µi in state Si when system transits directly to state Sj so that ⓢ/©: Symbol for Laplace Stieltjes convolution / Laplace convolution (desh) : Symbol for derivative of the function.

Simple probabilities considerations produce the following expressions for the non-zero elements

for the above transition probabilities, it can be verified that Let T denotes the time to system failure then the mean sojourn times ( and i) with particular values tetg)( and tetf)( in the state Si are given by

Let )(ti be the c.d.f. of the first passage time from regenerative state Si to a failed state Sj. Regarding the failed state as absorbing state, we have the following recursive relations for )(ti as follows: Taking Laplace Stieltjes transformation of relation (5) and solving for MTSF we get

Let )(tAi be the probability that the system is in up-state at instant ‗t‘ given that the system entered regenerative state Si at t = 0. The recursive relations for )(tAi are given as follows: )(tMi is the probability that the system is up initially in state ESi is up at time t without visiting to any other regenerative state where Now solving for availability0A, the steady state availability is given by

Evaluation of busy period of the server owing to repair of the failed unit Let )(tBi be the probability that the server is busy due to repair of the failed unit at instant t, given that the system entered the regenerative state iS at t = 0. The recursive relations for )(tBi are as follows: where )(tWi is the probability that the server is busy in state iS due to repairing of unit up to time t without making any transition to any other regenerative state or before returning to the same state via one or more non-regenerative states. The time period for which server is busy due to repair respectively is obtained by solving for )(*0RB we get

Evaluation of expected number of visits by the server owing to repair of the unit Let )(tRi be the expected number of visits by the server in (0, t], given that the system entered the regenerative state Si at t = 0. The recursive relations for )(tRi are as follows:

Evaluation of expected number of refreshments provided to server Let )(tTi be the expected number of treatments given to server in (0,t] such that the system entered the regenerative state at t = 0. The recursive relations for )(tTi are as follows: Taking Laplace Stieltjes transform of the relation (14), and solve for 0Twe have

Cost -Benefit evaluation of the system The profit appeared in the system model in steady state can be evaluated as Where K0 = 10000: Revenue per unit up- time of the system K1 = 600: Cost per unit time for which server is busy K2 = 500: Cost per unit visits by the server

DISCUSSION:

The system model illustrates such a system having two identical units in which one unit is absolutely needed to operate the system and other unit is kept as cold standby mode. The utility of the model can be seen in water supply boosting station. The entire system is examined by taking particular values of the various parameters like (, , λ and µ). The numerical behaviours of some reliability measures like mean time to system failure, availability and profit function have been examined with respect to repair rate () and treatment rate () as shown in the tables 2 to 4 respectively.

In this exploration, the effect of various parameters on performance measures of system model is envisaged. Table-2 reflects the facts and figures that MTSF having increasing trend with respect to increasing repair rate (θ). The second column of this table represents as the server failure rate (µ) decreases, values of MTSF in the table increase. In the third column reveals that whenever treatment rate () decreases, values of MTSF in the table decrease still having increasing pattern. But in the fourth column of the table, whenever the unit failure rate (λ) decreases, values of MTSF in sharply increase having increasing trend as compare with the first column of the table. Hence the effect of the parameters can be analysis form this table very easily or the numerical behaviour is very helpful to analysis the MSTF of such important system.

The table-3 clearly exhibits that availability keeps on increasing with respect to repair rate (θ). It also indicates that as server failure (µ) decrease, values in the table increase for fixed values of other parameters. So availability can be increased by keeping check on server failure. It is fascinating that as treatment rate () decreases, values in the table also decrease. It is also known from illustration that whenever rate of unit failure (λ) decreases from .55 to .45 then the values availability in the fourth column

Table-4 exhibits that profit is also increasing with increasing repair rate (θ) from 0.1 to 1.0. It is also observed from first and second columns that decrease in value of server failure rate (µ) causes increase in values of profit function. Third column of the table reveals that decrease in treatment rate () causes decrease in the values of profit but still having increasing trend and fourth column shows sharp increase in the values of profit whenever unit failure rate (λ) is reduced from .55 to .40.

CONCLUSION:

The idea to make refreshment available to the server (whenever required) enhances the efficiency of the server and having check on sever failure is more beneficial and economical for the productiveness of the system. It is also noteworthy during study that when MTSF, availability and profit are observed with respect to repair rate and with respect to treatment rate for fixed values of other parameters in the tables 2, 3 and 4. It means provision of refreshment is also significant to make the system profitable. Hence the sum and substance of this problem is that the system can be made more efficient and lucrative by increasing repair, having check on server failure, failure of unit and providing refreshment to the server at appropriate time. The study has its utility in water supply boosting station, power generation station with standby reservoirs etc. Keeping the above study in mind, there is possibility to make progress in the

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Research Scholar, School of Applied Science, Singhania University, Pechari Beri Jhunjhunu, Rajasthan Indurohilla24@gmail.com