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Cost- benefit analysis of stochastic models of A 2-out-of-3 redundant system with inspection

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  1. Abstract
  2. Introduction
  3. Notation
  4. Transition probabilities and mean sojourn times
  5. Analysis for model 1 and 2
    1. Mean Time to System Failure (MTSF)
    2. Steady state availability
    3. Busy period of the server
    4. Expected number of visits by the server
  6. Particular case
  7. References

In this paper two stochastic models for a 2-out-of-3 redundant system are developed and cost-benefit analysis is carried out by using semi-Markov and regenerative processes. There are three identical units in each model and the system is considered in up-state if 2-out-of-3 units are functioning. A single repair facility is available which plays the dual role of inspection and repair. In model1, server repairs the unit without inspection while in model 2, the unit is inspected at its failure to see the feasibility of repair. If repair of unit is not feasible, it is replaced by new one in order to increase the reliability of the system. The failure time distribution follows negative exponential while that of inspection and repair are taken as arbitrary. The expressions for MTSF, steady state availability, busy period and expected number of visits of the server are derived. The profit function of each model is also estimated. A particular case is considered to derive the results in parametric form. Keywords - 2-out-of-3 redundant system, inspection, regenerative process and cost-benefit analysis

[...] The possible transition states of both the system models are shown in the following table1 S0 Model I Model II S1 S2 N0, Fwr, Fur N N wi , FuI S3 S4 S5 N N Cs N N Cs N N Fur N N Fui N N Fur N N wi , FuR N FwI , Fur For model S0, For model II: S1, S2, S3} The possible transition states along with transition rates for model I and model II are shown in fig and fig respectively Transition Probabilities and Mean Sojourn Times Simple probabilistic considerations yield the following expressions for the non-zero elements, pij = Qij = q ij (t)dt as: For Model I p01=1, p10=g*(2?), p12=1-g*(2?), p 11.2 Clearly, p01=p10+p12=p10+p 11.2 For Model II p01 = p10 = bh* (2? p12 = 1 h * (2? p13 = ah *(2? p30 = (2? p34 = 1 (2? p 11.2 = b[1 h * (2? p 11.25 = a[1 h (2? p 31.4 = p34 It can be easily verified that * p01 = 1 = p10 + p12 = p30 + p34 = p30 + p 31.4 = p10 + p13 + p 11.2 + p 11.25 The unconditional mean time taken by the system to transit to any regenerative state Sj when it (time) is counted from epoch of entrance into that state Si, is given by mij = td{Qij = ij ' 0 and the mean Sojourn time in the state Si is given by µi = = P(T > t)dt 0 where T denotes the time to system failure. [...]


[...] Reliab. 655-659 (1985). Singh, S.K., Profit evaluation of a two unit cold standby system with random appearance and disappearance time of service facility, Microelectron Reliab., 29,705-709 (1989) Nakagawa, T., A replacement policy maximizing [...]


[...] Hence the total profit in is given by Gi ) = k1U i ) k2 µib ) k3 N i i = Where k1= Revenue per unit up time for the system k2= Cost per unit time for which the system is under repair k3= Cost per visit by the server Expected profit per unit time in steady state is given by Gi ) = lim t Gi ) = lim s 2G ( s i = t Or Gi = k1 Ai 0 k2 Bi 0 k3 N i Particular Case Let us take = ? e t , = ? e t , = ? ? t , we can obtain the following results For Model MTSF(T1)= N D11 Availability (A10) = N12 N , Busy period (B10) = D12 D12 Expected number of visits( N10) = For Model MTSF(T2)= N14 D12 N D 21 Availability (A20) = N 22 N , Busy period (B20) = D 22 D 22 Expected number of visits( N20) = N D 22 N 13 = where, N = (? + 4? ) N 12 = 2? ? N 14 = ? (? + 2? ) , D11 = 4? (? + 2? D12 = R0 S0 N 21 = ? + 4? , D21 = 4? N 22 = R 1 / S D 22 = R 2 / S N 23 = R 3 / S3 N 24 = R 4 / S4 R0 = ? (? + 2? ) + 4? R1 = (b? + 1)(? + 2? ) + 2a? , R 2 = [?? + 2? b? ]? (? + 2? ) + 2? (? + 2? + a? (? + 2? R 3 = + 2? ? + ? + a? ) 2??? + a? 2? (? + ? + 2? ) + a? 2 (? + ? + 2? )(2? ? ) R 4 = ? (? + 2b? ) S 0 = 2?? (? + 2? ) , S1 = (? + 2? + 2? ) , S2 = 2?? (? + 2? + 2? S3 = ?? (? + ? + 2? )(2? + ? + 2? ) , S4 = (? + 2? + 2? ) 8. References Chiang, D.T. and Niu, S.C., Reliability of Consecutive K-out-of-n:F system, IEEE Transactions on Reliability, R-30,87-89 (1981). Murari, K. and Goyal, V., Comparison of two-unit cold standby reliability models with three types of repair facilities, Microelectron. Reliab. 35-49 (1984). Goel, L.R., Sharma, G.C. and Gupta, R., Cost analysis of two-unit cold standby system under different weather conditions, Microelectron. [...]

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