Volume-5, Issue-, February-205 International Journal of Engineering and Management Research Page Number: 295-302 Stochastic Modeling of a Computer System with Software Redundancy V.J. Munday, S.C. Malik 2,2 Department of Statistics, M.D. University, Rohtak, Haryana, INDIA ABSTRACT The semi-markov process and regenerative point technique are adopted to obtain reliability measures of a computer system by providing software redundancy in cold standby. Initially, hardware and software work together in the system which may fail independently with some probability. There is a single server who repairs the system at hardware failure while software is up-graded as per requirements. The repair and up-gradation activities are performed perfectly and efficiently by the server. The time to hardware and software failures follows negative exponential distribution, whereas the distributions of hardware repair and software up-gradation times are taken as arbitrary with different probability density functions. Graphs are drawn to depict the behaviour of some important performance and economic measures of the system model for arbitrary values of various parameters and costs. Keywords- Computer System, Software Redundancy, Reliability Measures and Stochastic Model. I. INTRODUCTION The importance of computer systems cannot be denied in the corporate or business world, at the workplace and even in one s personnel life. They also serve as useful tools for communications and record keeping while saving tons of times of the organizations. But a computer system would not be able to function properly without software that empowers the computer to communicate the results. Therefore, there is a definite need to place emphasis on reliability of computer software. Several techniques have been suggested by the designers and engineers for performance improvement of the systems. The unit wise redundancy technique has been considered as one of these in the development of stochastic models for computer systems. Malik and Anand (200, 2), Malik and Sureria (202) and Kumar et al. (203) analyzed computers systems with cold standby redundancy under different failures and repair policies. Also, Malik and Munday (204) tried to establish a stochastic model for a computer system by providing hardware redundancy in cold standby. The basic interest of the authors in this paper is to evaluate reliability measures of a computer system with software redundancy in cold standby. For this purpose, a stochastic model is developed for a computer system in which hardware and software failures occur independently with some probability. There is a single server who repairs the system at hardware failure while software is up-graded as per requirements. The repair and up-gradation activities are performed perfectly and efficiently by the server. The time to hardware and software failures follows negative exponential distribution, whereas the distributions of hardware repair and software up-gradation times are taken as arbitrary with different probability density functions. The semi- Markov process and regenerative point technique are used to derive the expressions for transition probabilities, mean sojourn times, mean time to system failure (MTSF), availability, busy period of the server due to hardware repair and software up-gradation, expected number of hardware repairs and expected number of software up-gradations. Graphs are drawn to depict the behaviour of some important performance and economic measures of the system model for arbitrary values of various parameters and costs. II. NOTATIONS E : Set of regenerative states EE : Set of non-regenerative states O : Computer system is operative Scs : Software is in cold standby 295 Copyright 20-5. Vandana Publications. All Rights Reserved.
a/b : Probability that the system has hardware / software failure λλ /λλ 2 : Hardware/Software failure rate HFUr /HFWr : The hardware is failed and under repair/waiting for repair SFUg/SFWUg : The software is failed and under/waiting for up-gradation HFUR/HFWR : The hardware is failed and continuously under repair / waiting for repair from previous state SFUG/SFWUG : The software is failed and continuously under up-gradation /waiting for up- gradation from previous state g(t)/g(t) : pdf/cdf of hardware repair time f(t)/f(t) : pdf/cdf of software up-gradation time qq iiii (tt)/qq iiii (tt) : pdf / cdf of first passage time from regenerative state SS ii to a regenerative state SS jj or to a failed state SS jj without visiting any other regenerative state in (0, t] qq iiii.kk (tt)/qq iiii.kk (tt) : pdf/cdf of direct transition time from regenerative state SS ii to a regenerative state SS jj or to a failed state SS jj visiting state SS kk once in (0, t] MM ii (tt) : Probability that the system up initially in state SS ii εεεε is up at time t without visiting to any regenerative state WW ii (tt) : Probability that the server is busy in the state SS ii up to time t without making any transition to any other regenerative state or returning to the same state via one or more non-regenerative states. μμ ii : The mean sojourn time in state SS ii which is given by 0 μμ ii = EE(TT) = PP(TT > tt) dddd = jj mm iiii, where TT denotes the time to system failure. mm iiii : Contribution to mean sojourn time (μμ ii ) in state SS ii when system transits directly to state SS jj so that 0 μμ ii = jj mm iiii aaaaaa mm iiii = tttttt iiii (tt) = qq iiii (0) & : Symbol for Laplace-Stieltjes convolution/laplace convolution */** : Symbol for Laplace Transformation (LT)/Laplace Stieltjes Transformation (LST) 296 Copyright 20-5. Vandana Publications. All Rights Reserved.
O Scs State Transition Diagram S 0 S aaaa HFUr Scs g(t) bbbb 2 f(t) S 4 SFWUg SFUG f(t) bbbb 2 S 2 O SFUg f(t) aaaa HFWr SFUG Up-State Failed State Regenerative Point S 3 III. Fig. TRANSITION PROBABILITIES AND MEAN SOJOURN TIMES Simple probabilistic considerations yield the following expressions for the non-zero elements. pp iiii = QQ iiii ( ) = qq 0 iiii (tt)dddd pp 0 = aaaa, pp aaaa +bbλλ 02 = bbλλ 2, 2 aaaa +bbbb 2 pp 0 = gg (0) pp 20 = ff (aaaa + bbλλ 2 ) pp 23 = bbbb 2 { ff (aaaa aaaa +bbλλ + bbλλ 2 )} pp 24 = aaλλ { ff (aaaa 2 aaaa +bbλλ + bbbb 2 )}, 2 pp 3 = pp 42 = ff (0) For gg(tt) = ααee αααα aaaaaa ff(tt) = θθee θθθθ we have pp 2.3 = aaaa { ff (aaλλ aaaa +bbbb + bbbb 2 )} pp 22.4 = bbbb 2 { ff (aaλλ 2 aaaa +bbbb + bbbb 2 )} 2 But, ff (0) = gg (0) = aaaaaa pp + qq = It can be easily verified that pp 0 + pp 02 = pp 0 = pp 20 + pp 23 + pp 24 = pp 3 = pp 42 = pp 20 + pp 2.3 + pp 22.4 = The mean sojourn times (μμ ii ) is the state SS ii are μμ 0 = aaaa +bbλλ 2 μμ = αα μμ 2 = { ff (aaaa aaaa +bbbb + bbbb 2 )} = 2 Also aaaa +bbbb 2 +θθ mm 0 + mm 02 = μμ 0, mm 0 = μμ, mm 20 + mm 23 + mm 24 = μμ 2 297 Copyright 20-5. Vandana Publications. All Rights Reserved.
and mm 20 + mm 2.3 + mm 22.4 = μμ 2 = θθ IV. RELIABILITY AND MEAN TIME TO SYSTEM FAILURE (MTSF) Let φφ ii (tt) be the cdf of first passage time from regenerative state SS ii to a failed state. Regarding the failed state as absorbing state, we have the following recursive relations for φφ ii (tt), φφ 0 (tt) = QQ 02 (tt) & φφ 2 (tt) + QQ 0 (tt) φφ 2 (tt) = QQ 20 (tt) & φφ 0 (tt) + QQ 23 (tt) + QQ 24 (tt) () Taking LST of above relations () and solving for ϕ 0 (ss) We have RR (ss) = ϕ 0 (ss) ss The reliability of the system model can be obtained by taking Laplace inverse transform of the above equation. The mean time to system failure (MTSF) is given by MMMMMMMM = lim ss 0 ϕ 0 (ss) ss = NN DD (2) NN = μμ 0 + pp 02 μμ 2 aaaaaa DD = pp 02 pp 20 (3) V. STEADY STATE AVAILABILITY Let AA ii (tt) be the probability that the system is in up-state at instant t given that the system entered regenerative state SS ii aaaa tt = 0. The recursive relations for AA ii (tt) are given as: AA 0 (tt) = MM 0 (tt) + qq 0 (tt) AA (tt) + qq 02 (tt) AA 2 (tt) AA (tt) = qq 0 (tt) AA 0 (tt) AA 2 (tt) = MM 2 (tt) + qq 20 (tt) AA 0 (tt) + qq 2.3 (tt) AA (tt) + qq 22.4 (tt) AA 2 (tt) (4) where MM 0 (tt) = ee (aaλλ +bbλλ 2 )tt aaaaaa MM (tt) = ee (aaλλ +bbbb 2 )tt FF(tt) Taking LT of relations (4) and solving for AA 0 (ss), the steady state availability is given by AA 0 ( ) = lim ss 0 ss AA 0 (ss) = NN 2 (5) NN 2 = ( pp 22.4 )μμ 0 + pp 02 μμ 2 = ( pp 22.4 )μμ 0 + pp 02 μμ 2 + (pp 0 pp 20 μμ 2 + pp 2.3 )μμ (6) 298 Copyright 20-5. Vandana Publications. All Rights Reserved.
VI. BUSY PERIOD OF THE SERVER (a). Due to Hardware Repair Let BB HH ii (tt) be the probability that the server is busy in repairing the unit due to hardware failure at an instant t given that the system entered state SS ii aaaa tt = 0. The recursive relations for BB HH ii (tt) are as follows: BB HH 0 (tt) = qq 0 (tt) BB HH (tt) + qq 02 (tt) BB HH 2 (tt) BB HH (tt) = WW HH (tt) + qq 0 (tt) BB HH 0 (tt) BB HH 2 (tt) = qq 20 (tt) BB HH 0 (tt) + qq 2.3 (tt) BB HH (tt) + qq 22.4 (tt) BB HH 2 (tt) (7) where WW HH (tt) = GG(tt) dddd (b). Due to software Up-gradation Let BB SS ii (tt) be the probability that the server is busy due to replacement of the software at an instant t given that the system entered the regenerative state SS ii aaaa tt = 0. We have the following recursive relations for BB SS ii (tt): BB SS 0 (tt) = qq 0 (tt) BB SS (tt) + qq 02 (tt) BB SS 2 (tt) BB SS (tt) = qq 0 (tt) BB SS 0 (tt) BB SS 2 (tt) = WW SS 2 (tt) + qq 20 (tt) BB SS 0 (tt) + qq 2.3 (tt) BB SS (tt) + qq 22.4 (tt) BB SS 2 (tt) (8) where WW SS 2 (tt) = ee (aaaa +bbbb 2 )tt FF(tt) + aaλλ ee (aaλλ +bbbb 2 )tt FF(tt) + bbbb2 ee (aaaa +bbbb 2 )tt FF(tt) HH Taking LT of relations (7) & (8), solving for BB SS 0 (tt) aaaaaa BB 0 (tt). The time for which server is busy due to repair and replacements respectively are given by BB HH 0 (tt) HH = lim ss 0 ss BB 0 (tt) = NN 3 HH (9) BB SS 0 (tt) SS = lim ss 0 ss BB 0 (tt) = NN 3 SS (0) where NN HH HH 3 = pp 02 pp 2.3 WW (0) HH + pp 0 ( pp 22.4 )WW (0) NN SS SS 3 = pp 02 WW 2 (0)aaaaaa iiii aaaaaaaaaaaaaa mmmmmmmmmmmmmmmmmm. () VII. EXPECTED NUMBER OF HARDWARE REPAIRS Let NNNNNN ii (tt) be the expected number of hardware repairs by the server in (0, t] given that the system entered the regenerative state SS ii aaaa tt = 0. The recursive relations for NNNNNN ii (tt) are given as: NNNNNN 0 (tt) = QQ 0 (tt) & [ + NNNNNN (tt)] + QQ 02 (tt) & NNNNNN 2 (tt) NNNNNN (tt) = QQ 0 (tt) & NNNNNN 0 (tt) NNNNNN 2 (tt) = QQ 20 (tt) & NNNNNN 0 (tt) + QQ 2.3 (tt) & NNNNNN (tt) + QQ 22.4 (tt) & NNNNNN 2 (tt) (2) Taking LST of relations (2) and solving for NNNNNN 0 (ss). The expected number of hardware repair is given by NNNNNN 0 = lim ss 0 ssnnnnnn 0 (ss) = NN 4 (3) NN 4 = pp 0 ( pp 22.4 )aaaaaa iiii aaaaaaaaaaaaaa mmmmmmmmmmmmmmmmmm. (4) VIII. EXPECTED NUMBER OF SOFTWARE UP-GRADATIONS Let NNNNNN ii (tt) be the expected number of software up-gradations in (0, t] given that the system entered the regenerative state SS ii aaaa tt = 0. The recursive relations for NNNNNN ii (tt) are given as follows: NNNNNN 0 (tt) = QQ 0 (tt) & NNNNNN (tt) + QQ 02 (tt) & [ + NNNNNN 2 (tt)] NNNNNN (tt) = QQ 0 (tt) & NNNNNN 0 (tt) NNNNNN 2 (tt) = QQ 20 (tt) & NNNNNN 0 (tt) + QQ 2.3 (tt) & NNNNNN (tt) + QQ 22.4 (tt) & NNNNNN 2 (tt) (5) 299 Copyright 20-5. Vandana Publications. All Rights Reserved.
Taking LST of relations (5) and solving for NN 0 (ss). The expected numbers of software up-gradation are given by NNNNNN 0 ( ) = lim ss 0 ssnnnnnn 0 (ss) = NN 5 (6) NN 5 = pp 02 ( pp 22.4 ) aaaaaa iiii aaaaaaaaaaaaaa mmmmmmmmmmmmmmmmmm (7) IX. COST-BENEFIT ANALYSIS The profit incurred to the system model in steady state can be obtained as: PP = KK 0 AA 0 KK BB HH 0 KK 2 BB SS 0 KK 3 NNNNNN 0 KK 4 NNNNNN 0 (8) KK 0 = RRRRRRRRRRRRRR pppppp uuuuuuuu uuuu tttttttt oooo tthee ssssssssssss KK = CCCCCCCC pppppp uuuuuuuu tttttttt ffffff wwhiiiih ssssssssssss iiii bbbbbbbb dddddd tttt haaaaaaaaaaaaaa rrrrrrrrrrrr KK 2 = CCCCCCCC pppppp uuuuuuuu tttttttt ffffff wwhiiiih ssssssssssss iiii bbbbbbbb dddddd tttt ssssssssssssssss uuuu gggggggggggggggggg KK 3 = CCCCCCCC pppppp uuuuuuuu rrrrrrrrrrrr oooo tthee ffffffffffff haaaaaaaaaaaaaa KK 4 = CCCCCCCC pppppp uuuuuuuu uuuu gggggggggggggggggg oooo tthee ffffffffffff ssssssssssssssss aaaaaa AA 0, BB HH 0, BB SS 0, NNNNNN 0, NNNNNN 0 aaaaaa aaaaaaaaaaaaaa dddddddddddddd. X. PARTICULAR CASES Suppose gg(tt) = ααee αααα aaaaaa ff(tt) = θθee θθθθ We can obtain the following results: MMMMMMMM(TT 0 ) = NN DD AAAAAAAAAAAAAAAAAAAAAAAA(AA 0 ) = NN 2 BBBBBBBB PPPPPPPPPPPP dddddd tttt haaaaaaaaaaaaaa ffffffffffffff (BB 0 HH ) = NN 3 HH BBBBBBBB PPPPPPPPPPPP dddddd tttt ssssssssssssssss ffffffffffffff BB 0 SS = NN 3 SS EEEEEEEEEEEEEEEE nnnnnnnnnnnn oooo rrrrrrrrrrrr aaaa haaaaaaaaaaaaaa ffffffffffffff (NNNNNN 0 ) = NN 4 EEEEEEEEEEEEEEEE nnnnnnnnnnnn oooo uuuu gggggggggggggggggg aaaa ssssssssssssrrrr ffffffffffffff (NNNNNN 0 ) = NN 5 NN = NN 2 = NN 3 HH = NN 4 = aaλλ +2bbλλ 2 +θθ (aaλλ +bbλλ 2 )(aaλλ +bbλλ 2 +θθ) aaλλ +bbλλ 2 aaλλ (aaλλ +bbλλ 2 )αα aaλλ (aaλλ +θθ) (aaλλ +bbλλ 2 )(aaλλ +bbλλ 2 +θθ) DD = (aaλλ +bbλλ 2 )(aaλλ +bbλλ 2 +θθ) θθθθλλ 2 (aaλλ +bbλλ 2 )(aaλλ +bbλλ 2 +θθ) = αααα(aaλλ +θθ)+(θθθθλλ +ααααλλ 2 )(θθθθλλ +bbλλ 2 +θθ) αααα(aaλλ +bbλλ 2 )(aaλλ +bbλλ 2 +θθ) NN 3 SS = NN 5 = bbλλ (aaλλ +bbλλ 2 )θθ bbλλ 2 (aaλλ +θθ) (aaλλ +bbλλ 2 )(aaλλ +bbλλ 2 +θθ) 300 Copyright 20-5. Vandana Publications. All Rights Reserved.
XI. CONCLUSION REFERENCES The behaviour of some important performance measures such as MTSF, availability and profit with respect to hardware failure rate (λλ ) has been observed for arbitrary values of various parameters including K 0 = 5000, K 000, K 2 = 700, K 3 = 500, K 4 = 200 with a=0.6 and b=0.4 as shown respectively in figures 2, 3 and 4. It is revealed that these measures go on decreasing with the increase of hardware and software failure rates. But, their values increase with the increase of hardware repair rate (α) and up-gradation rate (θ). On the other hand, if the values of a and b are interchanged i.e. a=0.4 and b=0.6, than MTSF and availability of the system increase while profit declines. Hence the study reveals that a computer system in which software redundancy is provided in cold standby be more profitable if it has more chances of hardware failure may because of the less hardware repairable cost. [] Anand, Jyoti and Malik, S.C. (202): Analysis of a Computer System with Arbitrary Distributions for H/W and S/W Replacement Time and Priority to Repair Activities of H/W over Replacement of the S/W, International Journal of Systems Assurance Engineering and Management, Vol.3 (3), pp. 230-236. [2] Kumar, Ashish; Anand, Jyoti and Malik, S.C. (203): Stochastic Modeling of a Computer System with Priority to Up-gradation of Software over Hardware Repair Activities. International Journal of Agricultural and Statistical Sciences, Vol. 9(), pp. 7-26. [3] Malik, S.C. and Anand, Jyoti (200): Reliability and economic analysis of a computer system with independent hardware and software failures. Bulletin of Pure and Applied Sciences, Vol. 29E(0), pp.4-53. [4] Malik, S.C. and Munday, V.J. (204): Stochastic Modeling of a Computer System with Hardware Redundancy. International Journal of Computer Applications, Vol. 89(7), pp. 26-30. [5] Malik, S.C. and Sureria, J.K. (202): Profit Analysis of a Computer System with H/W Repair and S/W Replacement. International Journal of Computer Applications, Vol.47(), pp. 9-26. 30 Copyright 20-5. Vandana Publications. All Rights Reserved.
300 250 200 50 00 MTSF Vs H/w Failure Rate (λ) λ2=0.00,α=2,θ=5,a=0.6,b=0.4 λ2=0.002 θ=7 a=0.4,b=0.6 MTSF 50 0 0.00.020.030.040.050.060.070.080.09 0. Hardware Failure Rate (λ) Fig. 2 Availability.0 0.99 0.98 0.97 0.96 0.95 Availability Vs Hardware Failure rate (λ) λ2=0.00,α=2,θ=5,a=0.6,b=0.4 λ2=0.002 α=3 θ=7 a=0.4,b=0.6 0.00.020.030.040.050.060.070.080.09 0. Hardware Failure Rate (λ) Fig. 3 Profit 500 5000 4900 4800 4700 4600 4500 4400 4300 4200 400 Profit Vs Hardware Failure Rate (λ) λ2=0.00, α=2, θ=5, a=0.6, b=0.4 λ2=0.002 α=3 θ=7 a=0.4, b=0.6 0.0 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0. Hardware Failure Rate (λ) K 0 = 5000, K = 000, K 2 = 700, K 3 = 500, K 4 = 200 Fig. 4 302 Copyright 20-5. Vandana Publications. All Rights Reserved.