RELIABILITY MODELLING OF INTERNET DATA ...

International J. of Math. Sci. & Engg. Appls. (IJMSEA) ISSN 0973-9424, Vol. 3 No. IV (2009), pp. 231-244

RELIABILITY MODELLING OF INTERNET DATA CENTER V.V. SINGH, S. B. SINGH AND C. K. GOEL

Abstract This paper deals with the mathematically study of an Internet data center, which consists of a database main server, connected with a redundant server. By observing the different possibilities the system, it has been analyses to evaluate the various reliability characteristics of system. The system can fail completely due to database and redundant server failure, router failure and switch failure. Failure time is assumed to follow negative exponential distribution while repair follow general time distribution. By incorporating waiting time to repair the varies measures of reliability have been discussed such as reliability of system mean time to system failure steady state transition probabilities cost analyses. In last some particular cases are also highlighted and conclusion have been drown.

Introduction This paper deals with the study of an Internet data centre (IDC) with a redundant mail server. Internet data centre can have two types of failure namely partial and complete. The information technology enabled architecture of IDC handled by two switches L2 and L3 . The L3 switches, which a 6-part switch, is connected to a server via L2 switch. Whenever main mail server fails, redundant by comes into functioning automatically by switch over device. The switch over device is instantaneous and −−−−−−−−−−−−−−−−−−−−−−−−−−−−

Key Words : Database server failure, L2 , L3 Switch, Router failure, Redundant server failure etc. @2007 Ascent Publishing House

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automatic. The system as a whole can be fail due to: (1) failure of redundant server before the repair of main mail server, (2) switch failure and (3) router failure. The system will be in degraded state, when main mail server is in completely failed state and redundant mail server is in partial failure mode. The failure rates are assumed to follow exponential distribution however repair time follow general distribution. The system is analyzed by use of supplementary variable technique. Various measures of reliability such as reliability of systems mean time to failure, asymptotic, behaviour and profit functions have been discussed. Some particular cases are also taken to highlight various possibilities and the results obtained have been demonstrated by tables and graphs. The function of some technical terms attached with IDC is described below. Data base Server : It is a highly reliable and manageable facility. It maintains record of premises. It contents a buffer where each and every record is kept safely. Router : It provides route to the data packets and connected to a web space. L3 Switch : They are layer three switches generally used to transfer information packets to router. L2 Switch : These are layer two switches, which are most faithful switch, it takes information and transfers it to layer three switches for next transmission. They are also connected to the server.

Assumption The following assumptions are taken throughout the discussion of the model. (1) Initially the system is in good and working condition in state S1 . (2) System fails if redundant server fails before the repair of the main mail server. (3) Switch failure, router failure breakdown the system. (4) System is in degraded mode if main mail server has failed and redundant mail server is in partial failure mode. (5) The repaired system work like a new and repair don’t damage anything. (6) All failure rates are constant and follow negative exponential time distribution.

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(7) The system waits for repair if repair facility is not available. The waiting rate is also constant. (8) Switch over device is automatic and instantaneous. (9) All repairs are assumed to follow general time distribution.

Notations The following notations have been taken throughout study of the model. p0,s (t) : The main mail server is in operational mode and redundant mail server is in standby mode. Pf,0 (z, t) : Probability that main mail server has fail and redundant server is in operational mode. The main mail server is under repair, elapsed repair time is z, t. Pf,f (z, t) : Probability that the main as well as redundant mail server has fail, the system is under repair and elapse repair time is z, t. P0,f (z, t) : Probability that the main mail server is in operational and redundant server is in failure mode, the system is under repair and elapsed repair time is z, t. Pf,p (z, t) : Probability that the main mail server in failure mode and redundant mail server is in partial failure mode. The system is in degraded mode. The main mail server is under repair and elapsed repair is z, t. P0,p (z, t) : Probability that the main mail server is in operation mode and redundant mail server is in partial failure mode, the system is under repair and elapsed repair time is z, t. PR (y, t) : Probability that the system has fail due to router failure. The system is under repair and elapsed repair time is y, t. PL (x, t) : Probability that system is in completely failed due to switch failure. The system is under repair and elapsed repair time is x, t. λL /λR /λP : Switch failure/router failure /Partial failure rate λA /λw : Data base failure rate/ Waiting rate. vx /vy /vz /hv : Repair rates of switch / router / database server / partial failure of stand by server. P (s) : Laplace transform of state transition probability

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State Transition Diagram of Model

Formulation of Mathematical Model By probability of considerations and continuity arguments we can obtain the follow-

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ing set of difference differential equations governing the present mathematical model.   Z ∞ Z ∞ ∂ vx PL (x, t)dx vy PR (y, t)dy + + λA + λL + λR P0,s (t) = ∂t 0 0 Z ∞ Z ∞ Z ∞ vz Pw (z, t)dz (1) hv P0,p (v, t)dv + vz Pof (z, t)dz + + 0

0

0





∂ ∂ + + λA + λP Pf,0 (z, t) = 0 ∂t ∂z   ∂ ∂ + + λw + vz Pf,P (z, t) = 0 ∂t ∂z   ∂ ∂ + + hv P0,P (v, t) = 0 ∂t ∂v   ∂ ∂ + + hv Pw (v, t) = 0 ∂t ∂v   ∂ ∂ + + vy PR (y, t) = 0 ∂t ∂y   ∂ ∂ + + vx PL (x, t) = 0 ∂t ∂x   ∂ ∂ + + vz Pf,f = 0 ∂t ∂z   ∂ ∂ + + vz P0,f (z, t) = 0 ∂t ∂z

(2) (3) (4) (5) (6) (7) (8) (9)

Boundary Conditions Pf,0 (0, t) = λA P0,s (t)

(10)

Pf,P (0, t) = λP Pf,0 (0, t) Z ∞ P0,p (0, t) = vz Pf,p (z, t)dz

(11) (12)

0

Pw (0, t) = λw Pf,p (0, t)

(13)

Pf,f (0, t) = λA Pf,P (0, t) Z ∞ P0,f (0, t) = vz Pf,f (z, t)dz

(14) (15)

0

PR (0, t) = λR P0,s (t)

(16)

PL (0, t) = λL P0,s (t).

(17)

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Initials Conditions P0,s (0) = 1 and other state probabilities are zero at t = 0

(18)

Solution of the Model Taking Laplace transformation of equations (1) - (17) and using equation (18), we obtain  Z ∞ Z ∞ vy P R (y, s)dy + vx P L (x, s)dx [s + λA + λL + λP ]P 0,s (s) = 1 + 0 0  Z ∞ Z ∞ Z ∞ vz P w (z, s)dz + hv P 0,p (v, s)dv + vz P 0,f (z, s)dz + 0

0

(19)

0

  ∂ s+ + λA + λP P f,0 (z, s) = 0 ∂z   ∂ s+ + λw + vz P f,p (z, s) = 0 ∂z   ∂ s+ + hv P 0,p (v, s) = 0 ∂v   ∂ s+ + vz P w (z, s) = 0 ∂z   ∂ s+ + vy P R (y, s) = 0 ∂y   ∂ s+ + vx P L (x, s) = 0 ∂x   ∂ s+ + vz P f,f (z, s) = 0 ∂z   ∂ s+ + vz P 0,f (z, s) = 0 ∂z

(20) (21) (22) (23) (24) (25) (26) (27)

P f,0 (0, s) = λA P 0,s (s)

(28)

P f,p (0, s) = λp P f,0 (0, s)

(29)

P w (0, s) = λw P f,p (0, s)

(30)

P f,f (0, s) = λA P f,0 (0, s)

(31)

P R (0, s) = λR P 0,s (s)

(32)

P L (0, s) = λL P 0,s (s)

(33)

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∞

Z P0,P (0, s) =

vz P f,p (z, s)dz

(34)

vz P f,f (z, s)dz.

(35)

0

Z P0,f (0, s) =

∞

0

Solving (19)-(27) with the help of (28)-(35), one may get transition state probability of system. P 0,s (s) =

1 D(s)

λA P 0,s (s) (s + λA + λP )   1 − SV (s + λw ) P f,p (s) = λA λP P 0,S (s) s + λw   1 − SV (s) P 0,p (s) = λA λP Sv (s + λw ) P 0,s (s) s   1 − SV (s) 2 P f,f (s) = λA P 0,S (s) s   1 − SV (s) P L (s) = λL P 0,S (s) s   1 − SV (s) P R (s) = λR P 0,S (s) s   1 − SV (s) P w (s) = λw λp λA P 0,S (s) s   1 − SV (s) P 0,f (s) = λ2A Sv (s) P 0,S (s). s P f,0 (s) =

(36) (37) (38) (39) (40) (41) (42) (43) (44)

Here, D(s) = s + λA + λL + λR − {λL + λR + λ2A S v (s) + λA λP λw )S v (s). The laplace transformations of the probabilities that the system is in up (i.e. either good or degraded state) and failed state at any time are as follows: P up (s) = P 0,s (s) + P f,0 (s) + P 0,p (s) + P 0,f (s)   1 λA λA λP 1 − Sv (s) 2 = 1+ + Sv (s + λw ) + λA Sv (s) D(s) s + λA + λp s s P f ailed (s) = P f,p (s) + P f,f (s) + P w (s) + P L (s) + P R (s)

(45) (46)

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V. V. SINGH, S. B. SINGH & C. K. GOEL       1 − SV (s + λw ) 1 1 − SV (s) P f ailed (s) = λA λP + λ2A D(s) s + λw s       1 − SV (s) 1 − SV (s) 1 − SV (s) +λw λp λA + λL + λR . s s s

Asymptotic Behaviour of System Using Abel’s lemma in Laplace transformation the independent state transition probability can be find as, viz. Lts→0 {sf (s)} = ltt→∞ F (t) = F, provided limit of left hand side exists. The following time independent probabilities are obtained. P0,S =

1

(47)

D0 (0)

Pf,0 =

λA 1 0 λA + λP D (0)

(48)

Pf,P =

λA λp 1 λw + vz D0 (0)

(49)

P0,P =

λA λP 1 0 λw + hv D (0)

(50)

pf,f =

λ2A 1 vz D0 (0)

(51)

PL =

λL 1 vx D0 (0)

(52)

PR =

λR 1 vy D0 (0)

(53)

λw λp λA 1 vz D0 (0)

(54)

λ2A 1 vz D0 (0)

(55)

PW =

P0,f = where, D0 (0) =

d D(s), at s → 0. ds

Particular Cases When repair follows general distribution. Setting S v (s) = have

v s+v

in equation (45), we

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(i) Setting the values of different parameters as λA = 0.01, λP = 0.02, λL = 0.01, λR = 0.02, λw = 0.02, and v = 1, then taking inverse Laplace transform, one can obtain, Pup (t) := 0.06606474345e(−1.0333t) − 0.4752783528e(−0.03000t) −0.03301760963e(−1.026604t) − 0.003761281066e(−1.003795345t) +1.445992500e(0.9601166318t) .

(56)

Pf ailed (t) = 1 − Pup (t) For, t = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, · · · etc., one can get Table 1.

Table 1 : Time vs Avaliability Time(t) 0 1 2 3 4 5 7 8 9

Pup (t) 1.000 0.981 0.974 0.971 0.970 0.969 0.965 0.963 0.961

Figure - 1: Time vs Availability

(57)

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V. V. SINGH, S. B. SINGH & C. K. GOEL

Mean Time to Failure (MTTF) Where S v (s) =

v s+v

and taking all repairs to zero in equation (45). Taking limit as

s tends to zero one can obtain the MTTF as:

M.T.T.F. = lim P up (s) = s→0

(2λA + λP ) . (λA + λL + λR )

(58)

Setting λA = 0.01, λR = 0.02, λP = 0.02, and varying λL as 0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09 in (45) one may obtain variation of MTTF with respect to λL . Setting λL = 0.01, λR = 0.02, λP = 0.02 and varying λA as 0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09 in (45) one may obtain variation of MTTF with respect to λA . Setting λL = 0.01, λA = 0.01, λr = 0.02, and varying λP , as 0.10, 0.20, 0.30, 0.40, 0.50, 0.60, 0.70, 0.80, 0.90 in (45) one may obtain variation of MTTF with respect to λP . Setting λL = 0.01, λA = 0.01, λP = 0.02 and varying as 0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09 in (45) one may obtain variation of MTTF with respect to λR .

Table 2 Failure rate 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

MTTF λL 33.333 26.666 22.222 19.047 16.666 14.814 13.333 12.121 11.111

MTTF λA 43.010 36.585 31.372 27.322 24.144 21.604 19.536 17.821 16.380

MTTF λP 44.444 33.333 26.667 22.222 19.048 16.667 14.815 13.333 12.121

MTTF λR 36.500 33.333 31.250 30.000 29.166 28.571 28.125 27.777 27.500

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Figure - 2

Cost Analysis Let the failure rates of system be λA = 0.01, λP = 0.02, λL = 0.01, λw = 0.02, λR = 0.02 mean time to repair of v(x) = 1. Setting S v (s) =

v s+v

in equation (45) and taking

inverse Laplace transform one can obtain (45). Let the service facility be always available, then expected profit during the interval [0, t] is Z

t

Pup (t)dt − K2 t

Ep (t) = K1

(59)

0

where K1 and K2 revenue are service cost, per unit time. Hence Ep (t) = {K1 [(−14.9222212(e(−1.00000) − 1) + 13.66554338(e(−2.00000) − 1) −0.00286219827(e(−2.739887579) − 1) − 13.4258222(e(−0.1647641878t) − 1) −0.02045920153(e(−1.023648233t) − 1) − K2 t)]}

(60)

Setting K1 = 1 and K2 = 0.5, 0.25, 0.15, 0.10, 0.05, 0.02, and 0.01 respectively one get table 6 for, t = 0, 1, 2, 3, · · · etc.

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V. V. SINGH, S. B. SINGH & C. K. GOEL Table 3 : Time vs Ep (t)

Time (t) 0 1 2 3 4 5 6 7 8 9

Ep (t); K2 =0.5 0 0.531 1.024 1.503 1.976 2.446 2.915 3.383 3.849 4.313

Ep (t); K2 0.25 0 0.781 1.524 2.253 2.976 3.696 4.415 5.133 5.849 6.563

Ep (t); K2 0.15 0 0.881 1.724 2.553 3.376 4.196 5.015 5.833 6.649 7.463

Ep (t); K2 0.10 0 0.931 1.824 2.703 3.576 4.444 5.315 6.183 7.049 7.913

Ep (t); K2 0.05 0 1.026 2.014 2.988 3.956 4.921 5.885 6.848 7.809 8.768

Ep (t); K2 0.02 0 1.011 1.9845 2.943 3.896 4.846 5.795 6.743 7.689 8.634

Ep (t); K2 0.01 0 1.021 2.006 2.973 3.936 4.897 5.855 6.813 7.769 8.723

Figure - 3 : Time vs. Expected profit

Interpretation of the Result and Conclusion From Table 1 and Figure 1, one can observe the variation of availability with respect to change in time. When failure rates are fixed at lower values λL = 0.01, λA = 0.01, λP = 0.02, λR = 0.02, and λw = 0.02 we see that the availability of the system decreases and probability of failure increase, with passage of time. Hence one can safely predict the future behavior of complex system at any time for any given set of parametric values. From Tables 2 and Figure 2, we can observe variation of mean-time-to-failure (MTTF) of the system with respect to variation in λL , λA , λP and λR respectively

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when other parameters are kept at constant value. Study shows that the mean time to system failure decrease as failure rate increases. When revenue cost per unit time K1 fixed at 1, service cost K2 = 0.5, 0.25, 0.10, 0.05, 0.02, 0.01, profit has been calculated and result are demonstrated by graphs. It is also observed that as service cost decreases expected profit increases.

References [1] Govil, A. K., Operational behaviour of a complex system having shelf-life of the components under preemptive-resume repair discipline, Microelectronics and Reliab., 13 (1974), 97-101. [2] Lirong Cui and Haijun Li., Analytical method for reliability and MTTF assessment of coherent systems with dependent components, Reliability Engineering and System Safety (2007), 300-307. [3] Gupta, P. P., Complex system reliability with general repair time distributions under preemptiversum repair discipline, Microelectronics and Reliab., 12 (1973), 351-356. [4] Gupta, P. P. and Sharma, M. K., Reliability and M. T. T. F evaluation of a two duplex-unit standby system with two types of repair, Microelectron Reliab., 33(3) (1993), 291-295. [5] Goel, L. R. and Gupta, P. P., Analysis of two engine aeroplane model with two type of failure and.preventive maintanence Microelectron and Reliab., 24 (1984), 663-666. [6] Gupta, P. P. and Gupta, R. K., Costanalysis of an electronic repairable redundant system with crictical human error, Microelectron Reliab., 26 (1986), 417-421. [7] Gupta, P. P., Complex system reliability with general repair time distributions under preemptive-resume repair discipline, Microelectronics and Reliab., 12 (1973), 351-356. [8] Gupta, P.P. and Sharma, R. K., Operational Behaviour of a Three State Standby Redudant Electronic Equipment under Critical Human Errors, Microelectron Reliability, 26 (1986), 809-814.

V. V. Singh, Department of Mathematics NIET Gr. Noida, India E-mail:singh− [email protected]

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S. B. Singh, Department of Mathematics, Statistics and Computer Science, G. B. Pant University of Agriculture and Technology, Pantnagar, India E-mail: [email protected] C. K. Goel, Department of mathematics, C.C.SUniversity Meerut (India) E-mail: ck− [email protected]