Analysis of a Computer System with Software Redundancy Subject to Hardware Preventive Maintenance

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Analysis of a Computer System with Software Redundancy Subject to Hardware Preventive Maintenance

V. J. Munday*

*Department of Statistics, Ramjas College, University of Delhi,

Delhi 110007, India

Permila#

#Department of Statistics, Govt. College for Women, Rohtak 124001, India

S.C. Malik$

$Department of Statistics,

M.D. University, Rohtak 124001, (India)

Abstract :- The present work is devoted to the analysis of a computer system with software redundancy subject to hardware preventive maintenance. For this purpose, a system model has been developed by considering software redundancy. There is an independent hardware and software failure in the system model. The preventive maintenance of the hardware component has been conducted after a specific operation time t (called maximum operation time). A single server is provided immediately to the system for carrying out hardware repair, software up-gradation and hardware preventive maintenance as and when needed. The repair activities conducted by the server are perfect. The failure time of the hardware and software components follow negative exponential distribution whereas the distributions for hardware repair, software up-gradation and preventive maintenance times are taken as arbitrary with different probability density functions. Some important reliability characteristics of the system models have been examined stochastically in steady state for particular values of various parameters and costs by using semi-Markov process and regenerative point technique. Graphs are drawn to depict the behaviour of mean time to system failure (MTSF), availability and profit function under different sets of assumptions on the parameters. The profit of the present model has also been compared with the model as discussed in the research paper of [5].

Keywords – Computer System; Software Redundancy; Up- gradation; Preventive Maintenance, Profit Analysis and Stochastic Modelling

  1. INTRODUCTION

    In past few decades, the importance of computer widely increases in most of areas such as companies, medicals, banks, etc. In the previous research based on computer system, some stochastic models for a computer system have been developed and analyzed in steady state by considering the aspects of component wise redundancy in cold standby, priority in repair disciplines and maximum repair time to hardware. And, a computer system would be

    more profitable if redundancy is provided to the hardware along with maximum repair time rather than redundancy to the software. [1] examined a two-unit standby redundant system with preventive maintenance. [2] Analyzed a two dissimilar cold standby system with preventive maintenance and replacement of standby. On the other hand, the deterioration rate of a system can be reduced by conducting its preventive maintenance after a maximum operation time. [3] obtained reliability measures of a system under preventive maintenance. Hence, it becomes necessary to examine the effect of preventive maintenance on reliability measures of a computer system with the concepts of preventive maintenance and software redundancy. However, [4] tried to develop stochastic models of operating systems with preventive maintenance and priority subject to maximum operation and repair times. [6] established reliability measures of a computer system with hardware redundancy subject to preventive maintenance. The present work is devoted to the analysis of a computer system with software redundancy subject to hardware preventive maintenance. For this purpose, a system model has been developed by considering software redundancy in cold standby with the concept of hardware preventive maintenance. Hardware and software failures are independent in the system model. The preventive maintenance of the hardware component has been conducted after a specific operation time t (called maximum operation time). There is a single server who visits the system immediately for carrying out hardware repair, software up-gradation and hardware preventive maintenance as and when needed. The repair activities conducted by the server are perfect. The failure time of the hardware and software components follow negative exponential distribution whereas the distributions for hardware repair, software up-gradation and preventive maintenance times are taken as arbitrary with different probability density functions. Some important reliability characteristics of the system models have been examined

    stochastically in steady state for particular values of various

    = () = ( > ) = ,

    parameters and costs by using semi-Markov process and

    0

    regenerative point technique. Graphs are drawn to depict the behaviour of mean time to system failure (MTSF), availability and profit function under different sets of assumptions on the parameters. The profit of the present

    where denotes the time to system failure.

    mij : Contribution to mean sojourn time (i) in state Si when system transits directly to state Sj so that

    = = () = (0)

    model has also been compared with the model as discussed

    0

    in the research paper of [5].

  2. SOME IMPORTANT NOTATIONS O : Computer system is operative

    Scs : Software is in cold standby

    a/b : Probability that the system has hardware

    / software failure

    / : : Hardware/Software failure rate

    & : Symbol for Laplace – Stieltjes convolution

    / Laplace convolution

    */** : Symbol for Laplace Transformation (LT)

    /Laplace Stieltjes Transformation (LST) i(t) : cdf of first passage time from regenerative

    state Si to a failed state. Regarding the failed state as absorbing state

    1 2 Ai (t) : Probability that the system is in up-state at

    0 : The rate by which hardware component undergoes for preventive maintenance

    HFUr /HFWr : The hardware is failed and

    under/waiting for repair SFUg/SFWUg : The software is failed and

    under/waiting for up-gradation HFUPm /HFWPm : The hardware is under/waiting

    for preventive maintenance HFUR/HFWR : The hardware is failed and

    continuously under/ waiting for repair from previous state

    SFUG/SFWUG : The software is failed and

    continuously under /waiting for up- gradation from previous state

    HFUPM/HFWPM : The hardware is under/ waiting

    for preventive maintenance continuously from previous state

    g(t)/G(t) : pdf/cdf of hardware repair time f(t)/F(t) : pdf/cdf of software up-gradation time m(t)/M(t) : pdf/cdf of hardware preventive

    maintenance time

    qij(t)/ Qij(t) : pdf/cdf of first passage time from regenerative state Si to a regenerative state Sj or to a failed state Sj without visiting any other regenerative state in (0, t]

    qij.k (t)/Qij.k(t) : pdf/cdf of direct transition time

    from regenerative state Si to a regenerative state Sj or to a failed state Sj visiting state Sk once in (0, t]

    Mi(t) : Probability that the system up initially in

    state Si E is up at time t without visiting to any regenerative state

    Wi(t) : Probability that the server is busy in the state Si 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.

    i : The mean sojourn time in state which is given by

    instantt given that the system entered regenerative state Si at t=0.

    i

    i

    BH (t) : Probability that the server is busy in

    repairing the unit due to hardware failur at an instant t given that the system entered state Si at t = 0.

    i

    i

    BS (t) : Probability that the server is busy in up-

    gradation of the software at an instant t given that the system entered the regenerative state Si at t = 0.

    () : Probability that the server is busy in preventive maintenance of the hardware given that the system entered state at t=0.

    () : Expected number of hardware repairs by the server in (0, t] given that the system entered the regenerative state Si at t = 0.

    NSUi(t) : Expected number of software up- gradations in (0, t] given that the system entered the regenerative state Si at t = 0.

    ():Expected number of hardware preventive maintenances by the server in (0, t] given that the system

    entered the regenerative state = 0.

    K0 : Revenue per unit up time of the system

    K1 : Cost per unit time for which server is busy due hardware repair

    K2 : Cost per unit time for which server is busy due software Up-gradation

    K3 : Cost per unit repair of the failed hardware K4 : Cost per unit up-gradation of the failed

    software

    K5 : Cost per unit time for which server is busy due hardware preventive maintenance

    K6 : Cost per unit preventive maintenance of the failed hardware

    P : Profit of the present model

    P1 : Profit of the system model [5]

  3. ANALYSIS OF SYSTEM MODEL The state transition diagram is shown in the following figure:

    State Transition Diagram

    S6

    S6

    HFWr SFUG

    SFWUg SFUG

    SFWUg SFUG

    S

    1

    f(t)

    S0

    2

    O

    Scs

    O

    f(t)

    f(t)

    1

    g(t)

    m(t)

    0

    HFUr

    S1

    S1

    Scs

    HUPm

    Scs

    HUPm

    Scs

    S3

    5

    2

    SFUg

    S2

    0

    f(t)

    S4 HWPm SFUG

    Up-State Failed State Regenerative Point Fig. 1

    1. Transition Probabilities and Mean Sojourn Times

      Simple probabilistic considerations yield the

      The mean sojourn times (i) in the state Si are

      following expressions for the non-zero elements.

      = 1

      = 1

      = 1

      = () = ()

      0 1+2+0

      1

      2 1+2+0+

      0

      = 1

      = 1 ,

      = 2

      , =

      3

      01 1+2+0

      02 1+2+0 03

      Also

      0

      1+2+0

      0 = 01 + 02 + 03, 1 = 10,

      10 = (0), 20 = (1 + 2 + 0), 24 =

      2 = 20 + 24 + 25 + 26

      And = 20 + 21.6 + 22.5 + 23.4 (3)

      0 {1 (

      +

      + )}, 2

      1+2+0

      1 2 0

      = 2 {1 (

      +

      + )}, =

    2. Reliability and Mean Time to System Failure (MTSF)

      25 1+2+0

      1 2 0 26

      The recursive relations for () are as follows:

      1 {1 (

      +

      + )} , =

      + +

      1 2 0 30

      0(t) = Q01(t) + Q02(t) & 2(t) + Q03(t)

      1 2 0

      (0) , 43 = 52 = (0)

      2(t) = Q20(t) & 0(t) + Q24(t) + Q25(t) + Q26(t) (4)

      Taking LST of above relations (4) and solving for () ,

      For f (t) = e t , g (t) = et and m (t) =

      we have

      0

      1()

      we have

      () = 0

      = 0 {1 (

      +

      + )}()

      The reliability of the system model can be obtained by

      23.4

      1+2+0

      = 2 {1 (

      1

      +

      2 0

      + )}()

      taking Laplace inverse transform of the above equation. The mean time to system failure (MTSF) is given by

      22.5 1+2+0

      1

      1 2 0

      ( )

      1() 1

      = lim 0 =

      (5)

      21.6 = + +

      {1

      1 + 2 + 0 }()

      0 1

      1 2 0

      But, f*(0) = g*(0) = m*(0) =1 and a + b =1 (1)

      It can be easily verified that

      p01+p02+p03 = p10= p20+p24+p25+p26 = p20+p23.4+p22.5+p21.6 = p30 = p43=p52 =1 (2)

      Where 1 = 0 + 02 1 = 1 0220

      (6)

    3. Steady State Availability

    () = lim () = 3 (13)

    The recursive relations for () are given as:

    0 0 0

    2

    A0(t) = M0(t) + q01(t) A1(t) + q02(t) A2(t) + q03(t)

    () = lim () = 3

    (14)

    A3(t)

    0 0 0

    2

    A (t) = q

    (t) A (t)

    () = lim () = 3

    (15)

    1 10 0

    0 0 0

    2

    A2(t) = M2(t)+q20(t) A0(t) + q21.6 (t) A1 (t) + q22.5(t)

    where = (

    (1

    ) +

    )(0)

    A2(t) + q23.4(t) A3(t)

    =

    3 01 10

    22.5

    02 21.6 1

    3 02 2

    A3(t) = q30(t) A0(t)

    = (

    +

    (1 (7)

    )) (0)

    where

    3 02

    23.4 30

    03 30

    22.5 3

    M0 (t) =(1+2+0) ,

    2 . (16)

    M2 (t) =(1+2+0) ()

    (7)

    E. Expected Number of Hardware Repairs

    0

    0

    Taking LT of equation (7) and solving for A* (s) , the

    steady state availability is given by A () lim sA*(s)

    The recursive relations for () are given as:

    0() = 01() & 1() + 02() & 2() +

    0 s0 0

    03

    () & 3()

    N2

    (8)

    () =

    () & (1 + ())

    D

    1

    1

    0

    0

    2 () =

    () & ()+

    () & ()

    10

    10

    Where

    2 20

    0 21.6 1

    2 = 0 (1 p22.5) + p02 2

    2 = 0(1 22.5) + 1(01(1 22.5) +

    0221.6)+02 + 3(03(1 22.5) + 23.4) (9)

    + 22.5() & 2()

    +23.4() & 3()

    2 () = () & () (17)

    0

    0

    3 30 0

    D. Busy Period of the Server (a). Due to Hardware Repair

    Taking LST of relations (17) and solving for (). The expected number of hardware repairs are given by

    The recursive relations for () are as follows:

    = lim () = 4

    (18)

    0 0 0

    2

    0 () = 01()©1 () + 02()©2 ()

    3

    3

    + 03()©()

    0

    0

    1 1 10

    1 1 10

    () = () + ()©()

    2

    2

    0

    0

    1

    1

    () = 20()©() + 21.6()() +

    22.5() () + 23.4()©()

    where 4 = 01(10(1 22.5) +

    0221.6) 2 . (19)

    1. Expected Number of Software Up-gradations

      The recursive relations for NSU (t) are given as:

      2 3

      i

      3 () = 30()©0 () (10)

      NSU (t) = Q

      (t) & NSU (t) + Q

      (t) & NSU (t) +

      1

      1

      where () =

      ()

      0 01

      1 02 2

      1. Due to Software Up-gradation

        The recursive relations for () are as follows:

        () = 01()©() + 02()©()+03()©()

        Q03(t) & NSU3(t)

        NSU1(t) = Q10(t) & NSU0 (t)

        NSU2(t) = Q20(t) & (1 + NSU0(t)) + Q21.6(t) &

        0 1 2 3

        () = 10()©()

        1 0 (1 + NSU1(t)) + Q22.5(t) & (1 + NSU2(t)) + Q23.4(t) &

        () = () + 20()©() + 21.6()() +

        2 2 0 1

        2 23.4 3

        2 23.4 3

        22.5() () + ()©()

        () = 30()©() (11)

        (1 + NSU3(t))

        NSU3(t) = Q30(t) & NSU0(t) (20) Taking

        3 0

        2

        2

        where () = (1+2+0) () + (1(1+2+0)©1)() +

        LST of relations (20) and solving for NSU(s). The

        0

        0

        expected numbers of software up-gradations are given by

        ( + + )

        ( + + )

        NSU () = lim sNSU (s) = N5

        (21)

        (2 1

        0 ©1)() + (0 1

        2 0 ©1)()

        0 s0 0 D2

      2. Due to Preventative Maintenance

      The recursive relations for () are as follows:

      0 1

      0 1

      () = 01()©()

      + 02()© () + 03()© ()

      Where N5 = p02 and D2 is already mentioned. (22)

    2. Expected Number of Preventative Maintenance

    The recursive relations for () are given as:

    () = () & () + () &

    2 2

    () = 10()©()

    0 01

    1 02

    1 0 () + () & ()

    () = 20()©()+21.6() () +

    2 03 3

    2 0 1

    22.5()() + 23.4() ()

    () =

    () & ()

    1 1 1 10 0

    () = () + 30()©() (12)

    3 3 0

    2() = 20() & 0()

    where 3 () = ()

    Taking LT of relations (10), (11) & (12), solving

    + 21.6() & 1()

    for (), () (). The time for which server

    + () & ()

    0 0 0

    22.5 2

    is busy due to repairs, up-gradations and preventative maintenances respectively are given by

    + 23.4() & 3()

    3() = 30() & (1 + 0 ()) (23)

    Taking LST of relations (23) and solving for ().

    1

    1 + 2

    1 2 0

    1 2 0

    =

    =

    = ( + + + )

    0

    The expected numbers of hardware preventative

    1

    (1+2+0)(1+2+0+)2 ( + + )( + + +)

    maintenances are given by

    1 2 0

    1 2 0

    = lim () = 6

    (24)

    (1+2+0+)

    0 0

    0 2

    2 =

    where 6

    = 02

    23.4

    30

    + 03

    30

    (1

    (1+2+0)(1+2+0+)

    22.5) 2 . (25)

  4. PROFIT ANALYSIS

    The profit incurred to the system model in steady state can be obtained as:

    2

    (1+2+0+)((1+0+)+1(1+2+0+)+

    (0(1+0+)+0(1+2+0))

    = +2(2+(1+2+0)(1+2+0+2))

    (1+2+0)(1+2+0+)(1+2+0+)

    = 00 10 20 30 40

    0

    0

    5 60 (26)

    0, , , 0, 0, , 0

    3

    1((1 + 0 + )(1 + 2 + 0) + 12)

    1 2 0 1 2 0 1 2 0

    1 2 0 1 2 0 1 2 0

    = ( + + )( + + + )( + + )

    0 0 0

    = 2

    .

    3 (1+2+0)(1+2+0+)

    = 0(1+2+0+)

  5. PARTICULAR CASES

    Suppose () = , () =

    () =

    We can obtain the following results:

    3

    4

    (1+2+0)(1+2+0+)

    = 1((1+0+)(1+2+0)+12) (1+2+0)(1+2+0+)(1+2+0)

    2

    =

    ( ) = 1

    0 1

    , ( ) = 2

    0 2

    5

    6 =

    (1 + 2 + 0)

    0(1+2+0+)

    3

    3

    () =

    0 2

    (1+2+0)(1+2+0+)

  6. CONCLUSION

    The reliability measures of a computer system have been obtained by assuming () = , () =

    0

    0

    () =

    3

    2

    () =

    . And, graphs are drawn for the

    ( ) = 3

    0 2

    particular results of these measures as shown in figures 2 to

    4. It is analyzed that the mean time to system failure (MTSF), availability and profit go on decreasing with the increase of failure rates ( and ) and the rate ( ) by

    1 2 0

    4

    2

    2

    (0) =

    5

    2

    2

    (0) =

    6

    ( ) =

    which hardware undergoes for preventive maintenance after a pre specific operation timet while their values keep on moving up with the increase of hardware repair rate (), software up-gradation rate () and preventive maintenance rate () provided system has more chances of hardware failure. Further, it is interesting to note that a computer system has more values of these reliability measures when it has more chances of software failure than that of

    Where

    0 2

    hardware failure. The graphical presentation of the results related to these reliability measures obtained for the system model is shown in the figures 2 to 4.

  7. GRAPHICAL PRESENTATION OF RELIABILITY MEASURES FOR THE SYSTEM MODEL

    MTSF

    MTSF

    Fig. 2: MTSF Vs Hardware Failure Rate (1)

    250

    200

    150

    100

    2=0.001,=2,=5,a=0.6,b=0.4,=0.034,0=0.001

    2=0.002

    =3 =7

    a=0.4,b=0.6 =0.035 0=0.002

    250

    200

    150

    100

    2=0.001,=2,=5,a=0.6,b=0.4,=0.034,0=0.001

    2=0.002

    =3 =7

    a=0.4,b=0.6 =0.035 0=0.002

    50

    0

    50

    0

    0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

    Hardware Failure Rate (1)

    0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

    Hardware Failure Rate (1)

    Availability

    Availability

    Fig. 3: Availability Vs Hardware Failure Rate (1)

    0.98

    0.97

    0.96

    0.95

    0.94

    0.93

    0.92

    0.91

    0.9

    0.89

    2=0.001,=2,=5,a=0.6,b=0.4,=0.034,0=0.001

    2=0.002

    =3 =7

    a=0.4,b=0.6 =0.035 0=0.002

    0.98

    0.97

    0.96

    0.95

    0.94

    0.93

    0.92

    0.91

    0.9

    0.89

    2=0.001,=2,=5,a=0.6,b=0.4,=0.034,0=0.001

    2=0.002

    =3 =7

    a=0.4,b=0.6 =0.035 0=0.002

    0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

    Hardware Failure Rate (1)

    0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

    Hardware Failure Rate (1)

    Fig. 4: Profit (P) Vs Hardware Failure Rate (1)

    2=0.001,=2,=5,a=0.6,b=0.4,=0.034,0=0.001 2=0.002

    =3

    =7 a=0.4,b=0.6 =0.035

    14600

    K0=15000 K1= 1000

    K2 = 700

    14400

    14200

    Profit (P)

    Profit (P)

    14000

    13800

    13600

    13400

    13200

    13000

    0=0.002

    K3 = 1500

    K4 = 1200

    K5 = 300 K6 = 600

    0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

    Hardware Failure Rate (1)

  8. COMPARATIVE STUDY OF PROFIT OF SYSTEM MODELS

    The profit of the present model has been compared with that of the model discussed in the research paper Munday and Malik (2015). It is revealed that the present model is less profitable. And, hence we can say that

    the concept of hardware preventive maintenance in a computer system with software redundancy in cold standby is not much helpful in making the system more profitable. The behavior of the profit difference of the system model with respect to hardware failure rate (1) has been shown graphically in figure 5.

  9. GRAPHICAL PRESENTATION OF PROFIT DIFFERENCES

0

-100

-200

P-P1

P-P1

-300

-400

-500

-600

-700

Fig. 5: P P1 Vs Hardware Failure Rate (1)

Hardware Failure Rate (1)

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

01

01

K0=15000 K1= 1000

K2 = 700

K3 = 1500

K4 = 1200

2=0.001,=2,=5,a=0.6,b=0.4,=0.034,0=0.0 K7 = 300

2=0.002

=3 =7

a=0.4, b=0.6

K8 = 600

REFERENCES

[1]. S. Osaki, Reliability analysis of a two-unit standby redundant system with preventive maintenance, IEEE Transactions on Reliability, vol. R-21 (1), pp. 2 29, 1972.

[2]. M. Rander, S. Kumar and A. Kumar, Cost analysis of a two dissimilar cold standby system with preventive maintenance and replacement of standby, Microelectronics Reliability, vol. 34 (1), pp. 171 174, 1994.

[3]. S.C. Malik, P. Nandal and M.S. Barak, Reliability analysis of a system under preventive maintenance, Journal of Mathematics and System Sciences, vol. 5 (1), pp. 92 115,

2009.

[4]. S.C. Malik, Reliability modeling of a computer system with preventive maintenance and priority subject to maximum operation and repair times, International Journal of System Assurance Engineering and Management, vol. 4 (1), pp. 94-

100, 2013.

[5]. V.J. Munday and S.C. Malik, Stochastic modelling of a computer system with software redundancy, International Journal of Engineering and Management Research, vol. 5(1), pp.295-302, 2015.

[6]. V.J. Munday, Permila and S.C. Malik, Reliability modelling of a computer system with hardware redundancy subject to preventive maintenance, International Journal of Statistics and Reliability Engineering, vol. 3(2), pp. 176-189, 2016.

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