Reliability and Availability for Non-Repairable & Repairable Systems using Markov Modelling

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Reliability and Availability for Non-Repairable & Repairable Systems using Markov Modelling

1G.Saritha

Department of Mathematics Kakatiya University Warangal, TS, India

2M. Tirumala Devi and 3T. Sumathi Uma Maheswari

2,3Department of Mathematics Kakatiya University Warangal, TS, India

AbstractMarkov model provides great flexibility in modelling the timing of events. Markov analysis is a method of analysis that can be applied to both repairable and non- repairable types of systems. In this paper, Markov modelling technique is used to compute the reliability for non-repairable system and defined the mean time to failure of non-repairable systems with different failure rates. This technique is also used to compute the steady-state availability for repairable systems and to derived the mean time between failure of repairable systems with different failure rates and repair rates.

Keywords Markov model; Repairable and Non-repairable systems; MTTF; MTBF; Failure rate; Repair rate; Reliability;

INTRODUCTION

Markov analysis is the mathematical abstractions to model simple or complex concepts in quite easily computable form. The Markov analysis is also considered powerful modelling and analysis tool in solving reliability tribulations. Markov analysis is a tool for modelling complex system designs involving timings, sequence, repair, redundancy and fault- tolerance.

Mean Time To Failure (MTTF) describes the expected time to failure for a non-repairable system. MTTF is commonly refer to as the life time of any product or a device. MTTF can be mathematically calculated by

Mean Time Between Failure (MTBF) is the predicted time that passes between one previous failure of a mechanical/electrical system to the next failure during normal operation or, the time between one system breakdown and the next.

James Li [1] & [2] derived the reliability for a parallel redundant system with different failure rate & repair rate using Markov modelling and reliability comparative evaluation of active redundancy vs. Standby redundancy respectively. M.A. El-Damcese and N.S. Temraz [3] studied analysis for a parallel repairable system with different failure

al [6] estimated availability and reliability analysis of three elements parallel system with fuzzy failure and repair rate.

MATHEMATICAL MODEL:

  1. 3-component non-repairable system with different failure rates

    A non-repairable system has finite life time. The probability of a system failure will increase over time during system operation. A non-repairable system remains failed, after it failed once.

    3-component system can have 23 8 distinct states. The failure rates are 1 , 2 and 3 . The transition rate diagram is given below.

    State 2

    Component 1 fail

    Component 2 good

    Component 3 good

    State 5

    Component 1 fail

    Component 2 fail

    Component 3 good

    State 2

    Component 1 fail

    Component 2 good

    Component 3 good

    State 5

    Component 1 fail

    Component 2 fail

    Component 3 good

    State 1

    Component 1 good

    Component 2 good

    Component 3 good

    State 1

    Component 1 good

    Component 2 good

    Component 3 good

    State 3

    Component 1 good

    Component 2 fail

    Component 3 good

    State 3

    Component 1 good

    Component 2 fail

    Component 3 good

    State 8

    Component 1 fail

    Component 2 fail

    Component 3 fail

    State 8

    Component 1 fail

    Component 2 fail

    Component 3 fail

    State 6

    Component 1 good

    Component 2 fail

    Component 3 fail

    State 4

    Component 1 good

    Component 2 good

    Component 3 fail

    State 7

    Component 1 fail

    Component 2 good

    Component 3 fail

    State 4

    Component 1 good

    Component 2 good

    Component 3 fail

    State 7

    Component 1 fail

    Component 2 good

    Component 3 fail

    Figure 1: Non-repairable system state diagram

    Figure 1: Non-repairable system state diagram

    The probability of the system state one i.e., the probability of 3-component good at time is

    Taking the Limit as t 0 , and the differential equation is

    Solving the above equation by using the Laplace transformations, and taking assumption that the system good

    modes. Jacob Cherian et al [4] has described reliability of a

    standby system with repair. Garima Chopra [5] studied reliability measures of two dissimilar units parallel system using Gumbel-Hougaard family copula. M.A. El-Damcese et

    at initial time t = 0, hence

    then the following equation is obtained

    P1 0 1

    The probability of the system in state two at t t time is Taking the Limit as t 0 , and the differential equation is

    Taking Laplace transform, the equation is

    Using inverse Laplace transform, we get

    The probability of the system in state six at t t time

    Taking the Limit as

    t 0 , and the differential equation is

    Using inverse Laplace transform

    The probability of the system in state three at t t time is Taking the Limit as t 0 , and the differential equation is

    Taking Laplace transform, the equation is

    Using inverse Laplace transform, we get

    Taking Laplace transform, the equation is

    Using inverse Laplace transform, we get

    The probability of the system in state seven at t t time is

    The probability of the system in state four at t t time is

    Taking the Limit as t 0 , and the differential equation is

    Taking the Limit as

    t 0 , and the differential equation is

    Taking Laplace transform, the equation is

    Using inverse Laplace transform, we get

    The probability of being in state five at t t time

    Taking Laplace transform, the equation is

    Using inverse Laplace transform, we get

    Taking the Limit as

    t 0 , and the differential equation is

    Taking Laplace transform, the equation is

    The reliability of the 3-component system, if at least one component should operate

    Mean Time To Failure (MTTF) of the system

    The reliability function in the exponential case

    is failure rate and t is the period of time over which reliability is measured. The probability of failure is equal to .

    State

    Component 1

    Component 2

    Component 3

    System state

    1

    good

    good

    good

    success

    2

    failed

    good

    good

    success

    3

    good

    failed

    good

    success

    4

    good

    good

    failed

    success

    5

    failed

    failed

    good

    success

    6

    good

    failed

    failed

    success

    7

    failed

    good

    failed

    success

    8

    failed

    failed

    failed

    down

    Table.1

    If all the three components have the same failure rate, we get

  2. 3-component repairable system with different failure rates and repair rates

A repairable sytem is a system which, after failure, can be restored to a functioning condition by some maintenance action other than replacement of the entire system.

3-component system can have distinct states. The failure rates are and the repair rates are The transition rate diagram is given below.

State 2

State 5

Component 1 fail

Component 1 fail

Component 2 good

Component 2 fail

Component 3 good

Component 3 good

State 1

Component 1 good

Component 2 good

Component 3 good

State 3

State 6

Component 1 good

Component 1 good

Component 2 fail

Component 2 fail

Component 3 good

Component 3 fail

State 8

Component 1 fail

Component 2 fail

Component 3 fail

State 1

Component 1 good

Component 2 good

Component 3 good

State 3

State 6

Component 1 good

Component 1 good

Component 2 fail

Component 2 fail

Component 3 good

Component 3 fail

State 8

Component 1 fail

Component 2 fail

Component 3 fail

State 4

State 7

Component 1 good

Component 1 fail

Component 2 good

Component 2 good

Component 3 fail

Component 3 fail

Figure 2: Repairable system state diagram

Figure 2: Repairable system state diagram

The transition matrix is

State

Compon ent 1

Compone nt 2

Compone nt 3

System state

1

2

3

4

5

6

7

8

State

Compon ent 1

Compone nt 2

Compone nt 3

System state

1

2

3

4

5

6

7

8

1 2 3

1

2

2 1

0

2

0

3

0

0

3

0

0

1 3

0

0

1

0

0

2

0

0

3 2

0

0

0

0

0

0

0

0

0

0

0

0

1

0

0

2 1

0

0

0

0

0

0

0

0

1

1

2

2 1

0

2

0

3

0

0

3

0

0

1 3

0

0

1

0

0

2

0

0

3 2

0

0

0

0

0

0

0

0

0

0

0

0

1

0

0

2 1

0

0

0

0

0

0

0

0

1

1

2 3 0

0 0

3 2 0

0 0 0

0 0 0

3 1 3

From the transition matrix, the probability of the system in state one at t t time is

Table.2

From table.2 the probability of success is

The probability of the system in state two at t t time is

Taking the Limit as t 0 , and the differential equation is

Integrating above equation

and MTTF is

The boundary conditions are at all other conditions.

P1 0 1& P8 1 , zero

Where T1 ,T2 &T5

and five respectively.

are the expected time in state one, two

The probability of the system in state six at t t time is

The probability of the system in state three at t t time is Taking the Limit as t 0 , and the differential equation is

Integrating above equation

Where T1 ,T3 &T6 are the expected time in state one, three

Taking the Limit as t 0 , and the differential equation is

Integrating the above equation

Substituting T value in T we get

and six respectively. 6 3

The probability of the system in state four at t t time is Taking the Limit as t 0 , and the differential equation is

Integrating above equation

Substituting T3 value in T6 we get

The probability of the system in state seven at t t time is

Where T ,T &T

are the expected time in state one, four

Taking the Limit as t 0 , and the differential equation is

1 4 7

and seven respectively.

The probability of the system in state five at t t time is Taking the Limit as t 0 , and the differential equation is

Integrating above equation

Substituting T5 value in T2 we get

Substituting T2 value in T5 we get

Integrating the above equation

Substituting T7 value in T4

Substituting T4 value in T7 we get

The probability of the system in state eight at t t time is

Taking the Limit as t 0 , and the differential equation is

Substituting T5 ,T6 &T7 values in above equation

The MTBF is the sum of the expected time in state one, two, three, four, five, six and seven.

Substituting T1 value in above equation, we get

If all the 3-components have the same failure rate and repair rate, we get

0.02

0.06

0.001

40

0.987867

1041.975

0.02

0.06

40

0.903527

155.3059

0.02

0.06

0.017

40

0.838035

104.8716

0.02

0.06

0.025

40

0.787066

87.61905

0.02

0.06

0.033

40

0.747312

79.27058

0.02

0.06

0.041

40

0.716239

74.52798

0.02

0.06

0.049

40

0.691902

71.57095

0.02

0.06

0.057

40

0.672804

69.61199

0.02

0.06

0.065

40

0.657792

68.25818

0.02

0.06

0.001

40

0.987867

1041.975

0.02

0.06

0.009

40

0.903527

155.3059

0.02

0.06

0.017

40

0.838035

104.8716

0.02

0.06

0.025

40

0.787066

87.61905

0.02

0.06

0.033

40

0.747312

79.27058

0.02

0.06

0.041

40

0.716239

74.52798

0.02

0.06

0.049

40

0.691902

71.57095

0.02

0.06

0.057

40

0.672804

69.61199

0.02

0.06

0.065

40

0.657792

68.25818

Mean time to repair (MTTR) is a basic measure of the maintainability of repairable items. It represents the average time required to repair a failed component or device.

Where

If the repair rate t is constant and is equal to then

On simplification, we get

Numerical results:

  1. component non-repairable system with different failure rates

    0.001

    0.02

    0.05

    40

    0.988642

    1041.831

    0.009

    0.02

    0.05

    40

    0.909736

    155.7947

    0.017

    0.02

    0.05

    40

    0.848541

    105.835

    0.025

    0.02

    0.05

    40

    0.800984

    88.94737

    0.033

    0.02

    0.05

    40

    0.763956

    80.88555

    0.041

    0.02

    0.05

    40

    0.735073

    76.37223

    0.049

    0.02

    0.05

    40

    0.712504

    73.60144

    0.057

    0.02

    0.05

    40

    0.694843

    71.79583

    0.065

    0.02

    0.05

    40

    0.681005

    70.5698

    0.03

    0.001

    0.04

    40

    0.986559

    1030.164

    0.03

    0.009

    0.04

    40

    0.893242

    144.128

    0.03

    0.017

    0.04

    40

    0.820961

    94.16836

    0.03

    0.025

    0.04

    40

    0.764867

    77.2807

    0.03

    0.033

    0.04

    40

    0.721257

    69.21889

    0.03

    0.041

    0.04

    40

    0.687296

    64.70556

    0.03

    0.049

    0.04

    40

    0.660808

    61.93477

    0.03

    0.057

    0.04

    40

    0.640123

    60.12916

    0.03

    0.065

    0.04

    40

    0.623952

    58.90313

    In this case the repair rates

    1 2 3 , then

    1 , 2 & 3 are constants and

    The steady state availability is

    0.01

    0.02

    0.05

    10

    0.99638

    145

    0.01

    0.02

    0.05

    20

    0.978602

    145

    0.01

    0.02

    0.05

    30

    0.945621

    145

    0.01

    0.02

    0.05

    40

    0.901213

    145

    0.01

    0.02

    0.05

    50

    0.849698

    145

    0.01

    0.02

    0.05

    60

    0.794635

    145

    0.01

    0.02

    0.05

    70

    0.738605

    145

    0.01

    0.02

    0.05

    80

    0.683349

    145

    0.01

    0.02

    0.05

    90

    0.629983

    145

    0.01

    0.02

    0.05

    10

    0.99638

    145

    0.01

    0.02

    0.05

    20

    0.978602

    145

    0.01

    0.02

    0.05

    30

    0.945621

    145

    0.01

    0.02

    0.05

    40

    0.901213

    145

    0.01

    0.02

    0.05

    50

    0.849698

    145

    0.01

    0.02

    0.05

    60

    0.794635

    145

    0.01

    0.02

    0.05

    70

    0.738605

    145

    0.01

    0.02

    0.05

    80

    0.683349

    145

    0.01

    0.02

    0.05

    90

    0.629983

    145

    1. component repairable system with same failure rate and repair rates

      0.02

      0.1

      0.996169

      866.6667

      0.04

      0.1

      0.983087

      193.75

      0.06

      0.1

      0.964798

      91.35802

      0.08

      0.1

      0.944299

      56.51042

      0.1

      0.1

      0.923077

      40

      0.12

      0.1

      0.901863

      30.63272

      0.14

      0.1

      0.881027

      24.68416

      0.16

      0.1

      0.860756

      20.60547

      0.18

      0.1

      0.841142

      17.64975

      CONCLUSION:

      Reliability and MTTF have been derived for the non- repairable parallel redundant system with different failure rate and MTBF has been derived for repairable systems with different failure rate and repair rate. And also steady- state availability has been computed for repairable system with same failure rate and repair rate. It is observed from the computations that the reliability and MTTF decreases as

      1 , 2 & 3 increases and reliability decreases as t increases for the non repairable system and as

      increases steady state availability & MTBF decreases,

      as increases steady- state availability & MTBF increases for the repairable system.

      REFERENCES:

      1. James Li, Reliability calculation of a parallel redundant system with different failure rate & repair rate using Markov modelling, Journal of Reliability and Statistical Studies, Vol.9, Issue 1(2016):1-10.

      2. James Li, Reliability comparative evaluation of active redundancy vs. Standby redundancy, International Journal of Mathematical, Engineering and Management Sciences, Vol.1, No.3, 122-129, 2016.

      3. M.A. El-Damcese and N.S. Temraz, Analysis for a parallel repairable system with different failure modes, Journal of Reliability and Statistical Studies, Vol.5, Issue 1(2012):95-106.

      4. Jacob Cherian, Madhu Jain and G.C. Sharma, Reliability of a standby system with repair, Indian Journal Pure Applied Mathematics, 18(12):1061-1068, December 1987.

      5. Garima Chopra, Reliability Measures of two dissimilar units parallel system using Gumbel-Hougaard family copula, International Journal of Mathematical, Engineering and Management Science, Vol.4, No.1, 116-130, 2019.

        0.05

        0.1

        0.974359

        126.6667

        0.05

        0.12

        0.981706

        149.0667

        0.05

        0.14

        0.98647

        173.6

        0.05

        0.16

        0.989704

        200.2667

        0.05

        0.18

        0.99198

        229.0667

        0.05

        0.2

        0.993631

        260

        0.05

        0.22

        0.994857

        293.0667

        0.05

        0.24

        0.995787

        328.2667

        0.05

        0.26

        0.996506

        365.6

        0.05

        0.1

        0.974359

        126.6667

        0.05

        0.12

        0.981706

        149.0667

        0.05

        0.14

        0.98647

        173.6

        0.05

        0.16

        0.989704

        200.2667

        0.05

        0.18

        0.99198

        229.0667

        0.05

        0.2

        0.993631

        260

        0.05

        0.22

        0.994857

        293.0667

        0.05

        0.24

        0.995787

        328.2667

        0.05

        0.26

        0.996506

        365.6

      6. M.A. El-Damcese, F. Abbas and E. El-Ghamry, Availability and reliability analysis of three elements parallel system with fuzzy failure and repair rate, Journal of Reliability and Statistical Studies, Vol.75, Issue 1(2014):125-142.

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