 Open Access
 Authors : G. Saritha , M. Tirumala Devi , T. Sumathi Uma Maheswari
 Paper ID : IJERTV9IS030082
 Volume & Issue : Volume 09, Issue 03 (March 2020)
 Published (First Online): 16032020
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Reliability and Availability for NonRepairable & 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 nonrepairable system and defined the mean time to failure of nonrepairable systems with different failure rates. This technique is also used to compute the steadystate 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 Nonrepairable 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 nonrepairable 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. ElDamcese 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:

3component nonrepairable system with different failure rates
A nonrepairable system has finite life time. The probability of a system failure will increase over time during system operation. A nonrepairable system remains failed, after it failed once.
3component 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: Nonrepairable system state diagram
Figure 1: Nonrepairable system state diagram
The probability of the system state one i.e., the probability of 3component 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 GumbelHougaard family copula. M.A. ElDamcese 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 3component 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

3component 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.
3component 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 3components 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:

component nonrepairable 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

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:

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):110.

James Li, Reliability comparative evaluation of active redundancy vs. Standby redundancy, International Journal of Mathematical, Engineering and Management Sciences, Vol.1, No.3, 122129, 2016.

M.A. ElDamcese 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):95106.

Jacob Cherian, Madhu Jain and G.C. Sharma, Reliability of a standby system with repair, Indian Journal Pure Applied Mathematics, 18(12):10611068, December 1987.

Garima Chopra, Reliability Measures of two dissimilar units parallel system using GumbelHougaard family copula, International Journal of Mathematical, Engineering and Management Science, Vol.4, No.1, 116130, 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

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