Reliability Modeling and Analysis of a Parallel Unit System with Priority to Repair over Replacement Subject to Maximum Operation and Repair Times

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Reliability Modeling and Analysis of a Parallel Unit Priority to Repair o

Operation and Repair TimesReetu

1Amity Institute of Applied Science, AMITY University, Noida2Department of Statistics, M.

ABSTRACT This paper proposes the study of modelling and analysis of a two unit parallel system. A constant failure rate is considered for the units which are identical in nature. All repair activities like repair, replacement, preventive maintenance are mended immediately by a single server. The repair of the unit is done after its failure and if the fault is not rectified by the server within a given repair time, called maximum repair time, the unit replaced by new one. And, if there is no fault occurs up to a preoperation time, called maximum operation time, the unit undergoes for the preventive maintenance. The unit works as new after all repair activities done by the server. Priority to repair of one unit is given over the replacement of the other one. All random variables are statistically independent. The distribution for the failure, preventive maintenance and replacement rates are negative exponential whereas the distribution for all repair activities are taken as arbitrary with different probability density functions. Semi-Markov and regenerative point techniques are used to derive some reliability measures in steady state. The variation of MTSF, availability and profit function has been observed graphically for various parameters and costs. Keywords: Parallel system, Preventive MaiReplacement, Priority, Reliability Measures INTRODUCTION The general purpose of the modern world is to achieve the require performance level using the

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International Journal of Trend in Scientific Research and Development (IJTSRD)

International Open Access Journal

Reliability Modeling and Analysis of a Parallel Unit Priority to Repair over Replacement Subject to Maximum

Operation and Repair Timeseetu Rathee1, D. Pawar1, S. C. Malik2

1Assistant Professor, 2Professor Amity Institute of Applied Science, AMITY University, Noida, Uttar Pradesh

Department of Statistics, M. D. University, Rohtak, Haryana, India

modelling and analysis of a two unit parallel system. A constant failure rate is considered for the units which are identical in nature. All repair activities like repair, replacement, preventive maintenance are mended

repair of the unit is done after its failure and if the fault is not rectified by the server within a given repair time, called maximum repair time, the unit replaced by new one. And, if there is no fault occurs up to a pre-fixed

ximum operation time, the unit undergoes for the preventive maintenance. The unit works as new after all repair activities done by the server. Priority to repair of one unit is given over the replacement of the other one. All random

ally independent. The distribution for the failure, preventive maintenance and replacement rates are negative exponential whereas the distribution for all repair activities are taken as arbitrary with different probability density

d regenerative point techniques are used to derive some reliability measures in steady state. The variation of MTSF, availability and profit function has been observed graphically for various parameters and costs.

Parallel system, Preventive Maintenance, Measures

The general purpose of the modern world is to achieve the require performance level using the

lowest possible cost. And, the parallel system works not only for maximize the profit but the failure risk as well as cost.their practical applications, reliability models of parallel systems have been developed and analyzed stochastically by the researchers and reliability engineers. Kishan and Kumar (2009stochastically a parallel system using preventive maintenance. Further, kumar et al. (2010) and Malik and Gitanjali (2012) have analyzed costparallel system subject to degradation after repair and arrival time of the server respecenhance the profit of the system(2013) and Rathee and Chander (2014) developed parallel systems using the concept of priority. Also, the objective of the present paper is to determine the reliability measures by givingpriority to one repair activity over the other ones. A constant failure rate is considered for the units which are identical in nature. All repair activities are done immediately by a single server. The repair of the unit is conducted after its failure and if the fault is not rectified by the server within a given repair time, the unit replaced by new one. And, if there is no fault occurs up to a pre-fixed operation time, the preventive maintenance is conducted. The unit works as new after all repair activities done by the server. Priority to repair of one unit is given over the replacement of the other one. All random variables are statistically independent. The distribution for the failure, preventive maintenance and replacement rates are

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International Open Access Journal

Reliability Modeling and Analysis of a Parallel Unit System with Replacement Subject to Maximum

Uttar Pradesh, India , India

And, the parallel system works not only for maximize the profit but also for minimize the failure risk as well as cost. Keeping in view of their practical applications, reliability models of parallel systems have been developed and analyzed stochastically by the researchers and reliability

Kishan and Kumar (2009) evaluated stochastically a parallel system using preventive

kumar et al. (2010) and Malik and Gitanjali (2012) have analyzed cost-benefit of a

subject to degradation after repair and arrival time of the server respectively. However, to enhance the profit of the system Reetu and Malik (2013) and Rathee and Chander (2014) developed parallel systems using the concept of priority.

Also, the objective of the present paper is to determine the reliability measures by giving the priority to one repair activity over the other ones. A constant failure rate is considered for the units which are identical in nature. All repair activities are done immediately by a single server. The repair of the unit

e and if the fault is not rectified by the server within a given repair time, the unit replaced by new one. And, if there is no fault

fixed operation time, the preventive maintenance is conducted. The unit works as new

ctivities done by the server. Priority to repair of one unit is given over the replacement of the other one. All random variables are statistically independent. The distribution for the failure, preventive maintenance and replacement rates are

International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470

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negative exponential whereas the distribution for all repair activities are taken as arbitrary with different probability density functions. Semi-Markov and regenerative point techniques are used to derive some reliability measures in steady state. The variation of MTSF, availability and profit function has been observed graphically for various parameters and costs. NOTATIONS: E : Set of regenerative states 𝐸 : Set of non-regenerative states λ : Constant failure rate α0 : The rate by which system undergoes for preventive maintenance (called maximum constant rate of operation time) β0 : The rate by which system undergoes for replacement (called maximum constant rate of repair time) FUr /FWr : The unit is failed and under

repair/waiting for repair FURp/FWRp : The unit is failed and under

replacement/waiting for replacement UPm/WPm : The unit is under preventive

maintenance/waiting for preventive maintenance

FUR/FWR : The unit is failed and under repair / waiting for repair continuously from previous state

FURP/FWRP : The unit is failed and under /waiting for replacement continuously from previous state

UPM/WPM : The unit is under/waiting for preventive maintenance continuously from previous state

g(t)/G(t) : pdf/cdf of repair time of the unit f(t)/F(t) : pdf/cdf of preventive maintenance

time of the unit r(t)/R(t) : pdf/cdf of replacement time of the unit

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

qij.kr (t)/Qij.kr(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, Sr 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 𝑆 which is given by

𝜇 = 𝐸(𝑇) = 𝑃(𝑇 > 𝑡) 𝑑𝑡∞

= 𝑚 ,

where𝑇denotes the time to system failure. mij:Contribution to mean sojourn time (i) in state Si when system transits

directlyto state Sjso that i ijj

m and mij =

* '( ) (0)ij ijtdQ t q

& : Symbol for Laplace-Stieltjes convolution/Laplace convolution */** : Symbol for Laplace Transformation /LaplaceStieltjes Transformation The states S0, S1, S2, S4, S6and S7 are regenerative while the states S3,S5, S8,S9, S10, S11 and S12 are non-regenerative as shown in figure 1.



TRANSITION STATE DIAGRAM Up-State Failed-State Regenerative point

Fig. 1

Transition Probabilities and Mean Sojourn Times Simple probabilistic considerations yield the following expressions for the non-zero elements

0

)()( dttqQp ijijij as (1)

010

2

2p

, 002

02p

, *015 0 0

0 0

(1 ( )),( )

p g

*40 0( ),p r *

13 0 00 0

(1 ( ))( )

p g

, *

60 0( ),p f

*10 0 0( ),p g * *

17.3 0 0 00 0

(1 ( ))(1 ( ))( )

p g g

,

*06,12 66.12 0

0

(1 ( ))( )

p p f

, *

6,11 61.11 00

(1 ( ))( )

p p f

,

*049 46.9 0

0

(1 ( ))( )

p p r

, *

37 5,10 78 74.8 01 ( )p p p p g ,

*014 0 0

0 0

(1 ( ))( )

p g

, *

47 00

(1 ( ))( )

p r

,

*31 56 74 0( )p p p g , 26 84 96 10,6 11,1 12,6 1p p p p p p

It can be easily verify that

01 02 10 13 14 15 26 40 47 49 60 6,11 6,12 1p p p p p p p p p p p p p

10 14 11.3 16.5 16.5,10 17.3 40 47 46.9

60 61.11 66.12 74 74.8

1

1

p p p p p p p p p

p p p p p

βO

g(t) S10

S7

FUr FWRp FURp

WPM

S11

S12 S9 S8

S6

S5

S4

S3 S2 S1 SO

βO βO

βO

r(t)

r(t)

r(t)

r(t)

g(t)

g(t) g(t)

f(t) f(t)

f(t)

f(t)

λ λ

λ 2λ

αO αO

αO

αO

Upm WPm

FWr FUR

FUR WPm

FWr UPM

FURp FWRP

WPm FURP

UPM WPm

O O

O FUr

O FURp

O UPm



The mean sojourn times (𝝁𝒊) is in the state Si are

0 01 02 ,m m 1 10 13 14 15 ,m m m m 2 26 ,m 4 40 47 49 ,m m m 6 60 6,11 6,12 ,m m m '1 10 14 11.3 16.5 16.5,10 17.3 ,m m m m m m '4 40 47 46.9 ,m m m 7 74 74.8 ,m m

Reliability and Mean Time to System Failure (MTSF) Let i(t) be the cdf of first passage time from regenerative state Si to a failed state. Regarding thefailed state as

absorbing state, we have the following recursive relations for i(t):

0 01 1 02( ) ( ( )) ( )t Q t t Q t &

0 14 41 10 13 15( ) (( ) ( ) ( ) ( )) ( )t Q t t tQ Q tt t Q & &

0 48 44 0 94( ) ( ) ( ) ( ) ( )tt Q t Q tQ t & (2)

Taking LST of above relation (7.4) and solving for Ф∗∗(s), we have **

* 1 ( )( )

sR s

s

(3)

The reliability of the system model can be obtained by taking Inverse Laplace transform of (3). The mean time to system failure (MTSF) is given by

MTSF = **

0

1 ( )lims

s N

s D

(4)

Where

0 01 1 01 14 4N p p p and 01 10 01 14 401D p p p p p (5)

Steady State Availability Let Ai(t) be the probability that the system is in up-state at instant ‘t’ given that thesystementered regenerative state Si at t = 0.The recursive relations for ( )iA t are given as:

0 0 01 1 02 2( ) ( ) ( ) ( ) ( ) ( )A t M t q t A t q t A t

1 1 10 0 11.3 1 14 4

17.3 7 16.5 16.5,10 6

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ( ) ( )) ( )

A t M t q t A t q t A t q t A t

q t A t q t q t A t

2 26 6( ) ( ) ( )A t q t A t

4 4 40 0 47 7 46.9 6( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )A t M t q t A t q t A t q t A t

6 6 60 0 61.11 1 66.12 6( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )A t M t q t A t q t A t q t A t

7 74 74.8 4( ) ( ( ) ( )) ( )A t q t q t A t (6)

Where 0(2 )

0 ( ) ,tM t e 0 0( )1( ) ( ) ,tM t e G t

0( )4 ( ) ( ) ,tM t e R t 0( )

6 ( ) ( )tM t e F t (7)

Taking LT of above relation (6) and solving for A0*(s). The steady state availability is given by * 1

0 00 1

( ) lim ( )s

NA sA s

D

(8)

Where

1 0 47 60 11.3 61.11 10 61.11 40 14 17.3

6 47 01 16.5 16.5,10 02 11.3

01 46.9 14 17.3 1 47 01 66.12 02 61.11

4 14 17.3 01 60 61.11

[(1 ){ (1 ) } ( )]

[(1 ){ ( ) (1 )}

( )] (1 ){ (1 ) }

( ){ }

N p p p p p p p p p

p p p p p p

p p p p p p p p p

p p p p p

(9)



1 0 2 02 47 60 11.3 61.11 10 61.11 40 14 17.3

'6 47 01 16.5 16.5,10 02 11.3 01 46.9 14 17.3

' '01 60 61.11 4 14 17.3 7 14 47 17.3

'1 47 01

( )[(1 ){ (1 ) } ( )]

[(1 ){ ( ) (1 )} ( )]

( ){ ( ) ( )}

(1 ){ (

D p p p p p p p p p p

p p p p p p p p p p

p p p p p p p p

p p

66.12 02 61.111 ) }p p p

(10)

Busy Period Analysis for Server A. Due to Repair

Let ( )RiB t be the probability that the server is busy in repairof the unit at an instant‘t’ given that the system

entered regenerative state Si at t=0.The recursive relations for ( )RiB t are as follows:

0 01 1 02 2( ) ( ) ( ) ( ) ( )R R RB t q t B t q t B t

1 1 10 0 11.3 1 14 4

17.3 7 16.5 16.5,10 6

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ( ) ( )) ( )

R R R R

R R

B t W t q t B t q t B t q t B t

q t B t q t q t B t

2 26 6( ) ( ) ( )R RB t q t B t

4 40 0 47 7 46.9 6( ) ( ) ( ) ( ) ( ) ( ) ( )R R R RB t q t B t q t B t q t B t

6 60 0 61.11 1 66.12 6( ) ( ) ( ) ( ) ( ) ( ) ( )R R R RB t q t B t q t B t q t B t

7 7 74 74.8 4( ) ( ) ( ( ) ( )) ( )R RB t W t q t q t B t (11)

Where, 0 0 0 0 0 0( ) ( ) ( )

1 0( ) ( ) ( 1) ( ) ( 1) ( )t t tW t e G t e G t e G t

and 70( ) ( )tW t e G t (12)

Taking LT of above relation (11) and solving for *0 ( )RB s .The time for which serveris busydue to repair is given

by * 2

0 00 1

( ) ( )limR R

s

NB sB s

D

(13)

Where , * *

2 1 47 01 66.12 02 61.11 7 14 47 17.3 01 60 61.11(0)(1 ){ (1 ) } (0)( ){ }N W p p p p p W p p p p p p

and D1 is already mentioned. (14) B. Due to Replacement

Let ( )RpiB t be the probability that the server is busy in replacement of the unit at an instant ‘t’ given that the

system entered regenerative state Si at t=0.The recursive relations for ( )RpiB t are as follows:

01 020 1 2( ) ( ) ( ) ( ) ( )Rp Rp RpB t q t B t q t B t

10 11.3 141 0 1 4

17.3 16.5 16.5,107 6

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ( ) ( )) ( )

Rp Rp Rp Rp

Rp Rp

B t q t B t q t B t q t B t

q t B t q t q t B t

262 6( ) ( ) ( )Rp RpB t q t B t

4 40 47 46.974 0 6( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )Rp Rp Rp RpB t W t q t B t q t B t q t B t

60 61.11 66.126 0 1 6( ) ( ) ( ) ( ) ( ) ( ) ( )Rp Rp Rp RpB t q t B t q t B t q t B t

74 74.87 4( ) ( ( ) ( )) ( )Rp RpB t q t q t B t (15)

Where,



0 0( ) ( )4 0( ) ( ) ( 1) ( )t tW t e R t e R t (16)

Taking LT of above relation (15) and solving for *0 ( )RpB s .The time for which server is busy due to replacement

is given by * 3

0 00 1

( ) ( )limRp Rp

s

NB sB s

D

(17)

Where, *

3 4 14 17.3 01 60 61.11(0)( )( )N W p p p p p and D1 is already mentioned. (18)

C. Due to Preventive Maintenance

Let ( )PiB t be the probability that the server is busy in preventive maintenance of the unit at an instant‘t’ given

that the system entered regenerative state Si at t=0.The recursive relations for ( )PiB t are as follows:

0 01 1 02 2( ) ( ) ( ) ( ) ( )P P PB t q t B t q t B t

1 10 0 11.3 1 14 4

17.3 7 16.5 16.5,10 6

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ( ) ( )) ( )

P P P P

P P

B t q t B t q t B t q t B t

q t B t q t q t B t

2 2 26 6( ) ( ) ( ) ( )P PB t W t q t B t

4 40 0 47 7 46.9 6( ) ( ) ( ) ( ) ( ) ( ) ( )P P P PB t q t B t q t B t q t B t

6 6 60 0 61.11 1 66.12 6( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )P P P PB t W t q t B t q t B t q t B t

7 74 74.8 4( ) ( ( ) ( )) ( )P PB t q t q t B t (19)

Where, 0 0 0( ) ( ) ( )

6 0( ) ( ) ( 1) ( ) ( 1) ( )t t tW t e F t e F t e F t and 2( ) ( )W t F t (20)

Taking LT of above relation (19) and solving for *0 ( )PB s .The time for which server is busy due to preventive

maintenance is given by * 4

0 00 1

( ) ( )limP P

s

NB sB s

D

(21)

Where, *

4 2 02 47 60 11.3 61.11 10 61.11 40 14 17.3

*6 47 01 16.5 16.5,10 02 11.3 01 46.9 14 17.3

(0) [(1 ){ (1 ) } ( )]

(0)[(1 ){ ( ) (1 )} ( )]

N W p p p p p p p p p p

W p p p p p p p p p p

and D1 is already mentioned. (22) Expected Number of Repairs Let ( )iR t be the expected number of repairs by the server in (0, t] given that the system entered the regenerative

state Si at t = 0. The recursive relations for ( )iR t are given as:

0 01 1 02 2( ) ( ) ( ) ( ) ( )R t Q t R t Q t R t & &

1 10 0 11.3 1 14 4

17.3 7 16.5 6 16.5,10 6

( ) ( ) [1 ( )] ( ) [1 ( )] ( ) ( )

( ) ( ) ( ) [1 ( )] ( ) ( )

R t Q t R t Q t R t Q t R t

Q t R t Q t R t Q t R t

& & &

& & &

2 26 6( ) ( ) ( )R t Q t R t &

4 40 0 47 7 46.9 6( ) ( ) ( ) ( ) ( ) ( ) ( )R t Q t R t Q t R t Q t R t & & &

6 60 0 61.11 1 66.12 6( ) ( ) ( ) ( ) ( ) ( ) ( )R t Q t R t Q t R t Q t R t & & &

7 74 4 74.8 4( ) ( ) [1 ( )] ( ) ( )R t Q t R t Q t R t & & (23)



Taking LST of above relations (23) and solving for **0 ( )R s .The expected no. of repairs per unit time by the

server are giving by ** 5

0 00 1

( ) ( )lims

NR sR s

D

(24)

Where,

5 10 11.3 16.5 47 01 66.12 02 61.11

74 14 47 17.3 01 60 61.11

( )(1 ){ (1 ) }

( )( )

N p p p p p p p p

p p p p p p p

and D1 is already mentioned. (25) Expected Number of Replacements Let ( )iRp t be the expected number of replacements by the server in (0, t] given that the system entered the

regenerative state Si at t = 0. The recursive relations for ( )iRp t are given as:

0 01 1 02 2( ) ( ) ( ) ( ) ( )Rp t Q t Rp t Q t Rp t & &

1 10 0 11.3 1 14 4

17.3 7 16.5,10 6 16.5 6

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) [1 ( )] ( ) ( )

Rp t Q t Rp t Q t Rp t Q t Rp t

Q t Rp t Q t Rp t Q t Rp t

& & &

& & &

2 26 6( ) ( ) ( )Rp t Q t Rp t &

4 40 0 47 7 46.9 6( ) ( ) [1 ( )] ( ) ( ) ( ) [1 ( )]Rp t Q t Rp t Q t Rp t Q t Rp t & & &

6 60 0 61.11 1 66.12 6( ) ( ) ( ) ( ) ( ) ( ) ( )Rp t Q t Rp t Q t Rp t Q t Rp t & & &

7 74 4 74.8 4( ) ( ) ( )] ( ) [1 ( )]Rp t Q t Rp t Q t Rp t & & (26)

Taking LST of above relations (26) and solving for **0 ( )Rp s .The expected no. of replacements per unit time by

the server are giving by ** 6

0 00 1

( ) ( )lims

NRp sRp s

D

(27)

Where,

6 16.5,10 47 01 66.12 02 61.11

40 14 17.3 01 60 61.11

74.8 14 47 17.3 01 60 61.11

(1 ){ (1 ) }

( 46.9 )( ){ }

( )( )

N p p p p p p

p p p p p p p p

p p p p p p p

and D1 is already mentioned. (28) Expected Number of Preventive Maintenances Let ( )iP t be the expected number of preventive maintenance by the server in (0, t] given that the system entered

the regenerative state Siat t = 0. The recursive relations for ( )iP t are given as:

0 01 1 02 2( ) ( ) ( ) ( ) ( )P t Q t P t Q t P t & &

1 10 0 11.3 1 14 4

17.3 7 16.5 16.5,10 6

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) { ( ) ( )} ( )

P t Q t P t Q t P t Q t P t

Q t P t Q t Q t P t

& & &

& &

2 26 6( ) ( ) [1 ( )]P t Q t P t &

4 40 0 47 7 46.9 6( ) ( ) ( ) ( ) ( ) ( ) ( )P t Q t P t Q t P t Q t P t & & &

6 60 0 61.11 1 66.12 6( ) ( ) [1 ( )] ( ) [1 ( )] ( ) [1 ( )]P t Q t P t Q t P t Q t P t & & &

7 74 74.8 4( ) { ( ) ( )} ( )P t Q t Q t P t & (29)

Taking LST of above relations (29) and solving for **0 ( )P s .The expected no. of preventive maintenances per

unit time by the server are giving by



** 70 0

0 1

( ) lim ( )s

NP sP s

D

(30)

Where,

7 02 47 60 11.3 61.11 10 61.11 40 14 17.3

47 01 16.5 16.5,10 02 11.3 01 46.9 14 17.3

[(1 ){ (1 ) } ( )]

[(1 ){ ( ) (1 )} ( )]

N p p p p p p p p p p

p p p p p p p p p p

and D1 is already mentioned. (31) Profit Analysis The profit incurred to the system model in steady state can be obtained as

0 0 1 0 2 0 3 0 4 1 5 0 6 0R Rp PK A K B K B K B K R K R KP p P (32)

Where, P = Profit of the system model after reducing cost per unit time busy of the server and cost per repair activity per unit time

0K = Revenue per unit up-time of the system

1K =Cost per unit time for which server is busy due to repair

2K =Cost per unit time for which server is busy due to replacement

3K =Cost per unit time for which server is busy due to preventive maintenance

4K = Cost per repair per unit time

5K = Cost per replacement per unit time

6K = Cost per preventive maintenance per unit time

CONCLUSION After solving the equations of MTSF, availability and profit function for the particular case g(t) = 𝜃𝑒 , r(t) = 𝛽𝑒 and f(t) = 𝛼𝑒 , we conclude that These reliability measures are increasing as the repair rate θ, replacement rate β increases while their values

decline with the increase of failure rate λ and the rate 𝛼 by which the unit undergoes for preventive maintenance.

The MTSF and availability keep on upwards with the increase of the rate 𝛽 by which unit undergoes for replacement but profit decreases.

The system model becomes more profitable when we increase the preventive maintenance rate α. GRAPHS FOR PARTICULAR CASE

4.86

4.88

4.9

4.92

4.94

4.96

4.98

5

5.02

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

MT

SF

Failure Rate (λ)

MTSF Vs Failure Rate (λ)

α0=0.2,β0=3,θ=2.1,α=5,β=10β0=1θ=4.1α0=0.201α=7β=5

Fig.2



REFERENCES 1. R. Kishan and M. Kumar (2009), “Stochastic

analysis of a two-unit parallel system with preventive maintenance”, Journal of Reliability and Statistical Studies, vol. 22, pp. 31- 38.

2. Kumar, Jitender, Kadyan, M.S. and Malik, S.C. (2010): Cost-benefit analysis of a two-unit parallel system subject to degradation after repair. Journal of Applied Mathematical Sciences, Vol.4 (56), pp.2749-2758.

3. Malik S.C. and Gitanjali (2012). Cost-Benefit Analysis of a Parallel System with Arrival Time of the Server and Maximum Repair Time. International Journal of Computer Applications, 46 (5): 39-44.

4. Reetu and Malik S.C. (2013),A Parallel System

with Priority to Preventive Maintenance Subject to Maximum Operation and Repair Time. American journal of Mathematics and Statistics, Vol. 3(6), pp. 436-440.

5. Rathee R. and Chander S. (2014), A Parallel System with Priority to Repair over Preventive Maintenance Subject to Maximum Operation and Repair Time. International Journal of Statistics and Reliability Engineering, Vol. 1(1), pp. 57-68.

0.9550.9570.9590.9610.9630.9650.9670.9690.9710.973

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09Ava

ilab

ilit

y

Failure Rate (λ)

Availability Vs Failure Rate (λ)

α0=0.2,β0=3,θ=2.1,α=5,β=10β0=1θ=4.1α0=0.201α=7β=5

Fig.3

13050131501325013350134501355013650137501385013950

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

Pro

fit

(P)

Failure Rate (λ)

Profit Vs Failure Rate (λ)

α0=0.2,β0=3,θ=2.1,α=5,β=10β0=1θ=4.1α0=0.201α=7β=5

Fig.4

Education

Reliability Modeling and Analysis of a Parallel Unit System with Priority to Repair over Replacement Subject to Maximum Operation and Repair Times