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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 pre fixed operation 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. Reetu Rathee | S. C. Malik | D. Pawar "Reliability Modeling and Analysis of a Parallel Unit System with Priority to Repair over Replacement Subject to Maximum Operation and Repair Times" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-2 | Issue-5 , August 2018, URL: https://www.ijtsrd.com/papers/ijtsrd15655.pdf Paper URL: http://www.ijtsrd.com/mathemetics/statistics/15655/reliability-modeling-and-analysis-of-a-parallel-unit-system-with-priority-to-repair-over-replacement-subject-to-maximum-operation-and-repair-times/reetu-rathee
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ISSN No: 2456
InternationalResearch
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
@ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 5 | Jul-Aug 2018
ISSN No: 2456 - 6470 | www.ijtsrd.com | Volume
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
Aug 2018 Page: 350
6470 | www.ijtsrd.com | Volume - 2 | Issue – 5
Scientific (IJTSRD)
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
@ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 5 | Jul-Aug 2018 Page: 351
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.
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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
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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)
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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,
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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)
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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
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** 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
International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470
@ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 5 | Jul-Aug 2018 Page: 358
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
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