N0460284100

K. Uma Maheswari et al Int. Journal of Engineering Research and Applications www.ijera.com
ISSN : 2248-9622, Vol. 4, Issue 6( Version 2), June 2014, pp.84-100
www.ijera.com 84 | P a g e
Probabilistic Behaviour of A Redundant Complex System with Imperfect Switching, Environmental Common Cause and Human Error Effects under Head-Of-Line Repair Discipline K. Uma Maheswari#1, A. Mallikarjuna Reddy#1 and R. Bhuvana Vijaya#2 #1Department of Mathematics, Sri Krishnadevaraya University, Ananthapuramu-515003, A.P., India #2Department of Mathematics, J.N.T.U., Ananthapuramu-515001, A.P., India ABSTRACT Series-parallel systems are made up of combination several series & parallel configuration to obtain system reliability down into homogeneous subsystems. This is very simple to analyse a simple series-parallel system without any error effects, But very difficult to analyse a complex system with various error effects. This paper we presented mathematical analysis of a Redundant Complex System with Imperfect switching, Environmental and Common Cause and Human error effects under Head of Line Repair Discipline. Keywords: Imperfect switching, Reliability, Availability, MTTF, State Transition diagram.
I. INTRODUCTION
Kontoleon & Konoleon [2] had considered a system subject to partial and catastrophic failures to be repaired at a single service station and also assumed exponential distribution for repair. Gupta and Mittal [3] considered a standby redundant system and two types of failures to be repaired at a single service station with general repair time distribution under pre-emptive resume repair Discipline incorporating environmental effects. Further, Gupta and Agarwal [1] developed a model with two types of failure under general repair time distribution and under different repair discipline. Consequently Agarwal, Mittal et. al., [4] have solved the model of a Parallel redundant complex system with two types of failures with Environmental effect under preemptive Resume repair Discipline. Mittal, Gupta et. al., [1,3] have assumed constant repair rate due to environmental failure and perfect switching over device of a standby redundant system under preemptive resume and repeat discipline respectively. But it is not always possible that the switch is perfect. It may have some probability of failure. So the authors have initiated the study keeping in mind the practical aspect of imperfect switch. Also repair due to environmental effects is considered to follow general time distribution.
In this paper, we consider a complex system consisting of three independent, repairable subsystems A, B & C. Subsystem B is comprised of two identical units B1 & B2. The subsystem C is in standby redundancy to be switched into operation when both the identical units B1 & B2 of subsystem B fails, through an imperfect switching over device. For the smooth operation of the system, the subsystems B or C are of vital importance. The failure of the subsystem C results into non-operative state of the system. At the time of installation, the units of subsystem B have the same failure rate but due to adverse environmental effects, the failure rate of the subsystem C increases by the time, it is switched into operation. Subsystem A has minor failure and subsystem C has major and minor failures. Minor failure reduces the efficiency of the system causing a degraded state while major failure results into a non- operable state of the system. The whole complex system can also be in a total failure state due to Human error or Environmental effects like temperature, humidity etc. The repair rate due to Human error or Environmental failure is considered general. Laplace transforms of the time dependent probabilities of the system being in various states have been obtained by employing the techniques using supplementary variable under Head-of line Repair Discipline. The general solution of the problem is used to evaluate the ergodic behaviour and some particular case. At the end of the chapter some numerical illustrations are given and various reliability parameters have been calculated. The flow of states is depicted in the state transition diagram.
RESEARCH ARTICLE OPEN ACCESS

The following symbols are used in state transition diagram:
: Operable : Degarded
: Imperfect : : Failed
switch
Fig. 1. STATE TRANSITION DIAGRAM
II. NOTATIONS:
1 : Minor failure rate of the operating units B1
& B2 of subsystem B.
2(1<2): Minor failure rate of the
operating unit of subsystem C
m1  : Minor failure of subsystem A.
m2  : Major failure of subsystem B & C.
x: Elapsed repair time for both units of
subsystem B.
y: Elapsed minor repair time for the
subsystem A, B & C.
z: Elapsed major repair time for the
subsystem B & C.
v: Elapsed repair time for environmental
failure.
h: Elapsed repair time for human error
failure.
B(x) : Repair rate of subsystem B
( )
1
y m  : Repair rate of minor failure of
subsystems A, B & C.
( )
2
z m  : Repair rate of major failure of
subsystems B & C.
 : Failure due to environmental effects.
 : Failure due to human error.
 : Failure due to common cause.
(v) : Repair rate due to environmental
effects.
(h) : Repair rate due to Human error.
(c) : Repair rate due to common cause
b, R1 Probability of the successful operation
of the switching over device
and constant repair rate of
switching over device.
III. ASSUMPTIONS
(1) Initially, at time t = 0 the system is in operable
state, i.e., it operates in its normal efficiency
state
(2) The subsystem B consists of two identical units
B1 & B2.
(3) Switching device is imperfect.
(4) When the system starts functioning the
subsystems A & B will operate and C is in stand
by.
(5) When the system starts functioning both the
subsystems B & C have the same failure rate 1
as minor failure, but as the time passes due to the
adverse environmental effects, the failure rate of
the stand by unit i.e., subsystems C increase to 2
which is the minor failure of subsystem C by the
time it is needed to operate.
(6) Subsystem A has no major failures.
(7) During the degraded state of the system due to
minor failure in subsystems A, B & C, major
failures may also occur in subsystems B & C.
(8) All the failures are distributed exponentially.
(9) All the repair follows general time distribution.

(10) The system fails completely due to
environmental effects or due to major failures of
the subsystems B & C or due to the failure
caused by Human error.
(11) Repair of the subsystem C is taken only in failed
state.
(12) During the repair of the subsystem C in failed
state, the failed units of subsystem B are also
repaired.
STATE PROBABILITIES DESCRIPTION:
(i) ( ) ,0 P t a : Probability that the system is in
operable state at the time t, where a=0, 0, 1,
2.
(ii) ( , ) 3,0 P x t : Probability that the system is the
failed state and is under repair with elapsed
repair time in the interval (x, x+).
(iii) ( , )
, 1 P y t a m : Probability that the system is in
degraded state and is under repair with
elapsed repair time lying in the interval (y,
y+) where a = 0,1,2.
(iv) ( , )
, 2 P z t a m : Probability that the system is
the failed state and is under repair with
elapsed repair time in the interval (z, z+)
where a = 0,1,2.
(v) P (v, t) E : Probability that the system is in
failed state due to Environmental failure and
is under repair with elapsed Repair time in
the interval (v, v+).
(vi) P (t) r : Probability that the system is in
failed state due to failure of the switch in
switching the subsystem C when subsystem
B completely fails at any time t.
(vii) P (h, t) h : Probability of the system is in
failed state due to Human error and is under
repair with elapsed repair time in the
interval (h, h+).
(viii) P (u, t) v : Probability of by the system is in
failed state due to common cause error and
is under repair with elapsed repair time in
the interval (u, u+).
IV. FORMULATION OF THE
MATHEMATICAL MODEL:
Using continuity arguments and probability
considerations, we obtain the following difference
differential equations governing the stochastic
behaviour of the complex system, which is discrete is
space and continuous in time:

  
 


  
 


  
 
 



    
0
0 0
2,
0
0
0,
0
0,
0
1 0,0 3,0
( , ) ( )
( , ) ( ) ( , ) ( ) ( , ) ( )
( , ) ( ) ( , ) ( )
( ) ( , ) ( )
2 2
1 1 2 2
2 1
P u t u dx
P v t v dx P z t z dz P h t h dx
P y t y dy P z t z dz
P t P x t x dx
dt
d
C
E m m H
m m m m
m m B

  
 
     
(1)

  
 


  
 


  
 
 



    
0
0 0
2,
0
0
1,
0
1,
0
1 1,0 3,0
( , ) ( )
( , ) ( ) ( , ) ( ) ( , ) ( )
( , ) ( ) ( , ) ( )
( ) ( , ) ( )
2 2
1 1 2 2
2 1
P u t u dx
P y t y dy P z t z dx
P t P x t x dx
dt
d
C
E m m H
m m m m
m m B

  
 
     
(2)



 
  



     
0
1 2,
2 2,0 1 0,0 1 1,0
( ) ( , ) ( )
( ) ( ) ( )
1 1
1 2
R P t P y t y dy
P t b P t b P t
dt
d
r m m
m m

       
(3)



 





 



  
  








0
1,
1,0
0
0,0 1 0,0 0,
1
( , )
( ) ( ) ( , ) ( )
2 ( ) (1 )
1
1
1
P y t dy
P t b P t P y t dy P t
R P t b
dt
d
m
m
r


(4)
2 ( ) ( , ) 0 3,0  









x P x t
t x B  (5)
( ) ( , ) 0
2 1 1 0, 1  



    





y P y t
t y m m m      (6)
( ) ( , ) 0
2 1 1 1, 1  



    





y P y t
t y m m m      (7)
( ) ( , ) 0
2 2 1 2, 1  



    





y P y t
t y m m m      (8)
( ) ( , ) 0
2 0, 2  









z P z t
t z m m  (9)
( ) ( , ) 0
2 1, 2  









z P z t
t z m m  (10)
2 ( ) ( , ) 0
2 2, 2  









z P z t
t z m m  (11)
0 ) , ( ) ( 2  









h P h t
t h H  (12)
0 ) , ( ) ( 2  









v P v t
t v E  (13)
0 ) , ( ) ( 2  









u P u t
t u C  (14)
BOUNDARY CONDITIONS


 
0
3,0 2 2,0 2 2, (0, ) ( ) ( , )
1
P t P t P y t dy m   (15)
(0, ) ( ) 0, 1 1 0,0
P t P t m m   (16)
(0, ) ( ) 1, 1 1 1,0
P t P t m m   (17)
 
 
   
0
1 1, 1
0
2, 2,0 1 0, (0, ) ( ) ( , ) ( , ) ( )
1 2 1 1
P t P t b P y t dy b P y t dy R P t m m m m r    (18)


 
0
0, 0,0 0, (0, ) ( ) ( , )
2 2 2 1
P t P t P y t dy m m m m   (19)



 
0
1, 1,0 1, (0, ) ( ) ( , )
2 2 2 1
P t P t P y t dy m m m m   (20)


 
0
2, 2,0 2, (0, ) ( ) ( , )
2 2 2 1
P t P t P y t dy m m m m   (21)
 

 

 
   
0
2,
0
1,
0
0,0 1,0 2,0 0,
( , ) ( , )
(0, ) ( ) ( ) ( ) ( , )
1 1
1
P y t dy P y t dy
P t P t P t P t P y t dy
m m
E m
 
   
(22)
 

 

 
   
0
2,
0
1,
0
0,0 1,0 2,0 0,
( , ) ( , )
(0, ) ( ) ( ) ( ) ( , )
1 1
1
P y t dy P y t dy
m m
C m
 
   
(23)
(0, ) ( ) 2,0 P t P t H   (24)
INITIAL CONDITIONS
P0,0(0) = 1 and all other states probabilities are zero at time t = 0.
from 6
( ) ( , ) 0
2 1 1 0, 1  



    





y P y t
t y m m m     
Taking Laplace transforms on both sides
 


    


  


( )
( , )
( , )
2 1
1
1
1
0,
0,
y
y
S
P y s
P y s
y
m m
m
m
    
Integrating ‘y’ b/w the limits ‘0’ to ‘y’
     
 
y
s m y m y dy
m m P y s P s e 0
2 1 1
1 1
[ ] ( )
0, 0, ( , ) (0, )
    
from equation (16)
(0, ) ( ) 0, 1 1 0,0
P t P t m m  
     
 
y
s m y m y y
m m P y s P s e 0
2 1 1
1 1
[ ] ( )
0, 0,0 ( , ) ( )
    

On simplification
( , ) ( ) ( ) 0, 1 1 0,0 1
P y s P s V s m m   (25)
where
   
   
   
    

1
1
1
2
1 2
1 ( )
( )
m
m m
s
s s
V s
from equation (7)
( ) ( , ) 0
2 1 1 1, 1  



    





y P y t
t y m m m     

 ( )
( , )
( , )
2 1
1
1
1
1,
1,
s y
P y s
P y s
y
m m
m
m
      



Integrating ‘y’ between the limits ‘0’ to ‘y’
      
 
y
s m y m y dy
m m P y s P s e 0
2 1 1
1 1
[ ] ( )
1, 1, ( , ) (0, )
    
from equation 16
(0, ) ( ) 0, 1 1 1,0
P t P t m m 
(0, ) ( ) 1, 1 1 1,0
P s P s m m  
     
 
y
s m y m y dy
m m P y s P s e 0
2 1 1
1 1
[ ] ( )
1, 1,0 ( , ) ( )
    

( , ) ( ) ( ) 1, 1 1 1,0 1
P y s P s V s m m  (26)
From equation 4



 





 



 
  








0
1,
1,0
0
0,0 0,
1 1
( , )
( ) ( , ) ( )
2 ( ) (1 )
1
1
P y t dy
P t P y t dy P t
R P t b
dt
d
m
m
r 



 





 



 


 




0
1,
1,0
0
0,0 0,
1
1
( , )
( ) ( , ) ( )
2
(1 )
( )
1
1
P y s dy
P s P y s dy P s
s R
b
P s
m
m
r

[(1 ( ) ( ) (1 ( ) ( )]
2
(1 )
1 0,0 1 0,0
1
1
1 1
V s P s V s P s
s R
b
m m  

  



[(1 ( ))[ ]
2
(1 )
1 0 1
1
1
1
V s P P
s R
b
m  


 

( )[ ] 2 0 1 V s P  P (27)
where [(1 ( )
2
(1 )
( ) 1
1
1
2 1
V s
s R
b
V s m 





from equation 8
( ) ( , ) 0
2 2 1 2, 1  



    





y P y t
t y m m m     
 ( )
( , )
( , )
2 1
1
1
2
2,
2,
s y
P y s
P y s
y
m m
m
m
      



Integrating ‘y’ between the units ‘0’ to ‘y’
     
 
y
s m y m y dy
m m P y s P s e 0
2 2 1
1 1
[ ] ( )
2, 2, ( , ) (0, )
    
from equation 18

 
 
   
0
1 1, 1
0
2, 2,0 1 0, (0, ) ( ) ( , ) ( , ) ( )
1 1 1 1
P t P t b P y t dy b P y t dy R P t m m m m r   
( , ) ( ) ( ) ( )[ ( ) ( )] 2, 1 4 2,0 5 0,0 1,0
P y s V s P s V s P s P s m    
where
   
   
   
    

2
2
3
2
1 1
1 ( )
( )
m
m m
s
s s
V s (28)
1
( ) ( ) 4 3 m V s V s 
( ) ( ) ( ) 5 1 1 1 1 2 V s  b mV s  RV s
From equation (10)
( ) ( , ) 0
2 1, 2  









z P z t
t z m m 
 ( )
( , )
( , )
2
2
2
1,
1,
s z
P z s
P z s
y
m
m
m
  



Integrating ‘z’ between the limits ‘0’ to ‘z’
 

z
sz m z dz
m m P z s P s e 0
2
2 2
( )
1, 1, ( , ) (0, )

From equation 19


 
0
1, 1,0 1, (0, ) ( ) ( , )
2 2 2 1
P t P t P y t dy m m m m  
  
   
z
sz m z dz
m m m m P z s P s P y s dy e 0
2
2 2 2 1
( )
0
1, 1,0 1, ( , ) [ ( ) ( , ) ]

 
( ) ( )
( , ) [1 ( )] ( ) ( )
7 1,0
1, 2 2 1 1 6 1,0
V s P s
P z s V s V s P s m m m

 
(29)
where
s
s s
V s m 1 ( )
( ) 2
6


( ) [1 ( )] ( ) 7 2 1 1 6
V s V s V s m m   
From equation (12)
2 ( ) ( , ) 0
2 2, 2  









z P z t
t m m 

 2 ( )
( , )
( , )
2
2
2
2,
2,
s z
P z s
P z s
y
m
m
m
   



Integrating ‘z’ between the limits ‘0’ to ‘z’
 
 
z
sz m z dz
m m P z s P s e 0
2
2 2
( )
2, 2, ( , ) (0, )

From equation 21.



 
0
2, 2,0 2, (0, ) ( ) ( , )
2 2 2 2
P t P t P y t dy m m m m  
   




   
z
sz m z dz
m m m m P z s P s P y s dy e 0
2
2 2 2 1
2 ( )
0
2, 2,0 2, ( , ) ( ) ( , )

 
( ) ( ) ( )
[1 ( )] ( ) ( ) ( ) ( ) ( )
5 8 0,0
4 8 2,0 8 5 1,0
2
2 2
V s V s P s
V s V s P s V s V s P s
m
m m

 

  
( ) ( ) ( ) ( ) ( ) 9 2 10 1,0 0,0 V s P s V s P s  P s (30)
where
s
s s
V s m 1 2 ( )
( ) 2
8


( ) (1 ( )) ( ) 9 2 4 8
V s V s V s m  
( ) ( ) ( ) 10 2 8 5
V s V s V s m  
From equation (12)
0 ) , ( ) ( 2  









h P h t
t h H 
 2 ( )
( , )
( , )
s h
P h s
P h s
h
H
H
    


Integrating ‘h’ between the limits ‘0’ to ‘h’
  
 
h
sh h dh
H H P h s P s e 0
2 ( )
( , ) (0, )

From equation 22
(0, ) ( ) 2,0 P t P t H  
(0, ) ( ) 2,0 P t P t H  
After simplification
( , ) ( ) ( ) 11 2,0 P h s V s P s H   (31)
where
s
s s
V s h 1 2 ( )
( ) 11
 

from equation 13
0 ) , ( ) ( 2  









v P v t
t v E 
 2 ( )
( , )
( , )
s v
P v s
P v s
v
E
E
    


Integrating ‘v’ between the limits ‘0’ to ‘v’
  
 
v
sv v dv
E E P v s P s e 0
2 ( )
( , ) (0, )

From equation 14

0 ) , ( ) ( 2  









u P u t
t u C 
 ( )
( , )
( , )
s u
P u s
P u s
v
C
C
   


Integrating ‘u’ between the limits ‘0’ to ‘u’
  
 
u
su u du
C C P u s P s e 0
2 ( )
( , ) (0, )

From equation 22
 

 

 
   
0
2,
0
1,
0
0,0 1,0 2,0 0,
( , ) ( , )
(0, ) ( ) ( ) ( ) ( , )
1 1
1
P y t dy P y t dy
m m
E m
 
   
 

 

 
   
0
2,
0
1,
0
0,0 1,0 2,0 0,
( , ) ( , )
(0, ) ( ) ( ) ( ) ( , )
1 1
1
P y s dy P y s dy
P s P s P s P s P y s dy
m m
E m
 
   
[1 ( )] ( )) ( )
[1 ( ) ( )] ( ) ( )
( ) [1 ( ) ( )] ( ) ( )
4 12 2,0
1 5 12 1,0
1 5 12 0,0
1
1
V s V s P s
V s V s V s P s
P s V s V s V s P s
m
E m
 
  
  

 
 
( ) ( ) ( ) ( )[ ( ) ( )] 13 2,0 14 0,0 1,0 P s V s P s V s P s P s E     (32)
where
s
s s
V s
1 2 ( )
( ) 12
 

( ) [1 ( )] ( ) 13 4 12 V s  V s V s
( ) [1 ( ) ( )] ( ) 14 1 1 5 12
V s V s V s V s m   
From equation 23
 

 

 
   
0
2,
0
1,
0
0,0 1,0 2,0 0,
( , ) ( , )
(0, ) ( ) ( ) ( ) ( , )
1 1
1
P y t dy P y t dy
m m
C m
 
   
 

 

 
   
0
2,
0
1,
0
0,0 1,0 2,0 0,
( , ) ( , )
(0, ) ( ) ( ) ( ) ( , )
1 1
1
P y s dy P y s dy
P s P s P s P s P y s dy
m m
C m
 
   

[1 ( )] ( ) ( )
[1 ( ) ( )] ( ) ( )
( ) [1 ( ) ( )] ( ) ( )
4 12 2,0
1 5 12 1,0
1 5 12 0,0
1
1
V s V s P s
V s V s V s P s
P s V s V s V s P s
m
C m
 
  
  

 
 
( ) ( ) ( ) ( )[ ( ) ( )] 15 2,0 16 0,0 1,0 P s V s P s a V s P s P s C    (33)
where
s
s s
V s
1 2 ( )
( ) 12
 

( ) [1 ( )] ( ) 15 4 12 V s  V s V s
( ) [1 ( ) ( )] ( ) 16 1 1 5 12
V s V s V s V s m   
From equation 5
2 ( ) ( , ) 0 3,0  









x P x t
t x B 
 2 ( )
( , )
( , )
3,0
3,0
s x
P x s
P x s
x
B     


Integrating ‘x’ between the limits ‘0’ to ‘x’
  
 
x
sx B x dx
P x s P s e 0
2 ( )
3,0 3,0 ( , ) (0, )

From equation 14


 
0
3,0 2 2,0 2 2, (0, ) ( ) ( , )
1
P t P t P y t dy m  


 
0
3,0 2 2,0 2 2, (0, ) ( ) ( , )
1
P s P s P y s dy m  
Substituting the value of P3,0(0,s) in the above equation we get
( , ) ( ) ( ) ( )[ ( ) ( )] 3,0 18 2,0 19 1,0 0,0 P x s V s P s V s P s  P s (34)
where
s
s s
V s B 1 2 ( )
( ) 17


( ) [1 ( )] ( ) 18 2 4 17 V s   V s V s
( ) ( ) ( ) 19 2 5 17 V s  V s V s
From equation 3


 
  



     
0
1 2,
2 2,0 1 0,0 1 1,0
( ) ( , ) ( )
( ) ( ) ( )
1 1
1 2
R P t P y t y dy
P t b P t b P t
dt
d
r m m
m m

       
 


 
       
0
1 2,
2 2,0 1 0,0 1 1,0
( ) ( , ) ( )
( ) ( ) ( )
1 1
1 2
R P s P y s y dy
S P s b P s b P s
r m m
m m

       

Substituting the values of P (s) r , 2,m1 ( y,s) P in the above equation we get
( ) ( )[ ( ) ( )] 2,0 22 0,0 1,0 P s V s P s  P s (35)
( ) ( ) ( ) ( ) 20 1 1 2 5 1 2 1 2
           m m m V s b RV s V s S s
( )
( ) ( )
2
21 2
1 1 2 1
2 1
      
    
      
     
m m m m
m m
S s
V s s
( )
( )
( )
21
20
22 V s
V s
V s 
from equation 2

  
 


  
 


  
 
 



    
0
0 0
2,
0
0
1,
0
1,
0
1 1,0 3,0
( , ) ( )
( , ) ( ) ( , ) ( ) ( , ) ( )
( , ) ( ) ( , ) ( )
( ) ( , ) ( )
2 2
1 1 2 2
2 1
P u t u dx
P t P x t x dx
dt
d
C
E m m H
m m m m
m m B

  
 
     
 
  
  

  
  

  
  
     
0 0 0
2,
0 0
1,
0
1,
0
1 1,0 3,0
( , ) ( ) ( , ) ( ) ( , ) ( )
( , ) ( ) ( , ) ( ) ( , ) ( )
( ) ( , ) ( )
2 2
1 1 2 2
2 1
P z s z dz P h s h dx P u s u dx
P y s y dy P z s z dz P v s v dx
S P s P x s x dx
m H C
m m m m E
m m B
m
  
  
     
Substituting the values of
( , ) 3.0 P x s , ( , )
1. 1 P y s m , ( , )
1. 2 P z s m , ( , )
2. 2 P z s m , P (v, s) E , P (h, s) H , P (u, s) C
In the above equation we get
( ) ( ) ( ) ( ) ( ) 1,0 26 2,0 27 0,0 P s V s P s V s P s (36)
where
[(1 ( )] ( ) [(1 ( )] ( ) ( )
( ) (1 ( ) ( ) [(1 ( )] ( )
4 2 4 2
23 2 4 4
V s S s V s S s S s
V s V s S s V s S s
m m
B
    
  
    
   
( )] ( ) ( ) ( )
( ) ( ) ( ) [1 ( ) ( )] ( ) [1 ( )
2 2
1 1
5 8
24 2 5 1 5 1
V s S s V s S s
V s V s S s V s V s S s V s
m m
B m m
 
     
 
     
( ) ( ) ( ) ( ) [1 ( )] ( )
25 2 1 1 24 1 1 2 1 1 2 V s S V s m sm s V s S s m m m m m          
( )
( )
( )
25
23
26 V s
V s
V s 
( )
( )
( )
25
24
27 V s
V s
V s 
From equation 1


  
 


  
 


  
 
 



    
0
0 0
2,
0
0
0,
0
0,
0
1 0,0 3,0
( , ) ( )
( , ) ( ) ( , ) ( ) ( , ) ( )
( , ) ( ) ( , ) ( )
( ) ( , ) ( )
2 2
1 1 2 2
2 1
P u t u dx
P t P x t x dx
dt
d
C
E m m H
m m m m
m m B

  
 
     
 
  
  

  
  

  
  
     
0 0 0
2,
0 0
0,
0
0,
0
1 0,0 3,0
( , ) ( ) ( , ) ( ) ( , ) ( )
( , ) ( ) ( , ) ( ) ( , ) ( )
( ) ( , ) ( )
2 2
1 1 2 2
2 1
P z s z dz P h s h dx P u s u dx
P y s y dy P z s z dz P v s v dx
S P s P x s x dx
m H C
m m m m E
m m B
m
  
  
     
Substituting the values of
( , ) 3.0 P x s , ( , )
0. 1 P y s m , ( , )
0. 2 P z s m , ( , )
2. 2 P z s m , P (v, s) E , P (h, s) H , P (u, s) C in the above
equation we get
( ) ( ) ( ) ( ) ( ) ( ) 1 25 0,0 23 2,0 24 1,0 P s P s V s P s V s P s  (37)
From equation 34
( , ) ( )[ ( ) ( )] 2,0 22 0,0 1,0 P x s  V s P s  P s
Substitute the value of ( ) 1,0 P s in this equation we get
( ) ( ) ( ) ( ) ( ) P ( ) ( ) ( )P ( ) 2,0 22 0,0 22 26 2,0 22 27 0,0 P s V s P s V s V s s V s V s s
( )
1 ( ) ( )
( )[1 ( )]
( ) 0,0
22 26
22 27
2,0 P s
V s V s
V s V s
P s


 
( )
( )
( )
( ) 0,0
28
29
2,0 P s
V s
V s
 P s  (38)
where ( ) 1 ( ) ( ) 28 22 26 V s  V s V s
( ) ( )[1 ( )] 29 22 27 V s V s V s
From equation 36
( ) ( ) ( ) ( ) ( ) 1,0 26 2,0 27 0,0 P s V s P s V s P s
Substitute the vale of ( ) 2,0 P s in this equation we get
( )
1 ( ) ( )
( ) ( ) ( )
( ) 0,0
26 22
26 22 27
1,0 P s
V s V s
V s V s V s
P s


 
( )
( )
( )
( ) 0,0
28
30
1,0 P s
V s
V s
 P s 
where
( ) ( ) ( ) ( ) 30 26 22 27 V s V s V s V s
From equation 37
( ) ( ) ( ) ( ) ( ) ( ) 1 25 0,0 23 2,0 24 1,0 P s P s V s P s V s P s 

Substitute the value of ( ), ( ) 2,0 1,0 P s P s in this equation we get
( )
( )
( )
31
28
0,0 V s
V s
 P s  (39)
where ( ) ( )P ( ) ( ) ( ) ( ) ( ) 31 25 0,0 23 29 24 30 V s V s s V s V s V s V s
Substituting the value P ( ) 0,0 s in equation (37, 38) we get
( )
( )
( )
31
30
1,0 V s
V s
P s  (40)
( )
( )
( )
31
29
2,0 V s
V s
P s  (41)
Evaluation of operational Availability and Non-Availability
The Laplace transforms of the probability that the system is in operable up and down state at time ‘t’ can be
evaluated as follows.
[1 ( )] ( )
[1 ( ) ( )] ( ) [1 ( ) ( )] ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
4 2,0
m 1 5 0,0 m 1 5 1,0
0,0 1,0 2,0 0, 1, 2,
1 1
1 1 1
V s P s
V s V s P s V s V s P s
P s P s P s P s P s P s P s UP m m m
 
     
     
  (42)
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
3,0 0, 2 1, 2 2, 2 P s P s P s P s P s P s P s P s P s dawn m m m E r H C       
[ ( ) ( ) ( ) ( ) ( )]( ( )
[ ( ) ( ) ( ) ( ) ( )]( ( ) ( ))
9 11 13 15 18 2,0
7 10 14 16 19 0,0 1,0
V s V s V s V s V s P s
V s V s V s V s V s P s P s
    
     
(43)
ERGODIC BEHAVIOUR
Using Abel’s lemma is Laplace transform,
viz., ( ) ( )
0
( )
0
f t f say
t
lt
sf s
s
lt




, provided that the limit on the R.H.S exists, the time independent
up and down state probabilities are as follows.
(0)
(0)
[1 (0)]
(0)
(0)
[1 (0) (0)]
(0)
(0)
[1 (0) (0)]
( ) ( ) ( ) ( ) ( ) ( ) ( )
31
29
4
31
30
m 1 5
31
28
m 1 5
0,0 1,0 2,0 0, 1, 2,
1 1
1 1 1
V
V
V
V
V
V V
V
V
V V
P s P s P s P s P s P s P s UP m m m
 
     
     
 
(44)
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
3,0 0, 2 1, 2 2, 2 P s P s P s P s P s P s P s P s P s down m m m E r H C         




   
 


 


     
(0)
(0)
[ (0) (0) (0) (0) (0)]
(0)
(0)
(0)
(0)
[ (0) (0) (0) (0) (0)]
31
29
9 11 13 15 18
31
30
31
28
7 10 14 16 19
V
V
V V V V V
V
V
V
V
V V V V V
(45)
PARTICULAR CASE:- When repair follows exponential distribution setting
i
i
i s
S s



( )  and


 

s
S (s) where i = B, m1, m2 in results (25) to (34) (39) (40) (41) one may get the various probabilities as
follows:

( )
( )
( )
31
28
0,0 g s
g s
P s 
( )
( )
( )
31
30
1,0 g s
g s
P s 
( )
( )
( )
31
29
2,0 g s
g s
P s 
( ) ( ) ( ) ( )[ ( ) ( )] 3,0 18 2,0 19 0,0 1,0 P s  g s P s  g s P s  P s
( ) ( ) ( ) 0, 1 1 1 0,0
P s g s P s m m  
( ) ( ) ( ) 1, 1 1 1 1,0
P s g s P s m m  
( ) ( )[ ( ) ( )] 2 0,0 1,0 P s g s P s P s r  
( ) ( ) ( ) ( )[ ( ) ( )] 2, 1 4 2,0 5 0,0 1,0
P s g s P s g s P s P s m   
( ) ( ) ( ) 0, 2 7 0,0
P s g s P s m 
( ) ( ) ( ) 1, 2 7 1,0
P s g s P s m 
( ) ( ) ( ) ( )[ ( ) ( )] 2, 2 9 2,0 10 1,0 0,0
P s g s P s g s P s P s m   
( ) ( ) ( ) 11 2,0 P s g s P s H  
( ) ( ) ( ) ( )[ ( ) ( )] 13 2,0 14 0,0 1,0 P s g s P s g s P s P s E   
( ) ( ) ( ) ( )[ ( ) ( )] 15 2,0 16 0,0 1,0 P s g s P s g s P s P s C   
where
1
15 1 ( ) ( )
2 1
       m m g s S     
[1 ( )]
2
(1 )
( ) 1
1
1
2 1
g s
s R
b
g s m 





1
3 2 ( ) ( )
2 1
       m m g s S     
4 3 1 g (s)  g (s)m
( ) ( ) ( ) 5 1 1 1 1 2 g s  b m g s  R g s
1
6 ( ) ( )
2
   m g s s 
( ) [1 ( )] ( ) 7 2 1 1 6
g s g s g s m m  
( )
( )
2
2
8
m
m
s s
s
g s





( ) [1 ( )] ( ) 9 2 4 8
g s g s g s m   
( ) ( )] ( ) 10 2 5 8
g s g s g s m  
( )
( ) 11
h
h
s s
s
g s





( )
( ) 12 




s s
s
g s
( ) [1 ( )] ( ) 13 4 12 g s   g s g s
( ) [1 ( ) ( )] ( ) 14 1 1 5 12
g s g s g s g s m   
( ) [1 ( )] ( ) 15 4 12 g s   g s g s

( ) [1 ( ) ( )] ( ) 16 1 1 5 12
g s g s g s g s m   
( )
( ) 17
B
B
s s
s
g s





( ) [1 ( )] ( ) 18 2 4 17 g s   g s g s
( ) ( ) ( ) 19 2 5 17 g s  g s g s
1
20 1 1 2 5 2 ( ) ( ) ( ) ( )
1 2
           m m g s b R g s g s s
1
2
21 2
( )
( ) ( )
1 1 2 1
2 1
       
     
       
    
m m m m
m m
s
g s s
( )
( )
( )
21
20
22 g s
g s
g s 
1
1
4
1
4
1
4
1
23 2 4
( )
(1 ( )] ( ) [1 ( )] ( )
( ) (1 ( ) ( ) ) [1 ( )] ( )
2 2 2

 
 
 
     
       
 
     
     
s
g s s g s s
g s g s s g s s
m m m
B B
1
8
1
1 5
1
1 5
1
24 2 5
[1 ( ) ( )] ( ) ( ) ( )
( ) ( ) ( ) [1 ( ) ( )] ( )
1 2 2 2
1
 
 
     
      
m m m m
B B m
g s g s s g s s
g s g s s g s g s s
      
      
1
1
1
25 2 24
( ) [1 ( )] ( )
( ) ( ) ( )
1 1 1 2 1 2 2
2 1
      
      
m m m m m m m
m m
s g s s
g s s g s
      
    
( )
( )
( )
25
23
26 g s
g s
g s 
( )
( )
( )
25
24
27 g s
g s
g s 
( ) 1 ( ) ( ) 28 22 26 g s   g s g s
( ) ( )[1 ( )] 29 22 27 g s  g s  g s
( ) ( ) ( ) ( ) 30 26 22 27 g s  g s g s  g s
( ) ( ) ( ) ( ) ( ) ( ) ( ) 31 25 28 23 29 24 30 g s  g s g s  g s g s  g s g s
UP and Down state probabilities
The Laplace transforms of up and down state probabilities are as follows.
[1 ( )] ( )
[1 ( ) ( )] ( ) [1 ( ) ( )] ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
4 2,0
m 1 5 0,0 1 5 1,0
0,0 1,0 2,0 0, 1, 2,
1 1
1 1 1
g s P s
g s g s P s g s g s P s
P s P s P s P s P s P s P s
m
UP m m m
 
     
     
  (44)
[ ] ( )
[ ] ( )
[ ] ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
9 11 13 15 18 2,0
7 10 4 16 19 1,0
7 10 14 16 19 0,0
3,0 0, 2 1, 2 2, 2
g g g g g P s
g g g g g P s
g g g g g P s
P s P s P s P s P s P s P s P s P s down m m m E r H C
    
    
    
       
RELIABILITY:
Laplace transforms of the reliability of the system given as follows:





 









( )( )
1
1
( ) 1 1 1 1 1 1
s B s D
b
s B
b
s C
b
S A s C
R s m m m      
Taking inverse Laplace transforms of the above equation we get




 









 



 










 




 

 

( )( )
( ) ( )( ) ( ) ( )
( )( ) ( )
( ) 1
1
1 1
1 1 1
1
1 1 1 1
1 1 1
A B B D
b
e
C A
b
C A
m
e
A B D B
b
B A
b
e
C A
b
B A D A
b
B A
b
C A
R t e
Dt m
Bt m Ct m
At m m m
 
     
     
M.T.T.F
Mean time to system failure is the expected time to operate the system successfully which is given as
follows:



0
MT.T.F R(t)dt




 






 
( ) ( ) ( )( )
1
1 1 1 1 1 1 1
D A B A
b
C A
b
C A
m
B A
b m
A
m m      




 









 



 




( )( )
1
( ) ( )
1
( )( )
1
1
1 1 1
1
1 1 1
B D A D
b
D
A C
b
D B A B C C A
b
B A
b
B
m
m m m
 
     
where       m1 m2 1 A ,        m1 m2 1 B
     m2 1 C ,      m2 2 D
Numerical Illustrations:
Reliability Analysis:
0.01 1   , 0.02 2   , 0.011
1
 m  , 0.015
2
 m  ,  = 0.01,  = 0.02,  = 0.03, b = 0.96 and for
different values of t in the equation (60) one may obtain the reliability of the system as given in fig 2. The
reliability of the system decreases slowly as the time period increases. It also depicts the reliability of the system
for a long time period.
M.T.T.F
1 = 0.01, 2 = 0.02, 0.01
1
 m  , b = 0.96,  = 0.02 and taking different values of
m2  in the equation (61)
one may obtain the variations of M.T.T.F. of the system against the environmental failure rate  shown in figure
3.
The variations in M.T.T.F w.r.to Environmental failure rate as the major failure rate of the subsystem A
increases. The series of curves Represents that MTTF decreases as the environmental failure increases
apparently.

0
0.2
0.4
0.6
0.8
1
1.2
0 5 10 15 20 25
Time (t)
Reliability
Fig. 5.2.
Fig. 2 Fig. 3
Fig. 5.4
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0 0.02 0.04 0.06 0.08 0.1 0.12
Environmental Failure Rate
M.T.T.F.
Series1
Series2
Series3
Fig. 5.5.
Series 1 2 3
0.2
2
 m  0.4
2
 m  0.6
2
 m 
4 Series 1 2 3
0.2
1
 m  0.4
2
 m  0.6
3
 m 
Fig.5
Fig. 5.6.
6
V. DISCUSSION
In this paper we presented mathematical models
of imperfect switching, environmental, common
cause and human error effects, so probabilistic
behaviour considering various values of coefficient
of human error, common cause, environmental
effects respectively.
VI. CONCLUSIONS :
The Reliability, MTTF curves are plotted in figures (2),
(3). From these graphs we observe that
i) The Reliability of the system decreases slowly as
the time period increases.
ii) The MTTF decreases as the environmental
failure increases
REFERENCES
[1] Gupta, P.P. and Agarwal, S.C. : A parallel
redundant complex system with two types of
failure under different repair discipline, IEEE
Trans. Reliability, Vol R-32, 1983.
[2] Kontoleon, J.M. and Kontoleon, N., 1974.
Reliability Analysis of a system subjected to
partial & Catastrophic Failures, IEEE Trans.
On Rel., Vol. R-23, pp 277-278.
[3] Mittal, S.K., Gupta, Ritu and C.M. Batra,
2007. Probabilistic behavior of a Redundant
Complex System, With Imperfect Switching
and Environmental Effects under Head of Line
Repair Discipline, Bulletin of pure and
Applied Sciences, Vol. 6E(1), 2007, pp. no.43-
58.
[4] Mittal S.K., Agarwal, S.C., and Kumar Sachin,
2004. Operational Behaviour of a parallel
redundant complex system with two types of
failures with environmental effect under Pre-
Emptive Resume Repair Discipline, Bulletin of
Pure and Applied Sciences, Vol. 23E (No.2),
pp.253-263.

N0460284100

Recommended

Recommended

More Related Content

Similar to N0460284100

Similar to N0460284100 (20)

N0460284100