International Journal of Trend in Scientific
Research and Development (IJTSRD)
International Open Access Journal
ISSN No: 2456 - 6470 | www.ijtsrd.com | Volume - 2 | Issue – 2
Analysis of 1 out of 2 Main Units and
2 out
ut of 4 Subunits System
Praveen Gupta, Ruchi Yadav
School of Studies in Statistics Vikram Univesity,
Ujjain, Madhya Pradesh, India
ABSTRACT
Introduction of redundancy is one of the well
well-known
methods by which the reliability of a system can be
improved. A standby redundant system is one in
which one operating unit is followed by spare units
called standbys. On failure of operating unit, a
standby unit is in working mode. In the present paper
we investigate the probabilistic analysis of a two
two-main
unit and four subunit system. The system remain
operative if one mainn unit and two subunit are in
working mode. System is failed when the main unit
fails. A failed system is replaced by new one. The
failed sub-unit are repairable.
Keywords: Reliability, Availability, Mean time to
system failure, busy period analysis,, M
Markov renewal
process
Introduction
Two/three unit standby systems with two stages, that
is, working or failed, have been widely discussed by a
large number of researchers including (Mokaddis et.
Al. 1997, Taneja, 2005, Goel et. Al. 2010).
Introduction of redundancy is one of the well
well-known
methods by which the reliability of a system can be
improved. A standby redundant system is one in
which one operating unit is followed by spare units
called standbys. On failure of operating unit, a
standby unit is in working mode. Ritu Mittal (2006)
Analyzed
alyzed stochastic analysis of a compound
redundant system consisting three subsystems.
System Description and Assumption
Initially, the system starts operation with two main
units and four subunits. If one main unit and two
subunits work then system is in operative mode.
After failure of both the main units, it is replaced by
new – one. Failure time distributions of main unit and
sub-units
units are arbitrary functions of time. A single
repair facility is available to repair the sub-unit.
sub
The
repair rate of the sub-unit
unit is constant. The repair of
sub-unit
unit is as good as new. Using regenerative point
technique with the Markov-renewal
Markov
process, the
following reliability characteristics have been
obtained:
(i)
(ii)
Reliability of the system.
Steady state availability of the system.
Notation and States of the System
Constant
Constant failure rate of main unit
Constant
Constant failure rate of sub
sub-unit.
repair rate of sub-unit.
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International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470
f (.)
=
replacement time pdf of main unit.
S0 (M2S4)

2 main units and 4 sub-units are in normal mode.
S1(M1S4)

1 main unit and 4 sub-units are in working mode.
S2(M2S3)

2 main units and 3 sub-units are in working mode.
S3(M2S2)

2 main units and 4 sub-units are in working mode.
S4(M1S3)

1 main unit and 3 sub-units are in working mode.
S5(M1S2)

1 main unit and 2 sub-units are in working mode.
S6(M2S1)

System is in failed mode.
S7(M1S1)

System is in failed mode.
S8(M1S4)

System is in failed mode.
Figure 1: 1 out of 2 main units and 2 out of 4 subunits system
Analysis of Reliability and Mean Time to system Failure
Assuming that the failed states S7 and S8 to be the absorbing states and employing the arguments used in the
theory of regenerative process, the following relation among R i(t) be obtained :
R0(t) = Z0(t) + q01(t) © R1(t)+ q02(t) © R2(t)
R1(t) = Z1(t) + q14(t) © R4(t)
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R2(t) = Z2(t) + q20(t) © R0(t) + q24(t) © R4(t) + q23(t) © R3(t)
R3(t) = Z3(t) + q35(t) © A5(t) + q32(t) © A2(t)
R4(t) = Z4(t) + q41(t) © A1(t)
R5(t) = Z5(t) + q54(t) © A4(t)
(1-5)
where,
Z 0 (t)  e
Z2(t)  e
  t
 
 
Z4(t)  e
 t
,
Z1(t)  e
,
Z3(t)  e(),
,
Z5(t)  e().
,
For an illustration, the equation for R0(t) is the sum of the following mutually exclusive contingencies :
(i)
System sojourns in state S0up to time t. The probability of this contingency is
e
(ii)
  t
 z 0 (t), (Say).
System transits from state S0 to S1 during time (u, u+du); u t and then starting from S1, it survives for
the remaining time duration (t – u). The probability of this event is
t
q
01
(u)duR 1 (t  u)  q 01 (t) © R1(t).
0
(iii)
System transits from state S0 to S2 during time (u, u+du); u t and then starting from state S2, it survives
for the remaining time duration (t – u). The probability of this event is
t
q
02
(u)duR 2 (t  u)  q 02 (t) © R2(t).
0
Taking Laplace Transform of relations (1-5) and writing the solution of resulting set of algebraic equations in
the matrix form as follows
1
R*0   1 q*01 q*02 0
0
0 
 * 

*
1
0
0 q14 0 
R1   0
R*2  q*20 0
1 q*23 q*24 0 
 *

0 q*32 1
0 q*35 
R3   0
R*   0 q*
0
0
1 q*45 
4
41
 * 

*
0
0
0 q54
1 
R5   0
*
Z*0 
 *
 Z1 
Z*2 
 * .
Z3 
Z* 
 4* 
Z5 
(6)
*
*
The argument‘s’ has been omitted from Ri (s), qij(s) and Z1(s) for brevity. Computing the above matrix
*
equation for R 0 (s) , we get
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International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470
R *0 (s) 
N1(s)
D1 (s)
(7)
Where
N1  s 
D1  s  
Z*0 q*01 q*02
0
0
0
Z1*
1
0
0
*
q14
0
Z*2
0
0
q*23
0
0
Z*3
0
0
1
0
q*35
Z*4 q*41
0
0
1
0
Z5*
0
0
*
q54
1
0
1 q *01 q *02
0
0
0
0
1
0
0
*
q14
0
0
0
0
q *23
0
0
0
0
q *32
1
0
q *35
0 q *41
0
0
1
0
0
0
0
*
q 54
1
0
Taking the inverse Laplace transforms of the expression (7), we can get the reliability of the system when
system initially starts from state S0.
To get the MTSF, we have a well-known formula
E  T0    R0 (t)dt  lim R0* (S)
S0
*
So that, using q *ij(0)  pij and Zi (0)  i we get
E  T0  
 0  p011  p02 2  p 01p23 3  p02p14 4
.
1  p01p14p24  p02p35 p32p54
(8)
Availability Analysis
Ai (t) as the probability that the system is operative at epoch t due to both the units respectively. When initially
system starts from state SiE. Using the similar probabilistic arguments, point wise availability are satisfied
the recursive relations as follows:
A0(t)  Z0(t)  q01(t)© A1(t)  q02(t)© A2(t)
A1(t)  Z1(t)  q14(t)A4(t)  q18(t)A8(t)
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International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470
A6(t)  q63(t)A3(t)
A7(t)  q75(t)A5(t)
A8(t)  q80(t)A0(t).
(1-9)
*
Taking Laplace transform (L.T.) of the set of relations (1-9), the solution for A i (s) can be written in the
matrix form as follows*
Taking Laplace transform of relation (1-9), and solving for , A 0 (s) , we get
A *0 (s) 
N2 (s)
D2 (s)




*
*
*
*
*
*
*
*
*
*
*
Where N 2 (s)   q 01 q 23q14  q 32q 41  q 02 q 23q14  q 32q 41  Z1



* *
q*80 q*45q*32  q54
q01 Z3* .
(11)
and





*
* *
D2(s)  1  q*23q*24 1  q*01q10
 q*01 q*23  q14
q53 q*35




*
*
*
*
* *
*
*
q*02 q*24q*36  q*63q75
q32
 q03
q36
q14
q75  q14
q30




*
*
*
*
q*02 q*24  q*63q75
 q24
q75
 q58
q*45.
(12)
The steady state availability of the system due to both the units is given by
A0  lim A0(t)  lim A*0(s)
t 
 lim S
s 0
s 
N1(s)
.
D1(s)
Now,
D2(0)  1  p23p24  1  p10p01   p01  p23  p14p53  p35
p02  p24p36  p63p75  p32  p03p36  p14p75  p14 
p02  p24  p63p75   p24  p75  p58  p32
 p20 1  p01p10   p01p23  p01p14p53  p02p24p36p32
p03p75p32  p03p36p14p15  p03p36p14  p02p24
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International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470
p02p03p75  p24p75p45  p24p58p45
= 0.
Hence, by L. Hospital’s rule
N (s) N1(0)
A 0  lim 1'

s  0 D (s)
D1(0)
1
where,
N1(0)   p 01  p 23 p 24  p 32p 41   p 02  p 23 p 24  p 32p 41   1
+
 p45p32  p54p01  3
'
D1' (s)
D
(0)
s 0 as follows 1
In order to obtain
we collect the coefficient of mij in
Coefficient of
Coefficient of
m01  p23p24  p32p41
m02  p23p24  p32p41
Coefficient of
m23  p01p24  p02p24
Coefficient of
m24  p01p23  p02p23
Coefficient of
m32  p01p41  p02p41  p54
Coefficient of
 p41  p54
m41  p02p32  p01p32
Coefficient of
 p32
m45  p32
Coefficient of
 p24  p01  p02   p24
 p23  p01  p02   p23
  p01  p02  p41  p54
  p01  p02  p32
m54  p01.
Busy Period Analysis
Let Bi(t) be the probabilities that the repairman is busy in the repair of sub-unit at time t when system initially
starts from regenerative state Si using some the probabilistic arguments in respect to the definition of
have the following recursive relation –
B'i (t) , we
B00(t)  q 01(t)B1(t)  q02(t)B2(t)
B1k (t)  q 14 (t)B4 (t)  q18 (t)B8 (t)
B2(t)  W2q20(t)B0(t)  q24(t)B4(t)  q23(t)B3(t)
B3(t)  q35(t)B5(t)  q36(t)B6(t)  q32(t)B2(t)
B4(t)  W4  q41(t)B1(t)  q48(t)B8(t)  q45(t)B5(t)
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B5(t)  W5  q58(t)B8(t)  q57(t)B7(t)  q54(t)B4(t)
B6(t)  W6  q63(t)B3(t)
B7(t)  W7  q75(t)B5(t)
B8(t)  q80(t)B0(t)
where
W2  e
 t
W5  e
 
W4  e
;
  t
t
;
W7  et .
W6  e ;
;
(1-9)
B' (t)
For an illustration, 2 is the sum of the following mutually exclusive contingencies –
(i)
The repairman remains busy in state S2 continuously up to time t in the repair of sub-unit. The
probability of this event is -
e
 t
 W2(t).
(ii)
System transists from state S2 to S0 during time (u, u+du); u t and then repairman remains busy in the
repair of epoch t starting from state S0 at epoch u. The probability of this contingency is –
t
q
20
(u)duB '4 (t  u)  q 20 (t) B 0 (t)
.
(iii)
System
transits
from
state
S2
to
S4
during
the
time
(u, u+du); u t and then starting from state S4 at epoch u, the repairman may be observed to be busy in the
repair at epoch t. The probability of this contingency is
0
t
q
24
(u)duB '4 (t  u)  q 24 (t ) B '4 (t)
.
(iv)
Systemtransits
from
state
S2
to
S3
during
the
time
(u, u+du); u t and then starting from state S3 at epoch u, the repairman may be observed to be busy in the
repair at epoch t. The probability of this contingency is
0
t
q
23
(u)duB '3 (t  u)  q 23 (t ) B '3 (t)
.
0
Taking Laplace transform of relations (1-9) and solving the resulting set of the algebraic equation for
we get
B*0 (s) 
N3 (s)
.
D2 (s)
N 3 (s )  1  q q q q
*
24
*
41
*
14
*
57
q
*
35
W  q
*
3
*
01
q q q
*
02
*
23
*
58
B*0 (s) ,
(10)
 q 14* W4*  q 18* q 63* 
*
*
  q *02  q 14
q *45  q 25
W5*  q *24 q *75 W3*  q *80 q *75  .
In the long run, the expected fraction of the time for which the repairman is busy in the repair of the system, is
given by
B 0  lim B 0 (t)  lim sB 0* (s)
t

where
s0
N 3 (0)
D '2 (0)
N3 (0)  1  p24p41p14p57  p353   p01  p02p23p58   p144  p18 
  p02  p145  p255  p24p753  p75  .
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International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470
Conclusion
References
This paper describes an improvement over the Gupta,
R, Goel, C .K. and Tomar, (2010) Analysis of a two
unit standby system with correlated failure and repair
and random appearance and disappearance of
repairman. Using regenerative point technique
reliability analysis, availability analysis, busy period
analysis which shows that the proposed model is
better than Gupta, R, Goel, C .K. and Tomar,
(2010)because
we investigate the probabilistic
analysis of a two-main unit and four subunit system.
On failure of operating unit, a standby unit is in
working mode. The system remains operative if one
main unit and two subunit are in working mode.
1. Gupta, R., Goel, C.K. and Tomar, A., ‘Analysis of
a two unit standby system with correlated failure
and repair and random appearance and
disappearance of repairman’. Journal of
Reliability and Statistical Studies, Vol. 3, pp. 5361 (2010).
2. Goel, L. R. and Gupta, P., ‘Switch failure in a
two-unit standby system with better utilization of
units’. Microelectron Reliability. Vol. 24, pp. 439443 (1984).
3. Goel, L.R. and Praveen Gupta, ‘A two unit
deteriorating standby system with inspection’.
Microelectron and Reliability, Vol. 24, pp. 435438 (1984).
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