Research Journal of Applied Sciences, Engineering and Technology 7(16): 3280-3285, 2014
ISSN: 2040-7459; e-ISSN: 2040-7467
© Maxwell Scientific Organization, 2014
Submitted: August 05, 2013
Accepted: August 16, 2013
Published: April 25, 2014
Stochastic Analysis of a Parallel System under Maintenance and Repair Subject to
Random Shocks
1
A.K. Barak, 2S.K. Chhillar and 2S.C. Malik
Department of Mathematics, Manipal University, Jaipur-303007, Rajasthan, India
2
Department of Statistics, M.D. University, Rohtak-124001, Haryana, India
1
Abstract: This study presents the stochastic analysis of a two unit parallel system subject to random shocks. There
is a single server who visits immediately to do maintenance and repair of the system. The operative units in the
system undergo for maintenance if they are affected by the impact of shocks with some probabilities. However,
repair of the units is done when they fail due to some other reasons. The unit works as new after maintenance and
repair. The distributions of random shocks and failure times of the unit follow negative exponential while that of
maintenance and repair times are taken as arbitrary with different probability density functions. Using regenerative
point technique and semi-Markov process, several measures of system effectiveness are obtained in steady state. The
graphical behavior of MTSF, availability and profit function has been analyzed for particular values of various
parameters and costs. Profit of the present model has also been made with the profit of the system model proposed
by Malik and Chhillar (2012).
Keywords: Maintenance and repair, parallel system, random shocks
INTRODUCTION
The method of cold standby redundancy has widely
been adopted by the researchers including Murari and
Goyal (1984) and Singh (1989) while analyzing
systems of two or more units not only to attain better
reliability but also to reduce the down time of the
system. But sometimes cold standby redundancy is not
suggestive when shocks occur during operation of the
system. In such a situation, the units in the system may
be may be operated in parallel mode to share the impact
of shocks. It is a known fact that cold standby
redundancy is better than parallel redundancy so far as
reliability is concerned. But, no research study has been
written by the researchers on these types of systems so
far in the subject of reliability. However, a little work
has been carried out by the authors including Murari
and Al-Ali (1988) and Wu and Wu (2011) on the
reliability modeling of shock models with the concept
of maintenance and repairs. Shocks are the external
environmental conditions which cause perturbation to
the system, leading to its deterioration and consequent
failure. The shocks may be caused by external factors
such as fluctuation of unstable electric power, power
failure, change in climate conditions, change of
operator, etc., or due to internal factors such as stress
and strain. Many systems like power generation and
automotive industries are vulnerable to damage caused
by shock attacking that may occur over the service life.
Sometimes, a system may or may not be affected by the
impact of shocks and the system may fail due to
operation and/or due to random shocks.
Keeping the above observations and practical
situations in mind, here stochastic analysis of a two unit
parallel system subject to random shocks. There is a
single server who visits immediately to do maintenance
and repair of the system. The operative units in the
system undergo for maintenance if they are affected by
the impact of shocks with some probabilities. However,
repair of the units is done when they fail due to some
other reasons. The unit works as new after maintenance
and repair. The distributions of random shocks and
failure times of the unit follow negative exponential
while that of maintenance and repair times are taken as
arbitrary with different probability density functions.
Various reliability indices such as transition
probabilities, mean sojourn times, Mean Time to
System Failure (MTSF), availability, busy period of the
server due to repair and maintenance, expected number
of maintenance and repair and profit function are
evaluated in steady state using semi-Markov process
and regenerative point technique. The numerical results
giving particular values to various costs and parameters
are obtained for MTSF, availability and profit to depict
their graphical behavior with respect to shock rate.
MATERIALS AND METHODS
The following are the possible transition states of
the system:
S 0 = (O, O), S 1 = (SUm, O), S 2 = (SUM, SWm),
S 3 = (O, FUr), S 4 = (FUR, SWm), S 5 = (FUR,
FWr), S 6 = (SUM, FWr)
Corresponding Author: A.K. Barak, Department of Mathematics, Manipal University, Jaipur-303007, Rajasthan, India
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Res. J. App. Sci. Eng. Technol., 7(16): 3280-3285, 2014
The mean sojourn times µ i for m (t) = 𝜃𝜃𝑒𝑒 −𝜃𝜃𝜃𝜃 , g
(t) = 𝛼𝛼𝑒𝑒 −𝛼𝛼𝛼𝛼 are:
µ0 =
1
1
, µ1 =
,
λ + µ +θ
2q02 µ + 2λ + 4 p0 µ
µ3 = 1
, µ1' = λ + p0 µ + θ ,
θ (λ + µ + θ )
λ + µ +γ
'
p
λ
µ
γ
+
+
0
=
µ3
γ (λ + µ + γ )
(4)
where,
m00 + m01 + m03 =
µ0 , m10 + m11 + m12 + m16 =
µ1 ,
m30 + m33 + m34 + m35 =
µ3 ,
m10 + m11 + m11.2 + m13.6 =
µ1′ ,
Fig. 1: State transition diagram
m30 + m33 + m33.5 + m31.4 =
µ3′
The transition states S 0 , S 1 , S 3 , are regenerative
and states S 2 , S 4 , S 5 , S 6 are non-regenerative as shown
in Fig. 1.
Transition probabilities and mean sojourn times:
Simple probabilistic considerations yield the following
expressions for the non-zero elements:
∞
pij = Qij (∞) = ∫ qij (t )dt
0
RESULTS AND DISCUSSION
Reliability and Mean Time to System Failure
(MTSF): Let ϕ i (t) be the cdf of first passage time from
regenerative state i to a failed state. Regarding the
failed state as absorbing state, we have the following
recursive relations for ϕ i (t):
as:
p 03
ϕ 0 (t) = Q 00 (t) Ⓢϕ 0 (t) + Q 01 (t) Ⓢϕ 1 (t) + Q 03 (t)
Ⓢϕ 3 (t)
ϕ 1 (t) = Q 10 (t) Ⓢϕ 0 (t) + Q 11 (t) Ⓢϕ 1 (t) + Q 12 (t) +
Q 16 (t) ϕ 3 (t) = Q 30 (t) Ⓢϕ 0 (t) + Q 34 (t) + Q 35 (t) +
Q 33 (t) Ⓢϕ 3 (t)
(6)
4 p0 µ
2q02 µ
, p 01 =
,
2q02 µ + 2λ + 4 p0 µ
2q µ + 2λ + 4 p0 µ
2λ
=
, p 10 = θ
, p 11 = q0 µ
2
λ + µ +θ
λ + µ +θ
2q0 µ + 2λ + 4 p0 µ
p 00 =
2
0
,
p 12 =
p 30
p 34
p0 µ , p 16 =
λ
,
λ + µ +θ
λ + µ +θ
γ
=
, p 33 = q0 µ ,
λ + µ +γ
λ + µ +γ
p
µ
λ
0
=
, p 35 =
λ + µ +γ
λ + µ +γ
Taking LST of above relations (6) and solving for
~
φ0 ( s) :
We have:
~
R* (s) = 1 − φ0 ( s)
s
p 21 = m * ( 0 ) , p 41 = g * ( 0 ) ,
p 53 = g * ( 0 ) , p 63 = m * ( 0 )
~
MTSF = lim 1 − φ0 ( s ) = N1
s →o
s
D1
p 11.2 =
p 33.5
(7)
The reliability of the system model can be obtained
by taking Laplace inverse transform of (7).
The Mean Time to System Failure (MTSF) is given by:
(1)
For m (t) = 𝜃𝜃𝑒𝑒 −𝜃𝜃𝜃𝜃 and g (t) = 𝛼𝛼𝑒𝑒 −𝛼𝛼𝛼𝛼 we have:
p0 µ , p
λ
,
13.6 =
λ + µ +θ
λ + µ +θ
λ
p0 µ
=
, p 31.4 =
λ + µ +γ
λ + µ +γ
(5)
(8)
where,
(2)
N 1 = μ 0 (1-p 11 ) (1-p 33 ) + μ 1 p 01 (1-p 33 ) + μ 3 p 03 (1-p 11 )
It can be easily verified that:
and
p 00 + p 01 + p 03 = p 10 + p 11 + p 12 + p 16 = p 21 = p 30 +
p 33 + p 34 + p 35 = p 30 + p 33 + p 33.5 + p 31.4 = p 41 = p 53
= p 63 = p 10 + p 11.2 + p 11 + p 13.6 = 1
(3)
D 1 = (1-p 00 ) (1-p 11 ) (1-p 33 ) -p 01 p 10 (1-p 33 ) - p 03 p 30
(1-p 11 )
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Res. J. App. Sci. Eng. Technol., 7(16): 3280-3285, 2014
Steady state availability: Let A i (t) be the probability
that the system is in up-state at instant ‘t’ given that the
system entered regenerative state i at t = 0.
The recursive relations for A i (t) are given as:
A 0 (t) = M 0 (t) + q 00 (t) © A 0 (t) + q 01 (t) © A 1 (t)
+ q 03 (t) © A 3 (t)
A 1 (t) = M 1 (t) + q 10 (t) © A 0 (t) + [q 11 (t) + q 11.2
(t)] © A 1 (t) + q 13.6 (t) © A 1 (t)
A 3 (t) = M 3 (t) + q 30 (t) © A 0 (t) + [q 33.5 (t) + q 33
(t)] © A 3 (t) + q 31.4 (t) © A 1 (t)
(9)
where, M i (t) is the probability that the system is up
initially in state 𝑆𝑆𝑖𝑖 ϵ𝐸𝐸 is up at time t without visiting to
any other regenerative state, we have:
M 0 ( t ) = e − ( λ +µ ) t , M1 ( t ) = e − ( λ +µ ) t M ( t ) ,
M 3 ( t ) = e − ( λ +µ ) t G ( t )
(10)
Taking LT of above relations (9) and solving for
A 0 * (s), the steady state availability is given by:
A0 (∞) =lim sA0* ( s ) = N 2
s→0
Now taking L.T. of relations (12) and obtain the
value of BR 0 * (s) and by using this, the time for which
server is busy in steady state is given by:
BR 0 = N 3 /D 2
where,
N 3 = [p 10 p 13.6 + p 03 (1 - p 11.2 - p 11 )] w3∗ ( s ) and D 2 is
already defined
Due to maintenance: Let BM 0 (t) be the probability
that the server is busy in maintenance of the system at
instant t given that the system entered regenerative state
i at t = 0. The recursive relation for BR 0 (t) are as
follows:
BM 0 (t) = q 00 (t) © BM 0 (t) + q 01 (t) © BM 1 (t) + q 03
(t) © BM 3 (t)
BM 1 (t) = W 1 (t) + q 10 (t) © BM 0 (t) + [q 11 (t) +
q 11.2 (t)] © BM 1 (t) + q 13.6 (t) © BM 3 (t)
BM 3 (t) = q 30 (t) © BM 0 (t) + [q 33.5 (t) + q 33 (t)] ©
(13)
BM 3 (t) + q 31.4 (t) © BM 1 (t)
where,
(11)
W 1 (t) =
D2
e − ( λ + µ )t M (t ) + (λ e − ( λ + µ )t 1) M (t ) + ( p0 µ e − ( λ + µ )t 1) M (t ) + (q0 µ e − ( λ + µ )t 1) M (t )
where,
N 2 = [(1 - p 11.2 - p 11 ) (1 - p 33 - p 33.5 ) - p 31.4 p 13.6 ] μ 0
+ [p 01 (1 - p 33 - p 33.5 ) + p 31.4 p 03 ] μ 1 + [p 01 p 13.6 +
p 03 (1 - p 11.2 - p 11 )] µ 3
Now taking L.T. of relations (13) and obtain the
value of BM 0 * (s) and by using this, the time for which
server is busy in steady state is given by:
BM 0 = N 4 /D 2
and,
D 2 = [(1 - p 11.2 - p 11 ) (1 - p 33 - p 33.5 ) - p 31.4 p 13.6 ] μ 0
+ [p 01 (1 - p 33 - p 33.5 ) + p 31.4 p 03 ] µ1′ + [p 01 p 13.6 + p 03
where,
N 4 = [p 01 (1 - p 33 - p 33.5 ) + p 31.4 p 03 ] w1∗ ( s ) and D 2 is
(1 - p 11.2 - p 11 )] µ3′
already defined
Busy period analysis of the server:
Due to repair: Let BR i (t) be the probability that the
server is busy in repairing of the system at instant t
given that the system entered regenerative state i at
t = 0. The recursive relation for BR i (t) are as follows:
Expected number of maintenance: Let 𝑁𝑁𝑖𝑖𝑀𝑀 (𝑡𝑡) the
expected number of maintenance of the system by the
server in (0, t) given that the system entered the
regenerative state i at t = 0. The recursive relations for
N iM (t ) are given as:
NM 0 (t) = Q 00 (t) Ⓢ NM 0 (t) + Q 01 (t) Ⓢ NM 1 (t) +
Q 03 (t) Ⓢ NM 3 (t)
NM 1 (t) = Q 10 (t) Ⓢ (1 + NM 0 (t)) + Q 11 (t) ⓈNM 1
(t) + Q 11.2 (t) Ⓢ (1 + NM 1 (t)) + Q 13.6 (t) Ⓢ (1 +
NM 3 (t))
NM 3 (t) = Q 30 (t) Ⓢ NM 0 (t) + [Q 33.5 (t) + Q 33 (t)]
Ⓢ NM 3 (t) + Q 31.4 (t) Ⓢ NM 1 (t)
(14)
BR 0 (t) = q 00 (t) © BR 0 (t) + q 01 (t) © BR 1 (t) + q 03
(t) © BR 3 (t)
BR 1 (t) = q 10 (t) © BR 0 (t) + [q 11 (t) + q 11.2 (t)] ©
BR 1 (t) + q 13.6 (t) © BR 3 (t)
BR 3 (t) = W 3 (t) + q 30 (t) © BR 0 (t) + [q 33.5 (t) + q 33
(t)] © BR 3 (t) + q 31.4 (t) © BR 1 (t)
(12)
where,
W 3 (t) =
e − ( λ + µ )t G (t ) + (λ e − ( λ + µ )t 1)G (t ) + ( p0 µ e − ( λ + µ )t 1)G (t ) +
) + (q0 µ e − ( λ + µ ) t 1)G (t )
Now taking L.T. of relations (14) and obtain the
value of NM 0 * (s) and by using this, the time for which
server is busy in steady state is given by:
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Res. J. App. Sci. Eng. Technol., 7(16): 3280-3285, 2014
NM 0 = N 5 /D 2
P = K 0 A0 − K1 B0M − K 2 B0R − K3 N 0M − K 4 N 0R − K5
where,
N 5 = (p 10 + p 11.2 + p 13.6 ) [p 01 (1 - p 33 - p 33.5 ) + p 31.4 p 03 ]
and D 2 is already defined
Expected number of repairs: Let 𝑁𝑁𝑖𝑖𝑅𝑅 (𝑡𝑡) the expected
number of repairs of the system by the server in (0, t)
given that the system entered the regenerative state i at
t = 0. The recursive relations for 𝑁𝑁𝑖𝑖𝑅𝑅 (𝑡𝑡) are given as:
NR 0 (t) = Q 00 (t) Ⓢ NR 0 (t) + Q 01 (t) Ⓢ NR 1 (t) +
Q 03 (t) Ⓢ NR 3 (t)
NR 1 (t) = Q 10 (t) Ⓢ NR 0 (t) + Q 11 (t) Ⓢ NR 1 (t) +
Q 11.2 (t) Ⓢ NR 1 (t) + Q 13.6 (t) Ⓢ NR 3 (t)
NR 3 (t) = Q 30 (t) Ⓢ [1 + NR 0 (t)] + Q 33.5 (t) Ⓢ [1 +
NR 3 (t)] + Q 33 (t) Ⓢ NR 3 (t) + Q 31.4 (t) Ⓢ [1 + NR 1
(t)]
(15)
(16)
where,
K 0 = Revenue per unit up-time of the system
K 1 = Cost per unit time for which server is busy due to
maintenance
K 2 = Cost per unit time for which server is busy due to
repair
K 3 = Cost per unit maintenance of the shocked unit
K 4 = Cost per unit repair of the failed unit
K 5 = Total cost for the busy of the server and
𝐴𝐴0 , 𝐵𝐵0𝑀𝑀 , 𝐵𝐵0𝑅𝑅 , 𝑁𝑁0𝑀𝑀 , 𝑁𝑁0𝑅𝑅 are already defined
Particular cases:
Suppose g (t) = ge-gt, m (t) = θe-θt
We can obtain the following results:
Now taking L.T. of relations (15) and obtain the
value of NR 0 * (s) and by using this, the time for which
server is busy in steady state is given by:
MTSF (T 0 ) =
𝐷𝐷1
,
Availability (A 0 ) =
𝑁𝑁3
Busy period due to repair (𝐵𝐵0𝑅𝑅 ) =
NR 0 = N 6 /D 2
𝐷𝐷2
Busy period due to maintenance (𝐵𝐵0𝑀𝑀 ) =
where,
N 6 = (p 30 + p 33.5 + p 31.4 ) [p 03 (1 - p 11 - p 11.2 ) + p 01 p 13.6 ]
and D 2 is already defined
Profit analysis: The profit incurred to the system
model in steady state can be obtained as:
𝑁𝑁1
Expected number of maintenance
(𝑁𝑁0𝑀𝑀 )
Expected number of repairs (𝑁𝑁0𝑅𝑅 ) =
where,
[(λ + µ + γ )(λ + µ + θ ) − q0 µ (λ + µ + γ ) − q0 µ (λ + µ + θ ) + ( q0 µ ) + 4 p0 µ (λ + µ + γ )
2
N1 =
−4q0 p0 µ 2 + 2λ (λ + µ + θ ) − 2λ q0 µ ]
(2q02 µ + 2λ + 4 p0 µ )(λ + µ + γ )(λ + µ + θ )
[(λ + µ + γ )(λ + µ + θ )(2q02 µ + 2λ + 4 p0 µ ) − (λ + µ + γ )(λ + µ + θ )2q02 µ − q0 µ (λ + µ + γ )
(2q02 µ + 2λ + 4 p0 µ ) + 2q03 µ 2 (λ + µ + γ ) − 2q04 µ 3 (λ + µ + θ ) − 4 p0 µθ (λ + µ + γ ) + 4q0 p0 µ 2
− q0 µ (λ + µ + θ )(2q02 µ + 2λ + 4 p0 µ ) + 2q03 µ 2 (λ + µ + θ ) + (2q02 µ + 2λ + 4 p0 µ ) ( q0 µ )
D1 =
−2λγ (λ + µ + θ ) + 2λµ q0 ]
(2q02 µ + 2λ + 4 p0 µ )(λ + µ + γ )(λ + µ + θ )
[(λ + µ + γ )(λ + µ + θ ) − µ (λ + µ + γ ) − λ (λ + µ + θ ) − q0 µ (λ + µ + θ )
N2 =
+ q0 µ 2 + µλ − p0 µλ + 4 p0 µ (λ + µ + γ ) − 4q0 p0 µ 2 + 2λ (λ + µ + θ ) − 2λµ ]
(2q02 µ + 2λ + 4 p0 µ )(λ + µ + γ )(λ + µ + θ )
γθ [(λ + µ + γ )(λ + µ + θ ) − µ (λ + µ + γ ) − λ (λ + µ + θ ) − q0 µ (λ + µ + θ )
+ q0 µ 2 + µλ − p0 µλ ] + γ (λ + p0 µ + θ )[4 p0 µ (λ + µ + γ ) − 4q0 p0 µ 2 − 2 p0 µλ ]
D2 =
+θ (λ + p0 µ + γ )[2λ (λ + µ + θ ) − 2λµ + 2 p0 µλ ]
γθ (2q02 µ + 2λ + 4 p0 µ )(λ + µ + γ )(λ + µ + θ )
N 3 = [4 p0 µγ + 2λ (λ + µ + θ ) − 2λµ ]
γ (2q02 µ + 2λ + 4 p0 µ )(λ + µ + θ )
3283
2
𝑁𝑁6
𝐷𝐷2
𝑁𝑁2
𝐷𝐷2
𝑁𝑁4
𝐷𝐷2
𝑁𝑁5
=
𝐷𝐷2
(17)
Res. J. App. Sci. Eng. Technol., 7(16): 3280-3285, 2014
N 4 = [2q 2 0 µ (λ + µ + γ ) − 2q 30 µ 2 − 2q 2 0 µλ + 2 p0 µλ ]
θ (2q02 µ + 2λ + 4 p0 µ )(λ + µ + γ )
N5 =
[2q 2 0 µ (λ + µ + γ ) − 2q 30 µ 2 − 2q 2 0 µλ + 2 p0 µλ ] (λ + p0 µ + θ )
(2q02 µ + 2λ + 4 p0 µ )(λ + µ + γ )
(λ + µ + θ )
N 6 = [4 p0 µγ + 2λ (λ + µ + θ ) − 2λµ ] (λ + p0 µ + γ )
(2q02 µ + 2λ + 4 p0 µ )(λ + µ + θ ) (λ + µ + γ )
CONCLUSION
In the present study some important reliability
measures including MTSF, availability and profit for a
two- unit parallel system subject to random shocks have
been obtained numerically giving particular values to
various costs with p 0 = 0.6 and q 0 = 0.4. The graphical
behavior of these measures as shown in Fig. 2 to 4 go
on decline with increase of shock rate (µ) and failure
Fig. 2: MTSF vs. shock rate (μ)
Fig. 3: Availability vs. shock rate (μ)
Fig. 4: Profit vs. shock rate (μ)
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Res. J. App. Sci. Eng. Technol., 7(16): 3280-3285, 2014
Fig. 5: Profit difference vs. shock rate (μ)
rate (λ). However, there values increase with the
increase of repair and maintenance rate. Thus, the study
reveals that a parallel system in which operative units
subject to random shocks can be made more reliable
and profitable to use by doing repair and maintenance
of the system with higher rates.
Comparative study: As shown in Fig. 5, the profit of
the present model has been compared with the profit of
the model proposed by Malik and Chhillar (2012). It is
concluded that a two-unit cold standby system subject
to random shocks would have more profit as compare
to the two unit parallel system.
M i (t)
W i (t)
m ij
NOTATIONS
E
O
: Set of regenerative states
: The unit is operative and in normal
mode
: The probability that shock is effective
p0
: The probability that shock is not
q0
effective
µ
: Constant rate of the occurrence of a
shock
λ
: Constant failure rate of the unit
m (t) /M (t) : pdf/cdf of maintenance time of the unit
after the effect of a shock
FUr/FWr/FUR : The Unit is completely failed and under
repair/waiting
for
repair/under
Continuous repair from previous state
SUm/SUM : Shocked unit under maintenance and
under maintenance continuously from
previous state
SWm
: Shocked unit waiting for maintenance
g (t) /G (t)
: pdf/cdf of repair time of the completely
failed unit
q ij (t) /Q ij (t) : pdf and cdf of direct transition time
from a regenerative state i to a
regenerative state j without visiting any
other regenerative state
q ij.k (t) /Q ij.k (t) : pdf and cdf of first passage time from
a regenerative state i to a regenerative
(s) /©
~/*
/ (desh)
State j or to a failed state j visiting state
k once in (0, t)
: Probability that the system is up
initially in state S i ∈ E is up at time t
without visiting to any other
regenerative state
: Probability that the server is busy in
state S i up to time t without making
transition to any other regenerative state
or returning to the same via one or more
non regenerative states
: Contribution to mean sojourn time in
state S i when system transits directly to
state S j (S i , S j ϵ E) so that µ i = ∑𝑗𝑗 𝑚𝑚𝑖𝑖𝑖𝑖
where m ij = ∫ 𝑡𝑡d Q ij (t) = - q ij */ (0) and
µ i is the mean sojourn time in state S i ϵ
E
: Symbol for Stieltjes convolution/
Laplace convolution
: Symbol for Laplace Stieltjes Transform
(LST) /Laplace Transform (LT)
: Symbol for derivative of the function
REFERENCES
Malik, S.C. and S.K. Chhillar, 2012. Profit analysis of a
cold standby system under maintenance and repair
subject to random shocks. Int. J. Math. Arch.,
3(11): 3975-3981.
Murari, K. and V. Goyal, 1984. Comparison of two-unit
cold standby reliability models with three types of
repair facilities. J. Microelectron. Reliab., 24(1):
35-49.
Murari, K. and A.A. Al-Ali, 1988. One unit reliability
system subject to random shocks and preventive
maintenance. J. Microelectron. Reliab., 28(3):
373-377.
Singh, S.K., 1989. Profit evaluation of a two-unit cold
standby system with random appearance and
disappearance time of the service facility.
Microelectron. Reliab., 29: 705-709.
Wu, Q. and S. Wu, 2011. Reliability analysis of a twounit cold standby reparable systems under poison
shocks. Appl. Math. Comput., 218(1): 171-182.
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