Availability Measures of Two Unit Parallel
Redundant System with Replacement
1
Reena Grover,
2Praveen Saraswat,
3Viresh Sharma
12Assistant Professor, Department of Mathematics, SRM University NCR Campus Modinagar India. 3Assistant Professor, Department of Mathematics, N.A.S (P.G.) College Meerut India.
ABSTRACT:For the system which has two units in parallel redundancy, the reliability of repairable systems can be calculated as a result of the numerical solution of a simultaneous set of linear differential equations. The closed form solutions of the transient probabilities are used to obtain the reliability for non-repairable systems. The probabilities so obtained can be utilized to estimate the reliability and MTTF of the system.
KEYWORDS: Reliability, Multi-failure, Redundant system, Failure rate, Supplementary variable process.
INTRODUCTION
The system has two units in parallel redundancy. Since the system can fail partially or completely, a unit is said to be degraded when it fails partially, i.e., a degraded unit works, but not with its normal efficiency. A unit is said to be failed when it fails completely. Initially the system is in good state. There may be 3 states, good, degraded or failed, i.e., system is good when either or both units are in good state. The system is degraded when both units are neither good nor failed. The system fails when both units are failed. The failure and repair time for the system follows exponential distribution.
NOTATIONS
(i) G0, G1, G2 : Good state of the system, D1,D2, Degraded states of the system, F : Failed
(ii)
1
'
1 : Failure rates of a good unit to degraded state.(iii)
2,
'
2: Failure rates of a good unit to failure state, (iv)
3,
'
3: Failure rates of a degraded units to failed state.(v) Si(r),
i(
r
)
: Probability density function and hazard rates for repair time of the system.(vi) Pi (t) : Probability that the system in state i at time t i = G0, G1, G2, D1, D2, F
MATHEMATICAL ANALYSIS
,
)
,
(
)
(
)
(
2
2
0 2
1 hG
P
Gt
Fz
P
Fz
t
dt
(1)
0 1 1
1 1
2 1
3
'
'
P
(
t
)
2
P
(
t
)
P
(
x
,
t
)
(
x
)
dx
,
t
G
G D
Ddy
t
y
P
Y
D
)
D(
,
)
(
2 20
(2));
(
2
)
(
)
(
'
'
1 2P
2t
3P
1t
2P
t
t
G
G
G
(3)
;
0
)
,
(
)
(
'
2
3 1
1
t
x
P
x
x
t
D D (4)0
)
,
(
)
(
'
3 2
2
t
y
P
y
y
t
D D (5).
0
)
,
(
)
(
t
z
P
z
z
t
F F (6)Boundary conditions
);
(
'
)
,
(
1 11
o
t
P
t
P
D
G (7))
(
'
2
)
(
2
'
)
(
'
)
,
(
1 2 1 3 12
o
t
P
t
P
t
P
t
P
D
G
G
D (8)).
(
)
(
'
)
(
'
)
,
(
o
t
2P
2t
3P
2t
P
t
P
F
G
D
hG G (9)Initial conditions are
0
0
1
)
0
(
wheni
i
when
Pi
(10)
SOLUTIONS OF THE PROBLEM
Taking Laplace transform of above equations from (1) to (10) we obtain
dz
z
s
z
P
s
P
s
2
2
hG)
G(
)
1
F(
,
)
F(
)
(
*0 *
2
1
(11)dx
x
s
x
P
P
s
P
s
)
G(
)
2
G*
D(
,
)
D(
)
(
1 1
1
0 * 1 *
2 1
3
(12));
(
2
)
(
)
(
)
'
'
(
s
1
2P
*G2S
3P
*G1S
2P
*GS
(13)0
)
,
(
)
(
2
1 1
*
3
x
P
x
s
x
s
D D
0
)
,
(
)
(
2 2 *3
y
P
y
s
x
s
D D (15)0
)
,
(
)
(
*
z
P
z
s
z
s
F F (16)Boundary conditions (7) –(9) yield
);
(
)
,
0
(
1 1 * 1 *s
P
s
P
D
G (17));
(
2
)
(
)
(
)
,
0
(
2 1 12 * 3 * 2 *
*
s
P
s
P
s
P
s
P
D
G
G
D (18))
(
)
(
)
(
)
,
0
(
* * 3 * 2 * 22
s
P
s
P
s
P
s
P
F
G
D
hG G (19)Solving (11) to (16) and using (17) to (19) we obtain
)
'
2
(
)
(
)
(
,
'
'
)
(
2
)
(
)
(
2
*
3 * 1 * 2 * 2 * 3 1 11
P
s
P
s
Q
s
s
s
P
s
P
s
G
P
D G Di G G
2 1 1 2 * 2 2 1 1 3'
'
2
)
(
'
'
)
(
2
*
1
s
s
P
s
s
D
P
G
(
'
)
)
3
'
(
'
'
2
3 1 32
1
Q
DS
Q
Ds
(20)
3 2 1 1 3 3 2 3 2 1 3 2 * *'
(
'
'
)
(
)
(
)
(
21
s
Q
s
s
s
Q
s
P
s
P
F G F D
(
'
)
'
'
2
)
(
)
(
3 32 1 2 2 * 2
Q
s
s
s
Q
s
P
G F hG D (21).
2
2
)
(
)
,
(
1
)
(
2 1 * 0 * hG F F Gs
dz
z
s
z
P
s
P
(22)Now we get
3 2 1 1 3 2 3 2 1 3 2 * *'
(
'
'
'
'
'
'
'
'
'
)
(
)
(
)
(
21
s
Q
s
s
s
Q
s
P
s
P
G G F D)
'
(
'
'
'
2
)
(
)
(
3 32 1
2 2
*
2
Q
s
s
s
Q
s
P
F hG D (23)dz
z
s
x
P
s
P
s
2
2
hG)
G(
)
1
F(
,
)
F(
)
(
*0 *
2
1
)
(
2
)
(
1
2 / * 2 * 3 2 1s
Q
s
P
s
P
F GG
)
(
)
'
(
)
(
'
'
'
'
*3 *
2 2 1
1 3
3
P
1s
Q
1s
P
s
s
G D
G
2
'
'
(
2
'
)
(
'
)
'
'
2
3 3
1 3 2 1
1 2
2
1
s
Q
s
Q
s
D D)
(
*s
P
G hG
(24)which reduces to
)
'
(
'
)
(
'
'
'
2
2
2
)
(
3 32 1
2 2 2
1 *
2
Q
s
Q
s
s
s
s
P
G hG F D
Q
Ds
hGs
)
'
2
(
'
'
2
'
'
2
3 1
1 3 2 1
1 2
'
(
'
)
'
'
'
'
'
'
'
)
(
1
2 1 32 1
1 3 3 2
1 3 2
*
s
Q
s
s
Q
s
s
P
G F D (25)If we have
)
(
)
(
)
(
)
(
)
(
1 2 1 2* *
* *
*
s
P
s
P
s
P
s
P
s
P
up
G
G
D
D (26)
'
(
2
'
)
'
'
2
1
)
(
1 32 1
2 *
1
s
Q
s
s
P
G D
2
'
'
(
2
'
)
(
'
)
'
'
'
2
3 3
1 3 2 1
1 2
2
1
s
Q
s
Q
s
D D
'
(
'
)
'
'
'
'
'
1
)
(
2 32 1
1 3 2
1 3 *
1
1
s
Q
s
s
P
G D)
(
)
(
*
s
P
*s
P
down
F (27)We see that
1
)
(
)
(
**
s
P
s
P
up down (28)ERGODIC BEHAVIOUR OF THE SYSTEM
Using Abel's theorem
say
F
t
F
s
sF
t
s
.(
)
lim
(
)
lim
*0
we get
)
(
lim
*0
sP
s
P
ups
Here we use
sP
*(
s
)
P
GThus we have
1
2
'
1(
2
'
3)
2 1
2 *
1
D G
up
P
Q
P
2
'
'
(
2
'
)
'
'
'
2
)
'
(
3 1 32 1
1 2
3 1
2
DD
Q
Q
(
'
)
'
'
'
'
'
'
3
1
32 1
1 3 2
1
1
1
D
G
Q
P
(30)Similarly,
(
)
*
0sP
*s
P
down
s down
'
'
(
'
)
'
'
2
'
3
'
'
'
'
)
0
(
32 1
1 3 3 2
1 3 2
2
1
D F
G
Q
Q
P
'
(
'
)
'
2
)
0
(
3 32 1
1 2
2
1
hG DG
QF
Q
P
2
'
'
(
2
'
)
'
'
'
2
3 1
3 2 1
1 2
1
D
Q
(31)Here QF(0) is mean time to repair to the failed state of the system.
INTERPRETATIONS OF RESULTS
If we take particular values of
1,
2,
3,
'
1,
'
2,
'
3,
we see that results are similar i.e., availability of system decreases with time but as time progresses total decrease is low and expected profit v/s time will decrease rapidly.REFERENCES:
1. Gupta P.P., Singh S.B. and Goel C.K.: Stand by redundant system involving the concepts of multi-failure human failure under head of line repair policy. Bulletin of Pure &Applied Science, Volume 20, pp.345-351.(2001)
2. Mokaddis, G.S.; Tawfak, M.L.: 'Some characteristic of a two dissimilar unit cold stand by redundant with three modes." Micorelectron Reliability, Volume. 36(4), pp 497-504 (1996)
3. Singh, S.B., Goel, C.K. : "Stochastic Behavior of a Complex system involving major and minor failures, International Journal of Essential Sciences. Volume No. 1, (2008).
4. Widder, D.V., “The Laplace Transforms”, Princeton University Press, Princeton New Jersey (1946)
