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IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308
_______________________________________________________________________________________
Volume: 03 Issue: 01 | Jan-2014, Available @ http://www.ijret.org 295
RISK ASSESSMENT OF A HYDROELECTRIC DAM WITH PARALLEL
REDUNDANT TURBINE
Reena Garg
YMCA University, Faridabad
ABSTRACT
This paper deals with the risk assessment of a hydroelectric dam. Hydroelectric dam produces electric power with the help of
water collected in a pond. Here, in this paper, the author has been taken one parallel redundant turbine to improve system’s
overall performance. On failure of any one turbine, the whole system works in reduced efficiency state. The whole system can fail
due to failure of any of its subsystems. All failures follow exponential time distribution whereas all repairs follow general time
distribution. In order to risk assessment of the system, we have obtained reliability function, M.T.T.F. and availability function for
considered system. A particular case, when all repairs follow exponential time distribution, and long-run behavior of system have
also been computed to improve practical utility of the model. Graphical illustration followed by a numerical computation has also
been appended at the end to highlight important results of present study.
Key Words: Reliability, Mean time to failure, exponential time distribution, Inclusion of supplementary variables etc.
----------------------------------------------------------------***-------------------------------------------------------------------------
1. INTRODUCTION
The whole system consists of five major parts and these are
named here as:
(i) Subsystem A : Reservoir, intake gate
(ii) Subsystem B : Penstock
(iii) Subsystem C : Turbine
(iv) Subsystem D : Generator
(v) Subsystem E : Power house.
The system has been shown in fig -1(a). The subsystem A
takes the water and sends it to subsystem B (Penstock). By
the movement of this water, we rotate the turbine and by
rotation of turbine, generator produces electricity. The so
produced electric power can be stored at power house and
also can further distribute to various territories.
Since, the system under consideration in Non-Markovian,
the author has used supplementary variables to formulate
mathematical model of the system. This model has been
solved further by taking help of Laplace transform.
2. ASSUMPTIONS
This paper is based on following assumptions:
(1) Initially the whole system is good and operable.
(2) After failure, repair facilities can be provided
immediately.
(3)All failures follow exponential time distribution and are
S-independent.
(4) All repairs follow general time distribution and are
perfect.
(5) Nothing can fail from a failed state.
(6) On failure of any one turbine (subsystem C), the whole
system works in reduced efficiency
Fig-1(a): System Configuration of Hydro-electric Dam
3. LIST OF NOTATIONS USED:
x : Failure rates of subsystems A, B, C, D and E respectively.
 x1 etc. : First order probability that subsystem A can be repaired in the time interval
 xx, , conditioned that it was not repaired upto time x.
 u3,3 : First order probability that both the units of subsystem C can be repaired in
the time interval  uu, , conditioned that they were not repaired upto
time u.
 tP0
: Pr {at time t, whole system is operable}.
IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308
_______________________________________________________________________________________
Volume: 03 Issue: 01 | Jan-2014, Available @ http://www.ijret.org 296
 txP ,1 etc : Pr {at time t, system is failed due to failure of subsystem A}. Elapsed repair
time lies within  xx, .
 tzP ,3
Pr {at time t, system is degraded due to failure of any one component of
subsystem C}. Elapsed repair time lies within  zz, .
  etctxP ,1,3  Pr{at time t, system is failed due to failure of subsystem A while one
component of subsystem C has already failed}. Elapsed repair time of
subsystem A lies within  xx, .
 sP : Laplace transform (L.T.) of function P(t).
M.T.T.F. : Mean time to failure.
 jSi
:   djjj ii )(.exp)(  , ji and .
)( jDi
:   jijjSi and,/)(1  .
2 )(2 y
4 1
)(4 m )(1 x
5 1h
)(5 n )(kh
3 )(3 z
2h
2 1
)(2 y )(1 x
)(3,3 u )(5 n
3 5
)(4 m
4
Fig-1(b): Transition-state diagram
States:
),(2 tyP
),(4 tmP ),(1 txP
),( tkPh
),(5 tnP
),(1,3 txP
),(3,3 tuP
),(2,3 tyP
),(5,3 tnP
),(4,3 tmP
)(0 tP
),(3 tzP
Good Degraded Failed
IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308
_______________________________________________________________________________________
Volume: 03 Issue: 01 | Jan-2014, Available @ http://www.ijret.org 297
4. FORMULATIONOFMATHEMATICAL MODEL
Using continuity argument and limiting procedure, we obtain the following set of difference-differential equations, governing the
behaviour of considered system, discrete in space and continuous in time:
             









0
33
0
22
0
1101
5
1
,,, dzztzPdyytyPdxxtxPtPh
dt
d
i
i 
           


00
55
0
44 ,,, dkktkPdnntnPdmmtmP hh     


0
3,33,3 , duutuP 
(1)
0),()( 











tjPj
tj
ii , ly.respective,,,and5,4,2,1 nmyxji  …(2)
0),()( 332
5
1













tzPzh
tz i
i  …(3)
0),()( ,3 











tjPj
tj
ii , ly.respective,,,and5,4,2,1 nmyxji  …(4)
0),()( 3,33,3 









tuPu
tu

…(5)
0),()( 









tkPk
tk
hh
…(6)
Boundary conditions are:
)(),0( 0 tPtP ii  ,i = 1, 2, 4 and 5 …(7)
       


0
22,3
0
11,3033 ,,)(),0( dyytyPdxxtxPtPtP 
       


0
55,3
0
44,3 ,, dnntnPdmmtmP 
…(8)
)(),0( 3,3 tPtP ii  , i = 1, 2, 4 and 5 …(9)
)(),0( 333,3 tPtP  …(10)
)()(),0( 3201 tPhtPhtPh  …(11)
Initial conditions are:
1)0(0 P , otherwise zero. …(12)
5. SOLUTION OF THE MODEL
Taking Laplace transforms of equations (1) through (11) subjected to initial conditions (12) and then on solving them one by one,
we obtain the following probabilities of states depicted in fig-1(b):
)(
1
)(0
sB
sP 
…(13)
)(
)(
)(
sB
sD
sP ii
i

 for i = 1, 2, 4 and 5
…(14)
)(
)(
)(3
sB
sA
sP 
…(15)
)(
)()(
)(,3
sB
sDsA
sP ii
i

 for i = 1, 2, 4 and 5
…(16)
IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308
_______________________________________________________________________________________
Volume: 03 Issue: 01 | Jan-2014, Available @ http://www.ijret.org 298
)(
)()(
)(
3,33
3,3
sB
sDsA
sP


…(17)
  )()(
)(
1
)( 21 sDsAhh
sB
sP hh 
…(18)
where,

























2
5
1
3
5
1
33
2
5
1
33
)()(1
)(
hsDsSsS
hsD
sA
i
i
i
ii
i
i


…(19)
and
  1355442211 )()()()(1)( hsDsDsDsDssB  
  )()()()( 3,3321 sSsAsSsAhh h 
   





 
2
5
1
3554422113 )()()()()( hsSsSsSsSsSsA
i
i
…(20)
It is worth noticing that
Sum of equations (13) through (18) =
s
1
…(21)
6.LONG-RUN BEHAVIOR OF CONSIDERED SYSTEM:
Using final value theorem of Laplace transform, viz.,     PsPsLimtPLim
st

 0
(say), provided limit on left exists, we
obtain the following long-run behaviour of considered system from equations (13) through (18):
)0(
1
0
B
P


…(22)
)0(B
M
P ii
i



;for i = 1, 2, 4 and 5
…(23)
)0(
)0(
3
B
A
P


…(24)
)0(
)0(
,3
B
MA
P ii
i



;for i = 1, 2, 4 and 5
…(25)
)0(
)0( 3,33
3,3
B
MA
P


 …(26)
  hh MAhh
B
P )0(
)0(
1
21 


…(27)
where,































2
5
1
3
5
1
323
2
5
1
33 1
)0(
hSh
hS
A
i
i
i
i
i
i

 …(28)
0
)()0(






s
sB
ds
d
B
…(29)
and 

 )0(ii SM Mean time to repair
th
i failure.
…(30)
IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308
_______________________________________________________________________________________
Volume: 03 Issue: 01 | Jan-2014, Available @ http://www.ijret.org 299
7. PARTICULAR CASE:
When all repairs follow exponential time distribution
In this case, setting    ,
i
i
i
j
jS



 for all i and j, in equations (13) through (18), we obtain the following L.T. of
different state probabilities of fig-1(b):
)(
1
)(0
sE
sP 
…(31)
 i
i
i
ssE
sP




)(
)( ; For i = 1, 2, 4 and 5
…(32)
)(
)(
)(3
sE
sF
sP 
…(33)
 i
i
i
ssE
sF
sP




)(
)(
)(,3 ; For i = 1, 2, 4 and 5
…(34)
))((
)(
)(
3,3
3
3,3




ssE
sF
sP
…(35)
and  
)(
1
)(
)(
1
)( 21
h
h
s
sFhh
sE
sP


…(36)
where,

















 
5
1 3
33
32
5
1
3
)(
i i
ii
i
i
ss
hs
sF






…(37)
and 13
5
5
4
4
2
2
1
1
1)( h
ssss
ssE 












 








 
3,3
3,33
21
)(
)(








s
sF
s
sFhh
h
h
32
5
1
3
5
55
4
44
2
22
1
11
3 )(

































hs
ssss
sF
i
i
…(38)
8. RELIABILITY AND M.T.T.F. EVALUATION:
Form equation (13), we obtain the L.T. of reliability function
1
2
5
1
)(








  hssR
i
i
taking inverse Laplace transform, we get












 
thtR
i
i 1
5
1
.exp)(  (39)
Also, 


0
)(.... dttRFTTM
IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308
_______________________________________________________________________________________
Volume: 03 Issue: 01 | Jan-2014, Available @ http://www.ijret.org 300
1
5
1
1
h
i
i 



(40)
9. AVAILABILITY OF CONSIDERED SYSTEM:
Availability of considered system can be obtained from equations (13) and (15) as:
















 
2
5
1
3
1
5
1
1
1
)(
hshs
sP
i
i
i
i
up



takinginverseLaplacetransform,we
obta


































  
th
hh
th
hh
tP
i
i
i
iup 2
5
112
3
1
5
112
3
.exp.exp1)( 



(41)
Also, )(1)( tPtP updown  (42)
10. NUMERICAL ILLUSTRATION:
For a numerical illustration we consider the values:
009.0,008.0,15.0,08.0,06.0,001.0,002.0 2154321  hh
and .10,2,1,0 t Using these values in equations
(39), (40) and (41), we draw the graphs shown in fig-2, 3
and 4 respectively.
11. RESULTS AND DISCUSSION:
A critical examination of graphs, shown in fig-2, 3 and 4,
yields that reliability and availability of considered system
decreases smoothly with the increase in values of time t. It
should be noted that there are no sudden jumps in the values
of )(tR and )(tPup . Also M.T.T.F. of considered system
decreases catastrophically in the beginning but thereafter it
decreases approximately in constant manner.
Reliability Vs Time
0
0.2
0.4
0.6
0.8
1
1.2
1 2 3 4 5 6 7 8 9 10 11
t ---->
R(t)---->
M.T.T.F. Vs Human error
3.24
3.26
3.28
3.3
3.32
3.34
3.36
3.38
3.4
3.42
3.44
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01
h1---->
M.T.T.F.---->
Fig-2: Reliability Vs TimeFig-3:MTTF Vs Human Error
Availability Vs time
0
0.2
0.4
0.6
0.8
1
1.2
1 2 3 4 5 6 7 8 9 10 11
t ---->
Pup(t)---->
Fig-4: Availability Vs Time
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IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308
_______________________________________________________________________________________
Volume: 03 Issue: 01 | Jan-2014, Available @ http://www.ijret.org 301
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Risk assessment of a hydroelectric dam with parallel redundant turbine

  • 1. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 03 Issue: 01 | Jan-2014, Available @ http://www.ijret.org 295 RISK ASSESSMENT OF A HYDROELECTRIC DAM WITH PARALLEL REDUNDANT TURBINE Reena Garg YMCA University, Faridabad ABSTRACT This paper deals with the risk assessment of a hydroelectric dam. Hydroelectric dam produces electric power with the help of water collected in a pond. Here, in this paper, the author has been taken one parallel redundant turbine to improve system’s overall performance. On failure of any one turbine, the whole system works in reduced efficiency state. The whole system can fail due to failure of any of its subsystems. All failures follow exponential time distribution whereas all repairs follow general time distribution. In order to risk assessment of the system, we have obtained reliability function, M.T.T.F. and availability function for considered system. A particular case, when all repairs follow exponential time distribution, and long-run behavior of system have also been computed to improve practical utility of the model. Graphical illustration followed by a numerical computation has also been appended at the end to highlight important results of present study. Key Words: Reliability, Mean time to failure, exponential time distribution, Inclusion of supplementary variables etc. ----------------------------------------------------------------***------------------------------------------------------------------------- 1. INTRODUCTION The whole system consists of five major parts and these are named here as: (i) Subsystem A : Reservoir, intake gate (ii) Subsystem B : Penstock (iii) Subsystem C : Turbine (iv) Subsystem D : Generator (v) Subsystem E : Power house. The system has been shown in fig -1(a). The subsystem A takes the water and sends it to subsystem B (Penstock). By the movement of this water, we rotate the turbine and by rotation of turbine, generator produces electricity. The so produced electric power can be stored at power house and also can further distribute to various territories. Since, the system under consideration in Non-Markovian, the author has used supplementary variables to formulate mathematical model of the system. This model has been solved further by taking help of Laplace transform. 2. ASSUMPTIONS This paper is based on following assumptions: (1) Initially the whole system is good and operable. (2) After failure, repair facilities can be provided immediately. (3)All failures follow exponential time distribution and are S-independent. (4) All repairs follow general time distribution and are perfect. (5) Nothing can fail from a failed state. (6) On failure of any one turbine (subsystem C), the whole system works in reduced efficiency Fig-1(a): System Configuration of Hydro-electric Dam 3. LIST OF NOTATIONS USED: x : Failure rates of subsystems A, B, C, D and E respectively.  x1 etc. : First order probability that subsystem A can be repaired in the time interval  xx, , conditioned that it was not repaired upto time x.  u3,3 : First order probability that both the units of subsystem C can be repaired in the time interval  uu, , conditioned that they were not repaired upto time u.  tP0 : Pr {at time t, whole system is operable}.
  • 2. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 03 Issue: 01 | Jan-2014, Available @ http://www.ijret.org 296  txP ,1 etc : Pr {at time t, system is failed due to failure of subsystem A}. Elapsed repair time lies within  xx, .  tzP ,3 Pr {at time t, system is degraded due to failure of any one component of subsystem C}. Elapsed repair time lies within  zz, .   etctxP ,1,3  Pr{at time t, system is failed due to failure of subsystem A while one component of subsystem C has already failed}. Elapsed repair time of subsystem A lies within  xx, .  sP : Laplace transform (L.T.) of function P(t). M.T.T.F. : Mean time to failure.  jSi :   djjj ii )(.exp)(  , ji and . )( jDi :   jijjSi and,/)(1  . 2 )(2 y 4 1 )(4 m )(1 x 5 1h )(5 n )(kh 3 )(3 z 2h 2 1 )(2 y )(1 x )(3,3 u )(5 n 3 5 )(4 m 4 Fig-1(b): Transition-state diagram States: ),(2 tyP ),(4 tmP ),(1 txP ),( tkPh ),(5 tnP ),(1,3 txP ),(3,3 tuP ),(2,3 tyP ),(5,3 tnP ),(4,3 tmP )(0 tP ),(3 tzP Good Degraded Failed
  • 3. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 03 Issue: 01 | Jan-2014, Available @ http://www.ijret.org 297 4. FORMULATIONOFMATHEMATICAL MODEL Using continuity argument and limiting procedure, we obtain the following set of difference-differential equations, governing the behaviour of considered system, discrete in space and continuous in time:                        0 33 0 22 0 1101 5 1 ,,, dzztzPdyytyPdxxtxPtPh dt d i i                00 55 0 44 ,,, dkktkPdnntnPdmmtmP hh        0 3,33,3 , duutuP  (1) 0),()(             tjPj tj ii , ly.respective,,,and5,4,2,1 nmyxji  …(2) 0),()( 332 5 1              tzPzh tz i i  …(3) 0),()( ,3             tjPj tj ii , ly.respective,,,and5,4,2,1 nmyxji  …(4) 0),()( 3,33,3           tuPu tu  …(5) 0),()(           tkPk tk hh …(6) Boundary conditions are: )(),0( 0 tPtP ii  ,i = 1, 2, 4 and 5 …(7)           0 22,3 0 11,3033 ,,)(),0( dyytyPdxxtxPtPtP            0 55,3 0 44,3 ,, dnntnPdmmtmP  …(8) )(),0( 3,3 tPtP ii  , i = 1, 2, 4 and 5 …(9) )(),0( 333,3 tPtP  …(10) )()(),0( 3201 tPhtPhtPh  …(11) Initial conditions are: 1)0(0 P , otherwise zero. …(12) 5. SOLUTION OF THE MODEL Taking Laplace transforms of equations (1) through (11) subjected to initial conditions (12) and then on solving them one by one, we obtain the following probabilities of states depicted in fig-1(b): )( 1 )(0 sB sP  …(13) )( )( )( sB sD sP ii i   for i = 1, 2, 4 and 5 …(14) )( )( )(3 sB sA sP  …(15) )( )()( )(,3 sB sDsA sP ii i   for i = 1, 2, 4 and 5 …(16)
  • 4. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 03 Issue: 01 | Jan-2014, Available @ http://www.ijret.org 298 )( )()( )( 3,33 3,3 sB sDsA sP   …(17)   )()( )( 1 )( 21 sDsAhh sB sP hh  …(18) where,                          2 5 1 3 5 1 33 2 5 1 33 )()(1 )( hsDsSsS hsD sA i i i ii i i   …(19) and   1355442211 )()()()(1)( hsDsDsDsDssB     )()()()( 3,3321 sSsAsSsAhh h             2 5 1 3554422113 )()()()()( hsSsSsSsSsSsA i i …(20) It is worth noticing that Sum of equations (13) through (18) = s 1 …(21) 6.LONG-RUN BEHAVIOR OF CONSIDERED SYSTEM: Using final value theorem of Laplace transform, viz.,     PsPsLimtPLim st   0 (say), provided limit on left exists, we obtain the following long-run behaviour of considered system from equations (13) through (18): )0( 1 0 B P   …(22) )0(B M P ii i    ;for i = 1, 2, 4 and 5 …(23) )0( )0( 3 B A P   …(24) )0( )0( ,3 B MA P ii i    ;for i = 1, 2, 4 and 5 …(25) )0( )0( 3,33 3,3 B MA P    …(26)   hh MAhh B P )0( )0( 1 21    …(27) where,                                2 5 1 3 5 1 323 2 5 1 33 1 )0( hSh hS A i i i i i i   …(28) 0 )()0(       s sB ds d B …(29) and    )0(ii SM Mean time to repair th i failure. …(30)
  • 5. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 03 Issue: 01 | Jan-2014, Available @ http://www.ijret.org 299 7. PARTICULAR CASE: When all repairs follow exponential time distribution In this case, setting    , i i i j jS     for all i and j, in equations (13) through (18), we obtain the following L.T. of different state probabilities of fig-1(b): )( 1 )(0 sE sP  …(31)  i i i ssE sP     )( )( ; For i = 1, 2, 4 and 5 …(32) )( )( )(3 sE sF sP  …(33)  i i i ssE sF sP     )( )( )(,3 ; For i = 1, 2, 4 and 5 …(34) ))(( )( )( 3,3 3 3,3     ssE sF sP …(35) and   )( 1 )( )( 1 )( 21 h h s sFhh sE sP   …(36) where,                    5 1 3 33 32 5 1 3 )( i i ii i i ss hs sF       …(37) and 13 5 5 4 4 2 2 1 1 1)( h ssss ssE                          3,3 3,33 21 )( )(         s sF s sFhh h h 32 5 1 3 5 55 4 44 2 22 1 11 3 )(                                  hs ssss sF i i …(38) 8. RELIABILITY AND M.T.T.F. EVALUATION: Form equation (13), we obtain the L.T. of reliability function 1 2 5 1 )(           hssR i i taking inverse Laplace transform, we get               thtR i i 1 5 1 .exp)(  (39) Also,    0 )(.... dttRFTTM
  • 6. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 03 Issue: 01 | Jan-2014, Available @ http://www.ijret.org 300 1 5 1 1 h i i     (40) 9. AVAILABILITY OF CONSIDERED SYSTEM: Availability of considered system can be obtained from equations (13) and (15) as:                   2 5 1 3 1 5 1 1 1 )( hshs sP i i i i up    takinginverseLaplacetransform,we obta                                      th hh th hh tP i i i iup 2 5 112 3 1 5 112 3 .exp.exp1)(     (41) Also, )(1)( tPtP updown  (42) 10. NUMERICAL ILLUSTRATION: For a numerical illustration we consider the values: 009.0,008.0,15.0,08.0,06.0,001.0,002.0 2154321  hh and .10,2,1,0 t Using these values in equations (39), (40) and (41), we draw the graphs shown in fig-2, 3 and 4 respectively. 11. RESULTS AND DISCUSSION: A critical examination of graphs, shown in fig-2, 3 and 4, yields that reliability and availability of considered system decreases smoothly with the increase in values of time t. It should be noted that there are no sudden jumps in the values of )(tR and )(tPup . Also M.T.T.F. of considered system decreases catastrophically in the beginning but thereafter it decreases approximately in constant manner. Reliability Vs Time 0 0.2 0.4 0.6 0.8 1 1.2 1 2 3 4 5 6 7 8 9 10 11 t ----> R(t)----> M.T.T.F. Vs Human error 3.24 3.26 3.28 3.3 3.32 3.34 3.36 3.38 3.4 3.42 3.44 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 h1----> M.T.T.F.----> Fig-2: Reliability Vs TimeFig-3:MTTF Vs Human Error Availability Vs time 0 0.2 0.4 0.6 0.8 1 1.2 1 2 3 4 5 6 7 8 9 10 11 t ----> Pup(t)----> Fig-4: Availability Vs Time REFERENCES [1].Bala Krishnan N., Lin C.T., Ping-Shing C. “A Comparison of Two Simple Prediction Intervals for Exponential Distribution”, IEEE TR. on Reliability, Vol.54 (1), PP 27-33, 2005. [2]. Cassady C.R., Lyoob I.M., Schneider K., Pohi E. A“A Generic Model of Equipment Availability Under Imperfect Maintenance”, IEEE TR. on Reliability, Vol.54 (4), PP 564- 571, 2005 [3]. Fang Ming S., Xuemin S., Pin-Han H“Reliability Optimization of Distributed Access Network with
  • 7. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 03 Issue: 01 | Jan-2014, Available @ http://www.ijret.org 301 Constrained Total Cost”, IEEE TR. on Reliability, Vol.54 (3), PP 421-430, 2005. [4]. Gutierrez P.H., et. al“A Practical Method for Obtaining Prior Distribution in Reliability”, IEEE TR. on Reliability, Vol.54 (2), PP 262-269, 2005 [5]. Jen-Shyam W., Rong Jaye C“Efficient Algorithm for Reliability of a Circular Consecutive-K-out-of-N:F System”, IEEE TR. on Reliability, Vol.55 (1), PP 163-164, 2006 [6]. Jeon L.J“Sequence Confidence Interval Estimation for System Availability”, Quality and Reliability Engineering, Vol.22 (2), PP 165-176, 2006 [7]. Lai C.D., Xie M., Murthy D.N.P. “On Some Recent Modifications of Weibull Distribution”, IEEE TR. on Reliability, Vol.54 (4), PP 563-569, 2005. [8]. Myers, A. F. k-out-of-n:G System Reliability With Imperfect Fault Coverage”, IEEE Trans on Reliability, Vol.56, issue-3, PP 464-473, 2007. [9]. Marseguerra M., Zio E. Padofilline L., Coit D.W“Optimal Design of Reliability Network System in Presence of Uncertainty”, IEEE TR. on Reliability, Vol.54 (2), PP 243-253, 2005. [10]. Ramirez-Marquez J.E., Colt D.W. “Composite Importance Measure of Multi-State Systems with Multi- State Components”, IEEE TR. on Reliability, Vol.54 (3), PP 517-529, 2005. [11]. Sim S.H., Endrenyi J. “A Failure Repair Model with Minimal and Major Maintenance”, IEEE TR. on Reliability, Vol.55 (1), PP 134-140, 2006. [12]. Shey-Huei S., et.al“Optimal Policies with Decreasing Probability of Imperfect Maintenance”, IEEE TR. on Reliability, Vol.54 (2), PP 347-357, 2005. [13]. Wang F.K. “Quality Evaluation of a Manufactured Production with Multiple Characteristic”, Quality and Reliability Engineering, Vol.22 (2), PP 225-236, 2006. [14]. Yuegin W., Jiancheng G. “Repairable Consecutive K- out-of-N:G Systems with Repair”, IEEE TR. on Reliability, Vol.54 (2), PP 328-337, 2005. [15]. Young C., Jionghua J. “Quality Reliability Chain Modeling for System Reliability Analysis of Complex Manufacturing Processes”, IEEE TR. on Reliability, Vol.54 (3), PP 475-488, 2005. [16]. Zhimin He., Han T.L., Eng H.O“A Probabilistic Approach to Evaluate the Reliability of Piezoelectric Micro- Actuators”, IEEE TR. on Reliability, Vol.54 (1), PP 44-49, 2005. [17]. Way Kuo“Compatibility and Simplicity: The Fundamentals of Reliability”, IEEE TR. on Reliability, Vol.56 (4), PP 585-586, 2007. [18]. Wu, Y.P.; Hu, Q.P.; Xie, M.; Ng, S.H. “Modeling and Analysis of Software Fault Detection and Correction Process by Considering Time Dependency”, IEEE TR. on Reliability, Vol.56 (4), PP 629-642, 2007.