1. technology and computing
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International Journal of Latest Trends in Engineering and Technology (IJLTET)
Power System Reliability Evaluation
Dileep M John
Department of Electrical and Electronics Engineering
Amal Jyothi College of Engg., Kanjirappally, Kerala, India
Remyasree R
Department of Electrical and Electronics Engineering
Amal Jyothi College of Engg., Kanjirappally, Kerala, India
Jannet Teresa K Cyriac
Department of Electrical and Electronics Engineering
Amal Jyothi College of Engg., Kanjirappally, Kerala, India
Bobby George Joseph
Department of Electrical and Electronics Engineering
Amal Jyothi College of Engg., Kanjirappally, Kerala, India
Anoop C.P
Department of Electrical and Electronics Engineering
Amal Jyothi College of Engg., Kanjirappally, Kerala, India
Abstract- The basic function of a power distribution system is to supply customers with electrical energy as economically
as possible. The reliability performance of a power distribution system plays an important role to achieve this basic
function. The planning and operation of an electric power distribution system involves the accurate estimation of the
reliability of its constituent components. In order to have effective reliability performance estimation, a fair estimation of
the failure rates and failure distribution of these components need to be determined. Some basics concepts like general
reliability function, reliability indices and bathtub curve, which were required to proceed to the quantitative evaluation of
the power system reliability are discussed. This seminar also discusses about the different methods involved in the
computation of the reliability of various power system models. The models of Series, Parallel and Hybrid Systems and the
reliability evaluation in them are discussed. In addition, the concepts of Ties, Cuts, Decomposition, Path tracing method
and Event space method to compute reliability of systems that are not reducible to series or parallel systems are also
explained.
Keywords – Reliability, Reliability Variables, Reliability Indices, Bathtub curve, failure rates, failure distribution, Ties,
Cuts, Decomposition, Path Tracing method, Event space method
1. INTRODUCTION
Reliability engineering emphasis on the ability of a system or component to function under stated conditions for a
specified period of time. Reliability is defined through the mathematical concept of probability. Reliability is
theoretically defined as the probability of success for a system. Reliability plays a key role in cost-effectiveness of
systems. It is a general quality of an object – an ability to perform a desired function, sustaining the values of rated
operational indicators in given limits and time according to given technical conditions. Reliability is the probability
that an activity of an appliance in stated period and operating conditions will be adequate[1]
Electric power networks are prime examples of systems where a very high degree of reliability is expected. In
many power systems, the average duration of interruptions that a customer experiences is a total of 2-3 hours per year
and can be much less for stringent systems. The above translates into an availability of supply of about 0.99975. The
public has grown accustomed to such a reliable service and, by and large, would not accept much lower standards. A
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high degree of reliability is also essential for some industrial customers. For this reason, reliability is, and always has
been one of the major factors in the planning, design, operation, and maintenance of electric power systems. Power
systems are built and operated with the following goal to achieve a reliable and economic electric power supply. For
the consumer to have a reliable and economic electric power supply, a complex set of engineering analysis and design
solutions need to be implemented. The reliability of an electric power system is a measure of the probability of
providing the users with continuous service with frequency and voltage within tolerable limits. Reliability has the
following standard definitions.
Reliability of a power system refers to the probability of its satisfactory operation over the long run. It denotes the
ability to supply adequate electric service on a nearly continuous basis, with a few interruptions over an extended time
period.[2]
Reliability is a measure of the ability of a system, generally specified in numerical indices, to deliver power to all
points of utilisation within acceptable standards and in amounts desired. Power system reliability (comprising
generation, transmission and distribution facilities) can be described by two basic functional attributes: adequacy and
security. (CIGRE definition)
Reliability of a system is a function of time, defined as the conditional probability that the system will operate
correctly throughout the interval [t0, t], given that the system was performing correctly at the time t0.[3]
.
A. OVERVIEW OF RELIABILITY
Various indices have been introduced in reliability theory to facilitate reliability prediction; most of them falling
under the following categories[4]:
1) Probabilities, such as reliability or availability
2) Frequencies, such as average number of failures per unit time.
3) Mean durations, such as the mean time to first failure, mean time between failures and the mean
duration of failures.
4) Expectations such as the expected number of days in a year when a system failure occurs.
Consider the case where
R(t) = Reliability at any time t
λ(t) = Instantaneous hazard rate
 t
Then,
R(t) = e
(1)
Some of the Power system reliability indices are given below

SAIFI = System Average Interruption Frequency Index = Total number of customer interruptions / Total
number of customers served

SAIDI = System Average Interruption Duration Index = Customer interruption durations / Total number of
customers served

CAIFI = Customer Average Interruption Frequency Index = Total number of customer interruptions / Total
number of customers interrupted

CAIDI = Customer Average Interruption Duration Index= Customer interruption durations/ Total number of
customer interruptions = SAIDI/SAIFI

AENS = Average Energy Not Supplied = Total energy not supplied / Total number of customers served
All these indices are measured over a year. It is also presumed that these indices are applied to power systems.
Other applications will need other indices.[5]
Consider the following example. Table 1 shows all the five outages that were recorded on the 28 th of a month,
the duration of the outages and the customer-minutes. The utility has a total of 50,000 customers.
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As seen from Table 2.1, the first outage was at 9:53 in the morning and 10 customers where out of service for 90
minutes .Therefore, the customer minutes are 10*90 or 900. The customer-minutes are calculated for each outage and
then summed for a total of 21408 customer-minutes.
SAIDI = 21408/50000
SAIDI= 0.428 minutes
B. OVERVIEW OF RELIABILITY
Many years of experience of failure data of various units shows that failures, in general can be grouped into
different modes depending upon the nature of failure.The plot of failure rate with time is called as Bathtub Curve
and is shown in Figure 1.
The Bathtub curve has three distinctive phases[11]
1.
An “infant mortality” early life phase characterized by a decreasing failure rate (Phase one). Failure
occurrence during this period is not random in time but rather the result of substandard components with
gross defects and the lack of adequate controls in the manufacturing process. Components fail at a high but
decreasing rate.
2.
A “useful life” period where a relatively constant failure rate is caused by randomly occurring defects and
stresses (Phase two).This corresponds to a normal wear and tear period where failures are caused by
unexpected and sudden over stress conditions. During useful life, components exhibit a constant failure rate.
In exponential distribution the failure rate failure rate reduces to the constant λ for any time. Accordingly, the
reliability of a device can be modelled using an exponential distribution.
3.
A “wear out” period where the failure rate increases due to critical parts wearing out (Phase three). As they
wear out, it takes less stress to cause failure and the overall system failure rate increases. Accordingly failures
do not occur randomly in time.
The failure rate is represented by the height of the curve and is not related to the length of the curve. It is
therefore possible to have a long or short useful life period for a given failure rate.
Table 1: Calculation of SAIDI[6]
Calculation of Customer-Minutes
Date
Time
Customers
Duration
Customerminutes
28th
9:53
10
90
900
28th
11:02
1000
20
20000
28th
13:15
2
175
350
28th
20:48
1
120
120
28th
22:35
1
38
38
1014
443
21408
C. OVERVIEW OF RELIABILITY
Weibull Analysis can be used as a method of determining where a population of modules is on the bathtub curve.
The Weibull distribution is a 3-parameter distribution. The Weibull distribution is given by:
f(T) = [
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 
304
T

e [ ]
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where f(T) is the Probability Distribution Function of Weibull distribution and β,  , T are the three parameters.
Figure 1: Bathtub Curve[10]
The Weibull parameter β is the slope. It signifies the rate of failure.When β<1, the Weibull distribution models
early failures of parts. When β =1, the Weibull distribution models the exponential distribution. The exponential
distribution is the model for the useful life period, signifying that random failures are occurring. When β = 3, the
Weibull distribution models the normal distribution. This is the early wear out time. When = 10, rapid wear out is
occurring. The results of various β s on the Weibull analysis is shown in Figure 2 and it can be seen that if all curves
are combined, the resultant graph is similar to a bathtub curve.
Figure 2: Weibull Plot using various β values [6]
2. RELIABILITY OF SERIES,PARALLEL AND SERIES-PARALLEL STRUCTURES
A. Reliability Of Series Structures–
Components are in series if the failure of any one of them will result in system failure. Figure 3 shows series
structure with n components
Figure 3: : Series Structure with n components[7]
The following symbols are used in equations corresponding to reliability of series structures.
 S = Event that system is working

S = Event that system is not working
 X1= Event that the component 1 is working
 X2= Event that the component 2 is working

X 1 = Event that the component 1 is not working



X 2 = Event that the component 2 is not working
Ri = System reliability Index
R1 = Reliability index of component 1
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

R2 = Reliability index of component 2
Rn= Reliability index of component n
Thus for components 1 and 2 in series
S = X1∩ X2
(3)
S̅ = X1 U X2
(4)
For a pure series system, the system reliability is equal to the product of the reliabilities of its constituent
components
Rs=(R1)(R2)…(Rn)
(5)
R s = ∑𝑛𝑖=1 R i
(6)
Figure 4: Series System Reliability vs Component Reliability[5]
Example:[8]
Suppose three components are connected in series have reliabilities of 0.95, 0.98 and 0.97 respectively. Then the
overall reliability of the system is given by
Rs= (R1)(R2)(R3)
= (0.95)(0.98)(0.97)
= 90%
The Effect of the Number of Components on Series System Reliability and Component Reliability in a Series
System is shown in Figure 3.
B. Reliability Of Parallel Structures –
In a parallel system, the system will work as long as at least one component works.
Figure 5 shows parallel structure with n components
Using the same set of symbols used in analysis of series structures, for components 1 and 2 in parallel,
S= X1U X2
̅̅̅2
S̅ = ̅̅̅
X1 ∩ X
(7)
(8)
Figure 5: Parallel Structure with n components[4]
For a purely parallel system, the overall system reliability is equal to the product of the component unreliabilities.
Thus, the reliability of the parallel system is given by
Rs=1–[(1–R1)(1-R2)…(1-Rn)]
(9)
R s = 1 − ∏𝑛𝑖=1(1 − R i )
(10)
The Effect of the Number of Components on Parallel System Reliability and Component Reliability is shown in
Figure 6
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Figure 6: Parallel System Reliability vs Component Reliability[5]
Example: Suppose three components are connected in parallel.Component 1,2 and 3 have reliabilities of 0.95, 0.98
and 0.97 respectively. Then the overall reliability of the system is given by
Rs = 1 – [(1 – R1)(1 – R2)(1 – R3)]
(11)
= 1 – [(1 – 0.95)(1-0.98)(1-0.97)]
= 99.997 %
C. Reliability Of Of Series-Parallel Structures –
Consider a system with three components as in Figure 7. Units 1,2 and 3 have reliabilities of 0.99,0.98 and 0.97
respectively. Then the reliability of the system is computed as below:
Figure 7: Hybrid System[4]
Units 1 and 2 can be combined using formula for series combination as in Eqn. 6 and its resultant can be combined
with unit 3 using formula for parallel combination as in Eqn. 10. Hence the system reliability is given by
Rs = 1 – [(1 – (0.99)(0.98))(1 – (0.97)]
= 0.9991
3.
RELIABILITY OF NON SERIES-PARALLEL STRUCTURES
For systems which are not reducible to series or parallel, the methods of ties, cuts, path tracing method, event space
method or decomposition method may be used. In this section, a brief description of these methods are presented,
A. Concept of cuts and ties –
Ties are a set of components whose working will ensure system operation. Tie set is a set of units that form a
connection between input and output when traversed in the direction of arrows of the reliability block diagram. Cuts
are set of components whose failure will result in system failure. Cut sets are set of units that interrupt set of all
possible connections between input and output points.[9] A minimal tie is a subset of tie and it is the set remaining
after removal of any of its elements is no longer a tie. A minimal tie is path set containing the minimum number of
units needed to guarantee a connection between input and output points. A minimal cut is a subset of cut and it is the
set remaining after removal of any of its elements is no longer a cut. A minimal cut set is the smallest set of units
needed to guarantee an interruption of flow[10]. Consider a bridge circuit of 5 components- A,B,C,D and E as in
Figure 8.
For this particular bridge, there are 16 Ties, 4 Minimal Ties, 16 Cuts and 4 Minimal Cuts and they are listed
below.
TIES: AC, BD, AED, BEC, ABC, AEC, ADC, ABD, BED, BCD, ABED, ABCD, ABEC, AECD, BECD, ABCDE
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CUTS: AB, CD, AED, BEC, ABE, ABC, ABD, ACD, BCD, ECD, ABED, ABCD, ABEC, AEDC, BECD, ABCDE
MINIMAL TIES: AC, BD, AED, BEC
MINIMAL CUTS: AB, CD, AED, BEC
Figure 8: Bridge Circuit[12]
If a system has minimal paths denoted by T1, T2,...Tm , then the system reliability is given as
Rs(t) = Pr ( T1 U T2 U .. U Tm)
(12)
where each path sets Ti represents the event that units in the path set survive during the mission time t. This ensures
the success of the system.
Similarly system reliability can be expressed through minimal cuts sets. If the system has n minimal cuts denoted
by C1, C2,...Cn then the system reliability is obtained from
Rs(t) = 1- Pr (C1 U C2 U .. U Cn)
(13)
where Ci represent the event that units in the cut set fail some time before the mission time t. This ensures system
failure. Tie sets block representation of the bridge circuit is in Figure 9
Figure 9. Tie III.
Set E
block
representation
bridge circuit[12]
XPERIMENT
AND Rof
ESULT
Let
RS= System Reliability function
RA= Reliability Index of component A
RB= Reliability Index of component B
RC= Reliability Index of component C
RD= Reliability Index of component D
RE= Reliability Index of component E
The system reliability of the bridge circuit can be found out by the following expression
Rs = P(T1 ∪ T2 ∪ T3 ∪ T4)
Rs = P(T1)+P(T2)+P(T3)+P(T4)-P(T1∩T2)-P(T1∩T3)P(T1∩T4)-P(T2∩T3)-P(T2∩T4)-P(T3∩T4)+
P(T1∩T2∩T3)+P(T1∩T2∩T4)+
P(T1∩T3∩T4)+P(T2∩T3∩T4)P(T1∩T2∩T3∩T4)
P(T1) = RA .RC
P(T2)= RB .RD
P(T3)= RA.RD.RE
P(T4)= RB.RC.RE
P(T1∩T2)=P(T1).P(T2)= RA.RB.RC.RD
P(T1∩T3)=P(T1).P(T3)= RA.RC.RD.RE
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(16)
(17)
(18)
(19)
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P(T1∩T4)=P(T1).P(T4)= RA.RB.RC.RE
P(T2∩T3)= P(T2).P(T3)= RA.RB.RD.RE
P(T2∩T4)=P(T2).P(T4)= RB.RC.RD.RE
P(T3∩T4)= P(T1∩T2∩T3)= RA.RB.RC.RD.RE
P(T1∩T2∩T3) = P(T2).P(T3)= RA.RB.RC.RD.RE
P(T1∩T2∩T4)= RA.RB.RC.RD.RE
P(T1∩T3∩T4)= RA.RB.RC.RD.RE
P(T2∩T3∩T4)= RA.RB.RC.RD. RE
P(T1∩T2∩T3∩T4)=RA.RB.RC.RD. RE
Combining Eqn. 14 to Eqn. 30, the system reliability index is given by
Rs=RA.RC + RB .RD + RA.RD.RE+ RB.RC.RE - RA.RB.RC.RD RA.RB.RD.RE - RA.RB.RC.RERA.RC.RD.RE- RB.RC.RD.RE + 2 RA.RB.RC.RD. RE
If RA= RB= RC= RD= RE=R
Rs =2R2+2R3-5R4+2R5
If R = 0.99, Rs=0.99979805
B. Decomposition Method
(22)
(23)
(24)
(25)
(26)
(27)
(28)
(29)
(30)
(31)
(32)
Non-series-parallel structures can in many cases solved by a procedure in which a suitable component, X, is first
short-circuited (replaced by a component that never fails), then removed (considered a failure).If the resulting
structures after these changes are series-parallel, the network reliability will be given by.
Rs = P[system operates| X operates]P[x]
̅]
+ P[system operates| X fails] P[X
(33)
Where P[x] is probability that component x operates successfully.Consider Figure 10.
Let RA,RB,RC,RD and RE be the reliability index of components A,B,C,D and E respectively. Choosing E as the key
component, the system reliability index is computed as
̅]
Rs = P[system operates| E operates]P[E] + P[system operates| E fails]P[E
(34)
The network with component E short-circuited is shown in Figure 11 and with component E then removed is
shown in Figure 12. Since components A and B are parallel, the reliability of Section 1 can be computed using Eqn.
10.Reliability of Section 1 is given by
R(Section 1)=1-(1-RA)(1-RB )
(35)
Since components C and D are parallel, the reliability of Section 2 can be computed using Eqn. 10.Reliability of
Section 2 is given by
R(Section 2)=1-(1-RC)(1-RD)
(36)
Figure 10. Illustration of Decomposition Method [8]
Section 2
Section 1
Figure 11. Network obtained from Figure 9 with component E shorted[8]
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Figure 12. Network obtained from Figure 10 with component E removed[8]
Reliability of Figure 12 is given by the product of Eqn. 35 and Eqn.36
R=R(Section 1).R(section 2)
(37)
= RA RC+ RCRB+RBRD+RARD-RARBRC
-RBRCRD-RARBRD-RARCRD+RARBRCRD
If RA = RB = RC = RD = R
R =4R2-4R3+R4
(38)
Components A and C are in series. Also components B and D are in series. The resultant of the two are in parallel.
Combining Eqn. 6 and Eqn. 10,Reliability of Fig. 12 is given by
R=1-(1-RARC)(1- RBRD)
(39)
= RARC+ RBRD- RARCRBRD.
If RA = RB = RC = RD = R
R =2R2-R4
(40)
Combining Eqn. 34, Eqn. 38 and Eqn. 40
R=(4R2 -4R3+ R4)R+(2R2- R4)(1-R)
(41)
=2R5 -5R4+ 2R3+2R2
(42)
D. Event Space Method
The event space method is an application of the mutually exclusive events. The reliability of the system is simply
the probability of the union of all mutually exclusive events that yield a system success. Consider the system as in
Figure 11, where R1,R2,R3 are the reliabilities of unit1,unit2 and unit 3 respectively. Rs is the system reliability
index.
Let A,B and C be the event of unit 1, unit 2 and unit 3 success respectively. The mutually exclusive system events
are
X1 = ABC - all units succeed
̅BC -only unit 1 fails
X2 = A
̅C -only unit 2 fails
X3 = AB
̅ -only unit 3 fails
X4 = ABC
̅B
̅C -only unit 1 and 2 fails
X5 = A
̅BC
̅ - only unit 1and 3 fails
X6 = A
̅ -only unit 2 and 3 fails
̅C
X7 = AB
̅B
̅ -all unit fails
̅C
X8 = A
Figure 13. Illustration of the Event Space Method [8]
System events X6, X7 and X8 results in system failure. Thus the probability of failure of the system is
Pf = P(X6 U X7 U X8)
Since events X6,X7 and X8 are mutually exclusive,
Pf = P(X6) + P(X7) + P(X8)
The expansion of each of the terms are as below
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̅BC
̅ )= (1-R1)(R2)(1-R3)
P(X6) =P(A
̅ ) = (R1)(1- R2)( 1-R3)
̅
P(X7) =P(ABC
̅
̅ )= (1- R1)(1- R2)( 1-R3)
̅
P(X8) =P(ABC
(45)
(46)
(47)
Combining Eqn. 44 to Eqn. 47 yields,
Pf = 1- R1R2 - R3 + R1R2R3
But Rs= 1- Pf
Then Rs = R1R2 + R3 - R1R2R3
(48)
(49)
(50)
E. Path Tracing Method
With the path-tracing method, every path from a starting point to an ending point is considered. As long as at least
one path from the beginning to the end of the path is available, then the system has not failed. Consider Figure 14
for the illustration of the method.
The paths for this system are X1= 1,2 and X2= 3
The probability of success of the system is given by
P(X1 U X2 ) = P(X1)+P(X2)-P(X1 ∩ X2)
(51)
= P(1,2)+P(3)-P(1,2,3)
(52)
Rs = R1R2 + R3 - R1R2R3
(53)
which is same as that obtained in Eqn. 50 using event space method.
Figure 14. Illustration of the Path-Tracing Method [8]
F. Tie sets and Cut sets in large complex systems
For large complex systems, a computer program can be developed using Matrix Methods. The steps to be followed
are listed below
•
Obtain Reliability Network Model
•
Deduce Connection Matrix
•
Determine Minimal Paths (Tie Sets) from Connection Matrix multiplication
•
Build Incidence Matrix
•
Determine Minimal Cutsets from Incidence Matrix column operations
•
Use approximations and obtain Reliability index
The reliability network model of a bridge network is shown in the Figure 15.
Figure 15. Bridge Structure[12]
A Connection Matrix M is formed as below. The rows represents ‘From nodes’, columns represents ‘To nodes’ and
elements shows the connection between the nodes. The elements are marked 0 if there is no connection between the
nodes and are marked 1 if there is connection between the nodes.
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From Nodes
International Journal of Latest Trends in Engineering and Technology (IJLTET)
To Nodes
1 2 3 4
1 1 A B 0
2 0 1 E C
3
0
E
1
D
4
0
0
0
1
=M
(54)
Multiply the matrix by itself repeatedly, until the resulting matrix remains unchanged. M 2 was
obtained which was different from M. Hence M3 was calculated which was also found to be different from M.
1 A  BE B  AE AC  BD  BEC  AED
M3= 0
0
1
E
C  DE
E
1
EC  D
(55)
Minimal Paths
0
0
0
1
Then M4 was calculated which was found to be equal to M 3. From the Connection Matrix, the Minimal Paths (tie
sets) are AC,BD,BEC and AED. Now the Incident matrix is formed as below. Formulate a matrix that shows
components in each row of the minimal path.
Components
A B C D E
110 1 0 0
021 0 1 0
130 0 1 1
0 41 1 0 1
The steps to determine the minimal cutsets using matrix operations are as follows:
1. First step is to find the first order minimal cuts. If all the elements in a column are ‘1’, those
components form 1st order minimal cut.Here none of Columns A, B, C, D or E have all elements
marked ‘1’. So there are no 1st order minimal cut.
2. Second step is to find the second order minimal cuts.
a) Combine two columns at a time
b) All new set of columns having all elements ‘1’ forms 2 nd order cutsets
c) Eliminate cutsets containing 1st order cutsets
The new set of columns which have all elements ‘1’ are obtained when either columns A and B are combined or
when C and D are combined. Since there are no 1 st order cutsets to be eliminated, the result after Step 2 is AB, CD.
3. Third Step is to find the third order minimal cuts.
a. Combine three columns at a time
b. All new set of columns having all elements ‘1’ forms 3rd order cutsets
c. Eliminate cutsets containing 2 nd order cutsets.
The new set of columns which have all elements ‘1’ are obtained ABC, ABD, ABE, ACD, ADE, BCD, BCE, CDE.
Eliminating 2nd order cut sets AB and CD the result is
ABC, ABD, ABE, ACD, ADE, BCD, BCE, CDE
Thus the remaining 3rd order cut sets are ADE and BCE. Thus 1 st order cutsets are none, 2nd order cutsets are AB,
CD and 3rd order cutsets are ADE, BCE. Thus the cut sets of the bridge circuit are AB, CD, ADE and BCE. For
Higher order Min Cuts, just repeat the above method.
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IV. CONCLUSION
The paper attempts to illustrate the theoretical concepts and features of appropriate computational methodologies
that have been developed for analysing the overall reliability of power system networks. A single reliability formula
or technique which can be applied in all cases does not exist. The paper describes the different techniques to solve
reliability problems of various system models. Stress has been laid on network methods to determine system
reliability. These are fairly simple and fast, though requiring an essential logic diagram. Specific methods have been
analyzed and summarized.
Future work can lead to the following. It may be difficult to categorize general power system into
distinctive network models. In such cases, other methods such as frequency and duration methods, state space
methods may have to be resorted to. This can be achieved by analysis of failure data, leading to prediction of
relevant probability distributions.
V. REFERENCE
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ISSN: 2278-621X