RELIABILITY-BASED OPTIMIZATION OF DISTRIBUTION
SYSTEM PROTECTION
By
Yuting Tian

A THESIS
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
Electrical Engineering - Master of Science
2014

ABSTRACT
RELIABILITY-BASED OPTIMIZATION OF DISTRIBUTION SYSTEM PROTECTION
By
Yuting Tian
The power system is one of the most complex systems in existence. In any complex system, it is
impossible to avoid abnormal operations. In order to reduce the impact of abnormal operations
and to restore the system more quickly to normal states, many control and protection
mechanisms are established. Protection systems consist of sensing and isolation devices that
operate in the event of a fault to disconnect the part of the power system that is affected by the
fault. However, malfunctions can also occur within the protection system. Therefore, one
objective of this thesis is to develop approaches for analyzing and modeling the reliability of
power systems when the malfunctions of protection systems are considered. A methodology of
evaluating reliability of distribution systems are presented. This method uses matrices to
represent the relationship of devices in distribution systems and loads. Reliability indices such as
SAIFI and SAIDI of the distribution system of RBTS bus 2 are calculated using the proposed
methodology. Since power utilities are aim to provide reliable electric power with low cost to
customers, the investment of protective devices need to be taken into consideration. In this thesis,
an optimization problem is presented. The objective in this optimization problem is the cost of
disconnects and the constraint is the reliability indice, SAIDI. This reliability-based optimization
problem is solved for RBTS distribution system by Genetic Algorithm (GA).

Dedicated to:
My parents, Lifeng Tian and Jing Huang

iv

ACKNOWLEGEMENTS
I would like to thank my committee chair, Dr. Joydeep Mitra, for his guidance and support.
Without his help, I may not able to finish my master’s thesis. I would also like to thank my
committee members, Dr. Fang Z. Peng and Dr. Lixin Dong for their time, comments and
support.

Thanks also go to my friends and colleagues in ERISE lab, Dr. Niannian Cai, Ms. Nga Nguyen,
Mr. Mohammed Benidris, Mr. Samer Sulaeman, Mr. Salem Elsaiahs and Mr. Valdama
Johnsonthe. I would also thank all the faculty and staff of Michigan State University.

Finally, I want to express thanks to my parents for their love, care, understanding and support.

v

TABLE OF CONTENTS
LIST OF TABLES.....................................................................................................................vii
LIST OF FIGURES..................................................................................................................viii
CHAPTER 1 INTRODUCTION............................................................................................. 1
1.1 Introduction..............................................................................................................................1
1.2 Research Objectives.................................................................................................................2
1.3 Outline of Thesis......................................................................................................................4
CHAPTER 2 RELIABILITY EVALUATION OF DISTRIBUTION SYSTEM................6
2.1 Overview of Distribution System............................................................................................ 6
2.2 Previous Work......................................................................................................................... 7
2.3 Reliability Indices for Distribution System............................................................................. 8
CHAPTER 3 RELIABILITY EVALUATION OF DISTRIBUTION SYSTEM
CONSIDERING PROTECTION FAILURES........................................................................10
3.1 Methodology for the Distribution System Reliability Evaluation......................................... 10
3.2 Case Study............................................................................................................................. 12
3.2.1 Case 1................................................................................................................................14
3.2.2 Case 2................................................................................................................................16
3.2.3 Case 3................................................................................................................................19
3.2.4 Case 4................................................................................................................................21
3.2.5 Case 5................................................................................................................................24
3.2.6 Case 6................................................................................................................................26
3.3 Reliability Evaluation for RBTS Busbar 2............................................................................ 28
CHAPTER 4 OPTIMIZE ALLOCATION OF DISCONNECTS BY GENETIC
ALGORITHM............................................................................................................................37
4.1 Introduction............................................................................................................................37
4.2 Methodology.......................................................................................................................... 39
4.2.1 Coding...............................................................................................................................42
4.2.2 Fitness............................................................................................................................... 42
4.2.3 Reproduction.....................................................................................................................44
4.2.4 Crossover.......................................................................................................................... 44
4.2.5 Mutation............................................................................................................................45
4.3 Case Study............................................................................................................................. 45
4.3.1 Case 1................................................................................................................................45
4.3.2 Case 2................................................................................................................................49
CHAPTER 5 CONCLUSION................................................................................................51
5.1 Summary................................................................................................................................ 51
5.2 Future Work........................................................................................................................... 52

iii

BIBLIOGRAPHY...................................................................................................................... 54

iv

LIST OF TABLES
Table 3.1

Data of Transformer..................................................................................................13

Table 3.2

Data of Lines.............................................................................................................13

Table 3.3

Number of Customers Connected to Each Load...................................................... 13

Table 3.4

Feeder Type and Length........................................................................................... 28

Table 3.5

Customer Data.......................................................................................................... 29

Table 3.6

Loading Data.............................................................................................................29

Table 3.7

Reliability and System Data..................................................................................... 30

Table 3.8

Summary of Six Cases..............................................................................................32

Table 3.9

Result for Case 1.......................................................................................................32

Table 3.10

Result for Case 2.....................................................................................................33

Table 3.11

Result for Case 3.....................................................................................................33

Table 3.12

Result for Case 4.....................................................................................................34

Table 3.13

Result for Case 5.....................................................................................................34

Table 3.14

Result for Case 6.....................................................................................................35

Table 3.15

Comparison of Six Cases Results........................................................................... 35

Table 4.1

Optimum Solution for Case 1................................................................................... 47

Table 4.2

System Data for RBTS............................................................................................. 49

Table 4.3

Optimum Solution for Case 2................................................................................... 50

v

LIST OF FIGURES
Figure 3.1

Diagram for Basic Case...........................................................................................14

Figure 3.2

Diagram for Case 2..................................................................................................17

Figure 3.3

Diagram for Case 3..................................................................................................20

Figure 3.4

Diagram for Case 4..................................................................................................22

Figure 3.5

Diagram for Case 5..................................................................................................24

Figure 3.6

Distribution System of RBTS Busbar 2.................................................................. 31

Figure 4.1

Cost as a Function of Reliability............................................................................. 38

Figure 4.2

Flowchart.................................................................................................................48

vi

CHAPTER 1
INTRODUCTION
1.1 Introduction
The primary objective of the power grid is to provide reliable and economic electrical energy to
customers. However, the power system is one of the most complex systems in existence and it is
impossible to avoid abnormal operations in any complex system. Abnormal operation is usually
caused by system disturbances, which is defined as an undesired variable applied to a system that
tends to affect adversely the value of a controlled variable [1].The most common disturbances in
power systems are lightning, faults, large changes in load and failure of equipment due to
weather, fatigue, etc. In order to reduce the impact of abnormal operation and to restore the
system to normal status more quickly, many control and protection mechanisms are established.
Protection of power system is an extremely important aspect as the quality and scheme of the
protection decides system reliability, controllability and stability [2]. The performance of
protection systems affects the whole power system in many ways, including electrical facilities
and customers. Hence, there are always ongoing efforts toward improving the performance of
protection systems and the links between protection systems and power systems.

The electric power system is one of the largest and most complicated systems in the world, and
consists of a generation part, a transmission part and a distribution part. The research reported in

1

this thesis deals with protection issues encountered in distribution systems. The distribution
system is the portion of power systems which delivers electric energy from the transmission
system to the customer. The distribution system extends downstream from the distribution
substation to the customer meter. Often the initial overcurrent protection and voltage regulators
are within the substation fence and are considered to be part of the distribution system [3].

1.2 Research Objectives
The reliability of an electric power system is defined as the probability that the power system
will perform the function of delivering electric energy to customers on a continuous basis and
with acceptable service quality [4]. This character shows how well a system is performing its
intended function. There are several reliability indices that quantify different aspects of
distribution service interruptions, and some of the most commonly used indices are System
Average Interruption Frequency Index (SAIFI), System Average Interruption Duration Index
(SAIDI), Customer Average Interruption Duration Index (CAIDI), Customer Total Average
Interruption Duration Index (CTAIADI), and Customer Average Interruption Frequency Index
(CAIFI). These indices will be introduced in chapter two in detail. These indices help to evaluate
the customer satisfaction by representing the number of momentary and sustained interruptions,
duration of interruptions and number of customers interrupted. The significance of conducting
research of system reliability also plays roles in improving system performance. Besides, they
provide a basis for new or expanded system planning and maintenance scheduling, as well as
resource allocation.

2

It has been observed that protection system hidden failures commonly lead to multiple or
cascading outages, which consequently can cause large-scale power system blackouts [5]. The
protection system contains many protection devices, such as relays, breakers, fuses and
disconnects. Each component could fail to operate, which may lead to the failure of protection
systems and this would endanger the power grid in step. Thus, it is necessary to take the the
likelihood of protection failure into consideration when evaluating the reliability of the
distribution system.

Assessment of system performance is a valuable procedure for three important reasons. First of
all, it establishes the chronological changes in system performance and therefore helps to identify
weak areas and the need for reinforcement. Secondly, it establishes existing indices which serve
as a guide for acceptable values in future reliability assessments. Finally, it enables previous
predictions to be compared with actual operating experiences [6]. Therefore, as operating
conditions and technologies evolve in distribution systems, it is important to develop improved
and more sophisticated models and methods for reliability evaluation of these systems. One of
the objective of this thesis is to develop an improved model and method for distribution system
reliability that takes into account the role of protective devices.

The fundamental goal of an electric utility has always been to serve its customers with a reliable
and low cost power supply [7]. It is apparent that more the protective devices are installed in the
system, less the interruption duration is of the customer. However, the investment cost of utility
will be more with the increase of the number of protective devices. Therefore, an optimization
problem need to be solved to find the optimum number and placement for protective devices.

3

This is the second objective of this thesis. The objective function is the total cost of protective
devices, and reliability indice should be taken into consideration as a constraint. Genetic
Algorithm is applied in this thesis to solve the optimization problem, the detail is presented in
chapter four.

1.3 Outline of Thesis
The contents of this thesis are organized into five chapters. Following the chapter on introduction,
chapter two gives a brief overview of the distribution system. Also, the main indices to be used
in this research for assessing the reliability of the distribution system have been discussed.

Chapter three proposes a methodology based on relation matrix to evaluate the reliability indices
of distribution systems. Six cases are used to illustrate the method in detail. The variables in
these cases are fuse, disconnect, alternative supply and fuse failure. The system reliability and
individual load point reliability for six cases of the RBTS bus 2 distribution system are evaluated
by the methodology proposed in this thesis.

Chapter four presents a reliability-based optimization problem. This chapter proposes a
methodology of using Genetic Algorithm to solve the reliability-related optimization problem.
The optimum number and placement of disconnects are found for both the IEEE-RBTS bus 2
system and the whole RBTS system.

4

Chapter five is a summary of the whole thesis. It discusses the work that has been done and
makes suggestions for future research.

5

CHAPTER 2
RELIABILITY EVALUATION OF
DISTRIBUTION SYSTEM
2.1 Overview of Distribution System
Generally, an electric power system includes a generating, a transmission, and a distribution
system. The distribution system is the portion of power systems which delivers electric energy
from the transmission system to the customer. The distribution system extends downstream from
the distribution substation to the customer meter. Distribution lines are different from
transmission and subtransmission lines in that (1) they operate at lower voltages than
transmission lines, (2) they are usually radial, and (3) they usually have loads tapped all along
the line, not just at the terminals [8]. In the past, distribution system received less attention than
the other part of electric power system when considering reliability. The main reason for this are
that generating stations are individually very capital intensive and that generation inadequacy can
have widespread catastrophic consequences for both society and its environment [6]. However,
in 1977, [9] presented that analysis of the customer failure statistics of most utilities shows that
the distribution system makes the greatest contribution to the unavailability of supply of
customer. What’s more, in book [10], the author states that the distribution systems account for
up to 90% of all customer reliability problems. Therefore, it is essential to conduct research
about reliability evaluation of distribution systems.

6

2.2 Previous Work
After decades of effort in study of reliability of distribution systems, several methods have been
applied to this area. Basically, these methods can be divided into two parts, analytical methods
and simulation methods. Analytical methods are those that use system topology along with
mathematical expressions to calculate reliability indices, which include Markov modeling,
network reduction, fault tree analysis and cut-set analysis. Network reduction is useful for
systems consisting of series and parallel subsystems. This is a method that uses series-parallel
combinations to reduce network and then to determine load point indices and aggregate them to
calculate the system wide indices. When using fault tree analysis, the components that cause
interruptions to load, for each one should be determined first. Then the load point indices need to
be combined to get the system indices. As for the cut set method, it can be applied to systems
with simple as well as complex configurations and it is a very suitable technique for the
reliability analysis of power distribution system [11]. The first step is to determine the first and
second order minimal cut-sets that cause outages at each load point, then we can evaluate the
reliability indices.

When a system can be described by a set of discrete states, and the probability of moving to a
new state is only dependent upon the current state, the system can be described by a Markov
model [12]. Since protective devices and most components in power systems are repairable,
Markov modeling shows its advantage in evaluating the reliability of protective systems.
However, since distribution systems contain a large number of states, many simplifying
assumptions must be made to limit the Markov model to a manageable size [12].

7

Monte Carlo simulation is a way of simulating the conditions on the system by generating
system states of failure and repair randomly to compute the reliability indices. One advantage of
the simulation method is that this method is not restricted by the large size of the power grid.
Two types of simulation methods are often used: sequential Monte Carlo simulation and
non-sequential Monte Carlo simulation. The former one simulates the system operation by
generating an artificial history of failure and repair events in time sequence. The later type
evaluates the system’s response to a set of events and its order has no influence.

When evaluate the reliability of distribution system, the performance of protection system should
be taken into consideration. Hidden failures in the protection system are a main issue that affects
the reliability of the distribution system and decreases the satisfaction of customers. A large
amount of studies on reliability evaluation considering protection failures are conducted by
researchers [12-16].

2.3 Reliability Indices for Distribution System
System availability, estimated unsupplied energy, number of incidents and number of hours of
interruption are aspects that should be taken into consideration when evaluating the reliability of
the power grid. Reference [3] presents a set of terms and definitions which can be used to foster
uniformity in the development of distribution systems and several reliability indices are defined,
which are listed below.

The system average interruption frequency index (SAIFI) is defined as,

SAIFI=

 Total Number of Customers Interrupted
Total Number of Customers Served
8

(2.8)

This index indicates the average frequency of a sustained interruption that the customer
experiences. Sustained interruptions are those interruptions that last more than five minutes.

SAIDI refers to System Average Interruption Duration Index, which indicates the total duration
of interruptions for the average customer, and it is defined as,

SAIDI=

 Customer Minutes of Interruption
Total Number of Customers Served

(2.9)

CAIDI is Customer Average Interruption Duration Index, which indicates the average time
required to restore service, which is determined by,

CAIDI=

 Customer Minutes of Interruption
Total Number of Customers Interrupted

(2.10)

This index could also be calculated as below,

CAIDI=

SAIDI
SAIFI

(2.11)

CAIFI is the Customer Average Interruption Frequency Index, which represents the average
frequency of sustained interruptions for those customers experiencing sustained interruptions.
The equation is given below,

CAIFI=

 Total Number of Customer Interruptions
Total Number of Distinct Customers Interrupted

(2.12)

One thing that needs to be noticed is that the customer interrupted is counted once, regardless of
the the number of interruptions for this customer.

9

CHAPTER 3
RELIABILITY EVALUATION OF
DISTRIBUTION SYSTEM CONSIDERING
PROTECTION FAILURES
3.1 Methodology for the Distribution System
Reliability Evaluation
Distribution lines are different from transmission and subtransmission lines. Distribution lines
have characteristics such as lower voltage, radial and load tapped all along the line. If we assume
there are l lines, m transformers and n loads in a distribution system. We can use a matrix RL to
represent the relationship between loads and lines, the size of RL is n  l , and use RT

to

represent the relationship between loads and transformers, the size of RT is n  m . If load i
will be affected by line j , then RL (ij )  1 . If the fault on line j will not interrupt load i ,
RL (ij )  0 . Similarly, if load i will be affected by transformer j , then RT (ij )  1 , otherwise,
RT (ij )  0 . These could be represented as follows,

1, if load i will be affected by line j 1  i  n
RL (i, j )  
,
1 j  l
0, otherwise

10

(3.1)

1, if load i will be affected by transformer j 1  i  n
RT (i, j )  
,
1 j  m
0, otherwise

(3.2)

In this methodology, we use L to represent the failure rate of distribution lines. L is a 1  l
vector and L (i ) is the failure rate of line i , 1  i  l . I L and U L are used to represent the
number of interruptions and interruption duration of each load caused by distribution lines,
respectively. The size of I L is n  1 and the size of U L is n  l . The element of I L and U L
can be calculated as below,

I L (i )  RL (i )LT

(3.3)

U L (i, j )  tRL (i, j )L (1, j )

(3.4)

where t is the recovery time of load, which could be the fault clearing time or the isolation and
switching time. The value of t depends on situations, the detail will be illustrated later by six
cases in section 3.2.

Similarly, we use T

to represent the failure rate of transformers. I T

and U T represent the

number of interruptions and interruption duration of each load caused by transformers,
respectively. The size of I T

is n  1 and the size of U T

is n  m . I T

and U T

are

calculated as follows,

I T (i )  RT (i )T T
U T (i, j )  tRT (i, j )T (1, j )

(3.5)
(3.6)

Finally, the number of interruptions and interruption duration of each load can be calculated. We
use matrix I and U to represent the total number of interruptions and total interruption duration
of loads. The total number of interruptions is equal to the sum of the number of interruptions

11

caused by lines and the number of interruptions caused by transformers. Similarly, The total
interruption duration equals to the sum of interruption duration caused by lines and interruption
duration caused by transformers.
I (i )  I L (i )  I T (i )
l

m

j 1

j 1

(3.7)

U (i )   U L (i , j )   U T (i , j )

(3.8)

Then SAIFI and SAIDI can be calculated as

SAIFI=

 N    N I (i )
N
N
i i

i

i

SAIDI=

(3.9)

i

NU
N
i

i

i



 N U (i )
N
i

(3.10)

i

3.2 Case Study
Part of the distribution system of IEEE-RBTS bus 2 will be used as an example. The procedure
of evaluating reliability of this part by the methodology proposed in section 3.1 will be described.
Table 3.1 and 3.2 contain the reliability-related data of transformers and lines. Table 3.3 shows
the number of customer connected to each load. Six cases will be analyzed to show how to
calculate the reliability of distribution system according to the methodology proposed in this
thesis. Case 1 is the basic case, in this case the only protective device is a circuit breaker. In case
2, fuses are installed to protect the lateral lines. In case 3, disconnects are installed. In case 4,
both fuses and disconnects are installed. In case 5, fuses, disconnects and alternative supply are
all under consideration. Fuse failure is considered in case 6 based on case 5.

12

Table 3.1 Data of Transformer

Transformer

Failure rate (/yr)

Repair time (h)

1~7

0.015

200

Table 3.2 Data of Lines

Feeder type

Length (km)

1

0.60

2
3

Feeder section

Failure rate (/yr)

Repair time (h)

2, 6, 10

0.039

5

0.75

1, 4, 7, 9

0.04875

5

0.80

3, 5, 8, 11

0.052

5

numbers

Table 3.3 Number of Customers Connected to Each Load

Load

1

2

3

4

5

6

7

Number

210

210

210

1

1

10

10

13

3.2.1 Case 1
In this case, the reliability of a small branch will be evaluated, which does not contain fuse or
disconnect, only a circuit breaker is applied as protection in this basic case. The single line
diagram is shown in Figure 3.1.

Figure 3.1 Diagram

for Basic Case

The failure rate matrix of distribution lines is shown as below,

L  [0.0488 0.0390 0.0520 0.0488 0.0520 0.0390
0.0488

0.0520

0.0488

14

0.0390

0.0520]

(3.11)

Since there is no protective equipment at lateral lines, fault occurs on every line and each
transformer will trip the breaker and thus, interrupt all the customers. Hence, RL

and RT

should be all-ones matrices.

1
1

1

RL  1
1

1
1

1
1
1
1
1
1
1

1
1
1
1
1
1
1

1
1
1
1
1
1
1

1
1
1
1
1
1
1

1
1
1
1
1
1
1

1
1
1
1
1
1
1

1
1
1
1
1
1
1

1
1
1
1
1
1
1

1
1
1
1
1
1
1

1
1

1

1
1

1
1

1
1

1

RT  1
1

1
1

1
1
1
1
1
1
1

1
1
1
1
1
1
1

1
1
1
1
1
1
1

1
1
1
1
1
1
1

1
1
1
1
1
1
1

1
1

1

1
1

1
1

(3.12)

Then the number of interruptions and interruption duration can be calculated by the methodology
proposed.

The number of interruptions and interruption duration caused by lines are as follows,

RL (i )  1 1 1 1 1 1 1 1 1 1 1
I L (i )  RL (i )LT  0.52

(3.13)
(3.14)

Interruption duration of load i caused by distribution line j is

U L (i, j )  rL RL (i, j )L (1, j )

(3.15)

where, rL  5h .
In this case, the interruption duration for load i is
n

U
j 1

L

(i, j )  2.6 , for all i

(3.16)

Then the number of interruption and duration caused by transformers can be obtained.

T  [0.015 0.015 0.015 0.015 0.015 0.015 0.015]
RT (i )  1 1 1 1 1 1 1
15

(3.17)
(3.18)

I T (i )  RT (i )T T  0.105

(3.19)

The interruption duration for load i caused by transformer j is,

U T (i, j )  rT RT (i, j )T (1, j )  200  1  0.015  3, for all i, j

(3.20)

In this case,
7

U
j 1

T

(i, j )  21, for all i

(3.21)

Then the total number of interruption and interruption duration for each load can be calculated.

I (i )  I L (i )  I T (i )  0.52+0.105  0.625
l

m

j 1

j 1

U (i )  U L (i, j )  U T (i, j )  2.6  21  23.6

(3.22)
(3.23)

Hence, all the load will experience 0.625 times interruption in a year and the duration is 23.6
hours.
Therefore,

SAIFI=

 N    N I (i )  0.625
N
N

(3.24)

NU
N

(3.25)

i i

i

i

SAIDI=

i

i

i

i



 N U (i )  23.6
N
i

i

The units for SAIFI and SAIDI are interruption/customer.yr and hr/customer.yr, respectively.

3.2.2 Case 2
In this case, the transformer is assumed to be protected by a fuse in each lateral. When a short
circuit fault occurs, the corresponding fuse will blow and will disconnect that lateral. Hence, the

16

fault on that lateral will not affect loads on other laterals. The interruption duration for the
corresponding load is the repair time of failure.

Figure 3.2 Diagram

for Case 2

In this case the RL and RT matrix will not be all-ones matrix anymore. RL (i, j )  1 ，if j=1,4, 7,10
and if load i is connected to line j.

1
1

1

RL  1
1

1
1

1
0
0
0
0
0
0

0
1
0
0
0
0
0

1
1
1
1
1
1
1

0
0
1
0
0
0
0

0
0
0
1
0
0
0

17

1
1
1
1
1
1
1

0
0
0
0
1
0
0

0
0
0
0
0
1
0

1
1
1
1
1
1
1

0
0

0

0
0

0
1 

(3.26)

Now take load 1 as an example, the number of interruptions is,

I L (1)  RL (1)LT  0.2244

(3.27)

Interruption duration of load i caused by distribution line j is,

U L (i, j )  rL RL (i, j )L (1, j )

(3.28)

The interruption duration for load i is,
11

U L (1)  U L (1, j )  1.22

(3.29)

j 1

Now, we can evaluate the interruption caused by transformers. Load i will only be affected by
transformer j which connected directly to load i. Therefore,

1
0

0

RT  0
0

0
0

0
0

0

0
0

0 0 0 0 1 0
0 0 0 0 0 1 

0
1
0
0
0

0
0
1
0
0

0
0
0
1
0

0
0
0
0
1

0
0
0
0
0

( 2.30)

For load 1,

RT (1)  1 0 0 0 0 0 0

(3.31)

T  [0.015 0.015]

(3.32)

I T (1)  RT (1)T T  0.015

(3.33)

U T (1,1)  rT RT (1,1)T (1)  3

(3.34)

U T (1, j )  rT RT (1, j )T (1)  0, for 2  j  7

(3.35)

Thus,
7

U
j 1

T

(i, j )  3 for all i

18

(3.36)

Then we can calculate the total number of interruptions and interruption duration for each load.

I (1)  I L (1)  I T (1)  0.2244  0.015  0.2394
11

7

j 1

j 1

U (1)  U L (1, j )  U T (1, j )  1.122  3  4.122

(3.37)
(3.38)

Finally SAIFI and SAIDI can be calculated as,

SAIFI=

 N    N I (i )  0.248
N
N

(3.39)

NU
N

(3.40)

i i

i

i

SAIDI=

i

i

i

i



 N U (i )  4.165
N
i

i

The units for SAIFI and SAIDI are interruption/customer.yr and hr/customer.yr, respectively.

3.2.3 Case 3
In this case, disconnects are applied in the distribution system. When disconnect is added, faults
on feeders will still cause circuit breaker operates. Thus, RL and RT are all-ones matrices, and
the number of interruptions is the same as in case 1. However, loads between the circuit breaker
and disconnects will be recovered before fault is cleared in this case. The interruption duration
for those loads becomes the isolation and switching time, rather than the repair time.

19

Figure 3.3 Diagram for Case 3

Let the isolation and switching time s equals 1 hour in this case. Let Z i be the nearest line to load
i which installed disconnect. For example Z1  Z 2  4 , which means the nearest disconnect are
installed on line 4 for load 1 and load 2. Then the interruption duration could be calculated as
follows,

 r R (i, j )L (1, j ),
U L (i , j )   L L
 sRL (i, j )L (1, j ),

j  Zi
otherwise

(3.41)

For example,

U L (1,1)  rL RL (1,1)L (1,1)  5  1  0.0488  0.244

(3.42)

U L (1,5)  sRL (1,5)L (1,5)  1  1  0.0520  0.0520

(3.43)

After finding all the U L (1, j ) , then sum them up,

20

11

U
j 1

L

(1, j )  0.871

(3.44)

When calculating the duration caused by transformers, it is similar to the way of calculating
duration caused by lines,

 r R (i, j )T (1, j ),
U T (i , j )   T T
 sRT (i, j )T (1, j ),

j  Yi
otherwise

(3.45)

Where, Y (i ) is a set of transformers which are in the same zone of load i.
For instance, Y1  Y2  {1, 2}, Y7  {7} , then

U T (1,1)  U T (1, 2)  rT RT (1,1)T (1,1)  200  1  0.015  3

(3.46)

U T (1, j )  sRT (1, j )T (1, j )  1 1 0.015  0.015, for 3  j  7

(3.47)

After obtaining all the U T (1, j ) , we are able to calculate,
7

U
j 1

T

(1, j )  6.075

(3.48)

Thus,

U (1)  0.871  6.075  6.946

(3.49)

After finding all the U (i ) , it is easy to obtain that SAIDI is equal to 9.740, and SAIFI is the
same as in case 1, which is 0.625. The units for SAIFI and SAIDI are interruption/customer.yr
and hr/customer.yr, respectively.

3.2.4 Case 4
In this case, both fuses and disconnects are applied to protect the power grid. The number of
interruptions is the same as in case 2, since faults on line 1, 4, 7, 9 will still trip the circuit

21

breaker, though disconnects are installed. The way to calculate the interruption duration is the
same as in case 3. But the results are different, since RL and RT matrix are varied.

Figure 3.4 Diagram

for Case 4

In this case,

1
1

1

RL  1
1

1
1

1
0
0
0
0
0
0

0
1
0
0
0
0
0

1
1
1
1
1
1
1

0
0
1
0
0
0
0

0
0
0
1
0
0
0

1
1
1
1
1
1
1

0
0
0
0
1
0
0

 r R (i, j )L (1, j ),
U L (i , j )   L L
 sRL (i, j )L (1, j ),

22

0
0
0
0
0
1
0

1
1
1
1
1
1
1

0
0

0

0
0

0
1 

j  Zi
otherwise

(3.50)

(3.51)

Take load 1 as an example,

U L (1,1)  rL RL (1,1)L (1,1)  5  1  0.0488  0.244
U L (1,5)  sRL (1,5)L (1,5)  1  0  0.0520  0

(3.52)
(3.53)

Then,
11

U
j 1

L

(1, j )  0.5753

(3.54)

For transformers,

1
0

0

RT  0
0

0
0

0
1
0
0
0
0
0

0
0
1
0
0
0
0

0
0
0
1
0
0
0

0
0
0
0
1
0
0

0
0
0
0
0
1
0

 r R (i, j )T (1, j ),
U T (i , j )   T T
 sRT (i, j )T (1, j ),

0
0

0

0
0

0
1 

(3.55)

j  Yi
otherwise

(3.56)

U T (1,1)  rT RT (1,1)T (1,1)  200  1  0.015  3
U T (1, j )  sRT (1, j )T (1, j )  1 0  0.015  0, for 3  j  7
7

U
j 1

T

(1, j )  3

(3.57)
(3.58)
(3.59)

Thus,

U (1)  0.5753  3  3.5753

(3.60)

Finally, it is found that SAIDI is 3.697 hr/customer.yr. SAIFI is 0.248 interruption/customer.yr,
which is the same as in case 2.

23

3.2.5 Case 5
As shown in Figure 3.5 line 12 is connected to an alternative supply. In this case, the number of
interruption is the same as in case 2, so as SAIFI. But the interruption duration will be shortened.
For example, if there is a fault occurs on line 4, the breaker will trip and all the loads will be
affected. Then the disconnects on line 4 and line 7 will open which isolates the fault, and then the
breaker will reclosed. After this step, load 1 and load 2 are recovered. Later on, the normally
open disconnect on line 12 will close, load 5, 6, 7 will be recovered. Therefore, the interruption
duration for these loads, will change to the isolation and switching time, instead of the repair
time of fault in line 4 as in case 3.

Figure 3.5 Diagram for Case 5

The procedure to value the interruption duration is described below.
Let’s define a set Z to represent the protection zone sectioned by disconnects.

24

For example, Z1  {1, 2,3}, Z 3  {4,5,6} .

 r R (i, j )L (1, j ),
U L (i , j )   L L
 sRL (i, j )L (1, j ),

j  Zi
otherwise

(3.61)

Take load 1 as an example, since Z1  {1, 2,3} ,

U L (1,1)  rL RL (1,1)L (1,1)  5  1  0.0488  0.244

(3.62)

U L (1, 2)  rL RL (1, 2)L (1, 2)  5  0  0.0390  0

(3.63)

U L (1,3)  rL RL (1,3)L (1,3)  5  1  0.052  0.26

(3.64)

The rest U L (i, j )  sRL (i, j )L (i, j ) ,
For instance,

U L (1, 4)  sRL (1, 4)L (1, 4)  1  1  0.0488  0.0488

(3.65)

U L (1,5)  sRL (1,5)L (1,5)  1  0  0.052  0

(3.66)

After calculating all the U L (1, j ) , we are able to find the interruption duration caused by lines of
load 1, which is,
11

U
j 1

L

(1, j )  0.5753

(3.67)

The interruptions duration caused by transformers are the same as in case 4, which is 3h. Then,

U (1)  0.5753  3  3.5753

(3.68)

Finally, it is found that SAIFI is 0.248 interruption/customer.yr and SAIDI is 3.613
hr/customer.yr.

25

3.2.6 Case 6

Protective devices are not 100% reliable, so we need to consider
protection failures. In this case, we assume that fuse could operate when
it needed with probability 0.9 [6]. Then the number of interruptions and
duration will be different to case 5.

RL

and

RT

in this case is as shown

in 3.69 and 3.70.
1
1

1

RL  1
1

1
1

0.1
0.1

0.1

0.1
0.1 0.1 1 0.1 0.1 1 1 0.1 1 0.1

0.1 0.1 1 0.1 0.1 1 0.1 1 1 0.1
0.1 0.1 1 0.1 0.1 1 0.1 0.1 1 1 
1 0.1 1 0.1 0.1 1 0.1 0.1
0.1 1 1 0.1 0.1 1 0.1 0.1
0.1 0.1 1 1 0.1 1 0.1 0.1
0.1 0.1 1 0.1 1 1 0.1 0.1

1
0.1

0.1

RT  0.1
0.1

0.1
0.1

0.1
1
0.1
0.1
0.1
0.1
0.1

0.1
0.1
1
0.1
0.1
0.1
0.1

0.1
0.1
0.1
1
0.1
0.1
0.1

0.1
0.1
0.1
0.1
1
0.1
0.1

0.1
0.1
0.1
0.1
0.1
1
0.1

1
1
1
1

0.1
0.1

0.1

0.1
0.1

0.1
1 

Then the number of interruptions and duration caused by lines for load 1 are as below,

26

(3.69)

(3.70)

I L (1)  RL (1)LT

 1 1 0.1 1 0.1 0.1 1 0.1 0.1 1 0.1 LT

(3.71)

 0.2538
 r R (i, j )L (1, j ),
U L (i , j )   L L
 sRL (i, j )L (1, j ),

j  Zi
otherwise

(3.72)

For instance

U L (1,1)  rL RL (1,1)L (1,1)  5  1  0.0488  0.244

(3.73)

U L (1, 2)  rL RL (1, 2)L (1, 2)  5  0.1  0.0390  0.0195

(3.74)

U L (1,3)  rL RL (1,3)L (1,3)  5  1  0.052  0.26

(3.75)

U L (1, 4)  sRL (1, 4)L (1, 4)  1  1  0.0488  0.0488

(3.76)

U L (1,5)  sRL (1,5)L (1,5)  1  0.1  0.052  0.0052

(3.77)

Then,
11

U
j 1

L

(1, j )  0.5803

(3.78)

The analysis for transformer’s affect is similar.
Then the number of interruptions caused by transformers is,

I T (i )  RT (i )T T  0.024 for all i

(3.84)

In this case,

 r R (i, j )T (1, j ),
U T (i , j )   T T
 sRT (i, j )T (1, j ),

j  Yi
otherwise

(3.85)

For load 1,

U T (1,1)  rT RT (1,1)T (1,1)  200  1  0.015  3

(3.86)

U T (1, 2)  rT RT (1, 2)T (1, 2)  200  0.1  0.015  0.3

(3.87)

27

U T (1, j )  sRT (1, j )T (1, j )  1 0.1 0.015  0.0015, for 3  j  7

(3.88)

Therefore,
7

U
j 1

T

(1, j )  3.3075

(3.89)

Then we could calculate the total number of interruptions and duration for each load.
For example,

I (1)  I L (1)  I T (1)  0.2538  0.024  0.2778
11

7

j 1

j 1

U (1)  U L (1, j )  U T (1, j )  0.5803  3.3075  3.8878

(3.90)
(3.91)

Finally, SAIFI and SAIDI can be valued. In this case SAIFI is 0.284 interruption/customer.yr
and SIDAI is 3.888 hr/customer.yr.

3.3 Reliability Evaluation for RBTS Busbar 2
In this section the reliability of RBTS busbar 2 [17] will be evaluated. This system contains 36
lines, 22 loads and 1908 customers. The single line diagram for RBTS Busbar 2 is shown in
Figure 3.6. Customer data, feeder types and length, loading data, reliability and system data are
listed in Table 3.4~3.7.
Table 3.4 Feeder Type and Length
Feeder type

Length/km

Feeder section numbers

1

0.60

2, 6, 10, 14, 17, 21, 25, 28, 30, 34

2

0.75

1, 4, 7, 9, 12, 16, 19, 22, 24, 27, 29, 32, 35

3

0.80

3, 5, 8, 11, 13, 15, 18, 20, 23, 26, 31, 33, 36

28

Table 3.5 Customer Data

Number of
load points

Load points

Load level per load point,
MW

Customer
type

Average

Peak

Number of
customers

5

1-3, 10, 11

Residential

0.535

0.8668

210

4

12, 17-19

Residential

0.450

0.7291

200

1

8

Small user

1.00

1.6279

1

1

9

Small user

1.15

1.8721

1

6

4, 5, 13, 14, 20, 21

Govt/inst

0.566

0.9167

1

5

6, 7, 15, 16, 22

Commercial

0.454

0.7500

10

12.291

20.00

1908

Total

Table 3.6 Loading Data

Feeder load, MW

Feeder
number

Load
points

Average

Peak

Number of
customers

F1

1-7

3.645

5.934

652

F2

8-9

2.15

3.500

2

F3

10-15

3.106

5.057

632

F4

16-22

3.390

5.509

622

12.291

20.00

1908

Total

29

Table 3.7 Reliability and System Data
Component

P

T

 ''

T 33/11

0.015

0.05

1

11/0.415

0.015

B33

0.002

0.02

B11

0.006

L33

0.046

L11

0.065

Cable

0.040

r

rP

r ''

rC

s

15

120

0.083

1

200

10

1

0.5

4

96

8

0.083

1

0.06

1.0

4

72

8

0.083

1

0.06

0.5

8

8

8

0.083

2

5
30

Where,

P : permanent(total) failure rate (f/yr) (for lines/cables (f/yr.km))
T : temporary failure rate (f/yr) (for lines/cables (f/yr.km))
 '' : maintenance outage rate (out/yr)
r : repair time (hr)
rP : replace time by a spare (hr)
r '' : maintenance outage time (hr)
rC : reclosure time (hr)
s: switching time (hr)

30

3

Figure 3.6 Distribution System of RBTS Busbar 2

31

Reliability of six cases will be calculated and their characteristics are shown in Table 3.8. The
differences are the inclusion or not of disconnects in the main feeders, fuses in each lateral, an
alternative supply and fuses failures.

Table 3.8 Summary of Six Cases

Case

Fuse

Disconnect

Alternative
Supply

Repair of
Transformer

Failure of
Fuse

1
2

√

3

√
√

√
√

4

√

√

5

√

√

√

√

6

√

√

√

√

√

Table 3.9 Result for Case 1
CASE 1
SAIFI

SAIDI

F1

0.625

23.6

F2

0.192

0.959

F3

0.558

20.34

F4

0.625

23.6

SYSTEM

0.602

22.496

32

Table 3.10 Result for Case 2
CASE 2
SAIFI

SAIDI

F1

0.248

4.165

F2

0.14

0.699

F3

0.25

4.174

F4

0.247

4.16

SYSTEM

0.248

4.163

Table 3.11 Result for Case 3
CASE 3
SAIFI

SAIDI

F1

0.625

9.740

F2

0.192

0.777

F3

0.558

8.465

F4

0.625

11.66

SYSTEM

0.602

9.934

33

Table 3.12 Result for Case 4
CASE 4
SAIFI

SAIDI

F1

0.248

3.697

F2

0.14

0.621

F3

0.25

3.76

F4

0.247

3.75

SYSTEM

0.248

3.732

Table 3.13 Result for Case 5
CASE 5
SAIFI

SAIDI

F1

0.248

3.618

F2

0.14

0.523

F3

0.25

3.624

F4

0.247

3.605

SYSTEM

0.248

3.613

34

Table 3.14 Result for Case 6
CASE 6
SAIFI

SAIDI

F1

0.286

3.926

F2

0.145

0.529

F3

0.282

3.83

F4

0.285

3.917

SYSTEM

0.284

3.888

Table 3.15 Comparison of Six Cases Results

CASE

SAIFI

SAIDI

1

0.602

22.496

2

0.248

4.163

3

0.602

9.894

4

0.248

3.732

5

0.248

3.613

6

0.284

3.888

35

3.4 Summary
In this chapter, a methodology is proposed to calculate the reliability indices of distribution
system. This methodology is derived from R. Billiton’s book [6]. Reference [31] introduced a
methodology, zone branch reduction, which also uses matrix to calculate the reliability indices.
The comparison between cases demonstrates the effect of protective devices. It is obvious that
Case 5 is the most reliable one with lowest SAIFI and SAIDI, since it contains fuses, disconnects
and alternative supply and all the protective devices are assumed perfectly reliable. SAIDI in
case 1 is bigger than others, since it has no protection except the circuit breaker. The difference
of SAIFI between case 2 and case 4 is because of the effect of disconnects. Since, disconnects
are able to let the loads which are outside the fault zone recover more quickly. The interruption
duration for such loads are decreased from repair time as 5 hours to switching and isolation time
as 1 hour. Therefore, SAIDI in case 4 is smaller than case 2. SAIFI of case 2, 3 ,4 and 6 are the
same, since fuses are installed in laterals in these four cases. The effects of faults will be reduced
by fuses. Hence, SAIFI in these four cases are smaller than case 1 and case 5. Alternative supply
is also a very effective way to improve the system reliability. SAIDI is decreased from 3.732 as
in case 4 to 3.613 as in case 5.

36

CHAPTER 4
OPTIMIZATION OF ALLOCATION OF
DISCONNECTS USING GENETIC
ALGORITHM
4.1 Introduction
The fundamental goal of an electric utility has always been to serve its customers with a reliable
and low cost power supply [7]. From the cases in chapter 5, we find that disconnects are helpful
to reduce the interruption duration of customers, which is able to enhance the reliability of
distribution systems. It is apparent that more the disconnects are installed in the distribution
feeder, less the interruption duration is of the customer. However, the investment cost of utility
will be more with the increase of the number of disconnects. In recent years, electric utility
industry has confronted many challenges in the increasingly competitive market, which demands
them to serve its customers with higher reliability and lower cost power supply [20]. The
presence of inverse relation between economic constraint and reliability is obvious, as shown in
Figure 4.1 [21], so this causes complexity in management decisions. Hence, a method is needed

37

to find an optimum solution with lower investment cost and higher reliability. In this vein, this
chapter proposes an optimization method for planning adequate number and location of
disconnect in a distribution system with reliability consideration.

Figure 4.1 Cost as

a Function of Reliability

The relationship between allocation of protective devices and SAIDI is complex and the
constraint is non-differentiable. Thus, traditional analytical approaches such as linear and
nonlinear programming have difficulty in dealing with this optimization problem. Previous work,
such as [19][25-26], stated that the problem of location of protective devices naturally has binary
specification in distribution networks. Each protective device has two states: connected or
disconnected and can be model as 0 and 1, respectively. This specification causes using binary
programming method for optimal location of protective devices [20]. However, as the size of the

38

distribution system becomes larger, the possible position to install protective devices increases
rapidly and this lead to slower speed when solving this problem by binary programming.
Therefore, evolutionary methods are applied to solve this kind of problem [20][27-29]. In recent
past, nontraditional search and optimization methods are becoming popular in electrical
engineering, such as Genetic Algorithms (GA), Ant colony optimization (ACO) and Particle
Swarm Optimization (PSO). The method proposed in this chapter uses GA to find the optimum
solution. Genetic algorithms are computerized search and optimization algorithms based on the
mechanic of natural genetics and natural selection [23]. This method was first used by J. H.
Holland in mid-sixties [22]. The detail of this methodology is presented in the next section.

4.2 Methodology
The goal of system optimization is to minimize an objective function without violating any
constrains [18]. The objective of this problem is to minimize the cost of investment, which is
equal to the sum of the price of each disconnect.

Minimize F  X T C
Where,

F is the total cost of investment;
C is a vector of the unit price of disconnect, Ci is the price of disconnect installed on possible

39

position i;
X is a vector of decision of positions. xi is the decision for position i, if xi  1 , which means a
disconnect is installed on position i. Otherwise, if xi  0 , this means position i is not connected
to a disconnect. The constraint is reliability. In this model, we use SAIDI to evaluated the
reliability. Since the change of position and number of disconnects will also let the SAIDI varied.
Therefore, SAIDI should be updated each time. This step can be realized by the methodology
proposed in chapter 3. Using the methodology proposed in chapter 3, the interruption duration in
each feeder and the whole system can be obtained.
Then the constraint is as shown below,

S  Smax

(4.1)

Where S is the Sustained Average Interruption Duration Index (SAIDI), Smax is the maximum
SAIDI allowed.
As shown before [3],

SAIDI=

 Customer Minutes of Interruption
Total Number of Customers Served

(4.2)

Therefore, the mathematical model for this optimization problem can be stated as follows,

Minimize F  X T C
Subject to

(4.3)

S  Smax

40

This reliability-based optimization problem is solved by Genetic Algorithm (GA) and the steps
of a typical genetic algorithm is as follows [23],
Step 1 Choose a coding to represent problem parameters, a selection operator a crossover
operator, and a mutation operator. Choose population size n, crossover probability pc , and
mutation probability pm . Initialize a random population of strings of size l. Choose a maximum
allowable generation number tmax . Set t=0.
Step 2 Evaluate each string in the population.
Step 3 If t  tmax or other termination criteria is satisfied, Terminate.
Step 4 Perform reproduction on the population.
Step 5 perform crossover on random pairs of strings.
Step 6 Perform mutation on every string.
Step 7 Evaluate strings in the new population. Set t=t+1 and go to Step 3.
The detail of each step is presented below.

4.2.1 Coding

In order to solve the optimization problem by GAs, the variables are firstly coded in strings. GAs
work with a population of binary string (0 and 1), and because of the characteristic of this

41

problem, binary coding is used in this methodology. With the binary coding method, the decision
of possible positions would be coded as a binary string with length n, where n is the total number
of possible positions. The binary-coded string shows the decision for each possible position to
connect to a disconnect or not. In the string, “1” represents “connected” and “0” represents
“not-connected”. The string X can be shown as below,

X   x1

x2  xn 

（4.4）

Where,
n

Number of possible positions to install disconnect

xi Decision for position i, xi  0 or 1 and 1  i  n .

4.2.2 Fitness

When dealing with constrained optimization problems, a penalty-parameter-less constraint
handling approach is popular used, which is proposed by K. Deb in 2000 [24]. The concept is
simple. In a tournament selection operator comparing two population members, three
possibilities and corresponding selection strategy were suggested [23]:
(a) When one solution is feasible and the other is infeasible, the feasible solution is selected.
(b) When both solutions are feasible, the one with better function value is selected.

42

(c) When both solutions are infeasible, the one with smaller constraint violation is selected.
The definition of a Constraint Violation (CV) is determined in [24] . There are two steps to find
CV. First, normalize all constraints using the constant term in the constraint function. For
example, the constraint g j ( x )  b j  0 is normalized as g j ( x )  g j ( x ) / b j  1  0 [23]. Equality
constraints can also be normalized to h k ( x ) . Second, the constraint violation can be determined
as follows,
J

K

j 1

k 1

CV ( x )   g j ( x )   h j ( x )

（4.5）

A way to use the above penalty-parameter-less approach is to convert the problem into the
following unconstrained fitness function [23].


fitness( x )  

Where

f max

f ( x ),
if x is feasible
f max  CV ( x ), if x is infeasible

（4.6）

is the maximum objective function value among the feasible solutions or zero

when there is no feasible solution in a population. In this optimum allocation of disconnects
problem, the only constraint is the reliability requirement, therefore, CV can be expressed as
follows,

CV ( x )  S ( x ) / Smax  1, if x is infeasible

43

（4.7）

4.2.3 Reproduction

Reproduction, also known as selection operator. Reproduction selects good strings in a
population and forms a mating pool [23]. Roulette-wheel selection and tournament selection are
often used to form mating pool. In this work, tournament selection is applied. First, pick s
individuals from the population, then choose the best among them to the mating pool. A binary
tournament selection with s=2 is used in this work. After forming the mating pool, stings in
mating pool are going to the next step: crossover.

4.2.4 Crossover

In a crossover operator, new strings are created by exchanging information among strings of the
mating pool [23]. This step is mainly responsible for the search of new strings. In order to create
new strings, two strings are selected randomly from the mating pool, and these two strings are
called parent strings, the newly created strings are known as children strings. After randomly
selecting the parent strings from the mating pool, a single-point crossover is applied. First, a
crossing point is randomly chosen. Then exchange all the bits on the right side of the crossing
point, as shown below,

0 00 0 0 0 0 1 1 1

1 1 1 1 1
1 10 0 0
44

（4.8）

Not all the old strings are used in crossover, a crossover probability pc
preserve some of the good strings. Only 100 pc

is used in order to

percent of old strings are used in crossover

operation.

4.2.5 Mutation

Mutation of gene happens with low probability in nature. In GAs, a mutation operation is applied
to mimic this change. The mutation operator changes 1 to 0 and vice versa for each bit of every
string with a small mutation probability pm . This step allows the algorithm to a local search
around current solutions.

4.3 Case Study
In this section, two cases will be presented. One case is still the RBTS bus 2 [17], another is the
whole RBTS system. The information of the whole RBTS system can be found in [17] and [30].

4.3.1 Case 1

There are ten possible locations to install disconnect in this distribution system, which are on line
4, 7, 10, 14, 18, 21, 24, 29, 32 and 34. So we have 10 variables for this system. we can use

45

variables x1 to x10 to represent these possible positions. Then x1 represents the decision for
position 1, which is on line 4 and x10 represents the decision for position 10, which is on line 34.
For example, if the result shows that x1 is 1, this means a disconnect is installed on line 4.
Variables xi in the objective function are calculated using MATLAB to obtain the optimum
solution. X can be coded as below,

X   x1

x2  x10 

（4.9）

In this case, the mathematical model is as follows,

Minimize F  X T C
Subject to，

（4.10）

S  Smax

In this case we assume the price for each disconnect is 3000$ [20], that is Ci  3000 for all i.
Several optimum solutions for different maximum SAIDI are obtained by MATLAB. The result
is listed in Table 4.1. The flowchart for the methodology is as shown in Figure 4.1.

46

Table 4.1 Optimum Solution for Case 1
SAIDI_max

SAIDI

Position of disconnects

Number of
disconnects

Investment
cost ($)

3.66

3.6579

4, 18, 21, 29, 32

5

15000

3.63

3.6169

4, 7, 18, 21, 29, 32

6

18000

3.615

3.6150

4,7,10,18, 21, 29, 32

7

21000

3.614

3.6140

4, 7, 10, 18, 21, 24, 29, 32

8

24000

3.613

3.6128

4, 7, 10, 14, 18, 21, 24, 29, 32

9

27000

47

Figure 4.2 Flowchart

48

4.3.2 Case 2

In this case, the whole RBTS system is used. In this system, there are 287 lines, 170 loads and
18,289 customers in this test system, some of the system data are listed in Table 4.2 [17][30]. 69
possible places are chosen to install disconnects. The optimal solution is obtained by Genetic
Algorithm and the result is shown in Table 4.3.

Table 4.2 System Data for RBTS
Bus 2

Bus 3

Bus 4

Bus 5

Bus 6

Total

Lines

36

77

67

43

64

287

Loads

22

44

38

26

40

170

Customers

1,908

5,806

4,779

2,858

2,938

18,289

Positions

10

19

16

13

11

69

49

Table 4.3 Optimum Solution for Case 2
SAIDI_max

4.0

3.9

3.8

SAIDI

Position of disconnects

3.9959

Bus 3: 6, 23, 35, 45, 59
Bus 4: 5, 23, 36, 63
Bus 6: 7, 21

3.8897

Bus 2: 4
Bus 3: 3, 8, 23, 35, 38, 45,
Bus 4: 5, 23, 36, 60, 63
Bus 5: 7, 18, 39
Bus 6: 7, 21

3.7928

Bus 2: 4, 18, 21, 32
Bus 3: 3, 8, 21, 23, 35, 38, 45, 57, 59
Bus 4: 5, 7, 23, 26, 33, 36, 39, 60, 63
Bus 5: 4, 18, 36
Bus 6: 7, 17, 23

Number of
disconnects

Investment
cost ($)

11

33,000

18

54,000

28

84,000

47

141,000

59

Bus 2: 4, 7, 18, 21, 29, 32

3.7

3.6969

Bus 3: 3, 6, 8, 10, 21, 23, 26, 33, 35,
38, 40, 45, 50, 57, 59, 62
Bus 4: 3, 5, 7, 10, 21, 23, 26, 33, 36, 39,
58, 60, 63, 65
Bus 5: 4, 16, 20, 28, 36, 41
Bus 6: 5, 9, 17, 21, 23

50

CHAPTER 5
CONCLUSION
5.1 Summary
Protection systems play significant roles in power systems, which help to improve the reliability
of power grids. In this thesis, basic reliability indices, such as SAIFI and SAIDI are presented.
What’s more, this thesis proposes a methodology which is based on the methodology presented
in the book “Reliability Evaluation of Power Systems” by R. Billiton, but the methodology in
this thesis is easier to implement compared to the original one proposed by R. Billiton. To
calculate the number of interruptions and interruption duration for each loads, this method uses
matrices to represent the relationship of components, which is very straightforward and
self-explanatory. SAIFI and SAIDI of IEEE-RBTS bus 2 system are calculated in six different
cases. By comparing the results of the six cases, it is obvious to see the impact of protection
systems and protection failures.
Further, based on the methodology of evaluating reliability indices, a reliability-based
optimization problem is proposed and solved by Genetic Algorithm. This optimization problem

51

is aim to find the minimum cost of investment of disconnects with reliability constraint.
Optimum solutions have been found for RBTS bus 2 and the whole RBTS system. When I first
try to solve the optimization problem, the binary programming method is used as applied in
[25-26]. However, this method is slower compare to Genetic Algorithm, especially when the
system becomes larger. Therefore, GA is chosen to solve the optimization problem. It is suitable
for this allocation optimization problem and fast to find the optimum solution.

5.2 Future Work
A

multi-objective

reliability-based

optimization

model

would

be

established.

The

pareto-optimum can be found by performing a multi-objective optimization model. In addition,
disconnect is the only protective device taken into consideration in this work. Other protective
devices, such as reclosers, relays and fuses would also be considered in an optimization problem
in the future.

52

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53

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