Electrical Power and Energy Systems 62 (2014) 9094

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Electrical Power and Energy Systems

journal homepage: www.elsevier.com/locate/ijepes

Solving capacitor placement problem considering uncertainty in load

variation
M. Mukherjee , S.K. Goswami
Electrical Engineering Department, Power System Section, Jadavpur University, Kolkata 700032, India

a r t i c l e

i n f o

Article history:
Received 18 June 2013
Received in revised form 28 March 2014
Accepted 2 April 2014
Available online 13 May 2014
Keywords:
Capacitor placement problem
Load variation
Uncertainty
Interval arithmetic

a b s t r a c t
The paper reports on the solution of the capacitor placement problem in distribution system considering
uncertainty in the variation of loads. Solution techniques available in the literature generally consider
load variation as deterministic. In the present paper uncertainty in load variation is considered using
fuzzy interval arithmetic technique. Load variations are represented as lower and upper bounds around
base levels. Both xed and switchable capacitors have been considered and results for standard test systems are presented.
2014 Elsevier Ltd. All rights reserved.

Introduction
Shunt capacitors are used in distribution systems as a source of
reactive power. If they are connected with proper location and size,
load terminal voltage can be maintained within the acceptable
limit and the line loss and total system cost can be reduced. As
the load demand on distribution system may vary with time for
effective compensation, capacitors are to be of xed as well as
switchable in nature, where a minimum capacitor kvar is always
kept connected to the system (xed capacitor) and additional
capacitors are switched in or out as the load demand varies. Determination of the size, location and type of such capacitors for a distribution system is a complex optimization problem and requires
information regarding the load variation of the system with time.
Different solution techniques had been presented by many
researchers in the past for solving the problem of placing capacitor
in distribution system. Modied discrete PSO based solution was
proposed in [3,20]. In [4,5], the capacitor placement was formulated as a mixed integer non-linear problem. [6,16,17] proposes
Particle Swarm Optimization (PSO) based capacitor placement.
Loss saving equation based technique was proposed in [7]. In [8]
heuristics and greedy search technique based solution was proposed. Fuzzy reasoning based method was proposed in [9]. Simulated annealing was proposed in [15] and Genetic Algorithm
based solution has taken in [10,24] respectively. Interior point
based solution was proposed in [11,14]. Extended Dynamic Programming Approach was proposed in [12], Plant Growth
Corresponding author. Tel.: +91 9734248638.
E-mail address: manas202006@yahoo.com (M. Mukherjee).
http://dx.doi.org/10.1016/j.ijepes.2014.04.004
0142-0615/ 2014 Elsevier Ltd. All rights reserved.

Simulation Algorithm and using of loss sensitivity factor was

proposed in [13], heuristic search and node stability based method
was proposed in [18], and bacterial foraging solution was proposed
in [21]. Hybrid honey bee colony algorithm based solution was
proposed in [23] Uncertainty was taken into account in [19].
In all of the solution techniques load demand was assumed to
follow a denite pattern-represented by a number of xed load
levels. In reality however, the load demand is quite uncertain
and depends upon many factors in such a way that it is impossible
to predict the actual load before the actual occurrence. Load forecasts, based upon historic records of load variation can predict a
coarse picture of the probable situation. The actual scenario may
well deviate the predicted one by a considerable margin. Thus
instead of load representation by a number of denite load levels,
probabilistic variation of loads would be a better representation.
The capacitor placement decision based upon the xed pattern of
load variation thus may lead to an inferior solution than the solution where probability of load variation over the predicted one is
considered. The present paper thus proposes a method to take
uncertainty of the load variation in the capacitor placement
problem.
Problem formulation
For a distribution network, the loss associated with the reactive
components of branch currents can be written as

PLr

n
X
I2ri  Ri
i1

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M. Mukherjee, S.K. Goswami / Electrical Power and Energy Systems 62 (2014) 9094

Nomenclature
PLr

Iri
Inew
ri
Ri
Iril, Iriu

active power loss of the system associated with the

reactive components of branch currents for original system
reactive component of branch current of the ith branch
for original system
reactive component of branch current of the ith branch
for compensated system
resistance, respectively of the ith branch
lower and upper limit of Iri respectively

where Iri and Ri are the reactive component of branch current and
resistance, respectively of the ith branch.
But, actually Iri is not of xed value. Because, the load variation
in any power system cannot be truly represented by a single load
curve. Conventional way of representing load variation by a single
load curve basically represents the mean of load variation. A better
representation would be to use a curve like Fig. 1, where instead of
representing by a mean variation, the range of variation is shown.
So in the load duration curve, each load level is represented by a
range of load levels (like Fig. 2) rather than a single load. So it is
better to represent Iri as

Iri Iril ; Iriu 

where Iril and Iriu are lower and upper limit of Iri respectively.
Because of this variation in this pattern of the loads, the loss PLr
should be considered as an interval quantity instead of xed quantity. Therefore, in capacitor placement problem every quantity
should be considered as an interval quantity. For this purpose basic
operation of interval number is to be known which is described in
the next section.
Interval arithmetic
An interval number X = [xl, xu] is the set of real numbers x such
that xl 6 x 6 xu; xl and xu are known as the lower limit and upper
limit of the interval number, respectively. A rational number k is
represented as an interval number K = [k, k].
Let X = [xl, xu] and Y = [yl, yu] be the two interval numbers.
Then addition, subtraction, multiplication and division of these
two interval numbers are dened as below [22]:

X Y xl yl; xl yu

X  Y xl  yu; xu  yl

X  Y minxl  yl; xl  yu; xu  yl; xu  yu; maxxl  yl; xl

 yu; xu  yl; xu  yu

Fig. 1. Load curve considering load variation.

Ic
Pcom
Lr

S
Vm
k
Qc
Vc

reactive current drawn by the capacitor

active power loss of the system associated with the
reactive components of branch currents for compensated system
loss saving
magnitude of voltage of bus m before compensation
number of capacitor buses
capacitor size
voltage magnitude vector of capacitor bus

X  Y X  Y 1

where

Y 1 1=yu;1=ylif 0 R yl; yu

Also, the distance between these two interval numbers is

dened as [24]:

qX; Y maxjx1  y1j; jx2  y2j

For power system application, calculations involving complex

numbers, rather than real numbers are needed. Hence, in the next
sub-section, basic operations involving complex interval numbers
are presented.
Complex interval number
Any complex number Z = X + iY; where i is the complex operator, is said to be a complex interval number if both its real part
(X) and the imaginary part (Y) are interval numbers. Hence, X can
be represented as X = [x1, x2] and Y can be represented as
Y = [y1, y2], where, x1, y1 are the lower limits and x2, y2 are the
upper limits, respectively. The conjugate of a complex interval
number is given by Z = X  iY: Let Z1 = A1 + iB1 and Z2 = A2 + iB2
be two complex interval numbers. Then the addition, subtraction,
multiplication and division of these two complex interval numbers
are dened as [22]

Z 1 Z 2 A1 A2 iB1 B2

Z 1  Z 2 A1  A2 iB1  B2

Z 1  Z 2 A1  A2  B1  B2 iA1  B2 A2  B1

10

Z 1  Z 2 C iD

11

where C = (A1  A2 + B1  B2) 

(A22 + B22).

(A22

B22)

and D = (A2  B1  A1  B2) 

Fig. 2. Load duration curve considering load variation.

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M. Mukherjee, S.K. Goswami / Electrical Power and Energy Systems 62 (2014) 9094

It is to be noted that, Eqs. (8)(11) can be evaluated by applying

the fundamental operations as dened in Eqs. (2)(5).
Solution technique of the capacitor placement problem
The solution of the capacitor placement problem has two components the location of the capacitor and the size of capacitor at
each location. It is obvious that, location of buses must be a xed
number. So to nd the location, the method described in [7] is used
where only the base load will be considered. Once the location is
known, interval load will be used to nd the size of the required
capacitor.

ful when more than one capacitor is to be placed. So the size of multiple capacitors for optimal location is to be determined
simultaneously and the procedure of nding optimal sizes is
described in the following sections.
Size of capacitors for interval load
Here also the method described in [7] is used with a different
manner. First load ow solution is done taking interval load [1,2].
This will provide lower and upper limit of all the branch currents.
Let the followings are considered:
k = number of capacitor buses
Ic = k-dimensional vector consisting of capacitor currents.

Determination of location
A radial distribution system with n branches is considered here.
Let a capacitor C is placed at bus m (except the source bus) and a be
a set of branches connected between the source and capacitor
buses. The capacitor draws a reactive current Ic and as it is a radial
network it changes only the reactive component of current of
branch set a. The current of other branches (Ra) is remain
unchanged. Thus, the new reactive current Inew
of the ith branch
ri
is given by

Actually Ic = [Icl, Icu], where Icl and Icu are the lower an upper
limit of Ic respectively.

Inew
Iri Di  Ic
ri

Dij 1; if branch i 2 a 0; otherwise

12

(j = 1, 2, . . ., k)
D = a matrix of dimension n  k
The elements of D are considered as

When the capacitors are placed in the system, the new reactive
component of the branch current is given by,

where

Di 1 if branch i 2 a

Inew
ri  I r  D  Ic 

0 otherwise
Here Iri, is the reactive current of the ith branch in the original system obtained from the load ow solution. The loss Pcom
Lr , associated
with the reactive component of branch currents in the compensated
system (when the capacitor is connected) can be written as

Pcom

Lr

aj = set of branches from the source bus to the jth capacitor bus

n
X
Iri Di  Ic2  aRi

13

i1

As Ir and Ic are interval number,

will be also interval
number.
The loss Pcom
associated with the new reactive currents in the
Lr
compensated system is

Pcom

Lr

n
X

Iri

i1

The loss saving S is the difference between Eqs. (1) and (13) and
is given by
n
X
S PLr  Pcom
 2Di  Iri  Ic Di  Ic2  Ri
Lr

14

i1

n
X
@S
 2Di  Iri  Ic Di  Ic2  Ri 0
@Ic
i1

15

Thus the capacitor current for the maximum loss saving is

Ic

n
X
Di  Iri  Ri
i1

X
Iri  Ri
i2a

!,

!,

n
X

!
Di  Ri

i1

X
Ri

16

i2a

The corresponding capacitor size is Q c V m  Ic

17

Here Vm is the magnitude of voltage of bus m before compensation.

The above steps are repeated for all the buses (except the root bus)
to get the highest possible loss saving for a singly located capacitor.
The bus for which highest loss saving is obtained is termed as candidate bus. When the candidate bus is identied and compensated
using Eq. (17), the above technique is again used to identify the next
and subsequent candidate buses. That will provide only the locations where the capacitors are to be placed. Obviously capacitors
obtained from Eq. (17) are local optimal value. So they are not use-

k
X

!2
Dij  Icj

 Ri

19

j1

The loss saving S obtained by placing the capacitors is the difference between Eqs. (1) and (19) and is given by

"
!#
n
k
k
X
X
X
2
S
2Iri 
Dij  Icj
Dij  Icj
 Ri
i1

The capacitor current Ic, that provides the maximum loss saving
can be obtained from

18
Inew
ri

j1

20

j1

The optimal capacitor currents for the maximum loss saving can
be obtained by solving the following equations:

@S
0
@Ic1
@S
0
@Ic2
...
...
@S
0
@Ick

21

After some mathematical manipulations, Eq. (21) can be

expressed by a set of linear algebraic equations as follows:

A  Ic B

22

where A is a k x k square matrix and B is a k-dimensional vector. The

elements of A and B are given by

"
#
X X


Ajj Ajjl ; Ajju
Ri ;
Ri
i2aj

i2aj

23

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M. Mukherjee, S.K. Goswami / Electrical Power and Energy Systems 62 (2014) 9094

"
Ajm Ajml ; Ajmu 

Ri ;

i2aj\am

#
Ri

24

i2aj\am

Table 2
Possible choice of capacitor sizes and cost/kVAR.
Sl. No.
kVAR
($/kVAR)

1
150
.500

2
300
.350

3
450
.253

4
600
.220

5
750
.276

6
900
.183

Sl. No.
kVAR
($/kVAR)

7
1050
.228

8
1200
.170

9
1350
.207

10
1500
.201

11
1650
.193

12
1800
.187

25

Sl. No.
kVAR
($/kVAR)

13
1950
.211

14
2100
.176

15
2250
.197

16
2400
.170

17
2550
.189

18
2700
.187

where Bjl and Bju are lower and upper limit of Bj respectively. Also Iril
and Iriu are lower and upper limit of branch reactive current
respectively.
Only the branch resistances and reactive currents in the original
system are required to nd the elements of A and B. The capacitor
currents for the highest loss saving can be obtained as

Sl. No.
kVAR
($/kVAR)

19
2850
.183

20
3000
.180

21
3150
.195

22
3300
.174

23
3450
.188

24
3600
.170

Sl. No.
kVAR
($/kVAR)

25
3750
.183

26
3900
.182

27
4050
.179

where Ajjl and Ajju are the lower and upper limit of Ajj; and Ajml and
Ajmu are lower and upper limit of Ajm and both are same as branch
parameters are considered to be xed.

Bj Bjl ; Bju 

"
X
i2aj

Iril  Ri ;

X
Iriu  Ri

i2aj

Ic A1  B

26

As A and B are both interval quantity, R.H.S. of this equation is

interval quantity. So the value of Ic will be in the form [Icl Icu],
where Icl and Icu are lower and upper limit of capacitor current
respectively.
Once the capacitor currents are known, the optimal capacitor
sizes can be written as

Qc Vc  Ic

27

Here Vc is the voltage magnitude vector of capacitor buses, whose

value is like [Vcl, Vcu], where Vcl and Vcu are lower and upper limit
of Vc respectively. So Qc will be also an interval number which
would have a lower and upper limit.
Numerical results
The proposed method is tested for 10 bus system. The load and
bus data is given in Table 1 which is considered as base load. Cost
of energy is taken as 0.06 $/kW h and capacitor cost is obtained
from Table 2[9]. In this work the nearest value of the capacitor corresponding to Appendix-2 is taken, so that the cost of the capacitors can be calculated. For interval load, 5% of the base load is
taken.
Using Eqs. (14) and (17), corresponding loss saving is calculated
for each and every bus except the source bus i.e. Bus No. 0. Then it
is noticed that highest loss saving can be achieved if Bus No. 4 is
compensated with a capacitor of size 3998.5 kVAR and total loss
saving of 81 MW is obtained. So this bus is compensated with
the capacitor of 3998.5 kVAR. After that repeating the same process it is observed that highest loss saving is achieved if Bus No.
8 is compensated with capacitor of 852.09 kVAR and total loss saving of 14 MW is obtained. After that no such signicant loss saving

Table 1
Data for 10-Bus system.
From bus

0
1
2
3
4
5
6
7
8

To bus

1
2
3
4
5
6
7
8
9

Impedance

Load connected at to bus

R (ohm)

X (ohm)

kW

kVAR

0.1233
0.0140
0.7463
0.6984
1.9831
0.9053
2.0552
4.7953
5.3434

0.4127
0.6051
1.2050
0.6084
1.7276
0.7886
1.1640
2.7160
3.0264

1840
980
1790
1598
1610
780
1150
980
1640

460
340
446
1840
600
110
60
130
200

Bus 0 is the substation node, the voltage of which is xed at 23 kV.

can be achieved using this process. So it can be concluded that

optimal location is 4 and 8.
Then A matrix is formed as

0:0060

0:0060

0:0060 0:0428


p:u:

Solving interval load ow, lower and upper limits of B matrix is

obtained as


Blowerlimit

0:0140
0:0361


p:u: and Bupperlimit

0:113
0:0265


p:u:

So the lower and upper limit of required capacitor value nearer

to Appendix-2 is obtained as 2700 kVAR and 3300 kVAR for Bus No.
4 and 750 kVAR and 1050 kVAR for Bus No. 8 respectively. The
result is summarized in Tables 3 and 4.

Application to switched capacitor placement problem

considering load uncertainty
This method can be also applied to solve the switched capacitor
placement problem. It is assumed that the variation of the load is
conformal [4,5], the capacitor kVAR required in a particular load
level should be at least equal to that required at the immediate
lower load level. So the switched capacitor placement problem is
solved starting from the lowest load level and the capacitor
installed at a lower load level will be considered as xed capacitors
for all the higher load levels. The method is applied for the same 10
bus system. The load duration data are given in Table 5. It is
assumed that the substation voltage is 1.05 p.u. at peak load condition and 1.0 p.u. during the remaining periods [4]. The result is
summarized in Tables 6 and 7.
From the results interval of bus voltages, required capacitor
kVAR, system cost of the compensated are obtained by which all
these thing are whether in an acceptable limit or not can be
determined

Table 3
Required capacitor for 10-Bus system.
Bus no.

4
8

Capacitor size(kVAR)
Lower limit

Upper limit

2700
750

3300
1050

94

M. Mukherjee, S.K. Goswami / Electrical Power and Energy Systems 62 (2014) 9094

instead of sequential solution. This may be done by using robust

optimization technique which is being pursued by the authors at
present.

Table 4
Comparison between original system and compensated system (10 Bus).
Original system

Compensated system

Active power
loss (kW)

Lower
limit
629

Upper
limit
974

Active power
loss (kW)

Lower
limit
521

Upper
limit
941

Annual cost ($)

Lower
limit
105,672

Upper
limit
163,632

Annual cost ($)

Lower
limit
88,240

Upper
limit
158,900

Table 5
Load duration data for the test system.
Load level

Per unit load

Load duration (h)

0.3
1000

0.6
6760

1.1
1000

Table 6
Required capacitor for 10-Bus system for different load level.
Load level

Bus No.

4
8
4
8
4
8

2
3

Capacitor sizes(kVAR)
Lower limit

Upper limit

750
150
1650
450
3000
750

900
150
1800
450
3450
1050

Table 7
Comparison of system cost between original system and compensated system (10
Bus).
Original system

Compensated system

Lower limit

Upper limit

Lower limit

Upper limit

$95,809

$126,820

$88,900

$116,990

Conclusion
The present paper reports a new formulation of the capacitor
placement problem considering uncertainty in the variation of distribution system load. Unlike the conventional approaches of considering the load variation by simply using a number of load levels,
the present paper represents the load variation by upper and lower
bounds of the loads at different load levels and introduces interval
arithmetic to incorporate the effect of such variation in the solution of the capacitor placement problem. As the basic aim of the
paper is to introduce the interval arithmetic technique in the
capacitor placement problem, the solution approach had been
made simple by separating the problem of placing and sizing of
capacitors. The capacitor locations are rst selected based upon
the sensitivity of capacitor introduction at the location. Once locations are determined, the size is then found out. The problem however is not decoupled and should rather be solved simultaneously

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