This document discusses voltage sag analysis on a 14-bus power system network. It provides the Y-admittance and Z-impedance matrices of the network and calculates the voltage sag at buses 5 and 14 for various fault conditions. It finds that a three-phase fault at bus 5 results in the highest voltage sag of 1.57 p.u at bus 5, while a fault at bus 14 causes a sag of 1.39 p.u at bus 14. Opening certain lines or removing generators increases the voltage sag. Under expected fault rates, the estimated number of annual voltage sag events below 50% of nominal voltage is 4.65 for bus 5 and 6.55 for bus 14.

1. FACULTY OF ENGINEERING
(INDUSTRIAL ELECTRONIC AND CONTROL)
Semester II
Academic Session 2016
POWER QUALITY IN INDUSTRY
(KXGK6104)
ASSIGNMENT #1:
VOLTAGE SAGS EVALUATION STUDIES
KHAIRI AHMED ELRMALI
KGK1500011
SUBMITTED TO:DR. HAZLIE BIN MOKHLIS
SUBMITTED DATE: 4TH
JUNE 2016

2. Given below is an IEEE test system of 14-bus network:
Based on the data system given,figure 1 and 2 below are the Y-admittance matrix and Z
impedance matrix using MATLAB software. Here are some coding to obtain these matrices;
% Program to form Admittance And Impedance Bus Formation...
% Bus bus R X B/2 distance
% fr to p.u p.u p.u
linedata=[ 1 2 0.01938 0.05917 0.0528/2 12
1 5 0.05403 0.22304 0.0492/2 35
2 3 0.04699 0.19797 0.0438/2 20
2 4 0.05811 0.17632 0.0374/2 18
2 5 0.05695 0.17388 0.034/2 23
3 4 0.06701 0.17103 0.0346/2 19
4 5 0.01335 0.04211 0.0128/2 5
4 7 0 0.20912 0 0
4 9 0 0.55618 0 0
5 6 0 0.25202 0 0

3. 6 11 0.09498 0.1989 0 24
6 12 0.12291 0.25581 0 61
6 13 0.06615 0.13027 0 23
7 8 0 0.17615 0 0
7 9 0 0.11001 0 0
9 10 0.03181 0.0845 0 16
10 14 0.12711 0.27038 0 22
10 11 0.08205 0.19207 0 16
12 13 0.22092 0.19988 0 21
13 14 0.17093 0.34802 0 30];
%==============///### calculate voltage sag ###///======
% bus voltage angle
v_bus = [1 1.06 0
2 1.045 -4.98
3 1.01 -12.72
4 1.019 -10.33
5 1.02 -8.78
6 1.07 -14.22
7 1.062 -13.37
8 1.09 -13.36
9 1.056 -14.94
10 1.051 -15.1
11 1.057 -14.79
12 1.055 -15.07
13 1.05 -15.16
14 1.036 -16.04];
%==============### Generator data ###==========
% generator X capacity
gendata = [ 1 0.12 250
2 0.15 100
3 0.10 80
6 0.15 50
8 0.1 50 ];
base = 100;
% nl=linedata(:,1); % From bus number..
% nr=linedata(:,2); % To bus number...
% R=linedata(:,3); % Resistance, R...
% X=linedata(:,4); % Reactance, X...
nbranch=length(linedata(:,1)); % no. of branches...
nbus=max(v_bus(:,1)); % no. of buses...
Zline=R+j*X; % Z matrix...

4. y=ones(nbranch,1)./Zline; % To get inverse of each element...
Y=zeros(nbus,nbus); % Initialize YBus...
% Formation of the Off Diagonal Elements...
for k=1:nbranch;
Y(linedata(k,1),linedata(k,2))= -1/Zline(k);
Y(linedata(k,2),linedata(k,1))= -1/Zline(k);
end
aaa = zeros(nbus,1);
% Formation of Diagonal Elements....
for k=1:nbus
for l=1:nbranch
if((k==linedata(l,1))||(k==linedata(l,2)))
aaa(k,:)=1/(linedata(l,3)+linedata(l,4)*i+linedata(l,5)*i)+ aaa(k,:);
Y(k,k)=aaa(k,:);
end
end
end
YY = Y;
%========= To add generator sub-transient data into the Y matrix
for ia=1:nbus
for ib = 1:length(gendata)
if (ia == gendata(ib,1))
Y(ia,ia) = Y(ia,ia) + 1/(gendata(ib,2)*gendata(ib,3)/base)
end
end
end
Y; % Bus Admittance Matrix
Z = inv(Y); % Bus Impedance Matrix

5. Figure 1: Y-admittance matrix(data workspace)

6. Figure 2: Z-impedance matrix (data workspace)
Question1
1. Calculate voltage sag at bus 5 and 14 when three-phase-fault occurs at each bus in the
system.The pre-fault voltage (per unit) of all buses are given below:

7. Voltage sag formula:
Data for the pre-fault voltage (p.u) of
all the buses are convert into real and
imaginary form;
Figure 4 is obtain from these coding below;
% calculate angle
for ia = 1:nbus
ang = v_bus(ia,3)*3.142/180;
mag = v_bus(ia,2);
[a,b] = pol2cart(ang,mag);
v_bus_new(ia,:) = [a + b*i];
end
Figure 3: Cartesian form
Figure 4 is obtain from these coding below;
% voltage sag at 5
i_s = 5;
for i_f = 1:nbus;
V_val5(i_f,:) = v_bus_new(i_s,1) - v_bus_new(i_f,1)*Z(i_s,i_f)/Z(i_f,i_f);
end
V_val5
Figure 4 : voltage sag at 5
Figure5 is obtain from these coding below;
% voltage sag at 14
i_s = 14;
ff
fi
fii
Z
Z
VVVsag −=

8. for i_f = 1:nbus;
V_val14(i_f,:) = v_bus_new(i_s,1) - v_bus_new(i_f,1)*Z(i_s,i_f)/Z(i_f,i_f);
end
V_val14
Figure5: voltage sag at 14
% %% polar form
Figure 6 is obtain from these coding below;
for ia=1:nbus
[x1,y1] = cart2pol(real(V_val5(ia,1)),imag(V_val5(ia,1)));
plot_y(ia,1) = y1;
[x2,y2] = cart2pol(real(V_val14(ia,1)),imag(V_val14(ia,1)));
plot_y(ia,2) = y2;
end
%
Vsag1
4Vsag5
1 2 3 4 5 6 7 8 9 10 11 12 13 14
0
0.5
1
1.5
voltage sag at bus 5
line bus of system
voltagesagmagnitude
1 2 3 4 5 6 7 8 9 10 11 12 13 14
0
0.2
0.4
0.6
0.8
1
1.2
1.4
voltage sag at bus 14
line bus of system
voltagesagmagnitude

9. Figure 6: Bar graph for both voltage sag at bus 5 and bus 14
Question2
(a) Line 11 –10 and line 13 -14 is open.
Figure 7 and 8 are the Y-admittance and Z-impedance matrices when line 11-10 and line 13-14
are open

10. Figure 7: Y-admittance matrix when line 11-10 and line 13-14 is open
Figure 8: Z-impedance matrix when line 11-10 and line 13-14 is open
Figure 9 and 10 are the values of both voltage sag at bus 5 and 14 respectively when the line 11-
10 and line 13-14 are open.

11. Figure 9: Voltage sag at bus 5&14 when lines are open
Figure 10: bar graph of voltage sag at bus 5&14 when lines are open
1 2 3 4 5 6 7 8 9 10 11 12 13 14
0
0.5
1
1.5
system of 14-bus network
voltagesagmagnitude
Voltage sag at bus 5 line 11-10 & line 13-14 are open
1 2 3 4 5 6 7 8 9 10 11 12 13 14
0
0.2
0.4
0.6
0.8
1
1.2
1.4
system of 14 bus network
voltagesagmagnitude
voltage sag at 14 when line 10-11 and 13-14 are open
1 2 3 4 5 6 7 8 9 10 11 12 13 14
0
0.2
0.4
0.6
0.8
1
1.2
1.4
system 14 bus network
voltagemagnitude
comparison between voltage sag at 14 present lines & open lines
1 2 3 4 5 6 7 8 9 10 11 12 13 14
0
0.5
1
1.5
system of 14 bus network
voltagesagmagnitude
Comparison between valtage sag at 5 present lines & open lines

12. Figure 11 : comparison between voltage sag at bus 5&14
Q2 cont....
(b) Generator at bus 6 and 8 are taken out from the system.
Figure 12 and 13 are the Y-admittance and Z-impedance when generator at bus 6 and 8 are taken
out from the system.

13. Figure12: Y-admittance when generator at bus 6 and bus 8 are taken out from system

14. Figure13: Z-impedance when generator at bus 6 and bus 8 are taken out from system
Figure 14 and 15 are the values of voltage sag at bus 5 and 14 respectively when the generator 6
and 8 are taken out from the system.
Figure 14: voltage sag at bus 5 and 14 when generator are taken out
1 2 3 4 5 6 7 8 9 10 11 12 13 14
0
0.5
1
1.5
2
2.5
system of 14 bus network
voltagesagmagnitude
comparison between voltage sag at bus 5 present & generator taken out
1 2 3 4 5 6 7 8 9 10 11 12 13 14
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Comparison between voltage sag at 14 present & generator taken out
system of 14 bus network
voltagesagmagnitude

15. Figure 15 : comparison between voltage sag at bus 5&14
Question 3
*Short circuit
event = 5
events/100km/year
Figure 16 is obtain from these coding below;
for ia =1:nbranch
Event_data(ia,:) = [linedata(ia,:) linedata(ia,6)*5/100];
end
Event_data
eventcicuitShortc
100km
(km)Distance
frequencyVsagExpected ×=

16. Figure 16 :Expected V sag frequency
Estimation of voltage sag on a line can be calculated as the average of voltage sag between the
two connected busses
Figure 17 average of voltage sag between the two connected busses occurred at bus 5 & bus 14
for each lines.
is obtain from these coding below;
for ia = 1:nbranch
Vol_Event_data5(ia,1)= (V_val5(linedata(ia,1),1) + V_val5(linedata(ia,2),1))/2;
Vol_Event_data14(ia,1)= (V_val14(linedata(ia,1),1) + V_val14(linedata(ia,2),1))/2;
end
for ia = 1:nbranch

17. [x1,y1] = cart2pol (real(Vol_Event_data5(ia,1)),imag(Vol_Event_data5(ia,1)));
V_event5(ia,1) = y1;
[x1,y1] = cart2pol (real(Vol_Event_data14(ia,1)),imag(Vol_Event_data14(ia,1)));
V_event14(ia,1) = y1;
end
Event_val = [linedata(:,1) linedata(:,2) Event_data(:,7) V_event5(:,1) V_event14(:,1)];
Figure 17 average of voltage sag occurred at bus 5 & bus 14 for each lines.
Estimation of voltage event under 50% of the nominal 1.0 p.u a year for Bus 5: Voltage sag
< 0.5 :
To obtain from these coding below”
ib=1;
ic=1;
for ia=1:nbranch
if(Event_val(ia,4)<0.5) %VOLTAGE AS SEEN AT BUS 5
store5(ib,:) = [Event_val(ia,1:4)];

18. ib = ib +1;
end
Events_under_50_bus_5 = sum(store5(:,3))
Estimation of voltage event = 0.6 + 1.75 + 0.9 + 1.15 + 0.25 = 4.6500
Events_under_50_bus_5 =
4.6500
Estimation of voltage event under 50% of the nominal 1.0 p.u a year for Bus 14: Voltage
sag < 0.5 :
To obtain from these coding below”
ib=1;
ic=1;
for ia=1:nbranch
if(Event_val(ia,5)<0.5) %VOLTAGE AS SEEN ATBUS 14
store14(ic,:) = [Event_val(ia,1:3) Event_val(ia,5)];
ic = ic +1;
end
end
Events_under_50_bus_14= sum(store14(:,3)) % script file
Estimation of voltage event = 1.2 + 1.15 + 0.8 + 1.1 + 0.8 + 1.5 = 6.55
Events_under_50_bus_14 = % script file
6.5500
1 2 3 4 5 6
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
voltage event under 50% for Bus 5 & Bus 14
bus network
averagevoltagesag

19. Estimation of voltage event under 50%
Question 4
1. Based on the analysis the number of voltage sag at bus 5 and bus 14 according to the sag
magnitude level as in the following table (% nominal 1.0 pu).
%% less than 10 percent
ib=1;
ic=1;
for ia=1:nbranch
if(Event_val(ia,4)<0.1)
store5_10(ib,:) = [Event_val(ia,1:4)];
ib = ib +1;
end
if(Event_val(ia,5)<0.1)
store14_10(ic,:) = [Event_val(ia,1:3) Event_val(ia,5)];
ic = ic +1;
end
end
if (ib <2)
b5_10 = 0;
else
Vsag (%) 0-10 10-20 20 - 30 30- 40 40 - 50 50- 60 60 70 70- 80 80 - 90 90- 100
Number of
sag at
bus 5
3.50 1.15 0 0 0 0 0 1.0 0.95 0
Number of
sag at
bus 14
0 1.10 1.5 1.15 2.8 4.1 0.0 4.65 0.0 1.0

20. b5_10 = sum(store5_10(:,3));
end
if (ic <2 )
b14_10 = 0;
else
b14_10 = sum(store14_10(:,3));
End
Q5
By analyzing from the above graph we have seen that in Figure below on the estimation of
voltage event under 50% of the nominal a year for both bus 5 and 14, given that the value
estimation for bus 5 is lower than the value estimation of bus 14 which is 4.65 and 6.55
respectively.
To place a factory manufacturing electronic components, the place at bus 5 is more suitable place
rather than at bus 14 since it gives lower voltage sags under the voltage event.

21. Q6
The number of voltage sags that can occur at your facility depends on where you're located, the
characteristics of your utility's distribution system (underground vs. overhead, lengths of the
distribution feeder circuits, and number of feeders), lightning level in the area, number of trees
adjacent to the power lines, and several other factors.
 For improve lines and area in proper stability, we should arrange some methods and steps
thus system will back on its stability though fault is occurred in other lines. We have
introduces capacitor bank or Super Capacitor in the transmission lines
 Magnetic synthesizers, which are 3-phase devices that take advantage of their 3-phase
magnetics to provide improved voltage sag support and regulation.
Useful link
-http://ecmweb.com/content/dealing-voltage-sags-your-facility

22. clear all
clc
% Program to form Admittance And Impedance Bus Formation...
% Bus bus R X B/2 distance
% fr to p.u p.u p.u
linedata = [1 2 0.01938 0.05917 0.0528/2 12
1 5 0.05403 0.22304 0.0492/2 35
2 3 0.04699 0.19797 0.0438/2 20
2 4 0.05811 0.17632 0.0374/2 18
2 5 0.05695 0.17388 0.034/2 23
3 4 0.06701 0.17103 0.0346/2 19
4 5 0.01335 0.04211 0.0128/2 5
4 7 0 0.20912 0 0
4 9 0 0.55618 0 0
5 6 0 0.25202 0 0
6 11 0.09498 0.1989 0 24
6 12 0.12291 0.25581 0 61
6 13 0.06615 0.13027 0 23
7 8 0 0.17615 0 0
7 9 0 0.11001 0 0
9 10 0.03181 0.0845 0 16
10 14 0.12711 0.27038 0 22
10 11 0.08205 0.19207 0 16
12 13 0.22092 0.19988 0 21
13 14 0.17093 0.34802 0 30];

23. %==============///### calculate voltage sag ###///======
% bus voltage angle
v_bus = [1 1.06 0
2 1.045 -4.98
3 1.01 -12.72
4 1.019 -10.33
5 1.02 -8.78
6 1.07 -14.22
7 1.062 -13.37
8 1.09 -13.36
9 1.056 -14.94
10 1.051 -15.1
11 1.057 -14.79
12 1.055 -15.07
13 1.05 -15.16
14 1.036 -16.04];
%==============### Generator data ###==========
% generator X capacity
gendata = [1 0.12 250
2 0.15 100
3 0.10 80
6 0.15 50
8 0.1 50];
base = 100;
% nl=linedata(:,1); % From bus number..
% nr=linedata(:,2); % To bus number...
R=linedata(:,3); % Resistance, R...
X=linedata(:,4); % Reactance, X...
nbranch=length(linedata(:,1)); % no. of branches...
nbus=max(v_bus(:,1)); % no. of buses...
Zline=R+j*X; % Z matrix...
y=ones(nbranch,1)./Zline; % To get inverse of each element...
Y=zeros(nbus,nbus); % Initialise YBus...
% Formation of the Off Diagonal Elements...
for k=1:nbranch;
Y(linedata(k,1),linedata(k,2))= -1/Zline(k);
Y(linedata(k,2),linedata(k,1))= -1/Zline(k);
end
aaa = zeros(nbus,1);

24. % Formation of Diagonal Elements....
for k=1:nbus
for l=1:nbranch
if((k==linedata(l,1))||(k==linedata(l,2)))
aaa(k,:)=1/(linedata(l,3)+linedata(l,4)*i+linedata(l,5)*i)+ aaa(k,:);
Y(k,k)=aaa(k,:);
end
end
end
YY = Y;
%========= To add generator sub-transient data into the Y matrix
for ia=1:nbus
for ib = 1:length(gendata)
if (ia == gendata(ib,1))
Y(ia,ia) = Y(ia,ia) + 1/(gendata(ib,2)*gendata(ib,3)/base);
end
end
end
Y; % Bus Admittance Matrix
Z = inv(Y); % Bus Impedance Matrix
% calculate angle
for ia = 1:nbus
ang = v_bus(ia,3)*3.142/180;
mag = v_bus(ia,2);
[a,b] = pol2cart(ang,mag);
v_bus_new(ia,:) = [a + b*i];
end
%% Q 1 /// voltage sag calculation for faulted bus ////
% voltage sag at 5 and 14

25. i_s = 5;
for i_f = 1:nbus;
%V_val5(i_f,:) = v_bus(i_s,2) - v_bus(i_f,2)*Z(i_s,i_f)/Z(i_f,i_f);
V_val5(i_f,:) = v_bus_new(i_s,1) - v_bus_new(i_f,1)*Z(i_s,i_f)/Z(i_f,i_f);
end
V_val5;
i_s = 14;
for i_f = 1:nbus;
%V_val10(i_f,:) = v_bus(i_s,2) - v_bus(i_f,2)*Z(i_s,i_f)/Z(i_f,i_f);
V_val14(i_f,:) = v_bus_new(i_s,1) - v_bus_new(i_f,1)*Z(i_s,i_f)/Z(i_f,i_f);
end
V_val14;
%
% %% polar form
for ia=1:nbus
[x1,y1] = cart2pol(real(V_val5(ia,1)),imag(V_val5(ia,1)));
plot_y(ia,1) = y1;
[x2,y2] = cart2pol(real(V_val14(ia,1)),imag(V_val14(ia,1)));
plot_y(ia,2) = y2;
end
%
%% Question 3
for ia =1:nbranch
Event_data(ia,:) = [linedata(ia,:) linedata(ia,6)*5/100];
end
Event_data
for ia = 1:nbranch
Vol_Event_data5(ia,1)= (V_val5(linedata(ia,1),1) + V_val5(linedata(ia,2),1))/2;
Vol_Event_data14(ia,1)= (V_val14(linedata(ia,1),1) + V_val14(linedata(ia,2),1))/2;
end
for ia = 1:nbranch
[x1,y1] = cart2pol (real(Vol_Event_data5(ia,1)),imag(Vol_Event_data5(ia,1)));
V_event5(ia,1) = y1;
[x1,y1] = cart2pol (real(Vol_Event_data14(ia,1)),imag(Vol_Event_data14(ia,1)));
V_event14(ia,1) = y1;

26. end
Event_val = [linedata(:,1) linedata(:,2) Event_data(:,7) V_event5(:,1) V_event14(:,1)];
ib=1;
ic=1;
for ia=1:nbranch
if(Event_val(ia,4)<0.5) %VOLTAGE AS SEEN AT BUS 5
store5(ib,:) = [Event_val(ia,1:4)];
ib = ib +1;
end
if(Event_val(ia,5)<0.5) %VOLTAGE AS SEEN AT BUS 14
store14(ic,:) = [Event_val(ia,1:3) Event_val(ia,5)];
ic = ic +1;
end
end
Events_under_50_bus_5 = sum(store5(:,3))
Events_under_50_bus_14= sum(store14(:,3))
%
% %%Question 4
%% less than 10 percent
ib=1;
ic=1;
for ia=1:nbranch
if(Event_val(ia,4)<0.1)
store5_10(ib,:) = [Event_val(ia,1:4)];
ib = ib +1;
end
if(Event_val(ia,5)<0.1)
store14_10(ic,:) = [Event_val(ia,1:3) Event_val(ia,5)];
ic = ic +1;
end
end
if (ib <2)
b5_10 = 0;
else

27. b5_10 = sum(store5_10(:,3));
end
if (ic <2 )
b14_10 = 0;
else
b14_10 = sum(store14_10(:,3));
end
%% between 10 - 20 percent
ib=1;
ic=1;
for ia=1:nbranch
if(Event_val(ia,4)>0.1 && Event_val(ia,4)<0.2)
store5_20(ib,:) = [Event_val(ia,1:4)];
ib = ib +1;
end
if(Event_val(ia,5)>0.1 && Event_val(ia,5)<0.2)
store14_20(ic,:) = [Event_val(ia,1:3) Event_val(ia,5)];
ic = ic +1;
end
end
if (ib <2)
b5_20 = 0;
else
b5_20 = sum(store5_20(:,3));
end
if (ic <2 )
b14_20 = 0;
else
b14_20 = sum(store14_20(:,3));
end
%% between 20 - 30 percent
ib=1;
ic=1;
for ia=1:nbranch
if(Event_val(ia,4)>0.2 && Event_val(ia,4)<0.3)
store5_30(ib,:) = [Event_val(ia,1:4)];

28. ib = ib +1;
end
if(Event_val(ia,5)>0.2 && Event_val(ia,5)<0.3)
store14_30(ic,:) = [Event_val(ia,1:3) Event_val(ia,5)];
ic = ic +1;
end
end
if (ib <2)
b5_30 = 0;
else
b5_30 = sum(store5_30(:,3));
end
if (ic <2 )
b14_30 = 0;
else
b14_30 = sum(store14_30(:,3));
end
%% between 30 - 40 percent
ib=1;
ic=1;
for ia=1:nbranch
if(Event_val(ia,4)>0.3 && Event_val(ia,4)<0.4)
store5_40(ib,:) = [Event_val(ia,1:4)];
ib = ib +1;
end
if(Event_val(ia,5)>0.3 && Event_val(ia,5)<0.4)
store14_40(ic,:) = [Event_val(ia,1:3) Event_val(ia,5)];
ic = ic +1;
end
end
if (ib <2)
b5_40 = 0;
else
b5_40 = sum(store5_40(:,3));
end
if (ic <2 )

29. b14_40 = 0;
else
b14_40 = sum(store14_40(:,3));
end
%% between 40 - 50 percent
ib=1;
ic=1;
for ia=1:nbranch
if(Event_val(ia,4)>0.4 && Event_val(ia,4)<0.5)
store5_50(ib,:) = [Event_val(ia,1:4)];
ib = ib +1;
end
if(Event_val(ia,5)>0.4 && Event_val(ia,5)<0.5)
store14_50(ic,:) = [Event_val(ia,1:3) Event_val(ia,5)];
ic = ic +1;
end
end
if (ib <2)
b5_50 = 0;
else
b5_50 = sum(store5_50(:,3));
end
if (ic <2 )
b14_50 = 0;
else
b14_50 = sum(store14_50(:,3));
end
%% between 50 - 60 percent
ib=1;
ic=1;
for ia=1:nbranch
if(Event_val(ia,4)>0.5 && Event_val(ia,4)<0.6)
store5_60(ib,:) = [Event_val(ia,1:4)];
ib = ib +1;
end
if(Event_val(ia,5)>0.5 && Event_val(ia,5)<0.6)
store14_60(ic,:) = [Event_val(ia,1:3) Event_val(ia,5)];
ic = ic +1;
end
end

30. if (ib <2)
b5_60 = 0;
else
b5_60 = sum(store5_60(:,3));
end
if (ic <2 )
b14_60 = 0;
else
b14_60 = sum(store14_60(:,3));
end
%% between 60 - 70 percent
ib=1;
ic=1;
for ia=1:nbranch
if(Event_val(ia,4)>0.6 && Event_val(ia,4)<0.7)
store5_70(ib,:) = [Event_val(ia,1:4)];
ib = ib +1;
end
if(Event_val(ia,5)>0.6 && Event_val(ia,5)<0.7)
store14_70(ic,:) = [Event_val(ia,1:3) Event_val(ia,5)];
ic = ic +1;
end
end
if (ib <2)
b5_70 = 0;
else
b5_70 = sum(store5_70(:,3));
end
if (ic <2 )
b14_70 = 0;
else
b14_70 = sum(store14_70(:,3));
end
%% between 70 - 80 percent
ib=1;
ic=1;
for ia=1:nbranch

31. if(Event_val(ia,4)>0.7 && Event_val(ia,4)<0.8)
store5_80(ib,:) = [Event_val(ia,1:4)];
ib = ib +1;
end
if(Event_val(ia,5)>0.7 && Event_val(ia,5)<0.8)
store14_80(ic,:) = [Event_val(ia,1:3) Event_val(ia,5)];
ic = ic +1;
end
end
if (ib <2)
b5_80 = 0;
else
b5_80 = sum(store5_80(:,3));
end
if (ic <2 )
b14_80 = 0;
else
b14_80 = sum(store14_80(:,3));
end
%% between 80 - 90 percent
ib=1;
ic=1;
for ia=1:nbranch
if(Event_val(ia,4)>0.8 && Event_val(ia,4)<0.9)
store5_90(ib,:) = [Event_val(ia,1:4)];
ib = ib +1;
end
if(Event_val(ia,5)>0.8 && Event_val(ia,5)<0.9)
store14_90(ic,:) = [Event_val(ia,1:3) Event_val(ia,5)];
ic = ic +1;
end
end
if (ib <2)
b5_90 = 0;
else
b5_90 = sum(store5_90(:,3));
end

32. if (ic <2 )
b14_90 = 0;
else
b14_90 = sum(store14_90(:,3));
end
%% between 90 - 100 percent
ib=1;
ic=1;
for ia=1:nbranch
if(Event_val(ia,4)>0.9 && Event_val(ia,4)<1)
store5_100(ib,:) = [Event_val(ia,1:4)];
ib = ib +1;
end
if(Event_val(ia,5)>0.9 && Event_val(ia,5)<1)
store14_100(ic,:) = [Event_val(ia,1:3) Event_val(ia,5)];
ic = ic +1;
end
end
if (ib <2)
b5_100 = 0;
else
b5_100 = sum(store5_100(:,3));
end
if (ic <2 )
b14_100 = 0;
else
b14_100 = sum(store14_100(:,3));
end
bus_5_sags = [b5_10 b5_20 b5_30 b5_40 b5_50 b5_60 b5_70 b5_80 b5_90 b5_100]
bus_14_sags = [b14_10 b14_20 b14_30 b14_40 b14_50 b14_60 b14_70 b14_80 b14_90
b14_100]
%% display for question 4
fprintf ('| Bus number | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100
|n')
fprintf ('| Bus_5 | %.3f | %.3f | %.3f | %.3f | %.3f | %.3f | %.3f | %.3f | %.3f | %.3f
|n',bus_5_sags)
fprintf ('| Bus_14 | %.3f | %.3f | %.3f | %.3f | %.3f | %.3f | %.3f | %.3f | %.3f | %.3f

33. |n',bus_14_sags)