Forward and Backward swift Load flow solution MATLAB progra

Solution is correct and simply to understand...you can also verify this program taking different bus system....

Here, I took 33 bus data. In that program you have to input the data like what is the straight line bus,what is node where the another branch is there, what is starting bus, ending bus.
nbus=33;


basemva =100 ; accuracy = 0.0001; maxiter=100; basekv=12.66;

% Distribution System data (69 bus system)
% Bus Bus Voltage Angle ---Load---- -------Generator----- Injected
% No code Mag. Degree KW Kvar MW Mvar Qmin Qmax Mvar
busdata=[1 1 1 0 0 0 0 0 0 0 0 1
2 0 1 0 100 60 0 0 0 0 0 1
3 0 1 0 90 40 0 0 0 0 0 1
4 0 1 0 120 80 0 0 0 0 0 1
5 0 1 0 60 30 0 0 0 0 0 1
6 0 1 0 60 20 0 0 0 0 0 1
7 0 1 0 200 100 0 0 0 0 0 1
8 0 1 0 200 100 0 0 0 0 0 1
9 0 1 0 60 20 0 0 0 0 0 1
10 0 1 0 60 20 0 0 0 0 0 1
11 0 1 0 45 30 0 0 0 0 0 1
12 0 1 0 60 35 0 0 0 0 0 1
13 0 1 0 60 35 0 0 0 0 0 1
14 0 1 0 120 80 0 0 0 0 0 1
15 0 1 0 60 10 0 0 0 0 0 1
16 0 1 0 60 20 0 0 0 0 0 1
17 0 1 0 60 20 0 0 0 0 0 1
18 0 1 0 90 40 0 0 0 0 0 1
19 0 1 0 90 40 0 0 0 0 0 1
20 0 1 0 90 40 0 0 0 0 0 1
21 0 1 0 90 40 0 0 0 0 0 1
22 0 1 0 90 40 0 0 0 0 0 1
23 0 1 0 90 50 0 0 0 0 0 1
24 0 1 0 420 200 0 0 0 0 0 1
25 0 1 0 420 200 0 0 0 0 0 1
26 0 1 0 60 25 0 0 0 0 0 1
27 0 1 0 60 25 0 0 0 0 0 1
28 0 1 0 60 20 0 0 0 0 0 1
29 0 1 0 120 70 0 0 0 0 0 1
30 0 1 0 200 600 0 0 0 0 0 1
31 0 1 0 150 70 0 0 0 0 0 1
32 0 1 0 210 100 0 0 0 0 0 1
33 0 1 0 60 40 0 0 0 0 0 1];

% Line code
% Bus bus R X 1/2 B = 1 for lines
% nl nr Ohm Ohm p.u. > 1 or < 1 tr. tap at bus nl
linedata=[1 2 0.0922 0.047 0 1
2 3 0.493 0.2511 0 1
3 4 0.366 0.1864 0 1
4 5 0.3811 0.1941 0 1
5 6 0.819 0.707 0 1
6 7 0.1872 0.6188 0 1
7 8 0.7114 0.2351 0 1
8 9 1.03 0.74 0 1
9 10 1.044 0.74 0 1
10 11 0.1966 0.065 0 1
11 12 0.3744 0.1238 0 1
12 13 1.468 1.155 0 1
13 14 0.5416 0.7129 0 1
14 15 0.591 0.526 0 1
15 16 0.7463 0.545 0 1
16 17 1.289 1.721 0 1
17 18 0.732 0.574 0 1
2 19 0.164 0.1565 0 1
19 20 1.5042 1.3554 0 1
20 21 0.4095 0.4784 0 1
21 22 0.7089 0.9373 0 1
3 23 0.4512 0.3083 0 1
23 24 0.898 0.7091 0 1
24 25 0.896 0.7011 0 1
6 26 0.203 0.1034 0 1
26 27 0.2842 0.1447 0 1
27 28 1.059 0.9337 0 1
28 29 0.8042 0.7006 0 1
29 30 0.5075 0.2585 0 1
30 31 0.9744 0.963 0 1
31 32 0.3105 0.3619 0 1
32 33 0.341 0.5302 0 1];
This is the bus data....
in that like straight line bus is 1 to 18
Branch connected to which node=2
starting bus=19
ending bus=22
Branch connected to which node =3
Starting bus=23
Ending bus=25
Branch connected to which node =6
Starting bus=26
Ending bus=33
like data you have to put after that you will get results....