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//Book - Power System: Analysis & Design 5th Edition
//Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J. Overbye
//Chapter - 6 ; Example 6.8
//Scilab Version - 6.0.0 ; OS - Windows
clc;
clear;
y=[15;50]; //Constant coefficients in the equations
x=[4;9]; //Initial values for x1 and x2
err=1; //Initialization of error value
tol=1e-4; //Tolerance value for Newton Raphson method
iter=0; //Initialization of iteration value
while err>tol
temp=x;
f=[x(1)+x(2);x(1)*x(2)] //Function Value
dely=y-f;
J=[1 1;x(2) x(1)]; //Jacobian Matrix
//Reduction of Jacobian using Gauss elimination
Jg=[J(1,1) J(1,2);0 J(2,2)-J(2,1)/J(1,1)]
delyg=[dely(1);dely(2)-dely(1)*J(2,1)/J(1,1)]
//Solution using back substitution
delx2=delyg(2)/Jg(2,2);
delx1=(delyg(1)-Jg(1,2)*delx2)/Jg(1,1)
delx=[delx1;delx2]
x=x+delx
err=max(abs((x(1)-temp(1))/temp(1)),abs((x(2)-temp(2))/temp(2)));
iter=iter+1;
//Displaying first iteration results
if iter==1
printf('Values of x1 and x2 at the end of first iteration are:\n')
printf(' x1=%.4f and x2=%.4f\n\n',x(1),x(2))
end
end
printf('The convergence criterion is satisfied at the %dth iteration\n',iter)
printf('The solution is x1=%.4f and x2=%.4f',x(1),x(2))
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