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값은 =>\"+num2);
System.out.println(\"x[\"+i+\"]의 ea 값은 =>\"+num1);
}
System.out.println(\"---------------------------------------\");
}
}
2.결과값 도출 1. 문제 및 결과
2. 문제 및 결과
1. Gauss – seidel 법 소스코드 및 설명
2.결과값 도출
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Source
▶▶ Gauss - Seidel Method
#include <iostream.h>
#include <math.h>
void main()
{
double l_1,l_2,l_3,l_4;//the loading of chloride to each lakes
double c1,c2,c3,c4;//the resulting concentrations to each lakes
double ea1,ea2,ea3,ea4;//tolerance
double c1o,c2o,c3o,c4o;
int iter=0;
cout
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\\n\", k);
for(i=0;i<n;i++){
sum=0.;
for(j=0;j<n;j++){
if(i!=j)
sum=sum+A[i][j]*x[j];
}
next_x[i]=(A[i][n]-sum)/A[i][i];
printf(\"%lf \\t\", next_x[i]);
x[i]=next_x[i];
}
err = 0.;
sum = 0.;
for(i=0;i<n;i++){
for(j=0;j<n;j++)
sum = sum + pow(A[i][j],2);
}
err = pow(sum,0.5);
if(err <=
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n-1)
disp(x)
j=find(abs(x)==max(abs(x)));
c=x(j);
z=x/c;
P=R-c;
fprintf(\'\\n(원래 최대 lambda=24) lambda = %f=\',c);
disp(c);
fprintf(\' x(%.0f)\\n\',n);
disp(z(1));
disp(z(2));
disp(z(3));
fprintf(\'\\n\\n\');
n=n+1;
end 1.Gauss-Seidel법
2.Jacobi법
3.특성방정식
4.멱수법
5.소스코드
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seidel Method
double a,b,c,a0,b0,c0,ea;
b=0,c=0;
while (iter<=1000)
{
iter++;
a=(sumz-sumx*b-sumy*c)/n;
ea=fabs((a-a0)/a*100);
a0=a;
b=(sumxz-sumx*a0-sumxy*c)/sumx2;
b0=b;
c=(sumyz-sumy*a0-sumxy*b0)/sumy2;
c0=c;
if (ea<0.001)
{
//cout << \" \"<< iter << \"\\t\" << a &l
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