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+clear;

+clc;

+printf("\t\t\tExample Number 3.6\n\n\n");

+// Gauss-Seidal calculation

+// Example 3.6 (page no.-97-98)

+// solution

+

+// it is useful to think in terms of a resistance formulation for this problem because all the connecting resistances between the nodes in figure 3-6(page no.-83) are equal; that is 

+// R = dy/(k*dy) = dx/(k*dy) = 1/k                       (a)

+// therefore, when we apply equation(3-32) to each node, we obtain(qi = 0)

+// Ti = (sum Kj*Tj)/(sum Kj)                           (b)

+// because each node has four resistances connected to it and k is assumed constant,

+// sum Kj = 4*k

+// and

+// Ti = (1/4)*(sum Tj)                               (c)

+// we are now making four nadal equations for iteration

+// node 1 : T1 = (1/4)*(100+500+T2+T3)

+// node 2 : T2 = (1/4)*(500+100+T1+T4)

+// node 3 : T3 = (1/4)*(100+100+T1+T4)

+// node 3 : T4 = (1/4)*(T3+T2+100+100)

+// we now set up an iteration table as shown in output

+A=[4 -1 -1 0;-1 4 0 -1;-1 0 4 -1;0 -1 -1 4];

+b=[600;600;200;200];

+x=[300;300;200;200];

+NumIters=6;

+D=diag(A);

+A=A-diag(D);

+for i=1:4

+    D(i)=1/D(i);

+end

+n=length(x);

+x=x(:); 

+y=zeros(n,NumIters);

+for j=1:NumIters

+    for k=1:n

+        x(k)=(b(k)-A(k,:)*x)*D(k);

+    end

+    y(:,j)=x;

+end

+printf("the iteration table is shown as : \n\n");

+disp(y);

+printf("\n\n after five iterations the solution converges and the final temperatures are \n");

+disp(y(1,6),"T1=");

+disp(y(2,6),"T2=");

+disp(y(3,6),"T3=");

+disp(y(4,6),"T4=");

+