ps4solution - 13.002 Introduction to Numerical Methods for...

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Unformatted text preview: 13.002 Introduction to Numerical Methods for Engineers Solution to Problem Set 4 e'“ 1 .0 2:1 1 —1 e‘“ —1 2:; = e'“ 1 e‘“ ——2 :3 e'“ [-11 i :2” “[35 (f :2], ‘4 l [ 1.a=0 where Lc = b and Ux = c was used. For a —> oo 0 1 0 —1 0 —1 1 0 —-2 After swithching coulmns 1 and 2 and adding row3 to row2 1 0 0 $2 1 0 0 —1 —1 xx = 0 :1; = 1’ O 0 —3 1:3 0 O- 2. See attach. 3. For a large the soluton becomes ill-conditioned as can be seen from the results attached. 4. Partial pivoting reduced that problem. 5.F‘rom the original set of the equations by interchanging colums 1 and 2 1 6—“ 0 1:2 - l 6—“ —1 —1 21 = 6-“ . e'“ 1 —2 2:3 8"“ This particular rearrangement yields a stable solution due to the connection to the physical background of the problem. See attach. % Gaussian elimination clear % Examples %[a]=[2 1 0 0; ['b]=[2;l;4;8]=[1 -1 o; —i 2 —1; o —i 2] ,%L01=[1:0;0; %[a]=[0 1 0; -1 O -1; l 0 —2] *[b}=[1;0;0; 1 2 l 0; 0 1 2 1; 0 0 1 2] n=3; %alpha is p‘ [exp(—p) 1 0;-1 exp(-p) —1;1 exp(—p) —2] [1;exp(-p);eXP(-p)l %partial pivoting 1&2 row for j=l:n dum=a(2,j); 8(2.j)=a(1,j); a(1,j)=dum; end dum=b(2); b(2)=b(l); b(1)=dum; a b h . . . elimination c k=1:n for i=k+1:n m(i,k)=a(i,k)/a(k.k); for j=k:n a(i.j)=a(i.j)— m(i,k)*a(k,j); end ‘ ‘ b(i)=b(i)— m(i.k)*b(k); end end a b % Back substitution x(n)=b(n)/a(n,n): for i=n—1:-l:l sum=b(i) ; for k=i+1=n sum=sum—a(i,k)*x(k); end xti)=sum/a(i,i): _snd r ki)=sum/a(i,i) >> hwk4_3 %alpha is p p = O a = l l O -1 1 -1 l 1 -2 b = l 1 l l l O O 2 -1 O 0 -2 0 l 0 >> hwké_3 p = 5 a = 0.0067 1.0000 0 148.4l99 O O b = 1.0000 148.4199 0 -l.0000 -2.9999 >> hwk4_4 %alpha is p p = 0 a = l l 0 -1 l -1 l l -2 b = l l 1 % patial pivoting rows 1&2 a = -l l -l l l O l l -2 b = l l l % eliminacon a = -l l -1. 0 2 -1 O O —2 b : l 2 O % Back substitution x = O l 0 >> hwk4_4 ...
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