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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 illconditioned 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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 Spring '05
 HenrikSchmidt

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