EGR 511
NUMERICAL METHODS
_______________________
LAST NAME, FIRST
Problem set #2
1. Use Newton’s method with x(0) = 0 to compute x(2) for each of the following nonlinear
systems:
1
a. 4 x12 - 20x1 + x 22 + 8 = 0
b. sin(4x1x2) – 2x2 – x1 = 0
4
1
4  1 2 x 1
x1 x 22 + 2x1 – 5x2 + 8 = 0
(e
- e) + 4e x 22 - 2ex1 = 0
2
4
>> s2p1a
x = 0.4958936 1.9834235
>> s2p1b
x = -0.5131616 -0.0183762
2. Use the method of Steepest Descent with x(0) = 0 to approximate the solutions for
4 x12 - 20x1 +
1 2
x +8=0
4 2
1
x1 x 22 + 2x1 – 5x2 + 8 = 0
2
n
Iterating until g =
 f i2 < 0.05
i 1
% Steepest Descent Method
% Set 2, problem 2
%
f1='4*x(1)^2-20*x(1)+.25*x(2)^2+8';
f2='.5*x(1)*x(2)^2+2*x(1)-5*x(2)+8';
% Initial guess
%
x=[0 0];
f=[eval(f1) eval(f2)];
g1=f*f';
for i=1:50
Jt=[8*x(1)-20 .5*x(2)
.5*x(2)^2+2 x(1)*x(2)-5];
% Gradient vector
%
delg=2*Jt*f';
zo=sqrt(delg'*delg);
% Unit vector in the steepest descent
%
z=delg/zo;
xs=x;a3=1;x=x-a3*z';
f=[eval(f1) eval(f2)];
g3=f*f';
while g3>g1
a3=a3/2;
x=xs-a3*z';
f=[eval(f1) eval(f2)];
g3=f*f';
end
a2=a3/2;x=xs-a2*z';
fsave3=f;
f=[eval(f1) eval(f2)];
g2=f*f';h1=(g2-g1)/a2;h2=(g3-g2)/(a3-a2);h3=(h2-h1)/a3;
% Choose a0 so that g will be minimum in z direction
%
a0=.5*(a2-h1/h3);
x=xs-a0*z';
f=[eval(f1) eval(f2)];
g0=f*f';fsave0=f;
if g0<g3
alfa=a0;g=g0;f=fsave0;
else
alfa=a3;g=g3;f=fsave3;
end
x=xs-alfa*z';
fprintf('x = %9.6f %9.6f',x(1),x(2))
fprintf(' , g = %9.6f\n',g)
if abs(g)<0.05, break, end
g1=g;
end
>> s2p2
x = 0.398351
x = 0.688581
x = 0.423584
x = 0.490600
x = 0.409879
x = 0.459636
x = 0.545482
x = 0.481841
x = 0.490227
x = 0.487611
0.059753 , g = 72.674085
1.141483 , g = 29.558308
1.273358 , g = 8.398065
1.353318 , g = 7.249389
1.479892 , g = 4.538019
1.495889 , g = 3.863304
1.830068 , g = 1.507642
1.859146 , g = 0.275003
1.865165 , g = 0.258094
1.973402 , g = 0.033052
2.b Use the method of Steepest Descent with x(0) = 0 to approximate the solutions for
sin(4x1x2) – 2x2 – x1 = 0
4  1 2 x 1
(e
- e) + 4e x 22 - 2ex1 = 0
4
n
Iterating until g =
 f i2 < 0.05
i 1
% Steepest Descent Method
% Set 2, problem 2b
%
f1='sin(4*pi*x(1)*x(2))-2*x(2)-x(1)';
f2='(1-.25/pi)*(exp(2*x(1))-e)+4*e*x(2)^2-2*e*x(1)';
e=exp(1);
% Initial guess
%
x=[0 0];
f=[eval(f1) eval(f2)];
g1=f*f';
for i=1:50
con=4*pi*cos(4*pi*x(1)*x(2));
Jt=[x(2)*con-1
x(1)*con-2
(2-.5/pi)*exp(2*x(1))-2*e 8*e*x(2)]';
% Gradient vector
%
delg=2*Jt*f';
zo=sqrt(delg'*delg);
% Unit vector in the steepest descent
%
z=delg/zo;
xs=x;a3=1;x=x-a3*z';
f=[eval(f1) eval(f2)];
g3=f*f';
while g3>g1
a3=a3/2;
x=xs-a3*z';
f=[eval(f1) eval(f2)];
g3=f*f';
end
a2=a3/2;x=xs-a2*z';
fsave3=f;
f=[eval(f1) eval(f2)];
g2=f*f';h1=(g2-g1)/a2;h2=(g3-g2)/(a3-a2);h3=(h2-h1)/a3;
% Choose a0 so that g will be minimum in z direction
%
a0=.5*(a2-h1/h3);
x=xs-a0*z';
f=[eval(f1) eval(f2)];
g0=f*f';fsave0=f;
if g0<g3
alfa=a0;g=g0;f=fsave0;
else
alfa=a3;g=g3;f=fsave3;
end
x=xs-alfa*z';
fprintf('x = %9.6f %9.6f',x(1),x(2))
fprintf(' , g = %9.6f\n',g)
if abs(g)<0.05, break, end
g1=g;
end
>> s2p2b
x = -0.357926 0.000000 , g = 0.139390
x = -0.361009 0.057884 , g = 0.003316
3. Find the directional derivative of f(x, y) = 2x2 + y2 at x = 2 and y = 2 in the direction of h = 3i
+ 2j.
Ans: 8.875
4. Find the gradient vector and Hessian matrix for each of the following functions:
a) f(x, y) = 2xy2 + 3exy
b) f(x, y, z) = x2 + y2 + 2z2
c) f(x, y) = ln(x2 + 2xy + 3y2)
Ans:
2 y 2  3 ye xy 
(a) f = 
,H=
xy 
 4 xy  3xe 
2 x 
(b) f = 2 y  , H =
 
 4 z 
(c) f =

3 y 2 e xy

2 xy
xy
4 y  3 y e  3e
4 y  3 y 2 e xy  3e xy 

4 x  3x 2 e xy

 2 0 0
 0 2 0


0 0 4
2 x  2 y 
1
1
,H=

2
2 
x  2 xy  3 y 2 x  6 y 
x 2  2 xy  3 y 2 2
  2 x 2  2 y 2  4 xy  2 x 2  6 y 2  12 xy


2
2
2
2
 2 x  6 y  12 xy 2 x  18 y  12 xy 
5. a) Find the first two iterations of the Gauss-Seidel method for the following linear system using x(0)
= 0;
i) 3x1 - x2 + x3 = 1, ii) 10x1 - x2
= 9,
3x1 + 6x2 + 2x3
= 0,
- x1 + 10x2 - 2x3 = 7,
3x1 + 3x2 + 7x3
= 4.
- 2x2 + 10x3 = 6.
b) Repeat Exercise (a) using the SOR method with relaxation factor = 1.1.
Gauss Seidel
Gauss Seidel
0.1111111
0.9790000
-0.2222222
0.9495000
0.6190476
0.7899000
SOR
SOR
0.0541008
0.9876790
-0.2115435
0.9784934
0.6477159
0.7899328
method
method
6. Solve the system
x2 + y2 + z2
= 9,
xyz = 1,
x + y
-
z2
= 0
by Newton's method to obtain the solution near (2.5, 0.2, 1.6).
% Newton Method
%
f1='x(1)^2+x(2)^2+x(3)^2-9';
f2='x(1)*x(2)*x(3)-1';
f3='x(1)+x(2)-x(3)^2';
% Initial guess
%
x=[2.5 0.2 1.6];
for i=1:20
f=[eval(f1) eval(f2) eval(f3)];
Ja=[2*x(1)
2*x(2)
2*x(3)
x(2)*x(3) x(1)*x(3) x(1)*x(2)
1
1
-2*x(3)];
%
dx=Ja\f';
x=x-dx';
if max(abs(dx))<.00001, break, end
end
fprintf('Number of iteration = %g\n',i)
fprintf('x = ');
fprintf(' %10.7f %10.7f %10.7f\n',x(1),x(2),x(3))
Number of iteration = 3
x = 2.4913757 0.2427459 1.6535179
7. (P. 14.3 Chapra)
Given:
f(x, y) = 2xy + 1.5y – 1.25x2 – 2y2
Construct and solve a system of linear algebraic equations that maximizes f(x, y). Note that this is
done be setting the partial derivatives of f with respect to both x and y to zero.
Ans: x = 0.5, y = 0.625
8. (P. 14.4 Chapra)
(a) Start with an initial guess of x = 1 and y = 1 and apply two applications of the steepest ascent
method to the following function
f(x, y) = 2xy + 1.5y – 1.25x2 – 2y2
(b) Construct a plot from the results of (a) showing the path of the search.
Ans: x(2) = 0.543, y(2) = 0.657
9. (P. 14.7 Chapra)
Perform one iteration of the optimal gradient steepest descent method to locate the minimum of
f(x, y) = – 7x + 1.25x2 + 11y + 2y2 – 2xy
using initial guesses x = 0 and y = 0.
Ans: x(1) = 1.3013, y(1) =  2.0448