Difference between revisions of "Non-linear programming"
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[https://docs.google.com/document/d/1tHekhHdPWEPhPm_lGVP5cws_2-NsTw_6JpVhhLSlgmo Handout from class] | [https://docs.google.com/document/d/1tHekhHdPWEPhPm_lGVP5cws_2-NsTw_6JpVhhLSlgmo Handout from class] | ||
| [https://www.dropbox.com/s/0jls9rdtofpmbg1/2015-4G3-Class-10A.mp4?dl=1 Video] | |||
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| 18 March | | 18 March |
Revision as of 17:22, 18 March 2015
Class date(s): | 04 February 2015 | ||||
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Resources
Scroll down, if necessary, to see the resources.
Date | Class number | Topic | Slides/handouts for class | Video file | References and Notes |
---|---|---|---|---|---|
04 February | 05A |
|
Video | ||
09 February | 06A |
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Video | ||
11 February | 06B |
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Video | ||
16 to 27 February | 07 |
Reading week break and midterm | |||
02 March | 08A |
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Video | ||
04 March | 08B |
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Video |
Code used in class (see below) | |
09 March | 09A |
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Video | ||
11 March | 09B |
Guest lecture |
Video | ||
16 March | 10A |
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Video | ||
18 March | 10B |
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Taking full Newton's steps to solve the class example
clear all;
close all;
clc;
[X1,X2] = meshgrid(-0.5:0.1:6, 0:0.01:9);
Z = func(X1,X2);
contour(X1, X2, Z)
hold on
grid on
x = [1,3]';
plot(x(1), x(2), 'o')
text(x(1)+0.2, x(2), '0')
for k = 1:10
slope = -first_deriv(x)
step = hessian(x)\slope; % Solves the Ax=b problem, as x = A\b
x = x + step;
plot(x(1), x(2), '*')
text(x(1)+0.1, x(2), num2str(k))
end
func.m
function y = func(x1,x2)
y = 4.*x1.*x2 - 5.*(x1-2).^4 - 3.*(x2-5).^4;
first_deriv.m
function y = first_deriv(x)
y = [4*x(2) - 20*(x(1)-2)^3;
4*x(1) - 12*(x(2)-5)^3];
hessian.m
function y = hessian(x)
y = [-60*(x(1)-2)^2, 4;
4, -36*(x(2)-5)^2];