1 In The Last Homework You Were Asked To Outline The Process Of Bracketing The Minimum Of A Function Defined On Multipl 1 (61.52 KiB) Viewed 24 times
1 In The Last Homework You Were Asked To Outline The Process Of Bracketing The Minimum Of A Function Defined On Multipl 2 (64.74 KiB) Viewed 24 times
1 In The Last Homework You Were Asked To Outline The Process Of Bracketing The Minimum Of A Function Defined On Multipl 3 (49.29 KiB) Viewed 24 times
1. In the last homework you were asked to outline the process of bracketing the minimum of a function defined on multiple dimensions. In this question you will continue the process of finding the minimum of the three-dimensional function f(x, y, z) = 2x² + y² + z³-2xy + yz - 7y - 4z using the steepest descent method, using the process below. Starting at Po= (1, 1, 1) work towards the minimum using the steepest descent method as outlined below. a. [3 points] Use the method you devised in HW11 to bracket a minimum starting at Po b. [3 points] Once you have bracketed the minimum, write code to locate two interior points spaced at (1-r) and r (where r is the golden ratio) along it. c. [4 points] Now iteratively use the golden ratio search method to find the minimum of the function on the unimodal interval. d. [2 points] When you have found the minimum within the interval, call this P₁. What is the direction of steepest descent at P₁? c. [3 points] Starting from the unimodal interval you found in part a. compute the location of the minimum within the interval using quadratic minimization instead of golden ratio search. What is the direction of steepest descent at this point? BONUS [2 points] Extend your solution for problem 1 to locate a unimodal interval starting at P₁. [3 points] Once the new unimodal interval has been found, locate a minimum within it. You can use either the golden ratio search or quadratic minimization.
XX A %As an anonymous function, the function is f-@(x) p0= [1,1,1]; fo=f(p0); p=p0; % f0= -8 %% B GF (@(x) 4*x(1)-2*x (2); @(x) 2*x (2) -2*x (1)+x(3)-7; @(x) 3*x(3)^2+x(2)-4;}; s=-[GF {1} (p0), GF {2} (p0), GF {3} (p0)]; s=s/norm(s); %% D lambda=0.25; pstar-p0+s lambda/2; fstar=f(pstar) ; pstar2-p0+s*lambda; fstar2-f(pstar2) ; %% E max=100; solved=0; iter=1; lambda=0.25; x0-p0; f0-f(x0); while (solved && iter<max) x1=x0+lambda/2*s; f1=f(x1); x2=x0+lambda*s;f2=f(x2); if (f2>f1 && fØ>f1) solved=1; break; else 2*x(1)^2+x(2)^2+x(3)^3-2*x (1) *x (2)+x(2) *x (3)-7*x(2) -4*x(3); I
pstar-p0+s*lambda/2; fstar=f(pstar) ; pstar2-p0+s*lambda; fstar2-f(pstar2) ; %% E max=100; solved=0; iter=1; lambda=0.25; x0-p0; f0=f(x0); while (~solved && iter<max) x1=x0+lambda/2*s; f1=f(x1); x2=x0+lambda*s;f2=f(x2); if (f2>f1 && fØ>f1) solved=1; break; else if (f2<f1) % still decreasing lambda lambda*2; iter=iter+1; else %overshot lambda-lambda/3; iter=iter+1; end end end %located a unimodal interval when lambda=4 pstarfinal-p0+lambda/2*s; pstar2final-p0+lambda*s; d Window
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