How can find the maximum of f(x) by updating initial guesses in a loop?

As homework a 'simple grid search' must be coded to find the maxima of f(x) = 2*ln(x)-2*x+5.

The simple grid search takes two initial guesses (xa, xb) and devides them over 5 intervals of equal lenght (xa , (xa+xb)*(1/4) , (xa+xb)*(2/4) , (xa+xb)*(3/4) , xb). Then the code checks for three consecutive intervals in which a maximum is, and takes the outer two points to be iterated again as new initial guesses, this results in more narrow intervals each iteration until a chosen accuracy is acquired.

My code can find the intervals in which a maximum is because i leave it no choice. However, it fails to update the new initial guesses correctly, i think because i dont know how to properly start at the beginnen of the loop again. This is my code:

clear all

xa = 0;          %Initial guess xa < xb
xb = 10;         %Important: f(xa) < max < f(xb)
iter = 10;       
f = @(x) 2*log(x) - 2*x + 5;

for k = 1:iter
    
    x1 = (xa)        %Assuming a maximum is between xa and xb
    x2 = (xa+xb)*1/4 
    x3 = (xa+xb)*2/4 
    x4 = (xa+xb)*3/4
    x5 = (xb)
    
    if (f(x1) < f(x2))
        if (f(x2) < f(x3))
            if (f(x3) < f(x4))            
                if (f(x4) < f(x5))
                    xa = x3;
                    xb = x5;
                else
                    print('Choose better initial guesses')
                end               
            else
                xa = x2; %0.5
                xb = x4; %1.5               
            end            
        else
            xa = x1;
            xb = x3;           
        end                      
    else        
        print('Choose better initial guesses')
    end     
    if abs((x4-x3)) < 0.001
        break        
    end
end
    fprintf('The maximum is %f\nNumber of iterations: %d\n',x3,k)

Answer 1

[Divya Yerraguntla]

Hi Loic,

I dont see any issue with the updation of initial guesses in your code.

The initial guesses keep updating until the 3rd iteration and then they saturate because your maxima is already found at 3rd iteration itself. So you could change the value of 'iter' variable to 3 and still be able to find maxima.

Hope it helps!

Answer 2

[loic roure]

The problem was that the 5 intervals, x1 to x5, do not linearly increase. This is explained wrong in Numerical Methods by C. Woodford, C. Phillips on page 171.

To have a linearly increasing grid, set:

x1 = xa + (xb - xa)*0/4

x2 = xa + (xb - xa)*1/4

x3 = xa + (xb - xa)*2/4

x4 = xa + (xb - xa)*3/4

x5 = xa + (xb - xa)*4/4