Maximize a function that involves summation in MATLAB

Question

I want to maximize this function using fminbnd in MATLAB.

I am not able to correctly define the function in MATLAB. How can I create a loop for i to define the function accordingly and finally apply fminbnd ?. Here's my code ( which is actually wrong ):

k = 0.1;    
L = sum(k - (i/n)) + sum(((i/n) - k)*log(k - (i/n))) + k;

fminbnd(-L, [], []);

Any help will be very appreciated.

Answer 1

There are two things concerning your problem.

First : the thing with the loop

MATLAB is great at vectorizing stuff. The command sum adds-up a vector, so you need to create the vector i = 1:n (from 1 to n ... spaced by 1) yourself:

n = 10; % maximum in sum
i = 1:n; % line-space => use MATLAB's vetorization
in = i/n; % put this in a new variable for the sake of conciseness

k = 10; % some random value
L = sum( (k - in) + (in - k).*log(k - in) + k )

the output is:

 L = -17.7921

Second: the thing with optimizing a function

Handing over functions in MATLAB can be done with function handles . Essentially, this is an object pointing to a function. In your case, we in fact need an anonymous function handle :

L = @(k) sum( (k - in) + (in - k).*log(k - in) + k )

The variable in is "frozen". For the anonymous function handle it is as if in was hard coded. The @ indicates a function handle and the followed (k) denotes that the variable k should be considered as the first input of this new function (this is why it is an anonymous function handle. The variable k becomes and anonym input). So you could now call

L(10)

and would obtain

L = -17.7921

as before.

Now the optimization function can use this (note that you need to include the minus in the function handle, so we just wrap it in a new one:)

fnc = @(k) -L(k)

lb = 1+10-9; % lower bound
ub = 100; % upper bound
[k_opt,fval] = fminbnd(fnc ,lb,ub)

 k_opt = 3.2835 fval = -32.5310

Be aware that your function becomes complex for values k <= 1 , which is why I have restricted the minimum bound here to larger values.