MATLAB Code Performance Which of the following is slower

MATLAB Code Performance Which of the following is slower: A) x = zeros(1,1000000); for i = 1:1000000 x(i) = rand(1,1); end B) for i = 1:1000000 x(i) = rand(1,1); end Pick ONE option A is slower because there is an additional step of initializing 0’s B is slower because the memory allocated for ‘x’ keeps increasing A and B should take same amount of time

The Correct Answer and Explanation is:

Correct Answer: B is slower because the memory allocated for ‘x’ keeps increasing

Explanation:

In MATLAB, performance can vary significantly depending on how memory is managed during code execution. The difference between the two code snippets in this question relates to a concept called preallocation.

Code A:

matlabCopyEditx = zeros(1,1000000);
for i = 1:1000000
    x(i) = rand(1,1);
end

In this version, a vector x of size 1 by 1,000,000 is preallocated using the zeros function. This means that MATLAB sets aside enough memory for all the values of x before the loop begins. As a result, each assignment inside the loop simply fills an existing memory slot, which is efficient and avoids reallocating memory during execution.

Code B:

matlabCopyEditfor i = 1:1000000
    x(i) = rand(1,1);
end

In this version, the vector x is not preallocated. As the loop runs, MATLAB must dynamically increase the size of x with each new iteration. This involves repeated memory allocation and copying the current data into a larger memory block as x grows. This operation becomes increasingly expensive as the vector length increases.

Conclusion:

Code B is slower because the lack of preallocation forces MATLAB to reallocate memory repeatedly. This inefficiency can greatly increase runtime, especially for large arrays. Preallocating memory, as done in Code A, is a well-known best practice in MATLAB programming to improve performance.

Therefore, the correct choice is:

B is slower because the memory allocated for ‘x’ keeps increasing.