Compute the number of TLB misses with a replacement policy consisting of LRU

Question

Consider the following piece of code which multiplies two matrices:

int a[1024][1024], b[1024][1024], c[1024][1024];
multiply(){
    unsigned i,j ,k;
    for(i=0; i<1024; i++){
        for(j=0; j<1024; j++){
            for(k=0; k<1024; k++){
                c[i][j] += a[i][k] * b[k][j];
            }
        }
    }
}

Assume that the binary for executing this function fits in one page, and the stack also fits in one page. Assume further that an integer requires 4bytes for storage. Compute the number of TLB misses if the page size is 4096 and the TLB has 8 entries with a replacement policy consisting of LRU.

I tried to ask Chatgpt but he doesn't give me correct answers even the calculations not good. So after tried my best to solve this solution it comes like this:

• The page size is indeed 4,096 bytes (4KB), and each integer element requires 4 bytes.

• Each matrix (a, b, and c) has 1024x1024 elements, which is equivalent to 1,048,576 elements. Since each page can hold 1KB (1,024) elements (4,096 bytes / 4 bytes per element) => 2^10 (1,024) pages.

• The TLB has 8 entries.

• To access all three matrices, you have a total of 2^10 (1,024) * 3 = 3,072 pages to access at each time.

• For each page, there are 1024 memory accesses (since each page holds 1,024 elements).

Now, to calculate the total TLB misses:

Total TLB Misses = (Total Pages to Access) / (TLB Entries) Total TLB Misses = (3,072 pages) / (8 TLB entries) = 384 TLB misses