Question

A byte addressable direct-mapped cache with 1024 blocks/lines, and with each block are having 8 32-bit words. How many bits are required for block offset, assuming a 32-bit address?

Answer 1

Assuming that there is a memory of 2GWs where each of the individual frame consist of 8M frames or 256 words.

Each of the program consist of 4 bytes per word which equal to 203M words.

Each of the page has 256 words which means that there are 831488 pages, which is basically what the page table consist of.

Therefore the size of the page table becomes 831,488 * 3 bytes which equals 2,494,464 bytes.