0

Given a modern 64-bit architecture with 6-bytes used for each page pointer, 1 MB pages, and 16 GB installed memory:

a) How would I calculate how much memory can be addressed by the paging solution?
b) How would I calculate how large must each page table be?

I'm not looking for the answer. I would just appreciate a clear breakdown.

Improve this question
1

1 Answer 1

Reset to default
4

64 bit architecture allows 264 bytes to be addressed.

Six bytes is 48 bits, so the number of pages that can be addressed with a six byte page pointer is 248. Multiply that by a million to get the number of addressable bytes using 1 megabyte pages.

16 Gigabytes is approximately 230, and the paging solution already exceeds that capacity, even with a page size of 1 byte.

Improve this answer
4
  • So if I understand correctly the breakdown would look like this: 2^48 * 2^10 which translates into 2^58 byte giving you and addressable amount of 2^18 TB?
    – Stevenson
    Commented Oct 3, 2016 at 2:08
  • Where did you get 2^10 from?
    – Robert Harvey
    Commented Oct 3, 2016 at 2:10
  • Page size is 1024 bytes (2^10), right? or am I mistaken
    – Stevenson
    Commented Oct 3, 2016 at 2:19
  • 1
    It says in your question that page size is one megabyte.
    – Robert Harvey
    Commented Oct 3, 2016 at 2:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.