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In a paging system a virtual address consists of 24 bits, in which 16 bits are displacement and 8 bits for page number.

Calculate :

  1. Page size
  2. Maximum number of pages
  3. Maximum virtual address space
Ani Menon
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None
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    Great, do it. What's the problem? – harold May 27 '16 at 16:58
  • Well the the whole question is the problem, and I want it solved. Would you mind? – None May 27 '16 at 17:20
  • All three sub-questions have the form "compute some power of two", literally just put it in a calculator. – harold May 27 '16 at 17:22
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    @harold 'literally just put it in a calculator.' will just not do as a copypasta homework answer! – Martin James May 27 '16 at 18:39
  • @harold I'm confused on how to go about solving this, I think i'm not getting it right... would you mind solving it so i can compare it with my solution – None May 27 '16 at 20:02
  • I've got a better idea; why not show your solution and indicate where you have an issue with it? – Martin James May 28 '16 at 00:44
  • 16 bits are reserved for displacement, so the page size is 2^16 = 64KB 8 bits are reserved for page number, so each segment can contain 2^8= 256 pages – None May 28 '16 at 20:31
  • but maximum virtual address? – None May 28 '16 at 20:31
  • 'In a paging system a virtual address consists of 24 bits'. Anyway, why not ask your prof/TA instead of posting your homework here? One of your fellow students ratted this out as an assignment: http://stackoverflow.com/a/37502008/758133 – Martin James May 28 '16 at 21:05
  • got it maximum virtual address space=256*64K=16384K=16M ... thanks for your help @MartinJames – None May 28 '16 at 21:23

1 Answers1

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(1) 16 bits are reserved for displacement, so the page size is 2^16 = 64 KB

(2) 8 bits are reserved for page number, so each segment can contain 2^8= 256 pages

(3) maximum virtual address space=256*64KB=16384KB=16MB

None
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