baraka Posted August 11, 2011 Share Posted August 11, 2011 Hi im reading the Silberschatz book on OS concepts. I tried to do an exercise on memory management: physical address = 32 bit; physical address space is 4 times smaller than the logical address space; if a page is 2^22 byte, what is the number of the entries in the page table of a process? What is the number of frames? To solve this exercise i do this: physical address = number of frames + offset number of frames = physical address - offset = 32 - 22 = 10 now i know the number of frames that is 2^10. A frame has the same size of a page so the physical address space is 2^22 * 2^10 = 2^32 byte. The logical address space is 4 times bigger than the physical and 2^32 * 4 = 2^32 * 2^2 = 2^34 byte The number of entries of a page table of a process is 2^34/2^22 = 2 ^12 entries. The book doesn't do examples on how to solve this kind of exercises, it's only explain a lot of theory. I don't know if my solution is correct, what do you think about? Link to comment Share on other sites More sharing options...
galbotrix Posted August 30, 2011 Share Posted August 30, 2011 looks okay to me... Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now