Assume the computer is word addressable with 256K of addresses and is running a

Question

Assume the computer is word addressable with 256K of addresses and is running a process whose page table is given below. Answer the following questions.

a. How big is a page/frame in words?  

b. What is the physical address for the logical address 01100001111110?

c. What is the physical address for the logical address 10101010101010?

d. What is the physical address for the logical address 11011011011010?

e. How large is this program (approximately) in words?  

f. How much of the program is in memory?

Explanation / Answer

a) Here if we see the fist level page points to 1024 2nd level table, and each points to 256 3rd page table , where each is having 64 pages.
Now the program address space is having 1024 pages and hence we need 16 3rd level page tables and therefore there are 16 entries required in 2nd level page table and one entry in the first table.
Hence the size is 1024 entries for the 1st table, 256 entries for the 2nd table and 16 for 3rd level page table having 64 entries each. Assume we have two bytes per entry then space becomes 1024 * 2 + 256 * 2 + 16 * 64 * 2 = 4608 bytes.

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