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Q1. Assume that a virtual address is 24 bits: 14 bits for the virtual page numbe

ID: 3593072 • Letter: Q

Question

Q1. Assume that a virtual address is 24 bits: 14 bits for the virtual page number, and 10 bits for the offset. Also assume that each page table entry is 4 bytes in size.

If we use simple paging, what is the size of a page table, in bytes? (Make sure you get this right -- the following problems depend on it.) _____________

If we break the page table into page-sized chunks, how many chunks are there in the entire page table? ____________

If only 25% of the page-sized chunks of the page table are used, what is the total size, in bytes, of all the chunks that are used? _____________

Assume we move to multi-level paging, and split the VPN part of the virtual address into two parts of 6 bits and 8 bits, where the page directory index is the 6 bit part. Also assume that each page directory entry is 4 bytes in size. What is the total size of the page directory, in bytes? ____________

Following on from questions a-d above, how much memory is saved if we move from simple paging to multi-level paging? Express your answer as a decimal number between 0 and 1 that equals the number of bytes needed with multi-level paging divided by the number of bytes needed with simple paging. ___________.

Q2.

Using the LRU replacement policy, fill in the blanks in the table below. The cache size is 3.

                access                hit?                evict                resulting cache

                  0                no                 -                0

                  1                no                 -                0,1

                  1                yes                 -                0,1

                  2                no                 -                0,1,2

                  0                yes                 -                1,2,0

                  3                no                1                2,0,3

          1                ____                ____                ____

                  0                ____                ____                ____

                  2                ____                ____                ____

Using the FIFO replacement policy, fill in the blanks in the table below. The cache size is 3.

                access                hit?                evict                resulting cache

                  0                no                 -                0

                  1                no                 -                0,1

                  1                yes                 -                0,1

                  2                no                 -                0,1,2

                  0                yes                 -                0,1,2

                  3                no                0                1,2,3

          1                ____                ____                ____

                  0                ____                ____                ____

                  2                ____                ____                ____

Explanation / Answer

Q1)

A)size of the page table = 2^14 * 4 bytes = 2^16 bytes = 64kB

B)each page size = 2^10 = 1kB

Number of pages = 64kB/ 1kB =64 pages

C) size of 25% of page sized chunks = 64*25/100 * 1KB = 16kB

D) page directory has 6 bit index and hence each page directory contains 2^8 elements

page directory size = 2^ 8 * 4bytes = 1kB