Question

Consider a system with 64 bit words, 48 bit addresses, 32GB of real memory and 4kB pages.

a) How many virtual pages are there?

b) How many page frames (real memory) are there?

c) How many entries are needed in the page table?

d) If each entry in the table requires 2 words, how big is the table?

Answer 1

Answer:

Given, 64 bit words, 48 bit addresses, 32 GB of real memory and 4kB pages.

a)  How many virtual pages are there?

Total number of entries = number of virtual pages

Here word size = 64 bit = 2^6 , address size = 48 bit = 2^48

4kB = 2^2 x 2^10 = 2^12

virtual memory size = address size x word size = 2^48 x 2^6 = 2^54

Now, the number of virtual pages = virtual memory / page size = 2^54 / 2^12 x 2^3 (bytes) = 2^54 / 2^15 = 2^54-15 = 2^39 = 2^30 x 2 = 512G pages or simply 2^39 pages.

b) How many page frames (real memory) are there?

Page frames = real memory size/ page size = 32GB /4kB = 2^5 x 2^32 B / 2^2 x 2^10 = 2^35 -12 = 2^23 = 8 M page frames

c) How many entries are needed in the page table?

Number of entries = number of pages

We already got the number of pages = 2^39. Thus the number of entries = 2^39 or 512G

d)  If each entry in the table requires 2 words, how big is the table?

Size of entry = 2 x word = 2 x 64 = 128 bits

Total size becomes :

entry size x number of entries = 128 x 2^39 = 2^7 x 2^39 = 2^46 bits

Now 2^46 bits , convert it into bytes = 2^46 / 2^3 = 2^43 bytes = 2^3 x 2^40 = 8 x TB = 8TB

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