Question

1. Answer the questions based on the memory below: Page table register Virtual address 31 30 29 28 27........................15 14 13 12 11 10 9 8 ........3 2 1 0 Virtual page number Page offset 20 Valid Physical page number Page table J18 If O then page is not present in memory 29 28 27........ ..............-15 14 13 12 11 10 9 8.......3 2 1 0 Physical page number Page offset Physical address a) Page size in KB): ay offset. 212-22 210- TuKB b) Range of virtual page number: 0.720-17 c) Range of physical page number: [0.2 -) d) Physical memory size (in GB): virtual = 232 = 22 230 e) Page table size (in MB): virtual 132 133-4GB; physical => 230 - 163 220o (28+1) = 343 51864440 B/4210 ~2.375 33554560MB)

Need help with only Question E, how do I calculate the page table size? (2.3mb should be the answer).

Answer 1

E) Given that virtual page number is represented with 20 bits, And in paging there is an entry for each page in the page table.

Therefore the number of entries in the page table = 2²⁰ entries (because 20 bits for virtual page number).

And in the page table each entry contains 1 bit for valid and 18 bit for physical page number.

Therefore the size of each entry = 1 bit (valid) + 18 bits (physical page number) = 19 bits. (each entry).

Therefore total size of page table = number of entries * size of each entry = 2²⁰ entries * 19 bits

$\Rightarrow$2²⁰ * 19 bits = 19922944 bits = 19922944 bits / 8bits = 2490368 bytes (because 1 byte = 8 bits).

$\Rightarrow$ 2490368 bytes = 2490368 bytes / 1024 bytes = 2432 K bytes (because 1 Kilo byte = 1024 bytes).

$\Rightarrow$  2432 K bytes =  2432 K bytes / 1024 K bytes = 2.375 M bytes ((because 1 Mega byte = 1024 K bytes).

Therefore the size of the page table = 2.375 Mb.