ME MOCK

Question

Consider a computer system with 32 bit virtual addressing 36 bit physical addressing and page size of 8 KB. Each page table entry contain 2 valid bit 3 protection bit and 2 reference bit. The approximate size of page table if system uses single level paging ________ MB.

Comments

VAS=32 bits

offset = 13 bits (8KB)

bits for pages = 32 -13 = 19.

bits for frames = 36-19 = 23

entry of page table = 23 + 2 + 3 + 2 = 30 bits so we requires 4B

Size of page table = no. of entries in PT * size of each entry

                            = 2^19 * 4 B

                            = 2 MB