Question

consider a logical address space of 8 pages of 2028 words each, mapped onto a physical memory of 64 frames.
a) how many bits are there in the logical address?
B) how many bits are there in the physical address?

Answer 1

a) 1 word = 64 bytes => 2028*8 = 16224 bytes => 8*16224 bytes/page = 129792 bytes. We need 18 bits for offset ($2^{18}$=131072) and 3 bits which identify page number ($2^3=8$ pages). As a result, 18+3 = 21 bits.

b) 1 word = 64 bytes => 2028*8 = 16224 bytes => 64*16224 bytes/frame = 1038336 bytes. We need 20 bits for offset ($2^{20}$=1048576) and 6 bits which identify frame number ($2^6=64$ frames). As a result, 20+6 = 26 bits.