operating system - Hardware Support for Paging -

"the address consists of 16 bits, , page size 8kb. page table consists of 8 entries kept in fast registers." how total entries in page table 8?

according calculation should 1.

total entries in page table= ((2^16)/(2^3*2^10*2^3))=1. (the first 2^3 8 in 8kb, second 1 bytes bits conversion , 2^10 "kilo" in 8kb.)

thanks

memory byte-addressable hence, not need divide 2^3 bytes bit conversion.

explaining further, 16-bits address means processor generate memory addresses of length 16 bits used address byte or half-word or word present starting (or ending - depends on endianess of machine) @ 16-bit value.

now, page size total size of page in bits in case 2^16 bits. memory byte addressable, hence number of processor addresses in 1 page 2^16/2^3 i.e 2^13 addresses.

hence number of page table entries 2^16/2^13 = 8.