Question

A simple paging system has a memory size of 256 bytes and a page size of 16 bytes.
i. What is the size of the page table?
ii. How many bits exist for an address, assuming 1-byte incremental addressing?
iii. State p and d values (i.e. the page number and the offset).
iv. Perform address translation of 64 bytes to physical address space using the page table below.
0 8
1 6
2 3
3 11
4 7

Answer 1

is memory Size = 256 Page Size = 16 bytes bytes of entries in page Number table = memory size Page Size = 256 16 memory size = 256 bytes ligte 28 Encremental addressing mer size = 256 1 L > there will be 256 256 hence 8 lits exists addresses for an address chi الم پاک دا No = 16 =24 p=4 bits No of pages bits in address = 8 bits lits in offset = 8-4 = 4 bits d = 4 bits of of No

(IV) 64 luyter - 26 X8 = 29 Luis offset – Чы frame size = 9 -4 = 5 bits Ооооо) xxxx ) = | ooo доо | xxx x o 8 OVOX X X X DODOT XXX X ООП) xxxx 16 23 3 11 ч + оооо (x xxx7) Гооохххх — D Обо Хххх го)) xxxx Горіхххх

Please note that in part iv

XXXX represents the 4 bits of offset, these 4 bit remain same, rest 5 bits are converted to 4 bits of p (physical address) according to the page table given

E.g

5 bits representing 0 is changed to 4 bits representing 8

00000xxxx => 1000xxxx