Question

Suppose you have a hypothetical machine that has 4 MB of physical memory and a 24-bit address space.

Your hypothetical machine uses paging in a linear page table. Each page holds 512 bytes.

Each page still holds 512 bytes.

Given the following virtual address, partition the bits into a virtual page number and an offset.

0xC3D0E6

Answer 1

In linear page tables, the memory menagement unit(MMU) splits a virtual address into page number and page offset components. The page number is used to index into an array of page table entries.

page number

offset

phsical memory = 4MB=4 * 2²⁰ =2²² bytes

given 24 bit address space,

virtual address space = 2²⁴ Bytes (Total number of bits in virtual address=24 bits)

page size=512 Bytes =2⁹ Bytes

Number ofpages = virtual address space/(page size) =2²⁴/ 2⁹ =2¹⁵

So page number requies 15 bits.

number of bits in offset = (total no of bits in virtual address) -( number of bits in page number)

                                      =24-15=9 bits

So 24 bit virtual address is divided into as follows

15 bit page number

9 bit offset

(C3D0E6)₁₆ = (1100 0011 1101 0000 1110 0110)₂

[To convert hex to binary, write 4 bit binary of each hex bit (C-->1100, 3->0011 , D-->1101, 0-->0000 , E-->1110,

6-->0110)]

First 15 bit represents page number , 1100 0011 1101 000 =0x61E8

(110000111101000)₂ = ( 110 0001 1110 1000)_{2 =} (61E8)₁₆

(To convert binary to hex, we make group of 4 bits, starting from lsb and write the hex code for each group 1000->8, 1110-->E, 0001-->1, 0110-->6)

last 9 bits represents offset0 1110 0110

(011100110)₂ = (0 1110 0110)_{2 =}(0E6)₁₆ =(E6)₁₆

_So page number=(110000111101000)_{2 =}(61E8)_{16 ,} offset =(011100110)₂₌(E6)₁₆