Question

Paging Questions

1. A page is 1 KB in size. How many bits are required to store the page offset?

2. A page entry has 10 bits. What is the size of the page table?

3. A logical address is 32 bits long. The page size is 4 KB. Divide the address into its page number and offset.

4. The following hexadecimal addresses are used in a system with a 20-bit logical address where the page size is 256 bytes. What are the values of the page number and offset for each address? • 0x10fba • 0xabc28 5. Convert the following logical addresses (provided in hexadecimal) to their appropriate physical addresses using the provided page table. Assume logical addresses are 20 bits long with 12 bits for the page entry and 8 bits for the offset. List your answers in hexadecimal.

• 0x01127

• 0x019aa

• 0x03911

• 0x13dad

Entry Number(decimal) Page Frame(decimal)

0 99

... ....

17 320

... ...

25 24

... ...

45 425

.. ...

57 331

.. ...

300 57

... ...

317 200

.. ..

57 331

Answer 1

1.

Given,

Page size = 1 KB

Number of bits to store the offset = log ( 1 KB ) = 10 bits.

2.

Size of page entry = 10 bits

Since number of pages or logical address is not given,

Size of the page table = ( Number of pages x 10 ) / 8 Bytes.

3.

Logical address = 32 bits

Page size = 4 KB.

Number of bits to represent offset = log ( 4 KB ) = 12 bits

Number of bits used to represent the page number = Bits in Logical address - Number of bits to represent offset

= 32 - 12 = 20 bits

4.

Logical address = 20 bits

Page size = 256 Bytes

Number of bits to represent offset = log ( 256 B ) = 8 bits

Number of bits used to represent the page number = Bits in Logical address - Number of bits to represent offset

= 20 - 8 = 12 bits

So, the first 12 bits from the right of the logical address represent the page number and the next 8 bits represent the offset.

i. Logical address = 0x10fba = 00010000111110111010

Page number = 000100001111 = ( 271 )10.

Offset = 10111010 = ( 186 )10.

ii. Logical address = 0xabc28 = 10101011110000101000

Page number = 101010111100= ( 2748 )10.

Offset = 10111010 = ( 40 )10.

5.

Given,

Logical address = 20 bits

Offset = 8 bits

Bits to represent page number = 12 bits.

Page table entry = 12 bits.

Entry	Page Frame
0	99
17	320
25	24
45	425
57	331
300	57
317	200
57	331

i.

Logical address = 0x01127 = 00000001000100100111

Page number = 000000010001 = 17

Offset = 00100111

So, the page 17 maps to frame 320. =  101000000

Physical address = ( Frame Number ( ( Offset ) = 00010100000000100111 = 0x14027.

ii.

Logical address = 0x019aa = 00000001100110101010

Page number = 000000011001 = 25

Offset = 10101010

So, the page 25 maps to frame 24 = 000000011000

Offset = 10101010

Physical address = 00000001100010101010 = 0x18aa.

iii.

Logical address = 0x03911= 00000011100100010001

Page number = 000000111001= 57

Offset = 00010001

So, the page 57 maps to frame 331 = 000101001011

Offset = 00010001

Physical address = 00010100101100010001 = 0x14b11.

iv.

Logical address = 0x13dad= 00010011110110101101

Page number = 000100111101= 317

Offset = 10101101

So, the page 317 maps to frame 200 = 000011001000

Offset = 00010001

Physical address = 00001100100000010001 = 0xc811.