Consider a memory-management system based on paging. The total size of the physical memory is 2 GB, laid out over pages of size 8 KB. The logical address space of each process has been limited to 256 MB. a. Determine the total number of bits in the physical address. b. Determine the number of bits specifying page replacement and the number of bits for page frame number. c. Determine the number of page frames. d. Determine the logical address layout.

A. 31 bits

B. 13 bits, 18 bits

C. 256k

Explanation:

A.

We have total size of physical memory as 2gb

This means that 2¹ x 2³⁰

Using the law of indices,

= 2³¹

So physical address is equal to 31 bits

B.

We get page size = 8kb = 2³

1 kb = 2¹⁰

2³ * 2¹⁰ = 2¹³

So page replacement = 13 bits

Page frame number

= physical address - replacement bits

= 31 - 13

= 18 bits

C.

The number of frames is given as

2¹⁸ = 2⁸ x 2¹⁰ = 256K

So number of frames = 256K

D

Logical add space is 256K

Page size= physical pages = 8kb

Logical address layout= 28biys

Page number = 15 bits

Displacement = 12 bits

256k/8k = 32 pages in address space process

answer; i believe that the correct answer is a ( check first..good luck