Question

A 16 bit logical address is split into 5 bit page number and 11 bit page offset. Answer the following questions in the space provided:

a) The page size is ____ bytes

b) The page table size is ____

c) if the page table is as shown below, then the corresponding physical address for 0x174F is given by: ____.

Question 29 (3 points) Saved A 16 bit logical address is split into 5 bit page number and 11 bit page offset. Answer the following questions in the space provided: a) The page size is ____ bytes b) The page table size is c) if the page table is as shown below, then the corresponding physical address for 0x174F is given by: ___ (Format: use hexadecimal 16 bit number, if unknown write N/A) 0 Ox10 1 2 OxОА Ox1E OxOF 3

Answer 1

Logical address contains two Things

1. Page Number :--> the number of bits to Represent a Page

2. Page Offset :--> the number of bits to Represent a word in Page

Part A)

Page offset = 11 bits

So page size = 2^11 Bytes = 2 KB

Page size is 2 KB

Part B)

Size of page table is = Number of pages x size of each page table entry

Now Page table entry has frame number as in the given page table we can see that one frame need 8 bits to represent like 0x1C

So size of page entry = 8 bits = 1 Byte

To get number of pages we have page number = 5 bits So we have 2^5 = 32 pages  

Page Table Size = 32 B

Part C)

Now To write Physical address of the given logical address

Physical Address = Frame Number + Page/Frame Offset

As we see in page table one frame needs 8 bits and Page offset = 11 bits

Physical Address = 19 bits

Now In Given Logical Address we have 0x 174F

174F in binary = 00010 11101001111

First five bits represent the page number

SO we have page number = 00010 = 2

In page table at page 2  we have frame as 0x 1E = 0001 1110

Now Physical address (19 bits) = 00011110 11101001111

Hence Physical Address for 0x 174F is = 00011110 11101001111 = in hex ( 16 bit number ) 0x F74F

This is the complete solution for the given problem

I hope that you understood the answer and if u have any doubt do mention in comment .I will be happy to answer

Thank You