Explanation
1. Understanding the Components of a Logical Address
A logical address in a paging system is divided into two distinct parts:
Total Bits in Logical Address=Bits for Page Number (p)+Bits for Page Offset (d)
2. Finding the Bits for the Page Number (p)
The problem states that the logical address space consists of 256 pages.
Number of pages=256=28
To uniquely address 28 pages, we need exactly 8 bits.
Bits for Page Number (p)=8 bits
3. Finding the Bits for the Page Offset (d)
The size of each page is given as 210 bytes.
Page size=210 bytes
To address every individual byte within a page of size 210, we need exactly 10 bits.
Bits for Page Offset (d)=10 bits
4. Total Logical Address Size
Now, we add the bits calculated from both sections together:
Total Bits=8 bits+10 bits=18 bits
Correct Answer
The correct option is (d) 18 bits.
