Question

If there are $32$ segments, each of size $1$k byte, then the logical address should have

• A. $13$ bits
• B. $11$ bits
• C. $15$ bits
• D. $16$ bits

Answer 1

The correct option is C $15$ bits

If there are 32 segments, each of size 1k byte, then the logical address should have 15 bits.

To specify a particular segment, 5 bits are required. To select a particular byte after selecting a page, 10 more bits are required. Hence 15 bits are required.

To specify a particular segment, 5 bits are required (since 2^{^5} = 32). Having selected a page, to select a particular byte one needs 10 bits (since 2^{^10}= 1K byte). So, totally 5 + 10 =15 bits are needed.