Question

If there are 32 segments, each size 1 k bytes, then the logical address should have 
1. 13 bits
2. 14 bits
3. 15 bits
4. 16 bits

Answer 1

Correct Answer - Option 3 : 15 bits

Data:

Logical address size = L bytes

no of segments = n = 32

Size of each segment = s= 1 k byte = 1 × 2¹⁰ bytes

Formula:

Number of bits = log₂ L

Calculation:

L = n × s  bytes

L = 32 × 210  = 215 bytes

Number of bits = log2 215 = 15