If there are 32 segments, each size 1 k bytes, then the logical address should…
2015
If there are 32 segments, each size 1 k bytes, then the logical address should have
- A.
13 bits
- B.
14 bits
- C.
15 bits
- D.
16 bits
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Correct answer: C
In a segmented memory system, the logical address consists of a segment number and an offset. With 32 segments, we need log₂(32) = 5 bits for the segment number. Since each segment is 1 KB (1024 bytes), we need log₂(1024) = 10 bits for the offset. Therefore, the total logical address size is 5 + 10 = 15 bits.
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