101
If there are 64 segments, each of size 2KB. then how many bits should a logical address have ?
- A6 bits
- B11 bits
- C17 bits
- D5 bits
Answer not set
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CCommunity
64 segment means $ 64 = 2^6$
2KB means $ 2^{11}$
So in total we need $6 + 11 = 17$ bits for logical address
2KB means $ 2^{11}$
So in total we need $6 + 11 = 17$ bits for logical address