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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
in Operating System by Veteran (105k points) | 5.7k views

2 Answers

+15 votes
Best answer

Answer : 15 bits

To specify a particular segment, 5 bits are required. How 2^5 = 32 ===> so 5 bits

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

How 1 KByte ===> 2^10

So Total is 10+5 =15

by Boss (45.4k points)
selected by
0
with assumption of memory is byte addressable.
0
generally it is word addressable but here Byte.
+5
Generally it is byte addressable - it is more common.
+2 votes
Format of logical Address:       segment number           Block offset

32 segments, In order to represent these segments we require 5 bits

Block offset = 1 Kbytes = 2^10 bytes  =10 bits require for offset

5 + 10 =15 bits (C) will be answer
by Boss (42.4k points)
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