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DO bit 0 1 Q-2 (25p): Assume that logical addresses are 16-bit long: 3 bits for segment no and 13 bits for offset. Segment ta

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Answer #1

a. Since there are 3 bits for segment no in logical address, hence at max 00 segments are possible.

b. given logical address is 0100 1001 1101 1110, in this the most significant 3 bits represents the segment number.

which is 010 = 2. By looking at third row of segment table, base address of segment in main memory is

0010 0000 1000 0011

so physical address corresponding to given logical address = base address of segment + segment offset

physical address =

0010 0000 1000 0011

+   01001 1101 1110(offset of logical address)

-------------------------------------------

  0010 1010 0110 0001

c. since the present bit in third row of segment table is 1, means segment is present in main memory, hence no page fault.

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