The answer is not BEEFAC0H or 2DC9AFH. Question 15 6.6 A VM system uses segmenta

Question

The answer is not BEEFAC0H or 2DC9AFH. Question 15 6.6 A VM system uses segmentation with segment tables having maximum 256 entries and a maximum segment size of 4096 bytes. A section of the segment table is shown below, with all numbers expressed in hexadecimal (with the H suffix): mit Base ???? BH 320H42A40H CH 3FFH 9FAEH DH FAOH BEEFH EH 398H FABOOH What is the physical address expressed in base 16 for a hexadecimal virtual address of 2DACOH? ? 2DC9AFH ? 43500H O DOFFH O 399AFH ?9AFH No new data to save. Last checked at 6:06pm Su

Explanation / Answer

2DAC0, in this the first 8 bits are segment number i.e the segment number is 2D and the rest of the bits are offset i.e the offset here is AC0.

Therefore,

segment number = 2D

offset = AC0

Now if we map segment number 2D we see limit is FA0 .

Now by the rule of memory management offset size < limit.Here this is true as (AC0)10=2752 and (FA0)10 = 4000 and 2752 < 4000.

Now we add offset to the base to get the actual physical address = (AC0)10 + (BEEF)10 = 2752 + 48879 = 51631 = C9AFH

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