Solution
a)Determine number of lines in cache
in the questions, given
2 MB cache
1 KB per cache line
for easy calculation we will convert into 2 power
please note that
1 MB= 2 20
1 KB= 2 10
Number of lines in cache
=cache/line
cache=2 MB= 2* 2 20
line= 1 KB= 2 10
=2* 2 20/2 10
=2 21/2 10
=2 11
=2048
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b) Determine the number of address bits out of the processor
Number of bits required to address all bytes in the largest memory given as per the question is
=256 MB
=256 * 2 20
=2 8 * 2 20
= 2 28
so 28 bits is the answer
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c) Determine the number of bits needed for the block offset section of the address
Number of bits needed for the block offset section of the address
=size of the cache line
=1 KB
= 2 10
=10 bits
--
answered first 3 question, really sorry for that
please post remaining question separately, love to answer
all the best
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