Please help me understand these. Explanations would be helpful Thank You :) Supp

Question

Please help me understand these. Explanations would be helpful

Thank You :)

Suppose a computer using fully associative cache has 2^24 bytes of bvte-addressable main memory and a cache of 128 blocks, where each cache contains 64 bytes. How many blocks of main memory are there? What is the format of a memory' address seen by the cache; that is, what are the sizes of the tag and the offset field? To which cache block will the main memory address 0x01D872 map? Calculate the overall speedup of a system that spends 65% of its time on 1/0 with a disk upgrade that provides for 50% greater throughput.

Explanation / Answer

3)

i) Number of blocks of main memory are:

given 2^24 bytes and cache block contains 64 ==> 2^6

====> 2^24 / 2^6 == 2^18
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ii)
sizes of cache and tag fields are:
24 bit addresses with 18 bits in the tag field and 6 in the offset field
explaination:
--------------
cache 2^6 6 bits already given . So from total 24 bits ==>24-6 = 18 bits
which was calculated in above answer

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iii) Since it's associative cache, it can map anywhere

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4)
Amdahl's Law Equation

S = 1/(1-F) + F/K
where F = 0.65, K = 1.5

so S = 1.28 or 28% speedup

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