number 2 There are three small caches (direct mapped, 2-way set associative, and
ID: 3697675 • Letter: N
Question
number 2
Explanation / Answer
It is specified that there are 8 words per block.
This means 28=256 bytes per block. A 32 bit address gives 32-4=28 bytes to store the index and tag.
In the case of direct-mapped cache, the number of sets is the same as the number of blocks.
index=log2(1k)=10 bits
Total number of tag bits=(28-10)*1K sets=18*1=18Kbits
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