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4. (4pts) Assume that you have a direct-mapped cache with four-word blocks and a total size of 16 words that is initially empc). word address 23 (continuing from the previous step) Hit or Miss? What is the cache content at this point? Block number 06e) word address 43 (continuing from the previous step) Hit or Miss? What is the cache content at this point? Block number 010g). word address 25 (continuing from the previous step) Hit or Miss? What is the cache content at this point? Block number 0if there is an empty box please put none. If the image is too small please open it in a new tab for an enlarged version.

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Answer #1

Given that cache size is 16 words and block size is 4 words. So, total number of blocks in cache is = Cache Size/Block Size = 16/4 = 4.

Cache offset need log(4) = 2 bits because block size is 4 words. (Least significant 2 bits are used for block offset)

And number of blocks are 4, so number of bits are for block indexing is log(4) = 2 bits.

And the remaining most significant bits are for tag.

a). word address 21 = (010101). Tag: 01, Block: 01, Word Offset: 01 Hit or Miss ? = Miss in block 1 What is the cache content

c). word address 23 (continuing from the previous step) = (010111). Tag = 01, Block : 01, Word offset = 11 Hit or Miss? Hit i

e). word address 43 (continuing from the previous step) = (101011). Tag: 10, Block number = 10 Word offset = 11 Hit or Miss?

g). word address 25 (continuing from the previous step)= (011001. Tag = 01, Block number = 10, Word Offset = 01 Hit or Miss?

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