Made Easy Test Series cache Memory

Question

consider a 32 bit microprocessor which has 32K byte which is 4-way set associative cache.Block size of cache is 2 32-bit words.The set number to which wprd from memory location FAFEEBE1 wrapped ___.

I'm confused about 2 32-bit words as block size .Answer given was based on <Tag,Set,Word> =<19,9,4>

Answer 1

physical address

29

3

no of sets = n/ assocoativity 
        n= cache size/block size 
          =32Kbyte/2³

            =2¹⁵ / 2³ =2¹²

     no of sets = 2¹² /2²

                              = 2¹⁰ 

   set bits = 10 

19 TAG

10 set

3 block offset

1111 1010 1111 1110 1110 1011 1110 0001

blue color marked will give hexa decimal as : 0001 0111 1100

                                                                        : 17C (ANSWER)

Comments

Dont you think set bits should be 10?

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yes corrected now :)

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Here why we are considering byte offset in a block. It should be word offset. As it is mentioned the set number to which word from memory location.

Is it not so?

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Sir, in the question, it is given that "block size of cache is two 32 bit words". So, shouldn’t the block offset be taken in words instead of bytes? If taken in words, block offset will be 1 bit only.

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the amswer provided by madeasy is wrong i believe because block size bit denoted by number of words in it since 2 words hence block size should be 1 bit set size is 10 bit as solved in the answer and tag bit would be 21.

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I am also thinking the same but unable to verify this by any means. @focus _GATE sir, can you please verify this.

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let me try to clarify it to you.. we know memory is word addresable.. so if given microprocessor is 32 bit that means total 2^32 words are there... there is no point finding 1 byte in the memory.. if a word is greater than 1 byte..so we dont give significance to a byte here..because the least amount of memory that we can accees is the word ..so a block is also always represented as words..since there are only 2 words in a block you will only need 1 bit to identify them.. this is the basic logic that is very useful in all address to adress conversions

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Yes I am using the exact same logic. I also think that block offset should be 1 bit only. My doubt is “The accepted answer has taken 3 bits for block offset. Why?”