ACE TEST SERIES QUESTION

Question

Comments

12?

EDIT : 22

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I got 19  MiNiPanda

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Cache size =  2 ¹⁸ B

block size = 2 ¹⁰ B

given that , memory is word addressable

therefore ,  No of words in the block = 2 ¹⁰ / 2 ³  = 2⁷ B

No of blocks  = 2 ¹⁸ /  2 ¹⁰ =  2 ⁸

No of sets = 2 ⁸ / 2 ³ = 2 ⁵

therefore , tag bits + 5 + 7 = 34

                       tag bits = 22 bits

 

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@Magma

I got 12,15,7 for the 3 fields tag,set and word offset. What did you get?

And this is a virtual cache which is not very clear to me..i have tried reading several pdfs on it..Do we need to do some other calculation for this?

yes magma.22 should be right..i did calculation mistake

Answer 1

TAG	SET INDEX	WORD OFFSET
22	5	7

Cache Size = 256 KB

Block Size = 1KB

Word Length 64 bits.

8-way Set Associative Mapping 

#f blocks in Cache  = 256 KB /1KB = 2^8.

1 set acquiring 8 blocks.

Total number of sets possible  = 2^8/2^3=32=2^5

Word offset = words per block = 2^10*8bits/2^6 bits = 2^7.

TAg bits =22.