# made easy demo test #Q 23

Question

 my answer 5 plz check but answer given 4 

Comments

u  mean 5 bit  but answer given 4 ....so i confirm i  also  got 5 bit

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Let x be the set number

then x = 32-16-5 = 11

lines of the cache = 2^18/2^5 = 2^13

hence to match with set bits ,

2^11 = 2^13/k

hence k = 4

Answer 1

Answer given by them is absolutely right... I think the problem with your solution is that you are including the additional bits..

You must understand the question.. It is asking for additional bits in cache directory,not for the tag bits..

Therefore.. 16 bits will be for tag

5 bits for block size and 11 bits for sets..

#of cache lines are (2^8 * 2^10)/2^5= 2^13

#of sets =2^11=2^13/k

Then k should be equal to 4.

Comments

but these additional  bit is a part of TAG ... what is cache tag directory??

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I can understand your confusion.. Actually the cache directory is computed by multiplying the tag bits with cache lines.. So when you were asked in question to find the size of cache directory then you will add the additional bits with tag 

(16+2+1)*2^13

For the better understanding you can see the following discussion https://gateoverflow.in/157603/set-associative