ACE TESTSERIES

Question

WHAT WILL BE THE ANS I M GETTING 10  BITS IN TAG FIELD!!! IS IT CORRECT THEY SHOWS DIFFERENT ANS NOT 10!!

ANYONE TRY PLEASE AND LET ME KNOW THANKS IN ADVANCE!! 

Comments

Block size = 64 words

No of lines in cache  = 32 blocks

No of frames in MM = 256 blocks

BO : 6 bits

No of sets = 32 / 8 = 2²

2 bits required to represent no of of sets

 

MM size = 256 * 64 = 2 ¹⁴ words

14 - (6+2)

 = TAG BITS = 6 bits ?

 

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@magma they said 6

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Yes, Answer will be 6.
What's your problem??

Answer 1

Block size = 64 words = 2⁶

So, Block offset = 6 bits

No. of Blocks in Cache = 32 = 2⁵  = No. of Lines in cache

Main Memory size  = 256 Blocks = 256 * 2⁶  = 2^{14 .}

Set No = (No. of lines in the cache) / 8  [ Since 8-way set-associative]

= 2⁵ / 8 = 2²

So set no = 2 bits

Hence TAG = 14 - (6+2)

= 6 bits