Set Associative Mapping

Question

A byte addressable system with 16-bit address lines with a 2-way set associative, write back cache with perfect LRU replacement. Assume 1 valid bit and 1 dirty bit maintains for each block. The tag store requires a total of 4352 bits of storage. What is the block size of the cache? [in bytes]

Comments

i dont think without knowing size of cache we can solve this...

Answer 1

#Set=$2^{S}$

#blocks=$2\times 2^{S}$.

Tag size=Number of sets$\times$(associativity $\times$(tag bit+valid bit +dirty bit)+LRU bit).

$4352 =2^{S}\times(2\times(t+1+1)+1).$

$4352 =2^{S}\times (5+2t).$

After solving.

S=8 and t=6.

$t+s+w=16$

So $6+8+w=16$

$w=2$

Block size=$2^{2}=4B$

Comments

can u plz tell..how u solved 4352=  2^s x (5 +2t)2×(5+2t)

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Already explained check the previous comment.

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great work bro..this question demand "intuition"