set associative cache

Question

The physical address size on a machine is 36 bits.The number of tag bits in the physical address format in a 256 KB, 16 way block set associative cache is____bits.

What will be the ans?

Answer 1

Size of physical address, m = 36

$C = S \times E \times B$

where $C$ is cache size,

$S = 2^s$ is the number of sets, 

$E$ is the number of cache lines per set, and

$B = 2^b$ is the block size in bytes.

It is given that, $C = 256\ \rm{KB} = 2^{18}\ \rm{bytes}$ and $E = 16$.

So, $2^{18} = 2^s * 2^4 * 2^b$

or, $s + b = 14$.

Therefore, the number of tag bits will $m - (s+b) = 36 - 14 = 22$.

Answer 2

consider block size=cache line size=2^x bytes

then #cache line=2¹⁸/2^x=2^(18-x)

then #sets=2^(18-x)/2⁴  =2^(14-x)

physical address bits=36=#tagbits+#setbits+#blockoffsetbits

#tagbits+(14-x)+(x)=36

#tagbits=36-14=22bits

Comments

Thank u, I got it.