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Given,

cache size = 256KB

PA = 36 bits

set associativity = 16

So, assume block size is $2^{x}$

Then, no of cache blocks = $\frac{2^{18}}{2^{x}}$

no of sets in cache would be = $2^{18-x-4}$ 

So now the PA split up would be

t + 14-x + x =36 (Assume t bits for tag)

t = 22 bits.

Check this question for reference:

https://gateoverflow.in/39622/gate2016-2-32?show=39622#q39622

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