gate2018

Question

i think option b is correct

Comments

Bro from where u r getting this question ?? Plz share the link

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this one is the only sensible reply..the option is not present there.

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i am getting       P-N+logk

which is not matching with any option

Answer 1

Answer B

Yes true .. Simple formula ..

Tag bits in K-way = log( size of MM / size of CM ) + log(k)

here ... tag bits = log(2^p / 2^n ) + Log k  = log(2^(p-n) ) + log(k) = p - n + log(k)

Comments

The answer is not in option. Option B is P-M+ log(k).

Answer 2

Physical Address space is 2^P bytes i.e. P bits to represent total memory.
Cache Size 2^N i.e. bits to represent Cache memory.

K way associative.

(Size of Tag in bits) + (Cache Size/K in bits) = (Total memory in bits)
(Size of Tag in bits) + N-log(K) = P
(Size of Tag in bits) = P - N + log(K)

Answer 3

Yes b is correct