(A) 12 bits
(B) 15 bits
(C) 16 bits
(D) 20 bits
Answer c.
Explanation: Size of a page = 4KB = 212
Total number of bits needed to address a page frame is 32 – 12 = 20
If there are ‘n’ cache lines in a set, the cache placement is called n-way set associative. Since TLB is 4 way set associative and can hold total 64 (2^6) page table entries, number of sets in cache = 26/4 = 24. So 4 bits are needed to address a set, and 16 (20 – 4) bits are needed for tag.
Total number of bits needed to address a page frame is 32 – 12 = 20
If there are ‘n’ cache lines in a set, the cache placement is called n-way set associative. Since TLB is 4 way set associative and can hold total 64 (2^6) page table entries, number of sets in cache = 26/4 = 24. So 4 bits are needed to address a set, and 16 (20 – 4) bits are needed for tag.
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