Operating Systems | Memory Management | Question 5
A CPU generates 32-bit virtual addresses. The page size is 4 KB. The processor has a translation look-aside buffer (TLB) which can hold a total of 128 page table entries and is 4-way set associative. The minimum size of the TLB tag is:
(A) 11 bits
(B) 13 bits
(C) 15 bits
(D) 20 bits
Explanation: Size of a page = 4KB = 2^12
Total number of bits needed to address a page frame = 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 128 (2^7) page table entries, number of sets in cache = 2^7/4 = 2^5. So 5 bits are needed to address a set, and 15 (20 – 5) bits are needed for tag.
My Personal Notes arrow_drop_up
Please Login to comment...