# GATE | GATE-CS-2006 | Question 62

• Difficulty Level : Easy
• Last Updated : 28 Jun, 2021

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:
Virtual Memory would not be very effective if every memory address had to be translated by looking up the associated physical page in memory. The solution is to cache the recent translations in a Translation Lookaside Buffer (TLB). A TLB has a fixed number of slots that contain page table entries, which map virtual addresses to physical addresses.

Solution
Size of a page = 4KB = 2^12 means 12 offset bits CPU generates 32-bit virtual addresses
Total number of bits needed to address a page frame = 32 ā 12 = 20
If there are ānā 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 ā 5) bits are needed for tag.

Option (C) is the correct answer.

See Question 3 of https://www.geeksforgeeks.org/operating-systems-set-14/

This solution is contributed by Nitika Bansal

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