GATE 2015 CS (CS2) – Question 19
A computer has 40-bit virtual addresses, 8 KB pages, and a 128-entry TLB arranged as 32 sets with four ways each. The TLB stores no process ID. Find the minimum TLB-tag length in bits.
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Correct answer: 22
Explanation
An 8 KB page needs thirteen offset bits, leaving 40−13=27 virtual-page-number bits. Thirty-two TLB sets use five of those bits as the set index. The remaining 27−5=22 bits form the tag. Four-way associativity does not add tag bits.