GO Classes Test Series 2024 | Mock GATE | Test 13 | Question: 46

Answer 1

This question has been taken from page $7-$ https://www3.nd.edu/~mniemier/teaching/2011_B_Fall/lectures/2009_Final_Exam_Solutions.pdf

Coincidently GATE has also copied the same question from this university in year $2019$. Check -

https://gateoverflow.in/302815/gate-cse-2019-question-33

There are $128\left(2^7\right)$ entries in the TLB and there are $8192\left(2^{13}\right)$ entries per page. Therefore $2^7 \times 2^{13}$ implies $2^{20}$ (or $1 \mathrm{MB}$ ) addressable entries are covered by the TLB.

The answer would get better $b / c \;2^{13}$ would be larger.

Comments

What does the last line, “The answer would get better b/c 2^13 would be larger” mean?

---

The given page size is $2^{13}.$ So number of pages are $2^{40}/2^{13}=2^{27}$ pages and there are $128$ entries in $TLB.$

Since we use $TLB$ on top of paging, so number of addressable locations will be $2^{27}/2^7=2^{20}$.

Is this way of thinking correct? @Sachin Mittal 1