UGC NET CSE | December 2018 | Part 2 | Question: 19

Question

Consider a system with $2$ level cache. Access times of Level $1$ cache, Level $2$ cache and main memory are $0.5$ ns, $5$ ns and $100$ ns respectively. The hit rates of Level $1$ and Level $2$ caches are $0.7$ and $0.8$ respectively. What is the average access time of the system ignoring the search time within the cache?

• A. $35.20$ ns
• B. $7.55$ ns
• C. $20.75$ ns
• D. $24.35$ ns

Comments

Hi Arjun. The questions of UGC NET Dec-2018 were not asked in a particular order/format like the question number which you have posted here is numbered 19 whereas it was question number 90 in some other set. The better idea would be to post these questions by their unique question ID. The question ID of this question is : 91394310763 even better would be to use only the last three digits of this question ID i.e 763 which will be unique and in continuation and can be easily  searched by anyone in the format like UGCNET-DEC2018-II-763    UGCNET-DEC2018-II-764   UGCNET-DEC2018-II-765 etc

---

UGC answer key was published in this order rt?

---

Yes!! If you have considered the first answer key published then yes this question was published on number 19th in paper 2 section  but I was searching according to UNIQUE ID of question as a logical choice because question number is different in every paper in hand, so first logical choice came to me was to search using unique ID using the classical format of searching questions on Gate overflow i.e UGCNET-DEC2018(MONTH YEAR)-II(Paper Number)-763(Unique question ID).

---

Each question also have its question number - you can see the name attribute or file image name. That is the order we used and also for answer keys.

Answer 1

ignoring search time with in the cache ==> this statement tells us to use parallel access.

$T_{avg\;access} = H_1.T_1 + \color{red}{(}1-H_1 \color{red}{)} \color{red}{(} \color{green}{H_2 . T_2 + (1-H_2).T_M} \color{red}{)}$

$T_{avg\;access} = (0.7*0.5) + \color{red}{(}0.3\color{red}{)} \color{red}{(} \color{green}{ (0.8*5) + (0.2).100} \color{red}{)}$

$T_{avg\;access} = (0.35) + \color{red}{(}0.3\color{red}{)} \color{red}{(} \color{green}{ 4 + 20} \color{red}{)}$

$T_{avg\;access} = 0.35 + 7.2 = 7.55\;ns$

Comments

Hows search time related to "parallel access"?

---

sir, i mean to say " it is not hierarchical access ",

Due to when we miss in Cache$_1$, we doesn't add searching time of Cache$_1$ as miss penalty.

---

how we came to know that it is asking for parallel access not for hierarchal access.

Answer 2

[0.5+0.3(5+0.2*100)]

7.55

https://youtu.be/QgpdjU5K_N0

Comments

This one is giving 0.5+7.5=8.0

Answer 3

Average access time = level 1 hit rate( level 1 access time) + (level1 miss rate)(level 2 hit rate(level 2 access time)+ (level 1 miss rate)( level 2 miss rate) (main memory access time)
Average access time = 0.7(0.5) + 0.3(0.8)(5) + 0.3(0.2)(100)
Average access time = 7.55 ns