Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.3k views
in Computer by (106k points)
closed by
Consider a system with 2 level caches. Access times of Level 1 cache, Level 2 cache, and main memory are 1 ns, 10 ns, and 500 ns, respectively. The hit rates of Level 1 and Level 2 caches are 0.8 and 0.9, respectively. What is the average access time of the system ignoring the search time within the cache? 
1. 13.0 ns
2. 12.8 ns
3. 12.6 ns
4. 12.4 ns

1 Answer

0 votes
by (103k points)
selected by
 
Best answer
Correct Answer - Option 3 : 12.6 ns

Data:

T1 = 1 ns, T2 = 10 ns and T3 = 500 ns

H1 = 0.8, H2 = 0.9

Average access time = Tavg

Formula:

Tavg = H1T1 + (1 - H1) H2 (T1 + T2) + (1 - H1) (1 - H2) (T1 + T2 + T3)

Since search time within a cache is ignored

Tavg = H1T1 + (1 - H1) H2 (T2) + (1 - H1) (1 - H2) (T3)

Calculation:

Tavg =  0.8 × 1 + 0.2 × 0.9 × 10 + 0.2 × 0.1 × 500

Tavg = 0.8+ 0.18 +  10 = 12.6 ns

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...