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
240 views
in Computer by (101k points)
closed by
A computer system has a cache with access time 10 ns, a hit ratio of 80% and average memory access time is 20 ns. Then what is the access time for physical memory?
1. 50 ns
2. 40 ns
3. 30 ns
4. 20 ns

1 Answer

0 votes
by (108k points)
selected by
 
Best answer
Correct Answer - Option 1 : 50 ns

Concept:

TM = h × TC + (1 – h) × TP

TM is the average memory access time

TC is the cache access time

TP is the access time for physical memory

h is the hit ratio

Calculation:

Given: TC = 10 ns, TM = 20 ns, h = 80% = 0.8

TM = h × TC + (1 – h) × TP

20 = 0.8 × 10 + (1 – 0.8) × TP

20 = 8 + 0.2 TP

TP = 60 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

...