Question 52
52. In a paged memory, the page hit ratio is 0.40. The time required to access a page in secondary memory is equal to 120 ns. The time required to access a page in primary memory is 15 ns. The average time required to access a page is ______ .
Answer : (4) 78 | Previous | Next |
| UGC NET CS 2018 July - II Question 51 | UGC NET CS 2018 July - II Question 53 |
PAGE HIT = page present in ram or main memory
PAGE MISS = page fault so page swapped from the secondary memory
because we look for a page in secondary memory when the page is not present in the main memory, therefore, we need to add main memory lookup time.
Average access time
= hit ratio * primary memory access time + (1 – hit ratio) * secondary memory access time
Average access time = 0.4 * 15 + 0.6 * 120
Average access time = 6 + 72
Average access time = 78.
So, option (4) is correct answer.