Question 9946 – Computer-Organization
November 7, 2023Question 8084 – Operating-Systems
November 7, 2023Operating-Systems
| Question 50 |
If an instruction takes i microseconds and a page fault takes an additional j microseconds, the effective instruction time if on the average a page fault occurs every k instruction is:
| i + (j/k) | |
| i + (j*k) | |
| (i+j)/ k | |
| (i+j)*k |
= service time + page fault rate * page fault service time
= i + 1/k * j
= i + j/k
or
The effective instruction time, if on the average a page fault occurs every k instructions, is:
i + (j / k)
This formula accounts for the normal instruction time (i) and the additional time for a page fault (j), considering the average frequency of page faults (k instructions between page faults).
= service time + page fault rate * page fault service time
= i + 1/k * j
= i + j/k
or
The effective instruction time, if on the average a page fault occurs every k instructions, is:
i + (j / k)
This formula accounts for the normal instruction time (i) and the additional time for a page fault (j), considering the average frequency of page faults (k instructions between page faults).