Question 9946 – Computer-Organization
November 7, 2023Question 8084 – Operating-Systems
November 7, 2023Operating-Systems
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Question 50
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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:
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i + (j/k)
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i + (j*k)
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(i+j)/ k
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(i+j)*k
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= 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).