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If page fault service time is 50 milliseconds and memory access time is 100 nanoseconds, then what will be the effective access time, if the probability of page fault is p.
- a)(500000 + 100 p) nanoseconds
- b)(100 + 500000 x p) nanoseconds
- c)10-7 -10-7p (500000) seconds
- d)10-7 + 49.9 x 10-3 p seconds
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
If page fault service time is 50 milliseconds and memory access time i...
[(1- p)100 + p x 50 x 106] ns
Since, 1 ns = 10-9sec
=> 10-9[(1 -p)100 + p x 50 x 106] sec
=> [(1 - p) x 10-7 + p x 50 x 10-3] sec
=> 10-7 - 10-7p + p x 50 x 10-3
=> 10-7 + l0-3 x p f [50 - 10-4]
~ [10-7 + p x 10-3 x 49.9] sec
Since, 1 ns = 10-9sec
=> 10-9[(1 -p)100 + p x 50 x 106] sec
=> [(1 - p) x 10-7 + p x 50 x 10-3] sec
=> 10-7 - 10-7p + p x 50 x 10-3
=> 10-7 + l0-3 x p f [50 - 10-4]
~ [10-7 + p x 10-3 x 49.9] sec
Most Upvoted Answer
If page fault service time is 50 milliseconds and memory access time i...
Effective Access Time Calculation:
Given:
Page fault service time = 50 milliseconds = 50 * 10^6 nanoseconds
Memory access time = 100 nanoseconds
Probability of page fault = p
To calculate the effective access time, we need to consider two cases:
1. When a page fault occurs
2. When a page fault does not occur
1. When a page fault occurs:
In this case, the CPU needs to wait for the page fault service time before it can access the memory.
Effective access time = Page fault service time + Memory access time
= 50 * 10^6 nanoseconds + 100 nanoseconds
= (50 * 10^6 + 100) nanoseconds
2. When a page fault does not occur:
In this case, the CPU can directly access the memory without any delay.
Effective access time = Memory access time
= 100 nanoseconds
Now, we need to consider the probability of each case occurring. The probability of a page fault occurring is given as p.
Probability of page fault = p
Probability of no page fault = 1 - p
To calculate the overall effective access time, we need to consider the weighted average of the two cases based on their probabilities.
Effective access time = (Probability of page fault * Effective access time when page fault occurs) + (Probability of no page fault * Effective access time when no page fault occurs)
= (p * (50 * 10^6 + 100) nanoseconds) + ((1 - p) * 100 nanoseconds)
= (50 * 10^6p + 100p + 100 - 100p) nanoseconds
= (50 * 10^6p + 100) nanoseconds
Simplifying the expression, we get:
Effective access time = 50 * 10^6p + 100 nanoseconds
= 10^-7(500000p + 1) seconds
= 10^-7(500000p + 1) seconds
Therefore, the correct answer is option D) 10^-7(500000p + 1) seconds.
Given:
Page fault service time = 50 milliseconds = 50 * 10^6 nanoseconds
Memory access time = 100 nanoseconds
Probability of page fault = p
To calculate the effective access time, we need to consider two cases:
1. When a page fault occurs
2. When a page fault does not occur
1. When a page fault occurs:
In this case, the CPU needs to wait for the page fault service time before it can access the memory.
Effective access time = Page fault service time + Memory access time
= 50 * 10^6 nanoseconds + 100 nanoseconds
= (50 * 10^6 + 100) nanoseconds
2. When a page fault does not occur:
In this case, the CPU can directly access the memory without any delay.
Effective access time = Memory access time
= 100 nanoseconds
Now, we need to consider the probability of each case occurring. The probability of a page fault occurring is given as p.
Probability of page fault = p
Probability of no page fault = 1 - p
To calculate the overall effective access time, we need to consider the weighted average of the two cases based on their probabilities.
Effective access time = (Probability of page fault * Effective access time when page fault occurs) + (Probability of no page fault * Effective access time when no page fault occurs)
= (p * (50 * 10^6 + 100) nanoseconds) + ((1 - p) * 100 nanoseconds)
= (50 * 10^6p + 100p + 100 - 100p) nanoseconds
= (50 * 10^6p + 100) nanoseconds
Simplifying the expression, we get:
Effective access time = 50 * 10^6p + 100 nanoseconds
= 10^-7(500000p + 1) seconds
= 10^-7(500000p + 1) seconds
Therefore, the correct answer is option D) 10^-7(500000p + 1) seconds.
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If page fault service time is 50 milliseconds and memory access time is 100 nanoseconds, then what will be the effective access time, if the probability of page fault is p. a)(500000 + 100 p) nanosecondsb)(100 + 500000 x p) nanosecondsc)10-7 -10-7p (500000) secondsd)10-7 + 49.9 x 10-3 p secondsCorrect answer is option 'D'. Can you explain this answer?
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If page fault service time is 50 milliseconds and memory access time is 100 nanoseconds, then what will be the effective access time, if the probability of page fault is p. a)(500000 + 100 p) nanosecondsb)(100 + 500000 x p) nanosecondsc)10-7 -10-7p (500000) secondsd)10-7 + 49.9 x 10-3 p secondsCorrect answer is option 'D'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about If page fault service time is 50 milliseconds and memory access time is 100 nanoseconds, then what will be the effective access time, if the probability of page fault is p. a)(500000 + 100 p) nanosecondsb)(100 + 500000 x p) nanosecondsc)10-7 -10-7p (500000) secondsd)10-7 + 49.9 x 10-3 p secondsCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If page fault service time is 50 milliseconds and memory access time is 100 nanoseconds, then what will be the effective access time, if the probability of page fault is p. a)(500000 + 100 p) nanosecondsb)(100 + 500000 x p) nanosecondsc)10-7 -10-7p (500000) secondsd)10-7 + 49.9 x 10-3 p secondsCorrect answer is option 'D'. Can you explain this answer?.
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