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Let the page fault service time be 10ms in a computer with average memory access time being 20ns. If one page fault is generated for every 106 memory accesses, what is the effective access time for the memory?
- a)21ns
- b)30ns
- c)23ns
- d)35ns
Correct answer is option 'B'. Can you explain this answer?
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Let the page fault service time be 10ms in a computer with average mem...
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Let the page fault service time be 10ms in a computer with average mem...
Effective Access Time Calculation:
The effective access time (EAT) is calculated as the average time taken to access memory, taking into account both the time for a hit (when the data is found in the cache) and the time for a miss (when the data is not found in the cache and a page fault occurs).
Given:
Page fault service time = 10ms = 10,000,000ns
Average memory access time = 20ns
Page fault rate = 1 page fault / 10^6 memory accesses
To calculate the effective access time, we need to consider the probability of a hit and a miss.
Probability of a hit:
The probability of a hit is the complement of the page fault rate. In this case, the page fault rate is given as 1 page fault per 10^6 memory accesses. Therefore, the probability of a hit is 1 - 1/10^6 = 1 - 0.000001 = 0.999999.
Probability of a miss:
The probability of a miss is the page fault rate. In this case, the page fault rate is given as 1 page fault per 10^6 memory accesses. Therefore, the probability of a miss is 1/10^6 = 0.000001.
Calculation of Effective Access Time:
The effective access time can be calculated using the following formula:
EAT = (Probability of hit * Memory access time) + (Probability of miss * (Memory access time + Page fault service time))
Substituting the given values:
EAT = (0.999999 * 20ns) + (0.000001 * (20ns + 10,000,000ns))
= (19.99998ns) + (10ns)
= 29.99998ns
≈ 30ns
Therefore, the effective access time for the memory is approximately 30ns, which corresponds to option B.
The effective access time (EAT) is calculated as the average time taken to access memory, taking into account both the time for a hit (when the data is found in the cache) and the time for a miss (when the data is not found in the cache and a page fault occurs).
Given:
Page fault service time = 10ms = 10,000,000ns
Average memory access time = 20ns
Page fault rate = 1 page fault / 10^6 memory accesses
To calculate the effective access time, we need to consider the probability of a hit and a miss.
Probability of a hit:
The probability of a hit is the complement of the page fault rate. In this case, the page fault rate is given as 1 page fault per 10^6 memory accesses. Therefore, the probability of a hit is 1 - 1/10^6 = 1 - 0.000001 = 0.999999.
Probability of a miss:
The probability of a miss is the page fault rate. In this case, the page fault rate is given as 1 page fault per 10^6 memory accesses. Therefore, the probability of a miss is 1/10^6 = 0.000001.
Calculation of Effective Access Time:
The effective access time can be calculated using the following formula:
EAT = (Probability of hit * Memory access time) + (Probability of miss * (Memory access time + Page fault service time))
Substituting the given values:
EAT = (0.999999 * 20ns) + (0.000001 * (20ns + 10,000,000ns))
= (19.99998ns) + (10ns)
= 29.99998ns
≈ 30ns
Therefore, the effective access time for the memory is approximately 30ns, which corresponds to option B.
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Let the page fault service time be 10ms in a computer with average memory access time being 20ns. If one page fault is generated for every 106 memory accesses, what is the effective access time for the memory?a)21nsb)30nsc)23nsd)35nsCorrect answer is option 'B'. Can you explain this answer?
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