Suppose that the number of instructions executed between page faults is directly proportional to the number of page frames allocated to a program. If the available memory is doubled, the mean interval between page faults is also doubled. Further, consider that a normal instruction takes one micro second, but if a page fault occurs, it takes 2001 micro seconds. If a program takes 60 sec to run, during which time it gets 15000 page faults, how long would it take to run if twice as much memory were available? (A) 60 sec (B) 30 sec (C) 45 sec (D) 10 sec

Suppose that the number of instructions executed between page faults is directly proportional to the number of page frames allocated to a program. If the available memory is doubled, the mean interval between page faults is also doubled. Further, consider that a normal instruction takes one micro second, but if a page fault occurs, it takes 2001 micro seconds. If a program takes 60 sec to run, during which time it gets 15000 page faults, how long would it take to run if twice as much memory were available? (A) 60 sec (B) 30 sec (C) 45 sec (D) 10 sec
by

1 Answer

Answer :

Answer: C

Explanation:

T = Ninstr x 1µs + 15,000 x 2,000 µs = 60s

Ninstr x 1µs = 60,000,000 µs – 30,000,000 µs = 30,000,000 µs

Ninstr = 30,000,000

The number of instruction between two page faults is

Ninstr /NPageFaults = 30,000,000/15,000 = 2,000

If the mean interval between page faults is doubled, the number of instruction

between two page faults is also doubled and is 4,000. Now the number of page

faults is

30,000,000/4,000 = 7,500

T’ = 30,000,000 µs + 7,500 x 2,000 µs

= 30,000,000 µs + 15,000,000 µs = 45,000,000 µs = 45s

Doubling the memory, doesn’t mean that the program runs twice as fast as on

the first system. Here, the performance increase is of 25%.
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