. Suppose an instruction takes 1/2 microsecond to execute (on the average), and a page fault takes 250 microseconds of processor time to handle plus 10 milliseconds of disk time to read in the page. (a) How many pages a second can the disk transfer? (b) Suppose that 1/3 of the pages are dirty. It takes two page transfers to replace a dirty page. Compute the average number of instructions between page fault that would cause the system to saturate the disk with page traffic, that is, for the disk to be busy all the time doing page transfers.

The time required to execute (on the average) = 1/2 microsecond , a page fault takes of processor time to handle = 250 microseconds and the disk time to read in the page = 10 milliseconds.

Thus, the time taken by the processor to handle the page fault = 250 microseconds / 1000 = 0.25 milliseconds.

The execution time = [ 1/2 microseconds ]/ 1000 = 0.0005 milliseconds.

The number of Pages sent in a second by the disc = 1000/10 milliseconds = 100.

Assuming U = 1.

Hence, the disc transfer time = [2/3 × 1 } + [ 1/3 × 0.25 milliseconds + 15 ] × 2.

=0.667 + 15.083.

= 15.75 millisecond.

Average number of instruction = 15.75/0.0005 = 31500.

topeadeniran2 topeadeniran2 · Answer 2 · 2022-01-07T16:34:58+01:00

It should be noted that the number of pages in a second will be 100 pages.

From the information given, it was stated that the instruction takes 1/2 microsecond to execute and a page fault takes 250 microseconds of processor time to handle plus 10 milliseconds of disk time to read the page.

Therefore, the execution time will be:

= 0.5/1000

= 0.0005

Therefore, the number of pages will be:

= 1000/10

= 100

Also, the disc transfer time will be;

= (2/3 × 1) + (1/3 × 0.25 + 15) × 2

= 0.667 + 15.083

= 15.75

Therefore, the average number of instructions will be:

= 15.75/0.0005

= 31500

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