. 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.

a. 100.

b. 31500.

Explanation:

So, we are given the following data which is going to help in solving this particular question.

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.

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