Use proper words. Also, add interesting facts about the subject
I think Face book hope that helped XD
Answer:
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.
