CS 3733 Operating Systems, Fall 2005 Assignment 2 Comments

This assignment was graded on the basis of 30 points.

1. Some people did not understand what was meant by differ by a factor of at least 2.
2. The general idea of Part 3 is that if short and long processes are ready, the average waiting time is smallest when the short processes can finish before the long ones. This is essentially what SJF does.
3. For the first part, decreasing quantum produces increasing waiting time, you can start with several identical processes with constant CPU burst. As the quantum decreases from the CPU burst, the average waiting time will increase. It is difficult to get a factor of two difference, but you should be able to get close.
4. For the second part of 3b, decreasing quantum gives decreasing waiting time, start with one process with long CPU bursts followed by several with short CPU bursts. As the quantum decreases from the largest to the smallest, the shortest processes finish earlier and the waiting time decreases.
5. For part 3c, you should have found that the number of context switches increased between the largest and smallest quantum. For this part, you needed to compute the total waiting time, T, the number of context switches, n, and the load average, L, for each of the two experiments. If x is the context switch time, compute T + nxL for each of the experiments, set them equal, and solve for x.
6. If you did not get the correct answers for part 3c I will send you the correct answers if you send me an email with a line containing the following information separated by blanks:
   Average waiting time, run 1
   Average waiting time, run 2
   Total time for run 1
   Total time for run 2
   Number of CPU entries for run 1
   Number of CPU entries for run 2
   Number of processes for run 1
   Number of processes for run 2
   Remember that these 8 number must be on one line.