CS 3733 Operating Systems, Fall 2003 Assignment 2 Comments

Approximate point values are as follows:
Part 1: 5 points
Part 2: 4 + 1 + 5 + 5 = 15 points
Part 3: 10 points
If your experiment did not involve any I/O you lost 5 points.

Please remember that you are not allowed to work with others on these assignments. What you turn in must be representative of your work.

• The statement from Part 1 is a paraphrase of a statement from the SGG book on page 164.
• The following two statements are not the same:
  Machine A is 20 times slower than machine B.
  Machine A has a CPU that is 20 times slower than the one on machine B.
• If the quantum equals the CPU burst, this is a large quantum.
• If the duration does not exceed the CPU burst, no I/O will be done not matter how the I/O burst is set.
• You should know how to spell Gantt (upper case G, two t's)
• In Part 2, if you say the waiting time in 0 for one process, you should not then expect the waiting time to be the same when you have 20 processes.
• In Part 3, the experiment should confirm (or refute) the statement in Part 1.
• When there is I/O involved, running on a machine that has a CPU that is 20 times slower is not the same as saying the process takes 20 times as long to finish.

Some sample answers follow:

Part 1: This statement is only true if the processes do not do any I/O. It may be false if they do I/O and use the CPU at different times. In this case, one process may be using the CPU while the other processes are doing I/O.

Part 2:

1. The quantum should be small compared to the smallest CPU burst.
2. The duration and the CPU burst.
3. It is 0.
4. Turnaround time.

Part 3:
In your experiment, make the CPU bursts have a non-constant distribution so the processes do not all use the CPU at the same time. An example follows:

myexp.exp:

name myexp
comment For Assignment 2
run myrun1 algorithm FCFS key "FCFS 1 process"
run myrun2 algorithm RR 1 key "RR 1, 20 processes"
run myrun2 algorithm RR .1 key "RR .1, 20 processes"

myrun1.run:

name myrun1
comment For 1 process
algorithm FCFS
seed 5000
numprocs 1
firstarrival 0.0
interarrival constant 0.0
duration constant 2000
cpuburst uniform 20 40
ioburst constant 10
basepriority 1.0

myrun2.run

name myrun2
comment For 20 processes
algorithm RR 1
seed 5000
numprocs 20
firstarrival 0.0
interarrival constant 0.0
duration constant 100
cpuburst uniform 1 2
ioburst constant 10
basepriority 1.0