In Problem 1, Arrival Time 1 implies it has already been queued
at time 1. In the round-robin scheduling, the job that was
executed from time 2 to 3 will be enqueued after time 3.
In other words, it is not available for scheduling at time 3.
Normalized turnaround time is given by dividing the turnaround
time by the service time. It indicates the ratio between the
minimum and actual times for processing the job.
Exercise 5
October 31 pdf, Postscript.
Template files are at /home/course/os06s3/EX/ .
Exercise 6
November 21 pdf,
Postscript.
Template files are at /home/course/os06s3/EX6/ .
In principle, when calculating the throughput, the period
in which the disk is idle (i.e. no request is outstanding) must be
exlucded. However, you may assume that no such idle period exists
in the requqest sequence for this exercise.
Your program is given two names, ssf and elevator, and your
program determines the algorithm to schedule the request using
the name. You can copy the compiled binary (e.g. ssf) to
another name (elevator), but you can also do the following:
% gcc -o ssf ex12.c
% ln ssf elevator
In the template C file, the average throughput is in MB/Sec
but it should be KB/Sec for the disk parameters used in this
exercise.
While displaying the scheduling result of each request,
you also have to list the remaining requests when the request
is comleted. Therefore, the output should look like this:
Time 10.0 Track 25 Turn Around 10.0 Rem. Reqs 12, 30, 3, 5
Exercise 13
January 23, pdf,
Postscript.
Please read the instruction ("READ HERE FIRST") part before
start working.
Please assume that to access i-node, it takes the head to
seek 16 tracks.
Hitoshi Oi