# CSCI-GA.02250 Lab 2 (Short-term CPU) Scheduling solution

\$24.99

Category:

## Description

In this lab you will simulate scheduling in order to see how the time required depends on the scheduling algorithm
and the request patterns. The lab is due 2 March 2017.
A process is characterized by just four non-negative integers A, B, C, and M. A is the arrival time of the process
and C is the total CPU time needed. A process execution consists of computation alternating with I/O. I refer to
these as CPU bursts and I/O bursts. We make the simplifying assumption that, for each process, the CPU burst times
are uniformly distributed random integers (UDRIs) in the interval (0,B]. To obtain a UDRI t in some interval (0,U]
use the function randomOS(U) described below. If t, the value returned by randomOS(), is larger than the total CPU
time remaining, set t to the remaining time.
The I/O burst time for a process is its preceding CPU burst time multiplied by M.
You are to read a file describing n processes (i.e., n quadruples of numbers) and then simulate the n processes until
they all terminate. The way to do this is to keep track of the state of each process and advance time making any state
transitions needed. At the end of the run you first print an identification of the run including the scheduling algorithm used, any parameters (e.g. the quantum for RR), and the number of processes simulated. You then print for
each process
• A, B, C, and M.
• Finishing time.
• Turnaround time (i.e., finishing time – A).
• I/O time (i.e., time in Blocked state).
• Waiting time (i.e., time in Ready state).
Then print the following summary data.
• Finishing time (i.e., when all the processes have finished).
• CPU Utilization (i.e., percentage of time some job is running).
• I/O Utilization (i.e., percentage of time some job is blocked).
• Throughput, expressed in processes completed per hundred time units.
• Average turnaround time.
• Average waiting time.
You must simulate each of the following scheduling algorithms, assuming, for simplicity, that a context switch takes
zero time. You need only do calculations every time unit (e.g., you may assume nothing exciting happens at time
2.5).
• FCFS.
• RR with quantum 2.
• Uniprogrammed. Just one process active. When it is blocked, the system waits.
• SJF (This is not preemptive, but is not uniprogrammed, i.e., we do switch on I/O bursts). Recall that SJF is shortest job first, not shortest burst first. So the time you use to determine priority is the total time remaining (i.e., the
input value C minus the number of cycles this process has run).
For each scheduling algorithm there are several runs with different process mixes. A mix is a value of n followed by
n A, B,C, M quadruples. Here are the first two input sets. The comments are not part of the input.
1 0 151 about as easy as possible
2 0 151 0151 should alternate with FCFS
All the input sets, with their corresponding outputs are on the web.
The simple function randomOS(U), which you are to write, reads a random non-negative integer X from a file
named random-numbers (in the current directory) and returns the value 1 + (X mod U). I will supply a file with a
large number of random non-negative integers. The purpose of standardizing the random numbers is so that all correct programs will produce the same answers.
Professor Allan Gottlieb Lab 2 (Short-term CPU) Scheduling OS-2250
Breaking ties
There are two places where the above specification is not deterministic and different choices can lead to different
1. A running process can have up to three events occur during the same cycle.
i. It can terminate (remaining CPU time goes to zero).
ii. It can block (remaining CPU burst time goes to zero).
iii. It can be preempted (e.g., the RR quantum goes to zero).
They must be considered in the above order. For example if all three occur at one cycle, the process terminates.
2. Many jobs can have the same ‘‘priority’’. For example, in RR several jobs can become ready at the same cycle
and thus have the same priority. You must decide in what order they will subsequently be run. These ties are broken
by favoring the process with the earliest arrival time A. If the arrival times are the same for two processes with the
same priority, then favor the process that is listed earliest in the input. We break ties to standardize answers. We use
this particular rule for two reason. First, it does break all ties. Second, it is very simple to implement. It is not especially wonderful and is not used in practice. I refer to this method as the ‘‘lab 2 tie-breaking rule’’ and often use it
on exams for the same two reasons.