Assignment 1 – Priority Inversion solution


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The problem of priority inversion will be described in lectures but you should start this assignment
immediately. This assignment demonstrates the problem using a simple simulation of a scheduler.
You have to understand the scheduler and the resource controller, complete a little bit of scheduler
code and then implement a solution to the priority inversion problem. The source code you start
with is in the file
The scheduler
The Scheduler class implements a simple scheduler that runs on Linux (or any Unix). It is not a
real scheduler in the sense that it is invoked automatically by the operating system when a new
process needs to be scheduled. Instead, it is an emulation of a scheduler that runs once a second,
making scheduling decisions.
The scheduler works by sending signals to the processes it schedules. It uses the same job control
signals that the shell uses to suspend jobs and then resume them, e.g., by typing ctrl-Z in bash and
then typing fg. These signals are known as SIGSTOP and SIGCONT. When a process receives the
SIGSTOP signal it pauses, only to resume when it receives the SIGCONT signal. The code in the
scheduler is:
# Suspends the currently running process by sending it a STOP signal.
def suspend(process):
os.kill(, signal.SIGSTOP)

# Resumes a process by sending it a CONT signal.
def resume(process):
if # if the process has a pid it has started
os.kill(, signal.SIGCONT)
The resume also starts processes running for the first time by calling, a method in the
SimpleProcess class.
Every second the scheduler checks to see which runnable process has the highest priority. If the
current process still has the highest priority it continues. Otherwise the current process is su

If there are several processes with the same priority they are scheduled round-robin. e.g., If the
ready queue looks like this
process 1:priority 4 – process 2:priority 4 – process 3:priority 2
Process 1 will be scheduled first; then process 2. These two processes will take turns until they are
no longer runnable. Then process 3 can run.
Requesting the resource
In order to get priority inversion there must be a resource that is held by a lower priority process
when it is needed by a higher priority process. There has to be a way to allocate the resource and
control access to it. In this assignment the job is done by the Controller class.
A process requests a resource either by making a system call or by sending a message. In this
assignment the processes will send messages via pipes that are set up for you. The Controller thread
is actually the main thread of the program and is running in the same process as the Scheduler
thread. The processes that are to be scheduled each run in their own independent processes.
When a process wants to use the resource it sends a request message to the controller. It then
blocks until the controller sends a reply message back. After a process has finished with the
resource it must notify the controller with a release message. This is demonstrated by the
functions low_func and high_func which execute as SimpleProcesses in the file. e.g.,
def high_func(proc):
pid = os.getpid() # who am I?
print(‘High priority:’, pid, ‘- just about to request resource’)
controller_write.write(‘{0}:request\n’.format(pid)) # request
response = # wait for reply
print(‘High priority:’, pid, ‘- got resource’)
controller_write.write(‘{0}:release\n’.format(pid)) # release
print(‘High priority:’, pid, ‘- released resource’)
The request and release messages include the process id (pid) of the process making the request.
When the resource is being used by one process, all other requesting processes are kept in a FIFO
Part 1
You must complete the Scheduler class – add_process, remove_process, select_process in
the program. You may also modify the scheduler’s __init__ method (or constructor).
These methods add processes to the ready list, remove them from the ready list and select the
process that should be running next. The ready list is the queue of processes that are ready to run
(i.e. not waiting for any resource). The ready list must always be maintained in the order that
processes should be run, i.e. element 0 must be the currently executing process and element 1
should be the next process to execute etc.
Unless you use a lock (or similar synchronisation mechanism) there will be a race condition in your
code. Make sure you fix this. There are no race conditions within the controller because a single
pipe is shared by all processes to send messages to the controller and pipes are guaranteed to
preserve message integrity if the messages are small enough. The controller thread reads from this
COMPSCI 340 & SOFTENG 370 A1 2013
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one pipe. In order for the controller to send messages back to the correct process, each process has
its own pipe to read from.
When you have completed the Scheduler class the program should demonstrate priority
inversion when run.
The output from the program should look very much like this:
$ python3
Low priority: 99517 – just about to request resource
owner pid: 99517
Low priority: 99517 – got resource
High priority: 99528 – just about to request resource
owner pid: 99517
Mid priority: 1
Mid priority: 2
Mid priority: 3
Mid priority: 4
Mid priority: 5
Mid priority: 6
Mid priority: 7
Mid priority: 8
Mid priority: 9
Mid priority: 10
remove process 99551 from ready list
Low priority: 99517 – released resource
owner pid: 99528
High priority: 99528 – got resource
High priority: 99528 – released resource
owner pid: None
remove process 99528 from ready list
Low priority: 99517 – finished
remove process 99517 from ready list
The high priority process is blocked because the low priority process is holding the resource and the
low priority process cannot release the resource until the middle priority process has completed.
Even after the middle priority process is finished the high priority process has to wait but that isn’t
priority inversion it is just normal waiting for a resource to be released.
Part 2
Once you have that working, make a copy of your program and rename it
Then modify the Controller class to prevent priority inversion.
Also ensure that if more that one process is waiting for the resource, the process with the highest
priority will get the resource when it is released. If there are several waiting processes with the
same priority the resource is allocated to the one that has been waiting the longest.
The output from the program should be very much like this:
$ python3
Low priority: 181 – just about to request resource
owner pid: 181
Low priority: 181 – got resource
High priority: 191 – just about to request resource
owner pid: 181
Low priority: 181 – released resource
owner pid: 191
High priority: 191 – got resource
High priority: 191 – released resource
owner pid: None
COMPSCI 340 & SOFTENG 370 A1 2013
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remove process 191 from ready list
Mid priority: 1
Mid priority: 2
Mid priority: 3
Mid priority: 4
Mid priority: 5
Mid priority: 6
Mid priority: 7
Mid priority: 8
Mid priority: 9
Mid priority: 10
remove process 302 from ready list
Low priority: 181 – finished
remove process 181 from ready list
In this case the low priority process has its priority boosted to that of the high priority process when
the high priority process waits for the resource. This means that the low priority process releases the
resource quickly, allowing the high priority process to continue before the middle priority process
Your and programs will be tested as is. Then the markers will add extra
functions at the top of your program and create processes from them and add these to the
scheduler after the call to time.sleep near the bottom of the file to create a more complicated test.
The more complicated test will use different priority values (a higher number means a better
priority) and have a number of different processes running and accessing the resource. Several
processes may have the same priority and some processes may request the resource more than once.
If a process requests the resource while it already has access to it, your solution must count the
number of allocations and not free the resource until the process has released the resource the same
number of times it was allocated (similar to the way a Java thread can get repeated access to an
object lock).
1. Explain where the race condition could occur in, show and explain the code you
inserted to stop this happening.
2. In the Schedulers run method there is the following statement:
os.waitpid(, os.WNOHANG) != (0, 0)
What does this code do, and why is it necessary?
3. The Scheduler code is different from a true scheduler in very important ways – how the
scheduler code is invoked and how the processes are suspended and resumed. Briefly describe
how these things are done in a real scheduler.
4. The implementation of the Controller class depends on writing and reading from pipes.
Describe the purpose of each line in the SimpleProcess run method. Also explain what would
happen if the os._exit(0) call was not made, and why.
5. Extra for experts. There is at least one other possible race condition in the code (not counting
the one you fixed in Part 1. Describe what it is, how it could occur and what the consequences
would be in that case.
COMPSCI 340 & SOFTENG 370 A1 2013
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6. Extra for SOFTENG 370 students. The processes dictionary (hash map) variable has a
memory leak (or something similar to it). Describe how the leak occurs and give a solution so
that values which are no longer needed by the program can be released and eventually
garbage collected. You don’t need to provide the code to do this although some code would
probably help in explaining your solution.
Submitting the assignment
Make sure your name and upi is included in every file you submit. You may not get the marks
for these files if this is not done.
Use the assignment drop box to submit your, and files.
Submit your answers to the questions as a single text file, called a1Answers.txt.
Any work you submit must be your work and your work alone – see the assignment page on the
Operating System web site
Marking guide
Proper use of synchronization when required. (2 marks) works correctly (i.e. it shows priority inversion). (2 marks) works correctly (i.e. solves the problem of priority inversion). (4 marks) works correctly with more complicated schedules of processes and resource requests. (4
1. 2 marks
2. 2 marks
3. 2 marks
4. 2 marks
5. 1 mark (but you can’t go over 20 or 22 marks)
6. 2 marks
Total: 20 marks for COMPSCI 340, 22 marks for SOFTENG 370
Penalty for late submission: 1 mark per day up until 4 days late