Working with UNIX, C programming to do a project, the project is as follows:
Step 1: Populate the job Queue with 100 processes. For each process, auto increment
(starting from 1) the PID and randomly select Time (in the range 1-30). Display logging
message to the screen: “[Kernel] Process X created with Time = Y”
Step 2: Create the two threads for the two schedulers.
Step 3: Invoke the long-term scheduler. Every time long-term scheduler method is invoked,
display the logging message: “[Kernel] Long Term Scheduler Invoked” from the main
program. First, it will display a logging message to show the current content of the job queue
and ready queue as:
Winter 2019 LE/EECS 3221 2
“[LTS] Job Queue: [Process X1: Time Y1], ….”
“[LTS] Ready Queue: [Process X2: Time Y2], ….” or “[LTS] Ready Queue: EMPTY”
Now, Long-term scheduler method will dequeue an element (process) from the job queue
and enqueue it into the ready queue. Display logging message to the screen: “[LTS] Process
X removed from the Job Queue and inserted to the Ready Queue”. However, the ready queue
has maximum length 5; if the ready queue is full, display logging message to the screen:
“[LTS] Ready Queue is Full, cannot enter more”. Long term scheduler will try to insert
multiple processes to the queue if there are free spots. Now, Long-term scheduler will again
display the content of the two queues using same format as mentioned above and will pass
the control to short-term scheduler.
Step 4: Display the logging message: “[Kernel] Short Term Scheduler Invoked” from the
main program; this message will be displayed every time the short term scheduler is invoked.
Then, short term scheduler will display the content of two queues using the format discussed
earlier, change LTS to STS. Then, short-term scheduler method will dequeue an element
(process) from the ready queue and display the logging message to the screen: “[STS] Process
X now executing”. It will then reduce its time by two and enqueue it at the end of the ready
queue. Also display the logging message to the screen: “[STS] Process X with remaining time
Y enqueued to the Ready Queue”. If the message has consumed its entire time, then it will
not be enqueued to the ready queue and the logging message will be: “[STS] Process X
terminated”. Short term scheduler will repeat this action five times i.e. serve five processes.
Display the content of the two queues once again and then pass the control to the long-term
scheduler.
Hello! I'm an experienced C and UNIX programmer so I believe I could help you complete this project. If you have any further requirements concerning the deadline I may be able to accommodate. Your description has given me a good idea of the project, but if you would like to discuss the finer details please feel free to contact me via chat. Regards
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No problem!
I have read your description carefully and very interested in your project.
I am working on Desktop App with C/C++,C#,Python & Java for 7years.
I think i can do it perfectly.
If you hire me, you will get cool results.
i can work full-time in your time zone.
Best Regards
Hi. Dear
I read your job description in detail and feel I can help your project.
I have full experience and skills for C#, C++ and .NET Framework and Database Management etc.
I have done the many project as same as your project with Visual StudioS2019 .
I want to work for your project.
Best regards...