help2002

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help2002

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UWA week 39 (2nd semester, week 9) ↓

Project 2 processes (all 4)

10:16pm Thu 26th Sep, ANONYMOUS

For Project 2, once you get to the last page of a process, does it resolve so you should remove all the pages associated with that process from RAM? So for a sample of 0 0 0 0, should you end with 4 pages of the process in RAM (whole thing twice), or should the process be removed so RAM is empty? (relevant if the sample is something like 0 0 0 0 0 0 0, more pages of the process in RAM than it has, so its process table (which only has 4 entries) can't handle this). Either that, or the process tables need to be made more complex. Thanks in advance

UWA week 40 (2nd semester, week 10) ↓

2nd project Project 2 processes (all 4)

9:24pm Mon 30th Sep, Joshua N.

ANONYMOUS wrote:

> For Project 2, once you get to the last page of a process, does it resolve so you should remove all the pages associated with that process from RAM? So for a sample of 0 0 0 0, should you end with 4 pages of the process in RAM (whole thing twice),

You should have 4 pages in RAM.

> ....Either that, or the process tables need to be made more complex. > Thanks in advance

I'm not sure if I understand your question but if you exceed the number of pages for a process you likely would need to evict the first loaded page for that process and replace it with the new process. In the end this just ends up removing pg0 and then adding a new pg0 for that process. So, the only change would be an update to the load time of page 0. i.e. If you push 0 0 0 0 onto RAM, you have pg. 0,1,2,3 on RAM. When another zero is read RAM would be still be 0 0 0 0, except the first page's last load time is updated.

2nd project Project 2 processes (all 4)

11:05am Thu 3rd Oct, Kristiaan M.

so if the RAM has page numbers 0-7 from processes 0 and 1 all initially time step 0, and the disc has processes 2-3, and we receive the instruction '0' there is nothing to move in from the disc as all of process 0 is in RAM. We need to evict the first page loaded (page 0 in process 0 in RAM slot index 0) and gets moved into the disc now there are 7 slots loaded into the RAM and 9 slots loaded in the disc so now there is a page in disc that can be loaded into RAM from process 0 (the page that was evicted from RAM) and this page now gets moved into the empty slot in the RAM so basically the only thing that changes is the time step of the first page when being evicted and selected (increases by 2) is this correct?

2nd project Project 2 processes (all 4)

2:42pm Thu 3rd Oct, Joshua N.

"Kristiaan Maree" <24*1*7*3@s*u*e*t*u*a*e*u*a*> wrote:

> so if the RAM has page numbers 0-7 from processes 0 and 1 all initially time step 0, and the disc has processes 2-3, and we receive the instruction '0' there is nothing to move in from the disc as all of process 0 is in RAM. > We need to evict the first page loaded (page 0 in process 0 in RAM slot index 0) and gets moved into the disc now there are 7 slots loaded into the RAM and 9 slots loaded in the disc > so now there is a page in disc that can be loaded into RAM from process 0 (the page that was evicted from RAM) and this page now gets moved into the empty slot in the RAM > > so basically the only thing that changes is the time step of the first page when being evicted and selected (increases by 2) > is this correct?

Yes that seems right. just remember that Disk has all 16 pages in it (4 for each process) and disk is never changed. Also RAM has frames 0-7. Processes have pages 0-3.