help2002

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help2002

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UWA week 41 (2nd semester, week 11) ↓

2nd project Project 2 clarification (all 6)

1:54pm Mon 7th Oct, Joshua N.

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

> the way I understand the project is that it starts loaded in with the first 2 processes (page numbers 0-7) but because there are 16 pages in disc there is also a copy of process 0 and 1 in disc

RAM starts empty. You only load processes based on the file.

> so instruction 0: > bring in least recently used page into RAM from process 0 - currently in disc there are 4 pages in process 0 all with page num 99 and and time step 0 > since there is no space in RAM we evict the smallest time step (process 0, time step 0, page num 0)

At instruction 0, there is no process to evict. Since the RAM is empty.

> now we add (process 0, time step 0, page num 0) from the disc > now we increment the time step > > this would repeat until the 5th 0, I'm not quite sure what would happen here because in the given sample output they demonstrate that the time step can be greater than 1 but we are also not supposed to modify the disc so I am not sure how this is possible

Amitava said a process can only be loaded once. So, the fifth 0 would be ignored. I think that was a mistake on Amitava's part, you still need to change the last accessed time.