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 UWA week 40 (2nd semester, week 10) ↓
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"Kristiaan Maree" <24*1*7*3@s*u*e*t*u*a*e*u*a*> wrote:
> Ok that makes sense, > RAM and DISC are the values of the page table of the currently loaded processes so > RAM: 0,1,2,3,4,5,6,7 > DISK: 99,99,99,99,99,99,99,99 > instruction 3 > RAM: 0->99 (evicting process 1 (value 0) and replacing with process 3 (value 99)),1,2,3,4,5,6,7 > DISK: 99->0,99,99,99,99,99,99,99 > here initially processes 1 and 2 are in RAM (each with 4 pages for each page table) and a page from process 3 in disc swaps with the first page in ram after getting the '3' instruction. > Is this correct?
Isn't this your page table? process 1: 0,1,2,3 process 2: 4,5,6,7 process 3: 99,99,99,99 process 4: 99,99,99,99 It's hard to follow what you are doing to RAM and disk because it doesn't provide information as to which page of which process is in RAM. RAM: 0, 1, 2, 3, 4, 5, 6, 7 is just the frame numbers or the possible positions in RAM the page of a process could be loaded into.

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