It's UWAweek 42 (2nd semester, week 12)

help2002

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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:
> Hi, > if we load the page frames in sorted by page num the memory looks like this: > > process 1: 0,1,2,3 > process 2: 4,5,6,7 > process 3: 99,99,99,99 > process 4: 99,99,99,99 > > > RAM: 0,1,2,3,4,5,6,7 > DISK: 99,99,99,99,99,99,99 > > but if we get the instruction '3' there is no process 3 in main memory to swap to, > > instead are we supposed to put in 2 each to avoid this? > > e.g. > process 1: 0,1,99,99 > process 2: 2,3,99,99 > process 3: 4,5,99,99 > process 4: 6,7,99,99 > > RAM: 0,1,2,3,4,5,6,7 > DISK: 99,99,99,99,99,99,99,99 > > so now instruction 3 changes it to > > RAM: 0,1,2,99,4,5,6,7 > DISK: 99,99,99,99,99,99,3,99 > > or is there a better way?
Are you asking if there are any pages in RAM when the simulation starts? The RAM is empty there are no pages in RAM.


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1:23am Wed 2nd Oct, Kristiaan M.

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?


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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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11:56am Wed 2nd Oct, Kristiaan M.

yes that's the page table, with corrected indexes it looks like this: process 0: 0,1,2,3 process 1; 4,5,6,7 process 2: 99,99,99,99 process 3: 99,99,99,99 process 0 and 1 are loaded into ram with values 0-7 but now there is a new issue with instruction '0' 'the next page of process 0 has to be brought in from virtual memory to the RAM' however every page from process 0 is already loaded into RAM do I just switch the pages like so: process 0: 0,1,2,3 process 0: 1,0,2,3 (and update the time step for these 2 that change) or does something else need to happen?


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"Kristiaan Maree" <24*1*7*3@s*u*e*t*u*a*e*u*a*> wrote:
> yes that's the page table, with corrected indexes it looks like this: > process 0: 0,1,2,3 > process 1; 4,5,6,7 > process 2: 99,99,99,99 > process 3: 99,99,99,99 > > process 0 and 1 are loaded into ram with values 0-7 > but now there is a new issue with instruction '0' > > 'the next page of process 0 has to be brought in from virtual memory to the RAM' > however every page from process 0 is already loaded into RAM > do I just switch the pages like so: > process 0: 0,1,2,3 > process 0: 1,0,2,3 (and update the time step for these 2 that change) > > or does something else need to happen?
I've answered this question here: [help2002]

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