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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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