For a CPU with 12 bit addresses, demand paged virtual memory is used. Pages are 16 bytes each, it takes 1 millisec (10 s
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For a CPU with 12 bit addresses, demand paged virtual memory is used. Pages are 16 bytes each, it takes 1 millisec (10 s
questions may explicitly state othervise), there is 1 KB (2¹ KB) of real memory. the current page table contents that fit in the CPU are (remainder in memory): Time Loaded 16 16 20 21 22 23 Time Last Used 29 25 30 31 30 32 Mem Frane 7 4 1 6 3 2 Locked Page 0 0 0 0 1 1 Page Entry 5 1 2 4 3 6 Valid 0 1 1 0 1 1 (a) How large (in bytes) is total, addressable (logical) memory? (b) How large (in bytes), is one memory frame? (c) How many memory frames are there? (e) For the following physical addresses, please give the corresponding logical addresses in hex tif possible): (1) 011 Hex (2) 060 Hex (f) In this Virtual Memory, when does page replacement happen? (g) One of the page replacement algorithms (policy) is LRU Briefly (few words), which page would be replaced? (Please give page number if possible.) Dirty 1 1 0 (d) For each of the following logical (data pointer) addresses, please give the corresponding physical addresses (if possible): (1) 011 Hex (2) 555 Hex thi One of the page replacement algorithms (policy) is FIFO Briefly (few words), which page would be replaced? 1 1 1
For a CPU with 12 bit addresses, demand paged virtual memory is used. Pages are 16 bytes each, it takes 1 millisec (10 sec) to read or write a page to the swap flash, and it takes 2 microsecs (2 x 10") to read or write main memory (ignore the time to translate in page table if that table is in the CPU, some