a5 - CMPT 300 Assignment 5 Jake Wang 953022102 1 There are...

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CMPT 300 Assignment 5 Jake Wang 953022102 1. There are 2^32 / 2^7 = 2^25 memory addresses. Time to compact = (read + write) * access time * number of memory addresses = 2 x 10ns x 2^25 = 671 ms 2. First fit: (a) 20K (b) 10K (c) 18K Best fit: (a) 12K (b) 10K (c) 9K Worst fit: (a) 20K (b) 18K (c) 15K Next fit: (a) 20 KB (b) 18K (c) 9K 3a. There will be 12 bits page offset for 4KB pages and 13 bits for 8KB pages. Therefore: 2000010 = 01001110001000002. The page offset for both 4KB and 8KB pages is 3616. The VPN for 4KB is 4 (0100) while the VPN for 8KB page size is 2 (010). 3b. 3276810 = 10000000000000002. The page offset is 0 for both 4KB and 8KB pages. The VPN for 4KB is 8 and the VPN for 8KB page size is 4 . 3c. 6000010 = 11101010011000002. The page offset is 2656 for both 4KB and 8KB pages. The VPN for 4KB is 14 and the VPN for 8KB page size is 7. 4. For a one-level page table, there are 2^32/2^12 = 2^20 pages entries. For two-level paging, the main page table has 2^10 entries with each pointing to a second page table. Since only two entries are needed here, the total number of page table entries will be 3 (ie. 1 entry in first level table and 2 in each of the lower-level
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a5 - CMPT 300 Assignment 5 Jake Wang 953022102 1 There are...

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