sol-Tutorial5 - Question 1 Given address space = 22 bits...

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Question 1 Given: * address space = 22 bits * page size = 1 KiB = 1024 = 2 10 bytes a. No. of PTEs in one-level page table = = 2 22 /2 10 = 2 12 Not given in the question is the size of each PTE, usually it is of size 4 bytes (IA-32 architecture). The required space to store 2 12 PTEs = 2 12 × 4 = 2 14 bytes The number of page for storing the one-level page table = 2 14 /2 10 = 2 4 = 16 pages To minimize the number of pages used to store the page tables of a two-level table scheme, the only solution is to have one table in the first level, which points to tables in the second level. It is known that the second level has 2 12 entries and needs 16 pages to store these PTEs. This implies that one solution is to have the first level table consists of 16 entries; each entry points to a small table in one single page. Convert these information into the logical address format, we need to find v = (p, t, d). d = no. of byte locations in a page = 1024 = 10 bits t = no. of entries in one small table = 1024/4 = 2 10 /2 2 = 2 8 = 8 bits p = no. of entries in the first level table = 16 = 4 bits
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  • Spring '10
  • DR
  • Virtual memory, page table, Power of two, 3 second

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