Consider a logical-address of 2048 pages of 1024 words (4 bytes) each, mapped onto a physical memory of 64 frames. Please answer the following questions assuming that the smallest memory allocation unit is one byte (8 bits). (c)Assuming a 2-level page table is used and the first-level table has 32 entries, what is the minimal amount of memory (in bytes) required by the page tables? 可以计算出每个page=1kb 共有2048个pages 以1-level来计算 table size=2048*6(entries)bits 以上one-level的算法没错吧? (64 frames=2^6) 再来是2-level的所有table size p1=5 bit p2=6 bit 再算第一level的table size时 我就卡住了到底它的每个entry大小为何呢? 且如何计算第一level的table size大小呢呢? 谢谢 --

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