HW12-sol - to store the physical page number and an additional 4 bits for the valid protection dirty and use bits We round the 26 bits up to a full

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1 CSE 341: Computer Organization Spring 2011 Solution to HW #12 Problem 1: The total size is equal to the number of entries times the size of each entry. Each page is 16 KB, and thus, 14 bits of the virtual and physical address will be used as a page offset. The remaining 40 – 14 = 26 bits of the virtual address constitute the virtual page number, and there are thus 2 26 entries in the page table, one for each virtual page number. Each entry requires 36 – 14 = 22 bits
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Unformatted text preview: to store the physical page number and an additional 4 bits for the valid, protection, dirty, and use bits. We round the 26 bits up to a full word per entry, so this gives us a total size of 2 26 × 32 bits or 256 MB. Problem 2: The solution is very similar to the diagram given in page 505 (Figure 5.24). You may also use the figure from page 469 (Figure 5.9) and the figure from page 486 (Figure 5.17) for better understanding of this problem....
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This document was uploaded on 10/20/2011.

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