By John Minot Rice, William Woolsey Johnson
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Extra resources for An Elementary Treatise on the Differential Calculus Founded on the Method of Rates or Fluxions
1 a. 59 ns. ) Thus the 2-way takes 54% more time than the direct-mapped, while the 4-way takes 53% more time than the directmapped cache. 5 Chapter 5 Solutions ■ L-47 b. 8% greater for the 64 KB. c. 06 ns. The larger cache is 8 times larger but has a 43% larger access time. This growth is between square root and logarithmic, but closer to log. d. 88 ns. 90 ns. 2 a. 11 cycles. 26. b. 56 ns. 63, or 63% faster execution. c. 93! d. 79, or 21% less power. 3 a. 58 pipe stages. b. 14. 34. c. There are two banks.
B. The 0th chunk and the 11th, the 1st and 10th, the 2nd and 9th, and so on. c. 6 disks. There are 12 chunks, and each chunk seems to be on the same disk as one other chunk. Hence, 12 divided by 2 gives us 6 disks. d. Start at the left, numbering chunks 0, 1, 2, . . , 5 across disks; then go to the next row, and instead number 11, 10, 9, . . , 6. This pattern aligns the chunks correctly in a pairwise fashion. 7 Your graph should be 12 by 12, and the following chunk pairs would conflict (and hence have lighter shading): (0, 0), (1, 1), .
26 a. A 40 MB file requires one 20-byte checksum per 4 KB block and 1 parity block. Hence, the additional write traffic is dominated by the checksums. It is easy to calculate that 40 MB implies 10,240 blocks, which leads to 200 KB of checksums plus a 4 KB parity block, or 204 KB of extra traffic. 5%. b. Now the costs go up quite a bit. For each write, we must do a “small write” RAID-style update of the parity block, as well as update the checksum information. An update to a particular block requires us logically to read the old data that was in the block, the old parity, compute the new parity (using the difference between the old and new data to derive the new parity), and then write the new data and new parity.
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