# Quiz for Chapter 6 Storage and Other I/O Topics 3.10

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4 7. [7 points] Suppose we have two different I/O system A and B. A has data transfer rate: 5KB/s and has access delay: 5 sec. While B has data transfer rate: 3 KB/s and has access delay: 4 sec. Now we have a 3M I/O request, taking performance into consideration, which I/O system will you use? What about for a 3KB request? 3M request Case 1: t (in sec) = 5 + (3 * 1024 * 1024 ) / (5 * 1024) = = Case 2: t (in sec) = 4 + (3 * 1024 * 1024) / (3 * 1024) = = 1028 So system 1 will be chosen. 3K request Case 1: t (in sec) = 5 + (3 * 1024) / (5 * 1024) = 5.6 Case 2: t (in sec) = 4 + (3 * 1024) / (3 * 1024) = 5 So system 2 will be chosen. 8. [7 points] If a system contains 1,000 disk drives, and each of them has a 800,000 hour MTBF, how often a drive failure will occur in that disk system? Could you give some idea to improve that? And why will your idea work? MTBF (array) = MTBF (one disk) / Number of disks in array So, answer is 800,000 / 1000 = 800 hrs Set them up in a RAID 5 configuration which involves distributing the data and parity bits across the disk drives. This way even when one drive fails, the parity information enables the disk array to continue operation and rebuild the failed drive online once it has been replaced. Now, the likelihood that a second drive will fail before the failed drive is restored is pretty low. Hence the reliability can be shown to increase dramatically. To further improve the reliability, one can have a spare disk which is hot-swapped with the failed disk in order to reduce the restoration time. 9. [6 points] What is the average time to read a 512 byte sector for Seagate ST NS in Figure 6.5? What is the minimum time? Assume that the controller overhead is 0.2 ms, and the disk is idle so that there is no waiting time. Average = Average seek time + Average rotational delay + Transfer time + controller overhead = rotations/7200rpm + 0.5KB/105MBps = = ms Minimum: Minimum seek time + Minimum rotational delay + Transfer time + controller overhead = rotations + 0.5KB/105MBps = = 1.005ms Page 4 of 7

7 Now, for the second half, we compare the time to perform the read given our predictions are correct to the time required to perform the read. Consider that any number of reads m is possible, independent of our n-block read strategy. The total time required to perform an m-block read as a number of n- block reads, a la T_n(m) = ceil( m/n ) * T(n) And our expected read time is: E(n) = SUM from m=1 to inf of T_n(m) * P(exactly m blocks) Now, we choose n to minimize E(n). The optimal read strategy is 4. Page 7 of 7

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