08~solutions_for_chapter_8_exercises

# 08~solutions_for_chapter_8_exercises - Solutions for...

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Solutions for Chapter 8 Exercises 1 Solutions for Chapter 8 Exercises 8.1 Each transaction requires 10,000 × 50 = 50,000 instructions. CPU limit: 500M/50K = 10,000 transactions/second. The I/O limit for A is 1500/5 = 300 transactions/second. The I/O limit for B is 1000/5 = 200 transactions/second. These I/O limits limit the machine. 8.2 System A Thus system A can only support 9 transactions per second. System B—Frst 500 I/Os (Frst 100 transactions) Thus system B can support 11 transactions per second at Frst. 8.3 Time/file = 10 seconds + 40 MB * 1/(5/8) seconds/MB = 74 seconds Power/Fle = 10 seconds * 35 watts + (74 – 10) seconds * 40 watts = 2910 J Number of complete Fles transferred = 100,000 J/2910 J = 34 Fles 8.4 Time/file = 10 seconds + 0.02 seconds + 40 MB * 1/(5/8) seconds/MB = 74.02 seconds Hard disk spin time/Fle = 0.02 seconds + 40 MB * 1/50 seconds/MB = 0.82 sec- onds Power/Fle = 10 seconds * 33 watts + 0.02 seconds * 38 watts + 0.8 seconds * 43 watts + 63.2 seconds * 38 watts = 330 J + 0.76 J + 34.4 J + 2401.6 J = 2766.76 J Number of complete Fles transferred = 100000 J / 2766.76 J = 36 Fles Energy for all 100 Fles = 2766.76 * 100 = 276676 J transactions 9 compute 1 I/Os 45 latency 5 times 900 ms 100 us 100 ms exceeds 1 s transactions 9 compute 1 1 I/Os 45 latency 5 5 times 810 ms 100 us 90 ms 90 ms 990.1 ms

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2 Solutions for Chapter 8 Exercises 8.5 After reading sector 7, a seek is necessary to get to the track with sector 8 on it. This will take some time (on the order of a millisecond, typically), during which the disk will continue to revolve under the head assembly. Thus, in the version where sector 8 is in the same angular position as sector 0, sector 8 will have already revolved past the head by the time the seek is completed and some large fraction of an additional revolution time will be needed to wait for it to come back again. By skewing the sectors so that sector 8 starts later on the second track, the seek will have time to complete, and then the sector will soon thereafter appear under the head without the additional revolution. 8.6 No solution provided. 8.7 a. Number of heads = 15 b. Number of platters = 8 c. Rotational latency = 8.33 ms d. Head switch time = 1.4 ms e. Cylinder switch time = 2.1 ms 8.8 a. System A requires 10 + 10 = 20 terabytes. System B requires 10 + 10 * 1/4 = 12.5 terabytes. Additional storage: 20 – 12.5 = 7.5 terabytes. b. System A: 2 blocks written = 60 ms. System B: 2 blocks read and written = 120 ms. c. Yes. System A can potentially accommodate more failures since it has more redundant disks. System A has 20 data disks and 20 check disks. System B has 20 data disks and 5 check disks. However, two failures in the same group will cause a loss of data in both systems. 8.9 The power failure could result in a parity mismatch between the data and check blocks. This could be prevented if the writes to the two blocks are performed simultaneously. 8.10 20 meters time: 20 m * 1/(1.5 * 10 8 ) s/m = 133.3 ns 2,000,000 meters time: 2000000 m * 1/(1.5 * 10 8 ) s/m = 13.3 ms 8.11 20 m: 133.3 * 10 –9 s * 6 MB/sec = 0.8 bytes 2000000 m: 13.3 * 10 –3 s * 6 MB/sec = 80 KB
Solutions for Chapter 8 Exercises 3 8.12 4 KHz * 2 bytes/sample * 100 conversations = 800,000 bytes/sec Transmission time is 1 KB/5 MB/sec + 150 μs = 0.00035 seconds/KB Total time/KB = 800 * 0.00035 = 0.28 seconds for 1 second of monitoring There should be sufFcient bandwidth.

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## This note was uploaded on 04/20/2010 for the course IE ie200 taught by Professor . during the Spring '10 term at 카이스트, 한국과학기술원.

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08~solutions_for_chapter_8_exercises - Solutions for...

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