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Chapter 11 Homework Solutions

Chapter 11 Homework Solutions. 11.1 , 11.3, 11.7, 11.8, 11.9, 11.12. Problem 11.1. Consider a program that accesses a single I/O device and compare unbuffered I/O compared to the use of a buffer. Show that the use of the buffer can reduce the running time by a factor of two.

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Chapter 11 Homework Solutions

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  1. Chapter 11 Homework Solutions 11.1, 11.3, 11.7, 11.8, 11.9, 11.12

  2. Problem 11.1 • Consider a program that accesses a single I/O device and compare unbuffered I/O compared to the use of a buffer. Show that the use of the buffer can reduce the running time by a factor of two. • If the calculation time exactly equals the I/O time (which is the most favorable situation), both the processor and the peripheral device running simultaneously will take half as long as if they ran separately. • Formally, let C be the calculation time for the entire program and let T be the total I/O time required. • Then the best possible running time with buffering is max(C, T), while the running time without buffering is C + T • And, of course, ((C + T)/2) ≤ max(C, T) ≤ (C + T).

  3. Problem 11.3 • Perform the same kind of analysis as in the above table for the following sequence of disk track requests. 27, 129, 110, 186, 147, 41, 10, 64, 120 Assume the disk head is initially positioned over track 100 and is moving in the direction of decreasing addresses.

  4. Problem 11.3 Initial position over 100, moving downward address direction

  5. Problem 11.3 (b) Initial position over 100, moving upward address direction Only change is in SCAN and C-SCAN, shown above.

  6. Problem 11.7 • Calculate how much disk space (in sectors, tracks and surfaces) will be required to store 300,000 120-byte logical records if the disk is fixed-sector with 512 bytes/sector, with 96 sectors/track, 110 tracks per surface and 8 usable surfaces. Ignore any file header record(s) and track indexes and assume that records cannot span two sectors. • Each sector can hold 4 logical records. • The required number of sectors is 303,150/4 = 75,788 sectors. • This requires 75,788/96 = 790 tracks, • which in turn requires 790/110 = 8 surfaces.

  7. Problem 11.8 • Consider the disk system described in problem 11.7, and assume that the disk rotates at 360 rpm. A processor reads one sector from the disk using interrupt-driven I/O, with one interrupt per byte. If it takes 2.5 s to process each interrupt, what percentage of time will the processor spend handling I/O (disregard seek time)? • There are 512 bytes/sector. • Since each byte generates an interrupt, there are 512 interrupts. • Total interrupt processing time = 2.5 × 512 = 1280 μs. • The time to read one sector is: ((60 sec/min) / (360 rev/min)) / (96 sectors/track) = 0.001736 sec = 1736 μs • Percentage of time processor spends handling I/O: (100) × (1280/1736) = 74%

  8. Problem 11.9 • Repeat the preceding using DMA, and assume one interrupt per sector • With DMA, there is only one interrupt of 2.5 μs. • Therefore, the percentage of time the processor spends handling I/O is (100) × (2.5/1736) = 0.14%

  9. Problem 12 • Consider a 4-drive, 200 GB-per-drive RAID array. What is the available data storage capacity for each of RAID levels 0, 1, 3, 4, 5 and 6? • RAID 0: 800 GB • RAID 1: 400 GB • RAID 3: 600 GB • RAID 4: 600 GB • RAID 5: 600 GB • RAID 6: 400 GB Complete redundancy One disk used for parity Double parity,  two disks for parity

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