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With single buffering, the operating system dedicates a buffer to a process. This allows the process to work on one block while theI/O system reads another block into the buffer.
With double buffering, the operating system dedicates two buffers to a process. The process can then empty one buffer while the I/Osystem fills the other. This allows the process to process data as fast as the I/O system can read it in.
With a sequential file, the records are stored in order of a particular key, but searching requires going through all of therecords sequentially until a match is found. In an indexed sequential file, the index contains pointers to every K records in the file. For a file with Nrecords, searching for a record with a particular key requires searching only the N/K entries in the index, and then the K records in that section of thefile. Depending on the size of K, this can be considerably less than sequentially searching through all N entries.
Answer:
FIFO | SSTF | SCAN | C-SCAN | ||||
Next track accessed | Number of tracks traversed | Next track accessed | Number of tracks traversed | Next track accessed | Number of tracks traversed | Next track accessed | Number of tracks traversed |
27 | 73 | 110 | 10 | 64 | 36 | 64 | 36 |
129 | 102 | 120 | 10 | 41 | 23 | 41 | 23 |
110 | 19 | 129 | 9 | 27 | 14 | 27 | 14 |
186 | 76 | 147 | 18 | 10 | 17 | 10 | 17 |
147 | 39 | 186 | 39 | 110 | 100 | 186 | 176 |
41 | 106 | 64 | 122 | 120 | 10 | 147 | 39 |
10 | 31 | 41 | 23 | 129 | 9 | 129 | 18 |
64 | 54 | 27 | 14 | 147 | 18 | 120 | 9 |
120 | 56 | 10 | 17 | 186 | 39 | 110 | 10 |
Average | 61.8 | Average | 29.1 | Average | 29.6 | Average | 38 |
SCAN | C-SCAN | ||
Next track accessed | Number of tracks traversed | Next track accessed | Number of tracks traversed |
110 | 10 | 110 | 10 |
120 | 10 | 120 | 10 |
129 | 9 | 129 | 9 |
147 | 18 | 147 | 18 |
186 | 39 | 186 | 39 |
64 | 122 | 10 | 176 |
41 | 23 | 27 | 17 |
27 | 14 | 41 | 14 |
10 | 17 | 64 | 23 |
Average | 29.1 | Average | 35.1 |
Consider the disk system described in Problem 11.9 and assume the disk rotates at 360 rpm. A processor reads one sector from thedisk using interrupt-driven I/O, with one interrupt per byte. If it takes 2.5 μs to process each interrupt, what percentage of time will the processor spendhandling I/O (disregard seek time)? Repeat using DMA, and assume one interrupt per sector.
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