描述

Data-Intensive Computing Systems Data Access from Disks Shivnath Babu Outline Disks Data access from disks Software-based optimizations Prefetching blocks Choosing the right block size Focus on: “Typical Disk” Top View … Head assembly Sector Terms: Platter, Head, Cylinder, Track Sector (physical), Block (logical), Gap Gap Block Address: Physical Device Cylinder # Surface # Start sector # Disk Access Time (Latency) block X in memory I want block X ? Access Time = Seek Time + Rotational Delay + Transfer Time + Other Seek Time 3 or 5x Time x 1 N Cylinders Traveled Average value: 10 ms → 40 ms Rotational Delay Head Here Block I Want Average Rotational Delay R = 1/2 revolution Example: R = 8.33 ms (3600 RPM) Transfer Rate: t t: 1 → 100 MB/second transfer time: block size t Other Delays CPU time to issue I/O Contention for controller Contention for bus, memory “Typical” Value: 0 So far: Random Block Access What about: Reading “Next” block? If we do things right … Time to get = Block Size + Negligible next block t skip gap switch track once in a while, next cylinder Rule of Thumb Ex: Random I/O: Expensive Sequential I/O: Much less 1 KB Block » Random I/O:  20 ms. » Sequential I/O:  1 ms. Cost for Writing similar to Reading …. unless we want to verify! To Modify Block: (a) Read Block (b) Modify in Memory (c) Write Block [(d) Verify?] A Synthetic Example 3.5 in diameter disk 3600 RPM 1 surface 16 MB usable capacity (16 X 220) 128 cylinders seek time: average = 25 ms. adjacent cylinders = 5 ms. 1 KB blocks = sectors 10% overhead between sectors capacity = 16 MB = (220)16 = 224 bytes # cylinders = 128 = 27 bytes/cyl = 224/27 = 217 = 128 KB blocks/cyl = 128 KB / 1 KB = 128 3600 RPM 60 revolutions / sec 1 rev. = 16.66 msec. One track: ... Time over useful data:(16.66)(0.9)=14.99 ms. Time over gaps: (16.66)(0.1) = 1.66 ms. Transfer time 1 block = 14.99/128=0.117

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