a) To optimize the reading of the file, we may store adjacent blocks, e.g., B1,
ID: 668086 • Letter: A
Question
a) To optimize the reading of the file, we may store adjacent blocks, e.g., B1, B2, B3, …, on the same cylinder. Given the disk configuration above, how many blocks can be aligned on the cylinder position to be read by the disk arms at the same time?
b) Assume the file will be read sequentially, i.e., B1, B2, B3, ….until the last block Blast. Describe the best way to store these blocks on disk to speed up the sequential read. Then measure out the average time needed to read the file given this organization.
Hint: Make use of cylinders. Note that if Blocks B1 and B2 are aligned under each other in the same cylinder, then the transfer time to read B1, B2, or both is the same.
Given a disk with the following configurations: There are 5 (double-sided) platters, each surface has 8,000 tracks, 208 sectors per track -A sector holds 512 Bytes - Disk rotates at 5400 rpm -Seek time: A warm-up time is Ims, and then the arm movement covers 500 tracks per 1 ms Seek time: A warm-up time is 1ms, and then the arm movement covers 500 tracks per 1 ms - Reading one sector takes 0.05 ms - A disk block is of size 8K BytesExplanation / Answer
a)
The no.of double sided platters = 5
The no.of tracks = 8000 per surface
The no.of sectors =208 per track
The no.of bytes per sector = 512bytes
The disk rotates at 5400rpm
The seek time = the warm-up time (1ms) + 1ms to arm movement covers 500 tracks
Time to read one sector = 0.05 ms
The size of disk block =8K bytes =8*1024 =8192 byte block
The number of blocks = 8000*208*512/8192 = 104000 blocks
Only one block can be read/write by the disk arm at particular point of time. So, the no.of blocks that can be aligned on the cylinder position to be read by the disk arm at the same time is 1. It is independent of the disk configuration.
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