a): Given a file that consists of 100,000 records, and each record is 128 bytes.

Question

a): Given a file that consists of 100,000 records, and each record is 128 bytes. No record is allowed to span multiple blocks.

- How many records fit in one block?

- How many blocks are needed to store the file?

- How many sectors are needed to store the file?

b) Assume Blocks B1, B2, B3, …., B10, are stored on the same track (track number 100) and on adjacent sectors. If the disk head is positioned at track 0, what is the disk delay (time) needed to get read these 10 blocks? (Assume a half rotation is needed to reach the first block of these 10 blocks)

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 Bytes

Explanation / Answer

a)   How many records in a block?
8000/208=38.We can have at most 38 records in a block.

b)   How many blocks are required to store the file
100,000/38=2631 blocks are required to store the entire file

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