Problem 3: Files Consider a disk with a sector size of 512 bytes, 2,000 tracks p
ID: 3711504 • Letter: P
Question
Problem 3: Files Consider a disk with a sector size of 512 bytes, 2,000 tracks per surface, 50 sectors per track, five double-sided platters, and average seek time of 10 msec, the disk platters rotate at 5,400 rpm (revolutions per minute).
Suppose that a block size of 1,024 bytes is chosen. Suppose that a file F containing 100,000 records of 100 bytes each is to be stored on such a disk and no record is allowed to span two blocks.
How many records fit onto a block?
How many blocks are required to store the entire file F?
How many records of 100 bytes each can be stored using this disk?
What time is required to read the file F sequentially?
What time is required to read the file F randomly?
Explanation / Answer
How many records fit into a block ?
Answer :- five double sided platters = 5 * 2 = 10 surfaces
2000 tracks per surface = 2000 * 10 = 20000 tracks
50 sectors per track = 20000 * 50 = 1000000 sectors
512 bytes per sector = 1000000 * 512 = 512000000 bytes
block size= 1024 bytes= 512000000/1024= 500000 blocks
record size=100 bytes
block size=1024 bytes
no. of records/block = 1024/100 = 10.24 (Assume approx of 10)
How many blocks are required to store the entire file F?
Total no. of records in a file = 100000
per block we can have 10 records
so total blocks required = 100000/10 = 10000
So, Approximately 10000 blocks are required to store entire file
How many records of 100 bytes each can be stored using this disk?
As we have 50000 blocks , we can store upto 50000*10 = 500000 records of 100 bytes each
What time is required to read the file F sequentially?
10000 blocks in a file
transfer time of one track (50 sectors)= 1/5400 * 60 = 0.011 seconds
No. of tracks to read = 10 * 1000 * 2 = 20000
Time taken to read all blocks = 20000 * 0.011 = 220 seconds
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