Data Str & Algorithm HW Solutions 54

Data Str & Algorithm HW Solutions 54 - exercise as...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
8 File Processing and External Sorting 8.1 Clearly the prices continue to change. But, the principles remain the same. 8.2 The f rst question is How many tracks are required by the f le? A track holds 144 . 5 K =72 K . Thus, the f le requires 5 tracks. The time to read a track is seek time to the track + latency time + (interleaf factor × rotation time). Average seek time is de f ned to be 80 ms. Latency time is 0 . 5 16 . 7 ms, and track rotation time is 16.7 ms for a total time to read the f rst track of 80 + 4 . 5 16 . 7 155 ms . Seek time for the remaining four tracks is de f ned to be 20 ms (since they are adjacent), with identical latency and read times. Thus, the total f le read time is 155 + 4(20 + 4 . 5 16 . 7) 536 ms which is pretty slow by today’s standards. 8.3 The expected time to read one track at random was given in the previous
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: exercise as 80 + 4 . 5 ∗ 16 . 7 = 155 ms. The expected time to read one sector at random is seek time plus latency plus the time to read one sector (which takes up 1 / 144 of a track). Thus, the time required is 80 + . 5 ∗ 16 . 7 + 1 / 144 ∗ 16 . 7 ≈ 88 . 5 ms. To read one byte, we save the sector read time of 1 / 144 ∗ 16 . 7 which is about .1 ms., which is insigni f cant. 8.4 This is quite similar to Exercise 8.2, but with more modern equipment. One track holds 31.5K bytes, so the f le requires 4 tracks plus 4 sectors of a f fth track. Seek time to the f rst track is 3 ms + 2100 / 3 ∗ . 08 ms ≈ 59 ms . 54...
View Full Document

This note was uploaded on 12/27/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

Ask a homework question - tutors are online