Karar khudhair HW - fhrtyrtyryr ryryr PDF

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COMPUTER Engineering ‫خضي جابر‬ ‫كرار ر‬ A ‫المرحلة الرابعة‬

2021

Information Technology

H.W.1 :

Consider a hard disk with the following characteristics:

disk unit of 20 surfaces, 800 tracks/surface, 22 sectors/track, 512 bytes/sector, 3600 rpm, 7 msec. track-to-track seek time, 28 msec. avg. seek time, 50 msec. max. seek time, find: (a) Average latency. (b) Time to read one track. (c) The total storage capacity of the disk in Megabytes. (d) No. of revolutions to read the disk without interleaving. (e) No. of revolutions to read the disk with 3:1 interleaving. solurtion / (a) / Average latency Average latency (r) = (min + max) / 2 = max / 2 = time for ½ disk revolution Where: Min latency = 0 Max latency = Time for one disk revolution the disk spins at 3600 RPM, so it takes 1/ 3600 of a minute to make one revolution Equivalently (1000 ms/sec × 60 sec/minute) / 3600 RPM = 16.6666667 msec = ½ * 16.6666667 msec = 8.33333334 msec

(b) / Time to read one track Time to read one track = average seek time +average latency +transfer time Tranfer time = (1/22) *(3600*10 -3) msec = 0.0126262 msec average seek time= 28 msec , average latency = 8.33333334 msec Time to read one track =28 + 8.33333334 + 0.0126262 = 36.34595954 msec

(c) / The total storage capacity of the disk in Megabytes Capacity of disk pack = Total number of surfaces x Number of tracks per surface x Number of sectors per track x Number of bytes per sector = 20 x 800 x 22 x 512 bytes =180224000 bytes , since megabyte= 2 20

, 180224000 =2 20 * 171.875

The total storage capacity of the disk in Megabytes = 171.875 MB

(d) No. of revolutions to read the disk without interleaving 1 revolution

(e) No. of revolutions to read the disk with 3:1 interleaving 3 revolution...