GATE | GATE-CS-2015 (Set 1) | Question 65
Consider a disk pack with a seek time of 4 milliseconds and rotational speed of 10000 rotations per minute (RPM). It has 600 sectors per track and each sector can store 512 bytes of data. Consider a file stored in the disk. The file contains 2000 sectors. Assume that every sector access necessitates a seek, and the average rotational latency for accessing each sector is half of the time for one complete rotation. The total time (in milliseconds) needed to read the entire file is _________.
Seek time (given) = 4ms RPM = 10000 rotation in 1 min [60 sec] So, 1 rotation will be =60/10000 =6ms [rotation speed] Rotation latency= 1/2 * 6ms=3ms # To access a file, total time includes =seek time + rot. latency +transfer time TO calc. transfer time, find transfer rate Transfer rate = bytes on track /rotation speed so, transfer rate = 600*512/6ms =51200 B/ms transfer time= total bytes to be transferred/ transfer rate so, Transfer time =2000*512/51200 = 20ms Given as each sector requires seek tim + rot. latency = 4ms+3ms =7ms Total 2000 sector takes = 2000*7 ms =14000 ms To read entire file ,total time = 14000 + 20(transfer time) = 14020 ms
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