System Software and Operating System - File and Input/Output System

61. If a disk has a seek time of 20 ms, rotates 20 revolutions per second, has 100 words per block, and each track has capacity of 300 words. Then the total time required to access one block is

  • Option : D
  • Explanation : Time taken to access one block = seek time + rotational delay + block transfer time
    Seek time = 20 ms (given)
    Rotational delay = on an average taken to be the time to rotate by half = 1/2 X time for 1 rotation = 1/2 X 1/20 seconds = 1/40 s = 25 ms
    Block Transfer time = block size / transfer rate = 100 / transfer rate
    Now, transfer rate = Track capacity / rotation rate = 300 / ( 1/ 20) = 6000 word per sec = 6 word per ms
    Block Transfer time = block size / transfer rate = 100 / 6 ~ 16.67 ms per block Time taken to access one block = seek time + rotational delay + block transfer time= 20 + 25 + 16.67 = 61.67 ms
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62. An unpaged or read-ahead cache associates disk domains with the address of the read and continues for a specific length. The major disadvantage of unpaged cache is that

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63. The maximum amount of information that is available with one position of the disk access arm for a removal disk pack (without further movement of the arm with multiple heads) is

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64. Disk requests are received by a disk drive for cylinders 5, 25, 18, 3, 39, 8 and 35 in that order. A seek takes 5 m sec per cylinder moved. How much seek time is needed to serve these requests for a Shortest Seek First (SSF) algorithm ? Assume that the arm is at cylinder 20 when the last of these requests is made with none of the requests yet served

  • Option : B
  • Explanation : Shortest Seek Time First - minimizes arm movement
    Order to be followed is - (20) , 18,25,35,39, 8,5,3
    Seek = 2+5+7+10+4+31+3+2 = 59 cylinders , 295 ms
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65. Disk requests come to a disk driver for cylinders 10, 22, 20, 2, 40, 6 and 38, in that order at a time when the disk drive is reading from cylinder 20. The seek time is 6 ms per cylinder.
The total seek time, if the disk arm scheduling algorithm is first-come-first-served is

  • Option : B
  • Explanation : According to FCFS order serverd will be (20),10, 22, 20, 2, 40, 6 and 38
    Seek time = 10+12+2+18+38+34+32 = 146
    The disk drive has to traverse totally 146 cylinders (verify). So, seek time is 6 x 146 = 876 ms.
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