Operating System - I/o System

6:  

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

A.

360 ms

B.

876 ms

C.

850 ms

D.

900 ms

 
 

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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Kishore R said: (6:17pm on Friday 9th October 2015)
disk drive is reading from cylinder 20,fcfs ,20 2 40 6 38 10 22=138seek time=138x6=824 I had some confusion

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7:  

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.

If the scheduling algorithm is the closest cylinder next, then the total seek time will be

A.

360 ms

B.

876 ms

C.

850 ms

D.

900 ms

 
 

Option: A

Explanation :

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8:  

Disk sheduling involves deciding

A.

which disk should be accessed next

B.

the order in which disk access requests must be serviced

C.

the physical location where files should be accessed in the disk

D.

none of these

 
 

Option: B

Explanation :

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9:  

Poor response time are caused by

A.

Process or busy

B.

High I/O rate

C.

High paging rates

D.

All of these

 
 

Option: D

Explanation :

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10:  

A certain moving arm disk storage with one head has following specifications : Number of tracks I recording surface 100 Disk rotation speed 2400 rpm Track storage capacity= 62500 bits

The average latency time (assume the head can move from one track to another only by traversing the entire track) is

A.

2.5 s

B.

2.9 s

C.

3.1 s

D.

3.6 s

 
 

Option: A

Explanation :

 In 60 seconds, the disk rotates 2400 times. So it takes 25ms to rotate it one time.
There are 200 tracks. On average, 100 tracks need to be traveresed to reach a specific track.
So 25ms*100 tracks= 2.5s time is taken on average to reach a particular track 

 

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Syllabus covered in this section is-

  • Main functions of operating systems
  • Multiprogramming, multiprocessing and multitasking
  • Memory Management- Virtual memory, paging, fragmentation
  • Concurrent Processing: Mutual exclusion, Critical regions, locks and unlock.
  • Scheduling: CPU scheduling. I/O scheduling, Resource scheduling
  • Deadlock and scheduling algorithms.
  • Banker's algorithm for deadlock handling.

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