Classical

Operating System - I/o System

1:  

Access to moving head disks requires three periods of delay before information is brought into memory. The response that correctly lists the three time delays for the physical access of data in the order of the relative speed from the slowest to the fastest is

A.

latency time, cache overhead time, seek time

B.

transmission time, latency time, seek time

C.

seek time, latency time, transmission time

D.

cache overhead time, latency time, seek time

 
 

Option: C

Explanation :

Seek time is the time required to move the disk arm to the required track. Rotational delay or latency is the time it takes for the beginning of the required sector to reach the head. Sum of seek time (if any) and latency is the access time. Time taken to actually transfer a span of data is transfer time or transmission time.


2:  

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

A.

25

B.

30

C.

40

D.

60

 
 

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 blockTime taken to access one block = seek time + rotational delay + block transfer time= 20 + 25 + 16.67 = 61.67 ms


3:  

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

A.

it allows cache domain to contain redundant data

B.

it does not allow writes to be cached

C.

its access time is greater than that of paged caching

D.

read ahead cache domain blocks are necessarily fixed in size

 
 

Option: A

Explanation :


4:  

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

A.

a plate of data

B.

a cylinder of data

C.

a track of data

D.

a block of data

 
 

Option: D

Explanation :


5:  

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

A.

125 msec

B.

295msec

C.

575 msec

D.

750 msec

 
 

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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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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