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Test: Disk Scheduling Algorithm - Computer Science Engineering (CSE) MCQ


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10 Questions MCQ Test Operating System - Test: Disk Scheduling Algorithm

Test: Disk Scheduling Algorithm for Computer Science Engineering (CSE) 2024 is part of Operating System preparation. The Test: Disk Scheduling Algorithm questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Test: Disk Scheduling Algorithm MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Disk Scheduling Algorithm below.
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*Answer can only contain numeric values
Test: Disk Scheduling Algorithm - Question 1

Let's say there are seven requests: 16, 24, 43, 82, 140, 170, and 190. Additionally, the disc arm is supposed to travel "towards the larger number" while the read/write arm is at 50. What is the total seek time?


Detailed Solution for Test: Disk Scheduling Algorithm - Question 1

The given data,

"towards the larger number
" means it works like a scan algorithm.
Let's say there are seven requests: 16, 24, 43, 82, 140, 170, and 190.
The read/write arm is at 50.

The seek time is calculated as:
=(199-50)+(199-16)
=332

Hence the correct answer is 332.

*Answer can only contain numeric values
Test: Disk Scheduling Algorithm - Question 2

Consider a disk that has 209 cylinders, starting from index 0. At some time, the disk arm is at the 57th cylinder from the end moving from largest to innermost cylinder. There is a queue of disk access requests for cylinders 5, 40, 75, 195, 145, 203, 106 and 122. The R/W head of the disk uses the Circular Scan algorithm to reach the disk and access the data. The disk arm requires 1 msec to move between adjacent cylinders and 9 msec to move between the largest and innermost cylinder. The total seek time is _______.


Detailed Solution for Test: Disk Scheduling Algorithm - Question 2

Algorithm - CSCAN
Requests - 5, 40, 75, 195, 145, 203, 106, 122
Current Cylinder - (208 - 57 + 1) = 151

  • Here, we start from the current cylinder, that is, 151 because as the question states, we are starting from the 57th cylinder from the largest cylinder. So, as per the chart, 57th cylinder from the right, that is, 151.
  • The direction is from largest cylinder to innermost cylinder, that is, right to left.
  • Now R/W head will move from 151 to 150, 150 to 149 and so on till it reaches 145. It reaches there and access the data. Total seek time till now can be calculated by: (151 - 145) * 1 = 6 msec. Similarly, R/W head will move from 145 to 122, 122 to 106 and so on till 0.
  • Now when the R/W head reaches 0, it will jump to 208 and then continue moving in the same direction. Now, this jumping will take 9 msec. After that, it will move as is mentioned in (step - 3).

Calculation - 
{[(151 - 145) + (145 - 122) + (122 - 106) + (106 - 75) + (75 - 40) + (40 - 5) + (5 - 0)] * 1} + {(208 - 0) * 9} + {[(208 - 203) + (203 - 195)] * 1} = 203

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Test: Disk Scheduling Algorithm - Question 3

There are 200 tracks on a disk platter and the pending requests have come in the order - 36, 69, 167, 76, 42, 51, 126, 12, and 199, Assume the arm is located at the 100th track and moving towards track 200. If the sequence of disc access is 126, 167, 199, 12, 36, 42, 51, 69, and 76 then which disc access scheduling policy is used?

Detailed Solution for Test: Disk Scheduling Algorithm - Question 3

Option 1: 
The elevator algorithm is also known as the SCAN algorithm. 

In this algorithm, the disk moves in a particular direction servicing the requests coming in the way till the end & reverse the direction & service all the requests.

Option 2: 
In Shortest Seek Time First(SSTF), the algorithm selects disk I/O which requires the least disk arm movement from the current position

Option 3: 
In C-scan i.e. Circular Elevator, the disk moves in a particular direction servicing the requests coming in the way till the end & reverse the direction & again goes to the start point & starts servicing the requests once reach the start point. 

Option 4:
In First Come First Serve(FCFS), the disk service the request that comes first.
Hence, the correct answer is "option 3".

Test: Disk Scheduling Algorithm - Question 4

Consider a disk system having 60 cylinders. Disk requests are received by a disk drive for cylinders 10, 22, 20, 2, 40, 6, and 38, in that order. Assuming the disk head is currently at cylinder 20, what is the time taken to satisfy all the requests if it takes 2 milliseconds to move from one cylinder to an adjacent one and Shortest Seek Time First (SSTF) algorithm is used ?

Detailed Solution for Test: Disk Scheduling Algorithm - Question 4

Shortest Seek Time First (SSTF)

SSTF =(22-20+22-10+10-6+6-2+38-2+40-38)
          =2+12+4+4+36+2
          =60

It takes 2 milliseconds to move from one cylinder to adjacent one = 60x2  =120 milliseconds
∴ Hence the correct answer is 120 milliseconds.

Test: Disk Scheduling Algorithm - Question 5

Match the following.

Detailed Solution for Test: Disk Scheduling Algorithm - Question 5

Stack algorithm:
A page replacement algorithm is said to satisfy the inclusion property, or it is called as stack algorithm if the set of pages in a k frame memory is always a subset of the pages in a k+1 frame memory.

Elevator algorithm:
It is a disk scheduling algorithm, also known as scan algorithm in which head, starts from one end of disk and moves towards other end by servicing each request in between and reach other end.

Priority scheduling algorithm:
It is a CPU scheduling algorithm in which each process is assigned a priority and processes according to their priority.

Havender’s Algorithm:
It is related with deadlock. In this, all the resources required by process will be requested at once. 

*Answer can only contain numeric values
Test: Disk Scheduling Algorithm - Question 6

Consider a storage disk with 4 platters (numbered as 0, 1, 2 and 3), 200 cylinders (numbered as 0, 1, … , 199), and 256 sectors per track (numbered as 0, 1, … , 255). The following 6 disk requests of the form [sector number, cylinder number, platter number] are received by the disk controller at the same time: [120, 72, 2] , [180, 134, 1] , [60, 20, 0] , [212, 86, 3] , [56, 116, 2] , [118, 16, 1]  Currently the head is positioned at sector number 100 of cylinder 80, and is moving towards higher cylinder numbers. The average power dissipation in moving the head over 100 cylinders is 20 milliwatts and for reversing the direction of the head movement once is 15 milliwatts. Power dissipation associated with rotational latency and switching of head between different platters is negligible. The total power consumption in milliwatts to satisfy all of the above disk requests using the Shortest Seek Time First disk scheduling algorithm is _______.


Detailed Solution for Test: Disk Scheduling Algorithm - Question 6

Total head movements in SSTF= (86−80)+(86−72)+(134−72)+(134−16)=200
Power dissipated in moving 100 cylinder = 20mW
Power dissipated by 200 movements = 0.2∗200=40mW = P
Power dissipated in reversing head direction once= 15mW
Number of times head changes its direction= 3
Power dissipated in reversing head direction = 3∗15= 45mW = Q
​​​​​​​Total Power Consumption is P + Q = 85mW

Test: Disk Scheduling Algorithm - Question 7

Let disk request come to a disk driver for cylinders in the order 10, 22, 20, 2, 40, 6 and 38, at the time when the disk driver is reading from cylinder 20. If the seek time is 6 ms per cylinder, then the total seek time for the disk arm scheduling time algorithm (FCFS - First Come First Serve) is:

Detailed Solution for Test: Disk Scheduling Algorithm - Question 7

Data:
Seek time = t = 6 ms per cylinder
Head = 20

Concept:
Since, it is a first-come-first-serve scheduling requests will be served in the given sequence:
10, 22, 20, 2, 40, 6 and 38

Formula:
Total Seek time = total head movement × t

Calculation:
Total head movement = (20 - 10) + (22 - 10) + (22 - 2) + (40 - 2) + (40 - 6) + (38 - 6) =
= 10 + 12 + 20 + 38 + 34 + 32 = 146
Total Seek time = 146 x 6 = 876 ms.

Test: Disk Scheduling Algorithm - Question 8

In Magnetic Disks Structures, The surface of a platter is logically divided into circular______

Detailed Solution for Test: Disk Scheduling Algorithm - Question 8

Magnetic Disks:

Magnetic Disks provide the bulk of secondary storage for modern computer systems. Conceptually, disks are relatively simple. Each disk platter has a flat circular shape, like a CD. The two surfaces of a platter are covered with magnetic material. We store information by recording it magnetically on the platters.

A read-write head "flies" just above each surface of every platter. The heads are attached to a disk arm that moves all the heads as a unit. The surface of a platter is logically divided into circular tracks which are subdivided into sectors.

*Answer can only contain numeric values
Test: Disk Scheduling Algorithm - Question 9

Consider the process track request are 98, 183, 37, 122, 14, 124, 14, 124, 65, 67 and the initial seek is 53. What is the total number of seeks for the C-Look algorithm?


Detailed Solution for Test: Disk Scheduling Algorithm - Question 9

The given data,
The process track request are= 98, 183, 37, 122, 14, 124, 14, 124, 65, 67 
The initial seek = 53
C-Look algorithm:

Total number of seeks= ( 183-53 )+(183-14) +(37-14)
Total number of seeks= 322
Hence the correct answer is 322.

*Multiple options can be correct
Test: Disk Scheduling Algorithm - Question 10

Suppose that the rotating disk head is moving with 200 tracks numbered from 0 to 199 and is currently serving a request at track 65. Disk queue requests are as follows: 93, 173, 132, 20, 35, 150, 100, 175 and 55.Which of the following statements satisfies these requests for the SSTF (Shortest Seek Time First) algorithm?

Detailed Solution for Test: Disk Scheduling Algorithm - Question 10

Given disk queue requests 93, 173, 132, 20, 35, 150, 100, 175, 55

Total head movements = 10 + 20 + 15 + 73 + 7 + 32 + 18 + 23 + 2 
Total head movements = 200 cylinders
Hence the correct answer is option 2 and option 3.

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