Chapter 11, Problem 22E

Explanation of Solution

Performance:

• The speed of the CPU is a key factor that often decides the performance of the system.
• Disk access speed is considered to be another factor that is used to determine the performance.
• The read write operation that is performed on the disk (milliseconds) is considered to take significantly longer access time in comparison with that of the operation that is performed in a CPU (nanoseconds).
• To improve the performance of the disk access they are different ways present.
• The algorithm that determines the order in which the data access made in the disk will be beneficial is calculated.

Given:

It is know that there are 100 tracks that are ranged from 0 to 99.

At the beginning the disk arm is considered to be at the position 50 and it will getting move through and towards the lower numbered tracks.

It will require 500 nanoseconds to pass over the track.

The desired track is accessed in 2milliseconds.

The service request order is as follows 48,14,85,35,84,61,30, and 22.

Disk traversed using FCFS (First come First Serve):

The service request is being handled in the way they are being received:

Service request	Access made	Number of access
48	50-48	2
14	48-14	34
85	14-85	71
35	85-35	50
84	35-84	49
61	84-61	23
30	61-30	31
22	30-22	8
Total number of disk access		268
Total time taken		8*2 milliseconds+(268-8)*500 nanoseconds

Therefore, total number of disk access required is “268”.

Disk traversed using SSTF (Shortest Seek Time First):

• The service request that are from the nearest of the sector to that of the current position of the disk arm are being handled on priority.
• The list is processed to be generated in the way from the current position of the disk arm and the instances of the next nearest access are being obtained at every instance.

Service request	Access made	Number of access
48	50-48	2
61	48-61	13
84	61-84	23
85	84-85	1
35	85-35	50
30	35-30	5
22	30-22	8
14	22-14	8
Total number of disk access		110
Total time taken		8*2 milliseconds+(110-8)*500 nanoseconds

Therefore, total number of disk access required is “110”...