Disk requests come in to the disk driver for cylinders 10, 22, 20, 2, 40, 6, and 38, in that order. a seek takes 6 msec per cylinder moved. the arm is initially at cylinder 20.

how much seek time is needed if

a) fcfs

b) sstf

c) scan (elevator)

algorithm is employed?

For  FCFS = 876msec For SSTF = 360msec For SCAN(elevator) = 348msec

Explanation:

Considering FCFS algorithm.

In FCFS, the requests are addressed in the order they arrive in the disk queue, this means that the number of cylinders traveled becomes equal to the total of disk requests. With the arm initially at 20, the first request is to read cylinder 10.

Therefore the cylinders traversed for the first request = 20 – 10 = 10  

For the second request i.e. a movement from cylinder 10 to cylinder 22, the number of cylinders traversed is = 22 - 10 = 12.

Similarly, for the third request seek arm will return to 20 from 22 so, cylinders traversed through would be = 22-20 = 2.    

Also for the fourth request, cylinders traversed would be = 20 – 2 = 18.

For the fifth request, cylinders traversed = 40 – 2 = 38.

Now for the sixth request cylinders traversed = 40 – 6 = 34.

For the seventh and last request, cylinders traversed = 38 – 6 = 32.

So now to get the how much seek time is required for Disk scheduling algorithm  

First we would add the total cylinders traversed = 10 + 12 + 2+ 18+ 38 + 34 + 32

     = 146 cylinders  

So therefore the total seek time = number of cylinders traversed X seek time per cylinder

               = 146 X 6

   = 876msec

Considering SSTF algorithm.

In SSTF (Shortest Seek Time First), requests having shortest seek time are executed first. So, the seek time of every request is calculated in advance in the queue and then they are scheduled according to their calculated seek time. What this means is that the closest disk (cylinder) next to the position of the seek arm is attended to first. With   the arm at 20 initially, the first request is to read cylinder 22 (i.e. the closest cylinder to the seek arm)

Therefore the cylinders traversed for the first request = 22-20 = 2

For the second request, the disk to focus on is disk 20 and the cylinders traversed = 22-20 = 2

Similarly for the third request the seek arm will move to 10 from 20 so, cylinder traversed = 20-10 =10

For fourth request, cylinder traversed = 10 – 6 = 4

For the fifth request, cylinder traversed = 6 – 2 = 4

For sixth request, since all other disk request closer to it has been attended to the seek arm will move to disk 38 to attend to that disk request So, the cylinder traversed = 38 – 2 = 36

For the last request, cylinder traversed = 40 -38 = 2

So now to get the how much seek time is required for Disk scheduling algorithm  

First we would add the total cylinders traversed = 2 + 2 +10 + 4 + 4 + 36 + 2  

     = 60 cylinders

So therefore the total seek time = number of cylinders traversed X seek time per cylinder

     = 60 X 6 = 360msec

From Here we see that SSTF is better or an improvement to FCFS as it decrease the average response time (Average Response time is the response time of the all requests).

Considering SCAN (elevator) algorithm  

In SCAN algorithm the disk arm moves into a particular direction and services the requests coming in its path and after reaching the end of disk, it reverses its direction and again services the request arriving in its path. So, this algorithm works as an elevator and hence also known as elevator algorithm. Therefore the number of cylinder traveled becomes equal to the total of disk request. With the arm at 20 initially

The first request is to read cylinder 22 i.e. the first cylinder on the upward movement  

Therefore the cylinders traversed would be  =   20 – 22 = 2

For the second request is to read cylinder 38, and the cylinders traversed would be   = 38 – 22 =16

For the third request, seek arm will move to 40 So, the cylinders traversed would be = 40 – 38 = 2

For the fourth request, seek arm will return to 20 since from 40 since 40 is the highest in this upward elevator movement So, cylinders traversed would be = 40 -20 = 20  

For the fifth request, cylinder traversed would be = 20 – 10 = 10

For the sixth request, cylinder traversed would be   = 10 – 6 = 4

For the seventh and last request, cylinder traversed = 6 – 2 = 4

So now to get the how much seek time is required for Disk scheduling algorithm  

First we would add the total cylinders traversed = 2 + 16 + 2 + 20 +10+ 4 + 4 = 58 cylinders

So therefore the total seek time = number of cylinders traversed X seek time per cylinder

   = 58 X 6

          = 348msec

From Here we see that SCAN is better or an improvement to FCFS and SSTF as it decrease the average response time (Average Response time is the response time of the all requests).

answer; i believe the correct answer is 50 pounds;