Question

1. Suppose that a disk drive has 5,000 cylinders, numbered 0 to 4,999. The drive is currently serving a request at cylinder 2,150, and the previous request was at cylinder 1,805. The queue of pending requests, in FIFO order, is: 2,069 1,212 2,296 2,800 544 1,618 356 1,523 4,956 3,681 Starting from the current head position, what is the total distance (in cylinders) that the disk arm moves to satisfy all the pending requests for each of the following disk-scheduling algorithms? Record your answers in the table below.

	Question 1: Total Distance	Question 2 (a): Total Seek Time
FCFS
SSTF
SCAN
LOOK
C-SCAN
C-LOOK

Answer 1

a. In FCFS ( First come first serve) disk scheduling, the requests are served in the same arder thye arrived. So, the total distance will be Initial disk arem position = 2150 FIFO order = 2069, 1212, 2296, 2800, 544, 1618, 356, 1523, 4956, .3681 So, Total distance is calculated as a 12150 - 20691 + 12069-1212) +1212-2296Y+ 12296-2800| + 12800-544) + 1544–16181 +11618 - 3561 +356-15231 +11523 - 49561 +14956 -3681|


= 81 +857 +084 +504 +2256 + 1074 +1262 +1167 + 3433 + 1275 = 12993 considering Total seek seek time = 6msec ) cylinder time =12993 X6 msec. = 77,958 msec. = 77.958 sec,

In SSTF (shortest seek Time Firest) Algorithm the request with the shaitest seeking time from the current head position is served first. Hence 2069 is the shortest distance from current head. So, Total distance will be 2150 - 20691 + 12069-2296) + 12296-1800 + |2800 -3681+/3681-4956/+/4956–1618 +11618 – 15231 +|1523-12121 + 112/2544) + 1544 - 561 = 81 +227 +504 +881+1275+ 3338 +95 + 311+668 +188 - 7568 Considering seek time in 6 msee) cylinder, Total seek time = 7568x6 msee = 45, 408 msec. = 45 - 408 see

arm C. In SEAN disk scheduling , the disk. moves in one direction serving the requests that come accross. Once it reaches the end of the disk, it reverses the direction and serves until it reach and of the dick. Previous request was at cylinder 1805 At present disk arm is at 2150. so the disk arom is moving towards the end of disk. So, it serves in the order 2150, 2296, 2800, 3681,4956, Anodisk) Now it reverse the direction. 2069, 1618, 1523, 1212, 544, 356 So total distance is 2150 - 2296/+/2296-2800/ + /2800-36811+ 13681 -4956) + | 4956-4999) + 4999 - 2069) + 1 1 2069-1618) + | 1618 - 1523)+|1523-1212) + 11212-5441 +1544-356| 4999 = 146+504 + 881+1275+43+2930 +451+ 95+ 311 +668 +188 - 7492 7492 considering seek time 6 msec/ cylinder the Total seek time in 7492x6 msec = 44,952 msec. - 49-952 sec.

d. In Look disk scheduling algorithm, the disk aron moves in one direction serving the requests that it come across once it reaches the last request, it reverses the diretion and goes to the other end of the disk. So, the order will be as belove 2150, 2296, 2800, 3681,4956 Now Reverse the direction 2069, 1618, 1523, 1212, 544, 356 Total distance will be 2150 -22961+12296-2800] +12800-36811+ 13681-49561+14956-2069/+ 2069-1618/+ 11618-1523/+/1523-1212/+/1212-544/+ 1544-3561 - 146 +504 +881 + 1275 +2887 +451 +95 +331+668 +188 - 7406 considering seek time, 6 msec/cylinder, the total seek time in 74 06 x 6 msec = 44, 436 msec - 44.436 sec.

in C-SCAN disk scheduline alzarithen, the - serving the request that it come, acrotoradan lisk arm moves in one directione, till it reaches the end of the disk, it reseives it direction and moves toack to the opposite side & the disk weithout serving in any request. Then starting from the disk it serves the remaining request as it comes going to the opposite side of disk. so, the order of request will be as below. 2150, 2296, 2800, 368 1,4956,4999 (And of disk) Now reverse and goes to the starting of disk. Then 0,356, 544, 1212, 1523, 1618, 2069 so, Total distance will be 12150 - 2296) + (2296-2800| +)2800-3681) + |3681-4956) +14956 - 49991+14999-0/+10-356) + 1356 - 544/+)544-1212/+1212-1523)+ 11523-1618) +11618-20691 - 146+504 +881+ 1275 +43 +4999 +356 + 188 + 668 + 311 +95 + 451 = 9917 Total seek time 9917x6 msec. - 59,502 msec - 59.502 sec.

ou In C-Look disk scheduling algarithen, the disk arm moves in one direction till at es the last request in that direction towards one end of the disk. After seaving last request, it reverse its direction, and back to the farthest request in oppostie moves direthon. It then serves مهره ) request at then serves all remaining disk end, moving the the direction of So, the total distance will be First write down the order of request. 2150, 2296, 2800, 3681, 4956 Now reverse the direction and start from come to the opposite side to start from 356,544, 1212, 1523, 1618, 2069 Total distance = 12 150-2296 +/2296-28001 + /2800-3681) + 13681-4956]+4956-356) + |356-544) + 1544-12121 +11212-1523)+1 1523 -1618/ +/1618 - 20691 = 146 +504+881+12 75+4600+ 188 +668+314 +95 +451 = 9119 Total seek tinue = 91198 6msec - 54,7 14 msee - 54.714 sec