Suppose that a disk drive has 10,000 cylinders, numbered 0 to 10,000. The drive is currently serving a request at cylinder 7,423 and the previous request was at cylinder 8213. The queue of pending requests, in FIFO order, is: 9324, 6324, 7232, 1304, 5621, 105, 5382, 3241, 2425, 9891.
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?
a. FCFS
b. SSTF
c. SCAN
d. LOOK
Queue of pending request is: 9324, 6324, 7232, 1304, 5621, 105, 5382, 3241, 2425, 9891
Current position = 7423
The disk scheduling or I/O scheduling is done by the operating system. Computer hard disk is slower and more than one I/O request may come at the same time but only one request can be served at the same time.
a. FCFS
The requests are assigned in the same order as they arrived in the disk queue.
The order is: 7423, 9324, 6324, 7232, 1304, 5621, 105, 5382, 3241, 2425, 9891
Total Distance = |7423 - 9324| + |9423 - 6324| + |6324 - 7232| + |7232 - 1304| + |1304 - 5621| + |5621 - 105| + |105 - 5382| + |5382 - 3241| + |3241 - 2425| + |2425 - 9891|
= 2369 Cylinders
b. SSTF
The request having shortest seek time are assigned first.
The order is: 7423, 7232, 6324, 5621, 5382, 3241, 2425, 1304, 105, 9324, 9891
Total Distance = |7423 - 7232| + |7232 - 6324| + |6324 - 5621| + |5621 - 5382| + |5382 - 3241| + |3241 - 2425| + |2425 - 1304| + |1304 - 105| + |105 - 9324| + |9324 - 9891|
= 2468 Cylinders
c. SCAN
The disk arm moves in a particular direction till the end of the disk.
The order is: 7423, 7232, 6324, 5621, 5382, 3241, 2425,1304,105, 0, 9324, 9891
Total Distance = |7423 - 7232| + |7232 - 6324| + |6324 - 5621| + |5621 - 5382| + |5382 - 3241| + |3241 - 2425| + |2425 - 1304| + |1304 - 105| + |105 - 0| + |0 - 9324| + |9324 - 9891|
= 2468 Cylinders
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