Question: Suppose that a disk drive has 10000 cylinders (numbered 0-9999). It ha

Question

Question:

Suppose that a disk drive has 10000 cylinders (numbered 0-9999). It has just finished a request at cylinder 188, and the previous request was at 102. The queue of pending requests (in FIFO order) is: 12, 7650, 867, 233, 9878, 444, 3827, 7878, 587. Starting at the current position, calculate the total distance (in cylinders) that the disk arm must move to satisfy all the pending requests for each of the disk scheduling algorithms:

a. FCFS

b. SSTF

c. SCAN

d. LOOK

e. C-LOOK

f. C-SCAN


For FCFS, I got 49031 as the total distance, but after that I get lost with the pecking order
for the remaining scheduling algorithms. So sorry for this embarrassing question! Again, thank you very much for your help and time!

Explanation / Answer

FFCS:- DISTANCE=188-12+7650-12+7650-867+867-233+9878-233+9878-444+3827-444+7878-3827+7878-587. =49035. SSTF:- DISTANCE=233-188+444-233+587-444+867-587+867-12+3827-12+7650-3827+7878-7650+9878-7878. =11400. SCAN- DISTANCE=233-188+444-233+587-444+867-587+3827-867+7650-3827+7878-7650+9878-7878+9999-9878+9999-12. =19798. LOOK:- DISTANCE=233-188+444-233+587-444+867-587+3827-867+7650-3827+7878-7650+9878-7878+9878-12 =19556. C-SCAN:- DISTANCE=233-188+444-233+587-444+867-587+3827-867+7650-3827+7878-7650+9878-7878+9999-9878+9999-0+12-0 =19822. C-LOOK:- DISTANCE=233-188+444-233+587-444+867-587+3827-867+7650-3827+7878-7650+9878-7878+9878-12 =19556.

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