OPERATING SYSTEMS DISK SCHEDULING PLEASE SHOW YOUR WORK IM T

OPERATING SYSTEMS: DISK SCHEDULING

PLEASE SHOW YOUR WORK! I\'M TRYING TO UNDERSTAND HOW TO ACTUALLY DO THE PROBLEM.

2. (3 points) Suppose that a disk drive has 3000 cylinders, numbered 0 to 2999. The drive is currently serving a request at cylinder 140, and the previous request was at cylinder 125. The queue of pending requests, in FIFO order, is

                                66, 147, 913, 289, 202, 1750, 15

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

e. C-SCAN

3. Requests are not usually uniformly distributed. For example, a cylinder containing the file system FAT or inodes can be expected to be accessed more frequently than a cylinder that only contains files. Suppose you know that 50 percent of the requests are for a small, fixed number of cylinders.

Propose a disk-scheduling algorithm (one that we have discussed) that gives even better performance by taking advantage of this “hot spot” on the disk. Explain your answer and discuss any problems and how you might mitigate them.

Solution

The drive is currently serving a request at cylinder 140

request queue -- 66,147,913,289,202,1750,15

a) FCFS

66<--- 140

66---->147

147 ----> 913

289 <----- 913

202<---- 289

202 ------>1750

15 <------ 1750

total time = |66-140|+|66-147|+|147-913|+|289-913|+||202-289|+|202-1750|+|15-1750|

=74+81+766+624+87+1548+1735

=4915

b) SSTF

140 ----> 147

147 ------>202

202----->289

66<----289

15<----66

15----->913

913----->1750

total time=7+55+87+223+51+898+837

=2158

c) SCAN

140--->147---->202---->289----->913----->1750----->2999---

0<-----15<------66<-----------------------------------------------

total time = |140-147|+|147-202|+|202-289|+|289-913|+|913-1750|+|1750-2999|+|2999-66|+|66-15|+|15-0|

=7+55+87+624+837+1249+2933+51+15

=5858

d)LOOK

140--->147---->202---->289----->913----->1750---

15<------66<-----------------------------------------------

total time = |140-147|+|147-202|+|202-289|+|289-913|+|913-1750|+|1750-66|+|66-15|

=7+55+87+624+837+1684+51

=3345

e)C-SCAN

140--->147---->202---->289----->913----->1750----->2999---

66<-----15<------0<-----------------------------------------------

total time = |140-147|+|147-202|+|202-289|+|289-913|+|913-1750|+|1750-2999|+|2999-0|+|15-0|+|66-15|

=7+55+87+624+837+1249+2999+15+51

=5924

SSTF can be the best disk scheduling algorithm when data is close to each other.

We can have hot spot in the middle of the disk . Aging can be done on older request by increasing their priority