1. Disk Scheduling
Because Disk I/O is so important, it is worth our time to
Investigate some of the issues involved in disk I/O.
One of the biggest issues is disk performance.

2. seek time is the time
required for the read head to
move to the track containing
the data to be read.

3. rotational delay or
latency, is the time
required for the sector
to move under the
read head.

4. Performance Parameters
rotational
delay
data
transfer
Wait for device
seek
(latency)
Wait for
Channel
Device busy
Seek time is the time required to
move the disk arm to the specified track
Ts = # tracks * disk constant + startup time ~
Rotational delay is the time required for the data on
that track to come underneath the read heads. For
a hard drive rotating at 3600 rpm, the average rotational
delay will be 8.3ms.
Transfer Time
Tt = bytes / ( rotation_speed * bytes_on_track )

5. Data Organization vs. Performance
Consider a file where the data is stored as compactly as
possible, in this example the file occupies all of the sectors on
8 adjacent tracks (32 sectors x 8 tracks = 256 sectors total).
The time to read the first track will be
average seek time ms
rotational delay ms
read 32 sectors ms
45ms
Assuming that there is essentially no seek time on the remaining tracks,
each successive track can be read in ms = 25ms.
Total read time = 45ms + 7 * 25ms = 220ms = 0.22 seconds

6. If the data is randomly distributed across the disk:
For each sector we have
average seek time 20 ms
rotational delay 8.3 ms
read 1 sector ms
Total time = 256 sectors * 28.8 ms/sector = 7.73 seconds
28.8 ms

7. In the previous example, the biggest factor on performance is ?
Seek time!
To improve performance, we
need to reduce the average seek time.

8. The operating system keeps a queue of requests to read/write the disk. …
The operating system keeps a queue of
requests to read/write the disk.
In these exercises we assume that all
of the requests are on the queue.

9. Queue If requests are scheduled in random order,…
If requests are scheduled in random order,
then we would expect the disk tracks to be
visited in a random order.

10. First-come, First-served
Queue
First-come, First-served
Scheduling
Request …
If there are only a few processes competing
for the drive, then we can hope for good
performance.
If there are a large number of processes
competing for the drive, then performance
approaches the random scheduling case.

11. While at track 15, assume some random set
of read requests -- tracks 4, 40, 11, 35, 7 and 16
Head Path
Tracks Traveled
Track
15 to 4 11
4 to
40 to
11 to
35 to 7 28
7 to
137 tracks 40 30 20 10

12. Shortest Seek Time First
Queue
Shortest Seek Time First
Request …
Always select the request that requires
the shortest seek time from the current
position.

13. Shortest Seek Time First
While at track 15, assume some random set of read
requests -- tracks 4, 40, 11, 35, 7 and 16
Shortest Seek Time First
Track
Head Path
Tracks Traveled 40 30 20 10
In a heavily loaded system, incoming requests with a
shorter seek time will constantly push requests with
long seek times to the end of the queue. This results
In what is called “Starvation”.
Problem?

14. Shortest Seek Time First
While at track 15, assume some random set of read
requests -- tracks 4, 40, 11, 35, 7 and 16
Shortest Seek Time First
Track
Head Path
Tracks Traveled 40
15 –
16 –
11 –
7 –
4 –
35 – 30 20 10
49 tracks 50
100
In a heavily loaded system, incoming requests with a
shorter seek time will constantly push requests with
long seek times to the end of the queue. This results
In what is called “Starvation”.
Problem?

15. The elevator algorithm
Queue
The elevator algorithm
(scan-look)
Request …
Search for shortest seek time from the
current position only in one direction.
Continue in this direction until all requests
in this direction have been satisfied,
then go the opposite direction.
In the scan algorithm, the head moves all the
way to the first (or last) track before it changes
direction.

16. Scan-Look While at track 15, assume some random set of read requests
Track 4, 40, 11, 35, 7 and 16. Head is moving towards higher
numbered tracks.
Scan-Look
Track
Head Path
Tracks Traveled 40 30 20 10
Steps 50
100

17. Scan-Look While at track 15, assume some random set of read requests
Track 4, 40, 11, 35, 7 and 16. Head is moving towards higher
numbered tracks.
Scan-Look
Track
Head Path
Tracks Traveled 40
15 –
16 –
35 –
40 –
11 –
7 – 30 20 10
61 tracks
Steps 50
100

18. Which algorithm would you choose if you were
implementing an operating system? Issues to
consider when selecting a disk scheduling algorithm:
Performance is based on the number and types of requests.
What scheme is used to allocate unused disk blocks?
How and where are directories and i-nodes stored?
How does paging impact disk performance?
How does disk caching impact performance?

19. Disk Cache The disk cache holds a number of disk blocks in memory,
usually in RAM on the disk controller.
When an I/O request is made for a particular block,
the disk cache is checked. If the block is in the cache,
it is read. Otherwise, the required block (and often some
contiguous blocks) are read into the cache.

20. Replacement Strategies
Least Recently Used
replace the block that has been in the cache the
longest, without being referenced.
Least Frequently Used
replace the block that has been used the least

21. Redundant Array of Independent Disks
RAID
Redundant Array of Independent Disks
Push Performance
Add reliability

22. RAID Level 0: Striping Physical Physical Drive 1 Drive 2 A Stripe
o o o
o o o
strip 8
strip 9
strip 10
Disk Management
Software
strip 11
o o o
Logical Disk

23. RAID Level 1: Mirroring High Reliability Physical Physical Physical
strip 0
strip 1
Physical
Drive 1
Physical
Drive 2
Physical
Drive 3
Physical
Drive 4
strip 2
strip 3
strip 0
strip 0
strip 1
strip 1
strip 0
strip 0
strip 1
strip 1
strip 4
strip 2
strip 2
strip 3
strip 3
strip 2
strip 2
strip 3
strip 3
strip 5
strip 4
strip 4
strip 5
strip 5
strip 4
strip 4
strip 5
strip 5
strip 6
strip 6
strip 6
strip 7
strip 7
strip 6
strip 6
strip 7
strip 7
strip 7
o o o
o o o
o o o
o o o
o o o
o o o
o o o
o o o
strip 8
strip 9
strip 10
Disk Management
Software
Duplicate writes to drive 1
and drive 2 on these disks
strip 11
o o o
Logical Disk

24. RAID Level 3: Parity High Throughput Physical Physical Physical
strip 0
strip 1
Physical
Drive 1
Physical
Drive 2
Physical
Drive 3
Physical
Drive 4
strip 2
strip 3
strip 0
strip 0
strip 1
strip 1
strip 2
strip 0
strip 1
para
strip 4
strip 3
strip 2
strip 4
strip 3
strip 2
strip 5
strip 3
parb
strip 5
strip 6
strip 4
strip 7
strip 5
strip 8
strip 4
strip 5
parc
strip 6
strip 9
strip 6
strip 10
strip 7
strip 11
strip 6
strip 7
pard
strip 7
o o o
o o o
o o o
o o o
o o o
o o o
o o o
o o o
strip 8
parity
strip 9
strip 10
Disk Management
Software
strip 11
o o o
Logical Disk

25. Thinking about what
you have learned

26. Suppose that 3 processes, p1, p2, and p3 are attempting to concurrently use a machine with interrupt driven I/O. Assuming that no two processes can be using the cpu or the physical device at the same time, what is the minimum amount of time required to execute the three processes, given the following (ignore context switches):
Process Time compute Time device

27. Process Time compute Time device 1 10 50 2 30 10 3 15 35 p3 p2 P1 10 20 30 40 50 60 70 80 90
100
110
120
130
105

28. Consider the case where the device controller is double buffering I/O.
That is, while the process is reading
a character from one buffer, the
device is writing to the second.
Process
What is the effect on the running time of the process if the process is I/O bound and requests characters faster than the device can provide them? A
Device
Controller B
The process reads from buffer A.
It tries to read from buffer B, but
the device is still reading. The process
blocks until the data has been stored
in buffer B. The process wakes up and
reads the data, then tries to read Buffer A.
Double buffering has not helped performance.

29. Consider the case where the device controller is double buffering I/O.
That is, while the process is reading
a character from one buffer, the
device is writing to the second.
Process
What is the effect on the running time of the process if the process is Compute bound and requests characters much slower than the device can provide them? A
Device
Controller B
The process reads from buffer A.
It then computes for a long time.
Meanwhile, buffer B is filled. When
The process asks for the data it is
already there. The process does not
have to wait and performance improves.

30. Suppose that the read/write head is at track is at track 97, moving toward the highest numbered track on the disk, track 199. The disk request queue contains read/write requests for blocks on tracks 84, 155, 103, 96, and 197, respectively.
How many tracks must the head step across using
a FCFS strategy?

31. How many tracks must the head step across using a FCFS strategy?
Suppose that the read/write head is at track is at track 97, moving toward the highest numbered track on the disk, track 199. The disk request queue contains read/write requests for blocks on tracks 84, 155, 103, 96, and 197, respectively.
How many tracks must the head step across using
a FCFS strategy?
Track
97 to steps
84 to steps
155 to steps
103 to steps
96 to steps
244 steps
199
150
100 50
Steps
100
200

32. Suppose that the read/write head is at track is at track 97, moving toward the highest numbered track on the disk, track 199. The disk request queue contains read/write requests for blocks on tracks 84, 155, 103, 96, and 197, respectively.
How many tracks must the head step across using
an elevator strategy?

33. How many tracks must the head step across using an elevator strategy?
Suppose that the read/write head is at track is at track 97, moving toward the highest numbered track on the disk, track 199. The disk request queue contains read/write requests for blocks on tracks 84, 155, 103, 96, and 197, respectively.
How many tracks must the head step across using
an elevator strategy?
Track
97 to steps
103 to steps
155 to steps
197 to steps
199 to steps
96 to steps
217steps
199
150
100 50
Steps
100
200

34. In our class discussion on directories it was suggested
that directory entries are stored as a linear list. What
is the big disadvantage of storing directory entries
this way, and how could you address this problem?
Consider what happens when look up a file …
The directory must be searched in a linear way.

35. Which file allocation scheme discussed in class gives the
best performance? What are some of the concerns with
this approach?
Contiguous allocation schemes gives the best performance.
Two big problems are:
* Finding space for a new file (it must all fit in contiguous blocks)
* Allocating space when we don’t know how big the file will be,
or handling files that grow over time.

36. What is the difference between internal and external fragmentation?
Internal fragmentation occurs when only a portion of a
file block is used by a file.
External fragmentation occurs when the free space on a disk
does not contain enough space to hold a file.

37. Linked allocation of disk blocks solves many of the
problems of contiguous allocation, but it does not
work very well for random access files. Why not?
To access a random block on disk, you must walk
Through the entire list up to the block you need.

38. Linked allocation of disk blocks has a reliability
problem. What is it?
If a link breaks for any reason, the disk blocks after
The broken link are inaccessible.