1. operating
systems
Disk Management

2. Goals of I/O Software
operating
systems
Provide a common, abstract view of all devices to the application programmer (open, read, write)
Provide as much overlap as possible between the operation of I/O devices and the CPU.

3. I/O – Processor Overlap
operating
systems
Application programmers expect
serial execution semantics
read (device, “%d”, x);
y = f(x);
We expect that this statement will complete
before the assignment is executed.
To accomplish this, the OS blocks the process
until the I/O operation completes.

4. Without Blocking! operating systems The read is issued.
read(device, %D”, x);
y = f(x); …
operating
systems
The read is issued.
The read completes
and the value of x is
updated.
The read has not
Completed … but the
process continues to
execute.
READ
User Process
Device Independent Layer
CPU
Device Dependent Layer
Interrupt Handler
data status command
Device Controller x 5 5 12 12

5. In a multi-programming environment, another
operating
systems
In a multi-programming environment, another
application could use the cpu while the first
application waits for the I/O to complete.
Request I/O
operation
I/O
Complete
app2
done
app1
app2
I/O
controller

6. Performance Thread execution time can be broken into:
operating
systems
Performance
Thread execution time can be broken into:
Time compute The time the thread spends doing computations
Time device The time spent on I/O operations
Time overhead The time spent determining if I/O is complete
So, Time total = Time compute + Time device + Time overhead

7. Performance Time total = Time compute + Time device + Time overhead
operating
systems
Performance
When the device driver polls
Time total = Time compute + Time device + Time overhead
Time overhead = The period of time between the point where the device
completes the operation and the point where the polling
loop determines that the operation is complete. This is
generally just a few instruction times.
Note that when the device driver polls, no other
process can use the cpu. Polling consumes the cpu.

8. Are you done yet?

9. Performance Time total = Time compute + Time device + Time overhead
operating
systems
When the device driver uses interrupts
Time total = Time compute + Time device + Time overhead
When the device driver uses interrupts
Time overhead = Time handler + Time ready
Time handler is the time spent in the interrupt handler
Time ready is the time the process waits for the cpu
after it has completed its I/O, while another process
uses the CPU.

10. For simplicity’s sake assume processes of the
operating
systems
For simplicity’s sake assume processes of the
following form: Each process computes for a
long while and then writes its results to a file.
We will ignore the time taken to do a context switch.
Request an
I/O operation
Time compute
Time compute
process
Time device
I/O
controller

11. Polling Case operating systems Proc 1 Proc 2 Time overhead
Time device
Time device
Time compute
Time compute
Time compute
Proc 1
Time compute
Proc 2
Proc1 polls
Proc2 polls
In the polling case, the process starts the I/O operation,
and then continually loops, asking the device if it is done.

12. Interrupt Case operating systems Proc 1 Proc 2 Time overhead
Time compute
Time interrupt
handler
Time device
Time compute
Proc 1
Time compute
Proc 2
In the interrupt case, the process starts the I/O
operation, and then blocks. When the I/O is done,
the os will get an interrupt.
Time device

13. Which gives better system throughput?
* Polling
* Interrupts
Which gives better application performance?
* Polling
* Interrupts
If you were developing an operating system,
would you choose interrupts or polling?

14. Buffering Issues operating systems
User
space
Kernel
Read from the disk
Into user memory
Assume that you are using interrupts…
What problems Exist in this situation?

15. Buffering Issues operating systems
User
space
Kernel
Read from the disk
Into user memory
Assume that you are using interrupts…
What problems Exist in this situation?
The process cannot be completely swapped out of memory. At least the page containing the addresses into which the data is being written must remain in real memory.

16. Buffering Issues operating systems User space
Read from the disk into kernel buffer.
When the buffer is full, transfer to memory in user space.
Kernel

17. What problems Exist in this situation?
Buffering Issues
operating
systems
User
space
We can now swap the user processor out
While the I/O completes.
What problems Exist in this situation?
Kernel
1. The O/S has to carefully keep track of the assignment of system buffers to user processes.
2. There is a performance issue when the user process is not in memory and the O/S is ready to transfer its data to the user process. Also, the device must wait while data is being transferred.
3. The swapping logic is complicated when the swapping operation uses the same disk drive for paging that the data is being read from.

18. Buffering Issues operating systems User space
Some of the performance issues can be addressed by double buffering. While one buffer is being transferred to the user process, the device is reading data into a second buffer.
Kernel

19. Networking may involve many copies
operating
systems

20. operating
systems
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.

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

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

23. Performance Parameters
operating
systems
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 )

24. Data Organization vs. Performance
operating
systems
Consider a file where the data is stored as compactly as
possible, in this case 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

25. If the data is randomly distributed across the disk:
operating
systems
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
Random placement of data can be a problem when multiple
processes are accessing the same disk.

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

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

28. First-come, First-served
Queue
First-come, First-served
Scheduling
operating
systems
Request …
If there are few processes competing
for the drive, 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.

29. While at track 15, assume some random set
of read requests -- tracks 4, 40, 11, 35, 7 and 14
operating
systems
Head Path
Tracks Traveled
Track
15 to steps
4 to steps
40 to steps
11 to steps
35 to steps
7 to steps
135 steps 40 30 20 10
Steps 50
100

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

31. Shortest Seek Time First
While at track 15, assume some random set of read
requests -- tracks 4, 40, 11, 35, 7 and 14
operating
systems
Shortest Seek Time First
Track
Head Path
Tracks Traveled 40 30 20 10
Steps 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?

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

33. Scan-Look operating systems
While at track 15, assume some random set of read requests
Track 4, 40, 11, 35, 7 and 14. Head is moving towards higher
numbered tracks.
Scan-Look
Track
Head Path
Tracks Traveled 40 30 20 10
Steps 50
100

34. 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?

35. Disk Cache The disk cache holds a number of disk sectors in memory.
operating
systems
The disk cache holds a number of disk sectors in memory.
When an I/O request is made for a particular sector,
the disk cache is checked. If the sector is in the cache,
it is read. Otherwise, the sector is read into the cache.

36. Replacement Strategies
operating
systems
Least Recently Used
replace the sector that has been in the cache the
longest, without being referenced.
Least Frequently Used
replace the sector that has been used the least

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

38. RAID Level 0: Striping operating systems Physical Physical Drive 1
A Stripe
strip 4
strip 2
strip 3
strip 5
strip 4
strip 5
strip 6
strip 6
strip 7
strip 7
o o o
o o o
strip 8
strip 9
strip 10
Disk Management
Software
strip 11
o o o
Logical Disk

39. RAID Level 1: Mirroring operating systems High Reliability 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
strip 11
o o o
Logical Disk

40. RAID Level 3: Parity operating systems High Throughput 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 2
strip 1
para
strip 4
strip 3
strip 2
strip 3
strip 4
strip 5
strip 2
strip 3
parb
strip 5
strip 6
strip 4
strip 7
strip 5
strip 4
strip 8
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

41. Thinking About What You Have Learned

42. Process Time compute Time device 1 10 50 2 30 10 3 15 35
operating
systems
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

43. 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

44. 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.

45. 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.

46. 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?

47. 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

48. 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?

49. 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

50. 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 problems?
Consider what happens when look up a file …
The directory must be searched in a linear way.

51. 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.

52. 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.

53. 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.

54. 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.