Transcript Slide Set 10

operating
systems
Disk Management
1
operating
systems
Goals of I/O Software
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.
2
operating
systems
I/O – Processor Overlap
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.
3
read(device, %D”, x);
y = f(x);
…
operating
systems
User Process
READ
Device Independent Layer
Without Blocking!
The read completes
is
has
issued.
not
and
Completed
the value
… but
of xthe
is
updated.
process continues to
execute.
CPU
Device Dependent Layer
Interrupt Handler
data
status
command
Device Controller
12
x
5
4
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
5
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
6
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.
7
Are you done yet?
8
operating
systems
Performance
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.
9
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.
process
Time compute
Request an
I/O operation
I/O
controller
Time compute
Time device
10
Polling Case
Time overhead
operating
systems
Time device
Time overhead
Time device
Time compute
Time compute
Proc2 polls
Proc 2
Time compute
Proc1 polls
Proc 1
Time compute
In the polling case, the process starts the I/O operation,
and then continually loops, asking the device if it is done.
11
Interrupt Case
operating
systems
Time
Time overhead
interrupt
handler
Time compute
Time device
Proc 1
Proc 2
Time compute
Time compute
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
12
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?
13
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?
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?
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.
15
Buffering Issues
operating
systems
User
space
Kernel
Read from the disk into kernel buffer.
When the buffer is full, transfer to
memory in user space.
16
Buffering Issues
operating
systems
User
space
Kernel
We can now swap the user processor out
While the I/O completes.
What problems Exist in this situation?
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.
17
operating
systems
Buffering Issues
User
space
Kernel
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.
18
operating
systems
Networking may involve
many copies
19
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.
20
seek time is the time
required for the read head to
move to the track containing
the data to be read.
21
rotational delay or
latency, is the time
required for the sector
to move under the
read head.
22
operating
systems
Performance Parameters
seek
Wait for device
Wait for
Channel
rotational
delay
(latency)
data
transfer
Device busy
Seek time is the time required to
move
the disk
arm
to the
specified
track
Transfer
Time
Rotational
delay
is the
time
required
for the
data on
that track to come
~ underneath the read heads. For
T
=
#/ tracks
* disk
constant
+ startup
time)
( rotation_speed
* bytes_on_track
s
a hard T
drive
rotating
at 3600
rpm,
the
average
rotational
t = bytes
delay will be 8.3ms.
23
operating
systems
Data Organization vs.
Performance
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
rotational delay
read 32 sectors
20 ms
8.3 ms
16.7 ms
45ms
Assuming that there is essentially no seek time on the remaining tracks,
each successive track can be read in 8.3 + 16.7 ms = 25ms.
Total read time = 45ms + 7 * 25ms = 220ms = 0.22 seconds
24
operating
systems
If the data is randomly distributed across the disk:
For each sector we have
average seek time
rotational delay
read 1 sector
20 ms
8.3 ms
0.5 ms
28.8 ms
Total time = 256 sectors * 28.8 ms/sector = 7.73 seconds
Random placement of data can be a problem when multiple
processes are accessing the same disk.
25
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.
26
Queue
operating
systems
Request
Request
Request
Request
If requests are scheduled in random order,
then we would expect the disk tracks to be
visited in a random order.
…
27
Queue
operating
systems
Request
Request
Request
Request
…
First-come, First-served
Scheduling
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.
28
While at track 15, assume some random set
of read requests -- tracks 4, 40, 11, 35, 7 and 14
operating
systems
Track
40
30
Head Path
Tracks Traveled
15 to 4
4 to 40
40 to 11
11 to 35
35 to 7
7 to 14
11 steps
36 steps
29 steps
24 steps
28 steps
7 steps
135 steps
20
10
50
100
Steps
29
Queue
operating
systems
Shortest Seek Time First
Request
Request
Request
Request
…
Always select the request that requires
the shortest seek time from the current
position.
30
operating
systems
While at track 15, assume some random set of read
requests -- tracks 4, 40, 11, 35, 7 and 14
Shortest Seek Time First
Track
Head Path
Tracks Traveled
40
30
20
10
50
Problem?
100
Steps
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”.
31
Queue
operating
systems
Request
The elevator algorithm
(scan-look)
Request
Request
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.
32
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
50
100
Steps
33
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?
34
operating
systems
Disk Cache
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.
35
operating
systems
Replacement Strategies
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
36
RAID
Redundant Array of Independent Disks
• Push Performance
• Add reliability
37
operating
systems strip 0
RAID Level 0: Striping
strip 1
Physical
Drive 1
strip 2
strip 3
strip 4
strip 5
strip 6
strip 7
strip 8
Physical
Drive 2
strip 0
strip 1
strip 2
strip 3
strip 4
strip 5
strip 6
strip 7
ooo
ooo
A Stripe
strip 9
strip 10
strip 11
Disk Management
Software
ooo
Logical Disk
38
RAID Level 1: Mirroring
High Reliability
operating
systems strip 0
strip 1
strip 2
Physical
Drive 1
strip 3
strip 0
strip 1
strip 0
strip 1
strip 4
strip 2
strip 3
strip 2
strip 3
strip 5
strip 4
strip 5
strip 4
strip 5
strip 6
strip 6
strip 7
strip 6
strip 7
strip 7
ooo
ooo
ooo
ooo
Physical
Drive 2
Physical
Drive 3
Physical
Drive 4
strip 8
strip 9
strip 10
strip 11
Disk Management
Software
ooo
Logical Disk
39
RAID Level 3: Parity
High Throughput
operating
systems strip 0
strip 1
strip 2
Physical
Drive 1
strip 3
strip 0
strip 1
strip 2
0
strip
para 1
strip 4
strip 3
2
strip 4
3
strip 5
2
strip
parb3
strip 5
strip 6
4
strip 7
5
strip 8
4
strip
parc 5
strip 6
strip 9
6
strip
strip 10
7
strip
strip 11
6
strip
pard7
strip 7
ooo
ooo
ooo
ooo
strip 8
Physical
Drive 2
Physical
Drive 3
Physical
Drive 4
parity
strip 9
strip 10
strip 11
Disk Management
Software
ooo
Logical Disk
40
Thinking About What You
Have Learned
41
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
1
2
3
Time compute
10
30
15
Time device
50
10
35
42
Process
1
2
3
Time compute
10
30
15
Time device
50
10
35
p3
p2
P1
0
10
20
30
40
50
60
70
80
90
100 110 120 130
105
43
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.
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?
Process
A
B
Device
Controller
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.
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.
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?
Process
A
B
Device
Controller
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.
45
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?
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?
Track
97 to 84
84 to 155
155 to 103
103 to 96
96 to 197
199
150
13 steps
71 steps
52 steps
7 steps
101 steps
244 steps
100
50
100
200
Steps
47
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?
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?
Track
97 to 103
103 to 155
155 to 197
197 to 199
199 to 96
96 to 84
199
150
6 steps
52 steps
42 steps
2 steps
103 steps
12 steps
217steps
100
50
100
200
Steps
49
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.
50
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.
51
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.
52
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.
53
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.
54