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Disk SchedulingDisk Scheduling
Because Disk I/O is so important, it is worth our time toInvestigate some of the issues involved in disk I/O.
One of the biggest issues is disk performance.
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seek time is the timerequired for the read head tomove to the track containingthe data to be read.
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rotational delay orlatency, is the time required for the sectorto move under the read head.
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Performance ParametersPerformance Parameters
Wait for device
Wait forChannel
Device busy
seek
rotationaldelay data
transfer
Seek time is the time required tomove the disk arm to the specified track
Ts = # tracks * disk constant + startup time~
Rotational delay is the time required for the data onthat track to come underneath the read heads. Fora hard drive rotating at 3600 rpm, the average rotationaldelay will be 8.3ms.
Transfer Time
Tt = bytes / ( rotation_speed * bytes_on_track )
(latency)
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Data Organization vs. Performance
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 20 ms rotational delay 8.3 ms read 32 sectors 16.7 ms
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
45ms
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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 0.5 ms
Total time = 256 sectors * 28.8 ms/sector = 7.73 seconds
28.8 ms
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In the previous example, the biggest factoron performance is ?
Seek time!
To improve performance, weneed to reduce the average seek time.
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Queue
Request
Request
Request
Request
…
The operating system keeps a queue of requests to read/write the disk.
In these exercises we assume that allof the requests are on the queue.
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Queue
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.
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Queue
Request
Request
Request
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.
First-come, First-served Scheduling
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Track
10
20
30
40
15 to 4 114 to 40 3640 to 11 2911 to 35 2435 to 7 287 to 16 9
137 tracks
While at track 15, assume some random set of read requests -- tracks 4, 40, 11, 35, 7 and 16
Head Path Tracks Traveled
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Queue
Request
Request
Request
Request
…
Shortest Seek Time First
Always select the request that requires the shortest seek time from the current position.
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Track
10
20
30
40
While at track 15, assume some random set of read requests -- tracks 4, 40, 11, 35, 7 and 16
Shortest Seek Time First
Problem?
In a heavily loaded system, incoming requests with ashorter seek time will constantly push requests with long seek times to the end of the queue. This resultsIn what is called “Starvation”.
Head Path Tracks Traveled
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Track
10
20
30
40
50 100
While at track 15, assume some random set of read requests -- tracks 4, 40, 11, 35, 7 and 16
Shortest Seek Time First
Problem?
In a heavily loaded system, incoming requests with ashorter seek time will constantly push requests with long seek times to the end of the queue. This resultsIn what is called “Starvation”.
Head Path Tracks Traveled
15 – 16 116 – 11 511 – 7 47 – 4 34 – 35 3135 – 40 5
49 tracks
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Queue
Request
Request
Request
Request
…
The elevator algorithm (scan-look)
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 theway to the first (or last) track before it changesdirection.
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Track
10
20
30
40
Steps50 100
While at track 15, assume some random set of read requestsTrack 4, 40, 11, 35, 7 and 16. Head is moving towards highernumbered tracks.
Scan-Look
Head Path Tracks Traveled
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Track
10
20
30
40
Steps50 100
While at track 15, assume some random set of read requestsTrack 4, 40, 11, 35, 7 and 16. Head is moving towards highernumbered tracks.
Scan-Look
Head Path Tracks Traveled
15 – 16 116 – 35 1935 – 40 540 – 11 2911 – 7 47 – 4 3
61 tracks
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Which algorithm would you choose if you wereimplementing 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?
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Disk CacheDisk 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 somecontiguous blocks) are read into the cache.
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Replacement StrategiesReplacement 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
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RAIDRedundant Array of Independent Disks
• Push Performance
• Add reliability
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RAID Level 0: Striping
Logical Disk
strip 0
strip 1
strip 2
strip 3
strip 4
strip 5
strip 6
strip 7
strip 8
strip 9
strip 10
strip 11
o o o
strip 0
strip 2
strip 4
strip 6
o o o
strip 1
strip 3
strip 5
strip 7
o o o
PhysicalDrive 1
PhysicalDrive 2
Disk ManagementSoftware
A Stripe
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RAID Level 1: Mirroring
Logical Disk
strip 0
strip 1
strip 2
strip 3
strip 4
strip 5
strip 6
strip 7
strip 8
strip 9
strip 10
strip 11
o o o
strip 0
strip 2
strip 4
strip 6
o o o
strip 1
strip 3
strip 5
strip 7
o o o
PhysicalDrive 1
PhysicalDrive 2
Disk ManagementSoftware
strip 0
strip 2
strip 4
strip 6
o o o
strip 1
strip 3
strip 5
strip 7
o o o
strip 0
strip 2
strip 4
strip 6
o o o
strip 1
strip 3
strip 5
strip 7
o o o
PhysicalDrive 3
PhysicalDrive 4
strip 0
strip 2
strip 4
strip 6
o o o
strip 1
strip 3
strip 5
strip 7
o o o
High Reliability
Duplicate writes to drive 1 and drive 2 on these disks
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RAID Level 3: Parity
Logical Disk
strip 0
strip 1
strip 2
strip 3
strip 4
strip 5
strip 6
strip 7
strip 8
strip 9
strip 10
strip 11
o o o
strip 0
strip 2
strip 4
strip 6
o o o
strip 1
strip 3
strip 5
strip 7
o o o
PhysicalDrive 1
PhysicalDrive 2
Disk ManagementSoftware
strip 0
strip 3
strip 6
strip 9
o o o
strip 1
strip 4
strip 7
strip 10
o o o
strip 0
strip 2
strip 4
strip 6
o o o
strip 1
strip 3
strip 5
strip 7
o o o
PhysicalDrive 3
PhysicalDrive 4
strip 2
strip 5
strip 8
strip 11
o o o
para
parb
parc
pard
o o o
High Throughput
parity
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Thinking about whatyou have learned
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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
1 10 50 2 30 10 3 15 35
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Process Time compute Time device
1 10 50 2 30 10 3 15 35
0 10 20 30 40 50 60 70 80 90 100 110 120 130
P1p2
p3
105
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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
DeviceController
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?
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 storedin buffer B. The process wakes up and reads the data, then tries to read Buffer A.Double buffering has not helped performance.
BA
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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
DeviceController
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?
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 isalready there. The process does nothave to wait and performance improves.
A B
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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 usinga FCFS strategy?
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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 usinga FCFS strategy?
Track
50
100
150
199
Steps100 200
97 to 84 13 steps84 to 155 71 steps155 to 103 52 steps103 to 96 7 steps96 to 197 101 steps
244 steps
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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 usingan elevator strategy?
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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 usingan elevator strategy?
Track
50
100
150
199
Steps100 200
97 to 103 6 steps103 to 155 52 steps155 to 197 42 steps197 to 199 2 steps199 to 96 103 steps96 to 84 12 steps
217steps
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In our class discussion on directories it was suggestedthat directory entries are stored as a linear list. Whatis the big disadvantage of storing directory entriesthis 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.
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Which file allocation scheme discussed in class gives thebest 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.
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What is the difference between internal andexternal fragmentation?
Internal fragmentation occurs when only a portion of afile block is used by a file.
External fragmentation occurs when the free space on a diskdoes not contain enough space to hold a file.
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Linked allocation of disk blocks solves many of theproblems of contiguous allocation, but it does notwork very well for random access files. Why not?
To access a random block on disk, you must walkThrough the entire list up to the block you need.
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Linked allocation of disk blocks has a reliabilityproblem. What is it?
If a link breaks for any reason, the disk blocks afterThe broken link are inaccessible.
