Input Output Management in Operating System

1. I/O MANAGEMENT
Output
Input

2. I/O Devices
 I/O devices are roughly categorized into three categories:
2
I/O Devices
Human
Readable
Machine
Readable
Communicati
on
Suitable for
communicating
with computer
user
Suitable for
communicating
with electronic
equipment
Suitable for
communicating
with remote
devices

3. I/O Devices
 There are great differences across classes of I/O devices. Key
differences are:
 Data Rate:
 There may be differences of several orders of magnitude between the
data transfers.
 Application:
 The use to which a device is put has an influence on the software and
policies in the OS and supporting utilities.
 Complexity of control:
 E.g. a printer requires a relatively simple control interface while a disk is
much more complex.
 Unit of transfer:
 Data may be transferred as a stream of bytes or characters or in a large
block.
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4. I/O Devices
 There are great differences across classes of I/O devices. Key
differences are:
 Data Representation:
 Different data encoding schemes are used by different devices,
including differences in character code and parity conventions.
 Error conditions:
 The nature of errors, the way in which they are reported, their
consequences and the available range of responses differ widely from
one device to another
 This diversity makes it difficult to achieve a uniform and
consistent approach to I/O for both OS as well as for user
processes.
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5. Organization of the I/O Function
 There are three approaches for performing I/O:
5
I/O
Programm
ed I/O
Interrupt
Driven I/O
Direct
Memory
Access
(DMA)

6. Organization of the I/O Function
 Programmed I/O
 The processor issues an I/O command, on behalf of a process, to an I/O
module.
 The process then busy waits for the operation to be completed before
proceeding.
 Interrupt driven I/O
 The processor issues an I/O command, on behalf of a process.
 If the I/O instruction is non-blocking, the processor continues to execute
instructions from the process that issued the I/O command.
OR
 If the I/O instruction is blocking, then the next instruction that processor
executes is from OS, which will put current process in a blocked state and
schedule another process.
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7. Organization of the I/O Function
 Direct Memory Access (DMA)
 Controls the exchange of data between main memory and an I/O module.
 The processor sends a request for the transfer of a block of data to the
DMA module and is interrupted only after the entire block has been
transferred.
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No Interrupts Use of Interrupts
I/O-to-Memory
transfer through
Processor
Programmed I/O Interrupt driven I/O
Direct I/O-to-Memory
transfer
Direct Memory Access
I/O Techniques

8. Organization of the I/O Function
 Direct Memory Access (DMA)
 The DMA is capable of
mimicking the processor and,
of taking over control of system
bus just like a processor.
 It transfers data to and from
memory over the system bus.
 When the processor wishes to
read or write a block of data, it
issues a command to a DMA
module
8
Data
count
Data
register
Address
register
Control
Logic
Data lines
Address
lines
Request to
DMA
Acknowledgement from
DMA
Interrupt
Read
Write
DMA Block Diagram

9. Organization of the I/O Function
 Direct Memory Access (DMA)
 When the processor wishes to read or write a block of data, it
issues a command to a DMA module by sending following
information:
 Whether a read or write is requested, using the read or write control
line between the processor and the DMA module.
 The address of the I/O device involved, communicated on the data line.
 The starting location in memory to read from or write to,
communicated on data lines and stored by the DMA module in its
address register.
 The number of words to be read or written, again communicated via
the data lines and stored in the data count register.
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10. Organization of the I/O Function
 Direct Memory Access (DMA)
 The processor delegates I/O operation to the DMA module and
continues with other work.
 DMA module transfers the entire block of data, one word at a time,
directly to or from memory, without going to a processor.
 When the transfer is complete, the DMA module sends an interrupt
signal to the processor.
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11. Disk Organization
 Magnetic disks provide bulk of secondary storage for modern
computer systems.
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12. Disk Organization
 The two surfaces of a platter are covered with a magnetic
material.
 Information is recorded magnetically on the platters.
 A read -write head flies just above each surface of every platter.
 The heads are attached to a disk arm that moves all the heads as a
unit.
 The surface of a platter is logically divided into circular tracks,
which are subdivided into sectors.
 The set of tracks that are at one arm position makes up a cylinder.
 There may be thousands of concentric cylinders in a disk drive,
and each track may contain hundreds of sectors.
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13. Disk Organization
 Disk speed has two parts.
 The transfer rate is the rate at which data flow between the drive
and the computer.
 The positioning time, sometimes called the random-access time,
consists of the time necessary to move the disk arm to the desired
cylinder, called the seek time and the time necessary for the
desired sector to rotate to the disk head, called the rotational
latency.
 The disk head flies on an extremely thin cushion of air. Although
the disk platters are coated with a thin protective layer, the head
will sometimes damage the magnetic surface. This accident is
called a head crash.
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14. Disk Organization
 A disk drive is attached to a computer by a set of wires called an
I/O bus.
 Several kinds of buses are available
 ATA (Advanced Technology Attachment
 SATA (Serial ATA)
 eSATA
 USB (Universal Serial Bus)
 FC (Fibre Channel)
 The data transfers on a bus are carried out by special electronic
processors called controller.
 Host controller is at the computer end of the bus
 Disk controller is built into each disk drive
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15. 15

16. Magnetic Tape
 Magnetic tape was used as an early secondary-storage medium.
 Information is accessed sequentially.
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17. Magnetic Tape
 It can hold large quantities of data, its access time is slow
compared with that of main memory and magnetic disk.
 Random access to magnetic tape is about a thousand times slower
than random access to magnetic disk
 Tapes are used mainly for backup, for storage of infrequently
used information, and as a medium for transferring information
from one system to another.
 A tape is kept in a spool and is wound or rewound past a read-
write head.
 Tapes can store from 20GB to 200GB.
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18. Disk Scheduling
 The operating system need to use the hardware efficiently.
 Disk drive is expected to have fast access time and large disk
bandwidth.
 Seek time:
 The time for the disk arm to move the heads to the cylinder containing the
desired sector.
 Rotational Latency:
 The additional time for the disk to rotate the desired sector to the disk head.
 Disk Bandwidth:
 The total number of bytes transferred, divided by the total time between the
first request for service and the completion of the last transfer.
 We can improve both the access time and the bandwidth by
managing the order in which disk I/O requests are serviced.
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19. Disk Scheduling
 Whenever a process needs I/O to or from the disk, it issues a
system call to the operating system.
 The request specifies several pieces of information:
 Whether this operation is input or output
 What the disk address for the transfer is
 What the memory address for the transfer is
 What the number of sectors to be transferred is
 If the desired disk drive and controller are available, the request
can be serviced immediately.
 If the drive or controller is busy, any new requests for service will
be placed in the queue of pending requests for that drive.
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20. Disk Scheduling
 FCFS Disk Scheduling
 The simplest form of disk scheduling based in first-come, first-
served (FCFS) policy.
 The algorithm is intrinsically fair, but it generally does not
provide the fastest service.
 E.g., a disk queue with requests for I/0 to blocks on cylinders in
the following order
98, 183, 37, 122, 14, 124, 65, 67
suppose, currently head starts at 53
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21. Disk Scheduling
 FCFS Disk Scheduling
Queue = 98, 183, 37, 122, 14, 124, 65, 67 head
starts at 53
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0 14 37 53 65 67 98 122 124 183
199

22. Disk Scheduling
 FCFS Disk Scheduling
Queue = 98, 183, 37, 122, 14, 124, 65, 67 head starts at 53
22
0 14 37 53 65 67 98 122 124 183
199
45 85
14
6
85
10
8
11
0 59
2
A total head movement
of
640 cylinders.

23. Disk Scheduling
 SSTF Disk Scheduling
 Shortest Seek Time First (SSTF)
 It selects the request with the least seek time from the current head
position.
 Since seek time increases with the number of cylinders traversed by the
head, SSTF chooses the pending request closest to the current head
position.
 Hence, the seek time of every request is calculated in advance in the
queue and scheduled according to their seek time.
 E.g., a disk queue with requests for I/0 to blocks on cylinders in the
following order
98, 183, 37, 122, 14, 124, 65, 67
suppose, currently head starts at 53
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24. Disk Scheduling
 SSTF Disk Scheduling
Queue = 98, 183, 37, 122, 14, 124, 65, 67 head starts at 53
24
0 14 37 53 65 67 98 122 124 183
199
1
2
2
3
0
2
3
8
4
2
4
2
5
9
A total head movement
of
236 cylinders.

25. Disk Scheduling
 SSTF Disk Scheduling
 Gives a substantial improvement in performance in comparison with
FCS.
 SSTF scheduling is essentially a form of shortest-job-first (SJF)
scheduling; and like SJF scheduling, it may cause starvation of some
requests.
 While servicing a current request, if new request arrives which is
nearest to current one, new request will be processed after this.
Again if new request comes which is closer to the previous request,
that incoming request will be served. If this goes on continue… the
initial request which are far away will be starved.
 Although the SSTF algorithm is a substantial improvement over the
FCFS algorithm, it is not optimal.
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26. Disk Scheduling
 SCAN Disk Scheduling
 The disk arm starts at one end of the disk moves towards the other
end.
 It services the requests as it reaches each cylinder, until it gets to the
other end of the disk.
 At the other end, the direction of head movement is reversed, and
servicing continues.
 The head continuously scans back and forth across the disk.
 The SCAN algorithm is sometimes called the elevator algorithm, since
the disk arm behaves just like an elevator in a building.
 First it services all the requests going up and then reversing to service
requests the other way.
 E.g., a disk queue with requests for I/0 to blocks on cylinders in the
following order
98, 183, 37, 122, 14, 124, 65, 67
suppose, currently head starts at 53 and the disk arm is moving 26

27. Disk Scheduling
 SCAN Disk Scheduling
Queue = 98, 183, 37, 122, 14, 124, 65, 67
Head starts at 53, the disk arm is moving toward 0.
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0 14 37 53 65 67 98 122 124 183
199
1
6
2
3
1
4
2
6
5
3
1
2
4
2 5
9
A total head movement
of
236 cylinders.

28. Disk Scheduling
 SCAN Disk Scheduling
 If a request arrives in the queue just in front of the head, it will be
serviced almost immediately;
 A request arriving just behind the head will have to wait until the arm
moves to the end of the disk, reverses direction, and comes back.
 Assuming a uniform distribution of requests for cylinders, consider the
density of requests when the head reaches one end and reverses
direction.
 At this point, relatively few requests are immediately in front of the
head, since these cylinders have recently been serviced.
 The heaviest density of requests is at the other end of the disk. These
requests have also waited the longest so why not go there first?
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29. Disk Scheduling
 C-SCAN Disk Scheduling
 Circular SCAN is a variant of SCAN scheduling designed to provide a
more uniform wait time.
 Like SCAN, C-SCAN moves the head from one end of the disk to the
other, servicing requests along the way.
 When the head reaches the other end, however, it immediately returns
to the beginning of the disk without servicing any requests on the
return trip.
 The C-SCAN scheduling algorithm essentially treats the cylinders as a
circular list that wraps around from the final cylinder to the first one.
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30. Disk Scheduling
 C-SCAN Disk Scheduling
Queue = 98, 183, 37, 122, 14, 124, 65, 67 head starts at 53
30
0 14 37 53 65 67 98 122 124 183
199
2
2
2
3
1
2
4
2
5
9 1
6
199
14
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31. Disk Scheduling
 LOOK and C-LOOK Disk Scheduling
 Both SCAN and C-SCAN move the disk arm across the full width of the disk.
 In practice, neither algorithm is often implemented this way.
 More common, the arm goes only as far as the final request in each direction.
 Then, it reverses direction immediately without going all the way to the end of
the disk
 Versions of SCAN and C-SCAN that follow this pattern are called LOOK and C-
LOOK, because they look for a request before continuing to move in a given
direction.
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32. Disk Scheduling
 LOOK Disk Scheduling
Queue = 98, 183, 37, 122, 14, 124, 65, 67
Head starts at 53, the disk arm is moving toward 0.
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0 14 37 53 65 67 98 122 124 183
199
1
6
2
5
1
3
1
2
4
2 5
9
2
3

33. Disk Scheduling
 C-LOOK Disk Scheduling
Queue = 98, 183, 37, 122, 14, 124, 65, 67
Head starts at 53, the disk arm is moving toward 199.
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0 14 37 53 65 67 98 122 124 183
199
1
2
2
3
1
2
4
2
5
9
169
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34. Disk Scheduling
 How to select a disk scheduling algorithm?
 With any scheduling algorithm, the performance depends heavily on the
number and types of requests.
 Requests for disk service can be greatly influenced by the file-allocation
method.
 A program reading a contiguously allocated file will generate several requests that
are close together on the disk, resulting in limited head movement.
 A linked or indexed file in contrast may include blocks that are widely scattered on
the disk resulting in greater head movement.
 The location of directories and index blocks is also important.
 Because of these complexities, the disk-scheduling algorithm should be
written as a separate module of the operating system, so that it can be
replaced with a different algorithm if necessary.
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35. Disk Scheduling
 Disk Access Time
𝐷𝑖𝑠𝑘 𝐴𝑐𝑐𝑒𝑠𝑠 𝑇𝑖𝑚𝑒 = 𝑆𝑒𝑒𝑘 𝑇𝑖𝑚𝑒 + 𝑅𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝐿𝑎𝑡𝑒𝑛𝑐𝑦 + 𝑇𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑇𝑖𝑚𝑒
 Disk Response Time
 Average of time spent by a request waiting to perform it’s I/O operation.
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Disk
Queuing
Delay
Seek Time
Rotational
Latency
Transfer
Time
Disk Access
Time
Disk Response Time

36. Disk Scheduling Example
Suppose a disk has 201 cylinders, numbered from 0 to 200. At
some time, the disk arm is at cylinder 100, and there is a queue of
disk access requests for cylinders 30, 85, 90, 100, 105, 110, 135
and 145. If Shortest-Seek Time First (SSTF) is being used for
scheduling the disk access, the request for cylinder 90 is serviced
after servicing ____________ number of requests.
1. 1
2. 2
3. 3
4. 4
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37. Disk Scheduling Example
Suppose the following disk request sequence (track numbers) for a
disk with 100 tracks is given: 45, 20, 90, 10, 50, 60, 80, 25, 70.
Assume that the initial position of the R/W head is on track 50. The
additional distance that will be traversed by the R/W head when the
Shortest Seek Time First (SSTF) algorithm is used compared to the
SCAN (Elevator) algorithm (assuming that SCAN algorithm moves
towards 100 when it starts execution) is _________ tracks.
1. 8
2. 9
3. 10
4. 11
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38. Disk Scheduling Example
Data requests are received by a disk drive for cylinder 5, 25, 18, 3,
39, 8, 35 in that order. A seek takes 5msec per cylinder moved.
How much seek time is needed to serve these requests for a SSTF
algorithm? Assume that the arm at cylinder 20 when the last of
these request is made with none of the requests yet served.
1. 125
2. 295
3. 575
4. 750
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39. Disk Scheduling Example
Consider a disk pack with the following specifications- 16 surfaces,
128 tracks per surface, 256 sectors per track and 512 bytes per
sector.
1. What is the capacity of disk pack?
2. What is the number of bits required to address the sector?
3. If the disk is rotating at 3600 RPM, what is the data transfer rate?
4. If the disk system has rotational speed of 3000 RPM, what is the
average access time with a seek time of 11.5 msec?
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40. 40

41. Disk Scheduling Example
Consider a disk pack with the following specifications- 16 surfaces,
128 tracks per surface, 256 sectors per track and 512 bytes per
sector.
1. What is the capacity of disk pack?
256 MB
2. What is the number of bits required to address the sector?
19 bits
3. If the disk is rotating at 3600 RPM, what is the data transfer rate?
120 MBps
4. If the disk system has rotational speed of 3000 RPM, what is the
average access time with a seek time of 11.5 msec?
21.5 msec
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