Outline
Review
EEC-682/782
Computer Networks I
Data Link Layer Design Issues
Error Detection and Correction
Data link layer
Elementary Data Link Protocols
Sliding Window Protocols
Lecture 6
Wenbing Zhao
w.zhao1@csuohio.edu
(Lecture nodes are based on materials supplied by
Dr. Louise Moser at UCSB and Prentice-Hall)
Spring Semester 2005
Data Link Layer Design Issues
How does DL layer form frames?
Point-to-point, source-to-destination, “wirelike”
Framing
Insert time gapes between frames
Physical bit stream divided up into frames
Too risky, no timing guarantees, not used
Error Control
Acks and nacks
Sender has timer –within timeout get ack, or send again
If frame transmitted multiple times, receiver may get several
copies
Use sequence numbers
Character count
Flag bytes with byte stuffing
Starting and ending flags, with bit stuffing
Physical layer coding violations
Applies only if encoding on medium contain some redundancy
Example: encode 1 bit with 2 bits
Flow Control
Sender may transmit frames faster than receiver can receive
them
Throttle sender so sends no faster than receiver can receive
them
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Framing
Services Provided to the Network Layer
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1 => 10
0 => 01
Can use 00 or 11 to delimit frames
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Error-Correcting Codes
Error-Correcting Codes
N-bit codeword – an n-bit unit containing data and
check bits (m bits of data, r bits redundant/check
bits)
Given any two codewords, it is possible to determine
how many corresponding bits differ, using exclusive
OR and counting number of 1 bits in the result
Hamming distance – number of bit positions in which
two codewords differ
In general, all 2m possible data messages are legal, but not all 2n
possible codewords are used
Given the algorithm for computing the check bits, it is possible
to
Construct a complete list of legal codewords
Find two codewords whose Hamming distance is minimum
This distance is the Hamming distance of the complete code
To detect d errors, need a distance d+1 code
No way that d single-bit errors can change a valid codeword into
another valid codeword
If two codewords are a Hamming distance d apart, it will
require d single-bit errors to convert one into the other
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To correct d errors, need a distance 2d+1 code
Legal codewords are so far part that even with d changes, orginal
codeword is still closer than any other codeword, so it can be
uniquely determined
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Hamming Code
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Hamming Code - Example
The bits of codeword are numbered consecutively, starting with bit 1
at the left end, bit 2 to its immediate right and so on
The bits that are powers of 2 (1,2,4,8,16,etc) are check bits
Data (character ‘H’): 1001000
Create the data word, leaving spaces for the parity bits:
The rest are filled up with the m data bits
Each parity bit calculates the parity for some of the bits in the code
word. The position of the parity bit determines the sequence of bits
that it alternately checks and skips.
__1_001_000
Calculate the parity for each parity bit
Position 1 checks bits 1,3,5,7,9,11:
Position 1: check 1 bit, skip 1 bit, check 1 bit, skip 1 bit, etc.
(1,3,5,7,9,11,13,15,...)
Position 2: check 2 bits, skip 2 bits, check 2 bits, skip 2 bits, etc.
(2,3,6,7,10,11,14,15,...)
Position 4: check 4 bits, skip 4 bits, check 4 bits, skip 4 bits, etc.
(4,5,6,7,12,13,14,15,20,21,22,23,...)
Position 8: check 8 bits, skip 8 bits, check 8 bits, skip 8 bits, etc. (8-15,2431,40-47,...)
etc.
? _ 1 _ 0 0 1 _ 0 0 0. Even number of 1’s, set position 1 to 0: 0 _ 1 _ 0 0 1 _ 0 0 0
Position 2 checks bits 2,3,6,7,10,11:
0 ? 1 _ 0 0 1 _ 0 0 0. Even number of 1’s, set position 2 to 0: 0 0 1 _ 0 0 1 _ 0 0 0
Position 4 checks bits 4,5,6,7:
0 0 1 ? 0 0 1 _ 0 0 0. Odd number of 1’s, set position 4 to 1: 0 0 1 1 0 0 1 _ 0 0 0
Position 8 checks bits 8,9,10,11,12:
0 0 1 1 0 0 1 ? 0 0 0. Even number of 1’s, set position 8 to 0: 0 0 1 1 0 0 1 0 0 0 0
Code word: 00110010000.
Set a parity bit to 1 if the total number of ones in the positions it
checks is odd. Set a parity bit to 0 if the total number of ones in the
positions it checks is even (assuming even parity is used)
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Hamming Code
Cyclic Redundant Code
Sender and receiver agree on generate polynomial
G(x), with high and low bits = 1
To compute checksum for some frame with m bits
corresponding to M(x)
When a codeword arrives,
Receiver initializes a counter to 0,
It then examines each check bit, k, to see if it has the right
parity.
M > deg G(x) = r
If not, adds k to the counter.
Append checksum to end of frame so polynomial T(x)
corresponding to checksummed frame is divisible by
G(x)
When receiver gets checksummed frame, divides T(x)
by G(x)
If remainder R(x) != 0, then transmission error
If counter is 0 after all check bits, the codeword is
accepted as valid.
If counter is nonzero, it contains the number of the
incorrect bit
e.g., if 1, 2, 8 in error, bit in position 11 is inverted
Can only correct single-bit error
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Algorithm to Compute CRC Checksum
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Elementary Data Link Protocols
Let m = deg M(x), r = deg G(x)
Append r 0 bits to lower-order end of frame to
obtain corresponding polynomial xrM(x)
Divide bit string corresponding to xrM(x) by bit
string corresponding to G(x)
Subtract remainder R(x) from bit string
corresponding to xrM(x)
An Unrestricted Simplex Protocol
A Simplex Stop-and-Wait Protocol
A Simplex Protocol for a Noisy Channel
Result is checksummed frame
Let T(x) be its polynomial
xrM(x) = Q(x)G(x) + R(x)
xrM(x) – R(x) = Q(x)G(x) = T(x)
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Elementary Data Link Protocols
Protocol Definitions
Initial assumptions
Physical, data link, network layers are independent processes
Sender has infinite amount of data ready to send, supplied by
network layer
“wire-like” service: reliable, source ordered delivery
Packet is treated as pure data
When sender accepts packet from NL, it encapsulates in a frame
with a header and trailer
Receiver waits for arrived of frame, which generates an interrupt
When frame arrives at receiver, hardware computes checksum
Continued Æ
If error then DLL informed event = chksum_err
Else DLL informed event = frame_arrival
DLL acquires frame from physical layer, checks control info in header
If ok then passes packet to NL
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Unrestricted
Simplex
Protocol
Protocol
Definitions
Additional
assumptions
Processing time
insignificant
Infinite buffer
space
Communication
channel never loses
or damages frames
Uses no sequence
numbers or acks
Only event type
frame_arrival
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Simplex
Stop-andWait
Protocol
Simplex Stop-and-Wait Protocol
Drop assumption
Receiver processing time insignificant, or, equivalently, infinite
input buffer at receiver
Main problem
To prevent sender from overwhelming the receiver
Sender sends
frame, stop and
wait for ack
Alternates
between sender
and receiver
If receiver takes t time units to execute from physical layer to
network layer, sender must not transmit more than one frame
per time t
One solution – too conservative
Restrict sender so transmits so slowly that even if frame
undergoes max delay no buffer overflows
Better solution
Half-duplex
Receiver provides feedback to sender and gives sender permission
to send next frame
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A Simplex Protocol for a Noisy Channel
A Simplex Protocol for a Noisy Channel
Drop assumption
Frames not damaged or lost
Further considerations
Assumption
Timeout must be long enough so sender does not send
duplicate when ack is on its way
Timeout must allow
If frame is damaged, receiver will detect it when it computes the
checksum
Possible solution
Receiver sends ack if receives uncorrupted frame, discards frame
if damaged
Sender sends frame again if doesn’t receive ack before timeout
Frame to get to receiver
Receiver to process frame
Ack to get to sender
Problem
Acks need to be matched against frames
Receiver receives uncorrupted frame, sends ack
Sender times out before receiving ack, sends frame again
Receiver receives send copy uncorrupted – has duplicate copies
Sender remembers next_frame_to_send
Receiver remembers frame_expected
Positive acknowledgement with retransmission (PAR), or,
Automatic Repeat reQuest (ARQ)
Solution
Use sequence numbers: 1 bit suffices
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A Simplex Protocol for a Noisy Channel
A Simplex Protocol for a Noisy Channel
A positive
acknowledgement with
retransmission
protocol.
A positive acknowledgement with retransmission protocol.
Continued Æ
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Sliding Window Protocols
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Sliding Window Protocols
Efficiency improvements
A One-Bit Sliding Window Protocol
A Protocol Using Go Back N
A Protocol Using Selective Repeat
Use same circuit for data in both direction: full-duplex
Interleave data and ack frames
Use “kind” field to distinguish them
B piggybacks its ack for A’s frame onto B’s next frame
Savings of header in separate ack frame
If B sends data frames infrequently, use timeout to
determine when B should send ack in separate ack frame
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Sliding Window Protocols
Sliding Window Protocols
Each frame contains sequence number in ranges
0..2n-1
Sending window – list of consecutive sequence
numbers of frames that sender is permitted to send
Receiving window – list of consecutive sequence
numbers of frames that receiver is permitted to
accept
When frame with (seq num = lower edge of window) arrives
Frame is passed to NL
Ack is generated
Window slid down by 1 (remains same size as was initially)
When new packet arrives from NL, it is given next highest
sequence number, and upper edge of window is incremented
When ack arrives from receiver, lower edge of window is
incremented
Within sending window, frame sent but not acked
Sender must keep those frames for possible retransmission
If max window size = w, need w buffers
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Sliding Window Protocols
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One-Bit Sliding Window Protocol
Sender transmits frame, waits for ack before
sending next frame
Next_frame_to_send – which frame sender is trying to send
Frame_expected – which frame receiver is expecting
Sender initially
Gets packet from NL
constructs frame
sends frame
A sliding window of size 1, with a 3-bit sequence number.
(a) Initially.
(b) After the first frame has been sent.
(c) After the first frame has been received.
(d) After the first acknowledgement has been received.
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One-Bit Sliding Window Protocol
One-Bit Sliding Window Protocol
Receiver on arrival of frame
Checks if it is duplicate
If expected frame
Passes frame to NL
Slides its window up
Sends ack back to sender indicating last uncorrupted frame it
received
Sender on receipt of ack
If ack agrees with seq num of frame it is trying to send
Gets next packet frame NL
Constructs frame
Sends frame
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One-Bit Sliding Window Protocol
Continued Æ
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One-Bit Sliding Window Protocol
No duplicate, no omissions, no deadlock, but inefficient
Two scenarios for protocol 4. (a) Normal case. (b) Abnormal case. The
notation is (seq, ack, packet number). An asterisk indicates where a
network layer accepts a packet.
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A Protocol Using Go Back n
Sliding Window Protocols
Drop assumption
Transmission time is negligible from sending til receiving ack
Problem: sender blocks till receives acks (does not
send another frame)
Solution: pipelining
Allow sender to send up to w frames before blocking
Problem: if frame in middle is lost or damaged, what
to do with the frames following it
Solution: two strategies
Go Back n - all the frames following it are discarded
Selective repeat – nack the lost/damaged frame and
retransmit that frame
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Pipelining and error recovery. Effect on an error when
(a) Receiver’s window size is 1.
(b) Receiver’s window size is large.
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Go Back n
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Go Back n
Drop assumption
NL has infinite supply of packets
Network_layer_ready event added
Max seq frames outstanding at any instant
Needs multiple timers, on per outstanding frame
Sender
Stores all frames in output buffer
Must get acks in order in which frames are sent
Receiver
Discards all frames following lost or damaged frame
Works well
if transmission errors rare and few retransmissions
If lot of traffic in both directions
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Continued Æ
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Go Back n
Go Back n
Continued Æ
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Go Back n
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Simulation of multiple timers in software
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Selective Repeat
Selective Repeat
Problem: non-sequential receive
Receiver (rather than sender as in Go Back n) accepts and
buffers frames following lost or damaged frame
Both sender and receiver maintain windows
New range of valid sequence numbers for receiver can
overlap old range
Overlap can contain duplicates
Sender’s window starts at 0, grows to MAX SEQ
Receiver’s window fixed at MAX SEQ
Receiver has buffer reserved for each seq num in its window
When frame arrives, receiver checks if seq num in window
If so and if not already received, frame is accepted and stored
If all lower numbered frames delivered to NL, frame is delivered
to NL
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Problem for Non-Sequential Receive
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Problem for Non-Sequential Receive
Scenarios: n = 3bit seq num (8 possible numbers, 0 through 7, back
to 0)
Sender sends 0,1,2,3,4,5,6
Receiver
receives 0,1,2,3,4,5,6
sends ack but ack gets lost
Assumes sender gets ack and expects to get 7,0,1,2,3,4,5
Sender times out, retransmits 0,1,2,3,4,5,6
Receiver checks 0 is in new window, thinks new 0 because sent ack for
old 6, waits for 7
Sender receives ack for 0-6, sends 7
Receiver receives 7, delivers 7 and old 0 (as new 0 !)
Solution:
(a) Initial situation with a window size seven.
(b) After seven frames sent and received, but not acknowledged.
(c) Initial situation with a window size of four.
(d) After four frames sent and received, but not acknowledged.
make sure no overlap when receiver advances its window
Make window size w =1/2 range of seq numbers
Note: number buffers at receiver needs = w
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Selective Repeat
Selective Repeat
If no reverse traffic before timer goes off,
separate ack frame is sent
Interrupt due to timer is ack_timedout event
Solves problem of traffic in one direction only and blocking
of Go Back n
Further enhancements
When receiver suspects error, sends NAK back to sender
(request for retransmission)
Two circumstances that trigger NAK
Damaged frame arrives
Frame other than expected one arrives, suspect expected one
is lost
Receiver sends only one NAK for frame expected
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Selective Repeat
Selective Repeat
Continued Æ
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Continued Æ
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Selective Repeat
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