2002 Prentice Hall Chapter 9 Networking and Telecommunication.
CS412 Introduction to Computer Networking & Telecommunication
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Transcript of CS412 Introduction to Computer Networking & Telecommunication
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Chi-Cheng Lin, Winona State University
CS412 Introduction to Computer Networking &
Telecommunication
Data Link Layer Part II – Sliding Window
Protocols
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Part 2 - Topics Sliding Window Protocols
Go Back N Sliding Window Protocol
Selective Repeat Sliding Window Protocol
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Data Frame Transmission Unidirectional in previous
elementary protocols Not general
Full-duplex - approach 1 Two separate communication channels
Forward channel for dataReverse channel for acknowledgement
Problems: reverse channel bandwidth wasted
cost
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Data Frame Transmission Full-duplex - approach 2
Same circuit for both directionsData and acknowledgement are intermixedHow do we tell acknowledgement from data?
"kind" field telling data or acknowledgementCan it be improved?
Approach 3 Attaching acknowledgement to outgoing data
frames Piggybacking
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Piggybacking Temporarily delaying outgoing
acknowledgement so that they can be hooked onto the next outgoing data frame
Advantage: higher channel bandwidth utilization
Complication: How long to wait for a packet to piggyback?If longer than sender timeout period then
sender retransmit Purpose of acknowledgement is lost
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Piggybacking Solution for timing complexion
If a new packet arrives quickly Piggybacking
If no new packet arrives after a receiver timeout Sending a separate acknowledgement frame
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Sliding Window Protocol We are going to study three
bidirectional sliding window protocolsOne-bit sliding window protocolGo back NSelective repeat
Differ in efficiency, complexity, and buffer requirements
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Sliding Window Protocol Each outbound frame contains an n-bit
sequence numberRange: 0 - MAX_SEQ (MAX_SEQ = 2n - 1)For stop-and-wait, n = __. Why?
At any instance of timeSender maintains a set of sequence numbers
of frames permitted to sendThese frames fall within sending window
Receiver maintains a set of sequence numbers of frames permitted to acceptThese frames fall within receiving window
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Sliding Window Protocol Lower limit, upper limit, and size of
two windows need not be the same Fixed or variable size Requirements
Packets delivered to the receiver's network layer must be in the same order that they were passed to the data link layer on the sending machine
Frames must be delivered by the physical communication channel in the order in which they were sent
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Sending Window Contains frames have been sent or
can be sent but not yet acknowledged – outstanding frames
When a packet arrives from network layerNext highest sequence number
assignedUpper edge of window advanced by 1
When an acknowledgement arrivesLower edge of window advanced by 1
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Sending Window If the maximum window size is n,
n buffers is needed to hold unacknowledged frames
Window full (maximum window size reached) shut off network layer
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Receiving Window
Contains frames may be accepted Frame outside the window
discarded When a frame's sequence number
equals to lower edgePassed to the network layerAcknowledgement generatedWindow rotated by 1
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Receiving Window Contains frames may be accepted Always remains at initial size (different
from sending window) Size
=1 means frames only accepted in order>1 not so
Again, the order of packets fed to the receiver’s network layer must be the same as the order packets sent by the sender’s network layer
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Actually, 1-bit sequence number is enough for this example. The purpose of using 3-bit is to demonstrate the idea of sliding window.
In many books, an array of boxes are used to represent the window.
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 Sending window size = receiving
window size = 1 Stop-and-wait Refer to algorithm in Fig 3-14 Acknowledgement =
Sequence number of last frame received w/o error*
Problem of sender and receiver send simultaneously
*: some protocols define the acknowledgement to be the sequence number expected to receive as in Forouzan’s
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Try drawing the sending windows and receiving windows!
(a) Case 1: Normal case. (b) Case 7: Abnormal case. The notation is (seq, ack, packet number). An asterisk indicates where a network layer accepts a packet.
Case 7: simultaneous startCase 1: normal case
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One Bit Sliding Window Protocol Case 1: no
error A BTime (0,1,A0)
(0,0,B0)
Case 2: data lost A
BTimeX
Timeout
(1,0,A1)
(1,1,B1)
(0,1,A2)
(0,0,B2)
(0,1,A0)
(0,1,A0)
(0,0,B0)
Try drawing the sending windows and receiving windows!
*
*
*
*
*
*
*
*
Exp=0
Exp=1
Exp=0
Exp=1
Exp=0
Exp=1
Exp=0
Exp=0
Exp=1
Exp=0
Exp=1
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One Bit Sliding Window Protocol Case 4: ack. lost
A BTime
XTimeout
Case 3: data error
A B
Time
ErrorTimeout
(0,1,A0)
(0,1,A0)
(0,0,B0)
(0,1,A0)
(0,1,A0)
(0,0,B0)
(0,0,B0)
duplicate,discarded
Try drawing the sending windows and receiving windows!
*
*
* *
Exp=0
Exp=1
Exp=0Exp=0 Exp=0
Exp=1
Exp=1 Exp=1
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One Bit Sliding Window Protocol Case 6: outgoing
frame timeout A BTime
Timeout
Case 5: early timeout
A B
Time
Timeout(0,1,A0)
(0,1,A0)
(0,0,B0)0,1,A0
1,1,A1
0,1,B0Try drawing the sending windows and receiving windows!
duplicate,discarded
(0,0,B0)(1,0,A1)
duplicate,discarded
(1,1,B1)
ACK 0
Exp=0 Exp=0Exp=0 Exp=0
Exp=0*
*Exp=1
Exp=1*Exp=1*
Exp=1*
Exp=0*
Exp=0*
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Performance of Stop-and-Wait Protocol Assumption of previous protocols:
Transmission time is negligibleFalse, when transmission time is long
Example - satellite communication channel capacity: 50 kbps, frame size: 1kb
round-trip propagation delay: 500 msec Time:t=0 start to send 1st bit frame
t=20 msec frame sent completely t=270 msec frame arrives t=520 msec best case of ack. received
Sender blocked 500/520 = 96% of timeBandwidth utilization 20/520 = 4%
t 0 20
270
520
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If channel capacity = b, frame size = L, and round-trip propagation delay = R, then bandwidth utilization = _____
Conclusion: Long transit time + high bandwidth +
short frame length disaster
Performance of Stop-and-Wait Protocol
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Solution: PipeliningAllowing w frames sent before blocking
In our example, for 100% utilizationw = __, max window size = __sequence number = __ bits
Problem: errors Solutions
Go back n protocol (GNP)Selective repeat protocol (SRP)Acknowledge n means frames n,n-1,n-2,… are
acknowledged (i.e., received correctly)
Performance of Stop-and-Wait Protocol
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Go Back n Protocol Receiver discards all subsequent frames
following an error one, and send no acknowledgement for those discarded
Receiving window size = 1 (i.e., frames must be accepted in the order they were sent)
Sending window might get fullIf so, re-transmitting unacknowledged frames
Wasting a lot of bandwidth if error rate is high
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Go Back n Protocol
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11.8 Control variables
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11.9 Go-Back-N ARQ, normal operation
In Forouzan’s book, an ACK n means “I am expecting frame n,” which is one off from Tanenbaum’s. To be consistent, I modified examples adopted from Forouzan’s book. When you read Forouzan’s book, be aware of the difference!
1
2
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11.10 Go-Back-N ARQ, lost frame
1
2
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Go Back n Protocol What is the maximum sending window
size? Maximum sending window size of =
MAX_SEQ, not MAX_SEQ+1With n-bit sequence number, MAX_SEQ =
2n – 1, maximum sending window size = 2n - 1
e.g., for 3-bit window, MAX_SEQ = 7, so window size = 7 although max. size could be 8
Why?
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Go Back n Protocol - Window Size
Suppose 3-bit window is used and max sending window size = MAX_SEQ+1 = 8Sender sends frames 0 through 7Piggybacked ack 7 comes backSender sends anther 8 frames w/ sequence
numbers 0 through 7Another piggybacked ack 7 comes backQ: Did all second 8-frames arrive successfully or
did all of them get lost?Ack 7 for both cases Ambiguous
Max. window size = 7
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11.11 Go-Back-N ARQ: sender window size
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Go Back n Protocol Implementation
Sender has to buffer unacknowledged frames
Acknowledge n means frames n,n-1,n-2, ... are acknowledged (i.e., received correctly) and those buffers can be released
One timer for each outstanding frame in sending window
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Select Repeat Protocol Receiver stores correct frames following the bad
one Sender retransmits the bad one after noticing Receiver passes data to network layer and
acknowledge with the highest number Receiving window > 1 (i.e., any frame within
the window may be accepted and buffered until all the preceding one passed to the network layer
Might need large memory
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Negative Acknowledgement (NAK)
SRP is often combined with NAK When error is suspected by receiver,
receiver request retransmission of a frameArrival of a damaged frameArrival of a frame other than the expected
Does receiver keeps track of NAK? What if NAK gets lost? To nak, or not to nak: that is the question
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Selective Repeat with NAK
Nak 2,
lost
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11.12 Selective Repeat ARQ, sender and receiver windows
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11.13 Selective Repeat ARQ, lost frame
1
What’s the new window here?What’s the new window here?
What’s the ACK or NAK here?
0
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Select Repeat Protocol Implementation
Receiver has a buffer for each sequence number within receiving window
Each buffer is associated with an "arrived" bit
Check whether sequence number of an arriving frame within window or notIf so, accept and store
Maximum window size = ? Can it be MAX_SEQ ?
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Select Repeat Protocol - Window Size
Suppose 3-bit window is used and window size = MAX_SEQ = 7 sender receiver 0 1 2 3 4 5 6 sent 0 1 2 3 4 5 6 accepted 0 through 6 to network layer all acknowledgements lost0 retransmitted 0 acceptedacknowledgement 6 received7 sent 7 accepted 7 and 0 to network layer
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11.14 Selective Repeat ARQ, sender window size
0
0 1
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Select Repeat Protocol - Window Size
Problem is caused by new and old windows overlapped
SolutionWindow size=(MAX_SEQ+1)/2E.g., if 4-bit window is used, MAX_SEQ =
15 window size = (15+1)/2 = 8
Number of buffers needed = window size
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Select Repeat Protocol
(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.
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Acknowledgement Timer Problem
If the reverse traffic is light, effect?If there is no reverse traffic, effect?
SolutionAcknowledgement timer:
If no reverse traffic before timeout send separate acknowledgementEssential: ack timeout < data frame
timeout Why?