1. Ch 16 Protocols and Layering
Spring 2003

2. Motivation
Explains why hardware along does not solve all communication problems, and shows why protocol software is also need
Describes the concept of layering

3. The Need for Protocols
Communication hardware can transfer bits from point to another
Many problems can occur
Bits corrupted or destroyed
Entire packet lost
Packet duplicated
Packets delivered out of order

4. The Need for Protocols
Software handles low-level communication details and problems
Protocol is a set of rules to be used when exchanging messages
Most AP do not interact with net HW directly AP HW
Protocol

5. Protocol Specifies Format of messages Meaning of messages
Rules for exchange
Procedure for handling problems

6. Protocol Suites Each protocol solves part of communication problem
Work well together; known as protocol suite /w/
Make each protocol easier to design, analyze, implement, and test AP AP P1
Protocol … HW HW

7. A Plan for Protocol Design
Layering model
Vertically divides protocols into layers
Each layer devoted to one subproblem
To help protocol designers understand subparts of the communication problem
E.g.: ISO 7-layer reference model

8. The Seven Layers
LAN

9. The Seven Layers Layer 1: Physical Layer 2: Data Link Layer 3: Network
Gives the detailed spec. of underlying hardware
Layer 2: Data Link
Specify hardware frame format, medium access, and checksum computation
Layer 3: Network
Specify address assignment and packet forwarding
Layer 4: Transport
Specify how to handle details of reliable transfer

10. The Seven Layers Layer 5: Session Layer 6: Presentation
Specifications for session establishment and security details
Layer 6: Presentation
Data representation and the translation
Layer 7: Application
Define how application programs, running on different end systems, pass messages to each other

11. Stacks: Layered Software
Protocol software follows layering model
One software module per layer
Modules cooperate
Incoming or outgoing data passes from one module to another
Entire set of modules known as stack

12. Data Encapsulation

13. Data Encapsulation L7 L4 Segment L3 Datagram L2 Frame FTP HTTP Telnet
Port 21 80 25 L4
TCP
UDP
TCP
header
Segment
ICMP
IGMP L3 IP IP
header
Datagram
ARP
RARP L2
MAC
discard
Ethernet
header
Frame
Ethernet
header IP
TCP

14. Data Encapsulation Each layer
Places additional information in a header before sending data to a lower layer
Uses, and then removes, the header to process incoming packets (e.g., L2 checksum)

15. Data Encapsulation Port number
An AP can be uniquely identified by (IP address, port number) pair
Assigned ranges
Numbers below 255 are for public applications
Numbers from 255 to 1023 are assigned to companies for marketable applications
Numbers above 1023 are unregulated
Bind: creates mapping between an AP and a port number

16. Nested Packet Headers

17. Scientific Basis for Layering
Software implementing layer N at the destination receives exactly the message sent by software implementing layer N at the source

18. Layering Principle: peer-to-peer

19. Techniques Protocols Use
For bit corruption
Parity bit
Frame checksum
CRC
For out-of-order delivery
Sequence numbers
For lost packets
Positive acknowledgement (ACK) and retransmission for time-out
Max. number of try 4 3 2 1
Ack1

20. Techniques Protocols Use
Duplication
Sequence numbers
For replay (excessive delay)
A termination request arrives from an previous conversation may terminate a current conversation
Unique session ID
For data overrun
Flow control
Bye

21. Flow Control Needed because Related to buffering Two forms
Sending computer system faster than receiving computer
Sending application faster than receiving application
Related to buffering
Two forms
Stop-and-go
Sliding window

22. Stop-And-Go Sending Side Receiving side Inefficient
Transmits one packet
Waits for signal from receiver
Receiving side
Receives and consumes packets
Transmits signal to sender
Inefficient

23. Sliding Window Flow Control
1 car
v.s. 4 cars

24. Sliding Window Flow Control

25. Sliding Window Flow Control
Receiving side
Establishes multiple buffers and informs sender
Sending side
Transmits packets for all available buffers
Only waits if no signal arrives before transmission
Sends signals as packets arrive
Window tells how many packets can be sent
Window moves as acknowledgements arrive

26. Comparison of Flow Control
Stop-and-go
Slow
Useful only in special cases
Sliding window
Fast
Needed in high-speed network

27. Why Sliding Window? Simultaneously Speedup Tw = min(B, TG  W) where
Increase throughput
Control flow
Speedup
Tw = min(B, TG  W)
where
TW is sliding window throughput
B is underlying hardware bandwidth
TG is stop-and-go throughput
W is the window size

28. Congestion Fundamental problem in networks
Caused by traffic, not hardware failure
Analogous to congestion on a highway
Principal cause of delay

29. Illustration of Architecture That Can Experience Congestion
Multiple sources
Bottleneck

30. Congestion and Loss
Modern network hardware works well; most packet loss results from congestion, not from hardware failure
Rate control (e.g. temporarily reducing WS)

31. The Art of Protocol Design
Protocol Design is “nontrivial”
Lots of Tradeoff
Size of “Sequence number” field
Balance between sliding window and congestion control

32. Exercise
16.4

33. Two Common Implementations of Sliding Window Protocol
Go-Back-N Protocol
Selective Repeat Protocol
Source
Address
Destination
Frame Number (K bits)
ACK
Type of
Frame
..Data...
CRC
02K-1
Frame number of a frame being acknowledged
(piggybacking)
Data/ACK/NACK

34. Go-Back-N receiver does not acknowledge each received frame explicitly
Sender
Receiver
(0,1,2,3)
0(1,2,3)
0,1(2,3)
0,1,2(3)
0,1,2,3(-)
ACK 3
(4,5,6,7)
4(5,6,7)
ACK 5 1 2 3 4 5
4,5(6,7)
(6,7,0,1)
(0)
(1)
(2)
(3)
(4)
(5)
(6)
Sequence number: 0-7
Window size=4
Sender buffers all frames in the windows
frames are removed when ACKed

35. Go-Back-N (continued)
receiver always expects to receive frames in order
(2,3,4,5)
Sender
Receiver
(0,1,2,3)
0(1,2,3)
0,1(2,3)
0,1,2(3)
0,1,2,3(-)
ACK 1
2,3(4,5)
2,3,4(5) 1 2 3
(0)
(1)
(2) X
discard 4 5
2,3,4,5(-)
NAK 2
2(3,4,5)
Sequence number: 0-7
Window size=4

36. Go-Back-N (continued)
frame damaged
Sender
Receiver
(0,1,2,3)
0(1,2,3)
0,1(2,3)
0,1,2(3)
0,1,2,3(-)
ACK 1
2,3(4,5)
2,3,4(5) 1 2 3
3(4,5,6)
(0)
(1)
(2)
discard 4 5
3,4(5,6)
3,4,5(6)
NAK 3
(3,4,5,6)
Sequence number: 0-7
Window size=4

37. Go-Back-N (continued)
Frame timer
Sender
Receiver
(0,1,2,3)
0(1,2,3)
0,1(2,3)
0,1,2(3)
ACK 1 1 2
(0)
(1)
(2) X
Time out 3 4 5
(2,3,4,5)
2(3,4,5)
2,3(4,5)
2,3,4(5)
2,3,4,5(-)
Sequence number: 0-7
Window size=4

38. Go-Back-N (continued)
Out of order + time out
Sender
Receiver
(0,1,2,3)
0(1,2,3)
0,1(2,3)
0,1,2(3)
ACK 1 1 2
(0)
(1)
(2) X 3
discard
Time out 4
(2,3,4,5)
2(3,4,5)
2,3(4,5)
2,3,4(5)
NAK 2
Sequence number: 0-7
Window size=4

39. Go-Back-N (continued)
ACK timer
a frame arriving should be ACKed within a period of time
use piggyback approach to ACK whenever possible
send a separate ACK frame in lieu of the piggyback ACK

40. Go-Back-N (continued)
Window size should be  2K-1
Station A
Station B
Send frames 0 through 7. t1
Receive frames 0 through 7 in order and send an acknowledgment for frames 7 t2
Window size=2K
The piggyback acknowledgment gets delayed somewhere
Assume the frames were lost and resend frames 0 through 7.
Receive frames 0 through 7 Since the receiver was expecting these number,it accept them as new frames,not recognizing them as duplicate t3 t4

41. Go-Back-N (continued)
Window size should be  2K-1
Station A
Station B
Send frames 0 through 6. t1
Receive frames 0 through 6 in order and send an acknowledgment for frames 6 t2
Window size=2K-1
The piggyback acknowledgment gets delayed somewhere
Assume the frames were lost and resend frames 0 through 6.
Receive frames 0 through 6 Since the receiver was expecting frame 7 it ignores them t3 t4

42. Selective Repeat Similarities to go-back-n Frame format sending window
use piggyback ACK whenever possible
not acknowledge every frame explicitly
use NAKs for damaged frames and out-of-order frames
Frame timer
ACK timer

43. Selective Repeat (continued)
Differences to go-back-n
Buffer arriving frames in receiving window until all predecessors received
Only the timed-out frame is resent when a frame timer expires
When a NAK is received, resend just the frame specified by the NAK
A piggyback ACK doesn’t necessary acknowledge the frame just received
Receive sequential frames
Sending Window size=3
Send frames
Receiving Window size=3 7 7 6 1 6 1 5 2 5 2 4 3 4 3
Receive ACKs
Receive sequential frames

44. Selective Repeat (continued)
Frame lost
Sender
Receiver
(0,1,2,3)
0(1,2,3)
0,1(2,3)
0,1,2(3)
0,1,2,3(-)
ACK 1
2,3(4,5)
2,3,4(5)
NAK 2 1 2 3 4 5
2,3,4,5(-)
(1,2,3,4)
(2,3,4,5) X
3 (buffered)
4 (buffered)
5 (buffered)
(6,7,0,1)
ACK 5
Sending Window size=4
Receiving Window size=4

45. Selective Repeat (continued)
CRC error
(0,1,2,3)
(0,1,2,3)
Sending Window size=4
0(1,2,3)
Receiving Window size=4 1
0,1(2,3) 2
(1,2,3,4) 1
0,1,2(3)
(2,3,4,5) 3
2 (CRC error)
0,1,2,3(-)
ACK 1
3 (buffered) 4
NAK 2
2,3(4,5)
2,3,4(5) 2
4 (buffered)
2,3,4(5) 5 2
2,3,4,5(-)
(5,6,7,0) 5
(6,7,0,1)
Sender
Receiver

46. Selective Repeat (continued)
Receiving window size should be  2K-1
Station A
Station B
Receiving window
size>2K-1
Send frames 0 through3 t1
Receive frames 0 through 3;advance window to include frames 4,5,6,7,and 0. t2
The acknowledgement
gets lost.
Assume the frames were lost and resend frames 0 through 3 t3
Receive frames 0 and assume incorrectly it is a new frame. t4
time
time

47. Selective Repeat (continued)
Receiving window size should be  2K-1
Station A
Station B
Receiving window
size>2K-1
Receive frames 0 through 3 and advance window to include frames 4-7. Receive frame 4 and send an acknowledgment to A. Advance window again to include frames 5,6,7, and 0.
Send frames 0 through 4 t1 t2
The acknowledgement
gets lost.
Assume the frames were lost and resend frames 0 through 4 t3
Receive frames 0 and assume incorrectly it is a new frame. t4
time
time