Ch 16 Protocols and Layering Spring 2003

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

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

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

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

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

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

The Seven Layers LAN

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

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

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

Data Encapsulation

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

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)

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

Nested Packet Headers

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

Layering Principle: peer-to-peer

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

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

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

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

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

Sliding Window Flow Control

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

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

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

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

Illustration of Architecture That Can Experience Congestion Multiple sources Bottleneck

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)

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

Exercise 16.4

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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