CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1...

47
CompSci 356: Computer Network Architectures Lecture 8: Spanning Tree Algorithm and Basic Internetworking Ch 3.1.5 & 3.2 Xiaowei Yang [email protected]

Transcript of CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1...

Page 1: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

CompSci 356: Computer Network Architectures

Lecture 8: Spanning Tree Algorithm and Basic Internetworking

Ch 3.1.5 & 3.2

Xiaowei [email protected]

Page 2: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Review• Past lectures– Single link networks

• Point-to-point, shared media– Ethernet, token ring, wireless networks

• Encoding, framing, error detection, reliability– Delay-bandwidth product, sliding window, exponential backoff, carrier

sense collision detection, hidden/exposed terminals

– Packet switching: how to connect multiple links• Connectionless: Datagram• Connection-oriented: Virtual circuits• Source routing• Pros and cons• Ethernet switches

Page 3: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Today

• Spanning Tree Algorithm• Virtual LAN• New topic: how to connect different types of

networks–E.g., how to connect an Ethernet and an ATM

network

Page 4: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Learning Bridges• Overall design goal: complete transparency

• �Plug-and-play�• Self-configuring without hardware or software changes• Bridges should not impact operations of existing LANs

• Three parts to learning bridges:• (1) Forwarding of Frames• (2) Learning of Addresses• (3) Spanning Tree Algorithm

Page 5: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

• Assume a MAC frame arrives on port x.

(1) Frame Forwarding

Bridge 2Port A Port C

Port x

Port B

Is MAC address of destination in forwarding

table?

Forward the frame on theappropriate port

Flood the frame, i.e.,

send the frame on all ports except port x.

Found?Notfound ?

Page 6: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

• Consider the two LANs that are connected by two bridges.

• Assume host A is transmitting a frame F with a broadcast address

What is happening?• Bridges A and B flood the frame to

LAN 2.• Bridge B sees F on LAN 2, and

updates the port mapping of MAC_A, and copies the frame back to LAN 1

• Bridge A does the same. • The copying continuesWhere�s the problem? What�s the

solution ?

Danger of Loops

LAN 2

LAN 1

Bridge BBridge A

host A

F

F F

FF

F F

Page 7: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Spanning Tree Algorithm• A solution is the spanning

tree algorithm that prevents loops in the topology– By Radia Perlman at DEC

LAN 2

Bridge 2

LAN 5

LAN 3

LAN 1

LAN 4

Bridge 5

Bridge 4Bridge 3d

Bridge 1

Page 8: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Algorhyme (the spanning tree poem)• I think that I shall never see

A graph more lovely than a tree.A tree whose crucial propertyIs loop-free connectivity.A tree that must be sure to spanSo packets can reach every LAN.First, the root must be selected.By ID, it is elected.Least-cost paths from root are traced.In the tree, these paths are placed.A mesh is made by folks like me,Then bridges find a spanning tree.

• —Radia Perlman

Page 9: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Graph theory on spanning tree

• For any connected graph consisting of nodes and edges connecting pairs of nodes, a spanning tree of edges maintains the connectivity of the graph but contains no loops– n-node�s graph, n - 1 edges on a spanning tree– No redundancy

Page 10: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

The protocol

• IEEE 802.1d has an algorithm that organizes the bridges as spanning tree in a dynamic environment

• Bridges exchange messages to configure the bridge (Configuration Bridge Protocol Data Unit, Configuration BPDUs) to build the tree– Select ports they use to forward packets

Page 11: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Configuration BPDUs

time since root sent a message on

which this message is based

DestinationMAC address

Source MACaddress

Configuration Message

protocol identifier

version

message type

flags

root ID

Cost

bridge ID

port ID

message age

maximum agehello time

forward delay

Set to 0 Set to 0Set to 0

lowest bit is "topology change bit (TC bit)

ID of root Cost of the path from the bridge sending this

message to root bridge

ID of port from which message is sent

ID of bridge sending this message

Time between recalculations of the

spanning tree (default: 15 secs)

Time between BPDUs from the root

(default: 1sec)

Page 12: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

What do the BPDUs do? • Elect a single bridge as the root bridge

• Calculate the distance of the shortest path to the root bridge

• Each bridge can determine a root port, the port that gives the best path to the root

• Each LAN can determine a designated bridge, which is the bridge closest to the root. A LAN's designated bridge is the only bridge allowed to forward frames to and from the LAN for which it is the designated bridge.

• A LAN's designated port is the port that connects it to the designated bridge

• Select ports to be included in the spanning tree.

Page 13: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Terms• Each bridge has a unique identifier: Bridge ID

Bridge ID = {Priority : 2 bytes; Bridge MAC address: 6 bytes}• Priority is configured• Bridge MAC address is the lowest MAC addresses of all ports

• Each port within a bridge has a unique identifier (port ID)

• Root Bridge: The bridge with the lowest identifier is the root of the spanning tree

• Root Port: Each bridge has a root port which identifies the next hop from a bridge to the root

Page 14: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Terms• Root Path Cost: For each bridge, the cost of the

min-cost path to the root– Assume it is measured in #hops to the root

• Designated Bridge, Designated Port: Single bridge on a LAN that is closest to the root for this LAN:– If two bridges have the same cost, select the one with

the highest priority; if they have the same priority, select based on the bridge ID

– If the min-cost bridge has two or more ports on the LAN, select the port with the lowest identifier

Page 15: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Spanning Tree Algorithm• Each bridge is sending out BPDUs that contain the following

information:

• The transmission of BPDUs results in the distributed computation of a spanning tree

• The convergence of the algorithm is very fast

root bridge (what the sender thinks it is) root path cost for sending bridge

Identifies sending bridgeIdentifies the sending port

root ID cost bridge ID port ID

Page 16: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Ordering of Messages• We define an ordering of BPDU messages

(lexicographically)

We say M1 advertises a better path than M2 (�M1<<M2�) if

(R1 < R2), Or (R1 == R2) and (C1 < C2), Or (R1 == R2) and (C1 == C2) and (B1 < B2), Or (R1 == R2) and (C1 == C2) and (B1 == B2) and (P1 < P2)

ID R1 C1 ID B1 M1 M2

ID P1 ID R2 C2 ID B2 ID P2

Page 17: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

• Initially, all bridges assume they are the root bridge. • Each bridge B sends BPDUs of this form on its LANs

from each port P:

• Each bridge looks at the BPDUs received on all its ports and its own transmitted BPDUs.

• Root bridge is the one with the smallest received root ID that has been received so far – whenever a smaller ID arrives, the root is updated

Initializing the Spanning Tree Protocol

B 0 B P

Page 18: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

• Each bridge B looks on all its ports for BPDUs that are better than its own BPDUs

• Suppose a bridge with BPDU:

receives a �better� BPDU:

Then it will update the BPDU to:

• However, the new BPDU is not necessarily sent out• On each bridge, the port where the �best BPDU� (via relation �<�) was received

is the root port of the bridge– No need to send out updated BPDUs to root port

Spanning Tree Protocol

R1 C1 B1 P1M1

R2 C2 B2 P2M2

R2 C2+1 B1 P1

Page 19: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

• Say, B has generated a BPDU for each port x

• B will send this BPDU on port x only if its BPDU is better (via relation �<�) than any BPDU that B received from port x.

• In this case, B also assumes that it is the designated bridge for the LAN to which the port connects

• And port x is the designated port of that LAN

When to send a BPDU

R Cost B

Bridge BPort A Port C

Port x

Port B

x

Page 20: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Selecting the Ports for the Spanning Tree

• Each bridge makes a local decision which of its ports are part of the spanning tree

• Now B can decide which ports are in the spanning tree:

• B�s root port is part of the spanning tree• All designated ports are part of the spanning tree• All other ports are not part of the spanning tree

• B�s ports that are in the spanning tree will forward packets (=forwarding state)

• B�s ports that are not in the spanning tree will not forward packets (=blocking state)

Page 21: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

LAN 2

Bridge1

LAN 5

LAN 3

LAN 1

LAN 4

Bridge2

Bridge5Bridge4

•d

Bridge3

•D•D

•D

•R

•D•R •R

•R

•D

Building the Spanning Tree

• Consider the network on the right.• Assume that the bridges have

calculated the designated ports (D) and the root ports (R) as indicated.

• What is the spanning tree?– On each LAN, connect D ports to

the R ports on this LAN– Which bridge is the root bridge?

• Suppose a packet is originated in LAN 5. How is the packet flooded?

Page 22: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Example• Assume that all bridges send out their BPDU�s once per

second, and assume that all bridges send their BPDUs at the same time

• Bridge1 < Bridge2 < Bridge3 < Bridge4 < Bridge5• Assume that all bridges are turned on simultaneously at time

T=0 sec.

Brige2

LAN 1

LAN 2

LAN 3

LAN 4

Brige1

Brige5

Brige3 Brige4AB

A B

A B AB

AB

Page 23: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Example: BPDUs sentBridge1 Bridge2 Bridge3 Bridge4 Bridge5

T=1sec

Page 24: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Example: BPDUs sentBridge1 Bridge2 Bridge3 Bridge4 Bridge5

T=2sec

Page 25: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Example: BPDUs sentBridge1 Bridge2 Bridge3 Bridge4 Bridge5

T=3sec

Page 26: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Example: BPDUs sent

Bridge1 Bridge2 Bridge3 Bridge4 Bridge5T=1sec Send:

A: (B1,0,B1,A)B: (B1,0,B1,B)Recv:A:(B5,0,B5,A)(B2,0,B2,B)B:(B3,0,B3,B)(B4,0,B4,A)

Send:A: (B2,0,B2,A)B: (B2,0,B2,B)Recv:A:B: (B1,0,B1,A)(B5,0,B5,A)

Send:A:(B3,0,B3,A)B:(B3,0,B3,B)Recv:A: (B5,0,B5,B)(B4,0,B4,B)B: (B1,0,B1,B) (B4,0,B4,A)

Send:A:(B4,0,B4,A)B:(B4,0,B4,B)Recv:A: (B3,0,B3,B) (B1,0,B1,B)B: (B3,0,B3,A) (B5,0,B5,B)

Send:A:(B5,0,B5,A)B:(B5,0,B5,B)Recv:A: (B2,0,B2,B)(B1,0,B1,A)B: (B3,0,B3,A)(B4,0,B4,B)

Page 27: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Example: BPDU�s sent

Bridge1 Bridge2 Bridge3 Bridge4 Bridge5

T=2sec D-port: A,BSend:A: (B1,0,B1,A)B: (B1,0,B1,B)Recv:

R-port: BD-port: ASend:A: (B1,1,B2,A)Recv:A: B:(B1,0,B1,A)

R-port: BD-port: ASend:A: (B1,1,B3,A)Recv:A:(B1,1,B4,B)(B1,1,B5,B)B:(B1,0,B1,B)

R-port: AD-port: BSend:B: (B1,1,B4,B)Recv:A:(B1,0,B1,B)B:(B1,1,B3,A)(B1,1,B5,B)

R-port: AD-port: BSend:B: (B1,1,B5,B)Recv:A:(B1,0,B1,A)B:(B1,1,B3,A)(B1,1,B4,B)

Page 28: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Example: BPDU�s sent

Bridge 1 Bridge 2 Bridge 3 Bridge4 Bridge5

T=3sec D-port: A,BSend:A: (B1,0,B1,A)B: (B1,0,B1,B)Recv:

R-port: BD-port: ASend:A: (B1,1,B2,A)Recv:A: B:(B1,0,B1,A)

R-port: BD-port: ASend:A: (B1,1,B3,A)Recv:A:B:(B1,0,B1,B)

R-port: ABlocked: B

Recv:A:(B1,0,B1,B)B:(B1,1,B3,A)

R-port: ABlocked: B

Recv:A:(B1,0,B1,A)B:(B1,1,B3,A)

Page 29: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Example: the spanning treeBridge1 Bridge2 Bridge3 Bridge4 Bridge5

Root Port

Designatedbridge

Designatedports

Bridge2

LAN 1

LAN 2

LAN 3

LAN 4

Bridge1

Bridge5

Bridge3 Bridge4AB

A B

A B AB A B

A packet is sentfrom LAN2

Page 30: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Example: the spanning treeBridge1 Bridge2 Bridge3 Bridge4 Bridge5

Root Port B B A ADesignatedbridge LAN2,3 LAN1 LAN4Designatedports A,B A A

Bridge2

LAN 1

LAN 2

LAN 3

LAN 4

Bridge1

Bridge5

Bridge3 Bridge4AB

A B

A B AB A B

A packet is sentfrom LAN2

Page 31: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Limitations of bridges

• Scalability– Broadcast packets reach every host!

• Security– Every host can snoop

• Non-heterogeneity– Can�t connect ATM networks

Page 32: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Virtual LANs • To address the scalability and security issues

• A bridge�s port is configured to have a VLAN ID

• Each VLAN has a spanning tree

• A VLAN header is inserted to a packet

• Packets are flooded to ports with the same VLAN ID

Page 33: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

5 U 5 U

VLAN100 VLAN100

VLAN200 VLAN200

100

B1 B2

Page 34: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Today

• Spanning Tree Algorithm• Virtual LAN• New topic: how to connect different types of

networks–E.g., how to connect an Ethernet and an ATM

network

Page 35: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Inter-networking

• Routers interface different networks• Uniform addressing (IP)• Routers send packets to their destination IP addresses

Page 36: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

• IP (Internet Protocol) is a Network Layer Protocol• IP’s current version is Version 4 (IPv4). It is

specified in RFC 791.• IPv6 is also deployed

NetworkLayer

Link Layer

IP

ARP NetworkAccess

Media

ICMP IGMP

TransportLayer

TCP UDP

Internet Protocol

Page 37: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

IP: the thin waist of the hourglass

• IP is the waist of the hourglass of the Internet protocol architecture

• Multiple higher-layer protocols• Multiple lower-layer protocols• Only one protocol at the

network layer.• What is the advantage of this

architecture?– To avoid the N * M problem

App lica tions

HTTP FTP S MTP

TCP U D P

IP

Da ta l ink layer pro tocols

Phys ical layer technolog ies

Page 38: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Application protocol • Routers look at a packet’s IP header and link

layer header

Application

TCP

IPData Link

Application

TCP

IP

Application protocol

TCP protocol

IP protocol IP protocol

DataLink

DataLink

IPDataLink

DataLink

IPDataLink

DataLink

DataLink

IP protocol

RouterRouter HostHost

Data Link

Page 39: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

A simple network

Page 40: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

IP Service Model• Delivery service of IP is minimal

• IP provides an unreliable connectionless best effort service (also called: “datagram service”).– Unreliable: IP does not make an attempt to recover lost packets

– Connectionless: Each packet (“datagram”) is handled independently. IP is not aware that packets between hosts may be sent in a logical sequence

– Best effort: IP does not make guarantees on the service (no throughput guarantee, no delay guarantee,…)

• Consequences:

• Higher layer protocols have to deal with losses or with duplicate packets

• Packets may be delivered out-of-order

Page 41: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Basic IP router functions

• Things you need to understand to do lab2– Internet protocol• IP header• IP addressing• IP forwarding

– Address resolution protocol– Error reporting and control• Internet Control Message Protocol

Page 42: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Fields of the IP header• ToS (8-bit): specifies the

type of differentiated services for a packet

• HLen (4-bit): the length of header in 32-bit words

• Length (16-bit): packet length in bytes, including the header– 65535 bytes– Fragmentation and

reassembly

Page 43: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Fields of the IP Header• Identification (16 bits):

Unique identification of a datagram from a host. Incremented whenever a datagram is transmitted (in some OS)

• Flags (3 bits): – First bit always set to 0– DF bit (Do not fragment)– MF bit (More fragments) Will be explained laterà

Fragmentation

• Fragment offset (13 bits)

Page 44: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Fields of the IP Header• Time To Live (TTL)

(1byte):– Specifies the longest path

before a datagram is dropped– Role of TTL field: Ensure that

a packet is eventually dropped when a routing loop occurs

Used as follows:– Sender sets the value (e.g., 64)– Each router decrements the

value by 1– When the value reaches 0, the

datagram is dropped

Page 45: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Fields of the IP Header• Protocol (1 byte):

• Specifies the higher-layer protocol.• Used for demultiplexing to higher

layers.

• Header checksum (2 bytes): A simple 16-bit long checksum which is computed for the header of the datagram– Function?

IP

1 = ICMP 2 = IGMP

6 = TCP 17 = UDP

4 = IP-in-IPencapsulation

Page 46: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Fields of the IP Header• Options:

• Record Route: each router that processes the packet adds its IP address to the header.• Timestamp: each router that processes the packet adds its IP

address and time to the header.• (loose) Source Routing: specifies a list of routers that must be

traversed.• (strict) Source Routing: specifies a list of the only routers that

can be traversed.• IP options increase routers processing overhead. IPv6 does not have

the option field.

• Padding: Padding bytes are added to ensure that header ends on a 4-byte boundary

Page 47: CompSci 356: Computer Network Architectures Lecture 8 ... · Spanning Tree Protocol M1 R1 C1 B1 P1 M2 R2 C2 B2 P2 R2 C2+1 B1 P1 • Say, B has generated a BPDU for each port x •

Summary• Spanning Tree Algorithm• Virtual LAN• New topic: how to connect different types of

networks–E.g., how to connect an Ethernet and an ATM

network• Looking forward–More about IP