1. The Network Layer

2. Purpose of Network layer
Given a packet, send it across the network to destination
2 key issues:
Portability:
connect different technologies
Scalability
To the Internet scale
application
transport
network
data link
physical
network
data link
physical

3. What does it involve? Two important functions:
routing: determine path from source to dest.
forwarding: move packets from router’s input to output T3 T1 T3
Sts-1 T1

4. Network service model
Q: What service model for “channel” transporting packets from sender to receiver?
guaranteed bandwidth?
preservation of inter-packet timing (no jitter)?
loss-free delivery?
in-order delivery?
congestion feedback to sender?
The most important
abstraction provided
by network layer: ?
virtual circuit or
datagram? ? ?
service abstraction
Which things can be “faked” at the transport layer?

5. Two connection models Connectionless (or “datagram”):
each packet contains enough information that routers can decide how to get it to its final destination
Connection-oriented (or “virtual circuit”)
first set up a connection between two nodes
label it (called a virtual circuit identifier (VCI))
all packets carry label B A b C 1 1 1 B A C

6. Virtual circuits: signaling protocols
used to setup, maintain teardown VC
setup gives opportunity to reserve resources
used in ATM, frame-relay, X.25
not used in today’s Internet
application
transport
network
data link
physical
application
transport
network
data link
physical
5. Data flow begins
6. Receive data
4. Call connected
3. Accept call
1. Initiate call
2. incoming call

7. Virtual circuit switching
Forming a circuit:
send a connection request from A to B. Contains VCI + address of B
rule: VCI must be unique on the link its used on
switch creates an entry mapping input messages with VCI to output port
switch picks a new VCI unique between it and next switch a b 2 5 1 c

8. Virtual circuit forwarding
For each VCI switch has a table which maps input link to output link and gives the new VCI to use
if a’s messages come into switch 1 on link 2 and go out on link 3 then the table will be:
(Input link,VCI) (output link, new VCI)
(1, 2) (?, ?)
(1, 5) (?, ?)
Switch 1 2
Switch 2 c 1 5 1 2
Switch 3 2 1 b a

9. Virtual Circuits: Discussion
Plusses: easy to associate resources with VC
Easy to provide QoS guarantees (bandwidth, delay)
Very little state in packet
Minuses:
Not good in case of crashes
Requires explicit connect and teardown phases
What if teardown does not get to all routers?
What if one switch crashes?
Will have to teardown and rebuild route

10. Datagram networks no call setup at network layer
routers: no state about end-to-end connections
no network-level concept of “connection”
packets typically routed using destination host ID
packets between same source-dest pair may take different paths
Best effort: data corruption, packet drops, route loops
application
transport
network
data link
physical
application
transport
network
data link
physical
1. Send data
2. Receive data

11. Datagrams: Forwarding
How does packet get to the destination?
switch creates a “forwarding table”, mapping destinations to output port (ignores input ports)
when a packet with a destination address in the table arrives, it pushes it out on the appropriate output port
when a packet with a destination address not in the table arrives, it must find out more routing information (next problem) a b c 1 d 2 S1 S2 S3

12. Datagrams Plusses: Minuses
No round trip connection setup time
No explicit route teardown
No resource reservation  each flow could get max bandwidth
Easily handles switch failures; routes around it
Minuses
Difficult to provide resource guarantees
Higher per packet overhead
Internet uses datagrams: IP (Internet Protocol)

13. Datagrams Forwarding How to build forwarding tables?
Manually enter it
What if nodes crashed
What about scale?
The graph-theoretic routing problem
Given a graph, with vertices (switches), edges (links), and edge costs (cost of sending on that link)
Find the least cost path between any two nodes
Path cost =  (cost of edges in path)

14. Simple Routing Algorithm
Choose a central node
All nodes send their (nbr, cost) information to this node
Central node uses info to learn entire topology of the network
It then computes shortest paths between all pairs of nodes
Using All Pair Shortest Path Algorithm
Sends the new matrix to every node
Nice, simple, elegant!
What is the problem?
Scalability: centralization hurts scalability
Central node is “crushed” with traffic

15. Link State Routing Basic idea: Mechanisms required:
Every node propagates its (nbr, cost) information
This information at all nodes is enough to construct topology
Can use a graph algorithm to find the shortest routes
Mechanisms required:
Reliable flooding of link information
Method to calculate shortest route (Dijkstra’s algorithm)
Example link state update packet:
[node id, (nbr, cost) list, seq. no., ttl]
Seq. no. to identify latest updates, ttl specifies when to stop msg.

16. Reliable flooding receive(pkt)
If already have a copy of LSP from pkt.ID
if pkt’s sequence number <= copy’s
discard pkt
else
decrement pkt.TTL
replace copy with pkt
forward pkt to all links besides the
one that we received it on
# done every 10 minutes or so
gen_LSP()
increment node’s sequence # by one
recompute cost vector
send created LSP to all neighbors

17. Discussion: Link-State Routing
Plusses:
Simple, determines the optimal route most of the time
Used by OSPF
Minuses:
Might have oscillations
Avoid using load as cost metric, reduce herding effect A A D C B
2+e
1+e 1 A A D C B
2+e e
1+e 1 1
1+e
2+e D B D B e 1 C
1+e C 1 1 e
… recompute
Initially start with
almost equal routes
… recompute
Least loaded =>
Most loaded
… everyone goes with
least loaded

18. Is our routing algo scalable?
Route table size grows with size of network
Because our address structure is flat!
Solution: have a hierarchical structure
Used by OSPF
Divide the network into areas, each area has unique number
Nodes carry their area number in the address 1.A, 2.B, etc.
Nodes know complete topology in their area
Area border routers (ABR) know how to get to any other area

19. Hierarchical Addressing
Zone 2
2.a
1.b S1 1 1
2.b 2 S2 2 3
1.a
3.b 1 S3
Forwarding table for switch 1
Destination switch port
2. ?
3. ?
1.b ?
1.a ? 2
3.a
Zone 3

20. IP has 2-layer addressing
Each IP address is 32 bits
Network part: which network the host is on?
Host part: identifies the host.
All hosts on same network have the same network part
3 classes of addresses: A, B and C
network
host
32-bits
0 net host
bits
1 0 net host
net host
bits
bits

21. IP addressing The different classes:
Problems: inefficient, address space exhaustion
class to A
network
host
Unicast B to 10
network
host to C
110
network
host to
Multicast D
1110
multicast address to
Reserved E
1111
reserved
32 bits

22. IP addressing: CIDR Classless InterDomain Routing
network portion of address of arbitrary length
address format: a.b.c.d/x, where x is # bits in network portion
Examples:
Class A: /8
Class B: /16
Class C: /24
network
part
host
/23

23. Internet Protocol Datagram
IP protocol version
Number
32 bits
total datagram
length (bytes)
head.
len
type of
service
header length
ver
length
for
fragmentation/
reassembly
“type” of data
fragment
offset
16-bit identifier
flgs
max number
remaining hops
(decremented at
each router)
time to
live
upper
layer
Internet
checksum
32 bit source IP address
32 bit destination IP address
upper layer protocol
to deliver payload to
Options (if any)
E.g. timestamp,
record route
taken, pecify
list of routers
to visit.
data
(variable length,
typically a TCP
or UDP segment)

24. Datagram Portability
IP Goal: To create one logical network from multiple physical networks
All intermediate routers should understand IP
IP header information sufficient to carry the packet to destination
Goal: Run over anything!
Problem:
Physical networks have different MTUs
“max. transmission unit”: 1500 for Ethernet, 48 for ATM
Solution 1:
Fit everything in the MTU (!)

25. IP Fragmentation & Reassembly
Solution 2: (the one used)
If packet size > MTU of network, then fragment into pieces
Each fragment is less than MTU size
Each has IP headers + frag bit set + frag id + offset
Packets may get refragmented on the way to destination
Reassembly only done at the destination
What is a good initial packet size?
reassembly
fragmentation:
in: one large datagram
out: 3 smaller datagrams