1. EE 122: Intra-domain routing: Distance Vector
Computer Science Division
Department of Electrical Engineering and Computer Science
University of California, Berkeley,
Berkeley, CA
September 16, 2004
(* based partially on Kurose & Ross’ slides)

2. Problem
How is a packet transmitted from source to destination in the Internet?
Example: how does packet sent by B reaches D
Host C
Host D
Host A
Node 1
Node 2
Node 3
Node 5
Host B
Host E
Node 7
Node 6
Node 4

3. Forwarding vs. Routing
Forwarding: the process by which a router sends an arriving packet to the next router towards final destination
Routing: the process by which a router acquires the information to perform packet forwarding (e.g., routing tables)

4. IP Packet Forwarding
Each router maintains a mapping of address prefixes to output ports
Upon receiving a packet, a router performs the longest prefix matching on packet’s destination address and forwards it to corresponding output
Prefix
interface
xxx 1
12.82.xxx.xxx 3
xxx 2 … … 1 2

5. Internet Routing Internet organized as a two level hierarchy
First level – autonomous systems (AS’s)
AS – region of network under a single administrative domain
ASes run an intra-domain routing protocols
Distance Vector, e.g., Routing Information Protocol (RIP)
Link State, e.g., Open Shortest Path First (OSPF)
Between ASes runs inter-domain routing protocols, e.g., Border Gateway Routing (BGP)
De facto standard today, BGP-4

6. Example
Interior router
BGP router
AS-1
AS-3
AS-2

7. Intra-domain Routing Protocols
Based on UDP (unreliable datagram delivery)
Distance vector
Routing Information Protocol (RIP), based on Bellman-Ford
Each neighbor periodically exchange reachability information to its neighbors
Minimal communication overhead, but it takes long to converge, i.e., in proportion to the maximum path length
Link state
Open Shortest Path First (OSPF), based on Dijkstra
Each network periodically floods immediate reachability information to other routers
Fast convergence, but high communication and computation overhead

8. Routing
Goal: determine a “good” path through the network from source to destination
Good means usually the shortest path
Network modeled as a graph
Routers  nodes
Link edges
Edge cost: delay, congestion level,… A E D C B F 2 1 3 5

9. Distance Vector Routing Algorithm
Iterative: continues until no nodes exchange info
Asynchronous: nodes need not exchange info/iterate in lock steps
Distributed: each node communicates only with directly-attached neighbors
Each router maintains
Row for each possible destination
Column for each directly-attached neighbor to node
Entry in row Y and column Z of node X  best known distance from X to Y, via Z as next hop (remember this !)
Note: for simplicity in this lecture examples we show only the shortest distances to each destination

10. Distance Vector Routing
Each node:
wait for (change in local link cost or msg from neighbor)
recompute distance table
if least cost path to any dest has changed, notify neighbors
Each local iteration caused by:
Local link cost change
Message from neighbor: its least cost path change from neighbor to destination
Each node notifies neighbors only when its least cost path to any destination changes
Neighbors then notify their neighbors if necessary

11. Distance Vector Algorithm (cont’d)
1 Initialization:
2 for all neighbors V do
if V adjacent to A
D(A, V) = c(A,V);
5 else
D(A, V) = ∞;
7 loop:
8 wait (until A sees a link cost change to neighbor V
or until A receives update from neighbor V)
9 if (c(A,V) changes by d)
for all destinations Y through V do
D(A,Y) = D(A,Y) + d
12 else if (update D(V, Y) received from V)
for all destinations Y do
if (destination Y through V)
D(A,Y) = D(A,V) + D(V, Y);
else /* update D(A, Y) if there is a shorter path through V */
D(A, Y) = min(D(A, Y), D(A, V) + D(V, Y));
18 if (there is a new minimum for destination Y)
send D(A, Y) to all neighbors
20 forever

12. Example: Distance Vector Algorithm
Node A
Node B
Dest.
Cost
NextHop B 2 C 7 D ∞ -
Dest.
Cost
NextHop A 2 C 1 D 3 3 B D 2 1 1 A C 7
Node C
Node D
1 Initialization:
2 for all neighbors V do
if V adjacent to A
D(A, V) = c(A,V);
5 else
D(A, V) = ∞; …
Dest.
Cost
NextHop A 7 B 1 D
Dest.
Cost
NextHop A ∞ - B 3 C 1

13. Example: 1st Iteration (C  A)
Node A
Node B
Dest.
Cost
NextHop B 2 C 7 D ∞ -
Dest.
Cost
NextHop A 2 C 1 D 3 3 B D 2 1 1 A C 7
(D(C,A), D(C,B), D(C,D)) …
7 loop:
12 else if (update D(V, Y) received from V)
for all destinations Y do
if (destination Y through V)
D(A,Y) = D(A,V) + D(V, Y);
else
D(A, Y) = min(D(A, Y),
D(A, V) + D(V, Y));
18 if (there is a new minimum for dest. Y)
send D(A, Y) to all neighbors
20 forever
Node C
Node D
Dest.
Cost
NextHop A 7 B 1 D
Dest.
Cost
NextHop A ∞ - B 3 C 1

14. Example: 1st Iteration (C  A)
Node A
Node B
Dest.
Cost
NextHop B 2 C 7 D 8
Dest.
Cost
NextHop A 2 C 1 D 3 3 B D 2 1 1 A C 7
D(A, D) = min(D(A, D), D(A, C) + D(C,D)
= min(∞ , 7 + 1) = 8 …
7 loop:
12 else if (update D(V, Y) received from V)
for all destinations Y do
if (destination Y through V)
D(A,Y) = D(A,V) + D(V, Y);
else
D(A, Y) = min(D(A, Y),
D(A, V) + D(V, Y));
18 if (there is a new minimum for dest. Y)
send D(A, Y) to all neighbors
20 forever
(D(C,A), D(C,B), D(C,D))
Node C
Node D
Dest.
Cost
NextHop A 7 B 1 D
Dest.
Cost
NextHop A ∞ - B 3 C 1

15. Example: 1st Iteration (C  A)
Node A
Node B
Dest.
Cost
NextHop B 2 C 7 D 8
Dest.
Cost
NextHop A 2 C 1 D 3 3 B D 2 1 1 A C 7 …
7 loop:
12 else if (update D(V, Y) received from V)
for all destinations Y do
if (destination Y through V)
D(A,Y) = D(A,V) + D(V, Y);
else
D(A, Y) = min(D(A, Y),
D(A, V) + D(V, Y));
18 if (there is a new minimum for dest. Y)
send D(A, Y) to all neighbors
20 forever
Node C
Node D
Dest.
Cost
NextHop A 7 B 1 D
Dest.
Cost
NextHop A ∞ - B 3 C 1

16. Example: 1st Iteration (BA, CA)
Node A
Node B
Dest.
Cost
NextHop B 2 C 3 D 5
Dest.
Cost
NextHop A 2 C 1 D 3 3 B D 2 1 1 A C 7 …
7 loop:
12 else if (update D(V, Y) received from V)
for all destinations Y do
if (destination Y through V)
D(A,Y) = D(A,V) + D(V, Y);
else
D(A, Y) = min(D(A, Y),
D(A, V) + D(V, Y));
18 if (there is a new minimum for dest. Y)
send D(A, Y) to all neighbors
20 forever
D(A,D) = min(D(A,D), D(A,B) + D(B,D))
= min(8, 2 + 3) = 5
D(A,C) = min(D(A,C), D(A,B) + D(B,C))
= min(7, 2 + 1) = 3
Node C
Node D
Dest.
Cost
NextHop A 7 B 1 D
Dest.
Cost
NextHop A ∞ - B 3 C 1

17. Example: End of 1st Iteration
Node A
Node B
Dest.
Cost
NextHop B 2 C 3 D 5
Dest.
Cost
NextHop A 2 C 1 D 3 B D 2 1 1 A C 7 …
7 loop:
12 else if (update D(V, Y) received from V)
for all destinations Y do
if (destination Y through V)
D(A,Y) = D(A,V) + D(V, Y);
else
D(A, Y) = min(D(A, Y),
D(A, V) + D(V, Y));
18 if (there is a new minimum for dest. Y)
send D(A, Y) to all neighbors
20 forever
Node C
Node D
Dest.
Cost
NextHop A 3 B 1 D
Dest.
Cost
NextHop A 2 B 3 C 1

18. Example: End of 3nd Iteration
Node A
Node B
Dest.
Cost
NextHop B 2 C 3 D 4
Dest.
Cost
NextHop A 2 C 1 D 3 B D 2 1 1 A C 7 …
7 loop:
12 else if (update D(V, Y) received from V)
for all destinations Y do
if (destination Y through V)
D(A,Y) = D(A,V) + D(V, Y);
else
D(A, Y) = min(D(A, Y),
D(A, V) + D(V, Y));
18 if (there is a new minimum for dest. Y)
send D(A, Y) to all neighbors
20 forever
Node C
Node D
Dest.
Cost
NextHop A 3 B 1 D
Dest.
Cost
NextHop A 4 C B 2 1
Nothing changes  algorithm terminates

19. Link Cost Changes “good news travels fast” B A C 1 4 1 50 Node B D C N
7 loop:
8 wait (link cost update or update message)
9 if (c(A,V) changes by d)
for all destinations Y through V do
D(A,Y) = D(A,Y) + d
12 else if (update D(V, Y) received from V)
for all destinations Y do
if (destination Y through V)
D(A,Y) = D(A,V) + D(V, Y);
else
D(A, Y) = min(D(A, Y), D(A, V) + D(V, Y));
18 if (there is a new minimum for destination Y)
send D(A, Y) to all neighbors
20 forever 1 B 4 1 A C 50
Node B D C N A 4 1 B D C N A 1 B D C N A 1 B D C N A 1 B
“good
news
travels
fast”
Node C D C N A 5 B 1 D C N A 5 B 1 D C N A 2 B 1 D C N A 2 B 1
time
Link cost changes here
Algorithm terminates

20. Count-to-Infinity Problem
7 loop:
8 wait (link cost update or update message)
9 if (c(A,V) changes by d)
for all destinations Y through V do
D(A,Y) = D(A,Y) + d
12 else if (update D(V, Y) received from V)
for all destinations Y do
if (destination Y through V)
D(A,Y) = D(A,V) + D(V, Y);
else
D(A, Y) = min(D(A, Y), D(A, V) + D(V, Y));
18 if (there is a new minimum for destination Y)
send D(A, Y) to all neighbors
20 forever 60 B 4 1 A C 50
Node B D C N A 4 1 B D C N A 6 1 B D C N A 6 1 B D C N A 8 1 B
“bad
news
travels
slowly”
Node C D C N A 5 B 1 D C N A 5 B 1 D C N A 7 B 1 D C N A 2 B 1 …
time
Link cost changes here; recall that B also maintains shortest
distance to A through C, which is 6. Thus D(B, A) becomes 6 !

21. Split Horizon
Don’t send the routes learned from each neighbor to that neighbor A C 1 4 50 B 60
Node B D C N A 4 1 B D C N A 60 1 B D C N A 60 1 B D C N A 51 1 B D C N A 51 1 B
Node C D C N A 5 B 1 D C N A 5 B 1 D C N A 50 B 1 D C N A 50 B 1 D C N A 50 B 1
time
Link cost changes here; C won’t sent to B, D(C, A) = 5
since C has learned about this path from B itself. Thus
D(B,A) becomes 60.
Algorithm terminates

22. Poison Reverse
Stronger version of split horizon: if a node X has learned a route to Y from V, it sends to V, D(X, Y) = ∞  make sure that V won’t use X to reach Y!
Does this completely solve the count to infinity problem?

23. Is “Count To Infinity” Solved?60
No! Split horizon and poison reverse only work for loops involving two nodes. B 4 1 A C 1 50 2
Node B D D C N D 1 C 8 A N 2 B 5 A 4 A C 1 C D 1 D
Exercise: continue…
Note: you need also to
write C’s routing table
Node D D C N A 5 B B 1 B C 2 C
time
Link cost changes here; in a previous iteration D told B that it can reach A through B with cost 7 (split-horizon is not invoked!) so B concludes that it can reach A through D with cost 8!