1. Network Layer (contd.) Routing

2. Packet transfer in network
source
message M
application
transport
network
link
physical
segment Ht M Ht
datagram Ht Hn M Hn
frame Ht Hn Hl M
link
physical
Encapsulation
switch
destination
network
link
physical Ht Hn M
message
application
transport
network
link
physical Ht Hn Hl M M Ht Hn M Ht M Ht Hn M
router Ht Hn Hl M
Lecture 3

3. Network-Layer Functions
Forwarding: move packets from router’s input to router output
Image courtesy:
Network Layer

4. Routing Routing Goal: determine “good” paths
Image courtesy:
Routing
Goal: determine “good” paths
(sequences of routers) through
network from source to dest.
application
transport
network
data link
physical
application
transport
network
data link
physical ??
1. Send data
2. Receive data ?? ??

5. Routing: Graph formulation
Graph abstraction for the routing problem:
graph nodes are routers
graph edges are physical links
links have properties: delay, capacity, $ cost, policy u y x w v z 2 1 3 5
Graph: G = (N,E)
N = set of routers = { u, v, w, x, y, z }
E = set of links ={ (u,v), (u,x), (v,x), (v,w), (x,w), (x,y), (w,y), (w,z), (y,z) }

6. Graph abstraction: costsu y x w v z 2 1 3 5
c(x,x’) = cost of link (x,x’)
- e.g., c(w,z) = 5
cost could always be 1, or
inversely related to bandwidth,
or inversely related to
congestion
Cost of path (x1, x2, x3,…, xp) = c(x1,x2) + c(x2,x3) + … + c(xp-1,xp)
Network Layer

7. Key Desired Properties of a Routing Algorithm
Simplicity
Robustness
Optimality
find good path (for user/provider)
Robustness: a network is expected to run continuously for many years, with changes in topology and traffic
Optimality: depends on whose point of view: users: networks

8. Routing Design Space Routing has a large design space
- Robustness
- Optimality
- Simplicity
Routing has a large design space
who decides routing?
source routing: end hosts make decision
network routing: networks make decision
how many paths from source s to destination d?
multi-path routing
single path routing
will routing adapt to network traffic demand?
adaptive routing
static routing …

9. Routing Design Space: example network routing
- Robustness
- Optimality
- Simplicity
Routing has a large design space
who decides routing?
source routing: end hosts make decision
network routing: networks/routers make decision
- Based on information available from the network
how many paths from source s to destination d?
multi-path routing
single path routing
will routing adapt to network traffic demand?
adaptive routing
static routing …

10. Routing Algorithm design based on available information
Global or decentralized information?
Global:
all routers have complete topology, link cost info
“link state” algorithms
Decentralized:
router knows physically-connected neighbors, link costs to neighbors
iterative process of computation, exchange of info with neighbors
“distance vector” algorithms u y x w v z 2 1 3 5
Network Layer

11. Link state (LS) routing algorithm
Network Layer

12. Link-State Routing Algorithm
Network topology, link costs known to all nodes
accomplished via “link state broadcast”
all nodes have same info
Problem
Given a topology, link costs, and a source-destination (S-D) pair, determine a route from S to D so that the route has the minimum cost (i.e., is the shortest).
Solution: Dijkstra’s shortest-path algorithm u y x w v z 2 1 3 5
Network Layer

13. A Link-State Routing Algorithm: Dijkstra’s Algorithm
Parameters:
c(x,y): link cost from node x to y; = ∞ if not direct neighbors
D(v): current value of cost of path from source to dest. v
p(v): predecessor node along path from source to v
N': set of nodes whose least cost path definitively known
Network Layer

14. Dijkstra’s algorithm: exampleu y x w v z 2 1 3 5
Shortest paths:
u  x: 1,u
u  v: 2,u
u  y: 2,x
u  w: 3,y
u  z: 4,y
Step 1 2 3 4 5 N' u ux
uxy
uxyv
uxyvw
uxyvwz
D(v),p(v)
2,u
D(w),p(w)
5,u
4,x
3,y
D(x),p(x)
1,u
D(y),p(y) ∞
2,x
D(z),p(z) ∞
4,y
Network Layer

15. Dijkstra’s algorithm: example (2)
Resulting shortest-path tree from u:
Shortest paths:
u  x: 1,u
u  v: 2,u
u  y: 2,x
u  w: 3,y
u  z: 4,y u y x w v z
Resulting forwarding table in u: v x y w z
(u,v)
(u,x)
destination
link
Network Layer

16. Dijkstra’s Algorithm 1 Initialization: 2 N' = {u} 3 for all nodes v
if v adjacent to u
then D(v) = c(u,v)
else D(v) = ∞ 7
8 Loop
9 find w not in N' such that D(w) is a minimum
10 add w to N'
11 update D(v) for all v adjacent to w and not in N' :
D(v) = min( D(v), D(w) + c(w,v) )
13 /* new cost to v is either old cost to v or known
shortest path cost to w plus cost from w to v */
15 until all nodes in N'
Network Layer

17. Let’s solve this example!
Determine shortest paths from R1 to all other nodes with corresponding predecessor nodes!
Network Layer

18. Distance Vector routing algorithm
Network Layer

19. Distance Vector Routing
Practically entire network topology (each link state) may not be known accurately
Only the information to a direct neighbor may be known
Distance-vector routing algorithm operates by
Each router trusting its direct neighbors
Each router maintain a table (vector) estimating
the best known distance to each destination
which line to use to get there
Tables (vectors) are updated by exchanging information with the direct neighbors

20. Distance Vector Routing: Basic Idea
At node i, the basic update rule
where
- di denotes the distance estimation from i to the destination,
- N(i) is set of neighbors of node i, and
- dij is the distance of the direct link from i to j
- dj denotes the distance estimation feedback from j to i
Based on Bellman-Ford algorithm
destination j i

21. Distance Vector Algorithm
Each node:
wait for (change in local link cost or msg from neighbor)
recompute estimates
if DV to any dest has changed, notify neighbors
Iterative & Distributed
neighbors then notify their neighbors and so on…
Network Layer

22. D-V Algorithm example z y x node x table x y z x y z 0 2 7 ∞ from ∞
from
cost to
node y table
cost to x z 1 2 7 y
x y z x ∞ ∞ ∞ y
from z ∞ ∞ ∞
node z table
cost to
x y z x
∞ ∞ ∞
from y ∞ ∞ ∞ z 7 1
time
Initially:
Network Layer

23. z y x Dx(z) = min{c(x,y) + Dy(z), c(x,z) + Dz(z)} = min{2+1 , 7+0} = 3
Dx(y) = min{c(x,y) + Dy(y), c(x,z) + Dz(y)} = min{2+0 , 7+1} = 2
node x table
x y z x y z ∞
from
cost to
cost to
x y z x 2 3
from y z
node y table
cost to x z 1 2 7 y
x y z x ∞ ∞ ∞ y
from z ∞ ∞ ∞
node z table
cost to
x y z x
∞ ∞ ∞
from y ∞ ∞ ∞ z 7 1
time
Network Layer

24. z y x node x table x y z x y z 0 2 7 ∞ from cost to cost to x y z x ∞
from
cost to
cost to
x y z x
from y z
node y table
cost to
cost to x z 1 2 7 y
x y z
x y z x ∞ ∞ x ∞ y
from y
from z z ∞ ∞ ∞
node z table
cost to
cost to
x y z
x y z x x
∞ ∞ ∞
from y
from y ∞ ∞ ∞ z z 7 1
time
Network Layer

25. z y x node x table x y z x y z 0 2 7 ∞ from cost to cost to cost to ∞
from
cost to
cost to
cost to
x y z
x y z x x
from y
from y z z
node y table
cost to
cost to
cost to x z 1 2 7 y
x y z
x y z
x y z x ∞ ∞ x ∞ x y
from y
from
from y z z ∞ ∞ ∞ z
node z table
cost to
cost to
cost to
x y z
x y z
x y z x x x
∞ ∞ ∞
from y
from y
from y ∞ ∞ ∞ z z z 7 1
time
Network Layer

26. Distance Vector: link cost changes
node detects local link cost change
updates routing info, recalculates distance vector
if DV changes, notify neighbors x z 1 4 50 y
At time t0, y detects the link-cost change, updates its DV,
and informs its neighbors.
“good
news
travels
fast”
At time t1, z receives the update from y and updates its table.
It computes a new least cost to x and sends its neighbors its DV.
At time t2, y receives z’s update and updates its distance table.
y’s least costs do not change and hence y does not send any
message to z.
Network Layer

27. Distance-Vector Algorithms: Link cost changes
A goes down or
line between
A and B is down
Bad news propagate slowly
This is called the counting-to-infinity problem
Distances to A

28. Solution: Split Horizon Hack
Actual distance to a destination is not reported on the line on which packets to that destination are sent.
Instead these distances are reported as
“infinity: Poisoned reverse”
Will this completely solve count-to-infinity problem? B D A C
C tells D the truth
about its distance to
A, but lies to B and
says the distance is
infinity.

29. A topology where split horizon fails
Suppose that D becomes
unreachable from C.
A and B are reporting
infinite distances to C, but
they are reporting distances
of length 2 to each other.
A and B will count to infinity.

30. Comparison of LS and DV algorithms
Message complexity
LS: with n nodes, E links, O(nE) msgs sent
DV: exchange between neighbors only
convergence time varies
Speed of Convergence
LS: O(n2) algorithm requires O(nE) msgs
may have oscillations
DV: convergence time varies
may be routing loops
count-to-infinity problem
Robustness: what happens if router malfunctions?
LS:
node can advertise incorrect link cost
each node computes only its own table
DV:
DV node can advertise incorrect path cost
each node’s table used by others
error propagate through network
Network Layer