08-routing-1-slides.pdf

Internet Technology
08. Routing
Paul Krzyzanowski
Rutgers University
Spring 2016
1
March 21, 2016 CS 352 © 2013-2016 Paul Krzyzanowski

Routing algorithm goal
2
switch
LAN
LAN
router
router
Routing algorithm: given routers connected with links,
what is a good (best?) path from a source to a destination router
good = least cost
cost = time or money
1st hop router =
source router
last hop router =
destination router

Routing graphs, neighbors, and cost
3
2
1
3
3
1
1
2
5
5
u
v w
z
y
x
Graph G = (N, E) N = set of nodes (routers)
E = set of edges (links)
Each edge = pair of connected nodes in N
Node y is a neighbor of node x if (x, y) ∈ E
Cost Each edge has a value representing the cost of the link
c(x, y) = cost of edge between nodes x & y
if (x, y) ∉ E, then c(x, y) = ∞
We will assume c(x, y) = c(y, x)
2

Path cost, least-cost path, & shortest path
4
2
1
3
3
1
1
2
5
5
u
v w
z
y
x
A path in a graph G = (N, E) is a sequence of nodes (x1, x2, …, xp)
such that each of the pairs (x1, x2), (x2, x3), …, (xp-1, xp) are edges in E.
The cost of a path is the sum of edge costs: c(x1, x2), c(x2, x3), …, c(xp-1, xp)
There could be multiple paths between two nodes, each with a different cost.
One or more of these is a least-cost path.
Example: the least-cost path between u and w is (u, x, y, w) ⇒ c(u, x, y, w) = 3
If all edges have the same cost, then least-cost path = shortest path
2

Algorithm classifications
Global routing algorithms
– Compute the least-cost path using complete knowledge of the network
– The algorithm knows the connectivity between all nodes & costs
– Centralized algorithm
– These are link-state (LS) algorithms
Decentralized routing algorithms
– No node has complete information about the costs of all links
– A node initially knows only its direct links
– Iterative process: calculate & exchange info with neighbors
• Eventually calculate the least-cost path to a destination
– Distance-Vector (DV) algorithm
5

Additional algorithm classifications
• Static routing algorithms
– Routes change very slowly over time
• Dynamic routing algorithms
– Change routing paths as network traffic loads or topology change
• Load-sensitive algorithms
– Link costs vary to reflect the current level of congestion
• Load-insensitive algorithms
– Ignore current or recent levels of congestion
6

Link-State (LS): Dijkstra’s Algorithm
• Assumption:
Entire network topology & link costs are known
– Each node broadcasts link-state packets to all other nodes
– All nodes have an identical, complete view of the network
• Compute least-cost path from one node to all other nodes
in the network
• Iterative algorithm
– After k iterations, least-cost paths are known to k nodes
7

Dijkstra’s Algorithm
8
D(v):
cost of least-cost path from source to v
p(v):
previous node (neighbor of v) along the
least-cost path to v
N’:
subset of nodes for which we found the
least-cost path
2
1
3
3
1
1
2
5
5
u
v w
z
y
x
Initialize:
N’ = current node
N’ = { u }
for all nodes v
if v is a neighbor of u
D(v) = c(u, v)
else
D(v) = ∞
step N’ D(v), p(v) D(w), p(w) D(x),p(x) D(y),p(y) D(z),p(z)
0 u 2,u 5,u 1,u ∞ ∞
2

9
D(v):
p(v):
N’:
least-cost path
2
1
3
3
1
1
2
5
5
u
v w
z
y
x
Loop until N’ = N:
Find a node n not in N’ such that D(n) is
a minimum
→ Node x has minimum D(n)
add n to N’
N’ = {u, x}
for each neighbor m of n not in N’:
for each neighbor of node x
D(m) = min( D(m), D(n) + c(n, m) )
new cost = old cost or cost through x
if D(m) changed, set p(m) = n
0 u 2,u 5,u 1,u ∞ ∞
1 ux 2,u 4,x 2,x ∞
Skip: u ∈ N’
Cost to v is
not better
through x
2
Cost to w
is better
through x
We now
have a
path to y
Ignore x;
it is in N’
Least cost path

10
D(v):
p(v):
N’:
least-cost path
2
1
3
3
1
1
2
5
5
u
v w
z
y
x
find n not in N’ such that D(n) is a
minimum
→ Nodes v & y have minimum D(n)
Pick any one: we choose y
add n to N’
N’ = {u, x, y}
for each neighbor of node y
D(m) = min( D(m), D(n) + c(n, m) )
0 u 2,u 5,u 1,u ∞ ∞
1 ux 2,u 4,x 2,x ∞
2 uxy 2,u 3,y 4,y
2
Cost to w
is even
better
through y
We now
have a
path to z
Skip: x and y
are in N’
Least cost path

11
D(v):
p(v):
N’:
least-cost path
2
1
3
3
1
1
2
5
5
u
v w
z
y
x
minimum
→ Node v has minimum D(n)
add n to N’
N’ = {u, x, y, v}
for each neighbor of node v
D(m) = min( D(m), D(n) + c(n,m) )
0 u 2,u 5,u 1,u ∞ ∞
1 ux 2,u 4,x 2,x ∞
2 uxy 2,u 3,y 4,y
3 uxyv 3,y 4,y
2
No improvement
(2+3) ≮ 3
No change: z is
not a neighbor

12
D(v):
p(v):
N’:
least-cost path
2
1
3
3
1
1
2
5
5
u
v w
z
y
x
minimum
→ Node w has minimum D(n)
add n to N’
N’ = {u, x, y, v, w}
for each neighbor of node w
D(m) = min( D(m), D(n) + c(n,m) )
0 u 2,u 5,u 1,u ∞ ∞
1 ux 2,u 4,x 2,x ∞
2 uxy 2,u 3,y 4,y
3 uxyv 3,y 4,y
4 uxyvw 4,y
2
No improvement
(3+5) ≮ 4

13
D(v):
p(v):
N’:
least-cost path
2
1
3
3
1
1
2
5
5
u
v w
z
y
x
minimum
→ Node z is the only one left!
add n to N’
N’ = {u, x, y, v, w, z}
There are no neighbors not in N’!
We’re done
0 u 2,u 5,u 1,u ∞ ∞
1 ux 2,u 4,x 2,x ∞
2 uxy 2,u 3,y 4,y
3 uxyv 3,y 4,y
4 uxyvw 4,y
5 uxyvwz
2

14
D(v):
p(v):
N’:
least-cost path
2
1
3
3
1
1
2
5
5
u
v w
z
y
x
N’ = N:
All nodes are in N’
For each node, we have the total
cost from the source and the
predecessor along that path.
We can look up the predecessor to
find its predecessor
E.g., least-cost path from u → y
is u → x → y
2
0 u 2,u 5,u 1,u ∞ ∞
1 ux 2,u 4,x 2,x ∞
2 uxy 2,u 3,y 4,y
3 uxyv 3,y 4,y
4 uxyvw 4,y
5 uxyvwz (1) y is w’s
predecessor
(2) x is y’s
predecessor
(3) u is x’s
predecessor

15
D(v):
p(v):
N’:
least-cost path
2
1
3
3
1
1
2
5
5
u
v w
z
y
x
We can create a forwarding table
that stores the next hop on the
least-cost route
Forwarding
table for
node u
2
0 u 2,u 5,u 1,u ∞ ∞
1 ux 2,u 4,x 2,x ∞
2 uxy 2,u 3,y 4,y
3 uxyv 3,y 4,y
4 uxyvw 4,y
5 uxyvwz
Destination Link
v uv
w ux
x ux
y ux
z ux

Computational cost
– 1st iteration: search n nodes to find the minimum cost node
– 2nd iteration: search n-1 nodes
– 3rd iteration: search n-2 nodes
– nth iteration: search 1 node
– Total of n iterations = n + (n - 1) + (n - 2) + … 1 = (𝑛 − 𝑖)
𝑛
𝑖=0
• We need to search n(n+1)/2 nodes
– Complexity = O(n2)
16

If link cost = load carried on the link
• Link costs are not symmetric
– c(u, v) = c(v, u) only if the same
load flows in both directions
• Example loads
– Load of 1 comes into z for w
– Load of 1 comes into x for w
– Load of e comes into y for w
• When LS is run
– y determines (y→z→w) cost is 1
compared to (y→x→w) cost, which is 1+e
– x determines that x→y→z→w is a lower-cost path
Oscillations with congestion-based routing
17
0
z x
w
y
0
0
1 1+e
e
1
1
e
Initial routing

• After route updates, LS is run again
• x, y, and z detect 0-cost path counterclockwise
18
0
z x
w
y
0
1
0
1
2+e
1
1
1
e
Counterclockwise routing
1
z x
w
y
1+e
0
2+e 0
0
1
1
e
Clockwise routing

1
z x
w
y
1+e
0
2+e 0
0
1
1
e
Clockwise routing
0
z x
w
y
0
1
0 2+e
1
1
1
e
Counterclockwise routing
• After route updates, LS is run yet again
• x, y, and z now detect 0-cost path clockwise
19
1

Avoiding oscillations
• Ensure that not all routers run the LS algorithm at the
same time
– Avoid synchronized routers by randomizing the time when a router
advertises its link state
20

Distance-Vector Routing Algorithm
• Initial assumption
– Each router (node) knows the cost to reach its directly-connected neighbors
• Iterative, asynchronous, distributed algorithm
– Multiple iterations
• Each iteration caused by local link cost change or distance vector update message
from neighbor
– Asynchronous
• Does not require lockstep synchronization
– Distributed
• Each node receives information from one or more directly attached neighbors
• Notifies neighbors only when its distance-vector changes
21

Bellman-Ford Equation
• What it says
– If x is not directly connected to y, it needs to first hop to some neighbor v
– The lowest cost is
(the cost of the first hop to v) + (the lowest cost from v to y)
= c(x, v) + dv(y)
– the least cost path from x to y, dx(y), is the minimum of the lowest cost of
all of x’s neighbors
dx(y) = minv{ c(x, v) + dv(y) }
• The value of v that satisfies the equation is the forwarding table entry
in x’s router for destination y
22
x
y
v'
v’’
c(x,v’)
c(x,v’’)
dv’ (y)
dv’’ (y)

Distance-Vector Routing Algorithm
• At each node x we store:
– c(x, v) = cost for the direct link from x to v for each neighbor v
– Dx(y) = estimate of the cost of the least-cost path from x to y
– Distance Vector is the set of Dx(y) for all nodes y in N
Dx = [ Dx(y): y ∈ N ]
– Distance vectors received from its neighbors
Dv = [ Dv(y): y ∈ N ]
• Each node v periodically sends its distance vector, Dv to its neighbors
– When a node receives a distance vector, it saves it and updates its own distance
vector using the Bellman-Ford equation
Dx(y) = minv{ c(x, v) + Dv(y) } for each node y ∈ N
– If this results in a change to x’s DV, it sends the new DV to its neighbors
Each cost estimate Dx(y) converges to the actual least-cost Dx(y)
23
Least-cost estimates from x to all other nodes y
Set of least-cost estimates from each neighbor v to each node y

Distance-Vector Example
24
2 1
7
y
x z
cost to
from
x y z
x 0 2 7
y ∞ ∞ ∞
z ∞ ∞ ∞
Node x DV table
cost to
from
x y z
x ∞ ∞ ∞
y 2 0 1
z ∞ ∞ ∞
Node y DV table
cost to
from
x y z
x ∞ ∞ ∞
y ∞ ∞ ∞
z 7 1 0
Node z DV table

25
2 1
7
y
x z
cost to
from
x y z
x 0 2 7
y ∞ ∞ ∞
z ∞ ∞ ∞
Node x DV table
cost to
from
x y z
x ∞ ∞ ∞
y 2 0 1
z ∞ ∞ ∞
Node y DV table
cost to
from
x y z
x ∞ ∞ ∞
y ∞ ∞ ∞
z 7 1 0
Node z DV table
cost to
from
x y z
x 0 2 7
y ∞ ∞ ∞
z ∞ ∞ ∞
Node x DV table
cost to
from
x y z
x 0 2 7
y 2 0 1
z ∞ ∞ ∞
Node y DV table
cost to
from
x y z
x 0 2 7
y ∞ ∞ ∞
z 7 1 0
Node z DV table
Node x sends its DV {0, 2, 7} to nodes y and z
c(y, x) = 2 c(z, x) = 7

26
2 1
7
y
x z
cost to
from
x y z
x 0 2 7
y ∞ ∞ ∞
z ∞ ∞ ∞
Node x DV table
cost to
from
x y z
x 0 2 7
y 2 0 1
z ∞ ∞ ∞
Node y DV table
cost to
from
x y z
x 0 2 7
y ∞ ∞ ∞
z 7 1 0
Node z DV table
cost to
from
x y z
x 0 2 3
y 2 0 1
z 7 1 0
Node x DV table
cost to
from
x y z
x 0 2 7
y 2 0 1
z 7 1 0
Node y DV table
cost to
from
x y z
x 0 2 7
y 2 0 1
z 3 1 0
Node z DV table
Node y sends its DV {2, 0, 1} to nodes x and z
Node z sends its DV {7, 1, 0} to nodes x and y
c(x, y) = 2 c(z, y) = 1
From y: c(y,z) is 1
c(x, z) = c(x, y) + c(y,z) =
= 2 + 1 = 3
Less than old value, 7
From y: c(y,x) is 2
c(z, x) = c(z,y) + c(y,x) =
= 1 + 2 = 3
Less than old value, 7
Every update to a
node’s DV also
updates the
forwarding table

27
2 1
7
y
x z
cost to
from
x y z
x 0 2 3
y 2 0 1
z 7 1 0
Node x DV table
cost to
from
x y z
x 0 2 7
y 2 0 1
z 7 1 0
Node y DV table
cost to
from
x y z
x 0 2 7
y 2 0 1
z 3 1 0
Node z DV table
cost to
from
x y z
x 0 2 3
y 2 0 1
z 3 1 0
Node x DV table
cost to
from
x y z
x 0 2 3
y 2 0 1
z 3 1 0
Node y DV table
cost to
from
x y z
x 0 2 3
y 2 0 1
z 3 1 0
Node z DV table
Node x sends its DV {0, 2, 7} to nodes y and z
Node y’s vector did not change – it stays quiet
Node z sends its DV {2, 0, 1} to nodes x and y
We converged. Everyone has the same view of the network. Nobody has updates to send.

Link cost changes
• The DV algorithm remains quiet once it converges
– … until some link cost changes
• If a node detects link cost change between itself and a neighbor
– It updates its distance vector
– If there is a change in the cost of any least-cost path
it informs its neighbors of the new distance vector
– Each neighbor computes a new least cost
• If the value changed from its previous value, it sends its DV to its neighbors
• Recompute until values converge
March 21, 2016 CS 352 © 2013-2016 Paul Krzyzanowski 28

Link loss
29
Suppose we lose the link to C: c(B,C) = ∞
B will send an update to A but A thinks its cost to C is 3
B will think there is a route to C: B→A→C with a cost of (c(B,A) + 3) = 4
1 2
Distance to
C =3
Distance to
C =2
A B C
1 ∞
Distance to
C =3
Distance to
C =4
A B C
Update (A,C)=3
1 ∞
Distance to
C =5
Distance to
C =4
A B C
Update (B,C)=4
This continues ad infinitum!
We created a
Routing Loop

Mitigation: Poison Reverse
• If A routes through B to get to C
– A will advertise to B that its distance is infinity
– B will then never attempt to route through A
• This does not work with loops involving 3 or more nodes!
• Other approaches
– Limit size of network by setting a hop (cost) limit
– Send full path information in route advertisement
• Perform explicit queries for loops
30

The end
31

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