1 © Nokia 2015

Bell Labs

SDN Enabled Path Switching

• Gordon Wilfong Bell Labs

• Joint work with Adiseshu Hari (Bell Labs) and Urs Niesen (Qualcomm)

Bell Labs

2 © Nokia 2015

How is routing currently

done?

Destination-Based Forwarding Table

Destination Output Port

10.1.0.0/24

10.1.2.0/24

10.2.1.0/24

10.3.1.0/24

20.2.0.0/16

20.1.1.0/28

3

2

2

1

2

3

1

2

3

R:

R

20.1.1.0/28

10.2.1.0/24

10.3.1.0/24

3

• Destination-based tables grow as address blocks fragment and increase in number

• Does not allow for flexible policy-based routing

Destination-based Routing Disadvantages

Destination-based Routing Advantages

• Destination-based routing tables populated locally by distributed routing protocols (e.g., BGP or RIP or …)

4

Policy-Based Routing

firewall

codecS

D

5

Policy-Based Routing

firewall

codecS

D

6

Policy-Based Routing

firewall

codecS

D

7

Policy-Based Routing

firewall

codecS

D

8

Flow-Based Forwarding Table

Flow Output Port

Red

Blue

Purple

Green

Black

??

3

2

2

1

3

3

1

2

3

R:

R

9

• Flow-based tables grow as more and more complex policies are introduced

Flow-based Routing Disadvantages

Q: Is there a way to allow for unbounded growth in policy-complexity (i.e., growth in number of flows) while keeping forwarding table size fixed?

Flow-based Routing Advantages

• Allows for flexible policy-based routing

10

11 © Nokia 2015

Our

proposal

1S

• For each node, the outgoing ports are given a distinct binary label

• Each packet header includes path encoded as a sequence of port labels

Path Encoding

1S

1S

1S

10

11

11

1011

11

12

Path Encoding Table

Label Output Port

label1

label2

label3

1

2

3

1

2

3

R:

Rlabel1 …

13

Software defined network

SDN Controller

14

• Takes up space in packet headers

Path Encoding Disadvantages

Path Encoding Advantages

• Table size is bounded by the number of ports

• As flows or destinations increase, table size remains fixed

15

Fixed-length Labels

1S4S 0S 5S 8S

3S2S 7S 6S 9S

01

00

10

11

10 01

00

1

0

Longest path encoding from to a leaf is 5,

e.g., encoding 00000 of path 8510 ,,, SSSS

0S

In the next few slides, we consider paths from to all leaf nodes. 0S

A node with k output ports uses labels of length klog

16

Variable-length Binary Labels

1S4S 0S 5S 8S

3S2S 7S 6S 9S

100

0

101

10

11 10

0

1

0

Longest path encoding from to a leaf has length 3

e.g., encoding 000 of path .

How to unambiguously decode a path label?

8510 ,,, SSSS

0S

17

• Path labels may not be unique

• and both have label 01

Decoding Variable-length Binary Labels: Uniqueness

1S4S 0S 5S

3S2S 7S 6S

10

0

0

01

11 01

1

510 ,, SSS 40 ,SS

18

• Path labels may not be locally decodable even if encodings are unique• does not know that sending packet out port labeled 01

will not end at leaf

• is unique path to leaf with label 01 and so should send

packets labeled 01 on port labeled 0

• Nodes also should not have to know labels at other nodes to decode

Decoding Variable-length Binary Labels: Local

1S4S 0S 5S

3S2S 7S 6S

01

0

0

10

11 01

1

510 ,, SSS

0S

0S

19

Prefix-free Labeling (PFL)

1S4S 0S 5S 8S

3S2S 7S 6S 9S

100

0

101

11

11 10

0

1

0

PFL: At each node, no label is a prefix of another label at that node.

• PFL implies distinct path encodings for paths from same source

• PFL implies unambiguous decoding at each node with only local information

20

Computing a prefix-free labeling

21

Preliminaries

0S

1S 2S

3S

1a

},)(:{)( SatailAaSout

},,{)( 3210 aaaSout

Def:

2a

3a

),( AVD Directed graph:

VS

22

Arc Label Lengths

0S

1S 2S

3S

1aL2aL

3aL

Def: Length of binary label of arc a = aL

23

Kraft’s Inequality

12

1

1

ki

Li

A code with codewords having

lengths is said to satisfy Kraft’s Inequality if: kLLL ,...,, 21

kwww ,...,, 21

24

Tree Codes

0

10

0

0

0 0 01 1

1

1

1 1

= 110

= 01

Prefix-free if no two chosen nodes on the same path to the root

25

Prefix-free Code Kraft’s Inequality

Theorem: A prefix-free code with codewords having lengths

exists if and only if the lengths satisfy Kraft’s Inequality.

Proof: Assume wlog .

Suppose lengths satisfy Kraft’s Inequality.

Start with full binary tree of depth .

At step i: choose a node at depth to determine codeword and remove all descendants.

Must show this is always possible for steps i:

},...,,{ 21 kLLL

kLLL ...21

kL

iL

ki 1

26

Example

0

10

0

0

0 0 01 1

1

1

1 1

= 01

2iL

27

Example

0

10

0

0

0 01 1

1

1

1

= 01

Remove descendants of

28

Example

0

10

0

0

0 01 1

1

1

1

= 110

= 01

31 iL

29

Proof continued

We show that at step i there is at least one remaining leaf at depth .

Number of remaining leaves at step i is:

iL

0212212221

1

1

1

1

k

j

LLi

j

LLi

j

LLL jkjk

jkk

Kraft’s Inequality

30

Proof continued … again

1222111

k

i

LLk

i

LLk

i

iikikS

Kraft’s Inequality

• Suppose it is a prefix-free code.

• Consider a complete binary tree of depth .

• A length codeword has a set of leaves under it where

and ∅ for .

kL

ik LL

iS

2iL iS

ji SS ji

total number of leaves

31

Definition: A valid labeling is a labeling of the arcs in A where if is thelength of the label of arc a then for each the label lengths of outgoing arcs

satisfy Kraft’s Inequality

Definition: The length of a path P is

Definitions

12)(

vouta

La

aL

Pa

aLPL )(

Vv

32

Given a digraph D=(V,A) and a set of paths P in D.

Find a valid labeling of the arcs of A to minimize .

Problem statement

kiPi 1:

kiPL i 1:)(max

We call this the optimal path encoding problem.

33

Examples

v 2

1

31

8

1

8

1

4

1

2

1

3

v 2

2

21

4

1

4

1

4

1

4

1

2

34

Optimal Path Encoding is APX-hard

Theorem: Approximating the optimal path-encoding problem to within a factor better than 8/7 is NP-hard.

Proof: Reduction from SAT where each clause has 2 or 3 literals and eachvariable is involved in at most 3 clauses [Tovey84]. => Each literal in 1 or 2 clauses

__,,

_,_,

__,,

3

2

1

i

i

i

xC

xC

xC

__,,

__,,

_,_,

__,,

3

3

2

1

i

i

i

i

xC

xC

xC

xC

or

35

Gadget for variable

i id

ix

ix

1

it

2

it

1

if

2

if

ix

36

Gadget for clause

id

ix

ix

1

it

2

it

jf i 1

2

if

)xxx(C kji

kt j 1

jd

jx

jx

2

jt

1

jf

2

jf

i

kd

kx

kx

1

kt

2

kt

1

kf

2

kf

CP

37

Gadget for clause

id

ix

ix

1

it

2

it

jf i 1

2

if

)xxx(C kji

kt j 1

jd

jx

jx

2

jt

1

jf

2

jf

i

kd

kx

kx

1

kt

2

kt

1

kf

2

kf

1

12

11

1

C satisfied by 7,, Ckji Lfalsextruextruex

1

1

2

2

2

2

2

)(: CC PLengthL

38

Gadget for clause

id

ix

ix

1

it

2

it

jf i 1

2

if

)xxx(C kji

kt j 1

jd

jx

jx

2

jt

1

jf

2

jf

i

kd

kx

kx

1

kt

2

kt

1

kf

2

kf

1

12

12

1

C not satisfied

2

2

2

1 1

1

2

8,, Ckji Lfalsexfalsextruex

39

Dummy gadget for variable

'iy'iu

'iz'iv

ix

'i

40

Gadget for clause

id

ix

ix

1

it

2

it

jf i 1

2

if

)xx(C ji

'1 jt j

jd

jx

jx

2

jt

1

jf

2

jf

i

CP

'ju

'jv

'jy

'jz

41

Optimal Path Encoding: Convex Minimization Problem

AaZL

ZL

Vv

PLL

L

a

vouta

L

i

Pa

a

,

,12

,

:subject to

min

)(

a

i

P

AaLa ,0

I

a

I LL ,Optimal solution:

(Kraft)

42

Relaxed version

AaRL

RL

Vv

PLL

L

a

vouta

L

i

Pa

a

,

,12

,

:subject to

min

)(

a

i

P

** , aLLOptimal solution:

0))2ln(exp(ln)(

vouta

aLor

Geometric program(efficiently solved by interior point methods)

43

Rounding Relaxed Solution

Define R

aL *

aL Aa,

pa

R

ap

R LLP

max

44

Solutions

** , aLL

I

a

I LL ,

R

a

R LL ,

Optimal integer solution (what we want)

Optimal relaxed solution (what we can get)

Rounded optimal relaxed solution (settle for this)

45

If wlog we can assume

Useful Fact

Lemma:

Proof: only if a is the only outgoing arc at tail(a).

0I

aL 0* aL

0I

aL

0I

aL 0*

aLFact:

46

Rounding Gives a 2-Approx

Theorem:

Proof:

IRI LLL 2

122)()(

*

vouta

L

vouta

L aRa

(hence valid labeling)

Let p be a path where

I

pa

I

a

I

I

a

a

pa

*

a

pa

R

a

R

L

LL

LpaL

LpaL

LL

2

0:

0:

*

*

(By Lemma)

Rounded solution is valid and

47

Experimental Results

48

Gradient Descent Algorithm

• An entropy form of the dual can be obtained (see paper)• Dual used to derive “simple” projected gradient descent algorithm

Graph FixedLength

GDA Nodes Edges Paths

AT&T MPLS backbone network 15 bits 10 bits 25 224 600

AS 4323 (Rocketfuel database) 25 bits 17 bits 51 284 2550

49

Note 1: Finer Grained Model

v

v

0

0

0

0

0

50

Note 1: Finer Grained Model

v

v

10

0

0

0

1

1

0

1

12

2

51

Conclusions

• Introduced and formalized the optimal path-encoding problem

• Showed that optimal path encoding is APX-hard

• Presented a 2-approximation algorithm

• Experiments show > 30% reduction in longest path encoding in real networks

52

Future Work

• Close gap between 2-approx and 8/7 lower bound on approx.• What about multicast?

53

Thank you!

54