- Paper By Yunfeng Lin, Baochun Li, Ben Liang

OutlineMotivation

Constraints in DTN Network coding Vs Replication Binary Spraying Vs Epidemic routing

Protocol Principle Assumptions Description Example

Performance Average transmission / relay Transmission delay Relay buffer size Optimal Maximal Spray count Vs Identical Maximal spray

count

Motivation – Constraints in DTNOpportunistic connections between nodesNodes have limited transmission capabilitiesBuffer space limitationsBattery power limitedNodes might be mobileDelay in packet delivery will be largeNode density is sparse

Motivation – NC Vs ReplicationReplication – Transmitting copies of a packet.Network coding – combining packets and transmitting the resultant packet

Example:

K = 3, K – Number of source packets, namely, A, B and C.

Number of encoded packets, K’ = 3

Buffer Space of each relay, BS = 2

Number of nodes = 4, which includes one source, one destination. Number of relays is two.

Motivation – Replication Example

A

B

C

A

B

A

B

R1

R2

DESTINATION

Motivation – Replication ExampleR1

R2

A

B

C

A

C Replaces B

A

Here also C replaces B

DESTINATION

ONLY A AND C ARE RETRIEVED. FOR, B MORE DELAY IS

INCURRED

Motivation – NC Example

A + B + C

A + 2B + 3C

2A + 2B + C

A + B + C

A + 2B + 3C

A + B + C

A + 2B + 3C

R1

R2

DESTINATION

Motivation – NC ExampleR1

R2

A + B + C

A + 2B + 3C

2A + 2B + C

3A +3B +2C, x = 1

3A + 4B + 4C, x = 1

5A + 5B + 3C, x = 2

5A + 6B + 5C, x = 2

DESTINATION

ANY THREE PACKETS CAN HELP RETRIEVE A, B AND C

Motivation – Binary Spraying Vs ER

S, K = 2

R1

R2

R3

R4

D

Epidemic routing

Binary spraying

S, K = 2, L = 2

R1, L = 1

R2, L = 1

R3

D

R4R5

Protocol - PrincipleThis protocol called the E-NCR, is a combination of

Network coding and Binary spraying.

NCER Binary Spraying

E-NCR

ER

NCER – Network Coding based Epidemic routingER – Epidemic RoutingE-NCR - Efficient Network coding based routing

Protocol - AssumptionsThere is once source with ‘K’ packets to be transmitted, ‘n’

relay nodes and a destinationFor every opportunistic contact, only one packet can be

transmitted.The source ‘S’ and the Destination ‘D’ have enough buffer

space to hold all ‘K’ packets. But relay nodes have buffer space B, defined as 1 <= B <= K

While a packet transmission is made, no other back-ground traffic exists.

A packet in the buffer in an node is purged as soon as a ACK is received from ‘D’ or Time-to-live field is zero.

Protocol - DescriptionSOURCE-RELAY:

K' = K + some more encoded packets L = c * log k, where c is some constanti = 0;S = K';do{ if(detect any node and <i,l> not already there with that node) { send an encoded packet <i, L, co-efficients, packet>

i++; }

}while(S != i);

Protocol - DescriptionRELAY-RELAY, SENDER SIDE:

do{ if(detect any node X) {

get spray list of X; //list element is a tuple <i, l>,where i is index of packet, ‘l’ is the //remaining spray count

do {

compare this->spraylist with x->spraylist;if(any this->spraylist-><i, l> such that l >=0 and i does not exist in x->spraylist)

{send encoded packet <i, floor(l/2)> to node x;update tuple <i, l> to <i, ceil(l/2)>;}

}while(end of x->spraylist); }}while(true);

Protocol - DescriptionDESTINATION:

do{

if(got a packet){

add to packet listtry to decode list of

packets;if(decode possible){ exit loop;}

}

}while(true);

RELAY-RELAY, RECEIVER SIDE:

if(packet received){

if(buffer size == max_buffer_size){encode incoming packet with all packets in list;}else{place packet in free slot;}add <i,l> of incoming packet to spray list;

}

Protocol - ExampleIn the following example,

There are 7 nodes. One Source, One destination and 5 relays

The buffer space is 1 for the relay nodes and 3 for the source and destination

Number of packets to be transmitted from source is k = 3

L, the maximal spray count is set to log2(k) which is approximately 2 and the threshold is T = 1.

In the first level, l, the spray count reduces to L/2 which is 1. Since, this is the threshold, all future transmissions are one per packet.

Protocol - Example

E1, E2, E3

E2,

L = 2

E2

E3

E1,

L = 2

E1

E1 + (X) E2

L = 1

E3 + (Y) E1

L = 1

E1 + (X) E2 E3 + (Y) E1

E3 + (Z) E2

E3,

L = 2

Performance - Avg transmission per relay

Performance - Transmission delay

Performance - Relay buffer size

Performance – Optimal Vs Identical Maximal spray count

Some LimitationsDestination has to wait till minimum of ‘K’

encoded packets have to be receivedSome packets which have linear dependence

could arise during encoding at relays.

AcknowledgementsMy Audience for your comments, time and

patience.Prof. Chan Mun ChoonDr. A. V. BabuManjunathChen Bin BinThe Authors of the paper