Streaming in Peer-to-peer Networks 2002/1/2

Outline The problem Assumption Network model / Node Construction –Flood-broadcast –Tail-broadcast –Leave-broadcast Simulation Future work

The Problem Given a graph G = (V,E) with order- vertexes and a distinguished source vertex s, we find a way to connect all the vertexes without violating the order, which means we have to connect the smaller order vertexes before the larger ones. Constraints: –Minimize the maximum of order-difference –Minimize the sum of order-difference –Minimize the construction cost

An Example Step 1Step 2Step 3Step 4 MAX ： 2 SUM ： Order-difference: ID – Max(Children’s ID)

Assumption Without messages, all nodes knows nothing about other nodes’ condition except the paths to connect to its neighbor nodes. Only in-stream nodes can actively tell other nodes to join the stream. If a node receives a message, it can directly response the message or forward the message to its neighbors according to the message’s TTL attribute.

Network Model Distributing n nodes across a Cartesian coordinate grid. The edge probability function is,where d(u,v) is the Euclidean distance, L is the maximum possible distance between two nodes.

A Graph Example

Node A node has following attributes: –(x,y) –Neighbor nodes’ ID –ID (also used as joined order) –Stream parent ID –Stream child IDs (order-difference) –Message parent ID –Message TTL –The sent messages number

Construction The ways to connect all the nodes: –Flood-broadcast –Tail-broadcast –Leave-broadcast

Flood-broadcast Every node in the stream can broadcast messages to all its neighbor nodes

Tail-broadcast Only the lastly in-stream nodes can broadcast messages Tail-1 case Tail-2 case

Leaves-broadcast Only the non-child nodes can broadcast messages

Leaves-broadcast Vs. Tail-broadcast (1) Tail-broadcast has local problem and does not spread well. Leaves-broadcast has a trend to result in a tree graph but at the same time occurs larger maximum and sum of order-difference.

Leaves-broadcast Vs. Tail-broadcast (2) 1n-1n … i j n+1 Tail-broadcast 1n-1n … i j n+1 Leave-broadcast

How to Simulation (1) Main loop ： for i = 1 to NUM_OF_NODES do ResetMessageParent() SetBroadcastNodes(BROADCASTWAY) SetInStream()

How to Simulation (2) ResetMessageParent() ： –Reset every node’s message server to null (clear history) –To avoid broadcasting messages back

How to Simulation (3) SetBroadcastNodes(BROADCASTWAY) ： –To setup which nodes to broadcast messages to others. –BROADCASTWAY: Flood-broadcast Leaves-broadcast Tail-broadcast

How to Simulation (4) SetInStream() ： –To set the node which wants to join the movie into the chaining stream –Select a parent node to join

How to Simulation (5) Node’s function ： –BroadcastMessage() –ReceiveMessage() –ResponseMessage()

How to Simulation (6) BroadcastMessage() ： –If the node is set to broadcast message, the node call this function to broadcast messages to all its neighbor nodes.

How to Simulation (7) ReceiveMessage() ： –If a node receive a message, it first store the message server’s ID, and see if it want to join the movie. –If a node want to join the movie, it stores the stream parent ID. –If not, it looks the TTL field and compare the value received last time to determine if it need to broadcast messages forward.

How to Simulation (8) ResponseMessage() ： –A node can determine which movie server to connect with if it had received messages before.

Simulation Result (1) Flood-broadcast

Simulation Result (2) Leaves-broadcast

Simulation Result (3) Tail-3-broadcast

Simulation Result (4) In-stream node number (total 100 nodes) TTL1234 Flood Leaves Tail Messages cost TTL1234 Flood19,478273,9413,094,24927,166,605 Leaves9567,80127,260148,592 Tail-32,63420,51858,193108,506 Max/Sum of order-difference TTL1234 Flood19/25619/28319/2879/171 Leaves1/91/249/6419/262 Tail-33/243/6514/13012/139

Future Work Other construction ways Different graph, ex: hierarchical graph History messages