1. ECE 544 Project3 Kush Patel Siddharth Paradkar Ke Dong

2. Assumptions and Address Scheme Assumptions Network contains Routing Controllers Routing Controllers contain info of content files at each host Hosts connected to routers via Ethernet We will have multiple Routing Controllers serving multiple areas of the network Routing Controllers will “talk” to each other to update their tables Other assumptions such as 255 hosts, 255 content files etc. Naming scheme and eventual address scheme Routers: Identify them as 1,2,3… Content: Use ID for each File

3. Bootstrapping and Discovery Methods “CreateEntry()” Send all content id # stored to the Routing Controller upon boot Routing Controller creates a table entry with the end host that has the content file “GetContent()” Default dest. Routing Controller. If Routing Controller does not contain file, Routing Controller performs table search and finds closest host Routing Controller queries “GetContent()” from Host X UpdateContent() Routing Controller runs method to match host content files with table entries periodically

4. Bootstrapping and Discovery Routes Routers use Dijkstra to build table of routes and output ports Hosts have a default router to send requests to No 2 Routers connect to the same host for simplicity

5. Baseline Algorithm Content routing algorithm How are contents advertised? Hosts update content periodically How to route a content-request packet? Via the Routing Controller How is the content actually delivered? UDP Protocol

6. Baseline Algorithm Content routing algorithm How to choose the ‘best’, among multiple hosts having the same content? By default, Host will send request to Routing Controller with its address as destination The default router will check the hop count to the Routing Controller and send a GetContent() broadcast message on all the nodes with a TTL = Distance to Routing Controller – 1 The router will flood its connected hosts for a response, If a router that is closer than to the host than the Routing Controller, it will respond with the packet. The host will discard any duplicate packet it received

7. Baseline Algorithm Content routing algorithm Routing Table for every Router: Router will use the Hop Count of the Routing Controller and send a broadcast with a TTL = Hop Count - 1 IP AddressOutput Port #Hop Count Routing Controller 25

8. Packet Structure IP based packet structure Two separate packet structures based on protocol being utilized. Initially when all of the hosts are being added they will send a packet based on TCP packet structure, it will wait for a ACK from the central Routing Controller. Src AddDest AddrContent Count Payload C1, C2, C3, C4

9. Packet Structure When a transmission from host to host happens, they will employ a UDP scheme for the packet structure. Src AddDest AddrContent ID# Payload

10. Data Transfer and Reliability Message Forward Unicast – For sending contents to Routing Controllers Boradcast – For GetContent Queries() Use UDP for packet transfer – No need for TCP Control. If Packet fails, host requests retransmission TCP for CreateEntry() ARQ Scheme We will want to use a stop and wait ARQ method, where the number of frames being sent are determined by the sliding window size, and the receiver buffers any out of order packets. We believe that because we have focused on reducing the number of hops needed to a transmission, the # of retransmissions required will be very low.

11. Advantages Centralized Routing Controllers More robust. Every host knows to send requests to one place. “Brain” of the network hereby managing nodes and periodic updates. There is not a complete flooding of the network.

12. Disadvantages Centralized Routing Controllers bottleneck. Multiple request from multiple hosts can overload the Routing Controller scheduling Flooding can introduce a network overload

13. Network Architecture ARP Table C1 C6 C4 C5 C9 R3 H1 H2 H3 R1 R2 R4 R5 DHCP Server Routing Controller C3 C2 C7 H4

14. 2 Scenarios Host wants to get file that has a hop count greater than the routing controller. Host wants to get file from a node that is closer than the routing controller.

15. Scenario 1: H2 – GetContent (C3) ARP Table C1 C6 C4 C5 C9 R3 H1 H2 H3 R1 R2 R4 R5 DHCP Server Routing Controller C3 C2 C7 H4 C3 Hop Count to Routing Controller =3 Broadcast with TTL = 2 TTL = 2 TTL = 1 TTL = 0 C3

16. Scenario 2: H2 – GetContent (C4) ARP Table C1 C6 C4 C5 C9 R3 H1 H2 H3 R1 R2 R4 R5 DHCP Server Routing Controller C3 C2 C7 H4 Hop Count to Routing Controller =3 Broadcast with TTL = 2 TTL = 2 TTL = 1 TTL = 0 C4 C3C4