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Delay and Routing in Computer Networks

## Calculating the Minimum Delay to Transfer a 512KB File using Normal Routers

The minimum delay to transfer a 512KB file from A to B using normal routers will be: (1) 512KB file can produce (512X1024/512)=1024 packets of which each has a 512 bytes payload and 64 bytes header. Each packet size is 512+64=576 bytes.

(2) There are three routers in the shortest path, and the minimum delay to build a path between A and B is 0.15ms, and the bandwidth is 100Mbps.

(3) There are 3 routers between A and B, so the number of hops that the 1st packet jumps is 4.

(4) The delay to transmit the first packet from A to B via 4 hops will be Delay in transmission: 4X (576X8)/100X106=0.18432ms Delay in path setup: 0.15ms Then total delay will be 0.18432ms+0.15ms=0.334ms

(5) The delay to transmit the following 1023 packets from A to B will be: For each of them, delay in transmission for the last hop is

(576X8)/100X106=0.04608ms

(6) For each packet, the minimum delay to set up a path is 0.15ms Then total delay to transmit 1023 packets will be 1023X(0.04608ms+0.15ms)=200.589ms

(7) The minimum delay to transmit the 512KB file will be

a)

The cost of sending one packet to 4 receivers via multicast is 8. Following the reverse path forwarding mechanism, packets arrived at the right interface of a multicast router will be forwarded, otherwise will be dropped. In this way, in multicasting, a packet traverses a network link exactly once. There are 8 links, so the total cost is 8 units.

If the packet is sent over the network via unicast to all receivers, then each of the 4 receivers is sent an individual packet. Each packet must travel across 4 links from the source to any receiver. Hence the number of links is 4*4=16. As a result, the cost is 4*4=16.

b)

The RSVP requests of R1 (20kbs) and R2 (100kbps) merge at Router C, Router C needs to reserve a minimum of 100kbps which satisfies both R1 and R2 requests. The RSVP requests of R3 (3Mbps) and R4 (3Mbps) merge at Router D, Router D needs to reserve a minimum of 3Mbps which satisfies both R3 and R4 requests Router B receives RSVP requests from Router C (100kbps) and Router D (3Mbps), Router B needs to reserve a minimum of 3Mbps which satisfies both Router C and Router D requests.

Router A receives a request from Router B (3Mbps), Router A needs to reserve 3Mbps to satisfy Router B request.

a)

Explicit routing means that the path along which a packet is to be forwarded is determined at an ingress routing node, not by intermediate routing node. Once a path is determined, routing is fast. However, for IP networks, an explicit route has to be carried by a (large) number of IP headers which will incur large overhead. Normally IP networks only support hop-by-hop routing.

MPLS uses labels in forwarding IP packets, a label only has 32 bits. Compared with a 20 bytes IP header, MPLS labels are much smaller in size. So an explicit route can be represented with a number of labels.

b)

MPLS is aware of not just individual packets, but flows of packets in which each flow has certain QoS requirements. With MPLS, it is possible to set up routes on the basis of these individual flows.

Further, when congestion happens, MPLS paths can be rerouted dynamically Instead of simply changing the route on a packet-by-packet basis, with MPLS, the routes are changed on a flow-by-flow basis, taking advantage of the known traffic demands of each flow.

Effective use of traffic engineering can substantially increase usable network capacity as shown in the figure below, flows of IP2 and IP1 are aggregated dynamically through a communication tunnel receiving the same amount of bandwidth from the network, then these two flows of packets are diverted to different paths.