EE5527 Diffserve Model Assignment Sample

• Subject Code : EE5527
• Subject Name : IT Computer Science

Network Design and Management - Question 1

a) The First N packets don’t have any kind of delay in queue.

The second packet has a delay of queuing that is L/R seconds.

The n th packet has a delay of (n-1)L/R seconds.

So the average delay is

The period of time is denoted by ∆t and the number of packets are used is = ∆t x r +b.

If ∆t is large, the packets that produced are much more than the larger one of that the token generated from the bucket. Hence, ∆t x r >> b,

So the tokens number can be used to measure the value of ∆t x r. hence, b can be the limit of average rate.

Network Design and Management - Question 2

a. 512KBis the minimum delay that helps in transferring the file from A to B with the help of normal routers are;

512KB file can generate (512X1024/512) =1024 packets of that everyone has a 512 bytes payload and a header of 64 bytes. Each packet size is 512+64=576 bytes.

Three routers in the shortest path, and the minimum delay to create a path between A and B is 0.15ms, and the bandwidth is 100Mbps.

3 routers between A and B, and the number of hops that the 1st packet jumps is 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

total delay will be 0.18432ms+0.15ms=0.334ms

Delay to transmit the following 1023 packets from A to B will be: For every, delay in transmission for the ending hop is (576X8)/100X106=0.04608ms

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

 Minimum delay to transmit the 512KB file will be 200.589ms + 0.18432ms = 200.773ms

Network Design and Management - Question 3

a) the price of sending the packet which is one and this packet receives by the 4 receiver through the multicast that is 8. There are following reverse path that helps in forwarding mechanism and packers arrive at the correct interface of the multicast router that can be forwarded, on the other hand it will be terminated. In the similar way, the multicasting, the packets can be reversed throughout a network like the one exactly before. There are 8 links and the cost of these 8 links are 8 units. If all the packets can be sent through the network with the help of unicast and these all to the receivers then every of the 4 receiver can sent to the packet that is individual. Every packet can travel across these 4 links to receiver end from the source. There the number of packets are

4*4=16.

And according to this result, the cost will also be 16 that is 4*4=16.

b) RSVP make a request of R1 which is of 20 kbps and R2 that is of 100 kbps and these two merges at the router C, and this Router C requires to make a reservation and the reservation should of minimum of 100 kbps that help in satisfying both requests of R1 and R2.

The other two requests of RSVP that is R3 and it is of 1 Mbps and R4 which is of 3 Mbps and these two also merge at the router D, and this router D requires to make a reservation of the minimum 3 Mbps that help in satisfying the request of Router C and Router D. Router A receives the request from router B that is of 34 Mbps and router A requires to reserve 3 Mbps to make the request satisfy of router B.

Network Design and Management - Question 4

a). The meaning of the explicit routing is that that routing along with the packet is to be forwarded along the path as determined at an ingress node of routing, it is not determined by the routing node. If the path of routing is determined that the routing is very fast. Nevertheless, For the internet protocol network, an explicit route has to be carried by the large number of headers of Ip address and it is a explicit route that can be carried a large number of IP addresses that incurs the large overhead. AT the normal Ip address, it alos supports hip by hop routing. MPLS is helped in using the labels to forward the IP packets and a label has 32 bits. If compared with the 20 byte with IP header then MPLS labels are very much smaller than in size. Hence, the explicit route can be represented in the form of number of labels. 

b) It is the MPLS that does not aware of all the individual packets, that flows in the packets that every flow has fixed requirement of quality of service. With the help of MPLS the paths can be re routed but these paths should be rerouted dynamically rather than simply flow to flow basis, it also helps in taking the advantage of the traffic demands of every flow. The effective and efficient use of traffic engineering can be increased with the help of network capacity usability as shown in the figure which is given below, that flows of IP2 and IP! Are aggregated with the help of communication tunned that receives the same amount of bandwidth from the network, then IP1 and IP2 packets can be flows to different path with the help of packets diverted.

Network Design and Management - Question 5

a) Diffserve model replies On the protocol DSCP code that can carried the code of 6 bit in the header of the IP packet. Every DSCP has a pre described meaning in the terms of networking resources needed, if an Ip address or network is deployed over the network of ATM, the IP packets can further be categorised into smaller parts of the ATM cells and as an outcome, the DSCP can be carried in an IP header that will be disappeared in the network of ATM. Hence, IP lay quality of service by using the Diffserve model cannot be supported as a backbone of ATM networks.

b). As given in the diagram. It is the MPLS that can co related and coexist with the switches of ATM. The label swapping of MPLS paradigm is the similar to the mechanism that the switches of ATM is use to forward the component of the labelling the information that can be executed for the ATM header, particularly, the VPI and the VCI fields. The main responsibility of the MPLS here is the control the elements of the IP on the switches and routers that is LSR in ATM. If passed an IP packet that it will help in carrying the code that is DSCP code on to the network of MPLS, it is the DSCP code that helps in reflecting in the form of MPLS label and it can be reflected further in the ATM network in the form of VPI/VCI that help in identifying the virtual circuit. Hence, if add MPLS in the middle then Diffserve helps in facilitating the QoS that can still be reflects in the network of ATM.

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