Question

. Consider the scenario shown below, with four different servers connected to four different clients over four three-hop paths. The four pairs share a common middle hop with a transmission capacity of R = 200 Mbps. The four links from the servers to the shared link have a transmission capacity of RS = 70 Mbps. Each of the four links from the shared middle link to a client has a transmission capacity of RC = 30 Mbps per second.

a) What is the maximum achievable end-end throughput (in Mbps) for each of four clientto-server pairs, assuming that the middle link is fair-shared (i.e., divides its transmission rate equally among the four pairs)?

b) Which link is the bottleneck link for each session?

c) Assuming that the senders are sending at the maximum rate possible, what are the link utilizations for the sender links (RS), client links (RC), and the middle link (R)?

2. Consider the figure below, in which a single router is transmitting packets, each of length L bits, over a single link with transmission rate R Mbps to another router at the other end of the link. Suppose that the packet length is L= 12000 bits, and that the link transmission rate along the link to router on the right is R = 1000 Mbps.

a) What is the transmission delay (the time needed to transmit all of a packet's bits into the link)?

b) What is the maximum number of packets per second that can be transmitted by the link?

c) Suppose there are 5 routers between the two end devices called sender and receiver. Assume that the link between the routers and the end systems have same transmission rate R = 1000 Mbps. What will be the total transmission delay to send the packet from the sender to the receiver?

3. Consider the figure below, with three links, each with the specified transmission rate and link length.

a) Find the end-to-end delay (including the transmission delays and propagation delays on each of the three links, but ignoring queueing delays and processing delays) from when the left host begins transmitting the first bit of a packet to the time when the last bit of that packet is received at the server at the right. The speed of light propagation delay on each link is 3x10**8 m/sec. Note that the transmission rates are in Mbps and the link distances are in Km. Assume a packet length of 16000 bits. Give your answer in milliseconds.

b) If the processing delay in each router is 0.0001 msec/bit and queuing delay is 0.0 msec then what will be the total end-to-end delay.

4. Suppose Host A wants to send a large file to Host B. The path from Host A to Host B has 5 links, of rates R1 = 500 kbps, R2 = 400 kbps, and R3 = 1 Mbps, R4 = 750 kbps, and R5 = 1.2 Mbps.

a) Assuming no other traffic in the network, what is the throughput for the file transfer?

b) Suppose the file is 4 million bytes. Dividing the file size by the throughput, roughly how long will it take to transfer the file to Host B?

c) Repeat (a) and (b), but now with R1 reduced to 400 kbps.

5. Answer following questions.

a. How many layers in the OSI model and how many layers in the internet protocol stack?

b. What is MAC address, IP address, and Port No.? What do they identify?

c. Different layers have different names for the internet data. What are those names at different layers?

d. What is IXP? How does it make money?

e. Which layer exists in a router but does not exist in a switch?

f. What are the advantages of a packet switch over a circuit switch?

g. As you know, I = aL/R is the equation for the traffic intensity. Now, if the packet arrival rate is 5/s, transmission rate is 100 kbps, and packet length is 10 kB. Do you think there will be queuing delay? Why do you think so?

Answer 1

1b. bottleneck link is the one which is fully utilized the one with the max data rate is the one

2a. transmission delay = length/rate

12000/10^9 = 0.000012 seconds

2b.

2c. td =(N*L)/R

td is transmission delay

N is no of routers+1

R is rate

= no of routers + 1 is N that is 6

=( 6*12000)/10^9 = 0.000072 seconds

3a.

as the image is not given do substitute the values in answer

transmission delay =td

pd = propagation delay

td link 1= L/R = 16000/R mseconds

td link 2= L/R = 16000/R mseconds

td link 3= L/R = 16000/R mseconds

pd link 1= L/R = distance/ 3*10^8 mseconds

pd link 2= L/R = distance/ 3*10^8 mseconds

pd link 3= L/R = distance/ 3*10^8 mseconds

3b. total delay =no of hops*(1st packet delay)+transmissiondelayN - (td+pd)

4a. throughput is minimum of (r1,r2,r3,r4,r5)=400kbps

4b.4million bytes to bits is 32000000bits

throughput = 400000 bps

=file size/throughput

32000000/400000 = 80 seconds

4c.throughput will remain same as minimum is still 400kbps so and time taken is 80 seconds

5a. OSI has 7 layers

internet protocol stack has 4 layers

5b.mac address is unique identification code on network card of your device it is used to identify devices on network

ip is a label given to devices that uses internet protocol for communication it identifies your device in network it changes according to network you get connected it is assigned while connecting to device

port number is a logical address of each application or process which uses internet to communication on your pc it identifies a network based application on our computer

5c. layer 1 physical layer
layer 2 data link
layer 3 network
layer 4 transport
layer 5 session
layer 6 presentation
layer 7application

5d.ixp is data centers for peering routers it earns by charging isp that connect to ixp

5e.network layer

5f.efficency is the advantage packet switch can find its own path, in circuit switch network devices cant use channel untill voice communication is is terminated