Problem Set 2 of computer Networks, Assignments of Data Communication Systems and Computer Networks

Download Problem Set 2 of computer Networks and more Assignments Data Communication Systems and Computer Networks in PDF only on Docsity!

Problem solving

Q.No.

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.

Q.No.

Q1. Suppose that the packet length is L = 8000 bits, and that the link transmission rate along the link to router on the right is R = 1 Mbps.. What is the transmission delay? What is the maximum number of packets per second that can be transmitted by this link? HThe transmission delay = L/R = 8000 bits / 1000000 bps = 0.008 seconds The number of packets that can be transmitted in a second into the link = R / L = 1000000 bps / 8000 bits = 125 packets

Q6.A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network?

Q8. What are the propagation time and the transmission time for a 5- Mbyte message (an image) if the bandwidth of the network is 1 Mbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 10^8 m/s. Solution

Q9. Consider the two scenarios below: A circuit-switching scenario in which Ncs users, each requiring a bandwidth of 20 Mbps, must share a link of capacity 200 Mbps. A packet-switching scenario with Nps users(19) sharing a 200 Mbps link, where each user again requires 20 Mbps when transmitting, but only needs to transmit 20 percent of the time.

9b.Suppose packet switching is used. What is the

probability that a given (specific) user is transmitting, and

the remaining users are not transmitting? Nps=

The probability that a given user is transmitting is p=0. Probability that other user is not transmitting= (1-p)=0. Probability that all of the other Nps-1 users are not transmitting = (1-p)Nps-1. Thus the probability that one specific user is transmitting and the remaining users are not transmitting = p*(1-p)Nps- =.

3. 17 10.Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P2 is (1/2)P1. In this case, the attenuation (loss of power) can be calculated as A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.

3. 19