Consider the following network setup, where the Length represents the link dista

Question

Consider the following network setup, where the Length represents the link distance and R represents the transmission rates. Suppose that the packet length is L 12000 bits Server (A) Host (A) R 30 Mbps Length = 1 Km R- 50 Mbps Length = 2 Km R1 R2 R-30 Mbps Length = 20 Km R 70 Mbps Length- 3 Km 20 Mbps Length -2.5 Km Host (B) Server (B) (a) What is the transmission delay from R2 to Server A (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 from R2 to Server A? (c) Find the end-to-end delay (including the transmission delays and propagation delays on each of the three links, but ignoring queuing delays and processing delays) from when the host A begins transmitting the first bit of a packet to the

Explanation / Answer

a) Transmission delay = time taken to tramsit the packet = packet size / bandwidth

Transmission delay from R2 -> A = 12000 bits / (50 * 1000) bits/sec = 12/50 sec = 0.24 sec

b) Maximum packets per second = amount of data that can be transmitted / amount of data to be transmitted (packet size)

So, max packets per second = bandwidth / packet size;

= ( 50 * 1000 bits per second ) / (12000 bits)

= 50/12 = 4.17 packets / second

c) End to End delay from Host A to Server A = End to end Transmission delay + End to end propagation delay.

Since, we've already seen Transmission delay formula, lets see how to calculate propagation delay.

Propagation delay = distance / transmission speed = length / speed of light in this case.

= Td(HA->R1) + Td(R1->R2) + Td(R2->SA) + Pd(HA->R1) + Pd(R1->R2) + Pd(HA->R1)

= (12000/30000 + 12000/30000 + 120000/50000) + (1000/speed of light) + (20000/speed of light) + (2000/speed of light)

= 1.04 + (23000/ (3 x 108))

= 1.04 + 0.000077 = 1.040077 sec

d) By Applying the same logic as c

1.1710085 sec

e) Let's split the bandwidth of R1->R2 link by half for each of the packets (A & B) since fairshare was mentioned.

Then by applying above calculations, total delays for A = 1.714 sec and B = 1.57 sec, then

Max Throughput can be calculated as max data per second

12000/1.714 b/s = 7 Mbps for A

12000/1.57 = 7.65 Mbps for B

f) For packet transmission from A to A bottleneck is at HA to R1 due to low bandwidth than other 2. Similarly, for packet transmission from B to B bottleneck is at HB to R1 due to low bandwidth than other 2.

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