A. We wish to send a 150,000 byte message over a network with four hops, each of

Question

A. We wish to send a 150,000 byte message over a network with four hops, each of length 20km and running at 100 Mb/s. What is the end-to-end delay of the message? Use speed of light in copper = c = 2 * 10^8 m/s, and round your answer to the nearest integer millisecond.

B. We wish to send a message of size 150,000 bytes over the network. There are four hops, each of length 20km and running at 100 Mb/s. However, before sending we split the message into 1500 byte packets. What is the end-to-end delay of the message? Use speed of light in copper = c = 2 * 10^8 m/s, and round your answer to the nearest integer millisecond.

Explanation / Answer

(A)

Propagation delay over a link = 20km/20*10^8=0.2 milisecond

Transmission time= 150,000*8/100Mbps= 12 milisecond+ 0.2= 12.2 milisecond

(B)

Message is divided into 100 packet each of size 1500 bytes

Transmission time = 1500/100 =0.015 milisecond

Propagation delay over a link = 20km/20*10^8=0.2 milisecond

S----R1------R2------R3--------R4--------D

Since Router/switches are store and forward device we won’t consider ack time

below table shows packet1 and packet 2 reaches different hops

S       R1     R2     R3     R4     D

0        0.215 0.430 0.645 0.860 1.075

0.015 0.230 0.445 0.660 0.875 1.090

Now to reach 100th packet at D we need 1.075+99*0.015 =2.560 milisecond

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