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Problem 3 Consider a simple network that is composed of two computers (A & B) an

ID: 3890678 • Letter: P

Question

Problem 3

Consider a simple network that is composed of two computers (A & B) and a router X in the middle. Com- puter A is connected to X by a link that is 1 Mbps and has a propagation delay of 1 msec. Computer B is connected to X by a link that is 2 Mbps and has a propagation delay of 2 msec. Suppose that computer A generates two packets (destined to B) that are sent back-to-back via router X. Assume that each packet is 1000 Bytes in size. Answer the following questions:

(a) When would the first packet arrive fully at X? (b) When would the first packet arrive fully at B? (c) When would the second packet arrive fully at B?

(d) Would this “packet-pair” still arrive back-to-back at B? If your answer is yes, indicate why. If your answer is no, indicate the time gap between them (the time between when the last bit of the first packet was received and when the first bit of the second packet was received).

Explanation / Answer

Let's consider the network connected as below

A _____ X ______ B

A is connected to B via X and vice-versa.

Link speed between A & X =1 Mbps

propagation delay between A & X = 1 msec

Link Transmission delay for 100 Bytes(8000 bits) = 8000/1Mbps = 8 msec

Total transmission time of 1000 Bytes from A to X = 8+1 = 9msec (Answer to question a, i.e., total time to reach a packet from A to X)

Link speed between B & X =2 Mbps

propagation delay between B & X = 2 msec

Link Transmission delay for 100 Bytes(8000 bits) = 8000/2Mbps = 4 msec

Total transmission time of 1000 Bytes from A to X = 4+2 = 6msec

Total transmission time for a packet to reach from A to B = 9 + 6=15 msec (answer to question b)

question c) Time for the second packet to reach B.

At T=9msec , the first packet reached X. Now A starts sending second packet

T=18 msec , the second packet reaches X

T=18 + 6 =24msec, the second packet reaches B

D) The two packets will arrive back-to-back because by the time the second packet reaches the router X, the first packet was already reached B by 3 msec ahead.

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