There is a networl with two computers A & B with a router Z in the middle. A is
ID: 668521 • Letter: T
Question
There is a networl with two computers A & B with a router Z in the middle. A is connected to Z by a link that is 1 Mbps and has a propagation delay of 1 msec. Computer B is connected to Z 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 Z. Assume that each packet is 1000 Bytes in size. (a) When would the first packet arrive fully at Z? (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
As per given question A ------ Router Z ------ B
1Mbps 2Mbps
a)
Now A wants to send 2 packets of 1000 bytes size
that is 1000 bytes = 0.9765 MB
Since connectin spped is 1Mbps ==> To send packets it would take ~~ 1 Seconds
Adding propagation delay to result ==> 1 + 1 = 2 seconds.
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b) Now we have to calcualte packet that send in first qestion to destination B
It has spped link about 2 Mbps. So to send 0.97 MB it would take 0.5 sec
And delay is 2 sec...so toal 2.5 seconds.
Finally packet will deliverd to B in 2+2.5 = 4.5 secondsa.
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c) For second packet..of same size it would take same time ...as time taken taken
by first packet.
So totaltime = time to deliver to first packet + time taken to deliver to second packet
totaltime = 4.5 + 4.5 = 9 sec
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d)
No they dont seliver back to back at B.
Once reaching first packet to router from A ...then second packet will start deliver.
So time delay is...time taken by packet to deliver from A to router Z..
which was already calcualted as 4.5 seconds.
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