in Computer Networks
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1 vote
1 vote

please provide the detailed solution

in Computer Networks
682 views

4 Comments

this is what they have provided

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time starts at t=0, then 1st packet sent will leave A at t = 8ms

time starts at t=8, then 2nd packet sent will leave A at t = 16ms

time starts at t=16, then 3rd packet sent will leave A at t = 24ms

time starts at t=24, then 4th packet sent will leave A at t = 32ms

time starts at t=32, then 5th packet sent will leave A at t = 40ms

therefore at t = 40ms, All packets left A

at t= 40 + 20, last packet reach R1 ==> at t=60

Now last packet leave the R1 at t = 60+8 = 68

at t= 68 + 20, last packet reach R2 ==> at t=88

Now last packet leave the R2 at t = 88+8 = 96

at t= 96 + 20, last packet reach R1 ==> at t=116
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look at the answer below once..is it correct.?

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1 Answer

2 votes
2 votes
Propagation delay Tp = 20ms

Transmission delay for each packet = 1000*8/10^6 = 8ms

Now each packet needs to be transmitted thrice and has to pass through three links each have one way delay as 20ms.

Here we have 5000B/1000B = 5 packets.

So 1st packet will reach host B after 3*8 + 3*20 = 84ms

2nd packet will start getting transmitted 8ms after the 1st packet started its transmission.

So 2nd packet will reach host B after 84+8= 92ms

Similarly 3rd packet will reach host B after 92+8 = 100ms

4th packet will reach host B after 100+8= 108ms

And finally, 5th packet will reach host B after 108+8= 116ms

So total time required is 116ms.

4 Comments

Yes..that's why is said:

This formula is valid only when Tp>Tt because in this case, no time would be wasted as queuing delay.

this is true right?

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i hope so, but i am not sure !
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you can apply the above concept. This question is from packetization.

First packets will take ===> 3* ( 8 + 20) = 84.

===> After this one by one packets will be out. 4*8 ==> 32.

116.

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