Test by Bikram | Computer Networks | Test 1 | Question: 20

Question

Two hosts $A$ and $B$ are connected via a Router $R$ with two links  $L_1$ and $L_2.$ Each link is $125m$ long and has a bandwidth of $109 \;bps.$ If router queuing delay is $5$ microsec and propagation speed is $250,000$ km/sec  the total transfer delay of $1$ kB packet from $A$ to $B$  is _______ microsec

Comments

@Billy

read my below answer please,

transfer Delay from A to R + transfer Delay from R to B 

Thats why we have to consider transfer delay 2 times ..

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I did the same thing in test, but I got wrong ans. and it shows correct ans as 22. Should we always round the answer to integer value?

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some one edit the question

bandwidth of 109bps should be 10^9bps.

(it is also logically implies that how BW become so low but question should be corrected.)

Answer 1

See, there are two  Hosts A and B are connected via a Router R with two links Link l1 and l2

so it is like that  A------Link1--------------- R-----Link2----------------B 

                             125 m                              125 m 

now, Total transfer Delay = transfer Delay from A to Link 1 + Propagation delay from A to R + Queuing delay at R + transfer Delay from R to Link 2 + Propagation delay from R to B  

router queuing delay is 5 microsec is stated in the question.

now you have to calculate each one separately 

transfer Delay from A to Link 1 = transfer Delay from R to Link 2

= 1 kB/ 10^9 bps = 8.00 microsecond  [ 1 kB = 1000 Bytes and not 1024 as for 1KB ]

Propagation delay from A to R = Propagation delay from R to B

= 125 m / 250,000 km/ sec = 0.5 microsec

Hence total Transfer delay = ( 8.00 + 0.5 + 5 + 8.00 + 0.5 ) microsecond = 22.00 microsecond

Comments

you need to correct your calculation here as i showed. consider both k same! consider any example from any book, you will find the same.

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yes, i understand, i took 1 KBps = 1024 but as it is bandwidth it would be 1000 .

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@Bikram sir exact answer is 22.384, It should be also in answer range.As you said 1KB=1024 BYTE and 1Kbps=1000bps is correct.I request you to please correct the answer range.

Answer 2

@Bikram sir, please edit the question, it shows 109 bps rather than 10^9 bps. I got wrong ans due to this. 

Comments

https://gateoverflow.in/42677/ace-booklet-vol2-level2-q3-in-swp-sender-utilization

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https://gateoverflow.in/86578/test-by-bikram-2017-computer-networks-20?show=88161#c88161