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in Computer Networks by Boss (11.9k points) | 279 views
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1) minimum hamming distance is 1 between

10101 and 10111

 none option match
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2) #f packets = 100000/1000= 10 packets

size of each packet = 1000bits

TT = 1000bits/2kbps = 500ms

PT = 10msec

1st packet will take = 5*TT +5*PT = 5*500+5*10 = 2550msec

9packets will take = 9*500msec = 4500msec

total =7050msec
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Yes ! I missed the transmission time at each router.!
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2) i am getting 7050 ms

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2) #f packets = 100000/1000= 10 packets

size of each packet = 1000bits

TT = 1000bits/2kbps = 500ms

PT = 10msec

For 1st packet

For remaining packet:



use pipelining = (5 1st packet+ 9)500 = 7000 + 1st packet propagation time = 7000+50= 7050msec
 

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@Anu Sir,@Nitish Detail explanation please didn`t get it why we are considering the transmision time of each packet because its should be overlapped as 1st packet get to 1st hop in 10msec after spending 500 but as soon packet transmitted other starts transmitting overlapping is 10msec.
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no, we have not considered transmission time of each packet.

infact in packet switching, pipelining is used.

Tt = 0.5 s, Tp =50 ms ,now

1st packet will take all 5 transmission delay, and next 9 packets will take only 1 transmission delay.

1*5*0.5 s + 9*0.5 s + 50 ms = 7050 ms..
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So lets just take 3 hops and 3 packets Same $T_{t}=500ms \ and  \ T_{p}=10ms/hop$ so,
 

  R1 R2 des
P1 510 1020 1530
P2 520 1030 1540
P3 530 1040 1550


$for \ first \ packet = 3*500+3*10 + \ for \  2_{nd} \ and \ 3_{rd} \ = 2*10 = 1550$
Correct or i am missing something please clarify ? @ joshi_nitish @ Anu007 Sir?

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3∗500+3∗10+ for 2nd and 3rd =2∗500=2530

take transmission time into account..

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But where in this table i am lacking ! I am not able to absorb this thing .
@Manu Thakur Sir,@Arjun Sir,@ Habibkhan Sir need your comments!

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a) Minimum HD=1 and maximum HD=3

b) similar question https://gateoverflow.in/2153/gate2012_44

answer is 7050msec.

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Rupendra Choudhary In this previous year question also Why we consider transmission delay instead its should be propagation(though no one menotioned that whether it was propagation or transmission delay because both are same)! Please see my table and find out where i committed mistake i am unable to debug it !

+3

Hello saxena

Observe this what i'm doing.

that GATE question data is bit easy to first we will take that into consideration.

a) when let suppose TT=1ms and PT=1ms

Path for P1 (packet number 1): Time to transmit from S + time to propagate from S-R1 +time to transmit from R1+Time to propagate from R1-R2+Time to transmit from R2+Time to propagate from R2-D =1(TT)+1(PT)+1(TT)+1(PT)+1(TT)+1(PT)=6ms. so now at T=6 packet P1 has reached destination.Now can you tell me one thing? After what time exactly source can start to transmit packet number P2?I will tell you.When source has finished transmitting (at time T=1)and now packet P1 is ready to propagate from S to R1.then source can transmit P2. means at just after finishing TT for one packet next packet TT is starting. So from here we can observe P1 is 1ms ahead of P2. lets don't take care about in b/w processing because it's complicated. just remember all in b/w work will be done with pipelining. So we have to remember P2 is 1 ms behind P1. so in b/w both are doing their share of PT and TT so at T=6 p1 when reached Destination.Can you tell me now where is P2? we know it would be 1ms behind because as P1 it also didn't stop anywhere in b/w so it should be at R2 at that time and has already finished it's TT at R2 and now ready to start it's PT from R2 to D. so at T=6ms P2 is ready to finish it PT. Now can you tell me about third packet P3? when source can be able to transmit it? when P2 has finished it's TT at S and now ready to start it's PT from S-R1. S is ready to transmit P3 so simple observation. P3 is 1ms behind P2.

So at t=6ms P1 reached destination

at t=7ms P2 reached destination

at t=8ms P3 reached destination

...

....

at t=1005 ms P1000 reached destination...

Now this is the simplest way i find to answer such type of questions ..You are confuse about that 1ms is TT or PT...actually when you're thinking that's PT then you are observing the scenario when next packet is at R2 and ready to finish it's last PT and when you're thinking like that's TT then you're observing the scenario when packet finished it's first TT at source and ready to start it's first PT from S-R1.So you answer this question. why P2 is 1ms behind of P1 ? that's your 1ms.

+2

Now consider PT=0.5ms and TT=1ms for same GATE question.

P1 finished it's first TT=1ms and now ready to start it's first PT from S-R1 so source is free and now can transmit another packet P2. At T=1.5ms P1 has reached R1. now at this time P2 just finished it's half of first TT . When P1 has finished half if it's TT at R1 ,Phas finished it's first TT and now ready to travel from S-R1.hope you can observe this.

So Total time to reach D for P1 = 1+1/2+1+1/2+1+1/2=4.5ms , so at T=4.5 P1 reached

Now as P2 is 1ms behind P1 so time to reach D for P2=5.5 , at T=5.5 ms P2 reached D.

at T=6.5 P3 reached.....

.......at T=1003.5  P1000 reached D.


Now consider PT=2ms and TT=1ms

so here at T=1ms P1 finished it's first TT and now read to travel form S-R1 and now S is free to transmit next packet P2 . At T=2ms P2 finished it's first TT and at that time P1 has completed half of it's travel time b/w S-R1.so we time to reach D for P1=1+2+1+2+1+2=9ms

At T=9ms P2 has finished half of it's last PT and it's in b/w the path R2-D.

so time to reach D for P2= at T=10 ms

for P3.....T=11 ms

.........

....at T=1008 ms P1000 would reach D

+2

Now come to your original question

TT=500 ms PT=1ms

At T=500ms P1 has finished it's first TT and now ready to travel from S-R1. At time T=510ms when P1 finished it's first PT , P2 (which start transmitting at T=500 ms )has finished it's 10ms from first TT and yet to complete remaining 490 ms of TT. at time T=1000ms when T1 has finished 490ms of it's second TT (at R1) packet P2 has finished it's first TT and now ready to travel from S-R1. this is pipelining and which we don't have to take care , the thing which we must remember is how much behind P2 of P1 which is 1TT=500ms

total time to reach D for P1= 500+10+500+10+500+10+500+10+500+10=2550ms

at T=2550 ms , P2 has finished 10ms of TT at R4 and after finishing 490 ms of last TT , it will finish 10 ms of last PT and then at T=3050ms it will reach D

but we don't have to focus on pipelining , we only have to remember P2 is 500 ms behind P1

So at T=3050 ms P2 reached D

at T=3550ms P3 reached D

....

....

at T=7050 ms P10 reached D

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Thank you so much Rupendra! :) its clear now.
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Very much detailed explanation @rupendra. Thank you.
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It means we can say that every second packet is lagging behind the first packet by gap of transmission time one TT in pipelining concept
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@Kaluti  i guess even when tp is greater that tt then also the time gap is by tt only
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Generally TT is greater than TP to avoid Collision in which case TP would be greater can u tell
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@Kaluti right my bad was just thinking of this case didnt think of the collision avoidance :')
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@Rupendra Choudhary

Thank you for this clear explanation.

So point to be noted is at the initial step itself how much p2 is delayed wrt p1 at source node i.e transmission time

Nice explanation Sir.

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