switching

Question

A sends 2 frames of 1000 bit each to B via switch S.Bandwidth=10Mbps  propogation delay  over links=5us find time when second packet reaches B completely

Comments

is ans 305 ?

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similar Gate question

https://gateoverflow.in/8495/gate2015-3_36

Answer 1

Answer should be 310 micro-seconds.

Tt = 1000 / 10Mbps = 100 micro-seconds.

Tp = 5 micro-seconds.

Since The Store and Forward delay is not mentioned, we can assume it to be negligible.

It is a case of pipelining in Packet Switching, so, The Propogation delay will only be considered once for the first packet since the rest of the packet follow it !

Also, the Transmission delay will be considered = No. of hops for the first packet only, and for the subsequent packets, after each Transmission delay time, 1 packet will be delivered.

So,

First packet = 2*Tt + Tp( from A to S )  + Tp ( from S to B )  = 2*100 +5 + 5 = 210 micro-seconds.

Second packet = Tt = 100 micro-seconds.

Total time at which the second packet is completely delivered to the destination = 210 + 100 = 310 micro-seconds.

Comments

Tt is time taken by sender to place all the bits on the link

Tp is the time taken by signal to travel from source to destination

here A sends to switch and switch sends to B so, for the first packet, Tt=2*100 and then Tp(frame) =5,

so total time for first frame = 200 +5 =205us

due to pipelining we will consider only Tt for second frame

so 2nd frame will reach at B after 205 +2*100 =405us

why u have taken only 100us as Tt for 2nd packet??

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That's the power of Pipelining.

The second packet will start transmission onto the channel at the sender as soon as the last bit of the first packet completes its transmission onto the channel at the sender.

And so on.. When the first packet completes its journey and completely reaches the destination, the second packet is in the middle of its journey.

If there were 10 more packets then each packet would have been delivered to the destination after 1 Tt one by one.

So, only the first packet will take total time to reach the destination. Other packets are just following it.

Got it ?

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@bikram sir

please verify which is right answer.

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yes, 310 is correct answer.

two times Tp is to be consider , because we consider propagation time only for first packet not for first link ..so from A to S one time Tp is consider then again from S to B another time Tp need to be consider.

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for first packet shouldn't we consider Tp twice, once from A to S and then from S to B?
to put first packet on the link = 100 microsecs
from A to S = 5 Microsecs
S will store and forward, so again to put the packet on second link = 100 microsecs
and Tp from S to B = 5 Microsec
and for second packet we need total 100 microsecionds hence total time comes to 310 Microsecs
why is 205 microsecs taken for first packet?

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any response?

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@ manu00x 

This question is based on " pipelining in Packet Switching " 

Read 2nd benefit of packet switching here :

http://www.erg.abdn.ac.uk/users/gorry/course/intro-pages/ps.html

Transmission delay refers to the time it takes to transmit a whole packet over a link (  all the bits ) from one end to other end and is given by L/R, where L is the amount of data and R is the rate.  

Propagation delay refers to the time it takes for a single bit, once on the link, to reach the destination.

That means it is the time to transmit one bit over a link from one end to another end .

for first packet shouldn't we consider Tp twice, once from A to S and then from S to B?

Yes , two time Tp is to be consider , because we consider propagation time only for first packet not for first link ..so from A to S one time Tp is consider then again from S to B another time Tp need to be consider.

At the time packet 1 is sent from B to C, packet 2 is sent from A to B; packet 1 is sent from C to D while packet 2 is sent from B to C, and packet 3 is sent from A to B, and so forth.

This simultaneous use of communications links represents a gain in efficiency, the total delay for transmission across a packet network may be considerably less than for message switching .

So, this concept is used here .

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I agree to each point said by you @Bikram but see my analysis below:

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@manu00x you have done the correct analysis ...your answer is different from the selected answer only because you have taken 5microsec delay for each link while in above answer he has taken 5 microsecond for complete link

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@Vaibhav Thanks a lot! yes you're right!

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yes @ manu00x 

your diagram is correct and your point is correct too .. 

i understand , the main point here is :two time Tp is to be consider ,

because we consider propagation time only for first packet not for first link ..so from A to S one time Tp is consider then again from S to B another time Tp need to be consider.

@hem chandra joshi  see the edited answer, it is 310 correct answer due to that above point.

Answer 2

Frame 2 will be recieved at reciever after ≈ 310  µs ,if switch uses a store and forward mechanism.

http://inst.eecs.berkeley.edu/~ee122/fa09/notes/04-Performancex6.pdf