GATE2006-IT-66

Question

A router has two full-duplex Ethernet interfaces each operating at $100$ $\text{Mb/s}$. Ethernet frames are at least $84$ $\text{bytes}$ long (including the Preamble and the Inter-Packet-Gap). The maximum packet processing time at the router for wirespeed forwarding to be possible is (in micro­seconds)

• A. $0.01$
• B. $3.36$
• C. $6.72$
• D. $8$

Comments

what is final ans???

b or c??

Answer 1

Here we need at least enough speed that we are able to transmit packets in a speed we get them !

We have got $2$ Full duplex ports, each operating at $100$ $Mb/s$. So we require incoming packets with $200$ $Mbps$, so that we can sent out data at $200$ $Mbps$ over the two interfaces.

For each packet to come In router, you will need transmission. Time, in case of single $84$ $byte$ packet you will get it as $6.72$ microsecond.

Now, D is simply wrong. You take $8$ Microsecond to process, soon you will have pile of packets waiting (Processing > Transmission), and we are getting $2$ packets per $6.72$ micro seconds as input.

C is wrong, here we can get $6.72$ micro seconds packets & we are barely able to process $1$ packet in that time. So, every processing time our Queue will increase size by $1$ & get full and overflow.

B & A are okay.

A is best though as we are asked to give maximum,.

B is answer.

Assume that in B you got 2 packet at time 0, by time 3.36 you can start sending packet 1, by 6.72 packet 2. By 6.72 you got 2 more packet. By time you finish processing Packet no 3, first port where you started processing with 3.36 is free, so you can start sending Packet 3 and  so on !

Reference: https://en.wikipedia.org/wiki/Wire_speed

Comments

According to me, Full Duplex doesn't double the bandwith in this case.

Also, while accepting, apart from queueing delay, there's processing delay; while sending there's transmission delay. So, first packet comes, gets processed in 6.72, gets transmitted in next 6.72, too much! But while its getting transmitted, next packet can get processed, and so there's no additional delay after the first packet, except transmission delay_, ofcourse (No time gap between packets necessary, inter-packet gap already added).
What do you think?

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Full duplex does not double the bandwidth, but 2 links does mean that we have double the data transfer.

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Tt = 6.72 micosecond. So why 2 packets will be sent ??

Is it because of FULL duplex?

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As router has two full duplex wires then to calculate wirespeed forwarding both wires willbe considered separately??

I mean both Ethernets should be always busy or not??

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@Arjun Sir is 2 full duplex then packet will fragment and then transfer partial packet from one link and second from another one. so that it is taking up less than 6.72 sec.

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@arjun @akash sir so can we conclude that maximum processing time equals to the average time to receive a frame considering all ports ?

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@Shivansh Gupta

I don't think we can conclude that maximum processing time equals to the average time to receive a frame considering all ports.

But surely if that is the case then the frames would not be required to wait unnecessarily.

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"

So we require incoming packets with 200 Mbps so that we can sent out data at 200 Mbps over the two interfaces.

For each packet to come In router, you will need transmission TIme, in case of single 84 byte packet you will get it as 6.72 microsecond."

Are we assuming that input ports are different than the two full duplex ports mentioned in question?If its a full duplex than input and output can happen on same port.Also,i didnt get point that for packet to come in router it will take 6.72 microsecond. 6.72 is the transmission time on the link.How can we say that this time is the time by which input come to router

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@rahul sharma 5 I also have the same doubt. Since the router takes some time to transmit a packet on the wire, it surely will take some time to receive the packet from the wire, I searched it but couldn't get any clarity. @Bikram sir any thoughts on this..

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@ rahul sharma 5 and @shivansh

its a full duplex that means input and output can happen on same port at same time...

router takes some time to transmit a packet on the wire, it surely will take some time to receive the packet from the wire

Yes there is some time obviously but that time is negligible ..so not used in our calculation. That's why for packet to come in router it will take 6.72 microsecond. 

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@Bikram sir

Just by reading the language of the question how one can conclude that the transmission time of packet is considered as negligible.

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we have to calculate Transmission time first ..just looking at question it is impossible to guess..

it is coming as 6.72 micro seconds in case of single 84 byte packet .

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@Bikram sir

Not clear what you intended to convey :(

I mean even if we calculate TT first then also how we conclude that we have to ignore it.

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We have got 2 Full duplex ports, each operating at 100 Mb/s. So we require incoming packets with 200 Mbps so that we can sent out data at 200 Mbps over the two interfaces.

Forwarding packets to the same port from which it got ingress is not good in practical scenario as it may lead to deliberate routing loops. 

https://en.wikipedia.org/wiki/Routing_loop_problem

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Had there been only one interface, the answer would have been 6.72, right?

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Why are we assuming that there are 2 input ports ?

Full duplex between A and B means both can communicate and then there are two links between A and B on which A and B both can transmit.

But here it is said that 2 full duplex interfaces,so do we receive input on this or do we send output on these?If both happens at same time then collisions will happen.

In the answer it is assumed that we are receiving input on two ports and forwarding output on two ports.So are these same ports?can some one explain?

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@rahul sharma 5 

half-duplex – a port can send data only when it is not receiving data. In other words, it cannot send and receive data at the same time.

full-duplex – node can send and receive on its port at the same time(sending and receiving is through same port). There are no collisions in full-duplex mode. 

Here we have 2 full duplex interfaces of 100Mbps bandwidth. that means each them will receive at 100Mbps and send at 100Mbps at the same time. 

we can assume there is no forwarding to the same port that we have received the packet. but ports can forward to each other though. 

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@Sumalya23 Yes, had there been one interface it would've been 6.72 micro-seconds.

@rahul sharma 5 This is very genuine concern for the question Rahul, we need not necessarily assume they are input ports. Since they are full duplex, they can function as both input and output ports. You can assume any order of the interfaces though. As for your second question, the method is to have a processing time that would be the long enough to send out one packet by the time it receives the last bit second packet through the second link. Just make sure you understand the full-duplex concept properly to avoid further confusion.

Explained best here,

Assume that in option B you got 2 packet at time 0, by time 3.36 you can start sending packet 1, by 6.72 packet 2. By 6.72 you got 2 more packet. By time you finish processing Packet no 3, first port where you started processing with 3.36 is free, so you can start sending Packet 3 and  so on !

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Also,i didnt get point that for packet to come in router it will take 6.72 microsecond. 6.72 is the transmission time on the link.How can we say that this time is the time by which input come to router

@rahul sharma 5 I'm also having this doubt. If you got this cleared, can you please explain?

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It will be overlapped by the transmitting time.

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What is meaning of 

at the router for wirespeed forwarding

may be silly question but i am not getting this.

Answer 2

Although Akash's answer is correct , adding my answer just to elaborate bit more

Here router has two incoming/outgoing hardware which are full duplex which means we can send and receive at same time. 

Transmission time is 6.72(How to get this Transmission time = Packet size / Bandwidth) 

Transmission time is time for packet to get out/in of wire.

Also assume even though  router has two incoming/outgoing hardware but it has single processing unit.

Now assume that at t=0 router is free. 

at t=6.72 two packets p1 & p2 arrived from two incoming/outgoing hardware. 

Now next slot of two packets will arrive at 6.72+6.72 before that processing of this two packets must be completed so that there will not be any kind of delay.

hence we just have  6.72 maximum time to process each packet , hence for each packet we can devote maximum 3.36.

Hence answer is 3.36.

Now for further explanation I will show when first two packets are processed.

at t=0 both packets starts coming out of wire

at t=6.72 both packets completely coming out of wire , and we start processing of one of them (lets say p1)

              and next packets p3 , p4 starts coming out.

at t=6.72+3.36 processing of p1 done ,start transmitting it through port 1 

at t=6.72+ 6.72 processing of p2 done start transmitting it through port 2.

                         at this time next packets p3 & p4 arrived , and we starts processing one of them (lets say p3)

at t=6.72+6.72+3.36 p1 completely transmitted., processing of p3 done , start transmitting it though port 1 (as port 2 is busy in transmitting packet 2) 

at t=6.72+6.72+6.72 transmitting of p2 done. processing of p4 done starts transmitting it through port 2 , next packets p5 & p6 arrived 

.

.

.

continue 

Comments

@mehul vaidya Great answer!! :) Thanks 

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this should be the best answer.

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@mehul vaidya

Thanx buddy !!

This sentence is kind of "TLB hit " ;)

Also assume even though  router has two incoming/outgoing hardware but it has single processing unit.

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Great answer buddy! :)

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@mehul vaidya can router process some packets and transmit others on link at the same time?

Answer 3

For maintaining the speed of forwarding of wire, i.e. $100\text{ Mbps},$ processing
time should be at most same as minimum transmission time.
(Otherwise the packet will be delayed for transmission due to processing). 

i.e. $\dfrac{84\times 8\text{ bits}}{100\text{ Mbps}}=6.72\text{ micro seconds}$.

Ref: https://en.wikipedia.org/wiki/Wire_speed

PS: Wire speed doesn't imply a packet is sent without any waiting time.
It just means receiving and sending rates are the maximum possible. 

Comments

Could you please explain it in more detail ??

If have 6.72 processing time only, same you will take for transmission also.this adds to 13.5 approx.

So from the time a packet is received from one port and till it is put on the transmission line, you needed 6.72 to just process the packet.That would leave the line IDLE for 6.72 units. How it will equal the wirespeed forwarding then ??

According to me (A) should be correct. As processing time must be negligible to achieve the required rate because as soon as packet is received,it must be forwarded.

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A should not be answer. Even with B you are able to process fast enough ! Though C is clearly disaster, because you'll keep getting 2 pakcet per 6.72 sec & you will be sending out only 1 in 6.72 sec. !

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#source- https://www.safaribooksonline.com/library/view/ethernet-the-definitive/1565926609/ch04.html
 With full-duplex mode enabled, both stations can simultaneously transmit and receive, which doubles the aggregate capacity of the link. For example, a half-duplex Fast Ethernet twisted-pair segment provides a maximum of 100 Mbps of bandwidth. When operated in full-duplex, the same 100BASE-TX twisted-pair segment can provide a total bandwidth of 200 Mbps.
While full-duplex operation has the potential to double the bandwidth of an Ethernet link segment, it usually won't result in a large increase in performance on a link that connects to a user's computer. That's because most network protocols are designed to send data and then wait for an acknowledgment.
 

Answer 4

I think this should do it

Answer 5

wirespeed forwarding 

To operate at "wirespeed" means to operate at the bandwidth of the physical wire (or medium/link). This would require for other delays (processing, queuing etc) to not degrade the performance so much that we can't practically match the "wire speed"

To take an example, say we have a physical wire that goes from A to B, and the bandwidth of the wire is 100Mbps. But we practically observe that data is being transferred only at 80 Mbps.

This would mean that we're not functioning at wirespeed — the reason could be anything — receiver taking too much time to accept, or sender is taking too long to send, or packets are coming in so fast that the receiver's buffer overflowed.

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Now here, we have a Sender and a Receiver (the router is the receiver)

Sender's $T_t$ is $6.72 \mu s$ from one link. There are two such links from Sender to Receiver, hence sender can transmit two packets to the receiver in $6.72 \mu s$. ($T_p$ is not given, so we assume it is negligible)

Now, Receiver has received 2 packets in $6.72 \mu s$. Two more packets would arrive after additional $6.72 \mu s$ and so on.

So, to clear the two packets out, Receiver has got $6.72 \mu s$ maximum. Why? Because if Receiver forwards them in more time, then we are piling up packets in it's buffer, and achieving wirespeed wouldn't be possible.

Hence, per packet Receiver can take maximum $\frac{6.72 \mu s }{2}=3.36\mu s$.

Option B

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If the receiver takes no time i.e. $0 \mu s$ then it's great. But we can allow the receiver to take a maximum of $3.36 \mu s$ to process each packet — and that's what the question asks for.

Comments

Simply B.W would be twice (full duplex)

and there are 84 bytes

So 84/200=3.36 micro second

Answer 6

here at 6.72 microseconds,router will recieve 2 packets from both interfaces.

so for the maximum processing time for the router so that there is no delay for forwarding. for this since router will recive next two packet after next 6.72 microsec so  router should have to forward both the  packet within this time slot(6.72 microsec).As a single router there is a single processing unit means only one packet is processed at a time and we have to process 2 packets within 6.72 microsec and since all the packet is of equal sizes so 6.72/2=3.36 micro sec for each packet .

so correct option is b.