GATE IT 2006 | Question: 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??

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According to me, in full duplex, router has to process incoming as well as outgoing packets. Suppose if there is only one full  duplex link of bandwidth 100Mbps, then outflow and inflow can happen simultaneously. The router processes both incoming as well as the outgoing packets which it is transmitting also, isn’t it? So the router should be able to process at a speed of 200 Mbps for no delay. And since there are 4 such links(2 incoming and 2 outgoing) so router should process at 400 Mbps and so processing time=84B/400Mbps= 1.68microseconds. But here we are considering only the processing time of incoming packets.

Is this because of definition of processing time?-PACKET processing time (PPT) of a host is the time elapsed between the arrival of a packet in the host's input queue of the network interface card, NIC, (i.e., the data-link layer of the TCP/IP protocol stack) and the time the packet is processed at the application layer                 (source:https://web.njit.edu/~rojasces/publications/khoroSJ14.pdf)

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What is “wire speed forwarding”?

Text from Cisco community forum

The claim of 'wire-rate forwarding' implies that there is a non-blocking end-to-end data path and that the forwarding engine is capable of making enough forwarding decisions on packets to not congest that data path. 

Here in this question, the router should be able to process the frame before the next frame comes in so that the next frame will not wait for processing. 

Answer 1

Two full-duplex Ethernet interfaces operating at $100$ Mbps means the total data transfer rate to the router is $200$ Mbps. (Whether full duplex or half, we just need to consider the bandwidth as that determines the speed of data arrival to the router.)
    
Maximum packet processing time allowed for wirespeed (no delay) forwarding $= \frac{84\times 8}{200} \mu s = 3.36\mu s.$

(The router operating at wirespeed means it should be fast enough to process the incoming traffic without causing a delay for transmission)

Comments

Would there be any change if the interfaces were half-duplex?

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I dont think so, because the answer mentions we have to consider only the bandwidth. Here 200Mbps is due to two interfaces and not due to it being full duplex… I may not be right though

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This video by Kameswari (IIT B) mam will help. 

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what are the various things need to be consider when question says full duplex?

Answer 2

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

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

Transmission time is time for a packet to get out/in of wire. Also, assume even though the router has two incoming/outgoing hardware but it has a single processing unit.

Now assume that at $t=0$ router is free. At $t=6.72$ two packets, $p_{1}\;\&\; p_{2}$ arrived from two incoming/outgoing hardware.  Now next slot of two packets will arrive at $6.72+6.72$ before that processing of these two packets must be completed so that there will not be any kind of delay. Hence we just have  $6.72$ maximum times to process each packet, hence for each packet, we can devote a maximum of $3.36.$

Hence the answer is $3.36.$

Now for further explanation, I will show when the first two packets are processed. At $t=0$ both packets start coming out of wire, at $t=6.72$ both packets completely coming out of wire, and we start the processing of one of them (let's say $p_{1}$) and next packets $p_{3}, p_{4}$ starts coming out.

At $t=6.72+3.36$ processing of $p_{1}$ done, start transmitting it through port $1$. 

At $t=6.72+ 6.72$ processing of $p_{2}$ done start transmitting it through port $2$.

At this time next packets $p_{3} \;\&\; p_{4}$ arrived, and we start processing one of them (let's say $p_{3}$), at t=6.72+6.72+3.36 p1 completely transmitted., processing of p3 done, start transmitting it through port $1\;($as port $2$ is busy in transmitting packet $2)$ 

At $t=6.72+6.72+6.72$ transmitting of $p_{2}$ done. processing of $p_{4}$ done starts transmitting it through port $2,$ next packets $p_{5} \;\&\; p_{6}$ arrived.

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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?

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Do we take the time to receive the data by the router equal to transmission time too ?

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Great answer

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great explanation!👍

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So clear explanation. Must be the best answer! To Note : They have asked about the maximum packet processing time, not transmission time. 

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They chosen best answer of their own answer

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This ans deserves to be best ans !!

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Lets flag the best ans as low quality instead he he he :)

Answer 3

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

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.

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@Bikram @Arjun @mehul vaidya @Akash Kanase Sir,

Transmission time is time for a packet to get out/in of wire. Also, assume even though the router has two incoming/outgoing hardware but it has a single processing unit.

For each packet to come In router, you will need transmission.

  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 !
 

 

In a full duplex network does it mean that while the packet is being transmitted from the node, we can not receive packet at the same port??

Once the router receives the packet at time 0, why router will wait for 6.72 unit for getting a new packet ??

Answer 4

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.