GATE Overflow | Mock GATE | Test 1 | Question: 40

Question

You are working on a laptop connected to a $100 \text{Mbps}$ Ethernet LAN. You need a $2 \text{GB}$ file that is on the server in the same LAN. The entire file is also on your pen drive but you have left the pen drive in another room. You have a dog sitting beside you, that is trained to bring the pen drive to you. The average speed of the dog is $20 \text{km/hour}$. Up to what distance (in meters) does the dog have the higher data rate than $100 \text{Mbps}$ Ethernet? (Assume $1K=1000$, up to 2 decimal places)
(Note: The dog should be able to go and bring the pen drive, before the transfer on the LAN completes. Assume continuous data transmission on the LAN(no packetization required)).

Comments

Please provide some explanation for this?

---

Pretty simple but yet tricky one. This is one of the iitb and iitd assignment questions of networking subject. I found this really nice.

hint...
Compute how long does it take to transmit 2GB at 100 Mbps. This is the RTT for the dog.
You have the time for dog. You know the speed. Now just compute the distance.
The distance works out to 4/9 of a KM.

---

okay, tell  me one thing.This 2GB is in powers of 2 right?

---

Nopes. It is mentioned in the question to take 1k = 1000. So solve accordingly in powers of 10.

---

@Ruturaj Mohanty-See at 444.44, the dog will be able to bring in the pen drive exactly when data transfer will complete.

So, if we want distance such that dog brings pen drive before file transfer is complete then we should take $d \lt 444.44$ and not $d=444.44$, so don't you think 444 would be fine in that case?

---

Yes. In real gate you would have a range. Like 444.00 to 445.00. So don't worry. Just see how simple the question is but how well it is framed.

---

Yes it's a nice way of asking working of CSMA/CD :)

---

what is the answer to this question??

---

444.44 to be exact in calculation.

---

when calculated trasmission time comes to be 160sec,and if we take that as the rtt for the dog then total distance travelled by the dog would be 888.88 in total ,so why are we taking 444.44?,i know it might be a silly doubt or i might be missing something in the question @Ruturaj Mohanty

---

Read the statements properly again. You will get your answer. We are not calculating total distance travelled by dog. We are calculating upto what distance the dog can go and come back in time.

---

@Ruturaj Mohanty

Higher data rate ==> less time will be required is this concept we are using here ?

Here time for copying data from pen drive to laptop is not considered ..we define data rate is how fast is data transfered ..in ethernet case using 100Mbps we calculate time till we received all data on laptop

But in case of dog only till dog bring pen drive ?

---

@Ruturaj Mohanty Why have not we considered the propagation delay of the packet/bits from reaching the LAN server to the laptop?

---

It is not significant and hence not mentioned anything about it in the question. But yes if it would have been mentioned then you need to take it into account. In some questions transmission time might be ignored, in some propagatiom delay. By looking at the question you should be able to extract what the question is trying to say.

---

@Ruturaj Mohanty and @Ayush Upadhyaya

please tell me the calculation as i am getting 160 sec. for file to transfer completely. Now this 160 sec. will be RTT for

the dog. and speed of dog is given as 20 Km/hour so in 160 sec. dog will travel

= (160x20x1000)/60  meters

=53333.33 meters

please tell me where i am doing mistake.

---

convert kmph to $ms^{-1}$

---

@Ruturaj Mohanty , @Ayush Upadhyaya

already converted into m/sec thats why i wrote

=(160 sec.* 20 *1000 meter ) / 60 sec.

---

@shubham-Your calculation is in $m/min$ and moreover it is $2d\,m$. You have to divide final value by 2.

---

ok brother thanks a lot. :)

---

let the distance traveled by dog d then at 20km/h speed it take 18d/100 sec 

now to transfer 2GB at 100 Mbps it take 160 sec  (it is RTT for dog @Ruturaj Mohanty)

now then tp = 80

so 18d/100<80

d = 444.4

---

I got the answer as 888.88. Looking at these comments, I got that I needed to divide it by 2... I understood the calculations, but I just can't wrap my head around this line

Up to what distance (in meters) does the dog have the higher data rate than 100Mbps Ethernet?

444.44m would be the maximum distance the dog could go bring the pen drive. But what does that have to do with the dog having "higher data rate"? Lol. This might be a silly doubt, I realise, but can someone please help me get this line?

---

@JashanArora Could you eplain me how 160 secs will be the RTT for the dog? i am not able to get it.

---

@vikas999 Because the dog has to not just reach the pen-drive, but also be able to bring it back. That's a round-trip, not one-way.

Answer 1

Time taken by dog to go and comeback,

T(dog) = t(go) + t(come) =  2 * t(go) = 2 * d/ 20kmph = 18d/50 sec

Within this time 2GB data has to be transfered,

T (transfer) = data size/link bandwidwth = 2GB/100Mbps = 160 sec

Now,

T(dog) < T(transfer)

=> 18d/50 < 160

=> d = 160 * 50 /18

=> d = 444.444

or d = 444 m