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

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?

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

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@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