@7:45 , the propagation time is distance/speed = 100km/1000km per hour= 0.1 hour=6mins. the transmission time per packet or per car is 1min (given). So total time for one packet to get from A to B is 6mins + 1 min= 7mins. Previously it took 12 mins for all the packets to get from A to B but now in the new conditions it takes 7 mins for ONE packet to get from A to B. So the answer is YES, so packets will actually reach B before all of them even leave A
Yes, this is correct. Even I noticed the same now. Its 6 mins(time taken by 1 car to travel from 1st toll to 2nd toll) + 1 min (service time at a booth) which is 7 mins. But Mr. Kenan told it as 10 mins + 1 min which is 11 mins (not correct). I guess he told that by oversight. But this analogy is correct.
And, if so, how would you calculate the total bandwidth of the Client to Server connection? Would it be similar to throughput, where Rs would cause the total bandwidth to decrease?
Once you study computer science for a few years it becomes obvious just how many youtube tutorials actually have very little idea of what they are talking about. Either that or they are just copying from someone else.
Unfortunately I've taken this class twice for my Masters. Between two different professors, and this video, it's like the only way to teach this class is by basically just reading the book to you. I can't tell if its due to the density of the course or if it's due to laziness of the professors for this course, but I haven't found someone who actually teaches the concepts. They just regurgitate the material.
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Thank you so much for this! Watching a video like this is way better than just reading a textbook.
@7:45 , the propagation time is distance/speed = 100km/1000km per hour= 0.1 hour=6mins. the transmission time per packet or per car is 1min (given). So total time for one packet to get from A to B is 6mins + 1 min= 7mins. Previously it took 12 mins for all the packets to get from A to B but now in the new conditions it takes 7 mins for ONE packet to get from A to B. So the answer is YES, so packets will actually reach B before all of them even leave A
Yes, this is correct. Even I noticed the same now. Its 6 mins(time taken by 1 car to travel from 1st toll to 2nd toll) + 1 min (service time at a booth) which is 7 mins. But Mr. Kenan told it as 10 mins + 1 min which is 11 mins (not correct). I guess he told that by oversight. But this analogy is correct.
Thanks for this :) I couldn’t understand the discussion so i turned to the comments section haha.
14:56 So, just to clarify, Rs and Rc are basically the respective bandwidths of the Server to the Router and the Router to the client?
And, if so, how would you calculate the total bandwidth of the Client to Server connection? Would it be similar to throughput, where Rs would cause the total bandwidth to decrease?
@@JagjitBrawler It is min(Rc, Rs) which is just like water flows through pipe.
With the caravan analogy, where did 7 come from?
its actually 6
@@roshaanzafar8092 and there is one more minute from the toll booth service
idk if you still need it, it's 100 (length) / 1000 (speed) = 0.1 Hour = 0.1^60 = 6 minutes.
and 1 minute delay in the first booth so 7 minutes.
why are you unsure when you explain?
Once you study computer science for a few years it becomes obvious just how many youtube tutorials actually have very little idea of what they are talking about. Either that or they are just copying from someone else.
I love you
can i get download link plzz..?
this dude is just reading from a book or something
just mumbling and no actual explanation
Unfortunately I've taken this class twice for my Masters. Between two different professors, and this video, it's like the only way to teach this class is by basically just reading the book to you. I can't tell if its due to the density of the course or if it's due to laziness of the professors for this course, but I haven't found someone who actually teaches the concepts. They just regurgitate the material.
@@GiggaGMikeE 😢