I also have question about the " every minute 500 hours worth of videos" and the bandwidth calculation.
My conclusion is with ir without the bandwidth limitation in mind, we will both get the right answer.
If we assume 230 videos/sec (stated in the lecture) uploads, and then we can assume 5min/video and 50M/minute. The we will have
230(videos’/sec’) * 60(sec/min) * (5min/video) * (50M/min) = 3369 G/min = 56G/s.
As we know the video may not be evenly distributed across the day, the number may be much larger or smaller.
But the key thing to notice is, we may not finish uploading the whole video in a minute despite its size with the bandwidth limit for client to upload anything. So we assume a 10M/min upload bandwidth. In turns, this will affect the server’s ingress bandwidth.
So let’s analyze it with the bandwidth limitation in mind:
Each video is (5min/video) * (50M/min) 250 M in size on average, and the bandwith is 10M/min. Then we need 25 min to finish the upload of a video.
Each minute we have 230*60 = 13,800 videos
We have 13,800 video are being uploaded per minute, that is 13,800*10M = 134.7G.
This is the bandwith we need to start uploading these 13,800 videos and we need to keep the upload for 25 mintues.
And we can see from the graph below, each moment we will have new connections for new video upload and old connections disconnect, and the bandwidth need to be mutiply. And the cycle is 25 minutes.
So, we also need to multiply 25 to the bandwitdth.
134.7G*25 = 3369 G/min
This is exact the same as the analysis we have at the begining…
For simplicity, we have a cycle of 4 instead but the pattern is the same.
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