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Computer Networking : A Top-Down Approach

Computer Networking : A Top-down Approach

Authors: James F. Kurose, Keith W. Ross Chapter: Multimedia Networking
ISBN: 9780132856201 Exercise: Review
Edition: 6 Question: 1

 

Question

Reconstruct Table 7.1 for when Victor Video is watching a 4 Mbps video, Facebook Frank is looking at a new 100 Kbyte image every 20 seconds, and Martha Music is listening to 200 kbps audio stream.

Answer

Assume that 1kbyte is 8000bits.

So, 100kbytes is equal to (8000*100kbytes)/1kbytes = 800000bits  

The bit rate for 20secs is 800000bits  

The bit rate of Face book Frank is calculated by

Therefore, the bit rate = 800000bits/20sec

                                 =40000bps    

                                 =40kbps

The bytes transferred in 67mins = 40kbps *67*60

                                               = 160800 kbps

                                               =20100kbytes

                                               =20.1Mbytes

 

The bit rate of Victor video is 4 Mbps.

The bytes transferred in 67mins =4Mbps *67 mins*60

                                               =16080Mbps/8

                                               =2010Mbps

                                               =2Gbytes

The bit rate of Martha Music is 200 kbps

The bytes transferred in 67mins =200kbps *67 mins*60

                                               = 804000kbps

                                               = 100500kbytes

                                               =100Mbytes

 

 

Bit rate

Bytes transferred in 67mins

Facebook Frank

40kbps

20.1Mbytes

Martha Music

200kbps

100Mbytes

Victor Video

4Mbps

2Gbytes

 

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