Networking: A system uses statistical time division multiplexing, there are 100

Question

Networking:

A system uses statistical time division multiplexing, there are 100 channels (100 different signals) being combined into one multiplexed signal. Each channel carries frames that contain 30 bytes of header and 1170 bytes of data (a total of 1200 bytes). Each channel supports a data rate of 1 MB per second (1KB = 1024 bytes, 1MB = 1024 KB, 1GB = 1024 MB). Assume that each channel is used an average of 45% of the time. Assume that on average we wish to use 90% of the capacity of the line (the line carrying the multiplexed signals). What is the capacity of the line required to carry the multiplexed signal (in bytes per second, Bps)?

Explanation / Answer

Solution : Each channel is used an average = 1200*45%

=1200(45/100)= 540.

As only 90% of the capacity of the line is used, therefore the capacity of the line required to carry the multiplexed signal = 100*540*90/100=48600 bytes per second.

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