Transmission Rate and Value of h

Note: I don't want a direct answer to the problem.

Goal: understand how to do the problem.
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JCW2Asked:
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Steve JenningsIT ManagerCommented:
Not sure what your question is . . . "how do I find 'h'?". . . the value for "h" is stated as the number of bytes represented by the UDP and IP headers. The UDP header is a constant value (sort of) and you should be able to compute the IP header length given the UDP datagram it's carrying. And then RTP puts some data (the "special" header referred to above that includes a time stamp and sequence number and payload type, etc) on the audio chunk.

So for grins lets say the UDP header is 8 bytes and the IP header is 20+ bytes for each "chunk" sent. That would mean we are sending 160 bytes of RTP "data" plus 16 bytes of RTP header plus 8 bytes of UDP header plus 20+ bytes of IP header . . . or 204 bytes every 20ms.

I've deliberately left out some stuff in the calculation (UDP has a pseudo header for IP4 and IP6, IP headers use options and other attributes that effect the end size of the IP header . . . RTP header is pretty straightforward).

Good luck,
SteveJ
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JCW2Author Commented:
What about (a)?
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Steve JenningsIT ManagerCommented:
Hmmmm. So I "calculated" the size of the datagram to be 204 bytes every 20ms. There are 50 20ms chunks in a second, so that would mean (assuming 0 delay between transmitted chunks) 50 x 204 bytes or 652,800 bits per second or 81,600 bytes per scond.

No?
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JCW2Author Commented:
Thank you for your help.
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