Transmission Rate and Value of h

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

Goal: understand how to do the problem.
Who is Participating?

[Product update] Infrastructure Analysis Tool is now available with Business Accounts.Learn More

I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

Steve JenningsSr Manager Cloud Networking OpsCommented:
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,

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
JCW2Author Commented:
What about (a)?
Steve JenningsSr Manager Cloud Networking OpsCommented:
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.

JCW2Author Commented:
Thank you for your help.
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Networking Protocols

From novice to tech pro — start learning today.