MESSAGE
| DATE | 2016-05-04 |
| FROM | Ruben Safir
|
| SUBJECT | Re: [Hangout-NYLXS] UPD Packet in an IP Datagram
|
On 05/04/2016 08:25 PM, prmarino1-at-gmail.com wrote:
> Dude you are thinking too hard about this.
> You have a header with a known size in a packet of a known size then the rest is payload.
> You aren't dealing with an ISO layer cake packet with additional application layer stuff
> So its just header and payload.
> It's a simple math problem.
> Total transfer/(MTU - header)= total packets
That is not right, i think.
Right we need to know what the ultimate per packet data payload is.
"Assuming MTU = 1500, a 4,998,913 byte MP4 video is to be transmitted
encapsulated within UDP. Assuming standard IP header of 20 bytes and
standard UDP header of 8 bytes, how many IP datagrams are required to
transmit the entire video? Assume the payload of each UDP datagram is
such that the resulting IP datagram never exceeds the MTU and thus need
not be fragmented."
Payload on IP level = 1500 - 20
1480 bytes
that is the UDP packet size now
1472 = 1480 - 8 Data in the UDP payload
4998913/1472 = 3396.0007
that is 3397 packets
--
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