[ee122] hmwk problem Number 2.16
nescionomen at gmail.com
Thu Sep 20 17:43:09 PDT 2007
I'm not sure I understand. (in a real life problem) why wouldn't we need to
know the size of the frame header just because we are given the MTU?
Wouldn't we try to find the amount of 'real data' that could fit by doing
something along the lines of MTU - frame header size - IP header size -
Layer 4 header size?
On 9/20/07, Jonathan D. Ellithorpe <jde at berkeley.edu> wrote:
> I think the MTU is actually defined as being the maximum amount of data
> that a link layer frame can encapsulate. Thus, we don't need to consider
> the size of the frame, since we're just given the MTU of the link-layer.
> Jorge Ortiz wrote:
> > On 9/20/07, Nescio Nomen <nescionomen at gmail.com> wrote:
> >> For hmwk Problem #2, P16, do we know if the datagram will have a TCP
> >> inside? (Is it necessarily the case that a datagram always
> encapsulated a
> >> TCP header?)
> > If it's specified as only being a plane datagram (as it is in the
> > problem), we only need to include the IP header.
> >> Also, when we are transmitting into the link, how big is the
> >> frame header?
> > Treat the packet as having only an IP header and the data you wish to
> >> We haven't really talked about this layer yet. Does the size
> >> of the frame header depend on the technology, i.e. optical?
> > It may, yes. Different mediums may have different header definitions.
> > jorge
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