Hello George,
as already noted the link speeds in telecommunications are expressed in bit/s and not byte/s.
Also in telecommunications we dot not use a 1024 multiplier but 1000 like international metric system
1 Kbps = 1000 bps
1 Mbps = 1000000 bps
And so on.
Of course packet size is in bytes and 1 byte = 8 bit.
However, in telecommunications you need to take in account the so called protocol overhead, that depends on the OSI layer 2 technology in use.
example for ethernet:
if the WAN link is ethernet, the IP packet is carried within an ethernet frame.
An ethernet frame has a 14 bytes header (6 bytes for destination MAC, 6 bytes for source MAC, 2 bytes ethertype = protocol type) and at the end there is a 4 byte FCS (a checksum of the whole frame).
However, ethernet is a LAN protocol that misses the use of a clock.
So actually on the physical layer the equivalent of 8 bytes are sent as a pre-amble just to allow the receiver side to synchronize before the frame is sent.
In addition to this a silence is needed between two frames to allow the receiver to understand and detect the end of the previous frame.
The total of pre-amble and inter-frame gap is 20,1 bytes equivalent.
So given a packet size in IP like 674 bytes, on wire they count like 674+18+20,1 = 712,1 bytes.
RTT = Round Trip Time = two way delay and it is a measure performed with ICMP or UDP with timestamps inside.
What is important to understand is that the total one way delay includes different components:
-serialization component (the one that I have explained above for the ethernet case)
-time to wait in queue variable on the router interface itself.
-propagation delay : time for the signal to arrive at destination can be important on long distance WAN links.
TCP connection setup requires a three way handshake that requires exchange of at least three packets
TCP SYN ---->
<----- SYN, ACK
TCP ACK
So the time to setup a TCP connection is not equivalent to RTT, your understanding is correct.
Hope to help
Giuseppe
