If the transmission delay is the time it takes for a router to put a certain number of bits onto a link, why wouldn't it take the same amount of time to take them off the link? Thus making the total time to transmit equal to twice the transmission delay plus the propagation delay.

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    Yes, it takes the same amount of time to get bits off the wire as it does to put them on the wire, but bits can be pulled from the wire at the same time other bits are being put onto the wire. You are assuming a frame must be fully sent before it can be received, but that is not true. – Ron Maupin Sep 11 '19 at 14:48

In practice, there are three delays in a simple transmission:

  • serialization delay of the source
  • propagation delay between source and destination
  • serialization delay of the destination

Once serialized, the first bit starts travelling to the destination while source serialization is still running. So, destination serialization entirely overlaps with either source serialization or propagation (if you neglect the first bit or symbol) and you can simply ignore it.

From another perspective, the first symbol just needs the propagation delay to arrive. The last symbol is delayed by the serialization of all previous symbols and then by the propagation delay - that is effectively what you're looking for.

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