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Why is packet delivery time equal to transmission time + propagation delay, and not equal to 2*(transmission time) + propagation delay?

At the end of packet delivery time, doesn't that mean that those bits haven't been received?

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  • $\begingroup$ Why do you think it should be two transmission times? The packet is only transmitted once! $\endgroup$ – David Richerby Jan 21 at 18:32
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After the transmission time has passed, the last bit of the packet has just been put on the wire. That bit (just like all the others) will take an amount of time equal to the propagation delay to reach the receiver. When the last bit reaches the receiver, all the earlier bits will have reached it. So the total time taken is the transmission time plus the propagation delay.

It might be easier to visualize what's going on if you think about a train instead of a data packet. Imagine that we're standing by the track, some distance apart. We want to know how long it is from when the front of the train passes me (the time at which "I start sending the train to you") until the whole train has passed you (the time at which "you finish receiving the train"). This is equal to the time it takes the whole train to pass me (transmission time), plus the time it takes the back of the train to reach you (propagation delay). Or, if you prefer, it's equal to the time it takes the front of the train to reach you (propagation delay) plus the time it takes the rest of the train to pass you (transmission time).

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