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It is well know (see this question) and easily verifiable that end-to-end delay in case of trasmission of P packets of size L over N equal links (same transmission rate R) with zero propagation delay is given by the formula

N*L/R + (P-1)*L/R

I'd like to know how change this formula if we add propagation delay in the hypothesis that all the links are of the same distance D and speed C.

In my convinction the propagation delay does not affect the transfert delay, and every link add a propagation delay of D/C so the formula should be:

 N*L/R + (P-1)*L/R + N * D/C

I looked for confirmation in the literature but couldn't find any source on how to add the (simplified) propagation delay, so I'm asking here.

This is not an home work: I'm looking for a more complete formula. The base formula already use simplified assumptions, so why not add the propagation delay too ?

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    Why don't you just ask yourself what is happening in slow motion? You are also making assumptions that all forwarding is store-and-forward which is not always the case.
    – Zac67
    Commented Dec 12, 2023 at 4:46
  • thanks @Zac67 for your reply but I cannot understand your hint
    – frhack
    Commented Dec 12, 2023 at 10:29
  • Which one? Just picture each single bit leaving the source, traveling the length of the cable and being received on the other end.
    – Zac67
    Commented Dec 12, 2023 at 11:55
  • @Zac67 I end up concluding that the propagation delays sum to N * D/C Isn't it correct ?
    – frhack
    Commented Dec 12, 2023 at 13:09
  • If C refers to media-adjusted speed of light (c0*VF), then yes. Trivially, v=d/t means t=d/v with d =N*D.
    – Zac67
    Commented Dec 12, 2023 at 14:14

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