Why does the delay in sending a packet from source to destination doesn't depend upon the transmission rate and the packet length? If we look at the delay formula then it is, L/R + D/V.

  • Sometimes, people think one bit must arrive before the next bit is sent, but that is not the case, and you could have all the bits on the wire at the same time. – Ron Maupin May 21 '19 at 18:59
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    It depends which "delay" you are referring to. You can measure it in different ways depending on your goal. – Ron Trunk May 21 '19 at 19:04
  • Did any answer help you? If so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you can provide and accept your own answer. – Ron Maupin Dec 15 '19 at 3:07

It's a component, but in most WAN situations it's the smallest contributor.

Assume a 100 Mbit/s link speed and a 1500 byte packet. Serialisation delay is 1500 bytes x 8 bits/byte / 100,000,000 bits per second = .00012s.

Many servers do process switching 1000 times per second. This means a processing delay of 1ms (0.001s) is common, or eight times as long as the delay from packet length.

Speed of light in fibre is approximately 200,000 km/s, so the packet transmission delay matches 24km of length, and the processing delay is equivalent to 200 km. Unless you're talking to a data centre in your local city, speed of light delay is typically the largest factor.

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There are several delays:

  • propagation delay = distance / wave propagation velocity (D/V)
  • serialization delay = packet length / bitrate (L/R)
  • processing delay = device dependent
  • access delay (only for shared/multiple-access media) = depends on access method (CSMA/CD, CSMA/CA) and channel load

Add them together for total delay.

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