Signals in fibre travel at about two thirds of the speed of light, so a round trip over a great-circle distance d cannot take less than 2d ÷ (⅔c), which is 3d/c. New York to London is 55.889 ms and no engineering removes it. What a measurement has above that floor is the part that is queueing, serialisation, middleboxes and software — and it is the only part anyone can fix.
Core ideas
The hard idea is that one term of a latency measurement is not an engineering quantity at all. It is geography, and the only lever on it is moving the endpoints closer together.
Light in vacuum travels 299 792 458 m/s, exactly — the metre is defined from it, so c is not measured but fixed. Glass slows it to about two thirds of that. A round trip covers 2d, so the minimum is 2d ÷ (2c/3) = 3d/c, and every figure on this page follows from those two numbers alone.
Subtract the floor from a measurement and what is left is queueing at each hop, the time to clock the bits onto each link, the middleboxes, and your own software. New York to London measured at 76 ms leaves 20.111 ms. That is the whole budget an engineering effort has to work inside, and knowing it stops an argument about the other 55.889 ms.
Putting a copy of the content nearer the reader reduces the distance, which reduces the floor, which is a real and large win — for content that can be served from the copy. For a request that must reach the origin to read or change something only the origin knows, the distance is unchanged and so is the floor.
The material
An exact fraction of a millisecond, and the honest reading of a measurement that sits above it.
The light-speed floor for a route is the shortest possible round-trip time over it: twice the great-circle distance, divided by the propagation speed of a signal in fibre. This course takes that speed as ⅔c, so the floor is 2d ÷ (2c/3) = 3d/c. It assumes a perfectly straight cable and zero processing, and it is therefore a bound nothing beats rather than a prediction.
Because c is exact by definition, so is the floor: for New York to London it is the fraction 8 377 500 000 / 149 896 229 milliseconds, which is 55.889 ms to three places. The course quotes it as 56 ms, and the lab prints the fraction it came from so that the rounding is visible rather than assumed.
Set the lab’s measured round trip to 76 ms, which is a plausible figure for that route on a good day. The measurement is 1.360 times the floor, and the unexplained remainder is 20.111 ms. That remainder is the number to take into a meeting. Halving it — which would be an extremely good quarter of work — takes the route from 76 ms to about 66 ms, a saving of thirteen percent.
A 170 ms floor between London and Sydney is why a service with one region has a bad day in the other hemisphere, and it is why the fix is a second copy of the data rather than a faster anything. This is the honest case for a content network: it does not accelerate a packet, it shortens the route the packet has to take.
A reader in Sydney loads a page: the images, the stylesheet and the script come from a nearby cache, and their floor drops from 170.048 ms to whatever the local distance is. Then they post a comment. That write must reach the one place that owns the data, so its floor is 170.048 ms again, and so is the floor on the confirmation. No configuration of the cache changes either number. “Put it on a CDN” is an answer about the first group of requests offered as if it were an answer about the second.
The lab will also let you enter a measurement below the floor, and it prints a negative remainder when you do. That is not a fast network; it is a wrong distance. The usual cause is measuring to a nearby edge node while believing you measured to the origin, which is the same confusion as the example above with the sign the other way round.
Two cautions on the model. The ⅔c figure is a convention — real fibre runs at roughly 0.66 to 0.69 of c depending on the glass — so treat the floor as accurate to a few percent rather than exact in the physical sense, even though the arithmetic that produces it is exact. And real cables are not great circles: they follow coastlines, rights of way and existing ducts, so the true path is longer than d and the true floor is therefore somewhat higher than the one computed here.
Interactive
What the distance costs, and what is left over to blame on the system
Method
Three lines to the floor, and a fourth that decides whether the remainder is worth anyone’s quarter.
The floor is a lower bound, so it should use the shortest distance that exists, which is the great circle. A real cable is longer and its true floor is higher; using the great circle keeps the bound honest.
Twice the distance at two thirds of c is three times the distance at c. With d in metres and c in metres a second the answer is in seconds, so multiply by a thousand once, at the end.
The difference is the whole of the improvable part. Quote it as a number of milliseconds and as a ratio: 76 ms against a 55.889 ms floor is 20.111 ms of remainder at 1.360 times the floor.
If the remainder is a fifth of the measurement, then a heroic effort that halves it moves the user-visible number by a tenth. If the remainder is four fifths of the measurement, the same effort is transformative. The ratio tells you which world you are in.
One measurement, split into the part that is geography and the part that is somebody’s code.
The remainder is 26% of the measurement, so the very best possible outcome — removing every queue, every middlebox and all the software — is a 26% improvement. That is the ceiling on the whole category of work, and it is worth knowing before the work starts.
The same arithmetic run the other way is how you decide to place a second region. If the target is a 100 ms round trip, then 3d/c ≤ 100 ms gives d ≤ 9993 km, and every user further away than that cannot be served inside the target from this location no matter what is built. That is a siting decision falling out of one division.
For a faded rehearsal, take Frankfurt to Singapore at 10 263 km with a measured round trip of 165 ms. The supplied first move is the floor, which the lab prints as 102.701 ms. Compute the ratio and the remainder, then say what fraction of the measurement the best possible engineering effort could remove, and whether a 120 ms target is reachable from Frankfurt at all. Write the answer down before opening the quiz.
You should be able to compute 3d/c for a route, express a measurement as a multiple of that floor, state the remainder in milliseconds, and say which proposed changes can move it and which cannot.
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