Paths and directions

A crossing is a chain of decisions

Eleven rungs hand back a curve and stop, and nothing anybody flies is decided once. Re-planning is worth exactly nothing when the forecast turns out right — a sub-path of an optimal path is optimal, and the chain is the plan. When it turns out wrong it recovers 19 per cent of the cost at two decisions and 83 at eighteen, and never all of it.

Assumes The shortest route between two coasts.

Every rung of this anchor produces a curve and stops. Some of them produce it under a constraint — an obstacle, a curvature bound, a flow — and all of them produce it once, from a starting point that is not moving, against conditions that are not changing.

Nothing anybody crosses an ocean in is planned that way. A route is filed, and then it is re-filed: a new forecast arrives, the aircraft is somewhere else by then, and the plan is remade from where it is to where it is going. What gets flown is a chain of quickest paths from a sequence of moving starting points, and whether that chain is anywhere near the quickest path from the original start is a question no single optimisation answers.

The plan, the crossing that was flown, and the one a right forecast would have given. A crossing of NaN km planned against a jet forecast to sit on the fortieth parallel, flown through a jet that is actually 6 degrees further south, and re-planned six times on the way. Committing to the plan costs 167.81 hours. Re-planning costs 165.84. A vehicle that had known where the jet was would have taken 163.21. So the wrong forecast is worth 4.60 hours and re-planning recovers 1.97 of them.
Fig. 1 A crossing planned against a jet forecast to sit on the fortieth parallel and flown through a jet that is actually six degrees further south, re-planned six times on the way. The dashed track is the plan. The solid one is what was flown. The dotted one is what a vehicle that knew where the jet was would have flown. The two parallels are the forecast jet and the real one.

Half the question has an exact answer and it is zero

Before measuring what re-planning buys, it is worth being clear about when it buys nothing, because that case is exact and it is most of the intuition.

If the forecast is right, re-planning changes nothing at all. A sub-path of a quickest path is a quickest path between its own ends — Bellman’s principle, and it holds for any cost that accumulates along a path with no memory. So re-planning from a point already on the optimal track returns the remainder of that track, and the chain is the plan.

That is not an approximation or a tendency. It is an identity, and it is the refusal this rung is built on: a measurement in which re-planning under a correct forecast saves time is a measurement whose solver found a better path the second time round, which means the first answer was not optimal and every other number in the essay is measuring the solver instead of the world.

Running it: with the forecast right, the chain of six legs takes 132.83 hours against the single plan’s 132.41 — 0.32 per cent slower, at a grid of half a degree, and 0.04 per cent slower at a quarter. The sign is right and the magnitude is the chain’s own discretisation, since six separately-polished legs cannot beat one polished optimum. It is measured on every run and subtracted from everything below.

The first version of this machinery did not pass that check. It reported re-planning as saving 1.7 per cent under a correct forecast, which is impossible, and the cause was that the solver rounds its starting point to the nearest grid node — so a leg ending between nodes teleported the vehicle up to half a cell before the next solve, six times over. On a one-degree grid that is fifty-five kilometres of free travel per decision.

What a wrong forecast costs

The interesting case is the forecast that is wrong, and the error modelled here is the simplest one with a size: the jet is not where it was said to be.

Three ways to cross, in hours. The same crossing under three policies, with the jet 6 degrees south of where it was forecast. A vehicle that knew would take 163.21 hours; one that re-plans six times takes 165.84; one that commits takes 167.81. The whole spread is 4.60 hours on a crossing of 163, which is 2.8 per cent — and re-planning takes back 43 per cent of it for the cost of running the solver six times instead of once.
Fig. 2 The same crossing under three policies with the jet six degrees south of its forecast. A vehicle that knew takes 163.21 hours. One that re-plans six times takes 165.84. One that commits to its filed plan takes 167.81. The whole spread is 4.60 hours on a crossing of 163, which is 2.8 per cent, and re-planning takes back 43 per cent of it for the cost of running the solver six times instead of once.

Two and eight-tenths per cent is a real number for a real operation. On a transatlantic crossing it is the difference between arriving on schedule and not; on a fuel budget it is the reserve.

What it is not is large compared with the things the rest of this anchor measures. Why Mercator exists finds a rhumb line up to twenty-eight per cent longer than a great circle; the quickest route is not the shortest finds a quickest track arriving eleven per cent sooner than the shortest one. A wrong forecast, at this size of error, is a second-order correction on top of those.

Where the two tracks part company

The hero figure shows three curves and the useful reading of it is where they separate rather than how far apart they end up.

The plan and the flown track leave the departure point together, because the first leg is the plan — nothing has been observed yet. They part at the first decision, and from there the flown track heads south towards where the jet actually is while the plan continues towards where it was forecast. By the third decision the two are several hundred kilometres apart in latitude and the flown track has effectively given up on the forecast.

What the flown track does not do is reach the perfect one. It converges towards it and arrives late, because the first sixth of the crossing was flown in the wrong direction and the jet it eventually joins is entered further downstream than it need have been. That is what the seventeen per cent residual at eighteen decisions is made of, seen as a picture.

The three curves also make visible something the numbers hide. The plan and the perfect track are both smooth — each is a single optimisation — and the flown chain is not: it has a corner at every decision. Those corners are real and they cost something, which is a term the shortest route a vehicle can fly prices for a vehicle with a turning radius and this rung does not model.

How often is worth deciding

The number of decisions is the free parameter, and it is the one an operator actually controls.

What re-planning recovers, against how often it is done. The share of a wrong forecast's cost that re-planning gets back, against the number of decisions. One decision is committing and recovers nothing, by definition. Two recover 19.4 per cent, six recover 44.0, eighteen recover 83.0. It rises towards a hundred and does not reach it, and the residual is what was flown in the wrong direction before the error was known — a sunk cost that no later decision touches. Each point is corrected for the chain's own discretisation, which is measured separately by running the identical chain against a forecast that is right.
Fig. 3 The share of the wrong forecast’s cost that re-planning recovers, against how many decisions are made. One decision is committing and recovers nothing, by definition. Two recover 19.4 per cent, four recover 27.7, six recover 44.0, eighteen recover 83.0. Each point is corrected for the chain’s own discretisation, which is measured by running the identical chain against a forecast that is right; the uncorrected curve is drawn beside it and the two never differ by more than seven points.

The curve rises towards a hundred per cent and does not reach it, and the residual is the interesting part. What re-planning cannot recover is what was flown in the wrong direction before the error was known — a sunk cost that no later decision touches. At eighteen decisions the first leg is a eighteenth of the crossing, and seventeen per cent of the penalty survives.

That gives the shape of the trade. The first re-plan is worth about a fifth of the penalty; each subsequent one is worth less; and there is no point at which further decisions become free, because each one costs a solve and the recovery is asymptotic. An operator choosing a re-planning interval is choosing a point on a curve of diminishing returns whose slope is computable in advance.

And there is an error too small to be worth correcting

Sweeping the forecast error instead of the decision count produces the other half of the design rule, and it contains a sign change.

What the wrong forecast costs, and what re-planning gets back. Two quantities against how wrong the forecast was. The upper curve is what committing to the plan costs over what a perfect forecast would have given: nil at nought, 0.75 hours at two degrees, 7.19 at twelve. The lower curve is what re-planning six times gets back. At one degree the recovery is NEGATIVE — the chain's own overhead is larger than the error it is correcting — and above two degrees it is between two-fifths and nearly all of the penalty. There is a size of forecast error below which re-planning is not worth doing, and it is where the two curves cross.
Fig. 4 Two quantities against how wrong the forecast was. The upper curve is what committing costs over a perfect forecast: nil at nought, 0.75 hours at two degrees, 4.60 at six, 7.19 at twelve. The lower curve is what re-planning six times gets back. At one degree the recovery is negative — the chain’s own overhead exceeds the error it is correcting — and above two degrees it runs between two-fifths and nearly all of the penalty.

The crossing point is the practical result. Below about a degree and a half of jet displacement, re-planning a crossing six times makes it slower, because the chain of separately-optimised legs is a worse approximation to the true optimum than the single plan is. The forecast has to be wrong by enough to be worth chasing.

That is not a statement about numerical noise, though it is measured through it. It is a statement about a real cost: every re-plan is a decision made on partial information, and a policy of continuously re-optimising against the latest observation is a policy of chasing a signal that may be smaller than the noise in the decision procedure itself.

The same shape appears wherever a controller reacts to a measurement, and it has a name in that field. What is worth noticing here is that the crossing point is computable in advance from two things an operator already has — the accuracy of the forecast, and the granularity of the planner.

Why the recovery is not a fixed fraction

A reader looking at the two ladders might expect the recovered share to be a property of the situation — some fixed fraction of a penalty that re-planning always gets. It is not, and both ladders say so in different ways.

The decision-count ladder says it depends on how early the first correction happens, because everything before it is sunk. The forecast-error ladder says it depends on the size of the error relative to the chain’s own overhead, because a small correction is not worth the discretisation it costs.

Putting the two together gives the rule with both variables in it: re-planning recovers roughly (1 − 1/n) of the penalty, less a fixed overhead that does not shrink with the penalty. At six decisions that first factor is 83 per cent and the measured corrected recovery is 44 — so something else is losing half of the available saving, and it is that each leg is planned as though the currently observed flow held everywhere for the rest of the crossing, which is itself a forecast and is itself wrong.

A vehicle re-planning against a persistence forecast is correcting one error with another. That it recovers anything at all is the finding; that it recovers about half of what the sunk-cost bound allows is where the rest goes.

What the solver is, and what its grid costs

Everything here is Dijkstra on a latitude–longitude grid with eight neighbours per node, the step cost being the ground length divided by the speed made good along it — the first-order Zermelo cost, exact while the flow is slower than the vehicle, which it is throughout.

The grid step is a quarter of a degree, and the choice was forced by the refusal rather than by taste. At one degree the chain’s own overhead under a correct forecast is 0.99 per cent; at half a degree it is 0.32; at a quarter it is 0.04. The quantity being measured is a recovery of two to five per cent, so an overhead of one per cent is a third of the signal and an overhead of four hundredths is not.

That progression is worth reading as a convergence check rather than as a setting. The overhead is a pure artefact — it must be zero for an exact solver — so its fall by a factor of twenty-five across two halvings of the step says the discretisation is second-order, which is what a midpoint-sampled cost should be. Every reach set ever drawn is too small makes the same kind of check on a different instrument in the neighbouring anchor: a quantity known to be zero, measured, and used to size everything that is not.

The cost of the finer grid is about a second and a half per crossing and thirteen seconds for a ladder, which is nothing. The reason the coarser grid was there first is that nobody had asked the question the refusal asks.

What is being modelled and what is not

Every number above rests on a model of how a forecast fails, and the model is one line: the jet is where it was said to be, minus a stated number of degrees of latitude.

That is a real failure mode and it is not the only one, and being explicit about the difference is the whole of what the rung can claim.

It is a bias rather than a noise. The jet is displaced consistently in one direction, so the error does not average out along the crossing and re-planning has something systematic to correct. A forecast wrong in a way that varies randomly along the route would be corrected far less effectively, because each re-plan would be chasing a different error.

It is known perfectly once observed. The re-planning vehicle is assumed to see the true flow at its current position and to assume it holds everywhere for the rest of the crossing. That is optimistic in one direction — a real observation is local and noisy — and pessimistic in another, since a real operator gets a new forecast rather than a point measurement.

The decisions are evenly spaced along the route rather than triggered by anything. A real operator re-plans when a new forecast arrives, or when the observed conditions depart from the plan by some threshold, and both of those put the decisions where they are worth most. An evenly-spaced chain is the policy that needs no information about when to act, so it is a floor rather than a typical case — the tolerance decides the model is the essay about choosing a threshold and this rung deliberately has none.

And the flow does not change with time. Both the forecast and the truth are steady fields; what changes between them is which field is real. A genuinely time-varying flow is a different problem, with a time coordinate in the cost, and nothing in this collection solves it.

Those three together mean the recovered shares here are a middle case rather than a bound. A biased forecast observed perfectly is the situation re-planning helps most in; a noisy one observed noisily is where it helps least.

The larger error, and where the recovery goes

At six degrees of jet displacement the picture is a correction. At twelve it is a different crossing.

The plan, the crossing that was flown, and the one a right forecast would have given. A crossing of NaN km planned against a jet forecast to sit on the fortieth parallel, flown through a jet that is actually 12 degrees further south, and re-planned twelve times on the way. Committing to the plan costs 206.14 hours. Re-planning costs 199.38. A vehicle that had known where the jet was would have taken 198.96. So the wrong forecast is worth 7.19 hours and re-planning recovers 6.77 of them.
Fig. 5 The same crossing with the jet twelve degrees south of its forecast, re-planned twelve times. The plan and the flown track diverge almost immediately and the flown track ends up in a completely different part of the ocean. Committing costs 206.14 hours against a perfect 198.96 — 7.19 hours, or 3.6 per cent — and re-planning twelve times recovers 6.77 of them, which is 94.2 per cent. Six decisions recover 96.8, slightly more, which is the discretisation of the longer chain rather than a real reversal.

The recovered share is much higher at the larger error and the reason is arithmetic rather than anything about weather. The sunk cost of the first leg is a fixed fraction of the crossing, and what it costs in hours depends on how wrong the flow was over that leg — but the penalty being recovered grows faster than linearly with the displacement, because a jet twice as far away is not merely twice as unhelpful. So the same sunk fraction is a smaller share of a larger penalty.

Which gives the counter-intuitive practical statement: re-planning is most valuable exactly when the forecast is worst, and its value as a fraction of what is at stake rises with the size of the error rather than falling. An operator who re-plans only in settled conditions has it precisely backwards.

What the chain does to the anchor’s own curves

Eleven rungs produce curves and this one produces a chain, and it is worth asking which of the eleven the chain would change.

The ones about the metric are untouched. The rhumb line’s excess, the ellipsoid’s flattening cost, the great-circle vertex, the antipodal degeneracy — all are statements about the geometry of a curve between two fixed points, and no amount of re-planning alters them.

The ones about a constraint are partly touched. The shortest route a vehicle can fly prices a curvature bound and finds a fixed turning cost decided by the two headings. Re-planning changes the headings at each decision, so the turning cost is paid once per leg rather than once per crossing — which makes a chain of eighteen decisions carry eighteen turning costs and is a term this rung does not model.

And the ones about a flow are directly touched. The quickest route is not the shortest computes an optimal track in a known flow, which is exactly the plan this rung then discovers was made against the wrong flow. Every number in that essay is right about the flow it was given and says nothing about whether that flow was the one that turned up.

That last one is the honest summary of the anchor’s whole position on flows. A quickest route is a conditional statement, and this rung is the first to price what the condition is worth.

It touches the neighbouring anchor the same way. A partition under a directed cost has two versions divides a sphere by travel time under a flow and finds the outbound and inbound divisions disagreeing about half of it; every one of those travel times is computed against a flow that is assumed known. A division computed against a forecast that turns out six degrees wrong is a third division again, and the same sunk-cost argument applies to it — a catchment cannot be re-planned at all, because the ground does not move.

What the residual is made of, at every size

Three numbers now describe the residual and they do not agree, which is worth facing rather than smoothing.

At six degrees of displacement and six decisions, re-planning recovers 44 per cent. At six degrees and eighteen decisions, 83. At twelve degrees and six decisions, 97. The sunk-cost bound — one minus one over the number of decisions — is 83 per cent at six decisions and 94 at eighteen, so the six-decision figure at six degrees is far below its bound and the six-decision figure at twelve degrees is above it.

Being above the bound is not a contradiction, because the bound assumes the first leg is flown exactly as the perfect track would fly it and it is not: at a large displacement the plan sends the vehicle so far the wrong way that the first leg itself is worse than useless, and re-planning after it can still beat the committed track by more than the bound suggests, since the committed track keeps going the wrong way.

What all three have in common is that the recovery depends on the interaction between the size of the error and the timing of the first correction, and that no single fraction describes it. The honest statement is a range with its ends measured: between two-fifths and nearly all, on the cases here, with the low end where the error is small and the corrections are few.

Where the ladder goes next: more than one objective at a time

Twelve rungs price a route’s metric, its obstacles, its flow, its instrument, its vehicle, its endpoints and now its schedule. Every one of them optimises something, and every one of them takes the objective as given: shortest, or quickest, or shortest-subject-to-a-turning-radius.

What none of them asks is what happens when there is more than one objective and they disagree. A crossing that is quickest is not the one that is most fuel-efficient, nor the one that stays nearest to a diversion airfield, nor the one that spends least time over water — and those are not weights on a single number but genuinely different problems with genuinely different answers. The set of routes that cannot be improved in one objective without being made worse in another is a curve rather than a point, and no rung of this anchor has drawn one.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

ConvergenceCostDiscretisationEstimatorFlow fieldOptimal controlPurposeRoute planningShortest pathToleranceVerificationZermelo navigation