1Demonstration 1 of 4
Two measures, two slopes
Does the quoted delay rise with how full the hour is, and does taxi-out?
Taxi-out runs from pushback to wheels up and tracks runway queueing. Departure delay is lateness against a schedule, which rises through the clock day and can fall in loaded morning hours.
Load is an hour's scheduled departures over the airport's 95th percentile hourly count, so 1 is as full as the busiest one hour in twenty. Taxi-out and departure delay are minutes per departure.
Predict first. Between the lowest bin and the bin at load 1.15, does the 2023 departure delay rise or fall?
Choose an example
Constructed example on the committed public record: the bin means are computed from it, and the minimum bin size is a constructed variable.
Calculated values
- Measure
- Taxi-out
- Year
- 2023
- Smallest bin kept
- load 0.05, 22,576 departures
- Mean at load 0.05 (minutes)
- 15.91
- Mean at load 1.15 (minutes)
- 20.87
- Change (minutes)
- 4.96
- Bins kept
- 14
Between the bin at load 0.05 and the bin at load 1.15 the mean rises: 20.87 - 15.91 = 4.96 minutes. Taxi-out is a proxy for runway queueing, and departure delay measures lateness against a schedule, so their slopes differ because they cover different stages. The pooled curve crosses thirty airports and is no one airport's. Bins under 20,000 departures are dropped, and a constructed larger cutoff changes which bins remain.
Use the idea
Plot the physical waiting time against load beside the target based statistic before pricing a change.
Where the conclusion applies
Bin means are weighted by scheduled departures and pool thirty airports. The different slopes do not identify padding, since the measures cover different stages.
Check your understanding: If taxi-out is 15.9 minutes at load 0.05 and 20.9 at load 1.15, what is the change?
Chapter 39 source: section "The record read across airports and months". Demonstration C39-D01.
2Demonstration 2 of 4
A convex curve fitted to one year
How far does the fitted curve sit from the bins it was fitted to, and from the next year's?
The grid search chose the curvature that minimizes the mean absolute error over the 2023 bins. A different curvature raises that error, and the next year's bins sit a little above the curve.
Load is unitless. Taxi-out is minutes per departure. Base is 16.79 minutes and the knee 0.0, both from the fit, and curvature is minutes per unit of squared load.
Predict first. At the fitted curvature, will the error against 2024 be smaller or larger than against 2023?
Choose an example
Constructed example on the committed public record: the base and knee are inferred from it, and the other curvatures are constructed values.
Calculated values
- Curvature (minutes per unit of squared load)
- 3.09
- Year scored
- 2023
- Bins scored
- 14
- Mean absolute error (minutes)
- 0.54
- Curve at load 1.35 (minutes)
- 22.42
- Bin mean at load 1.35 (minutes)
- 19.37
The curve is 16.79 + 3.09 x load squared, with the knee at 0.0. At load 1.35 it gives 16.79 + 3.09 x 1.35 x 1.35 = 22.42 against a bin mean of 19.37, a miss of 3.05 minutes. The mean absolute error over the 14 bins is 0.54 minutes. This is the fitted curvature, inferred by the grid search from the 2023 bins. A convex form cannot bend down, so the thin top bins are where form and record part.
Use the idea
Hold out a second period and report its error beside the fit error.
Where the conclusion applies
A convex form, so it cannot follow the easing of the two thinnest top bins. The fit does not place a knee. The mechanism is not identified.
Check your understanding: If the curve gives 22.4 at load 1.35 and the bin mean is 19.4, what is the miss?
Chapter 39 source: section "Fit and holdout". Demonstration C39-D02.
3Demonstration 3 of 4
Pricing a cap needs two points on the curve
What does capping the busiest hours buy, and what does it cost?
The benefit is the curve at the load as filed minus the curve at the capped load. Both must sit on the curve, so the slope between them is what prices the cap.
Load is as filed. A cap replaces the load with a lower fixed one. Benefit is minutes of taxi-out above the lowest-load bin per peak departure. Movements lost are shares of scheduled departures.
Predict first. At the record's mean peak load, does the p90 cap buy more or less than twice what the p95 cap buys?
Choose an example
Constructed example on the committed public record: the curve is inferred from it, the caps are the chapter's constructed policies, and the loads are constructed values.
Calculated values
- Cap
- p95
- Load as filed
- 1.094
- Load after the cap
- 1.000
- Delay above baseline, as filed (minutes)
- 4.59
- Delay above baseline, capped (minutes)
- 3.97
- Benefit per peak departure (minutes)
- 0.61
- Departures kept by the cap (percent)
- 98.95
- Movements lost, model (percent)
- 2.26
Benefit = delay as filed - delay capped = 4.59 - 3.97 = 0.61 minutes per peak departure. Movements lost = 1 - 0.98945 x (1 - 0.0121) = 2.26 percent, which adds the cancellation slope to the departures above the cap. The curve is steeper at higher load, so the same cap buys more when the load as filed is higher. The pooled curve is inferred from the record and is no one airport's, and the loads shown are constructed.
Use the idea
Price a schedule change from the slope of the curve between two loads, not from an average.
Where the conclusion applies
The pooled curve is no one airport's. Departures above the cap are counted as lost, not moved. One representative peak hour a day. A tie appears when a cap does not bind.
Check your understanding: If delay as filed is 4.58 and capped is 3.97, what is the benefit?
Chapter 39 source: section "Why an average delay per flight cannot price a schedule change". Demonstration C39-D03.
4Demonstration 4 of 4
Padding hides the delay it responds to
How much of the realized delay does a schedule's padding hide after a year?
Padding rises toward realized delay at a rate set by the adjustment time. The faster it adjusts, the less of the delay a schedule reports.
Realized delay and padding are minutes per departure. The padding adjustment time is in days. Load is as in the other demonstrations.
Predict first. With a 30 day adjustment time, is reported delay at day 365 larger or smaller than with 180 days?
Choose an example
Constructed example on the committed public record: the curve is inferred from it, and the adjustment times and loads are constructed values.
Calculated values
- Padding adjustment time (days)
- 90
- Load
- 1.094
- Realized delay, day 365 (minutes)
- 4.59
- Padding, day 90 (minutes)
- 2.87
- Padding, day 365 (minutes)
- 4.51
- Reported delay, day 365 (minutes)
- 0.08
Reported delay = realized delay - padding = 4.59 - 4.51 = 0.08 minutes at day 365, and padding at day 90 is 2.87. A longer adjustment time lets the gap stay open longer, so less of the realized delay is hidden by the end of the year. The balancing loop closes the gap between the schedule and the delay, and it does not establish that padding caused the pattern in the record. Adjustment times and loads here are constructed settings.
Use the idea
To test a padding explanation, compare schedule buffers over time for comparable flights.
Where the conclusion applies
A model of one balancing loop, with adjustment times chosen here. The record does not establish a padding response, and other mechanisms can produce the same slopes.
Check your understanding: If realized delay is 4.59 and padding 4.51, what is reported delay?
Chapter 39 source: section "The loops". Demonstration C39-D04.