1Demonstration 1 of 4
Growth looks like S-shaped growth until the ceiling bites
While a stock is far below its ceiling, can the trace tell growth from S-shaped growth?
S-shaped growth multiplies the growth term by (1 - level / capacity). While the level is tiny against the capacity that factor is close to 1, so the two paths agree. As the level nears the ceiling the factor shrinks and the S-shaped path flattens.
Both paths start at 1 and use a rate of 0.3 per step. The S-shaped path also takes a capacity, the ceiling, in the same units as the stock. The level is a count of anything that accumulates.
Predict first. After 10 steps with a ceiling of 100, are the growth and S-shaped levels close or far apart?
Choose an example
Constructed example: the start, rate and ceiling of the chapter's neighbor-confusion figure, recomputed with the pack's generators.
Calculated values
- Steps shown
- 20
- Growth level
- 190.0
- S-shaped level
- 72.4
- S-shaped level as a fraction of the ceiling
- 0.72
- Growth level as a fraction of the ceiling
- 1.90
Step 1 is almost the same: growth 1 + 0.3 x 1 = 1.300, S-shaped 1 + 0.3 x 1 x (1 - 1/100) = 1.297. After 20 steps growth is 1 x 1.3^20 = 190.0 and the S-shaped path is 72.4, which is 72.4 / 100 = 0.72 of the ceiling. Growth has passed the ceiling that the S-shaped path is bending away from.
Use the idea
Before labeling a rising trace as growth, write down an estimate of the limit and where the current level sits against it, because the trace alone does not carry that.
Where the conclusion applies
A fixed ceiling, a start of 1 and a constant rate of 0.3. The comparison fails if the limit itself moves, which is the next demonstration's subject.
Check your understanding: With a ceiling of 50 and 15 steps shown, is the S-shaped level above or below half of the ceiling?
Chapter 3 source: section "Telling the neighbors apart". Demonstration C03-D01.
2Demonstration 2 of 4
Decay is goal seeking with the goal at zero
What separates a decay trace from a goal-seeking trace?
Each step closes 1 / adjustment_time of the remaining gap. When the goal is zero the gap is the stock, so the update is the decay update. Any other goal changes where the path ends, not the way it approaches.
The stock starts at 100. The goal is the level the path closes toward. The adjustment time is how many steps it takes to close the whole gap at the first step's pace; the decay rate is 1 divided by that time.
Predict first. With a goal of 40, does the stock keep falling toward zero or level off near 40?
Choose an example
Constructed example: the goal-seeking call printed in the chapter, run downward from a start of 100 and compared with the pack's decay generator.
Calculated values
- Goal
- 40
- Adjustment time (steps)
- 4
- Goal seeking after step 1
- 85.0
- Decay after step 1
- 75.0
- Goal seeking after step 30
- 40.01
- Decay after step 30
- 0.02
- Same curve
- no
Goal seeking, step 1: 100 + (40 - 100) / 4 = 85.0. Decay, step 1: 100 - 0.25 x 100 = 75.0. After 30 steps goal seeking is 40 + (100 - 40) x 0.75^30 = 40.01. The curves part: decay keeps heading for 0 while goal seeking levels off at 40. Same shape, different mechanism, so the goal is the thing to find out.
Use the idea
When a trace falls and flattens, ask what level it is flattening toward before calling it decay, since the label asserts a target of zero.
Where the conclusion applies
Discrete steps of length 1 and a goal that does not move. If the goal drifts, the trace shows a different shape.
Check your understanding: With a goal of 70 and an adjustment time of 4, what is the level after the first step from 100?
Chapter 3 source: section "Telling the neighbors apart". Demonstration C03-D02.
3Demonstration 3 of 4
One erosion rate turns a ceiling into a collapse
What does the stock do when it wears away the limit it is growing toward?
The ceiling falls by erosion_rate x stock every step. Once the stock grows past the shrinking ceiling, its growth term turns negative and the stock falls. With zero erosion the ceiling never moves and the path is S-shaped.
The stock starts at 1 against a ceiling of 100. The rate is the growth per step. The erosion rate is how much of the ceiling each unit of stock removes per step.
Predict first. At erosion rate 0.02 and growth rate 0.4, does the path settle at 100 or turn over?
Choose an example
Constructed example: the overshoot call printed in the chapter, varying its erosion and growth rates.
Calculated values
- Erosion rate
- 0.02
- Growth rate
- 0.4
- Highest level
- 87.6
- Turning point (step)
- 20
- Level at step 200
- 1.99
Step 1: 1 + 0.4 x 1 x (1 - 1/100) = 1.396, and the ceiling falls to 100 - 0.02 x 1 = 99.98. That small cut repeats every step, in proportion to the stock. The path peaks at 87.6, 12.4 below the original ceiling, and ends at 1.99: overshoot and collapse.
Use the idea
A flattening trace is ambiguous at the moment it flattens. Ask whether the limit is fixed or whether the stock consumes it, which usually comes from knowing the system.
Where the conclusion applies
A linear erosion of the ceiling, 200 steps and no recovery of the ceiling. A system whose limit regrows would not collapse this way.
Check your understanding: At erosion 0.02 and growth rate 0.4, by how much does the ceiling fall at the first step from a stock of 1?
Chapter 3 source: section "Generating the six shapes". Demonstration C03-D03.
4Demonstration 4 of 4
Noise hides a slow approach first
How much measurement noise can a goal-seeking path take before its shape is lost?
The clean path moves by (goal - level) / adjustment_time per step, a geometric sequence. A fast approach makes a few large moves early; a slow one makes many small moves that sit below the noise sooner.
The clean path closes toward a goal of 100 from 0. Each observation adds Gaussian error with standard deviation sd. The seed is fixed, so a run can be repeated exactly.
Predict first. At sd 20 and an adjustment time of 4, on how many of the 60 steps does the clean path move by more than the noise?
Choose an example
Constructed example: the chapter's noise experiment with seed 7, using the standard deviations 1, 5 and 20.
Calculated values
- Adjustment time (steps)
- 4
- Noise standard deviation
- 5
- Move at step 1 (clean)
- 25.00
- Move at step 10 (clean)
- 1.88
- Steps whose clean move exceeds the noise sd
- 6 of 60
The clean move at step 1 is (100 - 0) / 4 = 25.00. At step 10 it is (100 - 92.49) / 4 = 1.88. Against a noise standard deviation of 5, 6 of the 60 steps move the clean path by more than that. The seed is fixed at 7, so only the noise level and the adjustment time change the picture. Whether the shape survives is a property of the record, and a slower approach has fewer large moves.
Use the idea
Before claiming a pattern in a real series, ask what noise level the pattern would survive, and record the smoothing or exclusion choices in the treatment field.
Where the conclusion applies
Independent Gaussian error and one fixed seed. Real measurement error can be correlated or change over the window, which would make the picture look different.
Check your understanding: With an adjustment time of 4 and sd 5, does the clean move at step 1 exceed the noise sd?
Chapter 3 source: section "Generating the six shapes". Demonstration C03-D04.