HL Logistic Modelling

HL logistic modelling

P (t) = \frac{L}{1 + C e^{- k t}}

A logistic model describes growth that rises and then levels off.

L = the maximum value the model approachesC = ?k = ?

Today: work out what C and k actually do.

Mini investigation 1 — L

P (t) = \frac{L}{1 + 15 e^{- 0.6 t}}

L800

Predict What should happen to the upper level as you drag L?

L sets the limiting value — the horizontal level the model approaches.

Mini investigation 2 — C

P (t) = \frac{800}{1 + C e^{- 0.6 t}}

C15

P(0) = 50.0

L = 800

Drag C. Watch the starting point and the rapid-growth region move.

C controls the starting position relative to the limit, and therefore helps determine when the rapid-growth region occurs.

Mini investigation 3 — k

P (t) = \frac{800}{1 + 15 e^{- k t}}

k0.60

P(0) = 50

logistic midpoint: P = L/2 = 400
half of the limiting value — not halfway from 50 to 800

Predict Which direction gives a more rapid transition towards 800?

Increasing k makes the transition towards the limiting value more rapid.

Read the whole model

P (t) = \frac{L}{1 + C e^{- k t}}

LMaximum / limiting value the model approaches.

CStarting position relative to the limit; shifts when rapid growth occurs.

kHow rapidly the transition towards the limit happens.

Why this shape?

Exponential growth does not know when to stop.

If growth keeps depending only on how much is already present, the model keeps accelerating.

Real systems often have a limit: space, resources, customers, people, capacity.

Verhulst's idea

Let L be the maximum value the system can approach.

\frac{d P}{d t} = k P (1 - \frac{P}{L})

You do not need to manipulate this yet. Read it as a sentence:

P
how much is already present

1 − P/L
the fraction of capacity still available

Think When P gets close to L, what happens to the second factor?

It approaches 0, so growth slows. Solving this differential equation gives the logistic family we just investigated.

P (t) = \frac{L}{1 + C e^{- k t}}

Unfamiliar context — information spread

R (t) = \frac{1800}{1 + 2 e^{- 1.4 t}}

R(t) is the number of people who have received a message; t is time in hours.

State the limiting value.

Find the number of people at t = 0.

Describe what k = 1.4 tells you qualitatively.

Limit: 1800
R(0) = 600
k = 1.4 gives a relatively rapid transition towards the limiting value.

P (t) = \frac{L}{1 + C e^{- k t}}

Use the model — TI-84

TEXAS INSTRUMENTSTI-84 Plus C Silver Edition

NORMAL FLOAT AUTO REAL RADIAN MP

Y1=12000/(1+23e^(-.42X))

Try: HOME → 4 → ENTER, or GRAPH

N (t) = \frac{12000}{1 + 23 e^{- 0.42 t}}

N(t) = participants in a digital service
t = months after launch

Question

Use the model to predict participation after 4 months.

After 4 months, the model predicts about 2,270 participants.

N (t) = \frac{12000}{1 + 23 e^{- 0.42 t}}

Solve a threshold

When does participation first reach 9,000?

Use any valid technology method. Then write the answer in context.

t ≈ 10.08 months
Participation first reaches 9,000 about 10.1 months after launch.

9,000 is 75% of the limiting value 12,000.

N (t) = \frac{12000}{1 + 23 e^{- 0.42 t}}

Reason from the shape

8,000 → 9,000 versus 9,000 → 11,000

Should those increases take about the same additional time? Explain from the curve before calculating.

8,000 → 9,000: about 0.97 months

9,000 → 11,000: about 3.09 months

The model slows as it approaches 12,000.

P (t) = \frac{L}{1 + C e^{- k t}}

Final prediction challenge

f (x) = \frac{500}{1 + 19 e^{- 0.3 x}}

g (x) = \frac{500}{1 + 4 e^{- 0.3 x}}

h (x) = \frac{500}{1 + 19 e^{- 0.9 x}}

Which model has the largest initial value?

Which has the most rapid transition towards the limit?

Which pair differs only in the value of C?

Which pair differs only in the value of k?

Largest initial value: g

Most rapid transition: h

Only C differs: f and g

Only k differs: f and h

How do I know I learned it?

Without notes, can you do these three things?

1 · ReadGiven a logistic equation, identify L and explain what C and k do.

2 · UseUse technology to find a predicted value or solve when a target is reached.

3 · JudgeExplain why growth slows as the model approaches its limiting value.

If yes: you have the core idea. If one is shaky: that is your next retrieval target.

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