Long-term behaviour and steady states | Year 12

01 / LEARN

Understand the idea.

Equilibrium balances the flows

A steady state satisfies $T s = s$ : one transition leaves the category totals unchanged. Individuals can still move, but inflow and outflow balance. Use the total population as well as balance equations to determine the actual counts.

In a closed two-state model, one independent balance equation plus the total is enough; the other balance equation repeats the same information. A steady distribution need not divide the population equally, because the two transfer rates may differ.

Fixed does not always mean attracting

Finding a fixed distribution and proving convergence to it are different questions. Repeated multiplication can investigate a particular forecast, but matching rounded displays alone does not prove that every starting distribution approaches the same limit.

A swap matrix fixes an equal split while making unequal starting states alternate forever. The identity fixes every state. Solve the steady-state equations first, then separately examine the dynamics and whether constant transition assumptions remain appropriate.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Commuting choices

Rows are (0.7,0.2),(0.3,0.8). Find steady counts for 250 commuters.

Worked solution

1
Write the A balance equation.
$\begin{aligned} a & = 0.7 a + 0.2 b \\ 0.3 a & = 0.2 b \\ 3 a & = 2 b \end{aligned}$
At equilibrium, commuters leaving A balance those entering A from B.
2
Use the fixed population total.
$\begin{aligned} a + b & = 250 \\ a + \frac{3}{2} a & = 250 \\ a & = 100, b = 150 \end{aligned}$
The balance ratio $a : b = 2 : 3$ divides 250 commuters into five equal parts.
3
Verify both destination counts.
$(\begin{array}{cccccccccc} 0.7 (100) + 0.2 (150) \\ 0.3 (100) + 0.8 (150) \end{array}) = (\begin{array}{cccccccccc} 100 \\ 150 \end{array})$
The forecast equals the starting state and the counts total 250.

Answer

(100,150).

EXAMPLE 02

Three-state balance

Find a steady state totalling 200 for the displayed matrix.

Columns=current, rows=next; order A,B,C.
Next / current	A	B	C
A	0.5	0.25	0
B	0.5	0.5	0.5
C	0	0.25	0.5

Worked solution

1
Balance A using the first matrix row.
$\begin{aligned} a & = 0.5 a + 0.25 b \\ a & = 0.5 b \end{aligned}$
The zero C-to-A entry contributes nothing to the first balance equation.
2
Balance C using the third row.
$\begin{aligned} c & = 0.25 b + 0.5 c \\ c & = 0.5 b \end{aligned}$
The third row relates the steady C count to B.
3
Apply the total of 200.
$\begin{aligned} a + b + c & = 200 \\ 0.5 b + b + 0.5 b & = 200 \\ b & = 100, a = c = 50 \end{aligned}$
Both outside states contain half as many members as B.
4
Verify the complete three-state forecast.
$T (\begin{array}{cccccccccc} 50 \\ 100 \\ 50 \end{array}) = (\begin{array}{cccccccccc} 25 + 25 \\ 25 + 50 + 25 \\ 25 + 25 \end{array}) = (\begin{array}{cccccccccc} 50 \\ 100 \\ 50 \end{array})$
The middle balance equation also holds, so $(50, 100, 50)$ is a steady state totalling 200.

Answer

(50,100,50).

03 / YOUR TURN

Try it yourself.

Check your understanding

How many fixed distributions total 100 under the identity matrix?

Show answer & reasoning

Infinitely many.

The identity leaves every state unchanged: $I {(a, 100 - a)}^{𝖳} = {(a, 100 - a)}^{𝖳}$ . With non-negative real state entries, every $0 \leq a \leq 100$ is permitted, giving infinitely many fixed distributions.

Keep practising

6 worksheet questions · 19 marks

Open worksheet Worked solutions

Read the questions online

For rows $(\begin{array}{cccccccccc} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array})$ , find steady proportions.
3 marks
Use the same model for 350 members. Find steady counts.
3 marks
For rows $(\begin{array}{cccccccccc} 0.9 & 0.4 \\ 0.1 & 0.6 \end{array})$ , find steady proportions.
3 marks
For the swap matrix, find steady proportions and explain the behaviour from $(1, 0)$ .
3 marks
Find steady proportions for rows $(\begin{array}{cccccccccc} 0.5 & 0.25 & 0 \\ 0.5 & 0.5 & 0.5 \\ 0 & 0.25 & 0.5 \end{array})$ .
3 marks
A closed model has steady counts $(80, 120)$ out of 200.
4 marks
1. (a)
  State the steady proportions.
  1 mark
2. (b)
  Predict steady counts if the total doubles and the transition rule is unchanged.
  2 marks
3. (c)
  Give one reason observed long-run counts might differ.
  1 mark

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

For rows $(\begin{array}{cccccccccc} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array})$ , find steady proportions.
3 marks
$0.2 a = 0.3 b$ and $a + b = 1$ .
$(0.6, 0.4)$ .
Use the same model for 350 members. Find steady counts.
3 marks
Multiply $(0.6, 0.4)$ by 350.
$(210, 140)$ .
For rows $(\begin{array}{cccccccccc} 0.9 & 0.4 \\ 0.1 & 0.6 \end{array})$ , find steady proportions.
3 marks
$0.1 a = 0.4 b$ and $a + b = 1$ .
$(0.8, 0.2)$ .
For the swap matrix, find steady proportions and explain the behaviour from $(1, 0)$ .
3 marks
Balance gives $a = b$ .
Applying the swap repeatedly never approaches that balance from $(1, 0)$ .
Steady $(0.5, 0.5)$ ; the specified start alternates $(1, 0)$ , $(0, 1)$ .
Find steady proportions for rows $(\begin{array}{cccccccccc} 0.5 & 0.25 & 0 \\ 0.5 & 0.5 & 0.5 \\ 0 & 0.25 & 0.5 \end{array})$ .
3 marks
The first and third equations give $a = c = 0.5 b$ .
Normalise using $2 b = 1$ .
$(0.25, 0.5, 0.25)$ .
A closed model has steady counts $(80, 120)$ out of 200.
4 marks
a) Divide by 200.
b) Multiply the proportions by 400.
c) The equilibrium prediction is conditional on its assumptions.
1. (a)
  State the steady proportions.
  1 mark
  Divide by 200.
  $(0.4, 0.6)$ .
2. (b)
  Predict steady counts if the total doubles and the transition rule is unchanged.
  2 marks
  Multiply the proportions by 400.
  $(160, 240)$ .
3. (c)
  Give one reason observed long-run counts might differ.
  1 mark
  The equilibrium prediction is conditional on its assumptions.
  The system may open or transition behaviour may change.
a) $(0.4, 0.6)$ .; b) $(160, 240)$ .; c) The system may open or transition behaviour may change.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 General Mathematics matrix-models option. This is the offered fifth-topic option, alongside the four compulsory topics; a school-designed Open Topic is an alternative, not prescribed content in this chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

Equilibrium balances the flows

Fixed does not always mean attracting

See the method in action.

Commuting choices

Write the A balance equation.

Use the fixed population total.

Verify both destination counts.

Three-state balance

Balance A using the first matrix row.

Balance C using the third row.

Apply the total of 200.

Verify the complete three-state forecast.

Try it yourself.

Check your understanding

Keep practising