Construct a transition model with a specified equilibrium

At equilibrium, which idea should be used to find a missing transition rate?

Rows give the next state and columns give the current state, in bus then car order. Move both sliders and compare the four totals. Which totals stay at 1?

For every position of the sliders, which totals stayed at 1?

Construct an equilibrium model

An equilibrium state is unchanged by one transition. Find flows using source populations, balance the flows, complete each source column to 1 and check the result by multiplication

1. 1Read the source state A transition rate is multiplied by the population in the state it leaves \(\displaystyle t_{21}:S_1\to S_2\)
2. 2Balance the flows At equilibrium, the total flow into each state equals the total flow out of that state \(\displaystyle F_{1\to2}=F_{2\to1}\)
3. 3Find a missing rate from its flow Divide the balanced flow by the population in the source state of the missing rate \(\displaystyle t_{12}=\frac{F_{2\to1}}{S_2}\)
4. 4Complete each source column The entries in one current-state column must add to 1 for an ordinary transition model \(\displaystyle t_{11}+t_{21}=1\)

Build and investigate a fixed-total depot model

A hypothetical fleet has 200 reusable containers, each assigned to depot A or B at a weekly observation. A container currently at A has probability 0.20 of being at B next week; a container currently at B has probability 0.30 of being at A next week. Initially 160 are at A and 40 at B. No containers enter or leave. Assume these probabilities are constant and depend only on the current depot. Subsequent calculated entries are expected counts, not guaranteed literal counts

1. 1Translate all representations Define the weekly index with the initial column at n=0. Construct T with row and column order A,B, draw every directed arrow and loop with probability labels, and write the state recurrence. Check exclusivity/exhaustiveness of the states and every column total. Explain why row totals need not be 1.
2. 2Produce a full investigation Choose a matrix tool or spreadsheet. Record states from n=0 to n=7 without rounding intermediate values. Plot both state components at integer weeks with labelled axes and a justified window. Derive a candidate equilibrium and verify it independently. Find the first week for which each expected count is strictly within 0.5 of its equilibrium value; prove the adjacent-week boundary rather than relying on the graph
3. 3Save your independent construction Choose and justify your representation, technology and display window. Save the complete matrices, labelled directed diagram, state records, actual tool formulas or commands, independent checks and written conclusions before opening the model response. Full construction, external-tool operation and reporting are not automatically marked; review them yourself or with a teacher

Model response: Build and investigate a fixed-total depot model

Compare the complete representations and checks. A selected option or scalar answer alone is not the independent investigation

1. 1Matrix, arrows and timing Column A is [0.8,0.2] and column B is [0.3,0.7]. Draw A→A 0.8, A→B 0.2, B→A 0.3 and B→B 0.7, including both loops. The state vector is a column; one multiplication advances one week. Row totals are 1.1 and 0.9, not probabilities for one starting group \(\displaystyle T=\begin{bmatrix}0.8&0.3\\0.2&0.7\end{bmatrix},\quad S_0=\begin{bmatrix}160\\40\end{bmatrix},\quad S_{n+1}=TS_n,\quad S_n=T^nS_0\)
2. 2Complete state record States (A,B) at n=0,1,2,3,4,5,6,7 are (160,40), (140,60), (130,70), (125,75), (122.5,77.5), (121.25,78.75), (120.625,79.375), (120.3125,79.6875). Every total is 200. In a spreadsheet with n in A, expected A in B and expected B in C, enter 0,160,40 in row 2; use A3=A2+1, B3=0.8*B2+0.3*C2 and C3=0.2*B2+0.7*C2, then copy down. Check each row total with =B3+C3. Record actual tool outputs, not just these example formulas
3. 3Independent equilibrium and convergence At equilibrium, 0.2a=0.3b and a+b=200, giving (120,80). Both directed flows are 24 per week in expectation; movement continues. Direct multiplication gives T(120,80)^T=(120,80)^T. Every entry of T is positive, so this is a regular finite transition model. For this specific case, eliminating b=200−a gives a(n+1)=0.5a(n)+60 and a(n)−120=40(0.5)^n, an independent convergence proof \(\displaystyle S_n=\begin{bmatrix}120+40(1/2)^n\\80-40(1/2)^n\end{bmatrix}\)
4. 4Window, threshold and limits A window n=0 to 8 and expected count 0 to 200 contains both component sequences and states the conserved-total scale. Plot the integer-week values; a joining line is only a visual guide, not an extra half-week transition rule. At week 6 the error is 0.625 and at week 7 it is 0.3125, so week 7 first satisfies the strict 0.5 bound. Fractional entries are expectations. Actual counts can differ, and these probabilities were assumed rather than estimated from observations

Which entry represents the arrow from B to A in this one-week diagram?

Given information

\(\displaystyle S_0=(160,40)^T,\quad T=\begin{bmatrix}0.8&0.3\\0.2&0.7\end{bmatrix}\)

Find the expected number at depot A after one week

Given information

\(\displaystyle S_0=(160,40)^T,\quad T=\begin{bmatrix}0.8&0.3\\0.2&0.7\end{bmatrix}\)

Find the expected number at depot B after three weeks

Given information

\(\displaystyle S_*=(120,80)^T,\quad t_{21}=0.2\)

At the fixed state, find the expected number moving from A to B in one week

Given information

\(\displaystyle |a_n-120|=|b_n-80|=40(1/2)^n,\quad n\ge0\)

Find the first integer n for which both expected counts are strictly within 0.5 of equilibrium

Construct admissible models with a chosen equilibrium

Keep the total 200 and target equilibrium (120,80), but do not retain the previous movement probabilities. Let alpha be the one-week probability of moving from A to B. Construct all two-state column-stochastic matrices with this target equilibrium and parameter alpha

1. 1Derive, constrain and compare Derive the reverse probability and both diagonal entries from balance and column sums. Find the complete admissible alpha interval. Generate and compare examples alpha=0, 0.2, 0.4, 0.6 and 2/3, plus a rejected alpha=0.8. Starting from (160,40), compare at least three updates for the valid examples. Determine which models are regular and whether they share a convergence speed
2. 2Check identification and technology Explain why the equilibrium alone does not identify one matrix. Check the alpha=0.4 matrix’s determinant and discuss whether a next state uniquely identifies an earlier state. Use exact boundary calculations, not merely a rounded graph, to justify admissibility. Record the actual formulas or commands used in your parameter investigation
3. 3Save your independent construction Choose and justify your representation, technology and display window. Save the complete matrices, labelled directed diagram, state records, actual tool formulas or commands, independent checks and written conclusions before opening the model response. Full construction, external-tool operation and reporting are not automatically marked; review them yourself or with a teacher

Model response: Construct admissible models with a chosen equilibrium

Compare the complete representations and checks. A selected option or scalar answer alone is not the independent investigation

1. 1Complete family and domain Balance requires 120 alpha=80 beta, so beta=1.5 alpha. Non-negative entries at most 1 give 0≤alpha≤2/3. Alpha=0.8 would require beta=1.2 and a negative lower-right entry, so it is not a transition model \(\displaystyle T(\alpha)=\begin{bmatrix}1-\alpha&3\alpha/2\\\alpha&1-3\alpha/2\end{bmatrix},\quad0\le\alpha\le2/3\)
2. 2Different transient behaviour For fixed total 200, the first-component deviation updates by a(n+1)−120=(1−2.5 alpha)(a(n)−120). Alpha=0.2 halves the positive deviation. Alpha=0.6 multiplies it by −0.5, giving (100,100), (130,70), (115,85): it alternates across the same equilibrium while approaching it. Alpha=0.4 gives (120,80) in one step from any state of total 200. Alpha=0 is the identity and leaves (160,40) unchanged, so the selected target is not its limit
3. 3Boundary and regularity For 0<alpha<2/3, all entries are positive, so T itself proves regularity. At alpha=2/3, the displayed T has a zero but its displayed square is entirely positive. Thus that endpoint is also regular. The identity at alpha=0 never has all entries positive. Its many fixed states show why equilibrium alone does not guarantee a unique long-run outcome \(\displaystyle T(2/3)=\begin{bmatrix}1/3&1\\2/3&0\end{bmatrix},\qquad T(2/3)^2=\begin{bmatrix}7/9&1/3\\2/9&2/3\end{bmatrix}\)
4. 4Valid does not imply invertible At alpha=0.4 the columns are both [0.6,0.4], so the determinant is zero. It is a valid regular transition matrix but not invertible. Both (160,40) and (80,120) map to (120,80). A unique backward state cannot be recovered from that next state without additional information. For parameter trials, record alpha and all four matrix entries, check both column sums, then generate the same initial-state sequence at every trial

Given information

\(\displaystyle F_{A\to B}=120\alpha,\quad F_{B\to A}=80\beta,\quad\alpha=0.2\)

Find the probability of moving from B to A when alpha is 0.2 and the fixed state is (120,80)

Given information

\(\displaystyle T(\alpha)=\begin{bmatrix}1-\alpha&3\alpha/2\\\alpha&1-3\alpha/2\end{bmatrix}\)

Find the largest admissible alpha in this family

Given information

\(\displaystyle S_0=(160,40)^T,\quad T=\begin{bmatrix}0.4&0.9\\0.6&0.1\end{bmatrix}\)

For alpha=0.6, find the expected number at A after the first week

Given information

\(\displaystyle T=\begin{bmatrix}0.6&0.6\\0.4&0.4\end{bmatrix}\)

Find the determinant of the one-step-to-equilibrium matrix

The endpoint matrix has a zero entry. What establishes its regularity?

Construct a circulating three-state equilibrium

A separate fixed-total model has target expected state counts A=90, B=60, C=30. At equilibrium, six units per period move along each clockwise link A→B, B→C, C→A, and three move along each reverse link. Everyone else stays in the same state. These are expected movements in a hypothetical model

1. 1Construct and verify all representations Construct the full flow table with source columns and destination rows, including stay flows. Divide by each source population to construct every entry of T. Draw all nine directed arrows/loops with probability labels. Verify each column sum and the full equation TS=S, showing each destination’s incoming contributions
2. 2Explain balance without an equal split Compare A→B and B→A. Explain why their unequal flows do not contradict equilibrium. State whether the matrix is regular, and give a different regular matrix with the same target equilibrium to show that the target alone is not a unique model. Name the extra flow assumptions that identify this particular construction
3. 3Save your independent construction Choose and justify your representation, technology and display window. Save the complete matrices, labelled directed diagram, state records, actual tool formulas or commands, independent checks and written conclusions before opening the model response. Full construction, external-tool operation and reporting are not automatically marked; review them yourself or with a teacher

Model response: Construct a circulating three-state equilibrium

Compare the complete representations and checks. A selected option or scalar answer alone is not the independent investigation

1. 1Full flow table and matrix The stay flows are 81 at A, 51 at B and 21 at C because nine leave each state. With source columns A,B,C, the flow rows are [81,3,6], [6,51,3], [3,6,21]. Divide columns by 90,60,30 respectively, not by destination populations \(\displaystyle T=\begin{bmatrix}9/10&1/20&1/5\\1/15&17/20&1/10\\1/30&1/10&7/10\end{bmatrix}\)
2. 2All destination checks The incoming totals are A:81+3+6=90, B:6+51+3=60 and C:3+6+21=30. Each column totals 1 and every entry is positive, so the matrix is regular. Opposite A/B flows are 6 and 3, not equal; the rest of the cycle supplies the difference. Equilibrium requires total inflow equal total outflow at each state, not equality on every pair
3. 3Non-uniqueness without the flow restrictions Another regular matrix has each column equal to [1/2,1/3,1/6]. It sends any state of total 180 to (90,60,30) in one period and preserves that state. It is not the same movement model. The specified six-unit clockwise flow, three-unit reverse flow and stay rule distinguish the constructed matrix from this alternative

For this diagram, which calculation gives the expected C-to-A flow at state (90,60,30)?

Given information

\(\displaystyle S_*=(90,60,30)^T,\quad F_{A\to B}=6\)

Find the A-to-B transition probability in the circulating model

Given information

\(\displaystyle S_{*,B}=60,\quad F_{B\to C}=6,\quad F_{B\to A}=3\)

Find the B-to-B transition probability

Test a counterexample and write a qualified report

Someone claims that if every state can reach every other state, repeated transitions must settle at a fixed state. Test the displayed two-state swap matrix U from initial state (160,40), then compare it with the regular depot model and the parameter examples above

1. 1Construct a decisive counterexample Generate at least four states, find the powers U squared and U cubed, and identify a fixed state of total 200. Explain why finding that fixed state does not make the chosen initial sequence converge. Do not infer non-regularity merely because a short numerical search failed to find a positive power: use the exact repeating power pattern \(\displaystyle U=\begin{bmatrix}0&1\\1&0\end{bmatrix}\)
2. 2Report the model choice and limits Write a short report stating the selected depot model, source-column convention, fixed-total and memory assumptions, complete evidence, threshold result, parameter limits and distinction between a fixed state and a limit. Explain what observed data would be needed to estimate or check the probabilities. Do not treat a hypothetical expected-count model as a guarantee about actual individual movements
3. 3Save your independent construction Choose and justify your representation, technology and display window. Save the complete matrices, labelled directed diagram, state records, actual tool formulas or commands, independent checks and written conclusions before opening the model response. Full construction, external-tool operation and reporting are not automatically marked; review them yourself or with a teacher

Model response: Test a counterexample and write a qualified report

Compare the complete representations and checks. A selected option or scalar answer alone is not the independent investigation

1. 1Exact cycle The sequence alternates (160,40), (40,160), (160,40), (40,160), so it does not converge. U squared is I, and odd powers are U while even powers are I. Every power has zero entries, hence no positive power exists. Both states are reachable from each other, but the chain is periodic. The fixed state (100,100) exists and stays fixed if used initially; this does not make every other initial state approach it \(\displaystyle U^2=I,\quad U^{2k}=I,\quad U^{2k+1}=U,\quad U\begin{bmatrix}100\\100\end{bmatrix}=\begin{bmatrix}100\\100\end{bmatrix}\)
2. 2Example qualified conclusion Under fixed weekly probabilities 0.20 from A to B and 0.30 from B to A, the expected state approaches (120,80), with continuing balanced expected flows of 24 each way. From (160,40), week 7 is first strictly within 0.5 expected units per component. Full exact iteration, column totals, substitution and the scalar deviation formula agree. This conclusion is conditional on the two exclusive/exhaustive states, fixed total, stable probabilities and current-state assumption. Repeated observed transitions by source state would be needed to estimate probabilities and investigate stability; this lesson supplies no such empirical evidence. External additions/removals and population births/deaths require a different model and are the next investigation

Given information

\(\displaystyle U=\begin{bmatrix}0&1\\1&0\end{bmatrix},\quad S_0=(160,40)^T\)

Under the swap model, find the first component after two transitions

Which conclusion is supported by the swap counterexample?