Apply and reverse a state recurrence with external change

In the adjusted recurrence below, which operation comes first?

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?

Choose the model before calculating

The reused explorer above shows only a fixed-total transition and its source-column convention. It does not apply external changes, births, deaths or Leslie fertility. The investigations below require separate complete models. Every fractional population is an expected count in a hypothetical model, not a literal fraction of a plant

1. 1Transition before adjustment Calculate the usual transition product before adding the signed external-change column \(\displaystyle S_{n+1}=T S_n+B\)
2. 2Apply the adjustment each period Use the same external-change column after every transition when the rule is repeated \(\displaystyle S_2=T S_1+B\)
3. 3Keep the signs in B A positive entry adds to a state and a negative entry removes from that state \(\displaystyle B=\begin{pmatrix}4\\-2\end{pmatrix}\)
4. 4Undo in reverse order To recover an earlier state, subtract B first and then multiply on the left by T inverse \(\displaystyle S_n=T^{-1}\left(S_{n+1}-B\right)\)

Construct and compare culling and restocking rules

A hypothetical nursery tracks plants in holding beds A and B at weekly censuses. During a week, each A plant has probability 0.20 of moving to B and each B plant has probability 0.30 of moving to A. Otherwise it stays in its current bed; there are no births or natural losses in this first model. Initially A=160 and B=40. Immediately after every weekly movement, ten plants are restocked into A and ten are culled from B. Treat subsequent state columns as expected counts under these constant assumed rates

1. 1Construct and compare timing Define the census index and state order, construct T and signed B, and write the full repeated recurrence. Draw the movement diagram and show external additions/removals separately, not as transition probabilities. Generate states through week 3. Compare the different policy that applies the same signed change before each transition. Find and verify a fixed state of total 200 for each policy, and explain why zero net change does not mean no effect
2. 2Reverse and infer For the after-transition policy, recover the initial state from week 1 and infer B independently from consecutive states. Compare the result with the incorrect operation T inverse times the later state, then minus B. Verify your recovered state by a forward calculation. Use a spreadsheet or matrix tool and record its actual formulas or commands
3. 3Save an independent model and report Choose your representations and technology before opening the model response. Save complete labelled matrices, timing conventions, state records, diagrams, actual formulas or commands, exact checks, graphs with justified windows and a written conclusion with limitations. These independent constructions, external-tool operations and reports are not automatically marked; review them yourself or with a teacher

Model response: Construct and compare culling and restocking rules

Compare the entire model and evidence, not only the short checked answers

1. 1After-transition construction and full states Use A,B for rows and columns and take the initial census as n=0. Draw A→A 0.8, A→B 0.2, B→A 0.3 and B→B 0.7; put +10 into A and −10 out of B after that movement. The complete states are S0=(160,40), S1=(150,50), S2=(145,55), S3=(142.5,57.5). Totals stay 200, but the distribution changes. In spreadsheet columns n,A,B, enter row 2 as 0,160,40; use A3=A2+1, B3=0.8*B2+0.3*C2+10 and C3=0.2*B2+0.7*C2-10, then copy down \(\displaystyle T=\begin{bmatrix}0.8&0.3\\0.2&0.7\end{bmatrix},\quad B=\begin{bmatrix}10\\-10\end{bmatrix},\quad S_{n+1}=TS_n+B\)
2. 2Timing changes the fixed state At fixed total 200, the after-transition first component follows the rule below. Its fixed state is (140,60). Applying B before the movement instead gives a different adjustment, with TB=(5,−5), rather than B. Its states are (160,40), (145,55), (137.5,62.5), (133.75,66.25), approaching (130,70). Direct substitution verifies both fixed states. Neither policy is the unadjusted fixed state (120,80) \(\displaystyle a_{n+1}=0.5a_n+70,\quad a_n=140+20(0.5)^n,\quad T(S+B)=TS+TB\)
3. 3Undo in reverse order and independently check The determinant of T is 0.5 and its inverse is displayed. Subtract B from S1 to obtain (140,60), then left-multiply by T inverse to recover (160,40). In contrast, T inverse times (150,50), then minus B gives (170,30), which is wrong. Inference from observed model states gives B=S1−TS0=(10,−10). Forward substitution returns (150,50). A window n=0 to 8 and expected count 0 to 200 compares both policies without implying fractional-week transitions \(\displaystyle T^{-1}=\begin{bmatrix}1.4&-0.6\\-0.4&1.6\end{bmatrix}\)

Given information

\(\displaystyle S_0=(160,40)^T,\quad T=\begin{bmatrix}0.8&0.3\\0.2&0.7\end{bmatrix},\quad B=(10,-10)^T\)

Find the expected number in A after one complete movement-and-adjustment week

Given information

\(\displaystyle S_1=T(S_0+B),\quad S_0=(160,40)^T,\quad B=(10,-10)^T,\quad T=\begin{bmatrix}0.8&0.3\\0.2&0.7\end{bmatrix}\)

For the policy applying B before movement, find the first component after one complete week

Given information

\(\displaystyle S_1=(150,50)^T,\quad B=(10,-10)^T,\quad T^{-1}=\begin{bmatrix}1.4&-0.6\\-0.4&1.6\end{bmatrix}\)

Recover the first component of the preceding state for the after-transition policy

Given information

\(\displaystyle a_{n+1}=0.5a_n+70\)

Find the first component of the after-transition fixed state with total 200.

Find when a repeated removal becomes infeasible

Use the nursery transition matrix and initial state below, but now remove ten plants from B after each movement and do not restock A. Also consider a separate policy adding ten plants to A each week with no removal. Assume the same deterministic expected-count recurrence only while every planned removal is feasible

1. 1Compare total and component evidence Derive a formula for the total under each policy. Determine whether either can have a finite fixed state while its nonzero net change continues. For B=(0,−10), generate a full state table up to the first negative component and prove the first infeasible week using exact adjacent values. Explain why checking only the total, or rounding to whole plants first, gives a misleading conclusion
2. 2Stop or revise the model explicitly Choose and justify graph windows for the state components and total. State the last feasible census and the attempted step that must be stopped or revised. Explain why replacing a negative answer by zero without redefining the rule is not the same model. Do not prescribe a real culling policy from these hypothetical values
3. 3Save an independent model and report Choose your representations and technology before opening the model response. Save complete labelled matrices, timing conventions, state records, diagrams, actual formulas or commands, exact checks, graphs with justified windows and a written conclusion with limitations. These independent constructions, external-tool operations and reports are not automatically marked; review them yourself or with a teacher

Model response: Find when a repeated removal becomes infeasible

Compare the entire model and evidence, not only the short checked answers

1. 1Totals rule out a fixed state Because each T column totals 1, total(n+1)=total(n)+sum(B). The removal-only total is 200−10n, and the addition-only total is 200+10n. A fixed state would require sum(B)=0, so neither policy has a finite fixed state while its net change continues. Zero net change is a necessary condition here, not a general sufficient condition for a feasible or unique fixed state
2. 2Exact component formula and boundary For removal only, the component rules are shown below. Substitution checks both the recurrence and the initial state. The B component starts 40, 50, 53 and then decreases, as the difference rule shows. At week 16 the total is 40. At week 17 the attempted state has a negative B component despite a positive total of 30. Thus week 17 is the first infeasible step and week 16 is the last feasible census. Rounding the negative B component to zero would hide the failure \(\displaystyle a_n=132-6n+28(1/2)^n,\quad b_n=68-4n-28(1/2)^n,\quad b_{n+1}-b_n=-4+14(1/2)^n<0\ (n\geq2),\quad S_{16}=\begin{bmatrix}\frac{589831}{16384}\\\frac{65529}{16384}\end{bmatrix},\quad S_{17}=\begin{bmatrix}\frac{983047}{32768}\\-\frac{7}{32768}\end{bmatrix}\)
3. 3Reproducible table and qualified display In the earlier spreadsheet replace the +10 in the A formula by 0; retain −10 in the B formula. Store all unrounded values, add total =B3+C3 and a feasibility flag =MIN(B3,C3)>=0, and copy through n=17. Plot integer n from 0 to 17 with expected counts from −5 to 200; label the negative point as an infeasible extrapolation, not a physical population. Use a zoom around the B-component boundary together with the exact fractions because its negative value is visually tiny. A revised rule might stop or cap removal, but that changes B and requires a new recurrence and a new analysis

Given information

\(\displaystyle N_n=200-10n\)

Find the expected total at the last feasible census n=16 under removal only

Given information

\(\displaystyle b_n=68-4n-28(1/2)^n,\quad a_n=132-6n+28(1/2)^n,\quad n\ge0\)

Find the first infeasible integer week under removal only

What should happen when the computed B-component is −7/32768?

Construct and investigate an age-structured population model

A separate hypothetical plant population is counted annually in two equal-width age groups: Y aged 0 to under 1 year, and O aged 1 to under 2 years. A Y plant contributes an expected 0.7 new Y plants at the next census; an O plant contributes 1.2. Half the current Y plants survive into O. No current O plant survives to the next census. The fertility rates already count offspring surviving to the next census. All subsequent counts are expectations. Initially Y=120 and O=50.

1. 1Translate and iterate Construct the complete Leslie matrix with order Y,O, draw and label all nonzero life-cycle arrows, and explain the structural zero. Distinguish a fertility loop from the same individual remaining in a group. Generate S0 through S3, their totals, their proportions and each component growth factor. Explain why a fertility rate above 1 is allowed and why column sums need not be 1.
2. 2Test what the total hides Repeat from a different initial column (100,70) through S2. Compare total growth factors with component growth factors and stage proportions. Determine which initial state already has a stable profile. Prove any exact stable profile claim by substitution rather than by visual similarity. Record actual matrix or spreadsheet operations and labelled graphs
3. 3Save an independent model and report Choose your representations and technology before opening the model response. Save complete labelled matrices, timing conventions, state records, diagrams, actual formulas or commands, exact checks, graphs with justified windows and a written conclusion with limitations. These independent constructions, external-tool operations and reports are not automatically marked; review them yourself or with a teacher

Model response: Construct and investigate an age-structured population model

Compare the entire model and evidence, not only the short checked answers

1. 1Matrix and diagram The first row contains expected offspring rates and the first subdiagonal contains survival. Draw Y→Y fertility 0.7, O→Y fertility 1.2 and Y→O survival 0.5. There is no O→O survival arrow; the lower-right entry is zero. The Y loop produces new plants, not surviving Y individuals staying young. Fertility 1.2 is an expected offspring count per source plant, not a probability \(\displaystyle L=\begin{bmatrix}0.7&1.2\\0.5&0\end{bmatrix},\quad S_{n+1}=LS_n,\quad S_n=L^nS_0\)
2. 2Stable proportions but increasing counts The states are (120,50), (144,60), (172.8,72), (207.36,86.4), with totals 170, 204, 244.8, 293.76. Every component is multiplied by 1.2, and the profile remains 12:5, with the proportions shown below. Exact substitution gives the common growth factor and the rule for this initial profile. This is not a fixed population or fixed state \(\displaystyle \frac{12}{17},\quad\frac{5}{17},\qquad L\begin{bmatrix}12\\5\end{bmatrix}=1.2\begin{bmatrix}12\\5\end{bmatrix},\qquad S_n=1.2^n\begin{bmatrix}120\\50\end{bmatrix}\)
3. 3Counterexample to relying on total growth From (100,70), the next two states are (154,50) and (167.8,77), again with totals 170,204,244.8. Both columns of this L total 1.2, so every state has total multiplied by 1.2. Yet the first step’s component factors are 1.54 and 5/7, which are unequal. Its profile is not initially stable. The total alone therefore cannot identify a stable stage distribution. These finite records do not by themselves prove convergence from every initial profile
4. 4Technology and display For n,Y,O in columns A,B,C, use A3=A2+1, B3=0.7*B2+1.2*C2 and C3=0.5*B2. Add total =B3+C3, Y proportion =B3/(B3+C3), and growth factors =B3/B2 and =C3/C2 where denominators are nonzero. Copy from each initial row without rounding the states. A window n=0 to 3 and expected count 0 to 310 contains the first profile’s components and total; display proportions separately on 0 to 1, not on the count scale. Label integer census times and the initial state clearly

What does the Y-to-Y loop labelled fertility 0.7 represent?

Given information

\(\displaystyle L=\begin{bmatrix}0.7&1.2\\0.5&0\end{bmatrix},\quad S_0=(120,50)^T\)

Find the total expected population after two annual updates

Given information

\(\displaystyle L=\begin{bmatrix}0.7&1.2\\0.5&0\end{bmatrix},\quad v=(12,5)^T\)

Find the common growth factor for the profile (12,5)

Given information

\(\displaystyle L=\begin{bmatrix}0.7&1.2\\0.5&0\end{bmatrix},\quad S_0=(100,70)^T\)

From the different initial population, find the expected Y count after one year

Populations starting at (120,50) and (100,70) both have total growth factor 1.2 under the matrix below. Which conclusion follows?

Compare intervention rules and a cyclic population model

Return to the displayed L and initial (120,50). Compare two hypothetical after-update culling rules: remove one-sixth of each resulting group, or remove the fixed column (24,10). Then test the displayed unadjusted Leslie matrix C. These are mathematical comparisons, not recommendations for managing a real population

1. 1Distinguish proportional and fixed changes Construct both complete recurrences and verify their first two states. Repeat the first step from (240,100) to show why agreement at one population does not identify the same policy. State whether a signed constant B can represent the proportional rule at all states. For a separate one-step restocking goal (180,75) from (120,50) after L, derive the required signed adjustment and verify it \(\displaystyle L=\begin{bmatrix}0.7&1.2\\0.5&0\end{bmatrix},\qquad C=\begin{bmatrix}0&2\\0.5&0\end{bmatrix}\)
2. 2Prove a cycle and report the scope For C, calculate C squared and states through four years from (120,50). Find a fixed positive profile and explain why it does not make the other initial sequence converge. Write a report comparing fixed-total movement, externally adjusted movement, Leslie growth, proportional control and cycling. Include the census convention, exact checks, feasibility horizon, assumed rates and the empirical evidence missing from any real-world recommendation
3. 3Save an independent model and report Choose your representations and technology before opening the model response. Save complete labelled matrices, timing conventions, state records, diagrams, actual formulas or commands, exact checks, graphs with justified windows and a written conclusion with limitations. These independent constructions, external-tool operations and reports are not automatically marked; review them yourself or with a teacher

Model response: Compare intervention rules and a cyclic population model

Compare the entire model and evidence, not only the short checked answers

1. 1Two different interventions with one common fixed state Removing one-sixth after L gives A=(5/6)L and recurrence S(n+1)=AS(n). It maps (120,50) to itself because L first gives (144,60), then removes (24,10). The fixed-removal rule S(n+1)=LS(n)−(24,10)^T also fixes (120,50). Starting instead at (240,100), proportional removal returns (240,100), while fixed removal gives (264,110). The proportional change is −(1/6)LS(n), so it is state-dependent and cannot be the same constant B for every population. A common first result does not establish identical models \(\displaystyle A=\frac{5}{6}L=\begin{bmatrix}\frac{7}{12}&1\\\frac{5}{12}&0\end{bmatrix}\)
2. 2Construct the restocking column For the separate target (180,75) after one ordinary L update from (120,50), the required B is target−LS0=(180,75)−(144,60)=(36,15). Both entries are additions. Check L(120,50)^T+(36,15)^T=(180,75)^T. This one-step goal does not imply a fixed state if the same restocking is continued indefinitely
3. 3Exact cycle rather than a falsely claimed limit For the displayed C, C squared=I. Starting from (120,50), the states alternate (120,50), (100,60), (120,50), (100,60), (120,50), and totals alternate 170 and 160. A fixed profile is (100,50), since C(100,50)^T=(100,50)^T, but the chosen starting state never approaches it. The exact power identity proves the cycle; a finite numerical pattern alone would not establish a universal long-run conclusion \(\displaystyle C=\begin{bmatrix}0&2\\0.5&0\end{bmatrix},\qquad C^2=I_2\)
4. 4Example qualified report State order and census timing determine every matrix product. Signed external changes apply after movement unless a different rule is explicitly defined; reversing requires subtraction before the inverse. Constant net removal can become infeasible in one component before the total becomes negative, as week 17 demonstrates. Leslie fertility permits growth and rates above 1, while stable proportions do not imply fixed counts. Proportional and fixed culls can agree at one state but differ elsewhere. Exact cycling is possible even with an available fixed profile. All rates here are assumed; observed source-to-destination movements, age-specific offspring/survival records, intervention capacity and consequences would be needed to assess a real model. The calculations supply conditional mathematical evidence, not an empirical recommendation or proof of learner attainment

Given information

\(\displaystyle c=1/6\)

Find the fraction retained after removing one-sixth of each post-update group

Given information

\(\displaystyle S_0=(240,100)^T,\quad L=\begin{bmatrix}0.7&1.2\\0.5&0\end{bmatrix},\quad B=(-24,-10)^T\)

From the larger initial state, find Y after the fixed-removal rule

Given information

\(\displaystyle S_0=(120,50)^T,\quad L=\begin{bmatrix}0.7&1.2\\0.5&0\end{bmatrix},\quad S_1=(180,75)^T\)

Find the required Y addition for the one-step restocking target

Given information

\(\displaystyle C=\begin{bmatrix}0&2\\0.5&0\end{bmatrix},\quad S_0=(120,50)^T\)

Under the cyclic model, find Y after two annual updates