Applied recurrence modeling

5 In this question, give probabilities correct to 4 decimal places.
Alpha and Delta are two companies which compete for the ownership of insurance bonds. Boyles and Cayleys are companies which trade in these bonds. When a new bond becomes available, it is first acquired by either Boyles or Cayleys. After a certain amount of trading it is eventually owned by either Alpha or Delta. Change of ownership always takes place overnight, so that on any particular day the bond is owned by one of the four companies. The trading process is modelled as a Markov chain with four states, as follows. On the first day, the bond is owned by Boyles or Cayleys, with probabilities \(0.4,0.6\) respectively.
If the bond is owned by Boyles, then on the next day it could be owned by Alpha, Boyles or Cayleys, with probabilities \(0.07,0.8,0.13\) respectively. If the bond is owned by Cayleys, then on the next day it could be owned by Boyles, Cayleys or Delta, with probabilities \(0.15,0.75,0.1\) respectively. If the bond is owned by Alpha or Delta, then no further trading takes place, so on the next day it is owned by the same company.

1. Write down the transition matrix \(\mathbf { P }\).
2. Explain what is meant by an absorbing state of a Markov chain. Identify any absorbing states in this situation.
3. Find the probability that the bond is owned by Boyles on the 10th day.
4. Find the probability that on the 14th day the bond is owned by the same company as on the 10th day.
5. Find the day on which the probability that the bond is owned by Alpha or Delta exceeds 0.9 for the first time.
6. Find the limit of \(\mathbf { P } ^ { n }\) as \(n\) tends to infinity.
7. Find the probability that the bond is eventually owned by Alpha. The probabilities that Boyles and Cayleys own the bond on the first day are changed (but all the transition probabilities remain the same as before). The bond is now equally likely to be owned by Alpha or Delta at the end of the trading process.
8. Find the new probabilities for the ownership of the bond on the first day.

6 When \(10 ^ { 6 }\) of a certain type of bacteria are detected in a blood sample of an infected animal, a course of treatment is started. The long-term aim of the treatment is to reduce the number of bacteria in such a sample to under 10000 . At this level the animal's immune system can fight the infection for itself. Once treatment has started, if the number of bacteria in a sample is 10000 or more, then treatment either continues or restarts. The model suggested to predict the progress of the course of treatment is based on the recurrence system \(P _ { n + 1 } = \frac { 2 P _ { n } } { n + 1 } + \frac { n } { P _ { n } }\) for \(n \geqslant 0\), with \(P _ { 0 } = 1000\), where \(P _ { n }\) denotes the number of bacteria (in thousands) present in the animal's body \(n\) days after the treatment was started. The table below shows the values of \(P _ { n }\), for certain chosen values of \(n\). Each value has been given correct to 2 decimal places (where appropriate).

1. Find the value of \(P _ { 6 }\) correct to 2 decimal places.
2. Using the given values for \(P _ { 0 }\) to \(P _ { 9 }\), and assuming that the model is valid,
   1. describe the effects of this course of treatment during the first 9 days,
   2. state the number of days after treatment is started when the animal's own immune system is expected to be able to fight the infection for itself.
3. 1. Using information from the above table, suggest a function f such that, for \(n > 10 , \mathrm { f } ( n )\) is a suitable approximation for \(P _ { n }\).
   2. Use your suggested function to estimate the number of days after treatment is started when the animal may once again require medical intervention in order to help fight off this bacterial infection.
   3. Using information from the above table and the recurrence relation, verify or correct the estimate which you found in part (c)(ii).
4. One criticism of the system \(P _ { n + 1 } = \frac { 2 P _ { n } } { n + 1 } + \frac { n } { P _ { n } }\), with \(P _ { 0 } = 1000\), is that it gives non-integer
   values of \(P\). values of \(P _ { n }\). Suggest a modification that would correct this issue.

5. \begin{figure}[h] \end{figure} Figure 1 shows the first three stages of a pattern that is created by a recursive process.
The process starts with a square and proceeds as follows

• each square is replaced by 5 smaller squares each \(\frac { 1 } { 9 }\) th the size of the square being replaced
• the 5 smaller squares are the ones in each corner and the one in the centre
• once each of the squares has been replaced, the square immediately to the right and above the centre square of the pattern is then removed

Let \(u _ { n }\) be the number of squares in the pattern in stage \(n\), where stage 1 is the original square.

1. Explain why \(u _ { n }\) satisfies the recurrence system $$u _ { 1 } = 1 \quad u _ { n + 1 } = 5 u _ { n } - 1 \quad ( n = 1,2,3 , \ldots )$$
2. Solve this recurrence system. Given that the initial square has area 25
3. determine the total area of all the squares in stage 8 of the pattern, giving your answer to 2 significant figures.

1. A population of deer on a large estate is assumed to increase by \(10 \%\) during each year due to natural causes.

The population is controlled by removing a constant number, \(Q\), of the deer from the estate at the end of each year. At the start of the first year there are 5000 deer on the estate.
Let \(P _ { n }\) be the population of deer at the end of year \(n\).

1. Explain, in the context of the problem, the reason that the deer population is modelled by the recurrence relation $$P _ { n } = 1.1 P _ { n - 1 } - Q , \quad P _ { 0 } = 5000 , \quad n \in \mathbb { Z } ^ { + }$$
2. Prove by induction that \(P _ { n } = ( 1.1 ) ^ { n } ( 5000 - 10 Q ) + 10 Q , \quad n \geqslant 0\)
3. Explain how the long term behaviour of this population varies for different values of \(Q\).

2. In two-dimensional space, lines divide a plane into a number of different regions. \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} It is known that:

• One line divides a plane into 2 regions, as shown in Figure 1
• Two lines divide a plane into a maximum of 4 regions, as shown in Figure 2
• Three lines divide a plane into a maximum of 7 regions, as shown in Figure 3
• Four lines divide a plane into a maximum of 11 regions, as shown in Figure 4

1. The number of visits to a website, in any particular month, is modelled as the number of visits received in the previous month plus \(k\) times the number of visits received in the month before that, where \(k\) is a positive constant.

Given that \(V _ { n }\) is the number of visits to the website in month \(n\),

1. write down a general recurrence relation for \(V _ { n + 2 }\) in terms of \(V _ { n + 1 } , V _ { n }\) and \(k\). For a particular website you are given that
   • \(k = 0.24\)
   • In month 1 , there were 65 visits to the website.
   • In month 2 , there were 71 visits to the website.
   • Show that
   $$V _ { n } = 50 ( 1.2 ) ^ { n } - 25 ( - 0.2 ) ^ { n }$$ This model predicts that the number of visits to this website will exceed one million for the first time in month \(N\).
2. Find the value of \(N\).

3. \begin{figure}[h] \end{figure} There are three lily pads on a pond. A frog hops repeatedly from one lily pad to another.
The frog starts on lily pad A, as shown in Figure 1.
In a model, the frog hops from its position on one lily pad to either of the other two lily pads with equal probability. Let \(p _ { n }\) be the probability that the frog is on lily pad A after \(n\) hops.

1. Explain, with reference to the model, why \(p _ { 1 } = 0\) The probability \(p _ { n }\) satisfies the recurrence relation $$p _ { n + 1 } = \frac { 1 } { 2 } \left( 1 - p _ { n } \right) \quad n \geqslant 1 \quad \text { where } p _ { 1 } = 0$$
2. Prove by induction that, for \(n \geqslant 1\) $$p _ { n } = \frac { 2 } { 3 } \left( - \frac { 1 } { 2 } \right) ^ { n } + \frac { 1 } { 3 }$$
3. Use the result in part (b) to explain why, in the long term, the probability that the frog is on lily pad A is \(\frac { 1 } { 3 }\)

1. In a model for the number of subscribers to a new social media channel it is assumed that

• each week \(20 \%\) of the subscribers at the start of the week cancel their subscriptions
• between the start and end of week \(n\) the channel gains \(20 n\) new subscribers

Given that at the end of week 1 there were 25 subscribers,

1. explain why the number of subscribers at the end of week \(n , U _ { n }\), is modelled by the recurrence relation $$U _ { 1 } = 25 \quad U _ { n + 1 } = 0.8 U _ { n } + 20 ( n + 1 ) \quad n = 1,2,3 , \ldots$$
2. Prove by induction that for \(n \geqslant 1\) $$U _ { n } = 325 \left( \frac { 4 } { 5 } \right) ^ { n - 1 } + 100 n - 400$$ Given that 6 months after starting the channel there were approximately 1800 subscribers,
3. evaluate the model in the light of this information.

1. A staircase has \(n\) steps. A tourist moves from the bottom (step zero) to the top (step \(n\) ). At each move up the staircase she can go up either one step or two steps, and her overall climb up the staircase is a combination of such moves.

If \(u _ { n }\) is the number of ways that the tourist can climb up a staircase with \(n\) steps,

1. explain why \(u _ { n }\) satisfies the recurrence relation $$u _ { n } = u _ { n - 1 } + u _ { n - 2 } , \text { with } u _ { 1 } = 1 \text { and } u _ { 2 } = 2$$
2. Find the number of ways in which she can climb up a staircase when there are eight steps. A staircase at a certain tourist attraction has 400 steps.
3. Show that the number of ways in which she could climb up to the top of this staircase is given by $$\frac { 1 } { \sqrt { 5 } } \left[ \left( \frac { 1 + \sqrt { 5 } } { 2 } \right) ^ { 401 } - \left( \frac { 1 - \sqrt { 5 } } { 2 } \right) ^ { 401 } \right]$$

7. Martina decides to open a bank account to help her to save for a holiday. Each month she puts \(\pounds \mathrm { k }\) into the account and allows herself to spend one quarter of what was in the account at the end of the previous month. Let \(u _ { n }\) (where \(\mathrm { n } \geqslant 1\) ) represent the amount in the account at the end of month n .
Martina has \(\pounds \mathrm {~K}\) in the account at the end of the first month.

1. By setting up a first order recurrence relation for \(u _ { n + 1 }\) in terms of \(u _ { n }\), determine an expression for \(u _ { n }\) in terms of n and k At the end of the 8th month, Martina needs to have at least \(\pounds 1750\) in the account to pay for her holiday.
2. Determine, to the nearest penny, the minimum amount of money that Martina should put into the account each month.

7 In a long-running biochemical experiment, an initial amount of 1200 mg of an enzyme is placed into a mixture. The model for the amount of enzyme present in the mixture suggests that, at the end of each hour, one-eighth of the amount of enzyme that was present at the start of that hour is used up due to chemical reactions within the mixture. To compensate for this, at the end of each six-hour period of time, a further 500 mg of the enzyme is added to the mixture.

1. Let \(n\) be the number of six-hour periods that have elapsed since the experiment began. Explain how the amount of enzyme, \(\mathrm { E } _ { \mathrm { n } } \mathrm { mg }\), in the mixture is given by the recurrence system \(E _ { 0 } = 1200\) and \(E _ { n + 1 } = \left( \frac { 7 } { 8 } \right) ^ { 6 } E _ { n } + 500\) for \(n \geqslant 0\).
2. Solve the recurrence system given in part (a) to obtain an exact expression for \(\mathrm { E } _ { \mathrm { n } }\) in terms of \(n\).
3. Hence determine, in the long term, the amount of enzyme in the mixture. Give your answer correct to \(\mathbf { 3 }\) significant figures.
4. In this question you must show detailed reasoning. The long-running experiment is then repeated. This time a new requirement is added that the amount of enzyme present in the mixture must always be at least 500 mg . Show that the new requirement ceases to be satisfied before 12 hours have elapsed. \section*{END OF QUESTION PAPER} }{www.ocr.org.uk}) after the live examination series.
   If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity.
   For queries or further information please contact The OCR Copyright Team, The Triangle Building, Shaftesbury Road, Cambridge CB2 8EA.
   OCR is part of Cambridge University Press \& Assessment, which is itself a department of the University of Cambridge.

3 The sequence of positive numbers \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\) is such that \(u _ { 1 } < 3\) and, for \(n \geqslant 1\), $$u _ { n + 1 } = \frac { 4 u _ { n } + 9 } { u _ { n } + 4 }$$

1. By considering \(3 - u _ { n + 1 }\), or otherwise, prove by mathematical induction that \(u _ { n } < 3\) for all positive integers \(n\).
2. Show that \(u _ { n + 1 } > u _ { n }\) for \(n \geqslant 1\).

9 A new lake is stocked with fish. Let \(P _ { t }\) be the population of fish in the lake after \(t\) years. Two models using recurrence relations are proposed for \(P _ { t }\), with \(P _ { 0 } = 550\). $$\begin{aligned} & \text { Model } 1 : P _ { t } = 2 P _ { t - 1 } \mathrm { e } ^ { - 0.001 P _ { t - 1 } } \\ & \text { Model } 2 : P _ { t } = \frac { 1 } { 2 } P _ { t - 1 } \left( 7 - \frac { 1 } { 160 } P _ { t - 1 } \right) \end{aligned}$$

1. Evaluate the population predicted by each model when \(t = 3\).
2. Identify, with evidence, which one of the models predicts a stable population in the long term.
3. Describe the long term behaviour of the population for the other model.

A financial institution models the repayment of a loan to a client in the following way.

• An amount, \(£C\), is loaned to the client at the start of the repayment period.
• The amount owed \(n\) years after the start of the repayment period is \(£L_n\), so that \(L_0 = C\).
• At the end of each year, interest of \(\alpha\%\) (\(\alpha > 0\)) of the amount owed at the start of that year is added to the amount owed.
• Immediately after interest has been added to the amount owed a repayment of \(£R\) is made by the client.
• Once \(L_n\) becomes negative the repayment is finished and the overpayment is refunded to the client.

1. Show that during the repayment period, \(L_{n+1} = aL_n + b\), giving \(a\) and \(b\) in terms of \(\alpha\) and \(R\). [2]
2. Find the solution of the recurrence relation \(L_{n+1} = aL_n + b\) with \(L_0 = C\), giving your solution in terms of \(a\), \(b\), \(C\) and \(n\). [5]
3. Deduce from parts (a) and (b) that, for the repayment scheme to terminate, \(R > \frac{\alpha C}{100}\). [2]

A client takes out a £30000 loan at 8% interest and agrees to repay £3000 at the end of each year.

1. 1. Use an algebraic method to find the number of years it will take for the loan to be repaid. [3]
   2. Taking into account the refund of overpayment, find the total amount that the client repays over the lifetime of the loan. [3]

In order to rescue them from extinction, a particular species of ground-nesting birds is introduced into a nature reserve. The number of breeding pairs of these birds in the nature reserve, \(t\) years after their introduction, is denoted by \(N_t\). The initial number of breeding pairs is given by \(N_0\). An initial discrete population model is proposed for \(N_t\). Model I: \(N_{t+1} = \frac{6}{5}N_t\left(1 - \frac{1}{900}N_t\right)\)

1. 1. For Model I, show that the steady state values of the number of breeding pairs are 0 and 150. [3]
   2. Show that \(N_{t+1} - N_t < 150 - N_t\) when \(N_t\) lies between 0 and 150. [3]
   3. Hence determine the long-term behaviour of the number of breeding pairs of this species of birds in the nature reserve predicted by Model I when \(N_0 \in (0, 150)\). [2]
   An alternative discrete population model is proposed for \(N_t\). Model II: \(N_{t+1} = \text{INT}\left(\frac{6}{5}N_t\left(1 - \frac{1}{900}N_t\right)\right)\)
2. Given that \(N_0 = 8\), find the value of \(N_4\) for each of the two models and give a reason why Model II may be considered more suitable. [3]