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Recursion and Finance, GEN1 2023 VCAA 24 MC

The following recurrence relation models the value, \(P_n\), of a perpetuity after \(n\) time periods.

\(P_0=a, \quad P_{n+1}=R P_n-d\)

The value of \(R\) can be found by calculating

  1. \(a+d\)
  2. \(\dfrac{a+d}{a}\)
  3. \(\dfrac{a+d}{d}\)
  4. \(1+\dfrac{a+d}{a}\)
  5. \(1+\dfrac{a+d}{d}\)
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\(B\)

Show Worked Solution

\(P_0=a, \quad P_{n+1}=R P_n-d\)

\(P_1=R P_0-d\)

\(\text{Since}\ \ P_0=P_1\ \ \text{(perpetuities retain same value)} \)

\(a\) \(=Ra-d\)  
\(Ra\) \(=a+d\)  
\(R\) \(=\dfrac{a+d}{a} \)  

 
\(\Rightarrow B\)

CORE, FUR2 2021 VCAA 6

Sienna invests $420 000 in a perpetuity from which she will receive a regular monthly payment of $1890.

The perpetuity earns interest at the rate of 5.4% per annum.

  1. Determine the total amount, in dollars, that Sienna will receive after one year of monthly payments.  (1 mark)
  2. Write down the value of the perpetuity after Sienna has received one year of monthly payments.  (1 mark)
  3. Let `S_n` be the value of Sienna's perpetuity after `n` months.
  4. Complete the recurrence relation, in terms of `S_0`, `S_{n + 1}` and `S_n`, that would model the value of this perpetuity over time. Write your answers in the boxes provided.  (1 mark)

    `S_n =`
     
    		  ,          `S_{n+1} =`
     
    		 `xx S_n - 1890`
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  1. `$ 22 \ 680`
  2. `$ 420 \ 000`
  3. `S_n = 420 \ 000, \ \ S_{n+1} = 1.0045 xx S_n – 1890`
Show Worked Solution
a.   `text{Total amount}` `= 12 xx 1890`
    `= $ 22 \ 680`

 

b.  `text{Value} = $ 420 \ 00 \ text{(balance remains the same after each payment).}`

 

c.  `S_0 = 420 \ 000 \ => \ S_n = 420 \ 000`

`S_{n+1} = RS_n – 1890 \ \ text{where}\ \ R = 1 + r/100`

`r = 5.4/12 = 0.45`

`:. \ S_{n+1} = 1.0045 xx S_n – 1890`

CORE, FUR2 2020 VCAA 10

Samuel now invests $500 000 in an annuity from which he receives a regular monthly payment.

The balance of the annuity, in dollars, after  `n`  months,  `A_n` , can be modelled by a recurrence relation of the form

`A_0 = 500\ 000, qquad A_(n+1) = kA_n - 2000`

  1. Calculate the balance of this annuity after two months if  `k = 1.0024`.  (1 mark)
  2. Calculate the annual compound interest rate percentage for this annuity if  `k = 1.0024`.  (1 mark)
  3. For what value of  `k`  would this investment act as a simple perpetuity?  (1 mark)
Show Answers Only
  1. `$498\ 398.08`
  2. `2.88 text(%)`
  3. `1.004`
Show Worked Solution
a.   `A_1 = 1.0024 xx 500\ 000 – 2000 = $499\ 200`
  `A_2 = 1.0024 xx 499\ 200 – 2000 = $498\ 398.08`

 

♦ Mean mark 48%.
b.   `text(Monthly interest rate)` `= (1.0024 – 1) xx 100 = 0.24text(%)`
  `text(Annual interest rate)` `= 12 xx 0.24 = 2.88text(%)`

 

♦ Mean mark 36%.
c.   `text(Perpetuity would occur when)`
  `k xx 500\ 000 – 2000` `= 500\ 000`
  `k` `= (502\ 000)/(500\ 000)`
    `= 1.004`

CORE, FUR1 2018 VCAA 21 MC

Which one of the following recurrence relations could be used to model the value of a perpetuity investment, `P_n`, after `n` months?

  1. `P_0 = 120\ 000,qquadP_(n + 1) = 1.0029 × P_n - 356`
  2. `P_0 = 180\ 000,qquadP_(n + 1) = 1.0047 × P_n - 846`
  3. `P_0 = 210\ 000,qquadP_(n + 1) = 1.0071 × P_n - 1534`
  4. `P_0 = 240\ 000,qquadP_(n + 1) = 0.0047 × P_n - 2232`
  5. `P_0 = 250\ 000,qquadP_(n + 1) = 0.0085 × P_n - 2125`
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`B`

Show Worked Solution

`text(Perpetuity) => P_0 = P_1 = P_2 = … = P_n`

`text(By trial and error, consider option)\ B:`

`P_1` `= 1.0047 xx 180\ 000 – 846`
  `= 180\ 000`
  `= P_0`

 
`=> B`