smc-717-30-Annuity

Recursion and Finance, GEN2 2023 VCAA 6

Arthur invests $600 000 in an annuity that provides him with a monthly payment of $3973.00.

Interest is calculated monthly.

Three lines of the amortisation table for this annuity are shown below.

\begin{array} {|c|c|}
\hline
\textbf{Payment} & \textbf{Payment} & \textbf{Interest} & \textbf{Principal reduction} & \textbf{Balance} \\
\textbf{number} & \textbf{(\$) } & \textbf{(\$) } & \textbf{(\$) } & \textbf{(\$) }\\
\hline
\rule{0pt}{2.5ex} 0 \rule[-1ex]{0pt}{0pt} & 0.00 & 0.00 & 0.00 & 600\ 000.00 \\
\hline
\rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt} & 3973.00 & 2520.00 & 1453.00& 598\ 547.00\\
\hline
\rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt} & 3973.00 & 2513.90 & 1459.10 & 597\ 087.90 \\
\hline
\end{array}

The interest rate for the annuity is 0.42% per month.
Determine the interest rate per annum. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Using the values in the table, complete the next line of the amortisation table.
Write your answers in the spaces provided in the table below.
Round all values to the nearest cent. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---

Let $V_n$ be the balance of Arthur's annuity, in dollars, after $n$ months.
Write a recurrence relation in terms of $V_0, V_{n+1}$ and $V_n$ that can model the value of the annuity from month to month. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
The amortisation tables above show that the balance of the annuity reduces each month.
If the balance of an annuity remained constant from month to month, what name would be given to this type of annuity? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $I\% (\text{annual}) = 12 \times 0.42 = 5.04\% $

b. $\text{Row 3 calculations are as follows:}$

$\text{Payment}\ = \$3973.00\ \text{(remains constant)}$

$\text{Interest}\ = 597\ 087.90 \times 0.0042 = \$2507.77 $

$\text{Principal reduction}\ = 3973.00-2507.77 = \$1465.23 $

$\text{Balance}\ = 597\ 087.90-1465.23 = \$595\ 622.67$

c. $V_0 = 600\ 000$

$V_{n+1} = 1.0042 \times V_n-3973$

d. $\text{Perpetuity}$

Show Worked Solution

a. $I\% (\text{annual}) = 12 \times 0.42 = 5.04\% $

b. $\text{Row 3 calculations are as follows:}$

$\text{Payment}\ = \$3973.00\ \text{(remains constant)}$

$\text{Interest}\ = 597\ 087.90 \times 0.0042 = \$2507.77 $

$\text{Principal reduction}\ = 3973.00+2507.77 = \$1465.23 $

$\text{Balance}\ = 597\ 087.90-1465.23 = \$595\ 622.67$

♦♦ Mean mark (b) 40%.

c. $V_0 = 600\ 000$

$V_{n+1} = 1.0042 \times V_n-3973$

♦ Mean mark (c) 41%.

d. $\text{Perpetuity}$

CORE, FUR2 2021 VCAA 9

Sienna invests $152 431 into an annuity from which she will receive a regular monthly payment of $900 for 25 years. The interest rate for this annuity is 5.1% per annum, compounding monthly.

Let `V` be the balance of the annuity after `n` monthly payments. A recurrence relation written in terms of `V_0 , V_{n + 1}` and `V_n` can model the value of this annuity from month to month.
Showing recursive calculations, determine the value of the annuity after two months.
Round your answer to the nearest cent. (2 marks)
After two years, the interest rate for this annuity will fail to 4.6%.
To ensure that she will still receive the same number of $900 monthly payments, Sienna will add an extra one-pff amount into the annuity at this time.
Determine the value of this extra amount that Sienna will add.
Round your answer to the nearest cent. (1 mark)

Show Answers Only

`$151 \ 925.59`
`$7039.20`

Show Worked Solution

a. `text{Payments are monthly} \ => \ r = 5.1/12 = 0.425text(%) \ text{per month}`

`R`	`= 1 + r/100 = 1.00425`
`V_1`	`= RV_0 – text{payment}`
	`= 1.00425 xx 152 \ 431 – 900`
	`= $152\ 178.83`
`:.V_2`	`= 1.00425 xx 152\ 178.83 – 900`
	`= $ 151 \ 925.59`

b. `text{Find} \ V_24 \ text{(annuity value after 2 years) by TVM Solver:}`

`N`	`= 24`
`I text{(%)}`	`= 5.1`
`PV`	`= -152 \ 431`
`PMT`	`= 900`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=> FV = 146 \ 073.7405`

`text{Find}\ PV\ text{of annuity needed (by TVM solver):}`

`N`	`= 243 xx 12 = 276`
`Itext{(%)}`	`= 4.6`
`PV`	`= ?`
`PMT`	`= 900`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> PV = -153\ 112.9399`

`:. \ text{Amount to add}`	`= 153 \ 112.94 – 146 \ 073.74`
	`= $ 7039.20`

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`

Calculate the balance of this annuity after two months if `k = 1.0024`. (1 mark)
Calculate the annual compound interest rate percentage for this annuity if `k = 1.0024`. (1 mark)
For what value of `k` would this investment act as a simple perpetuity? (1 mark)

Show Answers Only

`$498\ 398.08`
`2.88 text(%)`
`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 17-18 MC

The value of an annuity investment, in dollars, after `n` years, `V_n` , can be modelled by the recurrence relation shown below.

`V_0 = 46\ 000, quadqquadV_(n + 1) = 1.0034V_n + 500`

Part 1

What is the value of the regular payment added to the principal of this annuity investment?

$34.00
$156.40
$466.00
$500.00
$656.40

Part 2

Between the second and third years, the increase in the value of this investment is closest to

$656
$658
$661
$1315
$1975

Show Answers Only

`text(Part 1:)\ D`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(Regular payment = $500)`

`=> D`

`text(Part 2)`

`V_1`	`= 1.0034 xx 46\ 000 + 500 = $46\ 656.40`
`V_2`	`= 1.0034 xx 46\ 656.40 + 500 = $47\ 315.03`
`V_3`	`= 1.0034 xx 47\ 315.03 + 500 = $47\ 975.90`

`:.\ text(Increase)`	`= 47\ 975.90 – 47\ 315.03`
	`= $660.87`

`=> C`

CORE, FUR1 SM-Bank VCE 1-2 MC

Part 1

Candy deposits $80 000 into a savings account that earns 5% interest per annum, compounded annually.

At the end of the first year, Candy withdraws $5600.

What is the balance of Candy's savings account at the start of the second year?

`$74\ 400`
`$77\ 600`
`$78\ 400`
`$85\ 600`
`$89\ 600`

Part 2

If `B_n` is the balance of Candy's account at the start of year `n`, a recurrence equation that can be used to model the balance of the account over time is

`B_(n + 1) = 1.5 B_n - 5600\ \ \ \ text(where)\ B_1 = 80\ 000`
`B_(n + 1) = 0.05 B_n - 5600\ \ \ \ text(where)\ B_1 = 80\ 000`
`B_(n + 1) = 1.05 (B_n - 5600)\ \ \ \ text(where)\ B_1 = 80\ 000`
`B_(n + 1) = 0.05 (B_n - 5600)\ \ \ \ text(where)\ B_1 = 80\ 000`
`B_(n + 1) = 1.05 B_n - 5600\ \ \ \ text(where)\ B_1 = 80\ 000`

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ E`

Show Worked Solution

`text(Part 1)`

`text(Balance at start of 2nd year)`

`= 80\ 000 xx 1.05 – 5600`

`= $78\ 400`

`=> C`

`text(Part 2)`

`B_2`	`= 80\ 000 xx 1.05 – 5600`
	`= 1.05B_1 – 5600`
`B_3`	`= 1.05B_2 – 5600`
`vdots`
`B_(n + 1)`	`= 1.05B_n – 5600`

`=> E`

CORE, FUR2 SM-Bank VCE 2

Spiro is saving for a car. He has an account with $3500 in it at the start of the year.

At the end of each month, Spiro adds another $180 to the account.

The account pays 3.6% interest per annum, compounded monthly.

i. What is the interest rate per month? (1 mark)
ii. Write a recurrence relation that models Spiro's investment, with `V_n` representing the balance of his account after `n` months. (1 mark)
What will be the balance of Spiro's account after 3 months?

Write your answer correct to the nearest cent. (1 mark)

Show Answers Only

1. `0.3text(% per month)`
2. `V_0 = 3500,qquadV_(n + 1) = V_n xx 1.003 + 180`
`$4073\ \ (text(nearest $))`

Show Worked Solution

a.i.	`text(Interest rate)`	`= 3.6/12`
		`= 0.3text(% per month)`

a.ii.	`V_0`	`= 3500`
	`V_1`	`= 3500 xx 1.003 + 180`
	`V_2`	`= V_1 xx 1.003 + 180`
	`vdots`
	`V_(n + 1)`	`= V_n xx 1.003 + 180`

`:.\ text(Recurrence relationship:)`

`V_0 = 3500,qquadV_(n + 1) = V_n xx 1.003 + 180`

b.	`V_1`	`= 3500 xx 1.003 + 180 = $3690.50`
	`V_2`	`= 3690.50 xx 1.003 + 180 = $3881.5715`

`V_3`	`= 3881.5715 xx 1.003 + 180`	`= $4073.216…`
		`= $4073.22\ \ text{(nearest cent)}`

CORE, FUR1 2016 VCAA 18 MC

The value of an annuity, `V_n`, after `n` monthly payments of $555 have been made, can be determined using the recurrence relation

`V_0 = 100\ 000,\ \ \ \ \ V_(n + 1) = 1.0025 V_n - 555`

The value of the annuity after five payments have been made is closest to

`$97\ 225`
`$98\ 158`
`$98\ 467`
`$98\ 775`
`$110\ 224`

Show Answers Only

`C`

Show Worked Solution

`V_0`	`= 100\ 000`
`V_1`	`= 1.0025 xx 100\ 000 – 555 = 99\ 695`
`V_2`	`= 1.0025 xx 99\ 695 – 555 = 99\ 389.23…`
`V_3`	`= 1.0025 xx 99\ 389.24 – 555 = 99\ 082.71…`
`V_4`	`= 1.0025 xx V_3 = 98\ 775.41…`
`V_5`	`= 1.0025 xx V_4 = 98\ 467.35…`

`=> C`