smc-603-70-Recurrence relation

Recursion and Finance, GEN2 2023 VCAA 7

Arthur takes out a new loan of $60 000 to pay for an overseas holiday.

Interest on this loan compounds weekly.

The balance of the loan, in dollars, after $n$ weeks, $V_n$, can be determined using a recurrence relation of the form

$V_0=60\ 000, \quad V_{n+1}=1.0015\,V_n-d$

Show that the interest rate for this loan is 7.8% per annum. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Determine the value of $d$ in the recurrence relation if
i. Arthur makes interest-only repayments (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
ii. Arthur fully repays the loan in five years. Round your answer to the nearest cent. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Arthur decides that the value of $d$ will be 300 for the first year of repayments.
If Arthur fully repays the loan with exactly three more years of repayments, what new value of $d$ will apply for these three years? Round your answer to the nearest cent. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
For what value of $d$ does the recurrence relation generate a geometric sequence? (1 mark)

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

Show Answers Only

Show Worked Solution

a. $\text{Weekly interest rate factor}\ = 1.0015-1 = 0.0015 = 0.15\% $

$I\%(\text{annual}) = 52 \times 0.15 = 7.8\%$

♦♦ Mean mark (a) 32%.

b.i. $d= 0.15\% \times 60\ 000 = \dfrac{0.15}{100} \times 60\ 000 = 90$

b.ii. $\text{By TVM solver:} $

$N$	$=5 \times 52 = 260$
$I\%$	$=7.8$
$PV$	$= -60\ 000$
$PMT$	$= ?$
$FV$	$=0$
$\text{P/Y}$	$=\ \text{C/Y}\ = 52$

$\Rightarrow PMT = d= $278.86$

♦ Mean mark (b)(i) 47%.
♦ Mean mark (b)(ii) 40%.

c. $d=300\ \text{for the 1st 52 weeks.}$

$\text{Find}\ FV\ \text{after 52 weeks:}$

$N$	$=52$
$I\%$	$=7.8$
$PV$	$= -60\ 000$
$PMT$	$= 300$
$FV$	$=?$
$\text{P/Y}$	$=\ \text{C/Y}\ = 52$

$\Rightarrow FV = $48\ 651.67$

$\text{Find}\ PMT\ \text{given}\ FV=0\ \text{after 156 more weeks:}$

$N$	$=156$
$I\%$	$=7.8$
$PV$	$= -48\ 651.67$
$PMT$	$= ?$
$FV$	$=0$
$\text{P/Y}$	$=\ \text{C/Y}\ = 52$

$\Rightarrow PMT = d= $350.01$

♦♦ Mean mark (c) 32%.

d. $\text{Geometric sequence when}\ \ d=0$

$V_0=60\ 000, V_1=60\ 000(1.0015), V_2=60\ 000(1.0015)^2, …$

♦♦♦ Mean mark (d) 11%.

Recursion and Finance, GEN2 2023 VCAA 5

Arthur borrowed $30 000 to buy a new motorcycle.

Interest on this loan is charged at the rate of 6.4% per annum, compounding quarterly.

Arthur will repay the loan in full with quarterly repayments over six years.

How many repayments, in total, will Arthur make? (1 mark)

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

The balance of the loan, in dollars, after $n$ quarters, $A_n$, can be modelled by the recurrence relation

$A_0=30\ 000, \quad A_{n+1}=1.016 A_n-1515.18$

Showing recursive calculations, determine the balance of the loan after two quarters.
Round your answer to the nearest cent. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
The final repayment required will differ slightly from all the earlier repayments of $1515.18
Determine the value of the final repayment.
Round your answer to the nearest cent. (1 mark)

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

Show Answers Only

a. $\text{Repayments}\ = 6 \times 4 = 24$

b. $A_1=1.016 \times 30\ 000-1515.18=$28\ 964.82 $

$A_2=1.016 \times 28\ 964.82-1515.18=$27\ 913.08 $

c. $ \text{Final repayment}\ = 1515.18-0.14=\$1515.04$

Show Worked Solution

a. $\text{Repayments}\ = 6 \times 4 = 24$

b. $A_1=1.016 \times 30\ 000-1515.18=$28\ 964.82 $

$A_2=1.016 \times 28\ 964.82-1515.18=$27\ 913.08 $

Mean mark (b) 51%.
♦♦ Mean mark (c) 25%

c. $\text{Solve for}\ N\ \text{using TMV calculator:}$

$N$	$=24$
$I\%$	$=6.4$
$PV$	$= -30\ 000$
$PMT$	$= 1515.18$
$FV$	$=?$
$\text{P/Y}$	$=\ \text{C/Y}\ = 4$

$FV = -0.14$

$\therefore \text{Final repayment}\ = 1515.18-0.14=\$1515.04$

Recursion and Finance, GEN1 2022 VCAA 18-19 MC

The balance of a loan, $V_n$, in dollars, after $n$ months is modelled by the recurrence relation

$V_0=400\ 000,\ \ \ V_{n+1}=1.003\,V_n-2024$

Question 18

The balance of the loan first falls below $398 000 after how many months?

Question 19

With a small change to the final payment, the loan is expected to be repaid in full in

25 years.
26 years.
28 years.
29 years.
30 years.

Show Answers Only

$\text{Question 18: C}$

$\text{Question 19: A}$

Show Worked Solution

$\text{Question 18}$

$V_1$	$=1.003 \times 400\ 000-2024 = $399 176$
$V_2$	$=1.003 \times 399\ 176-2024= $398 349.528$
$V_3$	$=1.003 \times 398\ 349.528-2024= $397 520.58$

$\Rightarrow C$

$\text{Question 19}$

$\text{Monthly interest rate = 0.3%}$

$\text{Annual interest rate}\ (r) = 12 \times 0.3 = 3.6\%$

$\text{Solve for}\ N\ \text{using TMV calculator:}$

$I\%$	$=3.6$
$PV$	$= 400\ 000$
$PMT$	$= -2024$
$FV$	$=0$
$\text{P/Y}$	$=12$
$\text{C/Y}$	$=12$

$N = 300.002\ \text{months}\ \approx \dfrac{300}{12} \approx 25\ \text{years}$

$\Rightarrow A$

♦ Mean mark (Q19) 47%.

CORE, FUR2 2021 VCAA 8

For renovations to the coffee shop, Sienna took out a reducing balance loan of $570 000 with interest calculated fortnightly.

The balance of the loan, in dollars, after `n` fortnights, `S_n` can be modelled by the recurrence relation

`S_0 = 570 \ 000,` `S_{n+1} = 1.001 S_n - 1193`

Calculate the balance of this loan after the first fortnightly repayment is made. (1 mark)
Show that the compound interest rate for this loan is 2.6% per annum. (1 mark)
For the loan to be fully repaid, to the nearest cent, Sienna's final repayment will be a larger amount.
Determine this final repayment amount.
Round your answer to the nearest cent. (1 mark)

Show Answers Only

`$ 569 \ 377`
`2.6text(%)`
`$ 1198.56`

Show Worked Solution

a.	`S_1`	`= 1.001 xx 570 \ 000 – 1193`
		`= $ 569 \ 377`

b. `text{Fortnights in 1 year} = 26`

`text{Rate per fortnight} = (1.001 – 1) xx 100text(%) = 0.1text(%)`

`:.\ text(Annual compound rate)`	`= 26 xx 0.1text(%)`
	`= 2.6text(%)`

c. `text{Find}\ N \ text{by TVM Solver:}`

`N`	`= ?`
`I text{(%)}`	`= 2.6`
`PV`	`= 570 \ 000`
`PMT`	`= -1193`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 26`

`=> N = 650.0046 …`

`text{Find} \ FV \ text{after exactly 650 payments:}`

`N`	`= 650`
`Itext{(%)}`	`= 2.6`
`PV`	`= 570 \ 000`
`PMT`	`= -1193`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 26`

`=> FV = -5.59`

`:. \ text{Final repayment}`	`= 1193 + 5.59`
	`= $ 1198.59`

CORE, FUR1 2021 VCAA 23 MC

Bimal has a reducing balance loan.

The balance, in dollars, of the loan from month, `B_n`, is modelled by the recurrence relation below.

`B_o = 450\ 000, \ B_{n+1} = RB_n - 2633`

Given that the loan will be fully repaid in 20 years, the value of `R` is closest to

1.003
1.0036
1.03
1.036
1.36

Show Answers Only

`A`

Show Worked Solution

`text{By TVM Solver:}`

♦♦ Mean mark 31%.

`N`	`= 240`
`Itext{(%)}`	`= ?`
`PV`	`= 450\ 000`
`PMT`	`= -2633`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`:. Itext{(%)}= 3.5999 …`

`r_text{monthly} = 3.6/12 = 0.3`

`R`	`= 1 + r/100`
	`= 1 + 0.3/100`
	`=1.003`

`=> A`

CORE, FUR2 2020 VCAA 8

Samuel has a reducing balance loan.

The first five lines of the amortisation table for Samuel’s loan are shown below.

Interest is calculated monthly and Samuel makes monthly payments of $1600.

Interest is charged on this loan at the rate of 3.6% per annum.

Using the values in the amortisation table
i. calculate the principal reduction associated with payment number 3 (1 mark)
ii. calculate the balance of the loan after payment number 4 is made.
Round your answer to the nearest cent. (1 mark)
Let `S_n` be the balance of Samuel’s loan after `n` months.
Write down a recurrence relation, in terms of `S_0, S_(n+1)` and `S_n`, that could be used to model the month-to-month balance of the loan. (1 mark)

Show Answers Only

i. `$643.85`
ii. `$317\ 428.45`
`text(See Worked Solutions)`

Show Worked Solution

a.i.	`text(Principal reduction)`	`=\ text(Payment – interest)`
		`= 1600 – 956.15`
		`= $643.85`

♦ Mean mark part a.ii. 39%.

a.ii.	`text(Interest)`	`= 318\ 074.23 xx (0.036/12)`
		`= $954.22`

`:.\ text(Balance after payment 4)`

`= 318\ 074.23 – 1600 + 954.22`

`= $317\ 428.45`

♦♦ Mean mark part b. 30%.

b.	`S_0 = 320\ 000,\ S_(n+1)`	`= S_n(1 + 0.036/12) – 1600`
	`S_(n+1)`	`= 1.003 S_n – 1600`

CORE, FUR2 2019 VCAA 9

Phil would like to purchase a block of land.

He will borrow $350 000 to make this purchase.

Interest on this loan will be charged at the rate of 4.9% per annum, compounding fortnightly.

After three years of equal fortnightly repayments, the balance of Phil’s loan will be $262 332.33

What is the value of each fortnightly repayment Phil will make?

Round your answer to the nearest cent. (1 mark)
What is the total interest Phil will have paid after three years?

Round your answer to the nearest cent.
Over the next four years of his loan, Phil will make monthly repayments of $3517.28 and will be charged interest at the rate of 4.8% per annum, compounding monthly.

Let `B_n` be the balance of the loan `n` months after these changes apply.

Write down a recurrence relation, in terms of `B_0, B_(n + 1)` and `B_n`, that could be used to model the balance of the loan over these four years. (2 marks)

Show Answers Only

`$1704.03`
`$45\ 246.67`
`B_0=262\ 332.33,\ \ \ B_(n+1) = 1.004 B_n – 3517.28`

Show Worked Solution

a. `text(Find)\ PMT \ text{(by TVM solver)}:`

`N`	`= 3 xx 26 = 78`
`I(%)`	`= 4.9`
`PV`	`= 350\ 000`
`PMT`	`= ?`
`FV`	`= -262332.33`
`text(P/Y)`	`= text(C/Y) = 26`

`=> PMT = -1704.030`

`:.\ text(Fortnightly payment) = $1704.03`

b.	`text(Total interest)`	`= text(Payments) – text(decrease in principal)`
		`= 78 xx 1704.03 – (300\ 000 – 262\ 332.33)`
		`= $45\ 246.67`

c. `B_0 = 262\ 332.33`

`text(Monthly interest) = 4.8/12 = 0.4%`

`text(Monthly payment) = 3517.28`

`:.\ text(Recurrence relation:)`

`B_0=262\ 332.33,\ \ \ B_(n+1) = 1.004 B_n – 3517.28`

CORE, FUR1 2017 VCAA 19-20 MC

Shirley would like to purchase a new home. She will establish a loan for $225 000 with interest charged at the rate of 3.6% per annum, compounding monthly.

Each month, Shirley will pay only the interest charged for that month.

Part 1

After three years, the amount that Shirley will owe is

$73 362
$170 752
$225 000
$239 605
$245 865

Part 2

Let `V_n` be the value of Shirley’s loan, in dollars, after `n` months.

A recurrence relation that models the value of `V_n` is

`V_0 = 225\ 000,qquadV_(n + 1) = 1.003 V_n`
`V_0 = 225\ 000,qquadV_(n + 1) = 1.036 V_n`
`V_0 = 225\ 000,qquadV_(n + 1) = 1.003 V_n - 8100`
`V_0 = 225\ 000,qquadV_(n + 1) = 1.003 V_n - 675`
`V_0 = 225\ 000,qquadV_(n + 1) = 1.036 V_n - 675`

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

`text(If the loan payments are interest only,)`

`text(the principal outstanding after 3 years)`

`text(remains $225 000.)`

`=> C`

`text(Part 2)`

`text(Monthly interest rate)`

`= 3.6/12 = 0.3text(%) = 0.003`

`text(Monthly payment)`

`= 225\ 000 xx 0.3text(%)`

`= $675`

`:.\ text(Recurrence Relation is)`

`V_(n + 1) = 1.003V_n – 675`

`=> D`

CORE, FUR1 2017 VCAA 17 MC

The value of a reducing balance loan, in dollars, after `n` months, `V_n`, can be modelled by the recurrence relation shown below.

`V_0 = 26\ 000,qquadV_(n + 1) = 1.003 V_n - 400`

What is the value of this loan after five months?

`$24\ 380.31`
`$24\ 706.19`
`$25\ 031.10`
`$25\ 355.03`
`$25\ 678.00`

Show Answers Only

`=> A`

Show Worked Solution

`V_1 = 1.003 xx 26\ 000 – 400 = 25\ 678`

`V_2 = 1.003 xx 25\ 678 – 400 = 25\ 355.03`

`V_3 = 1.003 xx 25\ 355.03 – 400 = 25\ 031.10`

`V_4 = 1.003 xx 25\ 031.10 – 400 = 24\ 706.19`

`V_5 = 1.003 xx 24\ 706.19 – 400 = 24\ 380.31`

`=> A`

\(N\)	\(=5 \times 52 = 260\)
\(I\%\)	\(=7.8\)
\(PV\)	\(= -60\ 000\)
\(PMT\)	\(= ?\)
\(FV\)	\(=0\)
\(\text{P/Y}\)	\(=\ \text{C/Y}\ = 52\)

\(N\)	\(=52\)
\(I\%\)	\(=7.8\)
\(PV\)	\(= -60\ 000\)
\(PMT\)	\(= 300\)
\(FV\)	\(=?\)
\(\text{P/Y}\)	\(=\ \text{C/Y}\ = 52\)

\(N\)	\(=156\)
\(I\%\)	\(=7.8\)
\(PV\)	\(= -48\ 651.67\)
\(PMT\)	\(= ?\)
\(FV\)	\(=0\)
\(\text{P/Y}\)	\(=\ \text{C/Y}\ = 52\)

\(N\)	\(=24\)
\(I\%\)	\(=6.4\)
\(PV\)	\(= -30\ 000\)
\(PMT\)	\(= 1515.18\)
\(FV\)	\(=?\)
\(\text{P/Y}\)	\(=\ \text{C/Y}\ = 4\)

\(V_1\)	\(=1.003 \times 400\ 000-2024 = $399 176\)
\(V_2\)	\(=1.003 \times 399\ 176-2024= $398 349.528\)
\(V_3\)	\(=1.003 \times 398\ 349.528-2024= $397 520.58\)