Financial Mathematics

Worked example 1 | Annual compounding

Question

$3000 is invested at \(4.5\%\) per year, compounded annually. Find (a) the value after five years and (b) the interest earned, both to the nearest cent.

Solution

(a) Use booklet formula 1.4 with \(k=1\). The annual multiplier is \(q=1.045\): \[u_5=3000(1.045)^5=3738.545812\ldots\approx\$3738.55.\] (b) Interest earned \(=u_5-u_0=738.545812\ldots\approx\$738.55\).

Worked example 2 | Monthly compounding

Question

$5000 is invested for three years at a nominal annual rate of \(4.8\%\), compounded monthly. Find: (a) the final balance and (b) the interest earned, both to the nearest cent.

Solution

(a) Use \(r=4.8\), \(k=12\) and \(n=3\). The monthly multiplier is \(q=1.004\); there are \(m=kn=12(3)=36\) months. \[\mathrm{FV}=5000\left(1+\frac{4.8}{100(12)}\right)^{36} =5772.762169\ldots\] The final balance is \(\boxed{\$5772.76}\). (b) Interest earned \(=5772.762169\ldots-5000\approx\$772.76\).

Worked example 3 | A car’s value

Question

A car costs $24000 and depreciates by \(15\%\) of its current value each year. Find:

1. its value after four years;

2. the total value lost.

Solution

(a) It retains \(85\%\) each year, so \(q=0.85\). Using \(u_n=u_0q^n\), \[u_4=24000(0.85)^4=\$12528.15.\] (b) Value lost \(=u_0-u_4=24000-12528.15=\$11471.85\).

Worked example 4 | Find a depreciation rate

Question

A machine falls from $20000 to $14580 over three years. Its annual percentage depreciation is constant. Find this percentage.

Solution

Use \(u_n=u_0q^n\), where \(q=1-p/100\): \[\begin{aligned} 14580&=20000q^3,\\ q^3&=\frac{14580}{20000}=0.729,\\ q&=\sqrt[3]{0.729}=0.9. \end{aligned}\] It retains \(90\%\), so it loses \(100\%-90\%=10\%\) each year.

The annual depreciation rate is \(10\%\).

Worked example 5 | Balance growth versus buying power

Question

$10000 earns \(4\%\) per year, compounded annually, for five years. Inflation is \(2.5\%\) per year. Find, to the nearest cent:

1. the nominal final balance;

2. its real value in today’s money;

3. the gain in buying power.

Solution

(a) Use \(u_n=u_0q^n\), with \(q=1.04\): \[u_5=10000(1.04)^5=12166.529024\approx\$12166.53.\]

(b) Adjust for inflation.

Prices have been multiplied by \(1.025^5\), so divide the nominal balance by this multiplier: \[\frac{10000(1.04)^5}{(1.025)^5}=10753.438843\ldots\] In Scratchpad (or a Calculator page), enter the fraction and press Enter. The screen rounds the display to \(10753.439\); to the nearest cent, the real value is \(\boxed{\$10753.44}\).

(c) The gain in buying power is \(10753.438843\ldots-10000\), or $753.44. The nominal gain of $2166.53 overstates this gain.

Real annual growth (not given in the booklet). The annual multiplier is \(1.04/1.025=1.014634146\ldots\), so buying power grows by about \(1.46\%\) per year. Subtracting \(4-2.5=1.5\%\) is only an approximation.

Open the Finance Solver

1. Open a Calculator page.

2. Press the Menu button.

3. Choose 8: Finance, then

   1: Finance Solver.

Use Tab to move through the fields below. Once all known values are entered, place the cursor inside the unknown field and press Enter to calculate it.

Worked example 2 revisited | Use the Finance Solver

Question

$5000 is invested for three years at a nominal annual interest rate of \(4.8\%\), compounded monthly. Use the Finance Solver to find, to the nearest cent:

1. the final balance;

2. the total interest earned.

Solution

This is the monthly investment from section 2, now solved using technology.

(a) The booklet values are \(r=4.8\), \(k=12\) and \(n=3\). Find FV as follows.

1. Enter the known values

N \(=12(3)=36\) months; I(%) \(=4.8\) is the annual rate. PV \(=-5000\) records our deposit; Pmt \(=0\) means no further payments. Set PpY \(=12\). FV still shows its default \(0\); it has not been solved.

2. Check the lower settings

Set CpY \(=12\): interest compounds monthly. Check that PpY is also \(12\), so N counts months. Leave PmtAt at END; with Pmt \(=0\), payment timing has no effect. N and I(%) are above the visible area.

3. Select the unknown: FV

After entering and checking all other values, use Tab to place the cursor inside the FV cell. Then press Enter to calculate FV. The selected \(0\) is a starting value, not the answer.

4. Read and interpret the result

FV \(=5772.7621691837\). Its positive sign represents money coming back to us. The balance is $5772.76 to the nearest cent. The message below confirms that the solver values have been stored.

Check (a): \(5000(1+4.8/1200)^{36}=5772.762169\ldots\), matching the solver.

(b) Interest earned \(=5772.762169\ldots-5000\approx\boxed{\$772.76}\).

Worked example 6 | Find the initial deposit

Question

An investment must grow to $8000 in five years. The nominal annual rate is \(3.6\%\), compounded quarterly. Find the initial deposit to the nearest cent.

Solution

Choose either method below. Both find the same deposit; you do not need to complete both unless asked.

Method 1 | By hand: rearrange the formula

There are \(m=kn=4(5)=20\) quarters. The multiplier is \(q=1+3.6/(100\times4)=1.009\). Using the geometric model \(u_m=u_0q^m\), \[8000=u_0(1.009)^{20} \quad\Rightarrow\quad u_0=\frac{8000}{1.009^{20}}=6687.544151\ldots\] Initial deposit: $6687.54. Booklet: this deposit expression is not given; rearrange formula 1.4.

Method 2 | Use the Finance Solver

Enter the known values, check the settings, then solve directly for PV.

1. Enter the known values.

N \(=20\) quarters; I(%) \(=3.6\) is the annual rate. Set Pmt \(=0\), FV \(=+8000\) and PpY \(=4\). The positive FV is money we will receive. PV is the unknown: the displayed \(0\) has not been solved.

2. Check the lower settings.

PpY \(=4\) and CpY \(=4\): N counts quarters and interest compounds quarterly. Leave PmtAt at END; with Pmt \(=0\), timing has no effect. Once all other entries are complete, place the cursor inside the PV cell and press Enter.

3. Interpret the result.

PV \(=-6687.5441512098\). Its negative sign means that the deposit leaves us to enter the account. The amount deposited is positive: \(\boxed{\$6687.54}\), agreeing with Method 1.

Worked example 7 | Find the nominal annual rate

Question

$6000 grows to $7000 in three years with monthly compounding. Find the nominal annual rate to three significant figures.

Solution

Choose either method. Both find the same annual percentage rate.

Method 1 | By hand: rearrange the formula

Use \(u_m=u_0q^m\), with \(m=12(3)=36\) months and \(q=1+r/1200\): \[7000=6000\left(1+\frac{r}{1200}\right)^{36} \quad\Rightarrow\quad 1+\frac{r}{1200}=\left(\frac{7000}{6000}\right)^{1/36}.\] \[r=1200\left[\left(\frac{7000}{6000}\right)^{1/36}-1\right]=5.149372839\ldots\] Nominal annual rate: \(5.15\%\) (3 s.f.). Booklet: this rate expression is not given; rearrange formula 1.4.

Method 2 | Use the Finance Solver

Enter the known values, check the settings, then solve directly for I(%).

1. Enter the known values.

N \(=36\) months; PV \(=-6000\) records our deposit; Pmt \(=0\); FV \(=+7000\) is the money we will receive. Set PpY \(=12\). I(%) is the unknown. The selected \(0\) is a starting value, not the calculated rate.

2. Check the settings; then solve.

Set PpY \(=12\) and CpY \(=12\) for monthly periods and compounding. Leave PmtAt at END. Once all other values are entered and checked, return to the I(%) cell, place the cursor inside it and press Enter.

3. Interpret the result.

I(%) \(=5.1493728391728\). This is the nominal annual percentage rate, even though N counts months. Report \(\boxed{5.15\%\text{ per year}}\) to three significant figures. No further multiplication by \(12\) or \(100\) is needed.

Worked example 8 | The first month a target is reached

Question

$6000 is invested at a nominal annual rate of \(4.5\%\), compounded monthly. Interest is credited at each month-end. Find the minimum number of complete months for the balance to reach at least $8000.

Solution

Choose either method below.

The monthly multiplier is \(q=1+0.045/12=1.00375\). We need the smallest integer \(m\) with \(6000(1.00375)^m\geq8000\). Solve the boundary equality first:

Method 1 | By hand using logarithms (once studied)

Booklet: the log power rule is given in 1.7; the time expression below is rearranged, not given.

\[\begin{gathered} (1.00375)^m=\frac{8000}{6000} \quad\Rightarrow\quad m\ln(1.00375)=\ln\left(\frac{8000}{6000}\right),\\[3pt] m=\frac{\ln(8000/6000)}{\ln(1.00375)}=76.85897062\ldots \end{gathered}\] Since \(q>1\), we need \(m\geq76.85897062\ldots\): at least 77 complete months.

Method 2 | Use the Finance Solver

1. Enter the known values.

I(%) \(=4.5\), PV \(=-6000\), Pmt \(=0\), FV \(=+8000\), PpY \(=12\). N is the unknown; its initial \(0\) is not the answer.

2. Check the settings; then solve.

Set CpY \(=12\); leave PmtAt at END. After checking all other fields, return to the N cell, place the cursor inside it and press Enter.

3. Interpret the output in months.

N \(=76.858970624218\). With PpY \(=12\), this counts months, not years. We need a whole month-end: month 76 is too early, so round up to \(\boxed{77\text{ complete months}}\), or \(6\) years and \(5\) months.

Check the neighbouring balances: \[\begin{aligned} u_{76}&=6000(1.00375)^{76}\approx\$7974.32<\$8000,\\ u_{77}&=6000(1.00375)^{77}\approx\$8004.22>\$8000. \end{aligned}\] Rounding rule. For the first complete period when an increasing balance reaches a target, round a non-integer boundary up, not to the nearest integer. Keep full precision before deciding.

Worked example 9 | Which account grows faster?

Question

$5000 is invested for three years. Account A pays \(4.85\%\) per year, compounded annually. Account B pays a nominal annual rate of \(4.8\%\), compounded monthly. There are no further transactions or charges.

1. Find each balance after one year, then each percentage gain over that year. Which account grows faster?

2. Find each final balance and the difference, to the nearest cent.

Solution

(a) Grow the deposit for one year. A multiplies by \(1.0485\) once; B multiplies by \(1+0.048/12=1.004\) twelve times: \[\begin{gathered}\text{A: }5000(1.0485)=\$5242.50,\\[4pt]\text{B: }5000(1.004)^{12}=5245.351037\ldots\approx\$5245.35.\end{gathered}\] Express each gain as a percentage of the deposit. Keep unrounded balances: \[\begin{gathered}\text{A: }\frac{5242.50-5000}{5000}\times100\%=4.85\%,\\[4pt]\text{B: }\frac{5245.351037\ldots-5000}{5000}\times100\%\approx4.90702\%.\end{gathered}\] These one-year percentage gains are the effective annual rates. B grows faster, despite its lower quoted rate: monthly interest itself earns further interest during the year.

(b) Grow the original deposit for three years: \[\begin{gathered}\mathrm{FV}_{A}=5000(1.0485)^3\approx\$5763.35,\\[4pt]\mathrm{FV}_{B}=5000(1.004)^{36}\approx\$5772.76.\end{gathered}\] Using unrounded balances, B exceeds A by \(5000(1.004)^{36}-5000(1.0485)^3\approx\boxed{\$9.41}\). B’s final balance matches worked example 2.

Arithmetic sum, starting from \(u_0\)

Using \(u_n=u_0+nd\) and sum = number of terms \(\times\) average of first and last, \[S_n=\frac{n+1}{2}(u_0+u_n) =\frac{n+1}{2}(2u_0+nd).\]

Booklet: not given in this form. Adapt section 1.2 to \(n+1\) terms starting at \(u_0\).

Question

Four annual fees are $200, $220, $240 and $260. Find the total paid.

Solution

These are \(u_0,u_1,u_2,u_3\): the final index is \(n=3\), so the total is \(S_3\). With \(u_0=200\) and \(d=20\), \[S_3=\frac{3+1}{2}\bigl(2(200)+3(20)\bigr)=\$920.\]

Geometric sum, starting from \(u_0\)

\[S_n=\frac{u_0(1-r^{n+1})}{1-r} =\frac{u_0(r^{n+1}-1)}{r-1}\quad(r\ne1).\]

Booklet: not given in this form. Adapt section 1.3 to \(n+1\) terms starting at \(u_0\).

If \(r=1\), \(\boxed{S_n=(n+1)u_0}\): count the equal terms (not given in the booklet).

Question

Four annual charges start at $100 and increase by \(10\%\) each year. Find their total.

Solution

The four charges run from \(u_0\) to \(u_3\). With \(u_0=100\), \(r=1.10\) and final index \(n=3\), \[S_3=\frac{100(1.10^{3+1}-1)}{1.10-1}=\$464.10.\] This is \(100+110+121+133.10\). If all four charges were $100, then \(r=1\) and \(S_3=(3+1)(100)=\$400\).