6 Solved Review Exercises

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6.1 Time-Value of Money

6.1.1 Nominal vs. Effective Rates

A bank advertises a nominal interest rate of $9\%$ convertible quarterly.

Determine the effective annual interest rate.
Find the accumulation function $a(t)$ for this rate.
Compute the future value at $t=5$ of an initial investment of $12\,000$.
A second bank offers an effective annual rate of $9.2\%$. Which institution provides the higher accumulation at $t=5$?

Solution

Effective annual interest rate

$i = \left(1 + \frac{j}{m}\right)^m - 1 = \left(1 + \frac{0.09}{4}\right)^4 - 1 = (1.0225)^4 - 1 \approx 0.0930833$

So the effective annual rate is about $\boxed{9.308\%}$

Accumulation function

For a nominal rate $j$ convertible $(m$-thly, in years:

$a(t) = \left(1 + \frac{j}{m}\right)^{mt}= (1.0225)^{4t},\qquad t \ge 0$

Future value at $t = 5$

$A(5) = 12{,}000 \\ a(5)= 12{,}000 (1.0225)^{20}\approx 12{,}000 \times 1.56051 \approx 18{,}726.11$

Comparison with $9.2\%$ effective

Second bank: $i_2 = 0.092$

$A_2(5) = 12{,}000 (1.092)^5 \approx 12{,}000 \times 1.55279 \approx 18{,}633.50$

Bank 1 yields $18,726.11$ vs. $18,633.50$ from Bank 2. Therefore $\boxed{\text{Bank 1 is better over 5 years}}$

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6.1.2 Force of Interest, Inflation, Real Rates

An investment earns a time-varying force of interest $\delta(t) = 0.03 + 0.002t,\quad 0 \le t \le 6$.

Inflation is constant at $2.5\%$ per year (effective).

Express the accumulation function $a(t)$ implied by $\delta(t)$.
Compute the effective annual interest rate equivalent to this force over the first 3 years.
Determine the real effective annual rate for years $0–3$.
Find the present value at $t=0$ of a payment of $20{,}000$ due at $t=6$.

Solution

Accumulation function

$a(t) = \exp\!\left(\int_0^t \delta(s)\,ds\right)= \exp\!\left(\int_0^t (0.03 + 0.002s)\,ds\right)$

$\int_0^t (0.03 + 0.002s)\,ds= [0.03 s + 0.001 s^2]_0^t= 0.03 t + 0.001t^2$

$a(t) = \exp(0.03 t + 0.001 t^2),\quad 0 \le t \le 6$

Equivalent annual rate 0–3

$a(3) = \exp(0.03 \cdot 3 + 0.001 \cdot 3^2) = \exp(0.09 + 0.009) = \exp(0.099) \approx 1.10407$

Let $i$ be the level effective annual rate:

$(1 + i)^3 = 1.10407\quad\Rightarrow\quad i = 1.10407^{1/3} - 1 \approx 0.03355$

So $i \approx 3.36\%$.

Real effective annual rate 0–3

Real rate $r$ satisfies $1 + r = \frac{1 + i}{1 + j},\quad j = 0.025$

$r = \frac{1.03355}{1.025} - 1 \approx 1.00834 - 1 \approx 0.00834$

Thus the real rate is about $\boxed{0.83\%}$ per year.

Present value at $t=0$ of $20{,}000$ at $t=6$

$a(6) = \exp(0.03 \cdot 6 + 0.001 \cdot 6^2)= \exp(0.18 + 0.036)= \exp(0.216) \approx 1.24110$

$PV = \large{\frac{20{,}000}{a(6)}} \approx \large{\frac{20{,}000}{1.24110}} \approx 16{,}114.71$

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6.1.3 Equation of Value, Simple vs. Compound Interest

A borrower agrees to repay a debt through the following schedule:

$3{,}000$ at time $t = 1$ under simple interest at 4% per year.
$2{,}000$ at time $t = 2$ under compound interest at 5% effective.
A final payment $X$ at time $t = 4$, also at 5% effective.

All payments must satisfy an equation of value at time $t = 0$.

Let $L$ denote the present value of the loan at $t=0$.

Compute the present value at $t=0$ of the first two payments.
Express the value of $X$ in terms of $L$ when the final payment is valued at 5% effective.
Express the value of $X$ in terms of $L$ if instead the final payment is valued using a discount rate of 6%.

Solution

Present value of first two payments

First payment:

$PV_1 = \frac{3{,}000}{1.04} \approx 2{,}884.62$

Second payment:

$PV_2 = \frac{2{,}000}{1.1025} \approx 1{,}814.06$

Total:

$PV_{1,2} = PV_1 + PV_2 \approx 4{,}698.67$

Final payment valued at 5% effective

Present value of $X$ at $t=0$:

$PV_3 = \frac{X}{(1.05)^4}$

Equation of value:

$L = PV_1 + PV_2 + \frac{X}{(1.05)^4}$

Thus,

$\frac{X}{(1.05)^4} = L - 4{,}698.67 \quad \Rightarrow \quad X = (L - 4{,}698.67)(1.05)^4$

Since $(1.05)^4 \approx 1.21550625$

$X \approx 1.21551\,L - 5{,}711.27$

Final payment valued with discount rate $d = 6\%$

With a discount rate $d$, the present value of $X$ at time 0 for time 4 is

$PV_3 = X(1 - d)^4 = X(0.94)^4$

Equation of value:

$L = PV_{1,2} + X(0.94)^4 = 4{,}698.67 + X(0.94)^4$

$X(0.94)^4 = L - 4{,}698.67 \quad \Rightarrow \quad X = \frac{L - 4{,}698.67}{(0.94)^4}$

Since $(0.94)^4 \approx 0.7831$

$X \approx 1.28082\,L - 6{,}018.16$

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6.2 Annuities

6.2.1 Level Annuity-immediate vs. Annuity-due (payable m-thly)

Consider a nominal annual interest rate of 8% convertible monthly.

An annuity-immediate pays 500 at the end of each month for 10 years. Compute its present value at time 0.
An annuity-due pays 500 at the beginning of each month for 10 years at the same interest basis. Compute its present value at time 0.
Suppose an investor wants a present value of 30,000 under the same nominal rate (8% convertible monthly) with level monthly payments of 500 at the end of each month. Determine approximately how many payments are required and express the term in years (to two decimals).

Solution

Let the nominal rate be $j = 0.08$ convertible monthly, so

$i^{(12)} = \frac{j}{12} = \frac{0.08}{12} \approx 0.0066667$ and $v = \frac{1}{1 + i^{(12)}}$.

There are $n = 10 \times 12 = 120$ monthly payments.

Present value of the annuity-immediate

For a level annuity-immediate with payment $R$, interest rate $i$ per period, and $n$ periods,

$a_{\overline{n}|i} = \frac{1 - (1+i)^{-n}}{i}$

Thus

$PV_{\text{imm}} = 500 \, a_{\overline{120}|\,0.0066667}= 500 \frac{1 - (1.0066667)^{-120}}{0.0066667} \approx 41\,210.74$

Present value of the annuity-due

An annuity-due is related by $\ddot{a}_{\overline{n}|i} = (1+i) a_{\overline{n}|i}$

Hence

$PV_{\text{due}} = 500 \, \ddot{a}_{\overline{120}|\,0.0066667}= 500 (1.0066667) a_{\overline{120}|\,0.0066667} \approx 41\,485.48$

So moving payments from end-of-month to beginning-of-month increases the present value by about

$41\,485.48 - 41\,210.74 \approx 274.74$

Number of payments for a present value of 30,000

Let $n$ be the number of monthly payments of 500 at the end of each month with rate $i^{(12)}$.

$30\,000 = 500 \, a_{\overline{n}|\,0.0066667}= 500 \frac{1 - (1+i^{(12)})^{-n}}{i^{(12)}}$

Divide both sides by $500$:

$60 = \frac{1 - (1+i^{(12)})^{-n}}{i^{(12)}}$

Multiply by $i^{(12)}$:

$60 i^{(12)} = 1 - (1+i^{(12)})^{-n}$

With $i^{(12)} = 0.0066667$:

$60 \cdot 0.0066667 \approx 0.4 \quad \Rightarrow \quad 1 - (1+i^{(12)})^{-n} = 0.4$

$(1+i^{(12)})^{-n} = 0.6$

Taking natural logs:

$-n \ln(1 + i^{(12)}) = \ln(0.6) \quad \Rightarrow \quad n = -\frac{\ln(0.6)}{\ln(1 + i^{(12)})}$

Numerically,

$n \approx 76.88 \text{ months}$

In years:

$\frac{n}{12} \approx \frac{76.88}{12} \approx 6.41 \text{ years}$

So approximately 77 payments are required.

6.2.2 Perpetuities, m-thly vs. Continuous

The nominal annual interest rate is 6% convertible quarterly. A perpetuity pays 300 at the end of each quarter forever. Compute its present value at time 0.
Assume instead that the force of interest is $ = (1.06)$. Find the level continuous payment rate $k$ per year such that a perpetual continuous payment stream of rate $k$ has the same present value as in part (1).

Solution (click to expand)

Quarterly perpetuity

Nominal rate $j = 0.06$ convertible quarterly ($m = 4$) gives a quarterly effective rate $i^{(4)} = \frac{j}{4} = \frac{0.06}{4} = 0.015$

For a perpetuity with payment $R$ each period, $PV = \frac{R}{i}$

Hence, $PV_{\text{quarterly}} = \frac{300}{0.015} = 20\,000$

Equivalent continuous perpetuity under a force of interest

For a continuous payment stream at rate $k$ per year, paid forever, the present value is $PV_{\text{cont}} = \int_0^\infty k e^{-\delta t} \, dt = \frac{k}{\delta}$

We want this to equal the present value from part (1):

$\frac{k}{\delta} = 20\,000 \quad \Rightarrow \quad k = 20\,000 \, \delta$

With $\delta = \ln(1.06) \approx 0.0582689$, $k \approx 20\,000 \times 0.0582689 \approx 1\,165.38$. Thus a continuous payment rate of approximately 1 165.38 per year produces the same present value.

6.2.3 Arithmetic and Geometric Increasing Annuities

An arithmetic increasing annuity-immediate pays at the end of each year for 5 years. The payment in year $k$ is $1\,000 k$ (that is, 1 000, 2 000, 3 000, 4 000, 5 000). The effective annual interest rate is 7%. Compute the present value at time 0.
A geometric increasing annuity-immediate pays at the end of each year for 10 years. The first payment is 800, and each subsequent payment increases by 3% per year. The effective annual interest rate is 6%. Compute the present value at time 0.

Solution

Arithmetic increasing annuity-immediate

$v = \frac{1}{1+i} = \frac{1}{1.07}$

Payment in year $k$ is $1\,000 k$, so the present value is $PV_{\text{arith}} = \sum_{k=1}^{5} 1\,000 k v^k= 1\,000 \sum_{k=1}^{5} k v^k$

While there is a closed form, it is often straightforward to compute term by term:

\[ \begin{aligned} v^1 &= 1.07^{-1},\\ v^2 &= 1.07^{-2},\\ &\dots \end{aligned} \]

Numerically,

$\sum_{k=1}^5 k v^k \approx 11.7469$

$PV_{\text{arith}} \approx 1\,000 \times 11.7469 \approx 11\,746.86$

Geometric increasing annuity-immediate

Interest rate $i = 0.06$, so $v = 1/1.06$. First payment is 800, and each payment increases at 3%:

Payment at time 1: $800$,
Payment at time 2: $800 (1.03)$,
Payment at time 3: $800 (1.03)^2$,
…
Payment at time $k$: $800 (1.03)^{k-1}$.

Present value:

$PV_{\text{geom}} = \sum_{k=1}^{10} 800 (1.03)^{k-1} v^k = 800 \sum_{k=1}^{10} (1.03)^{k-1} (1.06)^{-k}$

Each term discounts the geometrically increasing payment back to time 0. Numerically, $PV_{\text{geom}} \approx 6\,655.04$

So the present values are approximately $PV_{\text{arith}} \approx 11\,746.86 \qquad PV_{\text{geom}} \approx 6\,655.04$