1 Time-Value of Money

1.1 Suplementary Concepts

The accumulation function \(a(t)\) models the accumulated amount of an investment of \(1\) at time \(t\geq0\). The present value function is the multiplicative inverse of \(a(t)\), in other words, \(\large{\frac{1}{a(t)}}\).
\(a(t)\) has three properties:

\(a(0)=1\), according to the definition above.
\(a(t)\) is generally increasing; it is decreasing in case of negative interest rates. For example, \(a(t)=1+it\), \(t\in \{1, 2, 3, \cdots\}\) is an accumulation function under simple interest; and \(a(t)=(1+i)^t\), \(t\in \{1, 2, 3, \cdots\}\) is an accumulation function under compound interest.
\(a(t)\) is normally a continuous function, given that under normal circumstances, interest is assumed to accumulate continuously.

\(A(t)=k \times a(t)\) is the amount of money accumulated at time \(t\geq 0\) of an original investment of \(k\) at \(t=0\). Thus, \(A(0)=k\).
The amount of interest earned in the \(n^{th}\) period from the investment time is \(I_n=A(n)-A(n-1)\), \(n\in \{1, 2, 3, \cdots\}\).
The effective rate of interest is the amount that an original investment invested at \(t=0\) will earn during the period. In other words, \(i=a(1)-a(0)\), or alternatively, \(a(1)=1+i\).

\[ \boxed{ \begin{gathered} \\ \quad i=\frac{(1+i)-1}{1}=\frac{a(1)-a(0)}{a(0)}=\frac{A(1)-A(0)}{A(0)}=\frac{I_n}{A(0)} \quad \\ \\ i_n=\frac{A(n)-A(n-1)}{A(n-1)}=\frac{I_n}{A(n-1)} \quad {,} \quad n \in \{1, 2, 3, \cdots \} \quad\\ \\ \end{gathered} } \]

The following table includes examples of \(a(t)\) for the computation of present and future values.

	Simple interest	Compound interest	Simple discount	Compound discount
Present value	\((1+it)^{-1}\)	\((1+i)^{-t}\)	\((1-dt)\)	\((1-d)^t\)
Future value	\((1+it)\)	\((1+i)^t\)	\((1-dt)^{-1}\)	\((1-d)^{-t}\)

The effective discount rate can be defined as follows:

\[ \boxed{% \begin{gathered} \\ \quad d=\frac{(1+i)-1}{1}=\frac{a(1)-a(0)}{a(1)}=\frac{A(1)-A(0)}{A(1)}=\frac{I_n}{A(1)} \quad\\ \\ d_n=\frac{A(n)-A(n-1)}{A(n)}=\frac{I_n}{A(n)} \quad {,} \quad \small{n \in \{1, 2, 3, \cdots \}} \quad\\ \\ \end{gathered}% } \]

Note the analogies with the formulae of #5 above.

The force of interest \(\delta _t\) is the intensity of interest earned in each moment, i.e., in each infinitesimally small time interval.

\[ \boxed{ \begin{gathered} \\ \\ \quad \delta _t=\frac{A'(t)}{A(t)}= \small{\lim} \frac{A(t+h)-A(t)}{h \cdot A(t)}=\frac{d}{dt} \log A(t) \quad\\ \\ a(n)= \exp (\int _0 ^n \delta _r \, dr)\\ \\ \delta = \log (1+i)\\ \\ \end{gathered} } \] 9. The following formulae provide for the relationships between effective rates of interest and discount, and nominal rates of interest and discount convertible \(m\) times per year. \[ \boxed{ \begin{gathered} \\ \quad 1+i &= \left[ 1 + \frac{i^{(m)}}{m} \right ]^m\quad & \Leftrightarrow \quad i^{(m)} = m \left[ (1+i)^{1/m}-1 \right] \quad\\ \\ \quad 1-d &= \left[ 1 + \frac{d^{(m)}}{m} \right ]^{-m}\quad & \Leftrightarrow \quad d^{(m)} = m \left[ 1-v^{1/m} \right] \quad\\ \\ \end{gathered} } \]

1.2 Solved Exercises

Before opening the solutions, take a moment to work through each problem—-your mastery grows most when you engage actively with the mathematics.

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

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

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

Accumulation function

For a nominal rate \(i^{(m)}\) convertible \(m\)-thly, in years:

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

Future value at \(t = 5\)

\(A(5) = 12{,}000\)

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

Comparison with \(9.2\%\) effective

Second bank: \(i_2 = 0.092\)

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

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

\(\\\)

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 \(0–3\)

\(a(3) = \exp(0.03 \cdot 3 + 0.001 \cdot 3^2) = 1.10407\)

Let \(i\) be the level effective annual rate:

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

So \(i \approx 3.36\%\).

Real effective annual rate 0–3

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

\(r = \frac{1.03355}{1.025} - 1 = 0.00834\)

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

Present value at \(t=0\) of \(20{,}000\) at \(t=6\)

\(a(6) = \exp(0.03 \cdot 6 + 0.001 \cdot 6^2)= 1.24110\)

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

\(\\\)

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:

\(PV_1 = 3{,}000 \times 1.04^{-1} \approx 2{,}884.62\)

Second payment:

\(PV_2 = 2{,}000 \times 1.1025^{-1} \approx 1{,}814.06\)

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:

\(L = PV_1 + PV_2 + X \times (1.05)^{-4}\)

Thus,

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

Since \((1.05)^4 = 1.21550625\)

\(X \approx 1.21551\,L - 5{,}711.27\)

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:

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

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

Since \((0.94)^4 = 0.7831\)

\(X \approx 1.28082\,L - 6{,}018.16\)

\(\\\)

1.3 Supplementary Exercises

If \(A(t)=t^{2}+2t+3\), find \(a(4)\).

Let \(a(t)=at^{2}+b\).

If \(1\,000\) invested at \(t=0\) accumulates to \(1\,720\) at \(t=3\), find the accumulated value at \(t=10\) of \(1\,000\) invested at \(t=5\).

If \(A(4)=1\,000\) and \(i_{n}=0.01n\), find \(A(7)\).

If \(A(t)=t^{2}+2t+3\), find the amount of interest earned between times \(10\) and \(20\).

In how many years will \(500\) accumulate to \(630\) at \(7.8\%\) simple interest?

At a certain rate of simple interest \(1\,000\) will accumulate to \(1\,110\) after a certain period.

Find the accumulated value of \(500\) at a rate of simple interest three-fourths as great over twice as long a period.

One share of NVIDIA stock at the company’s Initial Public Offering (IPO) was worth USD \(0.04\).

Twenty-seven years later, the stock traded for USD \(140.14\).

Calculate

the compounded annual growth rate (CAGR) of the NVIDIA stock; and
the amount that an investor who bought USD \(10\,000\) worth of stock at the IPO date would have accumualted \(27\) years later.

On 31 December 2024, one Bitcoin (BTC) was worth USD \(93\,429.20\).

Over the past \(10\) years, the price of BTC has grown by a compounded annual growth rate of \(76.734\%\).

Calculate

the price of BTC \(10\) years earlier;
the hypothetical price of BTC today, if the value of the BTC ten years earlier (calculated above) had grown at a simple annual interest rate of \(76.734\%\).

Simple interest of \(i=4\%\) is being credited to a fund.

In which period is this rate equivalent to an effective rate of \(2.5\%\)?

It is known that \(600\) invested for two years will earn \(264\) in interest.

Find the accumulated value of \(2\,000\) invested at the same rate of compound interest for three years.

Find the present value at \(12\%\) simple discount of \(1\,000\) due in five months.

A bank charges \(11\%\) simple interest in advance on short-term loans.

Find the sum received by the borrower who requests \(9\,000\) for \(90\) days.

A bank charges \(12\%\) simple discount rate on short-term loans.

A borrower needs \(20\,000\) cash, to be repaid with interest in nine months.

How much interest will she pay?

A bank charges \(12\%\) simple discount rate on short-term loans.

A borrower needs \(20\,000\) cash, to be repaid with interest in nine months.

What rate of simple interest does the bank realize?

A company has a liability of \(1\,000\,000\) due in five months.

The rate of simple interest is \(12\%\).

Calculate the true discount.

If \(d^{(4)}=5\%\), find \(10^{6} \cdot |i^{(3)}-d^{(2)}|\).

Find the accumulated value of \(1\,000\) at the end of two years if the nominal rate of discount is \(6\%\) convertible once every four years.

Given that \(i^{(m)}=0.1844144\) and \(d^{(m)}=0.1802608\), find \(m\).

Find \(m\) given that

\(1+\large \frac{i^{(m)}}{m} = \Large \frac{1+\frac{i^{(4)}}{4}}{1+\frac{i^{(5)}}{5}}\)

If \(\large{\frac{i^{(4)}}{d^{(4)}}}=1.012272234\), find \(v\).

In account \(A\) money accumulates at a force of interest \(\delta_{t}=0.01t+0.1\), \(0≤t≤25\). In account \(B\) money accumulates at an annual effective rate of interest \(i\). An amount of \(1\,000\) is invested in each account for \(25\) years. The accumulated value of both accounts at the end of the investment period is identical.

Calculate the value of account \(B\) at the end of two years.

If \(\delta_{t}=0.01t\), \(0≤t≤2\), find the equivalent annual effective rate of interest over the interval \(0≤t≤2\).

Find the accumulated value of \(1\,000\) at the end of \(25\) years if \(\delta_{t}=1.2(1+t)^{-2}\)

Eighty days after borrowing money, a person pays back \(850\).

How much was borrowed if the \(850\) includes principal and simple interest at \(9.75\%\)?

Assume a \(365\)-day year.

An investment fund advertises that it will triple your money in \(10\) years.

What rate of interest compounded monthly is implied?

Payments of \(1\,000\), \(2\,000\), and \(5\,000\) are due at the ends of years two, three, and eight, respectively.

Assuming an effective rate of interest of \(5\%\) per annum, find the point in time at with a payment of \(8\,000\) would be equivalent.

Find the number of years necessary for \(1\,000\) to accumulate to \(1\,500\) if invested at \(6\%\) per annum compounded seminannually.

At what interest rate convertible semiannually would an investment of \(1\,000\) immediately and \(2\,000\) three years from now accumulate to \(5\,000\) \(10\) years from now?

The amount of interest earned on \(A\) for one year is \(336\), while the equivalent amount of discount is \(300\).

Find \(A\).

A bank pays \(7\%\) effective on deposits at the end of each year.

At the end of every three years a \(2\%\) bonus is paid on the balance at that time.

Find the effective rate of interest earned by an investor if the money is left on deposit for seven years.

An investor deposits \(10\,000\) in a bank.

During the first year the bank credits an annual effective rate of interest \(i\).

During the second year, the bank credits an annual effective rate of interest \(i-0.05\).

At the end of two years, the account balance is \(12\,093.75\).

What would the account balance have been at the end of three years, if the annual effective rate of interest were \(i+0.09\) for each of these three years?

Charlie signs a one-year promissory note for \(1\,000\) and receives \(920\) from the bank.

At the end of six months, Charlie makes a payment of \(288\).

Assuming simple discount, to what amount does this reduce the face amount of the note?

Let \(A\) be the accumulated value of \(2\,000\) invested for four years, if the rate of simple interest is \(8\%\) per annum, and let \(B\) be the accumulated value of \(2\,000\) invested for four years, if the rate of compounded interest is \(8\%\) per annum.

Find \(│A-B│\).

Find the accumulated value of \(2\,480\) at the end of \(12\) years if the nominal interest rate was \(2\%\) convertible monthly for the first three years, the nominal rate of discount was \(3\%\) convertible seminannually for the next two years, the nominal rate of interest was \(4.2\%\) convertible once every two years for the next four years, and the annual effective rate of discount was \(0.058\) for the last three years.

Find the present value of \(5\,000\) to be paid at the end of \(25\) months at a rate of discount of \(8\%\) convertible quarterly, assuming simple discount during the final fractional period.

Money is invested for five years in a savings account earning \(7\%\) effective.

If the rate of inflation is \(10\%\), find the percentage of purchasing power lost during the period of investment.

1.4 Code snippets

Definite integral:

The integral \(\int_0^{25} t^2e^{-0.05t}dt\) can be evaluated as follows:

f <- function(t) {t^2 * exp(-0.05 * t)}

integrate(f, 0, 25)$value

[1] 2104.517

Calculation of present value of cash flows:

Consider the following sequence of annual cash flows of \(2\,000, 3\,000, ..., 21\,000\). To compute the present value of these cash flows at \(5\%\) effective, we use:

(cf <- seq(2, 21))

 [1]  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21

(dsct <- 1.05 ^ -(1:20))

 [1] 0.9523810 0.9070295 0.8638376 0.8227025 0.7835262 0.7462154 0.7106813
 [8] 0.6768394 0.6446089 0.6139133 0.5846793 0.5568374 0.5303214 0.5050680
[15] 0.4810171 0.4581115 0.4362967 0.4155207 0.3957340 0.3768895

(sum(cf * dsct))

[1] 123.4128