Worked examples · Sheet 1

Accumulated values, present values and discounts

Theory used: Principal, interest and accumulated value Simple interest Compound interest Present value Discount and discount rate Comparison of the two regimes

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

A principal $C = 1500$ euros is invested at the annual rate of $3\%$ . Compute the accumulated value after $8$ months under simple interest, and the accumulated value and the interest after $5$ years under compound interest.

t = \frac{8}{12} = \frac{2}{3}, \qquad M = 1500 \left( 1 + 0.03 \cdot \frac{2}{3} \right) = 1500 \cdot 1.02 = 1530 \text{ euros}

M = 1500 \cdot 1.03^{5} \approx 1500 \cdot 1.159274 \approx 1738.91 \text{ euros}, \qquad I = M - C \approx 238.91 \text{ euros}

Time must always be expressed in years, because the rate is annual; $8$ months are therefore $\dfrac{8}{12} = \dfrac{2}{3}$ of a year. Under simple interest the accumulated value is $C\,(1 + i\,t)$ , and the interest for $8$ months is $1500 \cdot 0.02 = 30$ euros, therefore the accumulated value is $1530$ euros.

Under compound interest the principal is multiplied by $1.03$ at the end of each year, so after $5$ years it is multiplied by $1.03^{5}$ . The interest is the difference between accumulated value and principal, about $238.91$ euros. As a check, under simple interest the interest for $5$ years would be $1500 \cdot 0.03 \cdot 5 = 225$ euros; the compound value is larger, as the comparison of the two regimes predicts for durations longer than a year.

Exercise 2

A bill of exchange of $3600$ euros matures in $5$ months. Compute the present value and the discount with commercial discount at the rate $d = 6\%$ and with rational discount at the rate $i = 6\%$ .

t = \frac{5}{12}, \qquad V = 3600 \left( 1 - 0.06 \cdot \frac{5}{12} \right) = 3600 \cdot 0.975 = 3510 \text{ euros}, \qquad D = 90 \text{ euros}

V = \frac{3600}{1 + 0.06 \cdot \dfrac{5}{12}} = \frac{3600}{1.025} \approx 3512.20 \text{ euros}, \qquad D \approx 87.80 \text{ euros}

Under commercial discount the discount is proportional to the amount at maturity and to the time, $D = M\,d\,t = 3600 \cdot 0.06 \cdot \dfrac{5}{12} = 90$ euros, so the present value is $3600 - 90 = 3510$ euros. The formula can be used because $t = \dfrac{5}{12}$ is much less than $\dfrac{1}{d} \approx 16.7$ years.

Under rational discount the present value is the principal that under simple interest at $6\%$ has accumulated value $3600$ euros after $5$ months, therefore one divides by the accumulation factor $1.025$ . The rational discount is smaller than the commercial discount, because the interest is computed on the present value, which is less than $3600$ euros, and not on the amount at maturity. As a check, $3512.20 \cdot 1.025 \approx 3600$ .

Exercise 3

$2000$ euros are deposited today and another $3000$ euros in $2$ years into an account at $4\%$ per year compound. Compute the balance of the account after $5$ years and the present value of the two deposits, and verify that the two results are consistent.

\begin{aligned} M &= 2000 \cdot 1.04^{5} + 3000 \cdot 1.04^{3} \\[6pt] &\approx 2433.31 + 3374.59 \\[6pt] &\approx 5807.90 \text{ euros} \end{aligned}

V = 2000 + 3000 \cdot 1.04^{-2} \approx 2000 + 2773.67 = 4773.67 \text{ euros}

V \cdot 1.04^{5} \approx 4773.67 \cdot 1.216653 \approx 5807.90 \text{ euros}

Each deposit stays in the account for a different time, so the two accumulated values are computed separately and then added. The first deposit stays in the account for $5$ years and is multiplied by $1.04^{5}$ ; the second is made at time $2$ , stays in the account for $3$ years and is multiplied by $1.04^{3}$ . The balance after $5$ years is therefore about $5807.90$ euros.

The present value of the two deposits is obtained by bringing each back to time $0$ ; the first is already at time $0$ , the second is multiplied by $1.04^{-2}$ . Carrying the present value to time $5$ , that is, multiplying it by $1.04^{5}$ , gives back the balance, because under compound interest moving an amount from $0$ to $2$ and then from $2$ to $5$ is the same as moving it directly from $0$ to $5$ . The two results are therefore consistent.