Tasks 08-01 - Compound Interest & Geometric Sequences

Section 08: Financial Mathematics

A: Skills

Problem 1: Terms of Geometric Sequences (x)

Give amounts in EUR to the cent (two decimal places) and keep all intermediate results unrounded: type the exact rate per period, e.g. \(0.05/12\), into the calculator. Give interest rates in percent with two decimal places. A number of years, months or terms that must reach a target is rounded up to the next whole number, and you check the whole numbers on both sides. An amount that must be invested at least is rounded up to the cent. This rule holds for the whole sheet.

  1. Compute the requested term of each geometric sequence and state whether the sequence grows, decays toward zero or alternates.

(1) \(a_1 = 4\), \(r = 3\): \(a_6\)

(2) \(a_1 = 2048\), \(r = \frac{1}{4}\): \(a_5\)

(3) \(a_1 = 6\), \(r = -2\): \(a_7\)

(4) \(80, 120, 180, \ldots\): the common ratio \(r\) and \(a_6\)

  1. A geometric sequence has \(a_3 = 50\) and \(a_6 = 3.2\). Determine the common ratio \(r\) and the first term \(a_1\).

  2. Determine the first term of the sequence \(a_n = 250 \cdot 1.4^{n-1}\) that is greater than 10,000.

  1. Explicit formula \(a_n = a_1 \cdot r^{n-1}\):

    (1) \(a_6 = 4 \cdot 3^5 = 4 \cdot 243 = 972\). \(r = 3 > 1\), so the sequence grows.

    (2) \(a_5 = 2048 \cdot \left(\frac{1}{4}\right)^4 = \frac{2048}{256} = 8\). \(0 < r < 1\), so the sequence decays toward zero.

    (3) \(a_7 = 6 \cdot (-2)^6 = 6 \cdot 64 = 384\). \(r < 0\), so the signs alternate (\(6, -12, 24, \ldots\)), and the terms grow in size because \(|r| = 2 > 1\).

    (4) \(r = \frac{120}{80} = 1.5\) (check: \(\frac{180}{120} = 1.5\) \(\checkmark\)), so \(a_6 = 80 \cdot 1.5^5 = 80 \cdot 7.59375 = 607.5\). The sequence grows.

  2. From \(a_3\) to \(a_6\) the sequence is multiplied by \(r\) three times:

    \[a_6 = a_3 \cdot r^3 \quad \Rightarrow \quad r^3 = \frac{3.2}{50} = 0.064 \quad \Rightarrow \quad r = \sqrt[3]{0.064} = 0.4\]

    \[a_3 = a_1 \cdot r^2 \quad \Rightarrow \quad a_1 = \frac{50}{0.4^2} = \frac{50}{0.16} = 312.5\]

    Check: \(a_6 = 312.5 \cdot 0.4^5 = 312.5 \cdot 0.01024 = 3.2\) \(\checkmark\)

  3. \[250 \cdot 1.4^{n-1} > 10000 \iff 1.4^{n-1} > 40\]

    Take the logarithm on both sides. \(\ln(1.4) > 0\), so the inequality sign stays:

    \[(n - 1) \cdot \ln(1.4) > \ln(40) \iff n - 1 > \frac{\ln(40)}{\ln(1.4)} \approx 10.96\]

    So \(n - 1 \geq 11\), i.e. \(n = 12\). Check both neighbours: \(a_{11} = 250 \cdot 1.4^{10} \approx 7{,}231.37 < 10{,}000\) and \(a_{12} = 250 \cdot 1.4^{11} \approx 10{,}123.91 > 10{,}000\) \(\checkmark\). The 12th term is the first one above 10,000.

Typical mistake: in c), solving \(1.4^n > 40\) and answering \(n = 11\). The exponent of the \(n\)-th term is \(n - 1\), not \(n\).

Problem 2: Sums of Geometric Sequences (x)

For a geometric sequence with \(r \neq 1\), the sum of the first \(n\) terms is

\[S_n = a_1 \cdot \frac{r^n - 1}{r - 1}\]

  1. Compute the sum of the first 9 terms of \(5 + 10 + 20 + \ldots\)

  2. Compute the sum of the first 6 terms of \(729 + 243 + 81 + \ldots\) and check your result by adding the six terms.

  3. Compute the sum of the first 8 terms of \(3 - 6 + 12 - 24 + \ldots\)

  4. For \(4 + 12 + 36 + \ldots\), determine how many terms you need until the sum first exceeds 50,000.

  1. \(a_1 = 5\), \(r = 2\), \(n = 9\):

    \[S_9 = 5 \cdot \frac{2^9 - 1}{2 - 1} = 5 \cdot 511 = 2555\]

  2. \(a_1 = 729\), \(r = \frac{243}{729} = \frac{1}{3}\), \(n = 6\):

    \[S_6 = 729 \cdot \frac{\left(\frac{1}{3}\right)^6 - 1}{\frac{1}{3} - 1} = 729 \cdot \frac{\frac{1}{729} - 1}{-\frac{2}{3}} = \frac{1 - 729}{-\frac{2}{3}} = 728 \cdot \frac{3}{2} = 1092\]

    Check: \(729 + 243 + 81 + 27 + 9 + 3 = 1092\) \(\checkmark\)

  3. \(a_1 = 3\), \(r = \frac{-6}{3} = -2\), \(n = 8\):

    \[S_8 = 3 \cdot \frac{(-2)^8 - 1}{-2 - 1} = 3 \cdot \frac{256 - 1}{-3} = -255\]

    Check: \(3 - 6 + 12 - 24 + 48 - 96 + 192 - 384 = -255\) \(\checkmark\)

  4. \(a_1 = 4\), \(r = 3\), so \(S_n = 4 \cdot \frac{3^n - 1}{2} = 2 \cdot (3^n - 1)\):

    \[2 \cdot (3^n - 1) > 50000 \iff 3^n > 25001 \iff n > \frac{\ln(25001)}{\ln(3)} \approx 9.22\]

    So \(n = 10\) terms. Check: \(S_9 = 2 \cdot (19683 - 1) = 39{,}364 < 50{,}000\) and \(S_{10} = 2 \cdot (59049 - 1) = 118{,}096 > 50{,}000\) \(\checkmark\)

Typical mistake: in c), using \(r = 2\) because the sizes of the terms double. The signs alternate, so \(r = -2\), and \((-2)^8 = 256\) is positive.

Problem 3: Simple and Compound Interest (x)

  1. A company deposits 3,600 EUR for 9 years at 4.25 % per year. Compute the final amount with simple interest and with compound interest (compounded annually). Explain where the difference comes from.

  2. 7,250 EUR are invested for 7 years at a nominal rate (the annual rate the bank states) of 2.9 % per year. Compute the final amount if the interest is compounded annually, semi-annually, quarterly, monthly and daily. Describe what you observe.

  3. Determine after how many months the monthly compounded deposit from b) first exceeds 9,000 EUR.

  1. Simple interest, \(A = P(1 + rt)\):

    \[A = 3600 \cdot (1 + 0.0425 \cdot 9) = 3600 \cdot 1.3825 = 4{,}977.00 \text{ EUR}\]

    Compound interest, \(A = P(1 + r)^t\):

    \[A = 3600 \cdot 1.0425^9 \approx 5{,}235.85 \text{ EUR}\]

    The difference of 258.85 EUR is interest on interest. With simple interest, only the 3,600 EUR earn interest every year (153 EUR per year). With compound interest, the interest of each year is added to the balance and earns interest itself from the next year on.

  2. \(A = P\left(1 + \frac{r}{n}\right)^{nt}\) with \(P = 7250\), \(r = 0.029\), \(t = 7\):

    Compounding \(n\) Final amount
    annually 1 \(7250 \cdot 1.029^7 \approx 8{,}856.16\) EUR
    semi-annually 2 \(7250 \cdot 1.0145^{14} \approx 8{,}868.84\) EUR
    quarterly 4 \(7250 \cdot 1.00725^{28} \approx 8{,}875.27\) EUR
    monthly 12 \(7250 \cdot \left(1 + \frac{0.029}{12}\right)^{84} \approx 8{,}879.60\) EUR
    daily 365 \(7250 \cdot \left(1 + \frac{0.029}{365}\right)^{2555} \approx 8{,}881.70\) EUR

    The more often the interest is compounded, the larger the final amount, because interest starts to earn interest earlier. The gains get smaller: from monthly to daily, the amount grows by only 2.10 EUR.

  3. Rate per month \(\frac{0.029}{12}\), \(m\) = number of months:

    \[7250 \cdot \left(1 + \frac{0.029}{12}\right)^m > 9000 \iff \left(1 + \frac{0.029}{12}\right)^m > \frac{9000}{7250}\]

    \[m > \frac{\ln\left(\frac{9000}{7250}\right)}{\ln\left(1 + \frac{0.029}{12}\right)} \approx 89.58\]

    So \(m = 90\) months. Check both neighbours: after 89 months the balance is about 8,987.42 EUR \(< 9{,}000\) EUR, after 90 months about 9,009.14 EUR \(> 9{,}000\) EUR \(\checkmark\)

Typical mistake: in b), computing the monthly case as \(7250 \cdot \left(1 + \frac{0.029}{12}\right)^7\). The exponent counts the periods, \(n \cdot t = 12 \cdot 7 = 84\) months, not the years.

Problem 4: Effective Annual Rate and Continuous Compounding (xx)

  1. Compute the effective annual rate (EAR) for each offer:

(1) 5.2 % compounded semi-annually

(2) 4.6 % compounded quarterly

(3) 3.7 % compounded monthly

(4) 4.4 % compounded continuously

  1. 12,800 EUR are invested for 7 years at 4.4 % compounded continuously. Compute the final amount and check it with the effective annual rate from a) (4).

  2. Three banks offer: Bank X 4.75 % compounded annually, Bank Y 4.66 % compounded monthly and Bank Z 4.63 % compounded continuously. Rank the offers by their effective annual rates and explain why the nominal rates alone do not decide.

  1. \(r_{\text{eff}} = \left(1 + \frac{r}{n}\right)^n - 1\), and \(r_{\text{eff}} = e^r - 1\) for continuous compounding:

    (1) \(r_{\text{eff}} = 1.026^2 - 1 = 0.052676 \approx 5.27\,\%\)

    (2) \(r_{\text{eff}} = 1.0115^4 - 1 \approx 0.046800 \approx 4.68\,\%\)

    (3) \(r_{\text{eff}} = \left(1 + \frac{0.037}{12}\right)^{12} - 1 \approx 0.037634 \approx 3.76\,\%\)

    (4) \(r_{\text{eff}} = e^{0.044} - 1 \approx 0.044982 \approx 4.50\,\%\)

  2. \(A = Pe^{rt}\):

    \[A = 12800 \cdot e^{0.044 \cdot 7} = 12800 \cdot e^{0.308} \approx 17{,}416.97 \text{ EUR}\]

    Check with the unrounded EAR: \(12800 \cdot (1 + r_{\text{eff}})^7 = 12800 \cdot \left(e^{0.044}\right)^7 = 12800 \cdot e^{0.308}\) \(\checkmark\). With the rounded 4.50 %, you would get \(12800 \cdot 1.045^7 \approx 17{,}419.03\) EUR, about 2 EUR too much. This is why intermediate results stay unrounded.

  3. Effective annual rates:

    • Bank X: \(r_{\text{eff}} = 4.75\,\%\) (compounded once a year, so nominal = effective)
    • Bank Y: \(r_{\text{eff}} = \left(1 + \frac{0.0466}{12}\right)^{12} - 1 \approx 0.047608 \approx 4.76\,\%\)
    • Bank Z: \(r_{\text{eff}} = e^{0.0463} - 1 \approx 0.047389 \approx 4.74\,\%\)

    Ranking: Y (4.76 %) before X (4.75 %) before Z (4.74 %). Bank Y has a lower nominal rate than Bank X, but it compounds 12 times a year, so its interest earns interest during the year. Only the effective annual rates put all offers on the same basis: what one euro grows to in one year.

Typical mistake: in c), ranking by the nominal rates (X, Y, Z). A nominal rate says nothing about the growth in one year until you know how often the interest is compounded.

Problem 5: Present Value, Rule of 72 and Real Interest Rate (xx)

  1. Compute how much must be invested today to have

(1) 18,000 EUR in 7 years at 3.75 % compounded annually,

(2) 9,400 EUR in 4 years at 2.8 % compounded quarterly,

(3) 26,500 EUR in 12 years at 5.1 % compounded monthly.

  1. For each interest rate, estimate the doubling time with the Rule of 72. Then compute the exact doubling time (compounded annually, two decimal places) and compare: 2.25 %, 5.4 % and 18 % per year.

  2. A savings account pays 3.8 % per year, and the inflation rate is 2.3 % per year. Compute the approximate and the exact real interest rate and explain what the real interest rate tells a saver.

  1. \(PV = FV \cdot (1 + r)^{-t}\), with \(n\) periods per year \(PV = FV \cdot \left(1 + \frac{r}{n}\right)^{-nt}\). The amount needed today is a minimum, so round up to the cent:

    (1) \(PV = 18000 \cdot 1.0375^{-7} \approx 13{,}910.92\) EUR

    (2) \(PV = 9400 \cdot \left(1 + \frac{0.028}{4}\right)^{-16} = 9400 \cdot 1.007^{-16} \approx 8{,}407.30\) EUR

    (3) \(PV = 26500 \cdot \left(1 + \frac{0.051}{12}\right)^{-144} \approx 14{,}388.68\) EUR

    Check for (1): \(13{,}910.92 \cdot 1.0375^7 \approx 18{,}000.00\) EUR \(\checkmark\)

  2. Rule of 72: \(t \approx \frac{72}{r}\) with \(r\) in percent. Exact: \(2 = (1 + r)^t \iff t = \frac{\ln(2)}{\ln(1 + r)}\).

    Rate Rule of 72 Exact Difference
    2.25 % \(\frac{72}{2.25} = 32\) years \(\frac{\ln(2)}{\ln(1.0225)} \approx 31.15\) years 0.85 years
    5.4 % \(\frac{72}{5.4} \approx 13.33\) years \(\frac{\ln(2)}{\ln(1.054)} \approx 13.18\) years 0.15 years
    18 % \(\frac{72}{18} = 4\) years \(\frac{\ln(2)}{\ln(1.18)} \approx 4.19\) years 0.19 years

    The rule is a good estimate for medium rates. At 2.25 % it gives about 10 months too much, and at 18 % it gives too little: the estimate is about 4.5 % too low, the largest relative error of the three.

  3. Approximate: \(r_{\text{real}} \approx 3.8\,\% - 2.3\,\% = 1.5\,\%\)

    Exact:

    \[1 + r_{\text{real}} = \frac{1.038}{1.023} \approx 1.014663 \quad \Rightarrow \quad r_{\text{real}} \approx 1.47\,\%\]

    The real interest rate is the growth of the saver’s purchasing power (what the money can buy): after one year, the money buys about 1.47 % more goods than today, not 3.8 % more, because prices have risen by 2.3 %.

Typical mistake: in a), multiplying instead of dividing, e.g. \(18000 \cdot 1.0375^7\). This is the value of 18,000 EUR in 7 years, not the amount needed today. An amount needed today must be smaller than the target.

B: Set Up & Check

Problem 6: From Words to Formulas (xx)

Translate each statement into a formula and compute the result. Then answer the check question: it tells you whether your translation works.

  1. A start-up puts 4,800 EUR into a savings account for 5 years. The bank pays 2.7 % per year, compounded monthly. Write the rate per period, the number of periods and the formula for the final amount, and compute it.

    Check: A classmate writes \(4800 \cdot 1.027^{60}\). Explain what this formula assumes and why the result is far too large.

  2. A family-run hotel wants to have 22,000 EUR in 5 years to renew its breakfast room. Its bank pays 3.65 % per year, compounded annually. Write the equation for the amount the hotel must invest today and compute it.

    Check: A classmate gets about 26,318.99 EUR. Explain with a quick plausibility check why this cannot be the answer.

  3. An investment fund promises 5.8 % per year, compounded annually. A customer asks after how many full years the invested money has doubled. Write the equation and solve it.

    Check: The customer says: “You cannot answer without knowing how much I invest.” Explain why the amount does not matter, and check your answer with the Rule of 72.

  4. Two banks offer a savings account: Bank P pays 3.95 % compounded annually, Bank Q pays 3.9 % compounded quarterly. Decide which account is better.

    Check: A colleague chooses Bank P “because 3.95 % is more than 3.9 %”. Explain what the colleague has compared.

  5. A cleaning company charges a school 14,500 EUR for the first year of a cleaning contract, and the price grows by 3 % each year. Write the price in year \(n\) as a geometric sequence and compute the price in the 6th year.

    Check: A classmate writes \(14500 \cdot 1.03^6\) for the 6th year. Explain which year this price belongs to.

  1. Rate per period \(\frac{r}{n} = \frac{0.027}{12} = 0.00225\), number of periods \(n \cdot t = 12 \cdot 5 = 60\) months:

    \[A = 4800 \cdot \left(1 + \frac{0.027}{12}\right)^{60} \approx 5{,}492.94 \text{ EUR}\]

    Check: \(1.027^{60}\) adds 2.7 % interest in each of the 60 months, as if the bank paid 2.7 % per month. That is about 37.67 % per year instead of 2.7 %, so the classmate gets about 23,739.30 EUR, almost five times the deposit.

    Typical mistake: using the yearly rate for each month, or keeping the exponent 5 with the monthly rate, \(4800 \cdot 1.00225^5\). Rate and periods must fit together: a monthly rate needs the number of months.

  2. The hotel needs the present value of 22,000 EUR:

    \[PV = 22000 \cdot 1.0365^{-5} \approx 18{,}389.77 \text{ EUR}\]

    (rounded up, because the hotel must invest at least this amount)

    Check: invested money grows, so the amount needed today must be smaller than 22,000 EUR. 26,318.99 EUR is larger: it is \(22000 \cdot 1.0365^5\), the future value of 22,000 EUR in 5 years.

    Typical mistake: multiplying by \(1.0365^5\) instead of dividing. “How much must be invested today” asks for a present value.

  3. \(P\) = invested amount, \(t\) = years:

    \[2P = P \cdot 1.058^t \iff 2 = 1.058^t \iff t = \frac{\ln(2)}{\ln(1.058)} \approx 12.29\]

    So the money has doubled after 13 full years (\(1.058^{12} \approx 1.97 < 2\), \(1.058^{13} \approx 2.08 > 2\) \(\checkmark\)).

    Check: dividing both sides by \(P\) removes the amount, so every amount doubles in the same time. Rule of 72: \(\frac{72}{5.8} \approx 12.41\) years, close to 12.29 years \(\checkmark\)

    Typical mistake: computing \(\frac{1}{0.058} \approx 17.24\) years. This is the doubling time with simple interest, where only the first deposit earns interest.

  4. Compare the effective annual rates:

    \[\text{P: } r_{\text{eff}} = 3.95\,\%, \qquad \text{Q: } r_{\text{eff}} = \left(1 + \frac{0.039}{4}\right)^4 - 1 \approx 0.039574 \approx 3.96\,\%\]

    Bank Q is (slightly) better.

    Check: the colleague has compared the nominal rates. Bank Q adds interest every quarter, and this interest earns interest in the rest of the year. Only the effective annual rates show the growth in one year.

    Typical mistake: choosing by the nominal rate. Different compounding periods make nominal rates incomparable.

  5. Each year’s price is the previous one times \(1 + 0.03 = 1.03\), so \(a_1 = 14500\) and \(r = 1.03\):

    \[a_n = 14500 \cdot 1.03^{n-1}, \qquad a_6 = 14500 \cdot 1.03^5 \approx 16{,}809.47 \text{ EUR}\]

    Check: \(14500 \cdot 1.03^6\) contains six price increases, but the 6th year has only five, so it is the price of the 7th year (\(a_7 \approx 17{,}313.76\) EUR).

    Typical mistake: writing the common ratio as \(r = 0.03\) or \(r = 3\). A growth of 3 % means the factor \(1.03\).

Problem 7: One Paragraph, All Quantities (xx)

A photographer has saved 15,000 EUR and wants to buy new camera equipment that will cost 19,000 EUR in 4 years. The bank pays 3.4 % per year, compounded quarterly. The photographer wants to know whether the savings will have grown to 19,000 EUR after 4 years, after how many quarters the savings reach 19,000 EUR, and how much would have to be deposited today to have exactly 19,000 EUR after 4 years. The photographer also asks for the effective annual rate of the account. In general, prices rise by 2 % per year (inflation). Finally, the photographer wants to know by how much the purchasing power of the savings (the amount of goods they can buy) grows per year.

List every given amount, rate and period with its symbol. Then write every question of the photographer as an equation or a formula with the unknown. Do not compute anything.

Check: An assistant writes the growth of the purchasing power as \(3.4\,\% - 2\,\% = 1.4\,\%\). Explain what is missing and write the exact formula.

Given values:

  1. Savings today: \(P = 15000\) EUR
  2. Target: \(19000\) EUR after 4 years (price of the equipment)
  3. Nominal rate \(r = 0.034\), compounded quarterly: \(n = 4\), rate per quarter \(\frac{0.034}{4}\)
  4. Time \(t = 4\) years, i.e. \(n \cdot t = 16\) quarters
  5. Inflation rate \(r_{\text{infl}} = 0.02\) per year

Questions:

  1. Value after 4 years: \(A = 15000 \cdot \left(1 + \frac{0.034}{4}\right)^{16}\), compared with 19,000 EUR
  2. Number of quarters \(m\): \(15000 \cdot \left(1 + \frac{0.034}{4}\right)^m \geq 19000\), solved with logarithms and rounded up
  3. Deposit needed today: \(PV = 19000 \cdot \left(1 + \frac{0.034}{4}\right)^{-16}\)
  4. Effective annual rate: \(r_{\text{eff}} = \left(1 + \frac{0.034}{4}\right)^4 - 1\)
  5. Growth of the purchasing power (real interest rate): \(1 + r_{\text{real}} = \frac{1 + r_{\text{eff}}}{1 + 0.02}\)

Check: \(3.4\,\% - 2\,\%\) is only the approximation \(r_{\text{real}} \approx r_{\text{nom}} - r_{\text{infl}}\), and it uses the nominal rate. The exact real rate divides the growth factors of one year, \(\frac{1 + r_{\text{eff}}}{1.02} - 1\). Because the interest is compounded quarterly, the growth factor of the account is \(1 + r_{\text{eff}}\), not \(1.034\).

Typical mistake: dividing the rates, \(\frac{0.034}{0.02}\). The exact formula divides the factors \(1 + r\), not the rates.

C: Applied

Problem 8: A Main Sponsor for a Handball Club (xx)

A handball club is looking for a main sponsor, i.e. a company that pays the club for showing its logo on the shirts. A local building materials dealer offers a contract for up to 10 seasons: 24,000 EUR for the first season, and the payment grows by 4.5 % from season to season. The first payment is due at signing, and every further payment one year after the previous one.

  1. Show that the payments form a geometric sequence. Give \(a_1\), the common ratio \(r\) and a formula for the payment \(a_n\) in season \(n\). Compute the payments in seasons 2, 3 and 10.

(for verification: the payment for season 10 is about 35,666.28 EUR)

  1. The club wants to hire a full-time coach, which costs 30,000 EUR per season. Determine the first season in which the sponsor’s payment reaches at least 30,000 EUR.

  2. The sum of the first \(n\) terms of a geometric sequence with \(r \neq 1\) is \(S_n = a_1 \cdot \frac{r^n - 1}{r - 1}\). Compute the total of all payments over the 10 seasons.

(for verification: about 294,917.02 EUR)

  1. Instead of the yearly payments, the dealer also offers a one-off payment of 250,000 EUR at signing for the same 10 seasons. Compare it with the total from c). Determine the smallest number of seasons for which the yearly payments bring more money in total than 250,000 EUR.

  2. The club can invest money at 3.25 % per year, compounded annually. Compute the present value at signing of the payment for season 10 and explain why the comparison of the totals in d) favours the yearly payments.

a) Each payment is the previous one times \(1 + 0.045 = 1.045\), so the ratio of two consecutive payments is always the same:

\[\frac{a_{n+1}}{a_n} = 1.045\]

This is a geometric sequence with \(a_1 = 24000\) and \(r = 1.045\):

\[a_n = 24000 \cdot 1.045^{n-1}\]

\[a_2 = 24000 \cdot 1.045 = 25{,}080.00 \text{ EUR}, \qquad a_3 = 24000 \cdot 1.045^2 = 26{,}208.60 \text{ EUR}, \qquad a_{10} = 24000 \cdot 1.045^9 \approx 35{,}666.28 \text{ EUR}\]

b) \[24000 \cdot 1.045^{n-1} \geq 30000 \iff 1.045^{n-1} \geq 1.25 \iff n - 1 \geq \frac{\ln(1.25)}{\ln(1.045)} \approx 5.07\]

\(\ln(1.045) > 0\), so the inequality sign stays. The smallest whole number is \(n - 1 = 6\), i.e. season 7. Check both neighbours: \(a_6 = 24000 \cdot 1.045^5 \approx 29{,}908.37\) EUR \(< 30{,}000\) EUR, \(a_7 = 24000 \cdot 1.045^6 \approx 31{,}254.24\) EUR \(\geq 30{,}000\) EUR \(\checkmark\)

c) \(a_1 = 24000\), \(r = 1.045\), \(n = 10\):

\[S_{10} = 24000 \cdot \frac{1.045^{10} - 1}{1.045 - 1} = 24000 \cdot \frac{1.045^{10} - 1}{0.045} \approx 294{,}917.02 \text{ EUR}\]

(The exact value is 294,917.0249 EUR. If you type the rounded display value of \(1.045^{10}\), you may get 294,917.03 EUR. Keep the power unrounded.)

d) Over 10 seasons, the yearly payments bring about \(294{,}917.02 - 250{,}000 = 44{,}917.02\) EUR more in total.

Smallest number of seasons with \(S_n > 250000\):

\[24000 \cdot \frac{1.045^n - 1}{0.045} > 250000 \iff 1.045^n - 1 > \frac{250000 \cdot 0.045}{24000} = 0.46875 \iff 1.045^n > 1.46875\]

\[n > \frac{\ln(1.46875)}{\ln(1.045)} \approx 8.73\]

So 9 seasons. Check both neighbours: \(S_8 \approx 225{,}120.33\) EUR \(< 250{,}000\) EUR and \(S_9 \approx 259{,}250.74\) EUR \(> 250{,}000\) EUR \(\checkmark\)

e) The payment for season 10 is due 9 years after signing:

\[PV = 35{,}666.28 \cdot 1.0325^{-9} \approx 26{,}745.29 \text{ EUR}\]

(computed with the unrounded \(a_{10}\)). The total in c) adds euros that arrive at different times as if they had the same value. A euro received at signing can earn interest, so a euro received 9 years later is worth less today: the last payment of 35,666.28 EUR is worth only about 26,745.29 EUR at signing. The later a payment, the more it shrinks. A fair comparison discounts every yearly payment to the signing date and compares the sum of these present values with the 250,000 EUR. The difference of 44,917.02 EUR in d) is therefore too large.

Typical mistake: in b), solving \(1.045^n \geq 1.25\) and answering season 6. The payment of season \(n\) contains only \(n - 1\) increases, and \(a_6 \approx 29{,}908.37\) EUR is still below 30,000 EUR.

Problem 9: Reserve for a New Haul Rope (xx)

A small company runs a cable car up a mountain. The haul rope, the steel rope that pulls the cabins, has to be replaced from time to time, and a new one costs 260,000 EUR. The company keeps a reserve of 185,000 EUR for this and makes no further deposits. Two banks make an offer: a regional bank pays 3.3 % per year, compounded quarterly, and an online bank pays 3.32 % per year, compounded annually.

  1. Compute the effective annual rate of both offers and decide which bank the company should choose.

(for verification: the regional bank, effective annual rate about 3.34 %)

  1. Compute the value of the reserve after 6 years at the bank from a).

  2. Determine after how many quarters the reserve first reaches 260,000 EUR. State the time in years.

(for verification: 42 quarters)

  1. Estimate the doubling time of the reserve with the Rule of 72, using the effective annual rate, and compute the exact doubling time in years (two decimal places). Explain without computing why the time in c) must be shorter than the doubling time.

  2. After an inspection, the authority demands a new haul rope in 8 years. Determine how much money the company must add to the reserve today so that it reaches 260,000 EUR after 8 years.

a) \[\text{Regional bank: } r_{\text{eff}} = \left(1 + \frac{0.033}{4}\right)^4 - 1 = 1.00825^4 - 1 \approx 0.033411 \approx 3.34\,\%\]

\[\text{Online bank: } r_{\text{eff}} = 3.32\,\% \quad \text{(compounded once a year)}\]

The regional bank is better: its nominal rate is lower, but quarterly compounding gives the higher effective annual rate, \(3.34\,\% > 3.32\,\%\).

b) Rate per quarter \(\frac{0.033}{4} = 0.00825\), \(4 \cdot 6 = 24\) quarters:

\[A = 185000 \cdot 1.00825^{24} \approx 225{,}324.94 \text{ EUR}\]

c) \(m\) = number of quarters:

\[185000 \cdot 1.00825^m \geq 260000 \iff 1.00825^m \geq \frac{260000}{185000} \iff m \geq \frac{\ln\left(\frac{260000}{185000}\right)}{\ln(1.00825)} \approx 41.42\]

So 42 quarters, i.e. 10.5 years. Check both neighbours: after 41 quarters the reserve is about 259,101.06 EUR \(< 260{,}000\) EUR, after 42 quarters about 261,238.64 EUR \(\geq 260{,}000\) EUR \(\checkmark\)

d) Rule of 72 with \(r_{\text{eff}} \approx 3.34\,\%\): \(t \approx \frac{72}{3.34} \approx 21.56\) years.

Exact, with the unrounded effective annual rate:

\[2 = (1 + r_{\text{eff}})^t \iff t = \frac{\ln(2)}{\ln(1.00825^4)} = \frac{\ln(2)}{4 \cdot \ln(1.00825)} \approx 21.09 \text{ years}\]

The estimate is about half a year too long. In c), the reserve only has to grow by the factor \(\frac{260000}{185000} \approx 1.41\), which is less than 2. The reserve grows all the time, so it reaches the factor 1.41 before it doubles.

e) The reserve needs the present value of 260,000 EUR after \(4 \cdot 8 = 32\) quarters:

\[PV = 260000 \cdot 1.00825^{-32} \approx 199{,}889.4929 \text{ EUR}\]

The company must add \(PV - 185000 \approx 14{,}889.4929\) EUR, i.e. 14,889.50 EUR, rounded up because the reserve needs at least this amount. Check: \((185000 + 14889.50) \cdot 1.00825^{32} \approx 260{,}000.01\) EUR \(\geq 260{,}000\) EUR \(\checkmark\). With 14,889.49 EUR, the reserve would reach only about 259,999.996 EUR, a fraction of a cent short of 260,000 EUR.

Typical mistake: in c), solving with the yearly rate and \(t\) in years, \(185000 \cdot 1.033^t\). The bank compounds quarterly, so the growth factor per year is \(1.00825^4\), not \(1.033\).

Problem 10: Real Returns of an Orchestra Foundation (xxx)

A charitable foundation (an organisation that invests its capital and spends the returns on a good cause) supports a youth orchestra. It has invested its capital of 360,000 EUR in a fund that earns 4.8 % per year, compounded annually. The foundation expects an inflation rate (the yearly rise of prices) of 2.6 % per year.

  1. Compute the value of the capital after 10 years.

(for verification: about 575,327.76 EUR)

  1. Compute the approximate and the exact real interest rate. Compute the purchasing power of the capital after 10 years in today’s euros in two ways: with the exact real interest rate, and by discounting the value from a) with the inflation rate.

Hint: The real interest rate works for several years like an interest rate in compound interest, so the purchasing power grows by the factor \(1 + r_{\text{real}}\) each year.

(for verification: exact real interest rate about 2.14 %)

  1. The foundation’s statute (its founding rules) demands that the capital grows by at least 2.5 % per year in real terms. Determine the nominal return the fund needs at an inflation rate of 2.6 %, and decide whether the fund meets the rule.

  2. In 6 years, the orchestra needs new timpani and double basses (large orchestra instruments). They cost 45,000 EUR at today’s prices, and their prices rise with the inflation rate. Compute the price in 6 years and the amount the foundation must put into the fund today to pay for the instruments. Check your result with the exact real interest rate. The foundation has 38,000 EUR left from this year’s donations. Decide whether this is enough.

a) \[A = 360000 \cdot 1.048^{10} \approx 575{,}327.76 \text{ EUR}\]

b) Approximate: \(r_{\text{real}} \approx 4.8\,\% - 2.6\,\% = 2.2\,\%\)

Exact:

\[1 + r_{\text{real}} = \frac{1.048}{1.026} \approx 1.021442 \quad \Rightarrow \quad r_{\text{real}} \approx 2.14\,\%\]

Purchasing power with the real interest rate (unrounded):

\[360000 \cdot \left(\frac{1.048}{1.026}\right)^{10} \approx 445{,}083.73 \text{ EUR}\]

Discounting the value from a) with the inflation rate:

\[\frac{575{,}327.7569}{1.026^{10}} \approx 445{,}083.73 \text{ EUR} \;\checkmark\]

Both ways agree, because \(360000 \cdot \frac{1.048^{10}}{1.026^{10}} = 360000 \cdot \left(\frac{1.048}{1.026}\right)^{10}\). After 10 years, the capital buys about as much as 445,083.73 EUR buy today, although the account shows 575,327.76 EUR.

c) The real growth factor must be at least 1.025:

\[\frac{1 + r_{\text{nom}}}{1.026} \geq 1.025 \iff 1 + r_{\text{nom}} \geq 1.025 \cdot 1.026 = 1.05165 \iff r_{\text{nom}} \geq 5.165\,\%\]

The fund needs a nominal return of at least 5.17 % (a minimum, so round up). Check both neighbours: \(\frac{1.0517}{1.026} \approx 1.025049 \geq 1.025\) \(\checkmark\), but \(\frac{1.0516}{1.026} \approx 1.024951 < 1.025\).

The fund earns only 4.8 %, i.e. 2.14 % in real terms, so it does not meet the rule.

d) Price in 6 years:

\[45000 \cdot 1.026^6 \approx 52{,}492.43 \text{ EUR}\]

Amount needed today, the present value at the fund’s rate (rounded up):

\[PV = 52{,}492.4301 \cdot 1.048^{-6} \approx 39{,}621.33 \text{ EUR}\]

Check with the exact real interest rate: \(45000 \cdot \left(\frac{1.048}{1.026}\right)^{-6} \approx 39{,}621.33\) EUR \(\checkmark\). Both ways agree, because \(45000 \cdot 1.026^6 \cdot 1.048^{-6} = 45000 \cdot \left(\frac{1.048}{1.026}\right)^{-6}\).

The 38,000 EUR are not enough: the foundation needs about 1,621.33 EUR more today.

Typical mistake: in c), using the approximation \(2.5\,\% + 2.6\,\% = 5.1\,\%\). With 5.1 %, the exact real growth is only \(\frac{1.051}{1.026} - 1 \approx 2.44\,\% < 2.5\,\%\), so the rule would be missed.

D: FSP Problem

Work on this problem as in the exam: write down every step. The points show what each part is worth.

FSP Problem: Selling a Patent [20 pts. total]

An inventor has developed a quick-release coupling for garden hoses and holds a patent on it, i.e. the exclusive right to produce and sell the invention. A garden-tool manufacturer wants to buy the patent and offers two ways to pay:

  • Offer A: 140,000 EUR at signing.
  • Offer B: 35,000 EUR one year after signing and 117,600 EUR two years after signing.

The inventor will put all the money into a bank account and makes no withdrawals. Round amounts in EUR to the cent and interest rates in percent to two decimal places, and keep intermediate results unrounded.

Part a)

Two banks make an offer: a savings bank pays 3.6 % per year, compounded monthly, and an online bank pays 3.58 % per year, compounded continuously. Compute the effective annual rate of both offers and decide which bank the inventor should choose. [3 pts.]

Savings bank: rate per month \(\frac{0.036}{12} = 0.003\):

\[r_{\text{eff}} = \left(1 + \frac{0.036}{12}\right)^{12} - 1 = 1.003^{12} - 1 \approx 0.036600 \approx 3.66\,\%\]

Online bank: continuous compounding:

\[r_{\text{eff}} = e^{0.0358} - 1 \approx 0.036449 \approx 3.64\,\%\]

Decision: \(3.66\,\% > 3.64\,\%\), so the inventor should choose the savings bank. Continuous compounding is the most frequent compounding, but it cannot make up for the lower nominal rate here.

Grading: 1 Pt. effective annual rate of the savings bank, 1 Pt. effective annual rate of the online bank, 1 Pt. decision by the effective annual rates.

Part b)

From now on, the inventor uses the savings bank from part a). Compute the present value of Offer B at this bank and decide which offer the inventor should accept.

(for verification: the present value of Offer B is about 143,206.46 EUR) [4 pts.]

Each payment is discounted to the signing date with the rate per month \(\frac{0.036}{12} = 0.003\): the first payment over \(12\) months, the second over \(24\) months.

\[PV_1 = 35000 \cdot \left(1 + \frac{0.036}{12}\right)^{-12} = 35000 \cdot 1.003^{-12} \approx 33{,}764.23 \text{ EUR}\]

\[PV_2 = 117600 \cdot 1.003^{-24} \approx 109{,}442.23 \text{ EUR}\]

\[PV_B = PV_1 + PV_2 \approx 143{,}206.46 \text{ EUR}\]

(sum of the unrounded values). Offer A is paid at signing, so its present value is 140,000 EUR.

Decision: \(143{,}206.46 > 140{,}000\), so the inventor should accept Offer B. At signing, it is worth about 3,206.46 EUR more than Offer A.

Grading: 1 Pt. discounting over 12 and 24 months with the monthly rate, 1 Pt. \(PV_1\), 1 Pt. \(PV_2\), 1 Pt. \(PV_B\) and decision.

Part c)

The manufacturer says that Offer B is only the better choice as long as the inventor’s money earns less than a certain interest rate. Determine the effective annual interest rate \(r\) at which both offers have the same present value. Use \(q = 1 + r\). Interpret your result for the inventor’s decision.

Hint: Write the present value of Offer B with \(q\) and multiply the equation by \(q^2\).

(for verification: \(r = 5\,\%\)) [5 pts.]

Equation: with the effective annual rate \(r\), a payment after one year is discounted with \(q^{-1}\), a payment after two years with \(q^{-2}\):

\[140000 = \frac{35000}{q} + \frac{117600}{q^2}\]

Zero form: multiply by \(q^2 > 0\) and bring all terms to one side:

\[140000 q^2 - 35000 q - 117600 = 0 \quad \Big| : 1400\]

\[100 q^2 - 25 q - 84 = 0\]

Quadratic formula:

\[q_{1,2} = \frac{25 \pm \sqrt{(-25)^2 - 4 \cdot 100 \cdot (-84)}}{2 \cdot 100} = \frac{25 \pm \sqrt{625 + 33600}}{200} = \frac{25 \pm \sqrt{34225}}{200} = \frac{25 \pm 185}{200}\]

\[q_1 = \frac{210}{200} = 1.05, \qquad q_2 = \frac{-160}{200} = -0.8\]

A growth factor \(q = 1 + r\) must be positive, so \(q_2 = -0.8\) is rejected. With \(q = 1.05\), both offers have the same present value at \(r = 5\,\%\).

Check: \(\frac{35000}{1.05} + \frac{117600}{1.05^2} \approx 33{,}333.33 + 106{,}666.67 = 140{,}000\) \(\checkmark\)

Interpretation: at the savings bank’s 3.66 % (part b), Offer B is worth more. At a higher rate, the later payments of Offer B are discounted more strongly, e.g. at 6 %: \(\frac{35000}{1.06} + \frac{117600}{1.06^2} \approx 137{,}682.45 < 140{,}000\). So Offer B is better as long as the inventor’s money earns less than 5 % per year. Above 5 %, Offer A is better, because the inventor can invest the 140,000 EUR at signing for a higher return.

Grading: 1 Pt. equation of the present values, 1 Pt. zero form, 1 Pt. quadratic formula with both solutions, 1 Pt. negative solution rejected and \(r = 5\,\%\), 1 Pt. interpretation (below 5 % Offer B, above 5 % Offer A).

Part d)

The inventor accepts Offer B and puts both payments into the account at the savings bank. The inventor wants to buy a small workshop for 165,000 EUR. Compute the balance right after the second payment. Determine in which month after signing the balance first reaches 165,000 EUR.

(for verification: the balance right after the second payment is about 153,881.00 EUR) [5 pts.]

Balance after the second payment: the first payment earns interest for 12 months, the second payment is added at month 24:

\[35000 \cdot 1.003^{12} + 117600 \approx 36{,}281.00 + 117{,}600 = 153{,}881.00 \text{ EUR}\]

Months until 165,000 EUR: \(k\) = number of months after month 24:

\[153{,}881.00 \cdot 1.003^k \geq 165000 \iff 1.003^k \geq \frac{165000}{153{,}881.00}\]

Take the logarithm on both sides. \(\ln(1.003) > 0\), so the inequality sign stays:

\[k \geq \frac{\ln\left(\frac{165000}{153{,}881.00}\right)}{\ln(1.003)} \approx 23.29\]

So \(k = 24\) months after the second payment, i.e. in month \(24 + 24 = 48\) after signing (exactly 4 years).

Check both neighbours: \(153{,}881.00 \cdot 1.003^{23} \approx 164{,}856.65\) EUR \(< 165{,}000\) EUR and \(153{,}881.00 \cdot 1.003^{24} \approx 165{,}351.22\) EUR \(\geq 165{,}000\) EUR \(\checkmark\)

(computed with the unrounded balance)

Grading: 1 Pt. first payment with 12 months of interest, 1 Pt. balance 153,881.00 EUR, 1 Pt. inequality and logarithm, 1 Pt. \(k = 24\) and month 48, 1 Pt. check of both neighbours.

Part e)

A friend of the inventor says: “Offer B pays 152,600 EUR in total, that is 12,600 EUR more than Offer A. So Offer B is worth 9 % more.” Assess the statement with your results from parts b) and c). [3 pts.]

The statement is wrong.

  • The friend adds payments that arrive at different times as if one euro in two years were worth as much as one euro today. Money at signing can earn interest, so later payments must be discounted first.
  • At the savings bank’s rate, Offer B is worth 143,206.46 EUR at signing (part b), only about 3,206.46 EUR or 2.29 % more than Offer A, not 12,600 EUR or 9 %.
  • Whether Offer B is better at all depends on the interest rate (part c): above 5 % per year, Offer A is worth more, although the total of Offer B stays 152,600 EUR.

Grading: 1 Pt. the total ignores the dates of the payments, 1 Pt. true advantage from b), 1 Pt. dependence on the interest rate with the 5 % from c).