Tasks 08-02 - Annuities & Loan Amortization

Section 08: Financial Mathematics

A: Skills

Problem 1: Future and Present Value of Ordinary Annuities (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. \(0.036/12\), into the calculator. A number of payments that must pay off a loan or reach a target is rounded up to the next whole number, and you check the whole numbers on both sides. A payment that is needed to reach a target or to pay off a loan is a minimum, so round it up to the cent. The largest loan or payment that someone can afford is a maximum, so round it down to the cent. This rule holds for the whole sheet.

  1. Compute the future value of each ordinary annuity (payments at the end of each period). For (1), also compute the total of the deposits and the interest earned.

(1) 240 EUR per month for 8 years at 3.6 % per year, compounded monthly

(2) 1,175 EUR per quarter for 10 years at 4.8 % per year, compounded quarterly

(3) 6,400 EUR per year for 15 years at 2.75 % per year, compounded annually

  1. Compute the present value of each ordinary annuity.

(1) 520 EUR per month for 4 years at 6.6 % per year, compounded monthly

(2) 3,200 EUR per quarter for 6 years at 5.2 % per year, compounded quarterly

(3) 17,250 EUR per year for 12 years at 3.5 % per year, compounded annually

  1. Compare the present value from b) (1) with the total of the 48 payments and explain the difference.

  1. \(FV = PMT \cdot \frac{(1 + r)^n - 1}{r}\) with the rate per period \(r\) and the number of payments \(n\):

    (1) \(r = \frac{0.036}{12} = 0.003\), \(n = 12 \cdot 8 = 96\):

    \[FV = 240 \cdot \frac{1.003^{96} - 1}{0.003} \approx 26{,}654.59 \text{ EUR}\]

    Deposits: \(240 \cdot 96 = 23{,}040\) EUR. Interest earned: \(26{,}654.59 - 23{,}040 = 3{,}614.59\) EUR.

    (2) \(r = \frac{0.048}{4} = 0.012\), \(n = 4 \cdot 10 = 40\):

    \[FV = 1175 \cdot \frac{1.012^{40} - 1}{0.012} \approx 59{,}872.48 \text{ EUR}\]

    (3) \(r = 0.0275\), \(n = 15\):

    \[FV = 6400 \cdot \frac{1.0275^{15} - 1}{0.0275} \approx 116{,}875.40 \text{ EUR}\]

  2. \(PV = PMT \cdot \frac{1 - (1 + r)^{-n}}{r}\):

    (1) \(r = \frac{0.066}{12} = 0.0055\), \(n = 48\):

    \[PV = 520 \cdot \frac{1 - 1.0055^{-48}}{0.0055} \approx 21{,}884.51 \text{ EUR}\]

    (2) \(r = \frac{0.052}{4} = 0.013\), \(n = 24\):

    \[PV = 3200 \cdot \frac{1 - 1.013^{-24}}{0.013} \approx 65{,}611.14 \text{ EUR}\]

    (3) \(r = 0.035\), \(n = 12\):

    \[PV = 17250 \cdot \frac{1 - 1.035^{-12}}{0.035} \approx 166{,}692.52 \text{ EUR}\]

  3. The 48 payments add up to \(520 \cdot 48 = 24{,}960\) EUR, but their present value is only 21,884.51 EUR, i.e. 3,075.49 EUR less. Every payment is discounted to today, and the later a payment, the less it is worth today. Put differently: if you invest 21,884.51 EUR today at 6.6 % per year, compounded monthly, the account can pay out 520 EUR at the end of every month for 4 years. The interest earned on the remaining balance covers the difference.

Typical mistake: in a) (2), using the monthly or the yearly values: the payments are quarterly, so the rate per quarter is \(\frac{0.048}{4}\) and \(n\) counts the quarters, \(4 \cdot 10 = 40\).

Problem 2: Annuity Due and Ordinary Annuity (x)

  1. A craft business deposits 425 EUR per month for 5 years into a savings account at 2.4 % per year, compounded monthly. Compute the future value if the deposits are made at the end of each month (ordinary annuity) and if they are made at the beginning of each month (annuity due). Compute the difference.

  2. A shop pays a rent of 2,150 EUR at the beginning of each month. The landlord offers to accept one payment today instead of the rent for the next 3 years. Compute the present value of the rent payments at 4.2 % per year, compounded monthly. Compare it with the present value if the rent were due at the end of each month.

  3. A classmate says: “With an annuity due you make one payment more. That is why its value is larger.” Explain with a timeline of the first three months why this is wrong and where the factor \((1 + r)\) comes from.

  1. \(r = \frac{0.024}{12} = 0.002\), \(n = 60\):

    \[FV = 425 \cdot \frac{1.002^{60} - 1}{0.002} \approx 27{,}064.37 \text{ EUR (ordinary annuity)}\]

    \[FV_{\text{due}} = FV \cdot (1 + r) = 27{,}064.37 \cdot 1.002 \approx 27{,}118.50 \text{ EUR (annuity due)}\]

    (computed with the unrounded \(FV\)). Difference: \(FV \cdot r \approx 54.13\) EUR.

  2. \(r = \frac{0.042}{12} = 0.0035\), \(n = 36\):

    \[PV = 2150 \cdot \frac{1 - 1.0035^{-36}}{0.0035} \approx 72{,}603.16 \text{ EUR (rent at the end of each month)}\]

    \[PV_{\text{due}} = PV \cdot 1.0035 \approx 72{,}857.27 \text{ EUR (rent at the beginning of each month)}\]

    The rent is due at the beginning of each month, so the one-off payment worth the same is 72,857.27 EUR, about 254.11 EUR more than for rent at the end of each month. Both values are less than the total of \(2150 \cdot 36 = 77{,}400\) EUR.

  3. Timeline for the first three months:

    • Ordinary annuity: payments at the end of month 1, 2, 3, i.e. at the times 1, 2, 3.
    • Annuity due: payments at the beginning of month 1, 2, 3, i.e. at the times 0, 1, 2.

    Both annuities have the same number of payments. Each payment of the annuity due is made one period earlier, so it earns interest for one more period (or is discounted for one period less). Every term of the sum is multiplied by \(1 + r\), and so is the whole value: \(FV_{\text{due}} = FV \cdot (1 + r)\) and \(PV_{\text{due}} = PV \cdot (1 + r)\).

Typical mistake: computing an annuity due with \(n + 1\) payments. The number of payments stays the same, only the timing changes.

Problem 3: Required Payments (xx)

  1. Compute the deposit at the end of each period that reaches the target:

(1) 38,500 EUR in 6 years at 3 % per year, compounded monthly, with monthly deposits

(2) 89,000 EUR in 12 years at 4.4 % per year, compounded quarterly, with quarterly deposits

  1. Compute the monthly payment (at the end of each month) for each loan, with interest compounded monthly. For (1), also compute the total of all payments and the total interest.

(1) 14,600 EUR over 4 years at 7.8 % per year

(2) 63,500 EUR over 10 years at 5.1 % per year

(3) 235,000 EUR over 20 years at 3.45 % per year

  1. A start-up can afford a loan payment of at most 1,150 EUR at the end of each month for 6 years. The bank charges 5.9 % per year, compounded monthly. Determine the largest loan the start-up can take.

  1. Payment from a target value, \(PMT = \frac{FV \cdot r}{(1 + r)^n - 1}\), rounded up:

    (1) \(r = \frac{0.03}{12} = 0.0025\), \(n = 72\):

    \[PMT = \frac{38500 \cdot 0.0025}{1.0025^{72} - 1} \approx 488.7065 \quad \Rightarrow \quad 488.71 \text{ EUR}\]

    (2) \(r = \frac{0.044}{4} = 0.011\), \(n = 48\):

    \[PMT = \frac{89000 \cdot 0.011}{1.011^{48} - 1} \approx 1{,}417.4912 \quad \Rightarrow \quad 1{,}417.50 \text{ EUR}\]

    Here rounding up matters: with 1,417.49 EUR, the account would reach only \(1417.49 \cdot \frac{1.011^{48} - 1}{0.011} \approx 88{,}999.93\) EUR.

  2. Loan payment, \(PMT = PV \cdot \frac{r}{1 - (1 + r)^{-n}}\), rounded up:

    (1) \(r = \frac{0.078}{12} = 0.0065\), \(n = 48\):

    \[PMT = 14600 \cdot \frac{0.0065}{1 - 1.0065^{-48}} \approx 355.0596 \quad \Rightarrow \quad 355.06 \text{ EUR}\]

    Total of all payments: \(355.06 \cdot 48 = 17{,}042.88\) EUR. Total interest: \(17{,}042.88 - 14{,}600 = 2{,}442.88\) EUR.

    (2) \(r = \frac{0.051}{12} = 0.00425\), \(n = 120\):

    \[PMT = 63500 \cdot \frac{0.00425}{1 - 1.00425^{-120}} \approx 676.6241 \quad \Rightarrow \quad 676.63 \text{ EUR}\]

    (3) \(r = \frac{0.0345}{12} = 0.002875\), \(n = 240\):

    \[PMT = 235000 \cdot \frac{0.002875}{1 - 1.002875^{-240}} \approx 1{,}356.8752 \quad \Rightarrow \quad 1{,}356.88 \text{ EUR}\]

  3. The largest loan is the present value of the payments the start-up can afford, \(r = \frac{0.059}{12}\), \(n = 72\):

    \[PV = 1150 \cdot \frac{1 - \left(1 + \frac{0.059}{12}\right)^{-72}}{\frac{0.059}{12}} \approx 69{,}588.4780 \text{ EUR}\]

    This is a maximum, so round down: the start-up can borrow at most 69,588.47 EUR.

Typical mistake: in a), using the loan formula with the target, \(38500 \cdot \frac{r}{1 - (1 + r)^{-n}}\). A target that you reach at the end is a future value: use \(PMT = \frac{FV \cdot r}{(1 + r)^n - 1}\).

Problem 4: Number of Payments (xx)

  1. Determine how many monthly payments (at the end of each month) are needed to pay off each loan. Interest is compounded monthly.

(1) 9,600 EUR at 8.4 % per year with 325 EUR per month

(2) 27,400 EUR at 5.4 % per year with 520 EUR per month

  1. A business wants to pay off a loan of 15,000 EUR at 9 % per year, compounded monthly, with 110 EUR at the end of each month. Show that the loan is never paid off and explain why.

  2. A club saves 380 EUR at the end of each month at 3.3 % per year, compounded monthly. Determine after how many months the savings first reach 25,000 EUR.

Hint: Solve the future value formula for \((1 + r)^n\) first, then take the logarithm on both sides.

  1. From the present value formula: \(n = -\frac{\ln\left(1 - \frac{PV \cdot r}{PMT}\right)}{\ln(1 + r)}\), rounded up.

    (1) \(r = \frac{0.084}{12} = 0.007\):

    \[n = -\frac{\ln\left(1 - \frac{9600 \cdot 0.007}{325}\right)}{\ln(1.007)} = -\frac{\ln(0.793231)}{\ln(1.007)} \approx 33.21\]

    So 34 payments. Check both neighbours: 33 payments have a present value of \(325 \cdot \frac{1 - 1.007^{-33}}{0.007} \approx 9{,}546.71\) EUR \(< 9{,}600\) EUR, so after 33 payments a balance is left. 34 payments have a present value of about 9,803.09 EUR \(\geq 9{,}600\) EUR \(\checkmark\). The 34th payment is smaller than 325 EUR: it only pays off the rest.

    (2) \(r = \frac{0.054}{12} = 0.0045\):

    \[n = -\frac{\ln\left(1 - \frac{27400 \cdot 0.0045}{520}\right)}{\ln(1.0045)} \approx 60.28\]

    So 61 payments, although 60.28 is close to 60. Check: 60 payments are worth about 27,289.37 EUR \(< 27{,}400\) EUR, 61 payments about 27,684.79 EUR \(\geq 27{,}400\) EUR \(\checkmark\)

  2. The interest of the first month is \(15000 \cdot \frac{0.09}{12} = 15000 \cdot 0.0075 = 112.50\) EUR. The payment of 110 EUR does not even cover it, so the balance grows every month instead of falling. In the formula, \(1 - \frac{15000 \cdot 0.0075}{110} = 1 - 1.0227 < 0\), and the logarithm of a negative number does not exist, so there is no number of payments \(n\). A payment must be larger than the interest of the first period.

  3. \(FV = PMT \cdot \frac{(1 + r)^n - 1}{r}\) with \(r = \frac{0.033}{12} = 0.00275\):

    \[25000 = 380 \cdot \frac{1.00275^n - 1}{0.00275} \iff 1.00275^n = 1 + \frac{25000 \cdot 0.00275}{380} \approx 1.180921\]

    \[n = \frac{\ln(1.180921)}{\ln(1.00275)} \approx 60.55\]

    So after 61 months. Check: after 60 months the savings are about 24,751.97 EUR \(< 25{,}000\) EUR, after 61 months about 25,200.03 EUR \(\geq 25{,}000\) EUR \(\checkmark\)

Typical mistake: rounding 60.28 in a) (2) down to 60 payments, because it is “close to 60”. Then a balance is left after the 60th payment, and a 61st payment is needed.

Problem 5: Amortization Schedule, Outstanding Balance and Total Interest (xx)

A carpentry workshop takes out a loan of 16,800 EUR at 6.9 % per year, compounded monthly. The loan is repaid in 36 equal monthly payments at the end of each month.

  1. Compute the monthly payment.

  2. Complete the first four rows of the amortization schedule. Round each interest portion to the cent, as the bank does.

Month Payment Interest Principal Balance
0 16,800.00
1
2
3
4

  1. Compute the outstanding balance after one year (12 payments) with the formula \(B_k = PMT \cdot \frac{1 - (1 + r)^{-(n-k)}}{r}\).

  2. Compute the total interest over the life of the loan.

  3. Describe how the interest portion and the principal portion change from month to month and explain why.

  1. \(r = \frac{0.069}{12} = 0.00575\), \(n = 36\):

    \[PMT = 16800 \cdot \frac{0.00575}{1 - 1.00575^{-36}} \approx 517.9675 \quad \Rightarrow \quad 517.97 \text{ EUR}\]

  2. Interest = balance \(\cdot\, r\), principal = payment \(-\) interest, new balance = old balance \(-\) principal:

    Month Payment Interest Principal Balance
    0 16,800.00
    1 517.97 96.60 421.37 16,378.63
    2 517.97 94.18 423.79 15,954.84
    3 517.97 91.74 426.23 15,528.61
    4 517.97 89.29 428.68 15,099.93

    For example, month 2: \(16{,}378.63 \cdot 0.00575 \approx 94.18\), \(517.97 - 94.18 = 423.79\), \(16{,}378.63 - 423.79 = 15{,}954.84\).

  3. After \(k = 12\) payments, \(n - k = 24\) payments are left:

    \[B_{12} = 517.97 \cdot \frac{1 - 1.00575^{-24}}{0.00575} \approx 11{,}580.63 \text{ EUR}\]

    Note: The formula \(B_k = PV \cdot \frac{(1 + r)^n - (1 + r)^k}{(1 + r)^n - 1}\) gives 11,580.58 EUR. It works with the unrounded payment, so the results differ by a few cents.

  4. Total of all payments: \(517.97 \cdot 36 = 18{,}646.92\) EUR. Total interest: \(18{,}646.92 - 16{,}800 = 1{,}846.92\) EUR.

  5. The interest portion falls and the principal portion grows from month to month, while their sum stays 517.97 EUR. The interest is always computed on the outstanding balance. Every payment lowers the balance, so the next interest portion is smaller, and a larger part of the same payment is left for the principal. Early payments are mostly interest, late payments mostly principal.

Typical mistake: in b), computing the interest of every month on the original 16,800 EUR. The interest is charged only on what is still owed.

B: Set Up & Check

Problem 6: From Words to Formulas (xx)

Translate each statement into the right formula with \(PMT\), \(r\), \(n\) and \(PV\) or \(FV\), and compute the result. Then answer the check question: it tells you whether your translation works.

  1. A graphic designer deposits 650 EUR at the beginning of each month for 4 years into a savings account that pays 2.2 % per year, compounded monthly. Compute the balance after 4 years.

    Check: A classmate gets about 32,582.78 EUR. Explain which formula the classmate used and why the correct balance must be larger.

  2. A software company sells a licence for 1,800 EUR at the end of each quarter for 5 years. A customer wants to know what these payments are worth today. The company calculates with 5.6 % per year, compounded quarterly. Decide whether you need the present value or the future value formula and compute the value.

    Check: A classmate gets about 41,215.23 EUR. Explain with the total of the payments why this cannot be the answer.

  3. A company borrows 32,000 EUR at 7.5 % per year, compounded monthly, and repays the loan with equal monthly payments for 6 years. Write the rate per period and the number of payments, and compute the monthly payment and the total interest.

    Check: A colleague estimates the interest as \(32000 \cdot 0.075 \cdot 6 = 14{,}400\) EUR. Explain why the true total interest is much smaller.

  4. A loan of 7,800 EUR at 9.6 % per year, compounded monthly, is repaid with 260 EUR at the end of each month. Determine the number of payments.

    Check: A classmate gets \(n \approx 34.44\) and answers “34 payments”. Explain what happens after the 34th payment and how large the last payment is compared with the others.

  5. A company takes out a 5-year loan of 23,600 EUR at 4.7 % per year, compounded monthly. It pays 442.13 EUR at the end of each month. Write the formula for the outstanding balance after 2 years and compute it.

    Check: A colleague computes \(23600 - 24 \cdot 442.13 = 12{,}988.88\) EUR. Explain why the true balance is larger.

  1. “At the beginning of each month” means an annuity due: \(r = \frac{0.022}{12}\), \(n = 48\):

    \[FV_{\text{due}} = 650 \cdot \frac{\left(1 + \frac{0.022}{12}\right)^{48} - 1}{\frac{0.022}{12}} \cdot \left(1 + \frac{0.022}{12}\right) \approx 32{,}642.51 \text{ EUR}\]

    Check: 32,582.78 EUR is the ordinary annuity, as if every deposit were made at the end of the month. The designer deposits at the beginning, so every deposit earns interest for one month more, and the balance is larger by the factor \(1 + r\).

    Typical mistake: taking \(n = 49\) for an annuity due. The number of deposits stays 48, only their timing changes.

  2. “Worth today” asks for a present value: \(r = \frac{0.056}{4} = 0.014\), \(n = 20\) quarters:

    \[PV = 1800 \cdot \frac{1 - 1.014^{-20}}{0.014} \approx 31{,}210.35 \text{ EUR}\]

    Check: the 20 payments add up to \(1800 \cdot 20 = 36{,}000\) EUR. Their value today must be less than that, because later payments are worth less today. 41,215.23 EUR is more: it is the future value, i.e. the value of the payments at the end of the 5 years.

    Typical mistake: choosing the formula by the word “value”. “Save … to have” asks for a future value, “borrow” or “worth today” for a present value.

  3. Rate per month \(r = \frac{0.075}{12} = 0.00625\), number of payments \(n = 12 \cdot 6 = 72\):

    \[PMT = 32000 \cdot \frac{0.00625}{1 - 1.00625^{-72}} \approx 553.2836 \quad \Rightarrow \quad 553.29 \text{ EUR}\]

    Total of the payments: \(553.29 \cdot 72 = 39{,}836.88\) EUR. Total interest: \(39{,}836.88 - 32{,}000 = 7{,}836.88\) EUR.

    Check: 14,400 EUR would be simple interest on the full 32,000 EUR for all 6 years. But every payment lowers the balance, and the interest is charged only on what is still owed. On average, the company owes only about half the loan, so the interest is only about half as large.

    Typical mistake: keeping the yearly values, \(r = 0.075\) and \(n = 6\). This gives the payment of a loan with one payment per year, about 6,817.44 EUR.

  4. \(r = \frac{0.096}{12} = 0.008\):

    \[n = -\frac{\ln\left(1 - \frac{7800 \cdot 0.008}{260}\right)}{\ln(1.008)} = -\frac{\ln(0.76)}{\ln(1.008)} \approx 34.44\]

    So 35 payments.

    Check: 34 payments of 260 EUR have a present value of only \(260 \cdot \frac{1 - 1.008^{-34}}{0.008} \approx 7{,}712.93\) EUR \(< 7{,}800\) EUR. After the 34th payment, a balance is left, so a 35th payment is needed. It only pays off this rest, so it is smaller than 260 EUR.

    Typical mistake: rounding the number of payments to the nearest whole number. A number of payments that must pay off a loan is always rounded up.

  5. After \(k = 24\) payments, \(n - k = 60 - 24 = 36\) payments are left. The balance is their present value, with \(r = \frac{0.047}{12}\):

    \[B_{24} = 442.13 \cdot \frac{1 - \left(1 + \frac{0.047}{12}\right)^{-36}}{\frac{0.047}{12}} \approx 14{,}818.48 \text{ EUR}\]

    Check: the colleague counts every payment as repayment of the loan. But part of each payment is interest, and only the principal portion lowers the balance. So the true balance is larger than 12,988.88 EUR.

    Typical mistake: using the exponent \(-k = -24\) instead of \(-(n - k) = -36\). The formula needs the number of payments that are left.

Problem 7: One Paragraph, All Quantities (xx)

A goldsmith (a craftsman who makes and repairs jewellery) wants to buy a laser welding machine in 3 years. It will cost 46,000 EUR. From now on, the goldsmith deposits 850 EUR at the beginning of each month into a savings account that pays 2.6 % per year, compounded monthly. The rest of the price will be paid with a bank loan at 6.3 % per year, compounded monthly, repaid in equal payments at the end of each month over 4 years. The goldsmith wants to know how much will be saved after 3 years, how much must be borrowed, how large the monthly loan payment is, how much is still owed after the first year of the loan, and how much interest the loan costs in total. Finally, the goldsmith asks how many monthly payments would be needed with a payment of 400 EUR instead.

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

Check: An assistant writes the savings after 3 years as \(850 \cdot \frac{(1 + r)^{36} - 1}{r}\). Explain what is missing.

Given values:

  1. Deposit \(PMT_1 = 850\) EUR at the beginning of each month (annuity due)
  2. Savings rate \(r_1 = \frac{0.026}{12}\) per month, \(n_1 = 36\) deposits
  3. Price of the machine: 46,000 EUR in 3 years
  4. Loan rate \(r_2 = \frac{0.063}{12}\) per month, \(n_2 = 48\) payments at the end of each month (ordinary annuity)
  5. Alternative payment: 400 EUR per month

Questions:

  1. Savings after 3 years: \(FV_{\text{due}} = 850 \cdot \frac{(1 + r_1)^{36} - 1}{r_1} \cdot (1 + r_1)\)
  2. Loan: \(PV = 46000 - FV_{\text{due}}\)
  3. Monthly payment: \(PMT_2 = PV \cdot \frac{r_2}{1 - (1 + r_2)^{-48}}\), rounded up to the cent
  4. Balance after one year: \(B_{12} = PMT_2 \cdot \frac{1 - (1 + r_2)^{-36}}{r_2}\)
  5. Total interest: \(48 \cdot PMT_2 - PV\)
  6. Number of payments with 400 EUR: \(n = -\frac{\ln\left(1 - \frac{PV \cdot r_2}{400}\right)}{\ln(1 + r_2)}\), rounded up

Check: the goldsmith deposits at the beginning of each month, so the savings are an annuity due. The factor \((1 + r_1)\) is missing: every deposit earns interest for one month more.

Typical mistake: writing one rate \(r\) for both accounts. The savings account and the loan have different rates, so they need different symbols.

C: Applied

Problem 8: New Pinsetters for a Bowling Centre (xx)

A bowling centre with 12 lanes has to replace its pinsetters in 7 years. These are the machines behind the lanes that set up the pins again after each throw. The new machines will cost 168,000 EUR. The owner opens a sinking fund, i.e. a savings account into which the business pays regular amounts to replace equipment later. The bank pays 3.9 % per year, compounded monthly.

  1. Compute the deposit at the end of each month that reaches 168,000 EUR after 7 years. Compute the total of the deposits and the interest earned.

(for verification: the deposit is 1,742.64 EUR)

  1. The bank also allows deposits at the beginning of each month. Compute the monthly deposit for this case and the saving over the 7 years. Explain why the saving is so small.

  2. In the first months, the owner can only spare 1,500 EUR at the end of each month. Determine after how many months the fund would reach 168,000 EUR with this deposit, and decide whether this is enough.

Hint: Solve the future value formula for \((1 + r)^n\) first, then take the logarithm on both sides.

  1. The owner deposits the amount from a) at the end of each month. After 3 years (36 deposits), the manufacturer announces that the pinsetters will cost 181,000 EUR. Compute the balance of the fund after 3 years and the new monthly deposit for the remaining 4 years that reaches 181,000 EUR.

Hint: The balance after 3 years keeps earning interest for 4 more years, and the new deposits form an ordinary annuity over 48 months.

a) Payment from a target value with \(r = \frac{0.039}{12} = 0.00325\) and \(n = 12 \cdot 7 = 84\):

\[PMT = \frac{168000 \cdot 0.00325}{1.00325^{84} - 1} \approx 1{,}742.6343 \text{ EUR}\]

The fund must reach the target, so round up: 1,742.64 EUR per month.

Deposits: \(1742.64 \cdot 84 = 146{,}381.76\) EUR. With the rounded deposit, the fund reaches \(1742.64 \cdot \frac{1.00325^{84} - 1}{0.00325} \approx 168{,}000.55\) EUR, so the interest earned is \(168{,}000.55 - 146{,}381.76 = 21{,}618.79\) EUR. About 13 % of the price is paid by interest.

b) For an annuity due, \(FV_{\text{due}} = PMT \cdot \frac{(1 + r)^n - 1}{r} \cdot (1 + r)\). Solve for \(PMT\):

\[PMT = \frac{168000 \cdot 0.00325}{\left(1.00325^{84} - 1\right) \cdot 1.00325} \approx 1{,}736.9891 \quad \Rightarrow \quad 1{,}736.99 \text{ EUR}\]

The owner saves \(1742.64 - 1736.99 = 5.65\) EUR per month, i.e. \(5.65 \cdot 84 = 474.60\) EUR over the 7 years. Each deposit is made only one month earlier and earns one extra month of interest at 0.325 %, so the required deposit falls only by the factor \(1.00325\).

c) \(r = 0.00325\), \(PMT = 1500\):

\[168000 = 1500 \cdot \frac{1.00325^n - 1}{0.00325} \iff 1.00325^n = 1 + \frac{168000 \cdot 0.00325}{1500} = 1.364\]

\[n = \frac{\ln(1.364)}{\ln(1.00325)} \approx 95.67\]

So 96 months, i.e. 8 years. Check both neighbours: after 95 months the fund holds about 166,634.00 EUR \(< 168{,}000\) EUR, after 96 months about 168,675.57 EUR \(\geq 168{,}000\) EUR \(\checkmark\)

This is not enough: the pinsetters must be replaced after 84 months, a year earlier.

d) Balance after 36 deposits (future value of the first 36 deposits):

\[B = 1742.64 \cdot \frac{1.00325^{36} - 1}{0.00325} \approx 66{,}438.12 \text{ EUR}\]

In the remaining 48 months, this balance grows to \(66{,}438.12 \cdot 1.00325^{48} \approx 77{,}634.97\) EUR. The new deposits \(PMT_{\text{new}}\) must make up the rest:

\[77{,}634.97 + PMT_{\text{new}} \cdot \frac{1.00325^{48} - 1}{0.00325} = 181000\]

\[PMT_{\text{new}} = \frac{(181000 - 77{,}634.97) \cdot 0.00325}{1.00325^{48} - 1} \approx 1{,}993.3257 \quad \Rightarrow \quad 1{,}993.33 \text{ EUR}\]

(computed with unrounded intermediate results, rounded up because the fund must reach the price). The owner must raise the deposit by 250.69 EUR per month.

Typical mistake: in d), computing a new deposit for 181,000 EUR over 48 months as if the fund were empty, or subtracting the balance of 66,438.12 EUR from 181,000 EUR without its interest. The balance keeps earning interest for 4 more years.

Problem 9: A New Fit-Out for a Barber Shop (xx)

A barber (a hairdresser for men) gives the shop a new fit-out: new chairs, mirrors, wash basins and lighting. The bank lends 62,400 EUR at 7.2 % per year, compounded monthly. The loan is repaid in equal payments at the end of each month, and the barber can choose a term of 6 or 8 years.

  1. Compute the monthly payment for both terms and the total interest the barber pays over each term.

(for verification: 1,069.87 EUR for 6 years, 856.97 EUR for 8 years)

  1. The barber can spend at most 1,000 EUR per month on the loan. Decide which term the barber must choose and state how much the lower payment costs in extra interest.

  2. The barber takes the 8-year loan. After 5 years (60 payments), another bank offers refinancing, i.e. a new loan that pays off the old one. Compute the outstanding balance after 5 years.

(for verification: about 27,672.20 EUR)

  1. The new bank lends the outstanding balance plus a processing fee of 950 EUR at 4.8 % per year, compounded monthly, with monthly payments over the remaining 3 years. Compute the new monthly payment and decide whether the barber should accept. Explain why comparing the monthly payments is enough here.

a) \(r = \frac{0.072}{12} = 0.006\). For 6 years, \(n = 72\), for 8 years, \(n = 96\):

\[PMT_6 = 62400 \cdot \frac{0.006}{1 - 1.006^{-72}} \approx 1{,}069.8608 \quad \Rightarrow \quad 1{,}069.87 \text{ EUR}\]

\[PMT_8 = 62400 \cdot \frac{0.006}{1 - 1.006^{-96}} \approx 856.9678 \quad \Rightarrow \quad 856.97 \text{ EUR}\]

Total interest = total of the payments minus the loan:

  • 6 years: \(72 \cdot 1069.87 - 62400 = 77{,}030.64 - 62{,}400 = 14{,}630.64\) EUR
  • 8 years: \(96 \cdot 856.97 - 62400 = 82{,}269.12 - 62{,}400 = 19{,}869.12\) EUR

b) The 6-year payment of 1,069.87 EUR is above the limit of 1,000 EUR, so the barber must choose 8 years. The lower payment costs \(19{,}869.12 - 14{,}630.64 = 5{,}238.48\) EUR more interest, because the loan stays open for two more years.

c) After \(k = 60\) payments, \(96 - 60 = 36\) payments are left:

\[B_{60} = PMT \cdot \frac{1 - (1 + r)^{-(n - k)}}{r} = 856.97 \cdot \frac{1 - 1.006^{-36}}{0.006} \approx 27{,}672.20 \text{ EUR}\]

After 5 of 8 years, the barber still owes almost 28,000 EUR, more than 44 % of the loan, because the early payments were mostly interest.

d) New loan: \(27{,}672.20 + 950 = 28{,}622.20\) EUR at \(r = \frac{0.048}{12} = 0.004\) with \(n = 36\):

\[PMT_{\text{new}} = 28{,}622.20 \cdot \frac{0.004}{1 - 1.004^{-36}} \approx 855.2653 \quad \Rightarrow \quad 855.27 \text{ EUR}\]

(computed with the unrounded balance). The barber pays \(856.97 - 855.27 = 1.70\) EUR less per month, i.e. \(1.70 \cdot 36 = 61.20\) EUR less over the 3 years. The lower rate saves money, but the fee eats almost all of it. The barber can accept, but the advantage is very small.

Comparing the monthly payments is enough because both loans end at the same time: with or without refinancing, the barber pays once a month on the same 36 dates. The new payment is lower at every one of these dates, so no present values are needed.

Typical mistake: in c), putting the 60 payments already made into the exponent, \(856.97 \cdot \frac{1 - 1.006^{-60}}{0.006} \approx 43{,}073.14\) EUR. That is the balance after 3 years. The formula needs the payments that are left, \(n - k = 36\).

Problem 10: Entry Fee for a Bubble-Tea Franchise (xxx)

A young entrepreneur wants to open a bubble-tea shop (bubble tea is a sweet tea drink with small balls of tapioca) as a franchise, i.e. under the brand name and with the products of a large chain. The chain charges an entry fee and offers two ways to pay it:

  • Plan A: 43,500 EUR at signing.
  • Plan B: 36 monthly payments of 1,285 EUR at the beginning of each month, the first one at signing.

The entrepreneur has the money in a savings account that pays 3 % per year, compounded monthly. Money that is not paid to the chain stays in this account.

  1. Compute the present value of Plan B and decide which plan the entrepreneur should choose.

(for verification: about 44,297.07 EUR)

  1. The chain also offers Plan C: a down payment (an amount paid at signing) plus 36 monthly payments of 1,010 EUR at the end of each month. Determine the down payment at which Plan C has the same present value as Plan A. State for which down payments Plan C is the better choice.

  2. The chain says that Plan B “contains an interest rate”. The monthly rate \(x\) at which Plan B is worth exactly 43,500 EUR solves

\[43500 = 1285 \cdot \frac{1 - (1 + x)^{-36}}{x} \cdot (1 + x)\]

Solve the equation with SOLVE (SHIFT + CALC) and the start value \(0.003\). Give the corresponding nominal rate per year (the annual rate a bank states, like the 3 % per year of the savings account) and explain what it means for the decision in a).

Hint: As in Session 02-03, enter the difference of both sides with ALPHA \(x\). SOLVE finds the \(x\) that makes it zero.

a) Plan B is an annuity due with \(r = \frac{0.03}{12} = 0.0025\) and \(n = 36\):

\[PV_{\text{due}} = 1285 \cdot \frac{1 - 1.0025^{-36}}{0.0025} \cdot 1.0025 \approx 44{,}297.07 \text{ EUR}\]

Plan A costs 43,500 EUR at signing, which is already its present value. Plan B is worth about 797.07 EUR more today, so it is more expensive. The entrepreneur should choose Plan A and pay at signing.

b) Let \(D\) be the down payment. The 36 payments of Plan C form an ordinary annuity:

\[PV_C = D + 1010 \cdot \frac{1 - 1.0025^{-36}}{0.0025} \approx D + 34{,}730.33\]

Same present value as Plan A:

\[D + 34{,}730.33 = 43500 \iff D \approx 8{,}769.67 \text{ EUR}\]

(computed with the unrounded present value). Plan C is the better choice if its present value is smaller than 43,500 EUR:

\[D + 34{,}730.33 < 43500 \iff D < 8{,}769.67\]

So Plan C is better for every down payment below 8,769.67 EUR, and Plan A for every down payment above.

c) As in Session 02-03, enter the difference of both sides, \(1285 \cdot \frac{1 - (1 + x)^{-36}}{x} \cdot (1 + x) - 43500\), with ALPHA \(x\). SOLVE finds the \(x\) that makes it zero. Press SHIFT + CALC (SOLVE), type the start value \(0.003\) and press =:

\[x \approx 0.0035605\]

Check: \(1285 \cdot \frac{1 - 1.0035605^{-36}}{0.0035605} \cdot 1.0035605 \approx 43{,}500\) \(\checkmark\)

Nominal rate per year: \(12 \cdot 0.0035605 \approx 0.0427 = 4.27\,\%\), compounded monthly.

Meaning: paying in instalments with Plan B costs the entrepreneur as much as a loan at 4.27 % per year, compounded monthly. The savings account pays only 3 %, less than 4.27 %, so it is cheaper to take the money out of the account and pay at once. This agrees with a). If the account paid more than 4.27 %, Plan B would be the better choice.

Typical mistake: in a), comparing the totals, \(36 \cdot 1285 = 46{,}260\) EUR with 43,500 EUR. The total ignores when the payments are made. Here the decision is the same, but the true difference today is only 797.07 EUR, not 2,760 EUR.

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: A Second Workshop for a Canoe Builder [22 pts. total]

A small boatyard builds wooden canoes by hand and sells them to rental stations and private customers. It wants to convert a neighbouring barn into a second workshop and needs a bank loan for this. At an output of \(x\) canoes per month, its monthly profit is

\[P(x) = -0.002x^3 + 0.03x^2 + 1.2x - 14, \qquad 0 \leq x \leq 30,\]

with \(P\) in thousand EUR per month. The bank charges 5.4 % per year, compounded monthly, and the loan is repaid in equal payments at the end of each month over 8 years. Round amounts in EUR to the cent and keep intermediate results unrounded. A payment is rounded up to the cent, a largest loan down, and a number of payments up.

Part a)

Compute the maximum profit and prove it really is a maximum. Give the output at which it is reached.

(for verification: the maximum profit is 6,000 EUR per month) [5 pts.]

Necessary condition: \(P'(x) = 0\) with

\[P'(x) = -0.006x^2 + 0.06x + 1.2\]

\[-0.006x^2 + 0.06x + 1.2 = 0 \quad \Big| : (-0.006) \quad \iff \quad x^2 - 10x - 200 = 0\]

Quadratic formula:

\[x_{1,2} = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot (-200)}}{2} = \frac{10 \pm \sqrt{900}}{2} = \frac{10 \pm 30}{2}\]

\[x_1 = 20, \qquad x_2 = -10 \notin [0; 30]\]

Sufficient condition: \(P''(x) = -0.012x + 0.06\), so

\[P''(20) = -0.24 + 0.06 = -0.18 < 0\]

\(P\) is concave at \(x = 20\), so it has a (local) maximum there.

Maximum profit:

\[P(20) = -0.002 \cdot 8000 + 0.03 \cdot 400 + 1.2 \cdot 20 - 14 = -16 + 12 + 24 - 14 = 6\]

The maximum profit is 6,000 EUR per month at an output of 20 canoes per month.

Ends of the domain: \(P(0) = -14\) and \(P(30) = -54 + 27 + 36 - 14 = -5\). Both are below 6, so no other output in \([0; 30]\) gives a higher profit.

Grading: 1 Pt. \(P'(x)\), 1 Pt. zero form and quadratic formula with \(x = 20\) (\(x = -10\) rejected), 1 Pt. \(P''(20) < 0\) as proof, 1 Pt. \(P(20) = 6\), i.e. 6,000 EUR at 20 canoes, 1 Pt. comparison with the ends of the domain.

Part b)

The bank grants the loan only if the monthly payment is at most 40 % of the maximum monthly profit from part a). Determine the largest loan the boatyard can get. [3 pts.]

Largest payment: \(0.4 \cdot 6000 = 2{,}400\) EUR per month.

Largest loan: the present value of 96 payments of 2,400 EUR with \(r = \frac{0.054}{12} = 0.0045\) and \(n = 12 \cdot 8 = 96\):

\[PV = 2400 \cdot \frac{1 - 1.0045^{-96}}{0.0045} \approx 186{,}752.6259 \text{ EUR}\]

This is a maximum, so round down: the boatyard can borrow at most 186,752.62 EUR.

Grading: 1 Pt. largest payment 2,400 EUR, 1 Pt. present value formula with \(r = 0.0045\) and \(n = 96\), 1 Pt. 186,752.62 EUR rounded down.

Part c)

The conversion of the barn costs 170,000 EUR, and the boatyard borrows exactly this amount. Compute the monthly payment and complete the first three rows of the amortization schedule. Round each interest portion to the cent.

Month Payment Interest Principal Balance
0 170,000.00
1
2
3

(for verification: the monthly payment is 2,184.71 EUR)

[5 pts.]

Payment: \(r = 0.0045\), \(n = 96\):

\[PMT = 170000 \cdot \frac{0.0045}{1 - 1.0045^{-96}} \approx 2{,}184.7082 \quad \Rightarrow \quad 2{,}184.71 \text{ EUR}\]

\(2{,}184.71 \leq 2{,}400\), so the bank’s condition from part b) holds.

Schedule: interest = balance \(\cdot\, 0.0045\), principal = \(2{,}184.71 -\) interest, new balance = old balance \(-\) principal:

Month Payment Interest Principal Balance
0 170,000.00
1 2,184.71 765.00 1,419.71 168,580.29
2 2,184.71 758.61 1,426.10 167,154.19
3 2,184.71 752.19 1,432.52 165,721.67

Grading: 1 Pt. \(r\) and \(n\), 1 Pt. payment 2,184.71 EUR, 1 Pt. per row of the schedule.

Part d)

Compute the outstanding balance after 3 years and the interest the boatyard has paid in these 3 years.

Hint: The principal repaid so far is the loan minus the outstanding balance.

(for verification: the outstanding balance is about 114,652.61 EUR) [4 pts.]

Outstanding balance: after \(k = 36\) payments, \(96 - 36 = 60\) payments are left:

\[B_{36} = 2184.71 \cdot \frac{1 - 1.0045^{-60}}{0.0045} \approx 114{,}652.61 \text{ EUR}\]

Interest paid: in 3 years, the boatyard has paid \(36 \cdot 2184.71 = 78{,}649.56\) EUR. The principal repaid is the loan minus the balance, \(170{,}000 - 114{,}652.61 = 55{,}347.39\) EUR. The rest is interest:

\[78{,}649.56 - 55{,}347.39 = 23{,}302.17 \text{ EUR}\]

Grading: 1 Pt. \(n - k = 60\) remaining payments, 1 Pt. \(B_{36} \approx 114{,}652.61\) EUR, 1 Pt. principal repaid, 1 Pt. interest 23,302.17 EUR.

Part e)

After 3 years, the boatyard has 30,000 EUR left from a large order and uses it for an extra repayment, i.e. a one-off payment on top of the monthly payments that lowers the outstanding balance. The bank offers two options for the rest of the loan:

(1) The monthly payment from part c) stays the same, and the loan is paid off earlier.

(2) The loan still ends after 8 years in total, and the monthly payment is lowered.

Determine the number of remaining payments for option (1) and the new monthly payment for option (2). The owner wants to pay as little interest as possible. Decide which option the owner should choose.

Note: In option (1), the last payment is smaller than the others.

[5 pts.]

New balance: \(114{,}652.61 - 30{,}000 = 84{,}652.61\) EUR.

Option (1): number of payments from the present value formula with \(PMT = 2{,}184.71\) and \(r = 0.0045\):

\[n = -\frac{\ln\left(1 - \frac{84{,}652.61 \cdot 0.0045}{2184.71}\right)}{\ln(1.0045)} = -\frac{\ln(0.825635)}{\ln(1.0045)} \approx 42.67\]

So 43 more payments instead of 60: the loan is paid off 17 months earlier. Check both neighbours: 42 payments have a present value of about 83,437.69 EUR \(< 84{,}652.61\) EUR, 43 payments about 85,238.83 EUR \(\geq 84{,}652.61\) EUR \(\checkmark\)

Option (2): the remaining \(n = 60\) payments stay:

\[PMT = 84{,}652.61 \cdot \frac{0.0045}{1 - 1.0045^{-60}} \approx 1{,}613.0588 \quad \Rightarrow \quad 1{,}613.06 \text{ EUR}\]

Decision: the interest still to be paid is the total of the remaining payments minus the balance of 84,652.61 EUR.

  • Option (1): at most \(43 \cdot 2184.71 = 93{,}942.53\) EUR (the last payment is smaller), so at most 9,289.92 EUR of interest.
  • Option (2): \(60 \cdot 1613.06 = 96{,}783.60\) EUR, so 12,130.99 EUR of interest.

Option (1) saves at least 2,841.07 EUR of interest, because the balance falls faster and less interest is charged on it. The owner should choose option (1). Option (2) only makes sense if the boatyard needs the lower monthly payment.

Grading: 1 Pt. new balance, 1 Pt. \(n\) with the logarithm, 1 Pt. 43 payments with the check of both neighbours, 1 Pt. new payment 1,613.06 EUR, 1 Pt. decision by the interest still to be paid.