You invest €20 000 at a nominal rate of 3% per year, compounded monthly. Find the value FV of the investment after 2 years.
PV = €20 0003% / 12n = 2
FV = €
2.Time value of moneyHard
A loan of €25 000 is repaid with equal monthly payments over 3 years. The interest rate is 2.4% per year, compounded monthly. Find the monthly payment A.
P = €25 0002.4% / 12N = 36
A = €
3.NPV & IRRMedium
A project costs €60 000 today and returns €15 500 at the end of each year for 5 years. The cost of capital is 6%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
An investment of €40 000 returns €7 000 at the end of each year for 10 years. Find its internal rate of return IRR in percent.
C0 = €40 000CF = €7 000n = 10
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 9% per year. Find the price P of the bond.
F = €1 000coupon = 2%YTM = 9%n = 7
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 8 years. It sells for €1 131. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 7% (2×)n = 8P = €1 131
YTM = %
7.CAPM & WACCMedium
A company has a debt-to-equity ratio (at market values) of 0.5. Its cost of equity is 10%, its pre-tax cost of debt is 7% and the tax rate is 20%. Find the WACC.
D/E = 0.5re = 10%rd = 7%T = 20%
WACC = %
8.CAPM & WACCHard
A company has 2 000 000 shares at €38.00 each and 10 000 bonds trading at €1 020 each. The pre-tax cost of debt is 3.5% and the tax rate is 25%. The share’s beta is 1.2, the risk-free rate 3.5% and the market return 9.5%. Find the WACC.
β = 1.2rf = 3.5%rm = 9.5%rd = 3.5%T = 25%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 53, 57, 40, 25, 35, 41.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is above 92.5?
μ = 100σ = 15x = 92.5
P =
11.ProbabilityMedium
A bag holds 8 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
8 / 5
P =
12.ProbabilityHard
A game costs €5 to play. You win €25 with probability 0.25 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€25: p = 0.25€100: p = 0.05
E(G) = €
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0606. Find z to two decimals.
A = 0.0606
z =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 20% of them lie (the 20th percentile).
μ = 500σ = 40P = 20%
x = h
15.Confidence intervalsMedium
A sample of 30 items has mean 83 and standard deviation s = 15.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.
n = 30x̄ = 83s = 15.595%t(29) = 2.045
E =
16.Confidence intervalsHard
The population standard deviation is 31. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 4?
σ = 31E = 499%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 250; the population standard deviation is known to be 10. A random sample of 100 observations has a mean of 248.16. Test H₀: μ = 250 against H₁: μ < 250. Find the p-value p.
μ₀ = 250σ = 10n = 100x̄ = 248.16H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 200 against H₁: μ > 200 is done at α = 0.01 with a sample of 100; the standard deviation is known to be 10. In reality the mean is 201.9. Find the probability of a Type II error β in percent.
μ₀ = 200μ = 201.9σ = 10n = 100α = 0.01
β = %
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 42Σy = 149Σx² = 364Σy² = 4611Σxy = 1294
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.5.
n = 6x = 5.5
ŷ =
3Solutions
1.Time value of money
Answer: FV = 21235 €
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
Monthly rate and number of payments
i = 0.024/12 = 0.002
N = 12·3 = 36
Payment
The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.
(1 + i)−N = 0.9306
A = 25 000·0.002/(1 − 0.9306) = €720.44
Answer
A = €720.44
3.NPV & IRR
Answer: DPB = 4.543 years
Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.
Discounted and cumulative cash flows
year 1: 15 500/1.06 = 14 622.64 → Σ = 14 622.64
year 2: 15 500/1.124 = 13 794.94 → Σ = 28 417.59
year 3: 15 500/1.191 = 13 014.10 → Σ = 41 431.69
year 4: 15 500/1.262 = 12 277.45 → Σ = 53 709.14
year 5: 15 500/1.338 = 11 582.50 → Σ = 65 291.64
Fraction of the last year
DPB = 4 + 6 290.86/11 582.50 = 4.543
Answer
DPB = 4.543 years
The discounted payback is always longer than the simple payback (3.871 years).
4.NPV & IRR
Answer: IRR = 11.73 %
Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.
Required annuity factor
(1 − (1 + i)−10)/i = 40 000/7 000 = 5.714
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 11%: 5.889
i = 12%: 5.65
IRR = 11.73%
Answer
IRR = 11.73%
5.Bond pricing & yield
Answer: P = 642.2 €
Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.
Per period
C = €10.00, i = 0.045, N = 14
Present value of the coupons
€10.00·(1 − 0.54)/0.045 = €102.23
Present value of the face value
€1 000·0.54 = €539.97
Price
P = €642.20
Answer
P = €642.20
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 4.994 %
Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.
Per period
C = €35.00, N = 16
Solve numerically
i = 2.497% per half-year
YTM
YTM = 2·2.497% = 4.994%
Answer
YTM = 4.994%
7.CAPM & WACC
Answer: WACC = 8.533 %
Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).
Weights from D/E
E/V = 1/(1 + 0.5) = 0.6667
D/V = 0.5/(1 + 0.5) = 0.3333
WACC
WACC = 0.6667·10% + 0.3333·7%·(1 − 0.2) = 8.533%
Answer
WACC = 8.533%
8.CAPM & WACC
Answer: WACC = 9.744 %
Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Standardise
z = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915
Answer
P = 0.6915
11.Probability
Answer: P = 0.359
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
First and second draw
P = 8/13 · 7/12 = 56/156 = 0.359
Answer
P = 0.359
12.Probability
Answer: E(G) = 6.25 €
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Net outcomes
G: 20 (p = 0.25), 95 (p = 0.05), −5 (p = 0.7)
Expected value
E(G) = 20·0.25 + 95·0.05 − 5·0.7 = 6.25 €
Answer
E(G) = 6.25 €
A positive value means you win money on average.
13.Normal distribution
Answer: z = 1.55
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left of z
Φ(z) = 1 − 0.0606 = 0.9394
Read the z-table backwards
Find 0.9394 inside the table and read off z.
z = 1.55
Answer
z = 1.55
14.Normal distribution
Answer: x = 466.3 h
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z for the percentile
Φ(z) = 0.2 → z = −0.8416
Back to x
x = 500 + −0.8416·40 = 466.34 h
Answer
x = 466.34 h
15.Confidence intervals
Answer: E = 5.787
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Critical value
df = n − 1 = 29, t = 2.045
Margin of error
E = 2.045·15.5/√30 = 5.787
Answer
E = 5.787
16.Confidence intervals
Answer: n = 399
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Solve for n
n = (2.576·31/4)² = 398.6
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 399
Answer
n = 399
17.Hypothesis tests & confidence intervals
Answer: p = 0.03288
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Test statistic
SE = σ/√n = 10/√100 = 1
z = (248.16 − 250)/1 = −1.84
Tail area
Φ(1.84) = 0.9671
1 − Φ(1.84) = 0.0329
p-value
p = 0.03288
Answer
p = 0.03288
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: β = 66.49 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 10/√100 = 1
x̄c = 200 + 2.326·1 = 202.3
z under the true mean
z = (202.3 − 201.9)/1 = 0.426
Type II error
power = P(Z > 0.426) = 33.51%
β = 100% − 33.51% = 66.49%
Answer
β = 66.49%
19.Linear regression & correlation
Answer: r = 0.994
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Centred sums
Sxy = 1294 − 42·149/6 = 251
Sxx = 364 − 42²/6 = 70
Syy = 4611 − 149²/6 = 910.8
Correlation coefficient
r = 251/√(70·910.8) = 0.994
Answer
r = 0.994
20.Linear regression & correlation
Answer: ŷ = 84
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Means
x̄ = 33/6 = 5.5
ȳ = 504/6 = 84
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 105
Regression line
b = 105/17.5 = 6
a = 84 − 6·5.5 = 51
ŷ = 51 + 6·x
Prediction
ŷ = 51 + 6·5.5 = 84
Answer
ŷ = 84
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