A property is expected to earn net rent of €8 000 next year. The rent is expected to grow by 1.5% per year forever. The required return is 6.5% per year. Find the value PV of the property today.
C1 = €8 000g = 1.5%r = 6.5%
PV = €
2.Time value of moneyHard
A loan of €8 000 is repaid with equal monthly payments over 4 years. The interest rate is 4.2% per year, compounded monthly. Find the monthly payment A.
P = €8 0004.2% / 12N = 48
A = €
3.NPV & IRRMedium
A machine costs €30 000 today. It generates €6 500 at the end of each year for 5 years and can then be sold for €6 000. The cost of capital is 12%. Find the net present value NPV.
C0 = €30 000CF = €6 500S = €6 000n = 5r = 12%
NPV = €
4.NPV & IRRHard
A project costs €10 000 and returns €7 000, €5 500, €5 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −10 000CF1 = 7 000CF2 = 5 500CF3 = 5 500
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 5 years. The yield to maturity is 3% per year. Find the price P of the bond.
F = €1 000coupon = 2%YTM = 3%n = 5
P = €
6.Bond pricing & yieldHard
A bond trades at €1 055 with a yield to maturity of 7% (annual compounding) and a Macaulay duration of 11.3 years. Use the modified duration to estimate the new price P if the yield rises by 25 basis points.
P0 = €1 055y = 7%D = 11.3Δy = +25 bp
P = €
7.CAPM & WACCMedium
A company has equity worth €2 000 000 and debt worth €4 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 5% and the tax rate is 25.8%. Find the WACC.
A company has 2 000 000 shares at €15.00 each and 5 000 bonds trading at €960 each. The pre-tax cost of debt is 7% and the tax rate is 28%. The share’s beta is 1.5, the risk-free rate 2% and the market return 6%. Find the WACC.
β = 1.5rf = 2%rm = 6%rd = 7%T = 28%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 44, 42, 52, 48, 26, 30.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is below 176?
μ = 170σ = 10x = 176
P =
11.ProbabilityMedium
A fair die is rolled 7 times. What is the probability P of getting at least one six?
n = 7
P =
12.ProbabilityHard
A game costs €3 to play. You win €15 with probability 0.15 and €30 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€15: p = 0.15€30: p = 0.01
E(G) = €
13.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 93.25 and 127.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 93.25x2 = 127.75
P =
14.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points80%
x = points
15.Confidence intervalsMedium
A sample of 25 items has mean 54.5 and standard deviation s = 2. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
n = 25x̄ = 54.5s = 299%t(24) = 2.797
lower limit =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 10 plots, mean yield 11 kg, standard deviation 7 kg. Plot group 2: 6 plots, mean 5 kg, standard deviation 8 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
Two teaching methods are compared. Group 1: 9 students, mean score 21.5, standard deviation 11. Group 2: 17 students, mean score 14.9, standard deviation 6. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 9x̄₁ = 21.5s₁ = 11n₂ = 17x̄₂ = 14.9s₂ = 6
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 500 against H₁: μ < 500 is done at α = 0.01 with a sample of 64; the standard deviation is known to be 4. In reality the mean is 498.05. Find the power 1 − β of the test in percent.
μ₀ = 500μ = 498.05σ = 4n = 64α = 0.01
1 − β = %
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 12?
n = 6ŷ = 12
x =
3Solutions
1.Time value of money
Answer: PV = 160000 €
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 = 0.042/12 = 0.0035
N = 12·4 = 48
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 = 0.8456
A = 8 000·0.0035/(1 − 0.8456) = €181.35
Answer
A = €181.35
3.NPV & IRR
Answer: NPV = −3164 €
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.
Present value of the yearly cash flows
6 500·3.605 = 23 431.05
Present value of the salvage value
6 000/1.762 = 3 404.56
NPV
NPV = −30 000 + 23 431.05 + 3 404.56 = −€3 164.39
Answer
NPV = −€3 164.39
4.NPV & IRR
Answer: IRR = 38.43 %
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.
Bracket the IRR
NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.
NPV(35%) = 438.45
NPV(40%) = −189.50
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 35 + 5·438.45/(438.45 − (−189.50)) = 38.49%
exact: NPV(38.43%) = 0 → IRR = 38.43%
Answer
IRR = 38.43%
5.Bond pricing & yield
Answer: P = 953.9 €
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 = 0.015, N = 10
Present value of the coupons
€10.00·(1 − 0.8617)/0.015 = €92.22
Present value of the face value
€1 000·0.8617 = €861.67
Price
P = €953.89
Answer
P = €953.89
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: P = 1027 €
Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.
Modified duration
Dmod = 11.3/1.07 = 10.56
Estimated price change
%ΔP ≈ −10.56·(0.0025) = −2.64%
P ≈ 1055·(1 − 0.0264) = €1 027.15
Answer
P = €1 027.15
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: WACC = 5.14 %
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
V = €6 000 000, E/V = 0.3333, D/V = 0.6667
WACC
WACC = 0.3333·8% + 0.6667·5%·(1 − 0.258) = 5.14%
Answer
WACC = 5.14%
8.CAPM & WACC
Answer: WACC = 7.592 %
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).
Market values
E = 2 000 000·€15.00 = €30 000 000
D = 5 000·€960 = €4 800 000
Cost of equity (CAPM)
re = 2% + 1.5·4% = 8%
Weights
V = €34 800 000, E/V = 0.8621, D/V = 0.1379
WACC
WACC = 0.8621·8% + 0.1379·7%·(1 − 0.28) = 7.592%
Answer
WACC = 7.592%
9.Descriptive statistics
Answer: s = 10.23
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 0.7257
Answer
P = 0.7257
11.Probability
Answer: P = 0.7209
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.
Complement
P(no six) = (5/6)7 = 0.2791
P = 1 − 0.2791 = 0.7209
Answer
P = 0.7209
12.Probability
Answer: E(G) = −0.45 €
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: 12 (p = 0.15), 27 (p = 0.01), −3 (p = 0.84)
Expected value
E(G) = 12·0.15 + 27·0.01 − 3·0.84 = −0.45 €
Answer
E(G) = −0.45 €
A negative value means you lose money on average.
13.Normal distribution
Answer: P = 0.6415
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-scores
z₁ = (93.25 − 100)/15 = −0.45
z₂ = (127.75 − 100)/15 = 1.85
Area between
P = Φ(1.85) − Φ(−0.45) = 0.9678 − 0.3264 = 0.6415
Answer
P = 0.6415
14.Normal distribution
Answer: x = 56.6 points
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
Φ(z) = 0.2
z from the table
z ≈ −0.84
Back to the original units
x = 65 − 0.84·10 = 56.6 points
Answer
x = 56.6 points
15.Confidence intervals
Answer: lower limit = 53.38
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 = 24, t = 2.797
Margin of error
E = 2.797·2/√25 = 1.119
Interval
54.5 ± 1.119 → [53.38, 55.62]
Answer
lower limit = 53.38
16.Confidence intervals
Answer: lower limit = −0.7046 kg
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Pooled standard deviation
sp² = (9·7² + 5·8²)/14 = 54.36
sp = 7.373
Margin of error
E = 1.761·7.373·√(1/10 + 1/6) = 6.705
Interval
6 ± 6.705 → [−0.7046, 12.7]
Answer
lower limit = −0.7046 kg
17.Hypothesis tests & confidence intervals
Answer: t = 1.996
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.
Pooled variance
sp² = (8·11² + 16·6²)/24 = 64.33
sp = 8.021
Standard error
SE = 8.021·√(1/9 + 1/17) = 3.306
Test statistic
t = (21.5 − 14.9)/3.306 = 1.996
Answer
t = 1.996
With df = 24, the two-sided 5% critical value is 2.064; |t| is smaller, so H₀ is not rejected.
18.Hypothesis tests & confidence intervals
Answer: 1 − β = 94.23 %
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 = 4/√64 = 0.5
x̄c = 500 − 2.326·0.5 = 498.8
z under the true mean
z = (498.8 − 498.05)/0.5 = 1.574
Power
power = P(Z < 1.574) = 94.23%
1 − β = 94.23%
Answer
1 − β = 94.23%
19.Linear regression & correlation
Answer: r = 0.9869
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 = 147/6 = 24.5
ȳ = 698/6 = 116.3
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 810
Syy = Σ(y − ȳ)² = 4277
Correlation coefficient
r = 810/√(157.5·4277) = 0.9869
Answer
r = 0.9869
20.Linear regression & correlation
Answer: x = 3.807
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 = 21/6 = 3.5
ȳ = 78/6 = 13
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −57
Regression line
b = −57/17.5 = −3.257
a = 13 − (−3.257)·3.5 = 24.4
ŷ = 24.4 − 3.257·x
Solve for x
x = (12 − 24.4)/(−3.257) = 3.807
Answer
x = 3.807
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