A property is expected to earn net rent of €8 000 next year. The rent is expected to grow by 3% per year forever. The required return is 5% per year. Find the value PV of the property today.
C1 = €8 000g = 3%r = 5%
PV = €
2.Time value of moneyHard
A loan of €15 000 is repaid with equal monthly payments over 3 years. The interest rate is 6% per year, compounded monthly. Find the monthly payment A.
P = €15 0006% / 12N = 36
A = €
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 4: €20 000, €25 000, €15 000, €15 000. The cost of capital is 5%. Find the net present value NPV.
A project costs €25 000 and returns €10 500, €14 500, €9 000 at the end of years 1 to 3. Cash flows can be reinvested at the cost of capital of 6%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 pays an annual coupon of 3% and matures in 5 years. Its yield to maturity changes from 5.5% to 6%. By how much ΔP does the bond price change?
F = €1 000coupon = 3%n = 5y1 = 5.5%y2 = 6%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 6 years. The yield to maturity is 5.5%. Find the Macaulay duration D in years.
F = €1 000coupon = 5%n = 6YTM = 5.5%
D = years
7.CAPM & WACCMedium
A company has a debt-to-equity ratio (at market values) of 1. Its cost of equity is 10%, its pre-tax cost of debt is 3.5% and the tax rate is 25.8%. Find the WACC.
D/E = 1re = 10%rd = 3.5%T = 25.8%
WACC = %
8.CAPM & WACCHard
A company has 2 000 000 shares trading at €30.50; the book value of its equity is €36 600 000. It just paid a dividend of €1.00 per share, and dividends grow by 3% per year. Its debt has a market value of €48 800 000 and a pre-tax cost of 6%. The tax rate is 20%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 42, 55, 56, 57, 48, 20.
n = 6
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 100. Find the z-score z of a score of 420.
μ = 500σ = 100x = 420
z =
11.ProbabilityMedium
A fair die is rolled 6 times. What is the probability P of getting at least one six?
n = 6
P =
12.ProbabilityHard
A game costs €1 to play. You win €2 with probability 0.2 and €20 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€2: p = 0.2€20: p = 0.02
E(G) = €
13.Normal distributionMedium
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 110.1 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 110.1 s
P =
14.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.04 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
A sample of 8 items has mean 40.5 and standard deviation s = 4. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 8x̄ = 40.5s = 499%t(7) = 3.499
E =
16.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.05?
p = 0.5E = 0.0590%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 250; the population standard deviation is known to be 20. A random sample of 64 observations has a mean of 254.25. Test H₀: μ = 250 against H₁: μ > 250. Find the p-value p.
μ₀ = 250σ = 20n = 64x̄ = 254.25H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
In a trial, 24 of 120 patients on treatment A recover, and 24 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 24n₁ = 120x₂ = 24n₂ = 150
z =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 15Σy = 351Σx² = 55Σy² = 25067Σxy = 1117
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (9, 29) is made. Find its residual e = y − ŷ.
n = 6(9, 29)
e =
3Solutions
1.Time value of money
Answer: PV = 400000 €
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 = 0.06/12 = 0.005
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 = 0.8356
A = 15 000·0.005/(1 − 0.8356) = €456.33
Answer
A = €456.33
3.NPV & IRR
Answer: NPV = 7021 €
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 each cash flow
CF0 = −60 000
CF1/(1 + 0.05)1 = 20 000/1.05 = 19 047.62
CF2/(1 + 0.05)2 = 25 000/1.103 = 22 675.74
CF3/(1 + 0.05)3 = 15 000/1.158 = 12 957.56
CF4/(1 + 0.05)4 = 15 000/1.216 = 12 340.54
Add them up
NPV = €7 021.46
Answer
NPV = €7 021.46
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: MIRR = 13.1 %
Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.
Terminal value of the inflows
10 500·1.062 = 11 797.80
14 500·1.061 = 15 370.00
9 000·1.060 = 9 000.00
TV = 36 167.80
MIRR
MIRR = (36 167.80/25 000)1/3 − 1 = 13.1%
Answer
MIRR = 13.1%
5.Bond pricing & yield
Answer: ΔP = −19.61 €
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.
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.
Present value of each cash flow
t = 1: 50/1.055 = 47.39, t·PV = 47.39
t = 2: 50/1.113 = 44.92, t·PV = 89.85
t = 3: 50/1.174 = 42.58, t·PV = 127.74
t = 4: 50/1.239 = 40.36, t·PV = 161.44
t = 5: 50/1.307 = 38.26, t·PV = 191.28
t = 6: 1050/1.379 = 761.51, t·PV = 4 569.05
Duration
P = 975.02
D = 5 186.76/975.02 = 5.32
Answer
D = 5.32 years
7.CAPM & WACC
Answer: WACC = 6.298 %
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 + 1) = 0.5
D/V = 1/(1 + 1) = 0.5
WACC
WACC = 0.5·10% + 0.5·3.5%·(1 − 0.258) = 6.298%
Answer
WACC = 6.298%
8.CAPM & WACC
Answer: WACC = 5.676 %
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 = (420 − 500)/100 = −0.8
Answer
z = −0.8
11.Probability
Answer: P = 0.6651
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)6 = 0.3349
P = 1 − 0.3349 = 0.6651
Answer
P = 0.6651
12.Probability
Answer: E(G) = −0.2 €
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: 1 (p = 0.2), 19 (p = 0.02), −1 (p = 0.78)
Expected value
E(G) = 1·0.2 + 19·0.02 − 1·0.78 = −0.2 €
Answer
E(G) = −0.2 €
A negative value means you lose money on average.
13.Normal distribution
Answer: P = 0.2119
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the sum
μ = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (110.1 − 105)/6.403 = 0.8
Upper tail
P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119
Answer
P = 0.2119
14.Normal distribution
Answer: P = 0.9452
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the difference
μ = 25.04 − 25 = 0.04 mm, σ = 0.025 mm
Standardise
z = (0 − 0.04)/0.025 = −1.6
Probability
P = 1 − Φ(−1.6) = 0.9452
Answer
P = 0.9452
15.Confidence intervals
Answer: E = 4.948
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 = 7, t = 3.499
Margin of error
E = 3.499·4/√8 = 4.948
Answer
E = 4.948
16.Confidence intervals
Answer: n = 271
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 1.645²·0.5·0.5/0.05² = 270.6
Round up
n = 271
Answer
n = 271
17.Hypothesis tests & confidence intervals
Answer: p = 0.04457
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 = 20/√64 = 2.5
z = (254.25 − 250)/2.5 = 1.7
Tail area
Φ(1.7) = 0.9554
1 − Φ(1.7) = 0.0446
p-value
p = 0.04457
Answer
p = 0.04457
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: z = 0.8542
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.
Sample proportions
p̂₁ = 24/120 = 0.2
p̂₂ = 24/150 = 0.16
p̂ = 48/270 = 0.1778
Standard error under H₀
SE = √(0.1778·0.8222·(1/120 + 1/150)) = 0.04683
Test statistic
z = (0.2 − 0.16)/0.04683 = 0.8542
Answer
z = 0.8542
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9796
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 = 1117 − 15·351/5 = 64
Sxx = 55 − 15²/5 = 10
Syy = 25067 − 351²/5 = 426.8
Correlation coefficient
r = 64/√(10·426.8) = 0.9796
Answer
r = 0.9796
20.Linear regression & correlation
Answer: e = −1.667
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 = 42/6 = 7
ȳ = 148/6 = 24.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 210
Regression line
b = 210/70 = 3
a = 24.67 − 3·7 = 3.667
ŷ = 3.667 + 3·x
Predicted value
ŷ = 3.667 + 3·9 = 30.67
Residual
e = 29 − 30.67 = −1.667
Answer
e = −1.667
The point lies below the line.
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