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 8% per year. Find the value PV of the property today.
C1 = €8 000g = 3%r = 8%
PV = €
2.Time value of moneyHard
A loan of €15 000 is repaid with equal monthly payments over 5 years. The interest rate is 4.2% per year, compounded monthly. Find the monthly payment A.
P = €15 0004.2% / 12N = 60
A = €
3.NPV & IRRMedium
A machine costs €80 000 today. It generates €11 500 at the end of each year for 8 years and can then be sold for €8 000. The cost of capital is 6%. Find the net present value NPV.
C0 = €80 000CF = €11 500S = €8 000n = 8r = 6%
NPV = €
4.NPV & IRRHard
A project costs €20 000 and returns €12 500, €13 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 12 500CF2 = 13 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 9 years. The yield to maturity is 7% per year. Find the price P of the bond.
F = €1 000coupon = 8%YTM = 7%n = 9
P = €
6.Bond pricing & yieldHard
A bond with 18 years to maturity (face value €1 000, annual coupon 9%) sells for €1 261. The issuer can call it in 6 years at a call price of €1 090. Find the yield to call YTC in percent.
F = €1 000coupon = 9%P = €1 261call price = €1 090call in 6 years
YTC = %
7.CAPM & WACCMedium
A company has a debt-to-equity ratio (at market values) of 0.4. Its cost of equity is 11%, its pre-tax cost of debt is 5% and the tax rate is 25%. Find the WACC.
D/E = 0.4re = 11%rd = 5%T = 25%
WACC = %
8.CAPM & WACCHard
A company has 2 000 000 shares at €47.50 each and 40 000 bonds trading at €1 050 each. The pre-tax cost of debt is 6% and the tax rate is 25.8%. The share’s beta is 1.5, the risk-free rate 2% and the market return 8.5%. Find the WACC.
β = 1.5rf = 2%rm = 8.5%rd = 6%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 54, 15, 32, 50, 33, 11.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 5. What proportion P of values is below 66?
μ = 70σ = 5x = 66
P =
11.ProbabilityMedium
In how many ways N can a committee of 4 people be chosen from a group of 7? The order does not matter.
n = 7k = 4
N =
12.ProbabilityHard
X has the distribution shown, and Y = 10X − 3. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 10X − 3
σY =
13.Normal distributionMedium
The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 6 of them exceeds 70 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 6> 70 kg
P =
14.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 30% is to the left of it. Find the cut-off x.
μ = 178 cmσ = 7 cm30%
x = cm
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 30 phones, mean 94 h, known standard deviation 7 h. Model 2: 60 phones, mean 98 h, known standard deviation 9 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.
A pilot study suggests that about 0.7 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.04?
p ≈ 0.7E = 0.0495%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 201.5 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 201.5s = 5n = 49
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 30 of 150 patients on treatment A recover, and 24 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 30n₁ = 150x₂ = 24n₂ = 100
z =
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 = 127Σx² = 364Σy² = 3331Σxy = 1100
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 53?
n = 6ŷ = 53
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 = 12·5 = 60
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 = 0.8109
A = 15 000·0.0035/(1 − 0.8109) = €277.60
Answer
A = €277.60
3.NPV & IRR
Answer: NPV = −3568 €
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
11 500·6.21 = 71 412.63
Present value of the salvage value
8 000/1.594 = 5 019.30
NPV
NPV = −80 000 + 71 412.63 + 5 019.30 = −€3 568.07
Answer
NPV = −€3 568.07
4.NPV & IRR
Answer: IRR = 19.15 %
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(15%) = 1 077.50
NPV(20%) = −208.33
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
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 = €40.00, i = 0.035, N = 18
Present value of the coupons
€40.00·(1 − 0.5384)/0.035 = €527.59
Present value of the face value
€1 000·0.5384 = €538.36
Price
P = €1 065.95
Answer
P = €1 065.95
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTC = 5.151 %
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.
Equation for the yield
1261 = 90·(1 − (1 + y)−6)/y + 1090/(1 + y)6
Solve numerically
y = 5%: €1 270.19
y = 6%: €1 210.97
YTC = 5.151%
Answer
YTC = 5.151%
The yield to maturity is 6.498%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.
7.CAPM & WACC
Answer: WACC = 8.929 %
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 + 0.4) = 0.7143
D/V = 0.4/(1 + 0.4) = 0.2857
WACC
WACC = 0.7143·11% + 0.2857·5%·(1 − 0.25) = 8.929%
Answer
WACC = 8.929%
8.CAPM & WACC
Answer: WACC = 9.513 %
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 = −0.8
Standard normal table
Φ(−0.8) = 0.2119
P = 0.2119
Answer
P = 0.2119
11.Probability
Answer: N = 35
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.
Combinations
C(7, 4) = (7·6·5·4)/(4·3·2·1) = 35
Answer
N = 35
12.Probability
Answer: σY = 8.047
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.
Spread of X
μ = 1.05, Var(X) = 0.6475, σX = 0.8047
Linear change
σY = |10|·0.8047 = 8.047
Answer
σY = 8.047
The mean does move: E(Y) = 10·1.05 − 3 = 7.5.
13.Normal distribution
Answer: P = 0.8461
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 total
μT = 72 kg, σT = 0.8·√6 = 1.96 kg
Standardise
z = (70 − 72)/1.96 = −1.02
Probability
P = 1 − Φ(−1.02) = 0.8461
Answer
P = 0.8461
14.Normal distribution
Answer: x = 174.4 cm
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) = 0.3
z from the table
z ≈ −0.52
Back to the original units
x = 178 − 0.52·7 = 174.4 cm
Answer
x = 174.4 cm
15.Confidence intervals
Answer: E = 4.449 h
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.
Point estimate
x̄₁ − x̄₂ = 94 − 98 = −4
Standard error
SE = √(7²/30 + 9²/60) = 1.727
Margin of error
E = 2.576·1.727 = 4.449
Answer
E = 4.449 h
The interval contains 0, so the data do not show a significant difference.
16.Confidence intervals
Answer: n = 505
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 = 1.96²·0.7·0.3/0.04² = 504.2
Round up
n = 505
Answer
n = 505
17.Hypothesis tests & confidence intervals
Answer: t = 2.1
Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).
Standard error
SE = 5/√49 = 5/7 = 0.7143
Test statistic
t = (201.5 − 200)/0.7143 = 2.1
Answer
t = 2.1
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = −0.7529
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 = 30/150 = 0.2
p̂₂ = 24/100 = 0.24
p̂ = 54/250 = 0.216
Standard error under H₀
SE = √(0.216·0.784·(1/150 + 1/100)) = 0.05313
Test statistic
z = (0.2 − 0.24)/0.05313 = −0.7529
Answer
z = −0.7529
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9947
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 = 1100 − 42·127/6 = 211
Sxx = 364 − 42²/6 = 70
Syy = 3331 − 127²/6 = 642.8
Correlation coefficient
r = 211/√(70·642.8) = 0.9947
Answer
r = 0.9947
20.Linear regression & correlation
Answer: x = 8.959
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 = 274/6 = 45.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 262
Regression line
b = 262/70 = 3.743
a = 45.67 − 3.743·7 = 19.47
ŷ = 19.47 + 3.743·x
Solve for x
x = (53 − 19.47)/3.743 = 8.959
Answer
x = 8.959
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