A property is expected to earn net rent of €5 000 next year. The rent is expected to grow by 2.5% per year forever. The required return is 5.5% per year. Find the value PV of the property today.
C1 = €5 000g = 2.5%r = 5.5%
PV = €
2.Time value of moneyHard
A loan of €20 000 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 = €20 0002.4% / 12N = 36
A = €
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 3: €18 000, €15 000, €12 000. The cost of capital is 15%. Find the net present value NPV.
A project costs €20 000 and returns €12 000, €11 000, €12 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 12 000CF2 = 11 000CF3 = 12 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 14 years and pays annual coupons. It sells for €1 053 when the yield to maturity is 4%. Find the coupon rate c in percent.
F = €1 000n = 14P = €1 053YTM = 4%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 3% matures in 9 years and sells for €1 080. Find its yield to maturity YTM in percent.
F = €1 000coupon = 3%n = 9P = €1 080
YTM = %
7.CAPM & WACCMedium
A company has equity worth €3 000 000 and debt worth €5 000 000. Its cost of equity is 9%, its pre-tax cost of debt is 3% and the tax rate is 25%. Find the WACC.
E = €3 000 000D = €5 000 000re = 9%rd = 3%T = 25%
WACC = %
8.CAPM & WACCHard
A company has equity with a market value of €6 000 000 and a cost of equity of 11%. Its debt consists of 4 000 bonds (face value €1 000, annual coupon 4%, 6 years to maturity) that trade at €1 027 each. The tax rate is 28%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.
Find the sample standard deviation s of the data: 27, 29, 49, 25, 49, 28, 32.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 100. What proportion P of values is below 650?
μ = 500σ = 100x = 650
P =
11.ProbabilityMedium
A bag holds 6 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
6 / 8
P =
12.ProbabilityHard
A game costs €2 to play. You win €8 with probability 0.1 and €100 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €2€8: p = 0.1€100: p = 0.01
E(G) = €
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9846. Find z to two decimals.
A = 0.9846
z =
14.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 95% of them lie (the 95th percentile).
μ = 100σ = 15P = 95%
x = points
15.Confidence intervalsMedium
The blood pressure of 11 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.4 and standard deviation 3.1. Find the lower limit of the 99% confidence interval for the mean difference.
n = 11d̄ = 6.4sd = 3.199%t(10) = 3.169
lower limit = mmHg
16.Confidence intervalsHard
The diameters of 30 machined shafts have a sample standard deviation of 9.9 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.
n = 30s = 9.9 µmχ² = 16.047, 45.722
σ = µm
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 40 parts has mean length 87.5 mm, and the line's known standard deviation is 5 mm. Line 2: a sample of 80 parts has mean 90.4 mm, known standard deviation 5 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A sample of 81 measurements has mean 119.7. The population standard deviation is 10. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 119.7σ = 10n = 8199%
U =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 5
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 (11, 26) is made. Find its residual e = y − ŷ.
n = 6(11, 26)
e =
3Solutions
1.Time value of money
Answer: PV = 166667 €
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 = 20 000·0.002/(1 − 0.9306) = €576.35
Answer
A = €576.35
3.NPV & IRR
Answer: NPV = −25115 €
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 + 0.15)1 = 18 000/1.15 = 15 652.17
CF2/(1 + 0.15)2 = 15 000/1.322 = 11 342.16
CF3/(1 + 0.15)3 = 12 000/1.521 = 7 890.19
Add them up
NPV = €−25 115.48
Answer
NPV = €−25 115.48
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 35.02 %
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%) = 5.08
NPV(40%) = −1 260.93
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 35 + 5·5.08/(5.08 − (−1 260.93)) = 35.02%
exact: NPV(35.02%) = 0 → IRR = 35.02%
Answer
IRR = 35.02%
5.Bond pricing & yield
Answer: c = 4.502 %
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.
Present value of the face value
1000·(1 + 0.04)−14 = €577.48
Coupon
C = (1053 − 577.48)/10.56 = €45.02
Coupon rate
c = 45.02/1000 = 4.502%
Answer
c = 4.502%
6.Bond pricing & yield
Answer: YTM = 2.019 %
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
€1 080 = €30·(1 − (1 + y)⁻⁹)/y + €1 000/(1 + y)⁹
Solve numerically
P(2%) = €1 081.62, P(3%) = €1 000.00
YTM = 2.019%
Answer
YTM = 2.019%
7.CAPM & WACC
Answer: WACC = 4.781 %
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 = €8 000 000, E/V = 0.375, D/V = 0.625
WACC
WACC = 0.375·9% + 0.625·3%·(1 − 0.25) = 4.781%
Answer
WACC = 4.781%
8.CAPM & WACC
Answer: WACC = 7.552 %
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 0.9332
Answer
P = 0.9332
11.Probability
Answer: P = 0.1648
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 = 6/14 · 5/13 = 30/182 = 0.1648
Answer
P = 0.1648
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: 6 (p = 0.1), 98 (p = 0.01), −2 (p = 0.89)
Expected value
E(G) = 6·0.1 + 98·0.01 − 2·0.89 = −0.2 €
Answer
E(G) = −0.2 €
A negative value means you lose money on average.
13.Normal distribution
Answer: z = −2.16
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 − 0.9846 = 0.0154
Read the z-table backwards
Find 0.0154 inside the table and read off z.
z = −2.16
Answer
z = −2.16
14.Normal distribution
Answer: x = 124.7 points
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) = 0.95 → z = 1.645
Back to x
x = 100 + 1.645·15 = 124.67 points
Answer
x = 124.67 points
15.Confidence intervals
Answer: lower limit = 3.438 mmHg
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.
Critical value
df = 10, t = 3.169
Margin of error
E = 3.169·3.1/√11 = 2.962
Interval
6.4 ± 2.962 → [3.438, 9.362]
Answer
lower limit = 3.438 mmHg
16.Confidence intervals
Answer: σ = 7.884 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 29·9.9² = 2842
Limits for σ²
2842/45.722 = 62.16
2842/16.047 = 177.1
Limit for σ
σ = √62.16 = 7.884 µm
Answer
σ = 7.884 µm
95% interval for σ: [7.884, 13.31] µm.
17.Hypothesis tests & confidence intervals
Answer: z = −2.995
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.
Standard error of the difference
SE = √(5²/40 + 5²/80) = √0.9375 = 0.9682
Test statistic
z = (87.5 − 90.4)/0.9682 = −2.995
Answer
z = −2.995
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: U = 122.6
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).
Margin of error
E = 2.576·10/√81 = 2.862
Interval limit
U = 119.7 + 2.862 = 122.6
Answer
U = 122.6
19.Linear regression & correlation
Answer: r = 0.9507
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 = 30/5 = 6
ȳ = 44/5 = 8.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 52
Syy = Σ(y − ȳ)² = 74.8
Correlation coefficient
r = 52/√(40·74.8) = 0.9507
Answer
r = 0.9507
20.Linear regression & correlation
Answer: e = 3.2
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 = 84/6 = 14
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 154
Regression line
b = 154/70 = 2.2
a = 14 − 2.2·7 = −1.4
ŷ = −1.4 + 2.2·x
Predicted value
ŷ = −1.4 + 2.2·11 = 22.8
Residual
e = 26 − 22.8 = 3.2
Answer
e = 3.2
The point lies above the line.
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