A property is expected to earn net rent of €8 000 next year. The rent is expected to grow by 2.5% per year forever. The required return is 6.5% per year. Find the value PV of the property today.
C1 = €8 000g = 2.5%r = 6.5%
PV = €
2.Time value of moneyHard
You borrow €20 000 at 3% per year, compounded monthly, and repay €275 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.
P = €20 000A = €2753% / 12
N = payments
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 4: €25 000, €28 000, €14 000, €27 000. The cost of capital is 10%. Find the net present value NPV.
A project costs €20 000 and returns €11 500, €8 500, €10 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 11 500CF2 = 8 500CF3 = 10 000
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 11 years. The yield to maturity is 3% per year. Find the price P of the bond.
F = €1 000coupon = 2%YTM = 3%n = 11
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 5% matures in 9 years and sells for €1 075. Find its yield to maturity YTM in percent.
F = €1 000coupon = 5%n = 9P = €1 075
YTM = %
7.CAPM & WACCMedium
A company has a debt-to-equity ratio (at market values) of 1. Its cost of equity is 9%, its pre-tax cost of debt is 4% and the tax rate is 30%. Find the WACC.
D/E = 1re = 9%rd = 4%T = 30%
WACC = %
8.CAPM & WACCHard
A company has 2 500 000 shares at €49.50 each and 20 000 bonds trading at €980 each. The pre-tax cost of debt is 4.5% and the tax rate is 20%. The share’s beta is 0.8, the risk-free rate 1.5% and the market return 8%. Find the WACC.
β = 0.8rf = 1.5%rm = 8%rd = 4.5%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 35, 34, 51, 30, 39.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 12. Find the z-score z of a score of 194.
μ = 170σ = 12x = 194
z =
11.ProbabilityMedium
A fair die is rolled 3 times. What is the probability P of getting at least one six?
n = 3
P =
12.ProbabilityHard
9 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?
n = 9p = 0.1
P =
13.Normal distributionMedium
X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 28.3 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 28.3 min
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 30% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points30%
x = points
15.Confidence intervalsMedium
A sample of 21 items has mean 78 and standard deviation s = 14. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.
n = 21x̄ = 78s = 1495%t(20) = 2.086
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 0.025?
p = 0.5E = 0.02590%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 48.8 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 48.8s = 12n = 81
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 38 of 150 patients on treatment A recover, and 98 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 38n₁ = 150x₂ = 98n₂ = 250
z =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (17.5, 73) is made. Find its residual e = y − ŷ.
n = 6(17.5, 73)
e =
3Solutions
1.Time value of money
Answer: PV = 200000 €
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.
Rearrange the annuity formula
i = 0.0025
(1 + i)−N = 1 − 20 000·0.0025/275 = 0.8182
Solve with logarithms
N = −ln(0.8182)/ln(1.002) = 80.37
Round up to whole payments
80 full payments leave a small balance, which the last, smaller payment clears.
N = 81
Answer
N = 81 payments
3.NPV & IRR
Answer: NPV = 14828 €
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 + 0.1)1 = 25 000/1.1 = 22 727.27
CF2/(1 + 0.1)2 = 28 000/1.21 = 23 140.50
CF3/(1 + 0.1)3 = 14 000/1.331 = 10 518.41
CF4/(1 + 0.1)4 = 27 000/1.464 = 18 441.36
Add them up
NPV = €14 827.54
Answer
NPV = €14 827.54
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 24.16 %
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(20%) = 1 273.15
NPV(25%) = −240.00
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 = €10.00, i = 0.015, N = 22
Present value of the coupons
€10.00·(1 − 0.7207)/0.015 = €186.21
Present value of the face value
€1 000·0.7207 = €720.69
Price
P = €906.90
Answer
P = €906.90
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 3.992 %
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
€1 075 = €50·(1 − (1 + y)⁻⁹)/y + €1 000/(1 + y)⁹
Solve numerically
P(3%) = €1 155.72, P(4%) = €1 074.35
YTM = 3.992%
Answer
YTM = 3.992%
7.CAPM & WACC
Answer: WACC = 5.9 %
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 = 0.5·9% + 0.5·4%·(1 − 0.3) = 5.9%
Answer
WACC = 5.9%
8.CAPM & WACC
Answer: WACC = 6.276 %
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 = (194 − 170)/12 = 2
Answer
z = 2
11.Probability
Answer: P = 0.4213
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)3 = 0.5787
P = 1 − 0.5787 = 0.4213
Answer
P = 0.4213
12.Probability
Answer: P = 0.7748
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.
Zero and one failure
P(X = 0) = 0.99 = 0.3874
P(X = 1) = 9·0.1·0.98 = 0.3874
Add them
P = 0.3874 + 0.3874 = 0.7748
Answer
P = 0.7748
13.Normal distribution
Answer: P = 0.9306
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 = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (28.3 − 32)/2.5 = −1.48
Upper tail
P = 1 − Φ(−1.48) = 1 − 0.06944 = 0.9306
Answer
P = 0.9306
14.Normal distribution
Answer: x = 70.2 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) = 0.7
z from the table
z ≈ 0.52
Back to the original units
x = 65 + 0.52·10 = 70.2 points
Answer
x = 70.2 points
15.Confidence intervals
Answer: E = 6.373
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 = 20, t = 2.086
Margin of error
E = 2.086·14/√21 = 6.373
Answer
E = 6.373
16.Confidence intervals
Answer: n = 1083
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 = 1.645²·0.5·0.5/0.025² = 1082
Round up
n = 1083
Answer
n = 1083
17.Hypothesis tests & confidence intervals
Answer: t = −0.9
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 = 12/√81 = 12/9 = 1.333
Test statistic
t = (48.8 − 50)/1.333 = −0.9
Answer
t = −0.9
t0.025; df = 80 = 1.99
|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = −2.834
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 = 38/150 = 0.2533
p̂₂ = 98/250 = 0.392
p̂ = 136/400 = 0.34
Standard error under H₀
SE = √(0.34·0.66·(1/150 + 1/250)) = 0.04892
Test statistic
z = (0.2533 − 0.392)/0.04892 = −2.834
Answer
z = −2.834
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r² = 0.8974
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 = 138/6 = 23
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −35
Syy = Σ(y − ȳ)² = 78
Coefficient of determination
r² = −35²/(17.5·78) = 0.8974
Answer
r² = 0.8974
About 89.74% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = −2.124
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 = 141/6 = 23.5
ȳ = 616/6 = 102.7
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 723
Regression line
b = 723/157.5 = 4.59
a = 102.7 − 4.59·23.5 = −5.21
ŷ = −5.21 + 4.59·x
Predicted value
ŷ = −5.21 + 4.59·17.5 = 75.12
Residual
e = 73 − 75.12 = −2.124
Answer
e = −2.124
The point lies below the line.
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