A property is expected to earn net rent of €5 000 next year. The rent is expected to grow by 1% per year forever. The required return is 7% per year. Find the value PV of the property today.
C1 = €5 000g = 1%r = 7%
PV = €
2.Time value of moneyHard
A loan of €25 000 is repaid with equal monthly payments over 5 years at 3% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €25 0003% / 12N = 60
I = €
3.NPV & IRRMedium
A project costs €200 000 today. It generates €23 000 at the end of the first year, and the cash flow then grows by 1% per year forever. The cost of capital is 10%. Find the net present value NPV.
C0 = €200 000CF1 = €23 000g = 1%r = 10%
NPV = €
4.NPV & IRRHard
A project costs €50 000 and returns €24 000, €21 000, €19 000, €17 500 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 12%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 pays an annual coupon of 4.5% and matures in 8 years. Its yield to maturity changes from 7% to 5%. By how much ΔP does the bond price change?
F = €1 000coupon = 4.5%n = 8y1 = 7%y2 = 5%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 10 years. It sells for €740. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 5% (2×)n = 10P = €740
YTM = %
7.CAPM & WACCMedium
A company has equity worth €8 000 000 and debt worth €3 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 5% and the tax rate is 20%. Find the WACC.
A company has 2 000 000 shares at €44.50 each and 20 000 bonds trading at €1 000 each. The pre-tax cost of debt is 7% and the tax rate is 30%. The share’s beta is 1.5, the risk-free rate 1.5% and the market return 8%. Find the WACC.
β = 1.5rf = 1.5%rm = 8%rd = 7%T = 30%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 56, 47, 59, 60, 48, 46, 42.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 5. Find the z-score z of a score of 103.
μ = 100σ = 5x = 103
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
X has the distribution shown, and Y = 10X + 4. Find the standard deviation σY of Y.
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 23 and 41.3 min? Give P as a decimal.
μ = 32σ = 6x1 = 23x2 = 41.3
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 15% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points15%
x = points
15.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 5.1. What is the margin of error E₂ of a 95% interval from the same data?
99%n₁ = 50E₁ = 5.195%
E₂ =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 9 plots, mean yield 93.5 kg, standard deviation 5 kg. Plot group 2: 13 plots, mean 97 kg, standard deviation 6.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
A sample of 100 students has a mean score of 502.9 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 502.9s = 10n = 100
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 200 against H₁: μ > 200 is done at α = 0.10 with a sample of 36; the standard deviation is known to be 6. In reality the mean is 201.2. Find the power 1 − β of the test in percent.
μ₀ = 200μ = 201.2σ = 6n = 36α = 0.10
1 − β = %
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.
n = 5x = 2.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 83333 €
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 payment
i = 0.0025, N = 60
A = 25 000·0.0025/(1 − (1 + 0.0025)−60) = €449.22
Total interest
I = N·A − P = 60·449.2 − 25 000 = €1 953.04
Answer
I = €1 953.04
3.NPV & IRR
Answer: NPV = 55556 €
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.
Value of the cash flows
PV = 23 000/(0.1 − 0.01) = 255 555.56
NPV
NPV = 255 555.56 − 200 000 = €55 555.56
Answer
NPV = €55 555.56
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: MIRR = 18.57 %
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
24 000·1.123 = 33 718.27
21 000·1.122 = 26 342.40
19 000·1.121 = 21 280.00
17 500·1.120 = 17 500.00
TV = 98 840.67
MIRR
MIRR = (98 840.67/50 000)1/4 − 1 = 18.57%
Answer
MIRR = 18.57%
5.Bond pricing & yield
Answer: ΔP = 117 €
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.
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 = €25.00, N = 20
Solve numerically
i = 4.499% per half-year
YTM
YTM = 2·4.499% = 8.997%
Answer
YTM = 8.997%
7.CAPM & WACC
Answer: WACC = 8.364 %
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 = €11 000 000, E/V = 0.7273, D/V = 0.2727
WACC
WACC = 0.7273·10% + 0.2727·5%·(1 − 0.2) = 8.364%
Answer
WACC = 8.364%
8.CAPM & WACC
Answer: WACC = 10.08 %
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 = (103 − 100)/5 = 0.6
Answer
z = 0.6
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: σY = 10.2
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.
Spread of X
μ = 1.4, Var(X) = 1.04, σX = 1.02
Linear change
σY = |10|·1.02 = 10.2
Answer
σY = 10.2
The mean does move: E(Y) = 10·1.4 + 4 = 18.
13.Normal distribution
Answer: P = 0.8726
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 = (23 − 32)/6 = −1.5
z₂ = (41.3 − 32)/6 = 1.55
Area between
P = Φ(1.55) − Φ(−1.5) = 0.9394 − 0.06681 = 0.8726
Answer
P = 0.8726
14.Normal distribution
Answer: x = 75.4 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) = 0.85
z from the table
z ≈ 1.04
Back to the original units
x = 65 + 1.04·10 = 75.4 points
Answer
x = 75.4 points
15.Confidence intervals
Answer: E₂ = 3.88
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.
Ratio of margins
E₂/E₁ = (1.96/2.576)·√(50/50) = 0.7609
New margin
E₂ = 5.1·0.7609 = 3.88
Answer
E₂ = 3.88
16.Confidence intervals
Answer: lower limit = −7.947 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 = (8·5² + 12·6.5²)/20 = 35.35
sp = 5.946
Margin of error
E = 1.725·5.946·√(1/9 + 1/13) = 4.447
Interval
−3.5 ± 4.447 → [−7.947, 0.9474]
Answer
lower limit = −7.947 kg
17.Hypothesis tests & confidence intervals
Answer: t = 2.9
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 = 10/√100 = 10/10 = 1
Test statistic
t = (502.9 − 500)/1 = 2.9
Answer
t = 2.9
t0.025; df = 99 = 1.984
|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: 1 − β = 46.73 %
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 = 6/√36 = 1
x̄c = 200 + 1.282·1 = 201.3
z under the true mean
z = (201.3 − 201.2)/1 = 0.082
Power
power = P(Z > 0.082) = 46.73%
1 − β = 46.73%
Answer
1 − β = 46.73%
19.Linear regression & correlation
Answer: r² = 0.9757
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 = 36/6 = 6
ȳ = 415/6 = 69.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 233
Syy = Σ(y − ȳ)² = 794.8
Coefficient of determination
r² = 233²/(70·794.8) = 0.9757
Answer
r² = 0.9757
About 97.57% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 21.25
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 = 15/5 = 3
ȳ = 100/5 = 20
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −25
Regression line
b = −25/10 = −2.5
a = 20 − (−2.5)·3 = 27.5
ŷ = 27.5 − 2.5·x
Prediction
ŷ = 27.5 + (−2.5)·2.5 = 21.25
Answer
ŷ = 21.25
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