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 8.5% per year. Find the value PV of the property today.
C1 = €8 000g = 2.5%r = 8.5%
PV = €
2.Time value of moneyHard
You save €100 at the end of every month for 10 years. The account pays 4.8% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €1004.8% / 12N = 120
FV = €
3.NPV & IRRMedium
An investment of €30 000 gives these cash flows at the end of years 1 to 4: €9 000, €15 000, €13 000, €13 000. The cost of capital is 6%. Find the net present value NPV.
A project costs €20 000 and returns €13 500, €15 000, €8 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 13 500CF2 = 15 000CF3 = 8 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 3% annual coupon in two half-yearly payments and matures in 10 years. The yield to maturity is 3.5% per year. Find the price P of the bond.
F = €1 000coupon = 3%YTM = 3.5%n = 10
P = €
6.Bond pricing & yieldHard
A bond with 20 years to maturity (face value €1 000, annual coupon 6%) sells for €1 355. The issuer can call it in 3 years at a call price of €1 060. Find the yield to call YTC in percent.
F = €1 000coupon = 6%P = €1 355call price = €1 060call in 3 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €10 000 000 and debt worth €3 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 3% and the tax rate is 25.8%. Find the WACC.
A company has 1 000 000 shares trading at €40.50; the book value of its equity is €12 200 000. It just paid a dividend of €0.80 per share, and dividends grow by 5% per year. Its debt has a market value of €12 200 000 and a pre-tax cost of 5%. The tax rate is 20%. Find the WACC, using the dividend growth model for the cost of equity.
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 5 of them exceeds 61.8 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 5> 61.8 kg
P =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g40%
x = g
15.Confidence intervalsMedium
A sample of 12 items has mean 13 and standard deviation s = 9. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
n = 12x̄ = 13s = 999%t(11) = 3.106
lower limit =
16.Confidence intervalsHard
The diameters of 16 machined shafts have a sample standard deviation of 4.6 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.
n = 16s = 4.6 µmχ² = 6.262, 27.488
σ = µm
17.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 508.5 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 508.5s = 20n = 16
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 20 of 80 patients on treatment A recover, and 28 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 20n₁ = 80x₂ = 28n₂ = 200
z =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 21Σy = 105Σx² = 91Σy² = 2025Σxy = 313
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.
n = 5x = 5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 133333 €
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.
You paid in €12 000 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = 13053 €
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 = −30 000
CF1/(1 + 0.06)1 = 9 000/1.06 = 8 490.57
CF2/(1 + 0.06)2 = 15 000/1.124 = 13 349.95
CF3/(1 + 0.06)3 = 13 000/1.191 = 10 915.05
CF4/(1 + 0.06)4 = 13 000/1.262 = 10 297.22
Add them up
NPV = €13 052.78
Answer
NPV = €13 052.78
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 41.64 %
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(40%) = 393.59
NPV(45%) = −767.15
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 40 + 5·393.59/(393.59 − (−767.15)) = 41.7%
exact: NPV(41.64%) = 0 → IRR = 41.64%
Answer
IRR = 41.64%
5.Bond pricing & yield
Answer: P = 958.1 €
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 = €15.00, i = 0.0175, N = 20
Present value of the coupons
€15.00·(1 − 0.7068)/0.0175 = €251.29
Present value of the face value
€1 000·0.7068 = €706.82
Price
P = €958.12
Answer
P = €958.12
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTC = 50 %
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
1355 = 60·(1 − (1 + y)−3)/y + 1060/(1 + y)3
Solve numerically
y = 50%: €398.52
y = 50%: €398.52
YTC = 50%
Answer
YTC = 50%
The yield to maturity is 3.502%. 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 = 9.744 %
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).
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 = (32 − 50)/12 = −1.5
Answer
z = −1.5
11.Probability
Answer: Var(X) = 4.35
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.
Mean
μ = 0·0.15 + 2·0.25 + 4·0.3 + 6·0.3 = 3.5
Mean of the squares
E(X²) = 0·0.15 + 4·0.25 + 16·0.3 + 36·0.3 = 16.6
Variance
Var(X) = 16.6 − 3.5² = 4.35
Answer
Var(X) = 4.35
12.Probability
Answer: σY = 24.58
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 = 3.4, Var(X) = 6.04, σX = 2.458
Linear change
σY = |10|·2.458 = 24.58
Answer
σY = 24.58
The mean does move: E(Y) = 10·3.4 + 1 = 35.
13.Normal distribution
Answer: P = 0.1562
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 = 60 kg, σT = 0.8·√5 = 1.789 kg
Standardise
z = (61.8 − 60)/1.789 = 1.01
Probability
P = 1 − Φ(1.01) = 0.1562
Answer
P = 0.1562
14.Normal distribution
Answer: x = 251.3 g
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) = 0.6
z from the table
z ≈ 0.25
Back to the original units
x = 250 + 0.25·5 = 251.3 g
Answer
x = 251.3 g
15.Confidence intervals
Answer: lower limit = 4.93
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 = 11, t = 3.106
Margin of error
E = 3.106·9/√12 = 8.07
Interval
13 ± 8.07 → [4.93, 21.07]
Answer
lower limit = 4.93
16.Confidence intervals
Answer: σ = 3.398 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 = 15·4.6² = 317.4
Limits for σ²
317.4/27.488 = 11.55
317.4/6.262 = 50.69
Limit for σ
σ = √11.55 = 3.398 µm
Answer
σ = 3.398 µm
95% interval for σ: [3.398, 7.119] µm.
17.Hypothesis tests & confidence intervals
Answer: t = 1.7
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 = 20/√16 = 20/4 = 5
Test statistic
t = (508.5 − 500)/5 = 1.7
Answer
t = 1.7
t0.025; df = 15 = 2.131
|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = 2.206
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 = 20/80 = 0.25
p̂₂ = 28/200 = 0.14
p̂ = 48/280 = 0.1714
Standard error under H₀
SE = √(0.1714·0.8286·(1/80 + 1/200)) = 0.04986
Test statistic
z = (0.25 − 0.14)/0.04986 = 2.206
Answer
z = 2.206
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = −0.9514
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 = 313 − 21·105/6 = −54.5
Sxx = 91 − 21²/6 = 17.5
Syy = 2025 − 105²/6 = 187.5
Correlation coefficient
r = −54.5/√(17.5·187.5) = −0.9514
Answer
r = −0.9514
20.Linear regression & correlation
Answer: ŷ = 19.7
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 = 116/5 = 23.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 140
Regression line
b = 140/40 = 3.5
a = 23.2 − 3.5·6 = 2.2
ŷ = 2.2 + 3.5·x
Prediction
ŷ = 2.2 + 3.5·5 = 19.7
Answer
ŷ = 19.7
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