A property is expected to earn net rent of €10 000 next year. The rent is expected to grow by 4% per year forever. The required return is 10% per year. Find the value PV of the property today.
C1 = €10 000g = 4%r = 10%
PV = €
2.Time value of moneyHard
You borrow €8 000 at 4.2% per year, compounded monthly, and repay €200 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 = €8 000A = €2004.2% / 12
N = payments
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 4: €23 000, €20 000, €23 000, €18 000. The cost of capital is 12%. Find the net present value NPV.
An investment of €80 000 returns €11 000 at the end of each year for 10 years. Find its internal rate of return IRR in percent.
C0 = €80 000CF = €11 000n = 10
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 9 years. The yield to maturity is 9% per year. Find the price P of the bond.
F = €1 000coupon = 7%YTM = 9%n = 9
P = €
6.Bond pricing & yieldHard
A bond with 20 years to maturity (face value €1 000, annual coupon 8%) sells for €1 299. The issuer can call it in 4 years at a call price of €1 080. Find the yield to call YTC in percent.
F = €1 000coupon = 8%P = €1 299call price = €1 080call in 4 years
YTC = %
7.CAPM & WACCMedium
A share with a beta of 1.4 has a required return of 7.3%. The risk-free rate is 1%. What expected market return rm does CAPM imply?
β = 1.4re = 7.3%rf = 1%
rm = %
8.CAPM & WACCHard
A company's shares have a beta of 1.2 at its current debt-to-equity ratio of 0.5. It plans to raise its debt-to-equity ratio to 1.25. The tax rate is 28%, the risk-free rate 3% and the market risk premium 5%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 58, 18, 50, 46, 35, 22.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is above 40?
μ = 50σ = 5x = 40
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.25, 0.3, 0.35, 0.1
Var(X) =
12.ProbabilityHard
7 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?
n = 7p = 0.7
P =
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.8789. Find z to two decimals.
A = 0.8789
z =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 25% of them lie (the 25th percentile).
μ = 250σ = 5P = 25%
x = g
15.Confidence intervalsMedium
A sample of 25 items has mean 59 and standard deviation s = 17. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
n = 25x̄ = 59s = 1799%t(24) = 2.797
lower limit =
16.Confidence intervalsHard
The diameters of 30 machined shafts have a sample standard deviation of 2.5 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 = 2.5 µmχ² = 16.047, 45.722
σ = µm
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 50 parts has mean length 49 mm, and the line's known standard deviation is 17 mm. Line 2: a sample of 50 parts has mean 43.4 mm, known standard deviation 18 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A sample of 36 measurements has mean 66. The population standard deviation is 20. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 66σ = 20n = 3690%
L =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 6
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. A new observation (6.5, 42) is made. Find its residual e = y − ŷ.
n = 6(6.5, 42)
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.
Rearrange the annuity formula
i = 0.0035
(1 + i)−N = 1 − 8 000·0.0035/200 = 0.86
Solve with logarithms
N = −ln(0.86)/ln(1.004) = 43.17
Round up to whole payments
43 full payments leave a small balance, which the last, smaller payment clears.
N = 44
Answer
N = 44 payments
3.NPV & IRR
Answer: NPV = 4290 €
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 + 0.12)1 = 23 000/1.12 = 20 535.71
CF2/(1 + 0.12)2 = 20 000/1.254 = 15 943.88
CF3/(1 + 0.12)3 = 23 000/1.405 = 16 370.95
CF4/(1 + 0.12)4 = 18 000/1.574 = 11 439.33
Add them up
NPV = €4 289.86
Answer
NPV = €4 289.86
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 6.253 %
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.
Required annuity factor
(1 − (1 + i)−10)/i = 80 000/11 000 = 7.273
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 6%: 7.36
i = 7%: 7.024
IRR = 6.253%
Answer
IRR = 6.253%
5.Bond pricing & yield
Answer: P = 878.4 €
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 = €35.00, i = 0.045, N = 18
Present value of the coupons
€35.00·(1 − 0.4528)/0.045 = €425.60
Present value of the face value
€1 000·0.4528 = €452.80
Price
P = €878.40
Answer
P = €878.40
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTC = 2.073 %
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
1299 = 80·(1 − (1 + y)−4)/y + 1080/(1 + y)4
Solve numerically
y = 2%: €1 302.37
y = 3%: €1 256.93
YTC = 2.073%
Answer
YTC = 2.073%
The yield to maturity is 5.498%. 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: rm = 5.5 %
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).
Market risk premium
rm − rf = (7.3% − 1%)/1.4 = 4.5%
Market return
rm = 1% + 4.5% = 5.5%
Answer
rm = 5.5%
8.CAPM & WACC
Answer: re = 11.38 %
Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.
Unlever the beta
βU = 1.2/(1 + 0.72·0.5) = 0.8824
Relever at the new D/E
βL = 0.8824·(1 + 0.72·1.25) = 1.676
Cost of equity (CAPM)
re = 3% + 1.676·5% = 11.38%
Answer
re = 11.38%
9.Descriptive statistics
Answer: s = 15.96
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772
Answer
P = 0.9772
11.Probability
Answer: Var(X) = 3.64
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 = 0·0.25 + 2·0.3 + 4·0.35 + 6·0.1 = 2.6
Mean of the squares
E(X²) = 0·0.25 + 4·0.3 + 16·0.35 + 36·0.1 = 10.4
Variance
Var(X) = 10.4 − 2.6² = 3.64
Answer
Var(X) = 3.64
12.Probability
Answer: P = 0.003791
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) = 0.37 = 0.0002187
P(X = 1) = 7·0.7·0.36 = 0.003572
Add them
P = 0.0002187 + 0.003572 = 0.003791
Answer
P = 0.003791
13.Normal distribution
Answer: z = 1.55
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 + 0.8789)/2 = 0.9394
Read the z-table backwards
Find 0.9394 inside the table and read off z.
z = 1.55
Answer
z = 1.55
14.Normal distribution
Answer: x = 246.6 g
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) = 0.25 → z = −0.6745
Back to x
x = 250 + −0.6745·5 = 246.63 g
Answer
x = 246.63 g
15.Confidence intervals
Answer: lower limit = 49.49
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 = 24, t = 2.797
Margin of error
E = 2.797·17/√25 = 9.51
Interval
59 ± 9.51 → [49.49, 68.51]
Answer
lower limit = 49.49
16.Confidence intervals
Answer: σ = 1.991 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 = 29·2.5² = 181.3
Limits for σ²
181.3/45.722 = 3.964
181.3/16.047 = 11.29
Limit for σ
σ = √3.964 = 1.991 µm
Answer
σ = 1.991 µm
95% interval for σ: [1.991, 3.361] µm.
17.Hypothesis tests & confidence intervals
Answer: z = 1.599
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 = √(17²/50 + 18²/50) = √12.26 = 3.501
Test statistic
z = (49 − 43.4)/3.501 = 1.599
Answer
z = 1.599
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: L = 60.52
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 = 1.645·20/√36 = 5.483
Interval limit
L = 66 − 5.483 = 60.52
Answer
L = 60.52
19.Linear regression & correlation
Answer: r = 0.9764
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 = 126/6 = 21
ȳ = 290/6 = 48.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 200
Syy = Σ(y − ȳ)² = 599.3
Correlation coefficient
r = 200/√(70·599.3) = 0.9764
Answer
r = 0.9764
20.Linear regression & correlation
Answer: e = 3.995
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 = 33/6 = 5.5
ȳ = 209/6 = 34.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 55.5
Regression line
b = 55.5/17.5 = 3.171
a = 34.83 − 3.171·5.5 = 17.39
ŷ = 17.39 + 3.171·x
Predicted value
ŷ = 17.39 + 3.171·6.5 = 38
Residual
e = 42 − 38 = 3.995
Answer
e = 3.995
The point lies above the line.
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