A property is expected to earn net rent of €12 000 next year. The rent is expected to grow by 2.5% per year forever. The required return is 7.5% per year. Find the value PV of the property today.
C1 = €12 000g = 2.5%r = 7.5%
PV = €
2.Time value of moneyHard
You borrow €20 000 at 5.4% per year, compounded monthly, and repay €475 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 = €4755.4% / 12
N = payments
3.NPV & IRRMedium
A machine costs €20 000 today, lasts 4 years and costs €3 000 per year to run (paid at the end of each year). The cost of capital is 10%. Find its equivalent annual cost EAC.
C0 = €20 000M = €3 000n = 4r = 10%
EAC = €
4.NPV & IRRHard
An investment of €10 000 returns €1 500 at the end of each year for 8 years. Find its internal rate of return IRR in percent.
C0 = €10 000CF = €1 500n = 8
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 6 years. The yield to maturity is 5.5% per year. Find the price P of the bond.
F = €1 000coupon = 7%YTM = 5.5%n = 6
P = €
6.Bond pricing & yieldHard
A bond with 18 years to maturity (face value €1 000, annual coupon 5%) sells for €1 359. The issuer can call it in 5 years at a call price of €1 050. Find the yield to call YTC in percent.
F = €1 000coupon = 5%P = €1 359call price = €1 050call in 5 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €2 000 000 and debt worth €6 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 5% and the tax rate is 25%. Find the WACC.
A company has 5 000 000 shares trading at €58.00; the book value of its equity is €87 000 000. It just paid a dividend of €2.30 per share, and dividends grow by 4% per year. Its debt has a market value of €232 000 000 and a pre-tax cost of 3%. The tax rate is 25.8%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 43, 55, 47, 37, 35, 45.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 25. What proportion P of values is above 550?
μ = 500σ = 25x = 550
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3, 4p(x): 0.25, 0.35, 0.25, 0.05, 0.1
Var(X) =
12.ProbabilityHard
A game costs €5 to play. You win €20 with probability 0.15 and €250 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€20: p = 0.15€250: p = 0.01
E(G) = €
13.Normal distributionMedium
X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 200.6 mm. Round z to two decimals.
μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.6 mm
P =
14.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 95% of them lie (the 95th percentile).
μ = 178σ = 7P = 95%
x = cm
15.Confidence intervalsMedium
A sample of 8 items has mean 15 and standard deviation s = 13. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 8x̄ = 15s = 1399%t(7) = 3.499
upper limit =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 13 plots, mean yield 42 kg, standard deviation 10 kg. Plot group 2: 8 plots, mean 35.5 kg, standard deviation 9 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.
Two production lines make the same part. Line 1: a sample of 30 parts has mean length 115 mm, and the line's known standard deviation is 18 mm. Line 2: a sample of 30 parts has mean 107.7 mm, known standard deviation 11 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A sample of 16 measurements has mean 78.5. The population standard deviation is 10. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 78.5σ = 10n = 1690%
U =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
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. For which value x does the line predict ŷ = 51?
n = 6ŷ = 51
x =
3Solutions
1.Time value of money
Answer: PV = 240000 €
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 = 0.0045
(1 + i)−N = 1 − 20 000·0.0045/475 = 0.8105
Solve with logarithms
N = −ln(0.8105)/ln(1.004) = 46.79
Round up to whole payments
46 full payments leave a small balance, which the last, smaller payment clears.
N = 47
Answer
N = 47 payments
3.NPV & IRR
Answer: EAC = 9309 €
Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.
Annuity factor
(1 − 1.1−4)/0.1 = 3.17
Annualised purchase price
20 000/3.17 = 6 309.42
EAC
EAC = 6 309.42 + 3 000 = €9 309.42
Answer
EAC = €9 309.42
4.NPV & IRR
Answer: IRR = 4.239 %
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)−8)/i = 10 000/1 500 = 6.667
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 4%: 6.733
i = 5%: 6.463
IRR = 4.239%
Answer
IRR = 4.239%
5.Bond pricing & yield
Answer: P = 1076 €
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 = 0.0275, N = 12
Present value of the coupons
€35.00·(1 − 0.7221)/0.0275 = €353.65
Present value of the face value
€1 000·0.7221 = €722.13
Price
P = €1 075.78
Answer
P = €1 075.78
The yield is below the coupon rate, so the bond sells at a premium.
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
1359 = 50·(1 − (1 + y)−5)/y + 1050/(1 + y)5
Solve numerically
y = 50%: €225.10
y = 50%: €225.10
YTC = 50%
Answer
YTC = 50%
The yield to maturity is 2.499%. 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 = 5.813 %
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 = 0.25, D/V = 0.75
WACC
WACC = 0.25·12% + 0.75·5%·(1 − 0.25) = 5.813%
Answer
WACC = 5.813%
8.CAPM & WACC
Answer: WACC = 5.503 %
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 1 − 0.9772 = 0.02275
Answer
P = 0.02275
11.Probability
Answer: Var(X) = 1.44
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.
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: 15 (p = 0.15), 245 (p = 0.01), −5 (p = 0.84)
Expected value
E(G) = 15·0.15 + 245·0.01 − 5·0.84 = 0.5 €
Answer
E(G) = 0.5 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.1151
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 = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm
Standardise
z = (200.6 − 200)/0.5 = 1.2
Upper tail
P = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151
Answer
P = 0.1151
14.Normal distribution
Answer: x = 189.5 cm
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 = 178 + 1.645·7 = 189.51 cm
Answer
x = 189.51 cm
15.Confidence intervals
Answer: upper limit = 31.08
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 = 7, t = 3.499
Margin of error
E = 3.499·13/√8 = 16.08
Interval
15 ± 16.08 → [−1.082, 31.08]
Answer
upper limit = 31.08
16.Confidence intervals
Answer: E = 12.4 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 = (12·10² + 7·9²)/19 = 93
sp = 9.644
Margin of error
E = 2.861·9.644·√(1/13 + 1/8) = 12.4
Answer
E = 12.4 kg
17.Hypothesis tests & confidence intervals
Answer: z = 1.895
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 = √(18²/30 + 11²/30) = √14.83 = 3.851
Test statistic
z = (115 − 107.7)/3.851 = 1.895
Answer
z = 1.895
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: U = 82.61
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 = 1.645·10/√16 = 4.112
Interval limit
U = 78.5 + 4.112 = 82.61
Answer
U = 82.61
19.Linear regression & correlation
Answer: r² = 0.6207
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 = 25/5 = 5
ȳ = 276/5 = 55.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 24
Syy = Σ(y − ȳ)² = 92.8
Coefficient of determination
r² = 24²/(10·92.8) = 0.6207
Answer
r² = 0.6207
About 62.07% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: x = 7.532
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 = 275/6 = 45.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 44.5
Regression line
b = 44.5/17.5 = 2.543
a = 45.83 − 2.543·5.5 = 31.85
ŷ = 31.85 + 2.543·x
Solve for x
x = (51 − 31.85)/2.543 = 7.532
Answer
x = 7.532
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