A property is expected to earn net rent of €20 000 next year. The rent is expected to grow by 1.5% per year forever. The required return is 6.5% per year. Find the value PV of the property today.
C1 = €20 000g = 1.5%r = 6.5%
PV = €
2.Time value of moneyHard
A loan of €15 000 is repaid with equal monthly payments over 7 years at 3% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €15 0003% / 12N = 84
I = €
3.NPV & IRRMedium
A project costs €50 000 today and returns €19 000, €13 500, €20 500, €16 000, €11 500, €14 500 at the end of years 1 to 6. The cost of capital is 8%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €50 000 today. Its cash flow is €7 500 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 8. The cost of capital is 10%. Find the net present value NPV.
C0 = €50 000CF1 = €7 500g = 3%n = 8r = 10%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 7 years and pays annual coupons. It sells for €1 120 when the yield to maturity is 4%. Find the coupon rate c in percent.
F = €1 000n = 7P = €1 120YTM = 4%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 4% matures in 5 years and sells for €915. Find its yield to maturity YTM in percent.
F = €1 000coupon = 4%n = 5P = €915
YTM = %
7.CAPM & WACCMedium
A company has equity worth €4 000 000 and debt worth €2 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 4.5% and the tax rate is 30%. Find the WACC.
A company has 5 000 000 shares trading at €41.00; the book value of its equity is €123 000 000. It just paid a dividend of €0.90 per share, and dividends grow by 4% per year. Its debt has a market value of €205 000 000 and a pre-tax cost of 4%. The tax rate is 20%. Find the WACC, using the dividend growth model for the cost of equity.
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9297. Find z to two decimals.
A = 0.9297
z =
14.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 80% of them lie (the 80th percentile).
μ = 65σ = 10P = 80%
x = points
15.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 5.1. What is the margin of error E₂ of a 95% interval based on 16 observations?
90%n₁ = 64E₁ = 5.195%n₂ = 16
E₂ =
16.Confidence intervalsHard
The population standard deviation is 38. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 3?
σ = 38E = 395%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 120; the population standard deviation is known to be 15. A random sample of 36 observations has a mean of 124.55. Test H₀: μ = 120 against H₁: μ ≠ 120. Find the p-value p.
μ₀ = 120σ = 15n = 36x̄ = 124.55H₁: μ ≠ μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 22. Find the minimum sample size n.
The table gives 6 paired observations of x = advertising spend in k€ and y = sales 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 = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.
n = 6x = 4.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 400000 €
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 = 84
A = 15 000·0.0025/(1 − (1 + 0.0025)−84) = €198.20
Total interest
I = N·A − P = 84·198.2 − 15 000 = €1 648.76
Answer
I = €1 648.76
3.NPV & IRR
Answer: DPB = 3.388 years
Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.
Discounted and cumulative cash flows
year 1: 19 000/1.08 = 17 592.59 → Σ = 17 592.59
year 2: 13 500/1.166 = 11 574.07 → Σ = 29 166.67
year 3: 20 500/1.26 = 16 273.56 → Σ = 45 440.23
year 4: 16 000/1.36 = 11 760.48 → Σ = 57 200.71
Fraction of the last year
DPB = 3 + 4 559.77/11 760.48 = 3.388
Answer
DPB = 3.388 years
The discounted payback is always longer than the simple payback (2.854 years).
4.NPV & IRR
Answer: NPV = −6174 €
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 the growing annuity
((1.03)/(1.1))8 = 0.591
PV = 7 500/(0.1 − 0.03)·(1 − 0.591) = 43 825.97
NPV
NPV = 43 825.97 − 50 000 = −€6 174.03
Answer
NPV = −€6 174.03
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: c = 5.999 %
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.
Present value of the face value
1000·(1 + 0.04)−7 = €759.92
Coupon
C = (1120 − 759.92)/6.002 = €59.99
Coupon rate
c = 59.99/1000 = 5.999%
Answer
c = 5.999%
6.Bond pricing & yield
Answer: YTM = 6.019 %
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
€915 = €40·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵
Solve numerically
P(6%) = €915.75, P(7%) = €876.99
YTM = 6.019%
Answer
YTM = 6.019%
7.CAPM & WACC
Answer: WACC = 9.717 %
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).
Cost of equity
D1 = 0.9·1.04 = €0.94
re = 0.936/41 + 0.04 = 6.283%
Market values
E = 5 000 000·€41.00 = €205 000 000
V = €410 000 000, E/V = 0.5, D/V = 0.5
WACC
WACC = 0.5·6.283% + 0.5·4%·(1 − 0.2) = 4.741%
Answer
WACC = 4.741%
9.Descriptive statistics
Answer: s = 13.89
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 = (76 − 70)/10 = 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 = 0.522
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 = 1.1, Var(X) = 1.09, σX = 1.044
Linear change
σY = |0.5|·1.044 = 0.522
Answer
σY = 0.522
The mean does move: E(Y) = 0.5·1.1 − 3 = −2.45.
13.Normal distribution
Answer: z = 1.81
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 + 0.9297)/2 = 0.9649
Read the z-table backwards
Find 0.9649 inside the table and read off z.
z = 1.81
Answer
z = 1.81
14.Normal distribution
Answer: x = 73.42 points
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) = 0.8 → z = 0.8416
Back to x
x = 65 + 0.8416·10 = 73.42 points
Answer
x = 73.42 points
15.Confidence intervals
Answer: E₂ = 12.15
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 = (1.96/1.645)·√(64/16) = 2.383
New margin
E₂ = 5.1·2.383 = 12.15
Answer
E₂ = 12.15
16.Confidence intervals
Answer: n = 617
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.
Solve for n
n = (1.96·38/3)² = 616.4
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 617
Answer
n = 617
17.Hypothesis tests & confidence intervals
Answer: p = 0.06876
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.
Test statistic
SE = σ/√n = 15/√36 = 2.5
z = (124.55 − 120)/2.5 = 1.82
Tail area
Φ(1.82) = 0.9656
1 − Φ(1.82) = 0.0344
p-value
p = 2·0.0344 = 0.06876
Answer
p = 0.06876
p > 0.05, so H₀ is not rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: n = 188
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((1.645 + 0.842)·22/4)² = 187.1
Round up
n = 188
Answer
n = 188
19.Linear regression & correlation
Answer: r² = 0.807
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 = 201/6 = 33.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 99
Syy = Σ(y − ȳ)² = 173.5
Coefficient of determination
r² = 99²/(70·173.5) = 0.807
Answer
r² = 0.807
About 80.7% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 18.39
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 = 117/6 = 19.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −19.5
Regression line
b = −19.5/17.5 = −1.114
a = 19.5 − (−1.114)·3.5 = 23.4
ŷ = 23.4 − 1.114·x
Prediction
ŷ = 23.4 + (−1.114)·4.5 = 18.39
Answer
ŷ = 18.39
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