A credit card charges a nominal interest rate of 18% per year, compounded monthly. Find the effective annual rate EAR in percent.
r = 18%m = 12
EAR = %
2.Time value of moneyHard
A mortgage of €200 000 is repaid with equal monthly payments over 25 years at 2.4% per year, compounded monthly. How much is still owed just after the payment at the end of year 13? Find the outstanding balance B.
P = €200 0002.4% / 12N = 300k = 156
B = €
3.NPV & IRRMedium
A project costs €30 000 today and returns €6 500, €8 500, €13 000, €8 000 at the end of years 1 to 4. The cost of capital is 6%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
An investment of €10 000 returns €2 500 at the end of each year for 6 years. Find its internal rate of return IRR in percent.
C0 = €10 000CF = €2 500n = 6
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 6 years. The yield to maturity is 5% per year. Find the price P of the bond.
F = €1 000coupon = 8%YTM = 5%n = 6
P = €
6.Bond pricing & yieldHard
A bond trades at €935 with a yield to maturity of 7% (annual compounding) and a Macaulay duration of 7 years. Use the modified duration to estimate the new price P if the yield rises by 25 basis points.
P0 = €935y = 7%D = 7Δy = +25 bp
P = €
7.CAPM & WACCMedium
A share with a beta of 1.5 has a required return of 10%. The risk-free rate is 1%. What expected market return rm does CAPM imply?
β = 1.5re = 10%rf = 1%
rm = %
8.CAPM & WACCHard
A company has 1 000 000 shares trading at €28.50; the book value of its equity is €17 100 000. It just paid a dividend of €1.50 per share, and dividends grow by 3% per year. Its debt has a market value of €14 300 000 and a pre-tax cost of 3%. The tax rate is 30%. Find the WACC, using the dividend growth model for the cost of equity.
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 45.5 and 85 points? Give P as a decimal.
μ = 65σ = 10x1 = 45.5x2 = 85
P =
14.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 70% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points70%
x = points
15.Confidence intervalsMedium
A sample of 15 items has mean 98 and standard deviation s = 16. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 15x̄ = 98s = 1699%t(14) = 2.977
E =
16.Confidence intervalsHard
A pilot study suggests that about 0.3 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.04?
p ≈ 0.3E = 0.0499%
n =
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 100 parts has mean length 65 mm, and the line's known standard deviation is 17 mm. Line 2: a sample of 40 parts has mean 69.5 mm, known standard deviation 7 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A sample of 25 measurements has mean 116.4. The population standard deviation is 12. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 116.4σ = 12n = 2590%
U =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 5 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 = 47?
n = 5ŷ = 47
x =
3Solutions
1.Time value of money
Answer: EAR = 19.56 %
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.
Rate per period
r/m = 0.18/12 = 0.015
Compound over one year
EAR = (1 + 0.015)12 − 1 = 1.196 − 1 = 19.56%
Answer
EAR = 19.56%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: B = 110909 €
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 = 0.002, N = 300
A = 200 000·0.002/(1 − (1 + 0.002)−300) = €887.19
Present value of the remaining payments
Whatever is still owed must be exactly paid off by the remaining payments.
N − k = 300 − 156 = 144
B = 887.2·(1 − (1 + 0.002)−144)/0.002 = €110 909.40
Answer
B = €110 909.40
3.NPV & IRR
Answer: DPB = 3.85 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: 6 500/1.06 = 6 132.08 → Σ = 6 132.08
year 2: 8 500/1.124 = 7 564.97 → Σ = 13 697.05
year 3: 13 000/1.191 = 10 915.05 → Σ = 24 612.10
year 4: 8 000/1.262 = 6 336.75 → Σ = 30 948.85
Fraction of the last year
DPB = 3 + 5 387.90/6 336.75 = 3.85
Answer
DPB = 3.85 years
The discounted payback is always longer than the simple payback (3.25 years).
4.NPV & IRR
Answer: IRR = 12.98 %
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)−6)/i = 10 000/2 500 = 4
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 12%: 4.111
i = 13%: 3.998
IRR = 12.98%
Answer
IRR = 12.98%
5.Bond pricing & yield
Answer: P = 1154 €
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 = €40.00, i = 0.025, N = 12
Present value of the coupons
€40.00·(1 − 0.7436)/0.025 = €410.31
Present value of the face value
€1 000·0.7436 = €743.56
Price
P = €1 153.87
Answer
P = €1 153.87
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: P = 919.7 €
Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.
Modified duration
Dmod = 7/1.07 = 6.542
Estimated price change
%ΔP ≈ −6.542·(0.0025) = −1.636%
P ≈ 935·(1 − 0.01636) = €919.71
Answer
P = €919.71
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: rm = 7 %
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 = (10% − 1%)/1.5 = 6%
Market return
rm = 1% + 6% = 7%
Answer
rm = 7%
8.CAPM & WACC
Answer: WACC = 6.309 %
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743
Answer
P = 0.2743
11.Probability
Answer: σ = 1.244
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 = 0·0.3 + 1·0.2 + 2·0.15 + 3·0.35 = 1.55
Mean of the squares
E(X²) = 0·0.3 + 1·0.2 + 4·0.15 + 9·0.35 = 3.95
Variance
Var(X) = 3.95 − 1.55² = 1.548
σ = √1.548 = 1.244
Answer
σ = 1.244
12.Probability
Answer: σY = 0.6124
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, Var(X) = 1.5, σX = 1.225
Linear change
σY = |0.5|·1.225 = 0.6124
Answer
σY = 0.6124
The mean does move: E(Y) = 0.5·3 − 3 = −1.5.
13.Normal distribution
Answer: P = 0.9517
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 = (45.5 − 65)/10 = −1.95
z₂ = (85 − 65)/10 = 2
Area between
P = Φ(2) − Φ(−1.95) = 0.9772 − 0.02559 = 0.9517
Answer
P = 0.9517
14.Normal distribution
Answer: x = 107.8 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) = 0.7
z from the table
z ≈ 0.52
Back to the original units
x = 100 + 0.52·15 = 107.8 points
Answer
x = 107.8 points
15.Confidence intervals
Answer: E = 12.3
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 = 14, t = 2.977
Margin of error
E = 2.977·16/√15 = 12.3
Answer
E = 12.3
16.Confidence intervals
Answer: n = 871
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 2.576²·0.3·0.7/0.04² = 870.9
Round up
n = 871
Answer
n = 871
17.Hypothesis tests & confidence intervals
Answer: z = −2.218
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 = √(17²/100 + 7²/40) = √4.115 = 2.029
Test statistic
z = (65 − 69.5)/2.029 = −2.218
Answer
z = −2.218
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: U = 120.3
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 = 1.645·12/√25 = 3.948
Interval limit
U = 116.4 + 3.948 = 120.3
Answer
U = 120.3
19.Linear regression & correlation
Answer: r = 0.9509
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 = 265/5 = 53
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 46
Syy = Σ(y − ȳ)² = 234
Correlation coefficient
r = 46/√(10·234) = 0.9509
Answer
r = 0.9509
20.Linear regression & correlation
Answer: x = 5.077
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 = 259/5 = 51.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 208
Regression line
b = 208/40 = 5.2
a = 51.8 − 5.2·6 = 20.6
ŷ = 20.6 + 5.2·x
Solve for x
x = (47 − 20.6)/5.2 = 5.077
Answer
x = 5.077
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