A credit card charges a nominal interest rate of 16% per year, compounded monthly. Find the effective annual rate EAR in percent.
r = 16%m = 12
EAR = %
2.Time value of moneyHard
You borrow €10 000 at 6% per year, compounded monthly, and repay €450 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 = €10 000A = €4506% / 12
N = payments
3.NPV & IRRMedium
A machine costs €40 000 today, lasts 6 years and costs €2 000 per year to run (paid at the end of each year). The cost of capital is 8%. Find its equivalent annual cost EAC.
C0 = €40 000M = €2 000n = 6r = 8%
EAC = €
4.NPV & IRRHard
A project costs €10 000 and returns €7 000, €4 500, €7 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −10 000CF1 = 7 000CF2 = 4 500CF3 = 7 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 4% annual coupon in two half-yearly payments and matures in 3 years. The yield to maturity is 2.5% per year. Find the price P of the bond.
F = €1 000coupon = 4%YTM = 2.5%n = 3
P = €
6.Bond pricing & yieldHard
A bond trades at €945 with a yield to maturity of 5% (annual compounding) and a Macaulay duration of 5.6 years. Use the modified duration to estimate the new price P if the yield rises by 50 basis points.
P0 = €945y = 5%D = 5.6Δy = +50 bp
P = €
7.CAPM & WACCMedium
A company has equity worth €5 000 000 and debt worth €1 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 5% and the tax rate is 21%. Find the WACC.
A company has 4 000 000 shares trading at €34.50; the book value of its equity is €82 800 000. It just paid a dividend of €2.00 per share, and dividends grow by 2% per year. Its debt has a market value of €69 000 000 and a pre-tax cost of 4.5%. The tax rate is 28%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 39, 11, 20, 57, 33.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 10. What proportion P of values is above 75?
μ = 60σ = 10x = 75
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3p(x): 0.15, 0.25, 0.25, 0.35
σ =
12.ProbabilityHard
12 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?
n = 12p = 0.4
P =
13.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 12 of them exceeds 298.3 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 12> 298.3 kg
P =
14.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.04 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 11.4. What is the margin of error E₂ of a 90% interval from the same data?
99%n₁ = 36E₁ = 11.490%
E₂ =
16.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. 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 = 0.5E = 0.0499%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 120; the population standard deviation is known to be 10. A random sample of 25 observations has a mean of 116.28. Test H₀: μ = 120 against H₁: μ < 120. Find the p-value p.
μ₀ = 120σ = 10n = 25x̄ = 116.28H₁: 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 2.5 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 = temperature in °C and y = ice creams sold. 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. A new observation (4.5, 18) is made. Find its residual e = y − ŷ.
n = 6(4.5, 18)
e =
3Solutions
1.Time value of money
Answer: EAR = 17.23 %
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 = 0.16/12 = 0.01333
Compound over one year
EAR = (1 + 0.01333)12 − 1 = 1.172 − 1 = 17.23%
Answer
EAR = 17.23%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: N = 24 payments
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 = 0.005
(1 + i)−N = 1 − 10 000·0.005/450 = 0.8889
Solve with logarithms
N = −ln(0.8889)/ln(1.005) = 23.62
Round up to whole payments
23 full payments leave a small balance, which the last, smaller payment clears.
N = 24
Answer
N = 24 payments
3.NPV & IRR
Answer: EAC = 10653 €
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 − 1.08−6)/0.08 = 4.623
Annualised purchase price
40 000/4.623 = 8 652.62
EAC
EAC = 8 652.62 + 2 000 = €10 652.62
Answer
EAC = €10 652.62
4.NPV & IRR
Answer: IRR = 40.23 %
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%) = 29.15
NPV(45%) = −571.98
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 40 + 5·29.15/(29.15 − (−571.98)) = 40.24%
exact: NPV(40.23%) = 0 → IRR = 40.23%
Answer
IRR = 40.23%
5.Bond pricing & yield
Answer: P = 1043 €
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 = €20.00, i = 0.0125, N = 6
Present value of the coupons
€20.00·(1 − 0.9282)/0.0125 = €114.92
Present value of the face value
€1 000·0.9282 = €928.17
Price
P = €1 043.10
Answer
P = €1 043.10
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: P = 919.8 €
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 = 5.6/1.05 = 5.333
Estimated price change
%ΔP ≈ −5.333·(0.005) = −2.667%
P ≈ 945·(1 − 0.02667) = €919.80
Answer
P = €919.80
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: WACC = 10.66 %
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 = €6 000 000, E/V = 0.8333, D/V = 0.1667
WACC
WACC = 0.8333·12% + 0.1667·5%·(1 − 0.21) = 10.66%
Answer
WACC = 10.66%
8.CAPM & WACC
Answer: WACC = 6.355 %
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 1 − 0.9332 = 0.06681
Answer
P = 0.06681
11.Probability
Answer: σ = 1.077
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 + 1·0.25 + 2·0.25 + 3·0.35 = 1.8
Mean of the squares
E(X²) = 0·0.15 + 1·0.25 + 4·0.25 + 9·0.35 = 4.4
Variance
Var(X) = 4.4 − 1.8² = 1.16
σ = √1.16 = 1.077
Answer
σ = 1.077
12.Probability
Answer: P = 0.01959
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) = 0.612 = 0.002177
P(X = 1) = 12·0.4·0.611 = 0.01741
Add them
P = 0.002177 + 0.01741 = 0.01959
Answer
P = 0.01959
13.Normal distribution
Answer: P = 0.8365
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 = 300 kg, σT = 0.5·√12 = 1.732 kg
Standardise
z = (298.3 − 300)/1.732 = −0.98
Probability
P = 1 − Φ(−0.98) = 0.8365
Answer
P = 0.8365
14.Normal distribution
Answer: P = 0.9452
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 difference
μ = 25.04 − 25 = 0.04 mm, σ = 0.025 mm
Standardise
z = (0 − 0.04)/0.025 = −1.6
Probability
P = 1 − Φ(−1.6) = 0.9452
Answer
P = 0.9452
15.Confidence intervals
Answer: E₂ = 7.28
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 = (1.645/2.576)·√(36/36) = 0.6386
New margin
E₂ = 11.4·0.6386 = 7.28
Answer
E₂ = 7.28
16.Confidence intervals
Answer: n = 1037
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 = 2.576²·0.5·0.5/0.04² = 1037
Round up
n = 1037
Answer
n = 1037
17.Hypothesis tests & confidence intervals
Answer: p = 0.03144
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 = 10/√25 = 2
z = (116.28 − 120)/2 = −1.86
Tail area
Φ(1.86) = 0.9686
1 − Φ(1.86) = 0.0314
p-value
p = 0.03144
Answer
p = 0.03144
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: n = 479
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)·22/2.5)² = 479
Round up
n = 479
Answer
n = 479
19.Linear regression & correlation
Answer: r² = 0.9746
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 = 135/6 = 22.5
ȳ = 438/6 = 73
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 612
Syy = Σ(y − ȳ)² = 2440
Coefficient of determination
r² = 612²/(157.5·2440) = 0.9746
Answer
r² = 0.9746
About 97.46% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = 2.286
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.