A credit card charges a nominal interest rate of 25% per year, compounded semi-annually. Find the effective annual rate EAR in percent.
r = 25%m = 2
EAR = %
2.Time value of moneyHard
You want €20 000 in 18 years. You save the same amount at the end of every month in an account paying 5.4% per year, compounded monthly. Find the monthly deposit A.
FV = €20 0005.4% / 12N = 216
A = €
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 4: €10 000, €7 000, €10 000, €5 000. The cost of capital is 5%. Find the net present value NPV.
A project costs €20 000 and returns €14 500, €9 000, €9 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 14 500CF2 = 9 000CF3 = 9 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 9% per year. Find the price P of the bond.
F = €1 000coupon = 6%YTM = 9%n = 7
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 4 years. The yield to maturity is 3.5%. Find the Macaulay duration D in years.
F = €1 000coupon = 5%n = 4YTM = 3.5%
D = years
7.CAPM & WACCMedium
A company has equity worth €12 000 000 and debt worth €4 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 5% and the tax rate is 25%. Find the WACC.
A company has 5 000 000 shares at €21.50 each and 10 000 bonds trading at €960 each. The pre-tax cost of debt is 4% and the tax rate is 28%. The share’s beta is 1.1, the risk-free rate 2.5% and the market return 8%. Find the WACC.
β = 1.1rf = 2.5%rm = 8%rd = 4%T = 28%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 10, 30, 28, 24, 55, 26.
n = 6
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 12. Find the z-score z of a score of 88.
μ = 70σ = 12x = 88
z =
11.ProbabilityMedium
A bag holds 6 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
6 / 3
P =
12.ProbabilityHard
A game costs €3 to play. You win €12 with probability 0.2 and €150 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€12: p = 0.2€150: p = 0.01
E(G) = €
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7967. Find z to two decimals.
A = 0.7967
z =
14.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 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 = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
A sample of 30 items has mean 113 and standard deviation s = 11.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 30x̄ = 113s = 11.590%t(29) = 1.699
upper limit =
16.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.03?
p = 0.5E = 0.0390%
n =
17.Hypothesis tests & confidence intervalsMedium
12 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −0.53 and the standard deviation of the differences is 2.3. Find the paired test statistic t for H₀: μd = 0.
n = 12d̄ = −0.53sd = 2.3
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 5. Find the smallest sample size n.
σ = 5E = 199%
n =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. 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 = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (3, 15) is made. Find its residual e = y − ŷ.
n = 6(3, 15)
e =
3Solutions
1.Time value of money
Answer: EAR = 26.56 %
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 = 0.25/2 = 0.125
Compound over one year
EAR = (1 + 0.125)2 − 1 = 1.266 − 1 = 26.56%
Answer
EAR = 26.56%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: A = 54.96 €
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.
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 = −20 000
CF1/(1 + 0.05)1 = 10 000/1.05 = 9 523.81
CF2/(1 + 0.05)2 = 7 000/1.103 = 6 349.21
CF3/(1 + 0.05)3 = 10 000/1.158 = 8 638.38
CF4/(1 + 0.05)4 = 5 000/1.216 = 4 113.51
Add them up
NPV = €8 624.90
Answer
NPV = €8 624.90
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 33.11 %
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(30%) = 803.37
NPV(35%) = −459.79
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 30 + 5·803.37/(803.37 − (−459.79)) = 33.18%
exact: NPV(33.11%) = 0 → IRR = 33.11%
Answer
IRR = 33.11%
5.Bond pricing & yield
Answer: P = 846.7 €
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 = €30.00, i = 0.045, N = 14
Present value of the coupons
€30.00·(1 − 0.54)/0.045 = €306.68
Present value of the face value
€1 000·0.54 = €539.97
Price
P = €846.66
Answer
P = €846.66
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: D = 3.731 years
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.
Present value of each cash flow
t = 1: 50/1.035 = 48.31, t·PV = 48.31
t = 2: 50/1.071 = 46.68, t·PV = 93.35
t = 3: 50/1.109 = 45.10, t·PV = 135.29
t = 4: 1050/1.148 = 915.01, t·PV = 3 660.06
Duration
P = 1 055.10
D = 3 937.01/1 055.10 = 3.731
Answer
D = 3.731 years
7.CAPM & WACC
Answer: WACC = 8.438 %
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 = €16 000 000, E/V = 0.75, D/V = 0.25
WACC
WACC = 0.75·10% + 0.25·5%·(1 − 0.25) = 8.438%
Answer
WACC = 8.438%
8.CAPM & WACC
Answer: WACC = 8.085 %
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 = (88 − 70)/12 = 1.5
Answer
z = 1.5
11.Probability
Answer: P = 0.4167
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.
First and second draw
P = 6/9 · 5/8 = 30/72 = 0.4167
Answer
P = 0.4167
12.Probability
Answer: E(G) = 0.9 €
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: 9 (p = 0.2), 147 (p = 0.01), −3 (p = 0.79)
Expected value
E(G) = 9·0.2 + 147·0.01 − 3·0.79 = 0.9 €
Answer
E(G) = 0.9 €
A positive value means you win money on average.
13.Normal distribution
Answer: z = −0.83
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 − 0.7967 = 0.2033
Read the z-table backwards
Find 0.2033 inside the table and read off z.
z = −0.83
Answer
z = −0.83
14.Normal distribution
Answer: P = 0.9207
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 = 25.03 − 25 = 0.03 mm, σ = 0.02121 mm
Standardise
z = (0 − 0.03)/0.02121 = −1.41
Probability
P = 1 − Φ(−1.41) = 0.9207
Answer
P = 0.9207
15.Confidence intervals
Answer: upper limit = 116.6
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 = 29, t = 1.699
Margin of error
E = 1.699·11.5/√30 = 3.567
Interval
113 ± 3.567 → [109.4, 116.6]
Answer
upper limit = 116.6
16.Confidence intervals
Answer: n = 752
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 = 1.645²·0.5·0.5/0.03² = 751.7
Round up
n = 752
Answer
n = 752
17.Hypothesis tests & confidence intervals
Answer: t = −0.7982
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).
Standard error of the mean difference
SE = 2.3/√12 = 0.664
Test statistic
t = −0.53/0.664 = −0.7982
Answer
t = −0.7982
With df = 11, the two-sided 5% critical value is 2.201; |t| is smaller, so H₀ is not rejected.
18.Hypothesis tests & confidence intervals
Answer: n = 166
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).
Solve for n
n = (2.576·5/1)² = 165.9
Round up
n = 166
Answer
n = 166
19.Linear regression & correlation
Answer: r² = 0.9902
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 = 42/6 = 7
ȳ = 82/6 = 13.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 160
Syy = Σ(y − ȳ)² = 369.3
Coefficient of determination
r² = 160²/(70·369.3) = 0.9902
Answer
r² = 0.9902
About 99.02% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = 3.929
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 = 42/6 = 7
ȳ = 123/6 = 20.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 165
Regression line
b = 165/70 = 2.357
a = 20.5 − 2.357·7 = 4
ŷ = 4 + 2.357·x
Predicted value
ŷ = 4 + 2.357·3 = 11.07
Residual
e = 15 − 11.07 = 3.929
Answer
e = 3.929
The point lies above the line.
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