A credit card charges a nominal interest rate of 16% per year, compounded semi-annually. Find the effective annual rate EAR in percent.
r = 16%m = 2
EAR = %
2.Time value of moneyHard
A loan of €200 000 is repaid with equal monthly payments over 30 years. The interest rate is 2.4% per year, compounded monthly. Find the monthly payment A.
P = €200 0002.4% / 12N = 360
A = €
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 4: €34 000, €20 000, €18 000, €26 000. The cost of capital is 5%. Find the net present value NPV.
A project costs €200 000 today. Its cash flow is €36 500 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 6. The cost of capital is 8%. Find the net present value NPV.
C0 = €200 000CF1 = €36 500g = 3%n = 6r = 8%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 5 years. The yield to maturity is 7% per year. Find the price P of the bond.
F = €1 000coupon = 5%YTM = 7%n = 5
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 8% matures in 10 years and sells for €935. Find its yield to maturity YTM in percent.
F = €1 000coupon = 8%n = 10P = €935
YTM = %
7.CAPM & WACCMedium
A company has equity worth €10 000 000 and debt worth €2 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 3.5% and the tax rate is 25.8%. Find the WACC.
A company's business has an unlevered (asset) beta of 1.2. It plans a debt-to-equity ratio of 1.5. The tax rate is 20%, the risk-free rate is 2.5% and the market risk premium is 4.5%. Find the cost of equity re at the new capital structure.
βU = 1.2D/E = 1.5T = 20%rf = 2.5%rm − rf = 4.5%
re = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 60, 60, 28, 41, 13, 52.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is below 38?
μ = 50σ = 12x = 38
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3, 4p(x): 0.25, 0.2, 0.35, 0.15, 0.05
Var(X) =
12.ProbabilityHard
5 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?
n = 5p = 0.1
P =
13.Normal distributionMedium
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 113 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 113 s
P =
14.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 20% of them lie (the 20th percentile).
μ = 32σ = 6P = 20%
x = min
15.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 3.2. What is the margin of error E₂ if the same study uses 50 observations instead?
95%n₁ = 25E₁ = 3.2n₂ = 50
E₂ =
16.Confidence intervalsHard
The population standard deviation is 7. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?
σ = 7E = 1.590%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 101.7 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 101.7s = 10n = 100
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 199.8. The population standard deviation is 5. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 199.8σ = 5n = 6499%
L =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 30Σy = 76Σx² = 220Σy² = 1300Σxy = 530
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. Predict ŷ for x = 4.5.
n = 5x = 4.5
ŷ =
3Solutions
1.Time value of money
Answer: EAR = 16.64 %
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 = 0.16/2 = 0.08
Compound over one year
EAR = (1 + 0.08)2 − 1 = 1.166 − 1 = 16.64%
Answer
EAR = 16.64%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: A = 779.9 €
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 rate and number of payments
i = 0.024/12 = 0.002
N = 12·30 = 360
Payment
The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.
(1 + i)−N = 0.4871
A = 200 000·0.002/(1 − 0.4871) = €779.88
Answer
A = €779.88
3.NPV & IRR
Answer: NPV = 12461 €
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 = −75 000
CF1/(1 + 0.05)1 = 34 000/1.05 = 32 380.95
CF2/(1 + 0.05)2 = 20 000/1.103 = 18 140.59
CF3/(1 + 0.05)3 = 18 000/1.158 = 15 549.08
CF4/(1 + 0.05)4 = 26 000/1.216 = 21 390.26
Add them up
NPV = €12 460.88
Answer
NPV = €12 460.88
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: NPV = −19293 €
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.
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: P = 916.8 €
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 = €25.00, i = 0.035, N = 10
Present value of the coupons
€25.00·(1 − 0.7089)/0.035 = €207.92
Present value of the face value
€1 000·0.7089 = €708.92
Price
P = €916.83
Answer
P = €916.83
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 9.013 %
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
€935 = €80·(1 − (1 + y)⁻¹⁰)/y + €1 000/(1 + y)¹⁰
Solve numerically
P(9%) = €935.82, P(10%) = €877.11
YTM = 9.013%
Answer
YTM = 9.013%
7.CAPM & WACC
Answer: WACC = 9.6 %
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).
Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.
Levered beta
βL = 1.2·(1 + 0.8·1.5) = 2.64
Cost of equity (CAPM)
re = 2.5% + 2.64·4.5% = 14.38%
Answer
re = 14.38%
9.Descriptive statistics
Answer: s = 18.9
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
Standard normal table
Φ(−1) = 0.1587
P = 0.1587
Answer
P = 0.1587
11.Probability
Answer: Var(X) = 1.348
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.
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) = 0.95 = 0.5905
P(X = 1) = 5·0.1·0.94 = 0.3281
Add them
P = 0.5905 + 0.3281 = 0.9185
Answer
P = 0.9185
13.Normal distribution
Answer: P = 0.1056
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 = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (113 − 105)/6.403 = 1.25
Upper tail
P = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056
Answer
P = 0.1056
14.Normal distribution
Answer: x = 26.95 min
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) = 0.2 → z = −0.8416
Back to x
x = 32 + −0.8416·6 = 26.95 min
Answer
x = 26.95 min
15.Confidence intervals
Answer: E₂ = 2.263
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 = (1.96/1.96)·√(25/50) = 0.7071
New margin
E₂ = 3.2·0.7071 = 2.263
Answer
E₂ = 2.263
16.Confidence intervals
Answer: n = 59
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 = (1.645·7/1.5)² = 58.93
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 59
Answer
n = 59
17.Hypothesis tests & confidence intervals
Answer: t = 1.7
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
SE = 10/√100 = 10/10 = 1
Test statistic
t = (101.7 − 100)/1 = 1.7
Answer
t = 1.7
t0.025; df = 99 = 1.984
|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: L = 198.2
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 = 2.576·5/√64 = 1.61
Interval limit
L = 199.8 − 1.61 = 198.2
Answer
L = 198.2
19.Linear regression & correlation
Answer: r = 0.9723
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.
Centred sums
Sxy = 530 − 30·76/5 = 74
Sxx = 220 − 30²/5 = 40
Syy = 1300 − 76²/5 = 144.8
Correlation coefficient
r = 74/√(40·144.8) = 0.9723
Answer
r = 0.9723
20.Linear regression & correlation
Answer: ŷ = 34.25
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 = 20/5 = 4
ȳ = 169/5 = 33.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 9
Regression line
b = 9/10 = 0.9
a = 33.8 − 0.9·4 = 30.2
ŷ = 30.2 + 0.9·x
Prediction
ŷ = 30.2 + 0.9·4.5 = 34.25
Answer
ŷ = 34.25
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