You invest €2 000 at a nominal rate of 5% per year, compounded quarterly. Find the value FV of the investment after 8 years.
PV = €2 0005% / 4n = 8
FV = €
2.Time value of moneyHard
A loan of €25 000 is repaid with equal monthly payments over 4 years. The interest rate is 6% per year, compounded monthly. Find the monthly payment A.
P = €25 0006% / 12N = 48
A = €
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 3: €7 000, €4 000, €8 000. The cost of capital is 8%. Find the net present value NPV.
Project A costs €10 000 and returns €5 500, €4 000, €2 500 at the end of years 1 to 3. Project B costs €12 000 and returns €3 000, €5 000, €7 000. Find the crossover rate r* at which both projects have the same NPV, in percent.
You buy a bond with 11 years to maturity (face value €1 000, annual coupon 2%) when its yield to maturity is 3.5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 4%. Find your holding period return HPR in percent.
F = €1 000coupon = 2%n = 11y0 = 3.5%y1 = 4%
HPR = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 6 years. It sells for €1 134. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 6% (2×)n = 6P = €1 134
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €12 000 000, preferred shares worth €500 000 and debt worth €2 000 000. The cost of equity is 14%, the cost of preferred stock 6.5% and the pre-tax cost of debt 5%. The tax rate is 21%. Find the WACC.
A company's business has an unlevered (asset) beta of 1.1. It plans a debt-to-equity ratio of 0.75. The tax rate is 21%, the risk-free rate is 3% and the market risk premium is 4.5%. Find the cost of equity re at the new capital structure.
βU = 1.1D/E = 0.75T = 21%rf = 3%rm − rf = 4.5%
re = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 50, 53, 15, 60, 22, 24.
n = 6
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 8. Find the z-score z of a score of 60.
μ = 50σ = 8x = 60
z =
11.ProbabilityMedium
In how many ways N can a committee of 4 people be chosen from a group of 9? The order does not matter.
n = 9k = 4
N =
12.ProbabilityHard
A game costs €1 to play. You win €3 with probability 0.1 and €10 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€3: p = 0.1€10: p = 0.01
E(G) = €
13.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 65 and 73 points? Give P as a decimal.
μ = 65σ = 10x1 = 65x2 = 73
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 sample of 15 items has mean 58.5 and standard deviation s = 12. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 15x̄ = 58.5s = 1299%t(14) = 2.977
upper limit =
16.Confidence intervalsHard
The population standard deviation is 31. 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 = 31E = 1.590%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 78 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 78s = 5n = 16
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 50 against H₁: μ < 50 is done at α = 0.05 with a sample of 64; the standard deviation is known to be 12. In reality the mean is 46.85. Find the power 1 − β of the test in percent.
μ₀ = 50μ = 46.85σ = 12n = 64α = 0.05
1 − β = %
19.Linear regression & correlationMedium
The table gives 5 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 = 5
r² =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.
n = 5x = 5
ŷ =
3Solutions
1.Time value of money
Answer: FV = 2976 €
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 rate and number of payments
i = 0.06/12 = 0.005
N = 12·4 = 48
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 = 0.7871
A = 25 000·0.005/(1 − 0.7871) = €587.13
Answer
A = €587.13
3.NPV & IRR
Answer: NPV = −3739 €
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 + 0.08)1 = 7 000/1.08 = 6 481.48
CF2/(1 + 0.08)2 = 4 000/1.166 = 3 429.36
CF3/(1 + 0.08)3 = 8 000/1.26 = 6 350.66
Add them up
NPV = €−3 738.51
Answer
NPV = €−3 738.51
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: r* = 9.256 %
Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.
Difference in cash flows
ΔCF0 = −12 000 − (−10 000) = −2 000
ΔCF1 = 3 000 − 5 500 = −2 500
ΔCF2 = 5 000 − 4 000 = 1 000
ΔCF3 = 7 000 − 2 500 = 4 500
IRR of the difference
NPVΔ(9%) = 22.93
NPVΔ(10%) = −65.36
r* = 9.256%
Answer
r* = 9.256%
Check: at 9.256% both NPVs are €301.94.
5.Bond pricing & yield
Answer: HPR = −0.8318 %
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.
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, N = 12
Solve numerically
i = 1.752% per half-year
YTM
YTM = 2·1.752% = 3.504%
Answer
YTM = 3.504%
7.CAPM & WACC
Answer: WACC = 12.36 %
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 = 1.1·(1 + 0.79·0.75) = 1.752
Cost of equity (CAPM)
re = 3% + 1.752·4.5% = 10.88%
Answer
re = 10.88%
9.Descriptive statistics
Answer: s = 19.14
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 = (60 − 50)/8 = 1.25
Answer
z = 1.25
11.Probability
Answer: N = 126
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.
Combinations
C(9, 4) = (9·8·7·6)/(4·3·2·1) = 126
Answer
N = 126
12.Probability
Answer: E(G) = −0.6 €
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: 2 (p = 0.1), 9 (p = 0.01), −1 (p = 0.89)
Expected value
E(G) = 2·0.1 + 9·0.01 − 1·0.89 = −0.6 €
Answer
E(G) = −0.6 €
A negative value means you lose money on average.
13.Normal distribution
Answer: P = 0.2881
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 = (65 − 65)/10 = 0
z₂ = (73 − 65)/10 = 0.8
Area between
P = Φ(0.8) − Φ(0) = 0.7881 − 0.5 = 0.2881
Answer
P = 0.2881
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: upper limit = 67.72
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 = 2.977·12/√15 = 9.224
Interval
58.5 ± 9.224 → [49.28, 67.72]
Answer
upper limit = 67.72
16.Confidence intervals
Answer: n = 1156
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 = (1.645·31/1.5)² = 1156
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1156
Answer
n = 1156
17.Hypothesis tests & confidence intervals
Answer: t = 2.4
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 = 5/√16 = 5/4 = 1.25
Test statistic
t = (78 − 75)/1.25 = 2.4
Answer
t = 2.4
t0.025; df = 15 = 2.131
|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: 1 − β = 67.54 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 12/√64 = 1.5
x̄c = 50 − 1.645·1.5 = 47.53
z under the true mean
z = (47.53 − 46.85)/1.5 = 0.455
Power
power = P(Z < 0.455) = 67.54%
1 − β = 67.54%
Answer
1 − β = 67.54%
19.Linear regression & correlation
Answer: r² = 0.7937
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 = 172/5 = 34.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 33
Syy = Σ(y − ȳ)² = 137.2
Coefficient of determination
r² = 33²/(10·137.2) = 0.7937
Answer
r² = 0.7937
About 79.37% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 16.55
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.