You invest €2 000 at 4% per year with continuous compounding. Find the value FV after 18 years.
PV = €2 000r = 4%t = 18
FV = €
2.Time value of moneyHard
A loan of €150 000 is repaid with equal monthly payments over 30 years. The interest rate is 3% per year, compounded monthly. Find the monthly payment A.
P = €150 0003% / 12N = 360
A = €
3.NPV & IRRMedium
A project costs €500 000 today. It generates €36 000 at the end of the first year, and the cash flow then grows by 3% per year forever. The cost of capital is 9%. Find the net present value NPV.
C0 = €500 000CF1 = €36 000g = 3%r = 9%
NPV = €
4.NPV & IRRHard
A project costs €25 000 and returns €15 500, €17 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −25 000CF1 = 15 500CF2 = 17 000
IRR = %
5.Bond pricing & yieldMedium
You buy a bond with 8 years to maturity (face value €1 000, annual coupon 2%) when its yield to maturity is 6%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 5%. Find your holding period return HPR in percent.
F = €1 000coupon = 2%n = 8y0 = 6%y1 = 5%
HPR = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 6% matures in 4 years and sells for €920. Find its yield to maturity YTM in percent.
F = €1 000coupon = 6%n = 4P = €920
YTM = %
7.CAPM & WACCMedium
A company has equity worth €12 000 000 and debt worth €6 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 4.5% and the tax rate is 25%. Find the WACC.
A company's shares have a beta of 1.5 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 0.75. The tax rate is 25%, the risk-free rate 1% and the market risk premium 7%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 33, 25, 26, 34, 32.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 5. Find the z-score z of a score of 92.5.
μ = 100σ = 5x = 92.5
z =
11.ProbabilityMedium
A fair die is rolled 6 times. What is the probability P of getting at least one six?
n = 6
P =
12.ProbabilityHard
A game costs €1 to play. You win €2 with probability 0.25 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€2: p = 0.25€10: 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 0.1093. Find z to two decimals.
A = 0.1093
z =
14.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 20% of them lie (the 20th percentile).
μ = 100σ = 15P = 20%
x = points
15.Confidence intervalsMedium
A sample of 13 items has mean 65 and standard deviation s = 1. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 13x̄ = 65s = 190%t(12) = 1.782
upper limit =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 7 plots, mean yield 92 kg, standard deviation 6.5 kg. Plot group 2: 24 plots, mean 96 kg, standard deviation 1.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.
Two production lines make the same part. Line 1: a sample of 80 parts has mean length 126 mm, and the line's known standard deviation is 12 mm. Line 2: a sample of 30 parts has mean 131.9 mm, known standard deviation 12 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A sample of 16 measurements has mean 204.5. The population standard deviation is 15. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 204.5σ = 15n = 1695%
U =
19.Linear regression & correlationMedium
The table gives 6 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 = 6Σx = 42Σy = 151Σx² = 364Σy² = 4795Σxy = 1320
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 19.5.
n = 5x = 19.5
ŷ =
3Solutions
1.Time value of money
Answer: FV = 4109 €
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.
Exponent
r·t = 0.04·18 = 0.72
e0.72 = 2.054
Future value
FV = 2 000·2.054 = €4 108.87
Answer
FV = €4 108.87
2.Time value of money
Answer: A = 632.4 €
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 = 0.03/12 = 0.0025
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 = 0.407
A = 150 000·0.0025/(1 − 0.407) = €632.41
Answer
A = €632.41
3.NPV & IRR
Answer: NPV = 100000 €
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.
Value of the cash flows
PV = 36 000/(0.09 − 0.03) = 600 000.00
NPV
NPV = 600 000.00 − 500 000 = €100 000.00
Answer
NPV = €100 000.00
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 19.1 %
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(15%) = 1 332.70
NPV(20%) = −277.78
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
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.
Equation for the yield
€920 = €60·(1 − (1 + y)⁻⁴)/y + €1 000/(1 + y)⁴
Solve numerically
P(8%) = €933.76, P(9%) = €902.81
YTM = 8.439%
Answer
YTM = 8.439%
7.CAPM & WACC
Answer: WACC = 7.792 %
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.
Unlever the beta
βU = 1.5/(1 + 0.75·0.2) = 1.304
Relever at the new D/E
βL = 1.304·(1 + 0.75·0.75) = 2.038
Cost of equity (CAPM)
re = 1% + 2.038·7% = 15.27%
Answer
re = 15.27%
9.Descriptive statistics
Answer: s = 4.183
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.
Mean
x̄ = 150/5 = 30
Squared deviations
Σ(x − x̄)² = 3² + (−5)² + (−4)² + 4² + 2² = 70
Standard deviation
s = √(70/4) = 4.183
Answer
s = 4.183
10.Descriptive statistics
Answer: z = −1.5
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 = (92.5 − 100)/5 = −1.5
Answer
z = −1.5
11.Probability
Answer: P = 0.6651
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.
Complement
P(no six) = (5/6)6 = 0.3349
P = 1 − 0.3349 = 0.6651
Answer
P = 0.6651
12.Probability
Answer: E(G) = −0.4 €
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: 1 (p = 0.25), 9 (p = 0.01), −1 (p = 0.74)
Expected value
E(G) = 1·0.25 + 9·0.01 − 1·0.74 = −0.4 €
Answer
E(G) = −0.4 €
A negative value means you lose money on average.
13.Normal distribution
Answer: z = 1.23
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 − 0.1093 = 0.8907
Read the z-table backwards
Find 0.8907 inside the table and read off z.
z = 1.23
Answer
z = 1.23
14.Normal distribution
Answer: x = 87.38 points
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 = 100 + −0.8416·15 = 87.38 points
Answer
x = 87.38 points
15.Confidence intervals
Answer: upper limit = 65.49
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 = 12, t = 1.782
Margin of error
E = 1.782·1/√13 = 0.4942
Interval
65 ± 0.4942 → [64.51, 65.49]
Answer
upper limit = 65.49
16.Confidence intervals
Answer: E = 2.85 kg
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Pooled standard deviation
sp² = (6·6.5² + 23·1.5²)/29 = 10.53
sp = 3.244
Margin of error
E = 2.045·3.244·√(1/7 + 1/24) = 2.85
Answer
E = 2.85 kg
17.Hypothesis tests & confidence intervals
Answer: z = −2.297
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 = √(12²/80 + 12²/30) = √6.6 = 2.569
Test statistic
z = (126 − 131.9)/2.569 = −2.297
Answer
z = −2.297
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: U = 211.8
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 = 1.96·15/√16 = 7.35
Interval limit
U = 204.5 + 7.35 = 211.8
Answer
U = 211.8
19.Linear regression & correlation
Answer: r = 0.9966
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 = 1320 − 42·151/6 = 263
Sxx = 364 − 42²/6 = 70
Syy = 4795 − 151²/6 = 994.8
Correlation coefficient
r = 263/√(70·994.8) = 0.9966
Answer
r = 0.9966
20.Linear regression & correlation
Answer: ŷ = 87.7
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 = 105/5 = 21
ȳ = 474/5 = 94.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 426
Regression line
b = 426/90 = 4.733
a = 94.8 − 4.733·21 = −4.6
ŷ = −4.6 + 4.733·x
Prediction
ŷ = −4.6 + 4.733·19.5 = 87.7
Answer
ŷ = 87.7
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