You invest €20 000 at a nominal rate of 5% per year, compounded monthly. Find the value FV of the investment after 6 years.
PV = €20 0005% / 12n = 6
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
An investment of €20 000 gives these cash flows at the end of years 1 to 4: €6 000, €8 000, €7 000, €9 000. The cost of capital is 10%. Find the net present value NPV.
A project costs €10 000 and returns €5 000, €4 000, €5 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −10 000CF1 = 5 000CF2 = 4 000CF3 = 5 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 3 years and pays annual coupons. It sells for €895 when the yield to maturity is 7%. Find the coupon rate c in percent.
F = €1 000n = 3P = €895YTM = 7%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 4% matures in 6 years and sells for €1 110. Find its yield to maturity YTM in percent.
F = €1 000coupon = 4%n = 6P = €1 110
YTM = %
7.CAPM & WACCMedium
A company has equity worth €5 000 000 and debt worth €2 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 6% and the tax rate is 28%. Find the WACC.
A company's shares have a beta of 1.2 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 1. The tax rate is 30%, the risk-free rate 3% and the market risk premium 4.5%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 34, 10, 10, 19, 51, 54, 31.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 20. Find the z-score z of a score of 190.
μ = 170σ = 20x = 190
z =
11.ProbabilityMedium
A bag holds 4 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
4 / 4
P =
12.ProbabilityHard
A game costs €3 to play. You win €9 with probability 0.1 and €60 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€9: p = 0.1€60: p = 0.05
E(G) = €
13.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 460 and 520 h? Give P as a decimal.
μ = 500σ = 40x1 = 460x2 = 520
P =
14.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points95%
x = points
15.Confidence intervalsMedium
A sample of 8 items has mean 71.5 and standard deviation s = 5.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 8x̄ = 71.5s = 5.590%t(7) = 1.895
upper limit =
16.Confidence intervalsHard
The diameters of 30 machined shafts have a sample standard deviation of 6.8 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.
n = 30s = 6.8 µmχ² = 16.047, 45.722
σ = µm
17.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 48.5 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 48.5s = 10n = 36
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 8. Find the smallest sample size n.
σ = 8E = 2.590%
n =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.
n = 5x = 2.5
ŷ =
3Solutions
1.Time value of money
Answer: FV = 26980 €
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 = 0.03/12 = 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 = 3472 €
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 + 0.1)1 = 6 000/1.1 = 5 454.55
CF2/(1 + 0.1)2 = 8 000/1.21 = 6 611.57
CF3/(1 + 0.1)3 = 7 000/1.331 = 5 259.20
CF4/(1 + 0.1)4 = 9 000/1.464 = 6 147.12
Add them up
NPV = €3 472.44
Answer
NPV = €3 472.44
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 18.96 %
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%) = 659.98
NPV(20%) = −162.04
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 15 + 5·659.98/(659.98 − (−162.04)) = 19.01%
exact: NPV(18.96%) = 0 → IRR = 18.96%
Answer
IRR = 18.96%
5.Bond pricing & yield
Answer: c = 2.999 %
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.
Present value of the face value
1000·(1 + 0.07)−3 = €816.30
Coupon
C = (895 − 816.30)/2.624 = €29.99
Coupon rate
c = 29.99/1000 = 2.999%
Answer
c = 2.999%
6.Bond pricing & yield
Answer: YTM = 2.034 %
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
€1 110 = €40·(1 − (1 + y)⁻⁶)/y + €1 000/(1 + y)⁶
Solve numerically
P(2%) = €1 112.03, P(3%) = €1 054.17
YTM = 2.034%
Answer
YTM = 2.034%
7.CAPM & WACC
Answer: WACC = 11.23 %
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 = €7 000 000, E/V = 0.7143, D/V = 0.2857
WACC
WACC = 0.7143·14% + 0.2857·6%·(1 − 0.28) = 11.23%
Answer
WACC = 11.23%
8.CAPM & WACC
Answer: re = 11.05 %
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 = 1.2/(1 + 0.7·0.2) = 1.053
Relever at the new D/E
βL = 1.053·(1 + 0.7·1) = 1.789
Cost of equity (CAPM)
re = 3% + 1.789·4.5% = 11.05%
Answer
re = 11.05%
9.Descriptive statistics
Answer: s = 18.05
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 = (190 − 170)/20 = 1
Answer
z = 1
11.Probability
Answer: P = 0.2143
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 = 4/8 · 3/7 = 12/56 = 0.2143
Answer
P = 0.2143
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: 6 (p = 0.1), 57 (p = 0.05), −3 (p = 0.85)
Expected value
E(G) = 6·0.1 + 57·0.05 − 3·0.85 = 0.9 €
Answer
E(G) = 0.9 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.5328
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 = (460 − 500)/40 = −1
z₂ = (520 − 500)/40 = 0.5
Area between
P = Φ(0.5) − Φ(−1) = 0.6915 − 0.1587 = 0.5328
Answer
P = 0.5328
14.Normal distribution
Answer: x = 48.6 points
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
Φ(z) = 0.05
z from the table
z ≈ −1.64
Back to the original units
x = 65 − 1.64·10 = 48.6 points
Answer
x = 48.6 points
15.Confidence intervals
Answer: upper limit = 75.18
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 = 7, t = 1.895
Margin of error
E = 1.895·5.5/√8 = 3.685
Interval
71.5 ± 3.685 → [67.82, 75.18]
Answer
upper limit = 75.18
16.Confidence intervals
Answer: σ = 9.141 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 29·6.8² = 1341
Limits for σ²
1341/45.722 = 29.33
1341/16.047 = 83.56
Limit for σ
σ = √83.56 = 9.141 µm
Answer
σ = 9.141 µm
95% interval for σ: [5.416, 9.141] µm.
17.Hypothesis tests & confidence intervals
Answer: t = −0.9
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/√36 = 10/6 = 1.667
Test statistic
t = (48.5 − 50)/1.667 = −0.9
Answer
t = −0.9
t0.025; df = 35 = 2.03
|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 28
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 = (1.645·8/2.5)² = 27.71
Round up
n = 28
Answer
n = 28
19.Linear regression & correlation
Answer: r = 0.9951
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 = 4488 − 48·504/6 = 456
Sxx = 454 − 48²/6 = 70
Syy = 45336 − 504²/6 = 3000
Correlation coefficient
r = 456/√(70·3000) = 0.9951
Answer
r = 0.9951
20.Linear regression & correlation
Answer: ŷ = 57.15
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 = 15/5 = 3
ȳ = 296/5 = 59.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 41
Regression line
b = 41/10 = 4.1
a = 59.2 − 4.1·3 = 46.9
ŷ = 46.9 + 4.1·x
Prediction
ŷ = 46.9 + 4.1·2.5 = 57.15
Answer
ŷ = 57.15
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