You invest €3 000 at a nominal rate of 4% per year, compounded monthly. Find the value FV of the investment after 4 years.
PV = €3 0004% / 12n = 4
FV = €
2.Time value of moneyHard
You save €50 at the end of every month for 28 years. The account pays 3% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €503% / 12N = 336
FV = €
3.NPV & IRRMedium
A machine costs €30 000 today, lasts 5 years and costs €1 500 per year to run (paid at the end of each year). The cost of capital is 10%. Find its equivalent annual cost EAC.
C0 = €30 000M = €1 500n = 5r = 10%
EAC = €
4.NPV & IRRHard
A project costs €50 000 and returns €21 500, €36 000, €22 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −50 000CF1 = 21 500CF2 = 36 000CF3 = 22 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 6% and matures in 6 years. Its yield to maturity changes from 9% to 8%. By how much ΔP does the bond price change?
F = €1 000coupon = 6%n = 6y1 = 9%y2 = 8%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 4% matures in 5 years and sells for €1 045. Find its yield to maturity YTM in percent.
F = €1 000coupon = 4%n = 5P = €1 045
YTM = %
7.CAPM & WACCMedium
A company has equity worth €8 000 000 and debt worth €1 000 000. Its WACC is 7.52%, its pre-tax cost of debt is 5% and the tax rate is 25.8%. Find the cost of equity re.
A company's shares have a beta of 1.3 at its current debt-to-equity ratio of 0.4. It plans to raise its debt-to-equity ratio to 1.5. The tax rate is 28%, the risk-free rate 4% and the market risk premium 4%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 26, 42, 52, 38, 26, 28, 10.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 10. Find the z-score z of a score of 50.
μ = 70σ = 10x = 50
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3, 4p(x): 0.3, 0.35, 0.2, 0.1, 0.05
Var(X) =
12.ProbabilityHard
12 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?
n = 12p = 0.3
P =
13.Normal distributionMedium
X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 201 mm. Round z to two decimals.
μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 201 mm
P =
14.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.
μ = 178 cmσ = 7 cm90%
x = cm
15.Confidence intervalsMedium
A sample of 7 items has mean 18.5 and standard deviation s = 8. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
n = 7x̄ = 18.5s = 899%t(6) = 3.707
lower limit =
16.Confidence intervalsHard
The population standard deviation is 33. 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 2.5?
σ = 33E = 2.590%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 250; the population standard deviation is known to be 48. A random sample of 144 observations has a mean of 242.48. Test H₀: μ = 250 against H₁: μ < 250. Find the p-value p.
μ₀ = 250σ = 48n = 144x̄ = 242.48H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 95%. The standard deviation is 21. Find the minimum sample size n.
The table gives 5 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.
n = 5Σx = 20Σy = 263Σx² = 90Σy² = 14059Σxy = 1098
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 19?
n = 5ŷ = 19
x =
3Solutions
1.Time value of money
Answer: FV = 3520 €
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.
You paid in €16 800 yourself; the rest is interest.
3.NPV & IRR
Answer: EAC = 9414 €
Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.
Annuity factor
(1 − 1.1−5)/0.1 = 3.791
Annualised purchase price
30 000/3.791 = 7 913.92
EAC
EAC = 7 913.92 + 1 500 = €9 413.92
Answer
EAC = €9 413.92
4.NPV & IRR
Answer: IRR = 27.31 %
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(25%) = 1 760.00
NPV(30%) = −1 918.53
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
€1 045 = €40·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵
Solve numerically
P(3%) = €1 045.80, P(4%) = €1 000.00
YTM = 3.017%
Answer
YTM = 3.017%
7.CAPM & WACC
Answer: re = 7.996 %
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 = €9 000 000, E/V = 0.8889, D/V = 0.1111
Debt part of the WACC
0.1111·5%·(1 − 0.258) = 0.4122%
Cost of equity
re = (7.52% − 0.4122%)/0.8889 = 7.996%
Answer
re = 7.996%
8.CAPM & WACC
Answer: re = 12.4 %
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 = 1.3/(1 + 0.72·0.4) = 1.009
Relever at the new D/E
βL = 1.009·(1 + 0.72·1.5) = 2.099
Cost of equity (CAPM)
re = 4% + 2.099·4% = 12.4%
Answer
re = 12.4%
9.Descriptive statistics
Answer: s = 13.59
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 = (50 − 70)/10 = −2
Answer
z = −2
11.Probability
Answer: Var(X) = 1.287
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) = 0.712 = 0.01384
P(X = 1) = 12·0.3·0.711 = 0.07118
Add them
P = 0.01384 + 0.07118 = 0.08503
Answer
P = 0.08503
13.Normal distribution
Answer: P = 0.02275
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 = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm
Standardise
z = (201 − 200)/0.5 = 2
Upper tail
P = 1 − Φ(2) = 1 − 0.9772 = 0.02275
Answer
P = 0.02275
14.Normal distribution
Answer: x = 187 cm
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) = 0.9
z from the table
z ≈ 1.28
Back to the original units
x = 178 + 1.28·7 = 187 cm
Answer
x = 187 cm
15.Confidence intervals
Answer: lower limit = 7.291
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 = 6, t = 3.707
Margin of error
E = 3.707·8/√7 = 11.21
Interval
18.5 ± 11.21 → [7.291, 29.71]
Answer
lower limit = 7.291
16.Confidence intervals
Answer: n = 472
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 = (1.645·33/2.5)² = 471.5
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 472
Answer
n = 472
17.Hypothesis tests & confidence intervals
Answer: p = 0.03005
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.
Test statistic
SE = σ/√n = 48/√144 = 4
z = (242.48 − 250)/4 = −1.88
Tail area
Φ(1.88) = 0.9699
1 − Φ(1.88) = 0.0301
p-value
p = 0.03005
Answer
p = 0.03005
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: n = 435
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((2.326 + 1.645)·21/4)² = 434.6
Round up
n = 435
Answer
n = 435
19.Linear regression & correlation
Answer: r = 0.9693
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 = 1098 − 20·263/5 = 46
Sxx = 90 − 20²/5 = 10
Syy = 14059 − 263²/5 = 225.2
Correlation coefficient
r = 46/√(10·225.2) = 0.9693
Answer
r = 0.9693
20.Linear regression & correlation
Answer: x = 3.533
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 = 103/5 = 20.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −30
Regression line
b = −30/10 = −3
a = 20.6 − (−3)·3 = 29.6
ŷ = 29.6 − 3·x
Solve for x
x = (19 − 29.6)/(−3) = 3.533
Answer
x = 3.533
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