You invest €2 000 at a nominal rate of 2% per year, compounded quarterly. Find the value FV of the investment after 6 years.
PV = €2 0002% / 4n = 6
FV = €
2.Time value of moneyHard
A loan of €8 000 is repaid with equal monthly payments over 5 years. The interest rate is 3.6% per year, compounded monthly. Find the monthly payment A.
P = €8 0003.6% / 12N = 60
A = €
3.NPV & IRRMedium
A project costs €30 000 today and returns €9 000 at the end of each year for 6 years. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €50 000 and returns €28 000, €26 000, €18 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −50 000CF1 = 28 000CF2 = 26 000CF3 = 18 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 5 years. The yield to maturity is 3.5% per year. Find the price P of the bond.
F = €1 000coupon = 2%YTM = 3.5%n = 5
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 8% matures in 4 years and sells for €950. Find its yield to maturity YTM in percent.
F = €1 000coupon = 8%n = 4P = €950
YTM = %
7.CAPM & WACCMedium
A company has equity worth €12 000 000 and debt worth €2 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 3.5% and the tax rate is 25%. Find the WACC.
A company has 5 000 000 shares at €56.50 each and 20 000 bonds trading at €980 each. The pre-tax cost of debt is 4% and the tax rate is 21%. The share’s beta is 1.5, the risk-free rate 2.5% and the market return 9.5%. Find the WACC.
β = 1.5rf = 2.5%rm = 9.5%rd = 4%T = 21%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 28, 51, 19, 32, 46, 12, 56.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is below 440?
μ = 500σ = 50x = 440
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.15, 0.35, 0.3, 0.2
σ =
12.ProbabilityHard
A game costs €3 to play. You win €6 with probability 0.3 and €30 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€6: p = 0.3€30: 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 480 and 540 h? Give P as a decimal.
μ = 500σ = 40x1 = 480x2 = 540
P =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 60% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g60%
x = g
15.Confidence intervalsMedium
A sample of 11 items has mean 44.5 and standard deviation s = 10. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
n = 11x̄ = 44.5s = 1099%t(10) = 3.169
lower limit =
16.Confidence intervalsHard
A pilot study suggests that about 0.15 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.03?
p ≈ 0.15E = 0.0399%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 77.1 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 77.1s = 8n = 64
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 200 against H₁: μ > 200 is done at α = 0.10 with a sample of 64; the standard deviation is known to be 24. In reality the mean is 205.4. Find the probability of a Type II error β in percent.
μ₀ = 200μ = 205.4σ = 24n = 64α = 0.10
β = %
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.
n = 6
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 = 2254 €
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 = 0.036000000000000004/12 = 0.003
N = 12·5 = 60
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 = 0.8355
A = 8 000·0.003/(1 − 0.8355) = €145.89
Answer
A = €145.89
3.NPV & IRR
Answer: DPB = 4.522 years
Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.
Discounted and cumulative cash flows
year 1: 9 000/1.12 = 8 035.71 → Σ = 8 035.71
year 2: 9 000/1.254 = 7 174.74 → Σ = 15 210.46
year 3: 9 000/1.405 = 6 406.02 → Σ = 21 616.48
year 4: 9 000/1.574 = 5 719.66 → Σ = 27 336.14
year 5: 9 000/1.762 = 5 106.84 → Σ = 32 442.99
Fraction of the last year
DPB = 4 + 2 663.86/5 106.84 = 4.522
Answer
DPB = 4.522 years
The discounted payback is always longer than the simple payback (3.333 years).
4.NPV & IRR
Answer: IRR = 22.46 %
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(20%) = 1 805.56
NPV(25%) = −1 744.00
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.
Per period
C = €10.00, i = 0.0175, N = 10
Present value of the coupons
€10.00·(1 − 0.8407)/0.0175 = €91.01
Present value of the face value
€1 000·0.8407 = €840.73
Price
P = €931.74
Answer
P = €931.74
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 9.562 %
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
€950 = €80·(1 − (1 + y)⁻⁴)/y + €1 000/(1 + y)⁴
Solve numerically
P(9%) = €967.60, P(10%) = €936.60
YTM = 9.562%
Answer
YTM = 9.562%
7.CAPM & WACC
Answer: WACC = 12.38 %
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).
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).
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 = −1.2
Standard normal table
Φ(−1.2) = 0.1151
P = 0.1151
Answer
P = 0.1151
11.Probability
Answer: σ = 9.734
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.
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: 3 (p = 0.3), 27 (p = 0.05), −3 (p = 0.65)
Expected value
E(G) = 3·0.3 + 27·0.05 − 3·0.65 = 0.3 €
Answer
E(G) = 0.3 €
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 = (480 − 500)/40 = −0.5
z₂ = (540 − 500)/40 = 1
Area between
P = Φ(1) − Φ(−0.5) = 0.8413 − 0.3085 = 0.5328
Answer
P = 0.5328
14.Normal distribution
Answer: x = 248.8 g
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) = 0.4
z from the table
z ≈ −0.25
Back to the original units
x = 250 − 0.25·5 = 248.8 g
Answer
x = 248.8 g
15.Confidence intervals
Answer: lower limit = 34.95
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 = 10, t = 3.169
Margin of error
E = 3.169·10/√11 = 9.555
Interval
44.5 ± 9.555 → [34.95, 54.05]
Answer
lower limit = 34.95
16.Confidence intervals
Answer: n = 941
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 2.576²·0.15·0.85/0.03² = 940.1
Round up
n = 941
Answer
n = 941
17.Hypothesis tests & confidence intervals
Answer: t = 2.1
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 = 8/√64 = 8/8 = 1
Test statistic
t = (77.1 − 75)/1 = 2.1
Answer
t = 2.1
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: β = 30.22 %
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 = 24/√64 = 3
x̄c = 200 + 1.282·3 = 203.8
z under the true mean
z = (203.8 − 205.4)/3 = −0.518
Type II error
power = P(Z > −0.518) = 69.78%
β = 100% − 69.78% = 30.22%
Answer
β = 30.22%
19.Linear regression & correlation
Answer: r² = 0.9031
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 = 21/6 = 3.5
ȳ = 143/6 = 23.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −32.5
Syy = Σ(y − ȳ)² = 66.83
Coefficient of determination
r² = −32.5²/(17.5·66.83) = 0.9031
Answer
r² = 0.9031
About 90.31% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 12.2
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.