You invest €8 000 at a nominal rate of 4.8% per year, compounded monthly. Find the value FV of the investment after 6 years.
PV = €8 0004.8% / 12n = 6
FV = €
2.Time value of moneyHard
A mortgage of €150 000 is repaid with equal monthly payments over 30 years at 2.4% per year, compounded monthly. How much is still owed just after the payment at the end of year 10? Find the outstanding balance B.
P = €150 0002.4% / 12N = 360k = 120
B = €
3.NPV & IRRMedium
A project costs €250 000 today. It generates €27 000 at the end of the first year, and the cash flow then grows by 3% per year forever. The cost of capital is 12%. Find the net present value NPV.
C0 = €250 000CF1 = €27 000g = 3%r = 12%
NPV = €
4.NPV & IRRHard
A project costs €25 000 and returns €9 000, €11 000, €8 500, €9 000 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 8%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 pays an annual coupon of 4.5% and matures in 10 years. Its yield to maturity changes from 7% to 8%. By how much ΔP does the bond price change?
F = €1 000coupon = 4.5%n = 10y1 = 7%y2 = 8%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 5 years. The yield to maturity is 2.5%. Find the Macaulay duration D in years.
F = €1 000coupon = 5%n = 5YTM = 2.5%
D = years
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €5 000 000, preferred shares worth €2 000 000 and debt worth €5 000 000. The cost of equity is 13%, the cost of preferred stock 8% and the pre-tax cost of debt 5%. The tax rate is 20%. Find the WACC.
A company has 2 500 000 shares at €49.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 4.5% and the tax rate is 30%. The share’s beta is 0.6, the risk-free rate 1% and the market return 6.5%. Find the WACC.
β = 0.6rf = 1%rm = 6.5%rd = 4.5%T = 30%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 24, 27, 39, 55, 56, 57.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is below 160.4?
μ = 170σ = 12x = 160.4
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2p(x): 0.3, 0.35, 0.35
σ =
12.ProbabilityHard
A game costs €5 to play. You win €25 with probability 0.25 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€25: p = 0.25€100: 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 540 h? Give P as a decimal.
μ = 500σ = 40x1 = 460x2 = 540
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 9 items has mean 64 and standard deviation s = 14. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 9x̄ = 64s = 1490%t(8) = 1.86
lower limit =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 6 plots, mean yield 79.5 kg, standard deviation 8 kg. Plot group 2: 16 plots, mean 86.5 kg, standard deviation 4.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.
Two production lines make the same part. Line 1: a sample of 30 parts has mean length 60 mm, and the line's known standard deviation is 16 mm. Line 2: a sample of 40 parts has mean 54.1 mm, known standard deviation 3 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 10. Find the smallest sample size n.
σ = 10E = 0.599%
n =
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 = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 87?
n = 5ŷ = 87
x =
3Solutions
1.Time value of money
Answer: FV = 10664 €
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 payment
i = 0.002, N = 360
A = 150 000·0.002/(1 − (1 + 0.002)−360) = €584.91
Present value of the remaining payments
Whatever is still owed must be exactly paid off by the remaining payments.
N − k = 360 − 120 = 240
B = 584.9·(1 − (1 + 0.002)−240)/0.002 = €111 402.32
Answer
B = €111 402.32
3.NPV & IRR
Answer: NPV = 50000 €
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 = 27 000/(0.12 − 0.03) = 300 000.00
NPV
NPV = 300 000.00 − 250 000 = €50 000.00
Answer
NPV = €50 000.00
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: MIRR = 14.08 %
Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.
Terminal value of the inflows
9 000·1.083 = 11 337.41
11 000·1.082 = 12 830.40
8 500·1.081 = 9 180.00
9 000·1.080 = 9 000.00
TV = 42 347.81
MIRR
MIRR = (42 347.81/25 000)1/4 − 1 = 14.08%
Answer
MIRR = 14.08%
5.Bond pricing & yield
Answer: ΔP = −59.26 €
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.
Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.
Present value of each cash flow
t = 1: 50/1.025 = 48.78, t·PV = 48.78
t = 2: 50/1.051 = 47.59, t·PV = 95.18
t = 3: 50/1.077 = 46.43, t·PV = 139.29
t = 4: 50/1.104 = 45.30, t·PV = 181.19
t = 5: 1050/1.131 = 928.05, t·PV = 4 640.24
Duration
P = 1 116.15
D = 5 104.68/1 116.15 = 4.573
Answer
D = 4.573 years
7.CAPM & WACC
Answer: WACC = 8.417 %
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 = −0.8
Standard normal table
Φ(−0.8) = 0.2119
P = 0.2119
Answer
P = 0.2119
11.Probability
Answer: σ = 0.8047
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.
Mean
μ = 0·0.3 + 1·0.35 + 2·0.35 = 1.05
Mean of the squares
E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75
Variance
Var(X) = 1.75 − 1.05² = 0.6475
σ = √0.6475 = 0.8047
Answer
σ = 0.8047
12.Probability
Answer: E(G) = 6.25 €
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: 20 (p = 0.25), 95 (p = 0.05), −5 (p = 0.7)
Expected value
E(G) = 20·0.25 + 95·0.05 − 5·0.7 = 6.25 €
Answer
E(G) = 6.25 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.6827
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 = (540 − 500)/40 = 1
Area between
P = Φ(1) − Φ(−1) = 0.8413 − 0.1587 = 0.6827
Answer
P = 0.6827
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: lower limit = 55.32
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 = 8, t = 1.86
Margin of error
E = 1.86·14/√9 = 8.68
Interval
64 ± 8.68 → [55.32, 72.68]
Answer
lower limit = 55.32
16.Confidence intervals
Answer: upper limit = −1.423 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 = (5·8² + 15·4.5²)/20 = 31.19
sp = 5.585
Margin of error
E = 2.086·5.585·√(1/6 + 1/16) = 5.577
Interval
−7 ± 5.577 → [−12.58, −1.423]
Answer
upper limit = −1.423 kg
17.Hypothesis tests & confidence intervals
Answer: z = 1.994
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 = √(16²/30 + 3²/40) = √8.758 = 2.959
Test statistic
z = (60 − 54.1)/2.959 = 1.994
Answer
z = 1.994
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 2655
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 = (2.576·10/0.5)² = 2654
Round up
n = 2655
Answer
n = 2655
19.Linear regression & correlation
Answer: r² = 0.9059
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 = 106/6 = 17.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −47
Syy = Σ(y − ȳ)² = 139.3
Coefficient of determination
r² = −47²/(17.5·139.3) = 0.9059
Answer
r² = 0.9059
About 90.59% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: x = 19.55
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.