You need €25 000 in 8 years. An account pays 2% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €25 0002% / 4n = 8
PV = €
2.Time value of moneyHard
A loan of €15 000 is repaid with equal monthly payments over 6 years. The interest rate is 6% per year, compounded monthly. Find the monthly payment A.
P = €15 0006% / 12N = 72
A = €
3.NPV & IRRMedium
A project costs €40 000 today and returns €14 000 at the end of each year for 4 years. The cost of capital is 8%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €100 000 today. Its cash flow is €13 500 at the end of year 1 and grows by 2% per year until the last cash flow at the end of year 9. The cost of capital is 10%. Find the net present value NPV.
C0 = €100 000CF1 = €13 500g = 2%n = 9r = 10%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 10 years. The yield to maturity is 9% per year. Find the price P of the bond.
F = €1 000coupon = 8%YTM = 9%n = 10
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 8% and matures in 6 years. The yield to maturity is 6%. Find the Macaulay duration D in years.
F = €1 000coupon = 8%n = 6YTM = 6%
D = years
7.CAPM & WACCMedium
A company has equity worth €5 000 000 and debt worth €1 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 5% and the tax rate is 20%. Find the WACC.
E = €5 000 000D = €1 000 000re = 8%rd = 5%T = 20%
WACC = %
8.CAPM & WACCHard
A company has 5 000 000 shares at €13.50 each and 5 000 bonds trading at €1 000 each. The pre-tax cost of debt is 6% and the tax rate is 25%. The share’s beta is 0.6, the risk-free rate 4% and the market return 8.5%. Find the WACC.
β = 0.6rf = 4%rm = 8.5%rd = 6%T = 25%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 19, 15, 50, 13, 50, 46.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is below 460?
μ = 500σ = 50x = 460
P =
11.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 6? The order does not matter.
n = 6k = 2
N =
12.ProbabilityHard
A game costs €2 to play. You win €10 with probability 0.3 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 = €2€10: p = 0.3€100: p = 0.05
E(G) = €
13.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 242.5 and 254.25 g? Give P as a decimal.
μ = 250σ = 5x1 = 242.5x2 = 254.25
P =
14.Normal distributionHard
Hole diameters D are normal with mean 25.03 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 = 25.03 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
A sample of 9 items has mean 52 and standard deviation s = 15. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 9x̄ = 52s = 1590%t(8) = 1.86
E =
16.Confidence intervalsHard
The diameters of 20 machined shafts have a sample standard deviation of 7.4 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 19, 2.5% of the χ² distribution lies below 8.907 and 2.5% lies above 32.852.
n = 20s = 7.4 µmχ² = 8.907, 32.852
σ = µm
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 40 parts has mean length 107.5 mm, and the line's known standard deviation is 13 mm. Line 2: a sample of 80 parts has mean 109.7 mm, known standard deviation 11 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A one-sided z test of H₀: μ = 50 against H₁: μ < 50 is done at α = 0.01 with a sample of 36; the standard deviation is known to be 9. In reality the mean is 46.85. Find the probability of a Type II error β in percent.
μ₀ = 50μ = 46.85σ = 9n = 36α = 0.01
β = %
19.Linear regression & correlationMedium
The table gives 5 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 = 5
r² =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 1.5.
n = 6x = 1.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 21312 €
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.
Rate and number of periods
i = 0.02/4 = 0.005
N = 4·8 = 32
Present value
PV = FV/(1 + i)N = 25 000/1.173 = €21 312.09
Answer
PV = €21 312.09
2.Time value of money
Answer: A = 248.6 €
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 = 0.06/12 = 0.005
N = 12·6 = 72
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 = 0.6983
A = 15 000·0.005/(1 − 0.6983) = €248.59
Answer
A = €248.59
3.NPV & IRR
Answer: DPB = 3.381 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: 14 000/1.08 = 12 962.96 → Σ = 12 962.96
year 2: 14 000/1.166 = 12 002.74 → Σ = 24 965.71
year 3: 14 000/1.26 = 11 113.65 → Σ = 36 079.36
year 4: 14 000/1.36 = 10 290.42 → Σ = 46 369.78
Fraction of the last year
DPB = 3 + 3 920.64/10 290.42 = 3.381
Answer
DPB = 3.381 years
The discounted payback is always longer than the simple payback (2.857 years).
4.NPV & IRR
Answer: NPV = −16779 €
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 the growing annuity
((1.02)/(1.1))9 = 0.5068
PV = 13 500/(0.1 − 0.02)·(1 − 0.5068) = 83 221.44
NPV
NPV = 83 221.44 − 100 000 = −€16 778.56
Answer
NPV = −€16 778.56
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: P = 935 €
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 = €40.00, i = 0.045, N = 20
Present value of the coupons
€40.00·(1 − 0.4146)/0.045 = €520.32
Present value of the face value
€1 000·0.4146 = €414.64
Price
P = €934.96
Answer
P = €934.96
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: D = 5.044 years
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: 80/1.06 = 75.47, t·PV = 75.47
t = 2: 80/1.124 = 71.20, t·PV = 142.40
t = 3: 80/1.191 = 67.17, t·PV = 201.51
t = 4: 80/1.262 = 63.37, t·PV = 253.47
t = 5: 80/1.338 = 59.78, t·PV = 298.90
t = 6: 1080/1.419 = 761.36, t·PV = 4 568.14
Duration
P = 1 098.35
D = 5 539.90/1 098.35 = 5.044
Answer
D = 5.044 years
7.CAPM & WACC
Answer: WACC = 7.333 %
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 = €6 000 000, E/V = 0.8333, D/V = 0.1667
WACC
WACC = 0.8333·8% + 0.1667·5%·(1 − 0.2) = 7.333%
Answer
WACC = 7.333%
8.CAPM & WACC
Answer: WACC = 6.548 %
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: N = 15
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.
Combinations
C(6, 2) = (6·5)/(2·1) = 15
Answer
N = 15
12.Probability
Answer: E(G) = 6 €
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: 8 (p = 0.3), 98 (p = 0.05), −2 (p = 0.65)
Expected value
E(G) = 8·0.3 + 98·0.05 − 2·0.65 = 6 €
Answer
E(G) = 6 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.7355
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 = (242.5 − 250)/5 = −1.5
z₂ = (254.25 − 250)/5 = 0.85
Area between
P = Φ(0.85) − Φ(−1.5) = 0.8023 − 0.06681 = 0.7355
Answer
P = 0.7355
14.Normal distribution
Answer: P = 0.8849
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 = 25.03 − 25 = 0.03 mm, σ = 0.025 mm
Standardise
z = (0 − 0.03)/0.025 = −1.2
Probability
P = 1 − Φ(−1.2) = 0.8849
Answer
P = 0.8849
15.Confidence intervals
Answer: E = 9.3
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 = 1.86·15/√9 = 9.3
Answer
E = 9.3
16.Confidence intervals
Answer: σ = 5.628 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 = 19·7.4² = 1040
Limits for σ²
1040/32.852 = 31.67
1040/8.907 = 116.8
Limit for σ
σ = √31.67 = 5.628 µm
Answer
σ = 5.628 µm
95% interval for σ: [5.628, 10.81] µm.
17.Hypothesis tests & confidence intervals
Answer: z = −0.9185
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 = √(13²/40 + 11²/80) = √5.737 = 2.395
Test statistic
z = (107.5 − 109.7)/2.395 = −0.9185
Answer
z = −0.9185
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: β = 58.94 %
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 = 9/√36 = 1.5
x̄c = 50 − 2.326·1.5 = 46.51
z under the true mean
z = (46.51 − 46.85)/1.5 = −0.226
Type II error
power = P(Z < −0.226) = 41.06%
β = 100% − 41.06% = 58.94%
Answer
β = 58.94%
19.Linear regression & correlation
Answer: r² = 0.9221
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 = 93/5 = 18.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −29
Syy = Σ(y − ȳ)² = 91.2
Coefficient of determination
r² = −29²/(10·91.2) = 0.9221
Answer
r² = 0.9221
About 92.21% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 22.5
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 = 100/6 = 16.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −51
Regression line
b = −51/17.5 = −2.914
a = 16.67 − (−2.914)·3.5 = 26.87
ŷ = 26.87 − 2.914·x
Prediction
ŷ = 26.87 + (−2.914)·1.5 = 22.5
Answer
ŷ = 22.5
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