You invest €1 000 at a nominal rate of 3.6% per year, compounded monthly. Find the value FV of the investment after 4 years.
PV = €1 0003.6% / 12n = 4
FV = €
2.Time value of moneyHard
You save €50 at the end of every month for 28 years. The account pays 4.2% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €504.2% / 12N = 336
FV = €
3.NPV & IRRMedium
A machine costs €80 000 today. It generates €19 000 at the end of each year for 5 years and can then be sold for €20 000. The cost of capital is 8%. Find the net present value NPV.
C0 = €80 000CF = €19 000S = €20 000n = 5r = 8%
NPV = €
4.NPV & IRRHard
A project costs €50 000 and returns €29 500, €33 000, €36 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −50 000CF1 = 29 500CF2 = 33 000CF3 = 36 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 9 years. The yield to maturity is 9% per year. Find the price P of the bond.
F = €1 000coupon = 5%YTM = 9%n = 9
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 is financed with ordinary shares worth €6 000 000, preferred shares worth €1 500 000 and debt worth €6 000 000. The cost of equity is 14%, the cost of preferred stock 6% and the pre-tax cost of debt 4%. The tax rate is 25%. Find the WACC.
A company has 2 500 000 shares at €47.00 each and 20 000 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 1.5, the risk-free rate 3.5% and the market return 9.5%. Find the WACC.
β = 1.5rf = 3.5%rm = 9.5%rd = 6%T = 25%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 31, 27, 46, 57, 52, 47, 41.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 15. What proportion P of values is below 32?
μ = 50σ = 15x = 32
P =
11.ProbabilityMedium
In how many ways N can a committee of 5 people be chosen from a group of 15? The order does not matter.
n = 15k = 5
N =
12.ProbabilityHard
A game costs €1 to play. You win €4 with probability 0.3 and €50 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€4: p = 0.3€50: p = 0.02
E(G) = €
13.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 6 of them is below 151.8 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 6< 151.8 kg
P =
14.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.01 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 = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −5.6 and standard deviation 0.6. Find the upper limit of the 90% confidence interval for the mean difference.
n = 8d̄ = −5.6sd = 0.690%t(7) = 1.895
upper limit = mmHg
16.Confidence intervalsHard
The diameters of 20 machined shafts have a sample standard deviation of 3.9 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 = 3.9 µmχ² = 8.907, 32.852
σ = µm
17.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 102.4 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 102.4s = 20n = 100
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 49 measurements has mean 43.6. The population standard deviation is 15. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 43.6σ = 15n = 4995%
L =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
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 = 9.
n = 5x = 9
ŷ =
3Solutions
1.Time value of money
Answer: FV = 1155 €
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: NPV = 9473 €
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 yearly cash flows
19 000·3.993 = 75 861.49
Present value of the salvage value
20 000/1.469 = 13 611.66
NPV
NPV = −80 000 + 75 861.49 + 13 611.66 = €9 473.15
Answer
NPV = €9 473.15
4.NPV & IRR
Answer: IRR = 41.83 %
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(40%) = 1 209.91
NPV(45%) = −1 986.96
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 = €25.00, i = 0.045, N = 18
Present value of the coupons
€25.00·(1 − 0.4528)/0.045 = €304.00
Present value of the face value
€1 000·0.4528 = €452.80
Price
P = €756.80
Answer
P = €756.80
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 = 8.222 %
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: N = 3003
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(15, 5) = (15·14·13·12·11)/(5·4·3·2·1) = 3003
Answer
N = 3003
12.Probability
Answer: E(G) = 1.2 €
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), 49 (p = 0.02), −1 (p = 0.68)
Expected value
E(G) = 3·0.3 + 49·0.02 − 1·0.68 = 1.2 €
Answer
E(G) = 1.2 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.9292
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 total
μT = 150 kg, σT = 0.5·√6 = 1.225 kg
Standardise
z = (151.8 − 150)/1.225 = 1.47
Probability
P = Φ(1.47) = 0.9292
Answer
P = 0.9292
14.Normal distribution
Answer: P = 0.9868
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 = 0.01803 mm
Standardise
z = (0 − 0.04)/0.01803 = −2.22
Probability
P = 1 − Φ(−2.22) = 0.9868
Answer
P = 0.9868
15.Confidence intervals
Answer: upper limit = −5.198 mmHg
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.
Critical value
df = 7, t = 1.895
Margin of error
E = 1.895·0.6/√8 = 0.402
Interval
−5.6 ± 0.402 → [−6.002, −5.198]
Answer
upper limit = −5.198 mmHg
16.Confidence intervals
Answer: σ = 2.966 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 = 19·3.9² = 289
Limits for σ²
289/32.852 = 8.797
289/8.907 = 32.45
Limit for σ
σ = √8.797 = 2.966 µm
Answer
σ = 2.966 µm
95% interval for σ: [2.966, 5.696] µm.
17.Hypothesis tests & confidence intervals
Answer: t = 1.2
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 = 20/√100 = 20/10 = 2
Test statistic
t = (102.4 − 100)/2 = 1.2
Answer
t = 1.2
t0.025; df = 99 = 1.984
|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: L = 39.4
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).
Margin of error
E = 1.96·15/√49 = 4.2
Interval limit
L = 43.6 − 4.2 = 39.4
Answer
L = 39.4
19.Linear regression & correlation
Answer: r = 0.9933
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 = 2784 − 33·483/6 = 127.5
Sxx = 199 − 33²/6 = 17.5
Syy = 39823 − 483²/6 = 941.5
Correlation coefficient
r = 127.5/√(17.5·941.5) = 0.9933
Answer
r = 0.9933
20.Linear regression & correlation
Answer: ŷ = 32.75
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.