You need €25 000 in 7 years. An account pays 2% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €25 0002% / 12n = 7
PV = €
2.Time value of moneyHard
A loan of €15 000 is repaid with equal monthly payments over 6 years. The interest rate is 4.8% per year, compounded monthly. Find the monthly payment A.
P = €15 0004.8% / 12N = 72
A = €
3.NPV & IRRMedium
A machine costs €80 000 today, lasts 10 years and costs €4 000 per year to run (paid at the end of each year). The cost of capital is 5%. Find its equivalent annual cost EAC.
C0 = €80 000M = €4 000n = 10r = 5%
EAC = €
4.NPV & IRRHard
A project costs €20 000 and returns €13 500, €14 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 13 500CF2 = 14 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 3% annual coupon in two half-yearly payments and matures in 11 years. The yield to maturity is 5.5% per year. Find the price P of the bond.
F = €1 000coupon = 3%YTM = 5.5%n = 11
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 3% and matures in 4 years. The yield to maturity is 5%. Find the Macaulay duration D in years.
F = €1 000coupon = 3%n = 4YTM = 5%
D = years
7.CAPM & WACCMedium
A company has equity worth €6 000 000 and debt worth €6 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 4.5% and the tax rate is 28%. Find the WACC.
A company has 5 000 000 shares trading at €28.00; the book value of its equity is €70 000 000. It just paid a dividend of €2.10 per share, and dividends grow by 2% per year. Its debt has a market value of €42 000 000 and a pre-tax cost of 4.5%. The tax rate is 25.8%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 55, 50, 26, 26, 13, 24.
n = 6
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 475.
μ = 500σ = 25x = 475
z =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.1, 0.3, 0.1, 0.15, 0.35
σ =
12.ProbabilityHard
A game costs €10 to play. You win €20 with probability 0.2 and €500 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€20: p = 0.2€500: p = 0.05
E(G) = €
13.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 32 and 37.1 min? Give P as a decimal.
μ = 32σ = 6x1 = 32x2 = 37.1
P =
14.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 5% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points5%
x = points
15.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 4.3. What is the margin of error E₂ of a 99% interval based on 72 observations?
95%n₁ = 36E₁ = 4.399%n₂ = 72
E₂ =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 12 plots, mean yield 99.5 kg, standard deviation 10 kg. Plot group 2: 5 plots, mean 98.5 kg, standard deviation 7.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
A sample of 81 students has a mean score of 51.29 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 51.29s = 4n = 81
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 20. Find the smallest sample size n.
σ = 20E = 295%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (21, 64) is made. Find its residual e = y − ŷ.
n = 6(21, 64)
e =
3Solutions
1.Time value of money
Answer: PV = 21736 €
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 = 0.02/12 = 0.001667
N = 12·7 = 84
Present value
PV = FV/(1 + i)N = 25 000/1.15 = €21 736.49
Answer
PV = €21 736.49
2.Time value of money
Answer: A = 240.2 €
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 = 0.048/12 = 0.004
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 = 0.7502
A = 15 000·0.004/(1 − 0.7502) = €240.18
Answer
A = €240.18
3.NPV & IRR
Answer: EAC = 14360 €
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 − 1.05−10)/0.05 = 7.722
Annualised purchase price
80 000/7.722 = 10 360.37
EAC
EAC = 10 360.37 + 4 000 = €14 360.37
Answer
EAC = €14 360.37
4.NPV & IRR
Answer: IRR = 23.97 %
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%) = 972.22
NPV(25%) = −240.00
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 20 + 5·972.22/(972.22 − (−240.00)) = 24.01%
exact: NPV(23.97%) = 0 → IRR = 23.97%
Answer
IRR = 23.97%
5.Bond pricing & yield
Answer: P = 795.7 €
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 = €15.00, i = 0.0275, N = 22
Present value of the coupons
€15.00·(1 − 0.5506)/0.0275 = €245.15
Present value of the face value
€1 000·0.5506 = €550.55
Price
P = €795.71
Answer
P = €795.71
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: D = 3.821 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: 30/1.05 = 28.57, t·PV = 28.57
t = 2: 30/1.103 = 27.21, t·PV = 54.42
t = 3: 30/1.158 = 25.92, t·PV = 77.75
t = 4: 1030/1.216 = 847.38, t·PV = 3 389.53
Duration
P = 929.08
D = 3 550.27/929.08 = 3.821
Answer
D = 3.821 years
7.CAPM & WACC
Answer: WACC = 5.62 %
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 = €12 000 000, E/V = 0.5, D/V = 0.5
WACC
WACC = 0.5·8% + 0.5·4.5%·(1 − 0.28) = 5.62%
Answer
WACC = 5.62%
8.CAPM & WACC
Answer: WACC = 8.194 %
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 = (475 − 500)/25 = −1
Answer
z = −1
11.Probability
Answer: σ = 1.459
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: 10 (p = 0.2), 490 (p = 0.05), −10 (p = 0.75)
Expected value
E(G) = 10·0.2 + 490·0.05 − 10·0.75 = 19 €
Answer
E(G) = 19 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.3023
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 = (32 − 32)/6 = 0
z₂ = (37.1 − 32)/6 = 0.85
Area between
P = Φ(0.85) − Φ(0) = 0.8023 − 0.5 = 0.3023
Answer
P = 0.3023
14.Normal distribution
Answer: x = 124.6 points
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) = 0.95
z from the table
z ≈ 1.64
Back to the original units
x = 100 + 1.64·15 = 124.6 points
Answer
x = 124.6 points
15.Confidence intervals
Answer: E₂ = 3.996
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.
Ratio of margins
E₂/E₁ = (2.576/1.96)·√(36/72) = 0.9293
New margin
E₂ = 4.3·0.9293 = 3.996
Answer
E₂ = 3.996
16.Confidence intervals
Answer: upper limit = 9.77 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 = (11·10² + 4·7.5²)/15 = 88.33
sp = 9.399
Margin of error
E = 1.753·9.399·√(1/12 + 1/5) = 8.77
Interval
1 ± 8.77 → [−7.77, 9.77]
Answer
upper limit = 9.77 kg
17.Hypothesis tests & confidence intervals
Answer: t = 2.902
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 = 4/√81 = 4/9 = 0.4444
Test statistic
t = (51.29 − 50)/0.4444 = 2.902
Answer
t = 2.902
t0.025; df = 80 = 1.99
|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 385
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 = (1.96·20/2)² = 384.2
Round up
n = 385
Answer
n = 385
19.Linear regression & correlation
Answer: r = 0.992
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 = 3724 − 100·181/5 = 104
Sxx = 2040 − 100²/5 = 40
Syy = 6827 − 181²/5 = 274.8
Correlation coefficient
r = 104/√(40·274.8) = 0.992
Answer
r = 0.992
20.Linear regression & correlation
Answer: e = −3.833
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 = 126/6 = 21
ȳ = 407/6 = 67.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 261
Regression line
b = 261/70 = 3.729
a = 67.83 − 3.729·21 = −10.47
ŷ = −10.47 + 3.729·x
Predicted value
ŷ = −10.47 + 3.729·21 = 67.83
Residual
e = 64 − 67.83 = −3.833
Answer
e = −3.833
The point lies below the line.
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