You need €20 000 in 6 years. An account pays 2% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €20 0002% / 4n = 6
PV = €
2.Time value of moneyHard
You deposit €3 000 at the start of every year for 29 years, the first deposit today. The account pays 5% per year. How much is in the account at the end of year 29? Find FV.
A = €3 000i = 5%n = 29
FV = €
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 3: €30 000, €25 000, €24 000. The cost of capital is 9%. Find the net present value NPV.
A project costs €25 000 and returns €13 000, €10 000, €13 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −25 000CF1 = 13 000CF2 = 10 000CF3 = 13 500
IRR = %
5.Bond pricing & yieldMedium
You buy a bond with 8 years to maturity (face value €1 000, annual coupon 4.5%) when its yield to maturity is 6%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 5%. Find your holding period return HPR in percent.
F = €1 000coupon = 4.5%n = 8y0 = 6%y1 = 5%
HPR = %
6.Bond pricing & yieldHard
A bond trades at €950 with a yield to maturity of 6% (annual compounding) and a Macaulay duration of 11.1 years. Use the modified duration to estimate the new price P if the yield rises by 75 basis points.
P0 = €950y = 6%D = 11.1Δy = +75 bp
P = €
7.CAPM & WACCMedium
A company has equity worth €3 000 000 and debt worth €5 000 000. Its WACC is 6.73%, its pre-tax cost of debt is 4% and the tax rate is 25.8%. Find the cost of equity re.
A company has 4 000 000 shares at €12.50 each and 10 000 bonds trading at €1 020 each. The pre-tax cost of debt is 4.5% and the tax rate is 21%. The share’s beta is 0.8, the risk-free rate 1% and the market return 5.5%. Find the WACC.
β = 0.8rf = 1%rm = 5.5%rd = 4.5%T = 21%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 45, 30, 30, 23, 16, 44.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 58?
μ = 70σ = 8x = 58
P =
11.ProbabilityMedium
A bag holds 5 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
5 / 7
P =
12.ProbabilityHard
A game costs €5 to play. You win €15 with probability 0.25 and €50 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€15: p = 0.25€50: p = 0.05
E(G) = €
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8186. Find z to two decimals.
A = 0.8186
z =
14.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.02 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
15.Confidence intervalsMedium
A sample of 30 items has mean 61 and standard deviation s = 9.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.
n = 30x̄ = 61s = 9.595%t(29) = 2.045
upper limit =
16.Confidence intervalsHard
The diameters of 15 machined shafts have a sample standard deviation of 3.8 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.
n = 15s = 3.8 µmχ² = 5.629, 26.119
σ = µm
17.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 8 students, mean score 42, standard deviation 10.5. Group 2: 9 students, mean score 47.6, standard deviation 10.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 8x̄₁ = 42s₁ = 10.5n₂ = 9x̄₂ = 47.6s₂ = 10.5
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 16 measurements has mean 505.4. The population standard deviation is 8. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 505.4σ = 8n = 1690%
L =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 21Σy = 133Σx² = 91Σy² = 2999Σxy = 437
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (4.5, 53) is made. Find its residual e = y − ŷ.
n = 5(4.5, 53)
e =
3Solutions
1.Time value of money
Answer: PV = 17744 €
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 = 4·6 = 24
Present value
PV = FV/(1 + i)N = 20 000/1.127 = €17 743.71
Answer
PV = €17 743.71
2.Time value of money
Answer: FV = 196317 €
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.
Ordinary annuity factor
((1 + 0.05)29 − 1)/0.05 = 62.32
Annuity due
Each deposit is invested one year longer.
FV = 3 000·62.32·1.05 = €196 316.54
Answer
FV = €196 316.54
3.NPV & IRR
Answer: NPV = −7903 €
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 each cash flow
CF0 = −75 000
CF1/(1 + 0.09)1 = 30 000/1.09 = 27 522.94
CF2/(1 + 0.09)2 = 25 000/1.188 = 21 042.00
CF3/(1 + 0.09)3 = 24 000/1.295 = 18 532.40
Add them up
NPV = €−7 902.66
Answer
NPV = €−7 902.66
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 21.5 %
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%) = 590.28
NPV(25%) = −1 288.00
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.
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.
Modified duration
Dmod = 11.1/1.06 = 10.47
Estimated price change
%ΔP ≈ −10.47·(0.0075) = −7.854%
P ≈ 950·(1 − 0.07854) = €875.39
Answer
P = €875.39
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: re = 13 %
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 = €8 000 000, E/V = 0.375, D/V = 0.625
Debt part of the WACC
0.625·4%·(1 − 0.258) = 1.855%
Cost of equity
re = (6.73% − 1.855%)/0.375 = 13%
Answer
re = 13%
8.CAPM & WACC
Answer: WACC = 4.423 %
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 = −1.5
Standard normal table
Φ(−1.5) = 0.06681
P = 0.06681
Answer
P = 0.06681
11.Probability
Answer: P = 0.1515
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.
First and second draw
P = 5/12 · 4/11 = 20/132 = 0.1515
Answer
P = 0.1515
12.Probability
Answer: E(G) = 1.25 €
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: 10 (p = 0.25), 45 (p = 0.05), −5 (p = 0.7)
Expected value
E(G) = 10·0.25 + 45·0.05 − 5·0.7 = 1.25 €
Answer
E(G) = 1.25 €
A positive value means you win money on average.
13.Normal distribution
Answer: z = −0.91
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 of z
Φ(z) = 1 − 0.8186 = 0.1814
Read the z-table backwards
Find 0.1814 inside the table and read off z.
z = −0.91
Answer
z = −0.91
14.Normal distribution
Answer: P = 0.8133
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 = 25.02 − 25 = 0.02 mm, σ = 0.02236 mm
Standardise
z = (0 − 0.02)/0.02236 = −0.89
Probability
P = 1 − Φ(−0.89) = 0.8133
Answer
P = 0.8133
15.Confidence intervals
Answer: upper limit = 64.55
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 = 29, t = 2.045
Margin of error
E = 2.045·9.5/√30 = 3.547
Interval
61 ± 3.547 → [57.45, 64.55]
Answer
upper limit = 64.55
16.Confidence intervals
Answer: σ = 5.993 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 = 14·3.8² = 202.2
Limits for σ²
202.2/26.119 = 7.74
202.2/5.629 = 35.91
Limit for σ
σ = √35.91 = 5.993 µm
Answer
σ = 5.993 µm
95% interval for σ: [2.782, 5.993] µm.
17.Hypothesis tests & confidence intervals
Answer: t = −1.098
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.
Pooled variance
sp² = (7·10.5² + 8·10.5²)/15 = 110.3
sp = 10.5
Standard error
SE = 10.5·√(1/8 + 1/9) = 5.102
Test statistic
t = (42 − 47.6)/5.102 = −1.098
Answer
t = −1.098
With df = 15, the two-sided 5% critical value is 2.131; |t| is smaller, so H₀ is not rejected.
18.Hypothesis tests & confidence intervals
Answer: L = 502.1
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).
Margin of error
E = 1.645·8/√16 = 3.29
Interval limit
L = 505.4 − 3.29 = 502.1
Answer
L = 502.1
19.Linear regression & correlation
Answer: r = −0.9555
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 = 437 − 21·133/6 = −28.5
Sxx = 91 − 21²/6 = 17.5
Syy = 2999 − 133²/6 = 50.83
Correlation coefficient
r = −28.5/√(17.5·50.83) = −0.9555
Answer
r = −0.9555
20.Linear regression & correlation
Answer: e = 2.1
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 = 20/5 = 4
ȳ = 243/5 = 48.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 46
Regression line
b = 46/10 = 4.6
a = 48.6 − 4.6·4 = 30.2
ŷ = 30.2 + 4.6·x
Predicted value
ŷ = 30.2 + 4.6·4.5 = 50.9
Residual
e = 53 − 50.9 = 2.1
Answer
e = 2.1
The point lies above the line.
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