A property is expected to earn net rent of €20 000 next year. The rent is expected to grow by 4% per year forever. The required return is 9% per year. Find the value PV of the property today.
C1 = €20 000g = 4%r = 9%
PV = €
2.Time value of moneyHard
You save €50 at the end of every month for 11 years. The account pays 6% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €506% / 12N = 132
FV = €
3.NPV & IRRMedium
An investment of €45 000 gives these cash flows at the end of years 1 to 4: €14 000, €17 000, €12 000, €11 000. The cost of capital is 9%. Find the net present value NPV.
A project costs €10 000 and returns €5 000, €4 000, €3 500, €4 500 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 matures in 3 years and pays annual coupons. It sells for €899 when the yield to maturity is 9%. Find the coupon rate c in percent.
F = €1 000n = 3P = €899YTM = 9%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 4% matures in 9 years and sells for €1 120. Find its yield to maturity YTM in percent.
F = €1 000coupon = 4%n = 9P = €1 120
YTM = %
7.CAPM & WACCMedium
A company has equity worth €4 000 000 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 25%. Find the WACC.
E = €4 000 000D = €1 000 000re = 8%rd = 5%T = 25%
WACC = %
8.CAPM & WACCHard
A company has 2 000 000 shares trading at €56.50; the book value of its equity is €67 800 000. It just paid a dividend of €2.70 per share, and dividends grow by 4% per year. Its debt has a market value of €56 500 000 and a pre-tax cost of 3%. 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: 28, 18, 26, 24, 45.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 5. Find the z-score z of a score of 96.
μ = 100σ = 5x = 96
z =
11.ProbabilityMedium
In how many ways N can a committee of 3 people be chosen from a group of 6? The order does not matter.
n = 6k = 3
N =
12.ProbabilityHard
A game costs €5 to play. You win €25 with probability 0.15 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 = €5€25: p = 0.15€50: p = 0.02
E(G) = €
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9788. Find z to two decimals.
A = 0.9788
z =
14.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
15.Confidence intervalsMedium
A sample of 7 items has mean 72.5 and standard deviation s = 8.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 7x̄ = 72.5s = 8.595%t(6) = 2.447
lower limit =
16.Confidence intervalsHard
The diameters of 16 machined shafts have a sample standard deviation of 5.8 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.
n = 16s = 5.8 µmχ² = 6.262, 27.488
σ = µm
17.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 197 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 197s = 8n = 49
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 16 measurements has mean 103.4. The population standard deviation is 5. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 103.4σ = 5n = 1699%
L =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
20.Linear regression & correlationHard
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 = 5.
n = 5x = 5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 400000 €
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 €6 600 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = −788.5 €
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.
Present value of each cash flow
CF0 = −45 000
CF1/(1 + 0.09)1 = 14 000/1.09 = 12 844.04
CF2/(1 + 0.09)2 = 17 000/1.188 = 14 308.56
CF3/(1 + 0.09)3 = 12 000/1.295 = 9 266.20
CF4/(1 + 0.09)4 = 11 000/1.412 = 7 792.68
Add them up
NPV = €−788.52
Answer
NPV = €−788.52
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: MIRR = 17.78 %
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
5 000·1.083 = 6 298.56
4 000·1.082 = 4 665.60
3 500·1.081 = 3 780.00
4 500·1.080 = 4 500.00
TV = 19 244.16
MIRR
MIRR = (19 244.16/10 000)1/4 − 1 = 17.78%
Answer
MIRR = 17.78%
5.Bond pricing & yield
Answer: c = 5.01 %
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.
Present value of the face value
1000·(1 + 0.09)−3 = €772.18
Coupon
C = (899 − 772.18)/2.531 = €50.10
Coupon rate
c = 50.1/1000 = 5.01%
Answer
c = 5.01%
6.Bond pricing & yield
Answer: YTM = 2.495 %
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
€1 120 = €40·(1 − (1 + y)⁻⁹)/y + €1 000/(1 + y)⁹
Solve numerically
P(2%) = €1 163.24, P(3%) = €1 077.86
YTM = 2.495%
Answer
YTM = 2.495%
7.CAPM & WACC
Answer: WACC = 7.15 %
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 = €5 000 000, E/V = 0.8, D/V = 0.2
WACC
WACC = 0.8·8% + 0.2·5%·(1 − 0.25) = 7.15%
Answer
WACC = 7.15%
8.CAPM & WACC
Answer: WACC = 6.722 %
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 = (96 − 100)/5 = −0.8
Answer
z = −0.8
11.Probability
Answer: N = 20
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, 3) = (6·5·4)/(3·2·1) = 20
Answer
N = 20
12.Probability
Answer: E(G) = −0.25 €
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 = 0.15), 45 (p = 0.02), −5 (p = 0.83)
Expected value
E(G) = 20·0.15 + 45·0.02 − 5·0.83 = −0.25 €
Answer
E(G) = −0.25 €
A negative value means you lose money on average.
13.Normal distribution
Answer: z = −2.03
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 − 0.9788 = 0.0212
Read the z-table backwards
Find 0.0212 inside the table and read off z.
z = −2.03
Answer
z = −2.03
14.Normal distribution
Answer: P = 0.8413
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 = 500 − 490 = 10 g, σ = 10 g
Standardise
z = (0 − 10)/10 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
15.Confidence intervals
Answer: lower limit = 64.64
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 = 6, t = 2.447
Margin of error
E = 2.447·8.5/√7 = 7.861
Interval
72.5 ± 7.861 → [64.64, 80.36]
Answer
lower limit = 64.64
16.Confidence intervals
Answer: σ = 8.977 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 = 15·5.8² = 504.6
Limits for σ²
504.6/27.488 = 18.36
504.6/6.262 = 80.58
Limit for σ
σ = √80.58 = 8.977 µm
Answer
σ = 8.977 µm
95% interval for σ: [4.285, 8.977] µm.
17.Hypothesis tests & confidence intervals
Answer: t = −2.625
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 = 8/√49 = 8/7 = 1.143
Test statistic
t = (197 − 200)/1.143 = −2.625
Answer
t = −2.625
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: L = 100.2
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).
Margin of error
E = 2.576·5/√16 = 3.22
Interval limit
L = 103.4 − 3.22 = 100.2
Answer
L = 100.2
19.Linear regression & correlation
Answer: r² = 0.8738
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 = 35/5 = 7
ȳ = 239/5 = 47.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 94
Syy = Σ(y − ȳ)² = 252.8
Coefficient of determination
r² = 94²/(40·252.8) = 0.8738
Answer
r² = 0.8738
About 87.38% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 21.1
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 = 30/5 = 6
ȳ = 125/5 = 25
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 156
Regression line
b = 156/40 = 3.9
a = 25 − 3.9·6 = 1.6
ŷ = 1.6 + 3.9·x
Prediction
ŷ = 1.6 + 3.9·5 = 21.1
Answer
ŷ = 21.1
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