A property is expected to earn net rent of €8 000 next year. The rent is expected to grow by 1.5% per year forever. The required return is 6.5% per year. Find the value PV of the property today.
C1 = €8 000g = 1.5%r = 6.5%
PV = €
2.Time value of moneyHard
You deposit €2 500 at the start of every year for 28 years, the first deposit today. The account pays 7% per year. How much is in the account at the end of year 28? Find FV.
A = €2 500i = 7%n = 28
FV = €
3.NPV & IRRMedium
A project costs €30 000 today and returns €12 000, €7 000, €8 500, €11 000, €13 000 at the end of years 1 to 5. The cost of capital is 5%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
An investment of €25 000 returns €12 000 at the end of each year for 3 years. Find its internal rate of return IRR in percent.
C0 = €25 000CF = €12 000n = 3
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 5 years. The yield to maturity is 5% per year. Find the price P of the bond.
F = €1 000coupon = 7%YTM = 5%n = 5
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 7% and matures in 3 years. The yield to maturity is 8%. Find the Macaulay duration D in years.
F = €1 000coupon = 7%n = 3YTM = 8%
D = years
7.CAPM & WACCMedium
A company has equity worth €4 000 000 and debt worth €5 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 3% and the tax rate is 25.8%. Find the WACC.
A company has 1 000 000 shares at €46.00 each and 10 000 bonds trading at €1 020 each. The pre-tax cost of debt is 5% and the tax rate is 21%. The share’s beta is 1.8, the risk-free rate 1.5% and the market return 8%. Find the WACC.
β = 1.8rf = 1.5%rm = 8%rd = 5%T = 21%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 35, 30, 14, 42, 45, 44, 48.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 20. What proportion P of values is above 200?
μ = 170σ = 20x = 200
P =
11.ProbabilityMedium
A fair die is rolled 6 times. What is the probability P of getting at least one six?
n = 6
P =
12.ProbabilityHard
A game costs €5 to play. You win €15 with probability 0.25 and €250 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€250: 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 0.0778. Find z to two decimals.
A = 0.0778
z =
14.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 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 = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 5.3. What is the margin of error E₂ if the same study uses 16 observations instead?
95%n₁ = 64E₁ = 5.3n₂ = 16
E₂ =
16.Confidence intervalsHard
A pilot study suggests that about 0.6 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.01?
p ≈ 0.6E = 0.0190%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 200.9 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 200.9s = 5n = 49
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 90 of 150 patients on treatment A recover, and 38 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 90n₁ = 150x₂ = 38n₂ = 80
z =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 26?
n = 6ŷ = 26
x =
3Solutions
1.Time value of money
Answer: PV = 160000 €
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.
Ordinary annuity factor
((1 + 0.07)28 − 1)/0.07 = 80.7
Annuity due
Each deposit is invested one year longer.
FV = 2 500·80.7·1.07 = €215 866.32
Answer
FV = €215 866.32
3.NPV & IRR
Answer: DPB = 3.539 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: 12 000/1.05 = 11 428.57 → Σ = 11 428.57
year 2: 7 000/1.103 = 6 349.21 → Σ = 17 777.78
year 3: 8 500/1.158 = 7 342.62 → Σ = 25 120.40
year 4: 11 000/1.216 = 9 049.73 → Σ = 34 170.12
Fraction of the last year
DPB = 3 + 4 879.60/9 049.73 = 3.539
Answer
DPB = 3.539 years
The discounted payback is always longer than the simple payback (3.227 years).
4.NPV & IRR
Answer: IRR = 20.71 %
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.
Required annuity factor
(1 − (1 + i)−3)/i = 25 000/12 000 = 2.083
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 20%: 2.106
i = 21%: 2.074
IRR = 20.71%
Answer
IRR = 20.71%
5.Bond pricing & yield
Answer: P = 1088 €
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 = €35.00, i = 0.025, N = 10
Present value of the coupons
€35.00·(1 − 0.7812)/0.025 = €306.32
Present value of the face value
€1 000·0.7812 = €781.20
Price
P = €1 087.52
Answer
P = €1 087.52
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: D = 2.805 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: 70/1.08 = 64.81, t·PV = 64.81
t = 2: 70/1.166 = 60.01, t·PV = 120.03
t = 3: 1070/1.26 = 849.40, t·PV = 2 548.20
Duration
P = 974.23
D = 2 733.04/974.23 = 2.805
Answer
D = 2.805 years
7.CAPM & WACC
Answer: WACC = 7.014 %
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 1 − 0.9332 = 0.06681
Answer
P = 0.06681
11.Probability
Answer: P = 0.6651
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.
Complement
P(no six) = (5/6)6 = 0.3349
P = 1 − 0.3349 = 0.6651
Answer
P = 0.6651
12.Probability
Answer: E(G) = 11.25 €
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), 245 (p = 0.05), −5 (p = 0.7)
Expected value
E(G) = 10·0.25 + 245·0.05 − 5·0.7 = 11.25 €
Answer
E(G) = 11.25 €
A positive value means you win money on average.
13.Normal distribution
Answer: z = 1.42
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 − 0.0778 = 0.9222
Read the z-table backwards
Find 0.9222 inside the table and read off z.
z = 1.42
Answer
z = 1.42
14.Normal distribution
Answer: P = 0.9207
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 = 0.02121 mm
Standardise
z = (0 − 0.03)/0.02121 = −1.41
Probability
P = 1 − Φ(−1.41) = 0.9207
Answer
P = 0.9207
15.Confidence intervals
Answer: E₂ = 10.6
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 = (1.96/1.96)·√(64/16) = 2
New margin
E₂ = 5.3·2 = 10.6
Answer
E₂ = 10.6
16.Confidence intervals
Answer: n = 6495
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 1.645²·0.6·0.4/0.01² = 6494
Round up
n = 6495
Answer
n = 6495
17.Hypothesis tests & confidence intervals
Answer: t = 1.26
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 = 5/√49 = 5/7 = 0.7143
Test statistic
t = (200.9 − 200)/0.7143 = 1.26
Answer
t = 1.26
t0.025; df = 48 = 2.011
|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = 1.817
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.
Sample proportions
p̂₁ = 90/150 = 0.6
p̂₂ = 38/80 = 0.475
p̂ = 128/230 = 0.5565
Standard error under H₀
SE = √(0.5565·0.4435·(1/150 + 1/80)) = 0.06878
Test statistic
z = (0.6 − 0.475)/0.06878 = 1.817
Answer
z = 1.817
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9899
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 = 120/5 = 24
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 150
Syy = Σ(y − ȳ)² = 574
Correlation coefficient
r = 150/√(40·574) = 0.9899
Answer
r = 0.9899
20.Linear regression & correlation
Answer: x = 11.14
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 = 42/6 = 7
ȳ = 102/6 = 17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 152
Regression line
b = 152/70 = 2.171
a = 17 − 2.171·7 = 1.8
ŷ = 1.8 + 2.171·x
Solve for x
x = (26 − 1.8)/2.171 = 11.14
Answer
x = 11.14
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