You invest €10 000 at 3% per year with continuous compounding. Find the value FV after 9 years.
PV = €10 000r = 3%t = 9
FV = €
2.Time value of moneyHard
You deposit €2 000 at the start of every year for 28 years, the first deposit today. The account pays 8% per year. How much is in the account at the end of year 28? Find FV.
A = €2 000i = 8%n = 28
FV = €
3.NPV & IRRMedium
A project costs €250 000 today. It generates €22 000 at the end of the first year, and the cash flow then grows by 2% per year forever. The cost of capital is 10%. Find the net present value NPV.
C0 = €250 000CF1 = €22 000g = 2%r = 10%
NPV = €
4.NPV & IRRHard
A project costs €10 000 and returns €6 000, €7 000, €4 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −10 000CF1 = 6 000CF2 = 7 000CF3 = 4 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 11 years. Its yield to maturity changes from 9% to 8%. By how much ΔP does the bond price change?
F = €1 000coupon = 5%n = 11y1 = 9%y2 = 8%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 11 years. It sells for €1 239. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 5% (2×)n = 11P = €1 239
YTM = %
7.CAPM & WACCMedium
A company has equity worth €12 000 000 and debt worth €3 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 5% and the tax rate is 21%. Find the WACC.
A company has 4 000 000 shares at €32.50 each and 5 000 bonds trading at €1 020 each. The pre-tax cost of debt is 3% and the tax rate is 28%. The share’s beta is 1.3, the risk-free rate 3% and the market return 9%. Find the WACC.
β = 1.3rf = 3%rm = 9%rd = 3%T = 28%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 26, 23, 20, 23, 19.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 20. Find the z-score z of a score of 60.
μ = 100σ = 20x = 60
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.3, 0.15, 0.25, 0.3
Var(X) =
12.ProbabilityHard
A game costs €2 to play. You win €8 with probability 0.15 and €20 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €2€8: p = 0.15€20: 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 0.9082. Find z to two decimals.
A = 0.9082
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 12 items has mean 101 and standard deviation s = 12.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 12x̄ = 101s = 12.599%t(11) = 3.106
upper limit =
16.Confidence intervalsHard
The population standard deviation is 30. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?
σ = 30E = 1.590%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 72.75 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 72.75s = 5n = 16
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 200 against H₁: μ < 200 is done at α = 0.10 with a sample of 100; the standard deviation is known to be 10. In reality the mean is 197.9. Find the power 1 − β of the test in percent.
μ₀ = 200μ = 197.9σ = 10n = 100α = 0.10
1 − β = %
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. 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. For which value x does the line predict ŷ = 10?
n = 5ŷ = 10
x =
3Solutions
1.Time value of money
Answer: FV = 13100 €
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.
Exponent
r·t = 0.03·9 = 0.27
e0.27 = 1.31
Future value
FV = 10 000·1.31 = €13 099.64
Answer
FV = €13 099.64
2.Time value of money
Answer: FV = 205932 €
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 + 0.08)28 − 1)/0.08 = 95.34
Annuity due
Each deposit is invested one year longer.
FV = 2 000·95.34·1.08 = €205 931.87
Answer
FV = €205 931.87
3.NPV & IRR
Answer: NPV = 25000 €
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.
Value of the cash flows
PV = 22 000/(0.1 − 0.02) = 275 000.00
NPV
NPV = 275 000.00 − 250 000 = €25 000.00
Answer
NPV = €25 000.00
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 35.88 %
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(35%) = 114.31
NPV(40%) = −502.92
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 35 + 5·114.31/(114.31 − (−502.92)) = 35.93%
exact: NPV(35.88%) = 0 → IRR = 35.88%
Answer
IRR = 35.88%
5.Bond pricing & yield
Answer: ΔP = 58.04 €
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.
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, N = 22
Solve numerically
i = 1.251% per half-year
YTM
YTM = 2·1.251% = 2.501%
Answer
YTM = 2.501%
7.CAPM & WACC
Answer: WACC = 9.59 %
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 = €15 000 000, E/V = 0.8, D/V = 0.2
WACC
WACC = 0.8·11% + 0.2·5%·(1 − 0.21) = 9.59%
Answer
WACC = 9.59%
8.CAPM & WACC
Answer: WACC = 10.47 %
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 = (60 − 100)/20 = −2
Answer
z = −2
11.Probability
Answer: Var(X) = 1.448
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.
Mean
μ = 0·0.3 + 1·0.15 + 2·0.25 + 3·0.3 = 1.55
Mean of the squares
E(X²) = 0·0.3 + 1·0.15 + 4·0.25 + 9·0.3 = 3.85
Variance
Var(X) = 3.85 − 1.55² = 1.448
Answer
Var(X) = 1.448
12.Probability
Answer: E(G) = 0.2 €
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: 6 (p = 0.15), 18 (p = 0.05), −2 (p = 0.8)
Expected value
E(G) = 6·0.15 + 18·0.05 − 2·0.8 = 0.2 €
Answer
E(G) = 0.2 €
A positive value means you win money on average.
13.Normal distribution
Answer: z = −1.33
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 − 0.9082 = 0.0918
Read the z-table backwards
Find 0.0918 inside the table and read off z.
z = −1.33
Answer
z = −1.33
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 = 112.2
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 = 11, t = 3.106
Margin of error
E = 3.106·12.5/√12 = 11.21
Interval
101 ± 11.21 → [89.79, 112.2]
Answer
upper limit = 112.2
16.Confidence intervals
Answer: n = 1083
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.
Solve for n
n = (1.645·30/1.5)² = 1082
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1083
Answer
n = 1083
17.Hypothesis tests & confidence intervals
Answer: t = −1.8
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/√16 = 5/4 = 1.25
Test statistic
t = (72.75 − 75)/1.25 = −1.8
Answer
t = −1.8
t0.025; df = 15 = 2.131
|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: 1 − β = 79.33 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 10/√100 = 1
x̄c = 200 − 1.282·1 = 198.7
z under the true mean
z = (198.7 − 197.9)/1 = 0.818
Power
power = P(Z < 0.818) = 79.33%
1 − β = 79.33%
Answer
1 − β = 79.33%
19.Linear regression & correlation
Answer: r = 0.9975
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 = 3676 − 48·412/6 = 380
Sxx = 454 − 48²/6 = 70
Syy = 30364 − 412²/6 = 2073
Correlation coefficient
r = 380/√(70·2073) = 0.9975
Answer
r = 0.9975
20.Linear regression & correlation
Answer: x = 4.811
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 = 61/5 = 12.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 74
Regression line
b = 74/40 = 1.85
a = 12.2 − 1.85·6 = 1.1
ŷ = 1.1 + 1.85·x
Solve for x
x = (10 − 1.1)/1.85 = 4.811
Answer
x = 4.811
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