You invest €20 000 at 6% per year with continuous compounding. Find the value FV after 4 years.
PV = €20 000r = 6%t = 4
FV = €
2.Time value of moneyHard
You borrow €25 000 at 5.4% per year, compounded monthly, and repay €425 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.
P = €25 000A = €4255.4% / 12
N = payments
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 3: €25 000, €20 000, €21 000. The cost of capital is 6%. Find the net present value NPV.
An investment of €25 000 returns €5 000 at the end of each year for 6 years. Find its internal rate of return IRR in percent.
C0 = €25 000CF = €5 000n = 6
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 11 years. The yield to maturity is 3.5% per year. Find the price P of the bond.
F = €1 000coupon = 8%YTM = 3.5%n = 11
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 7% matures in 10 years and sells for €1 155. Find its yield to maturity YTM in percent.
F = €1 000coupon = 7%n = 10P = €1 155
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €10 000 000, preferred shares worth €2 000 000 and debt worth €6 000 000. The cost of equity is 12%, the cost of preferred stock 6% and the pre-tax cost of debt 3%. The tax rate is 20%. Find the WACC.
A company has 5 000 000 shares at €8.00 each and 40 000 bonds trading at €980 each. The pre-tax cost of debt is 4% and the tax rate is 30%. The share’s beta is 0.6, the risk-free rate 3% and the market return 10%. Find the WACC.
β = 0.6rf = 3%rm = 10%rd = 4%T = 30%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 16, 18, 10, 32, 32, 23, 35.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is below 55?
μ = 50σ = 5x = 55
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4p(x): 0.35, 0.3, 0.25, 0.1
σ =
12.ProbabilityHard
A game costs €5 to play. You win €20 with probability 0.25 and €250 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€20: p = 0.25€250: 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 0.0256. Find z to two decimals.
A = 0.0256
z =
14.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 30% is to the left of it. Find the cut-off x.
μ = 65 pointsσ = 10 points30%
x = points
15.Confidence intervalsMedium
A sample of 10 items has mean 34 and standard deviation s = 9.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 10x̄ = 34s = 9.599%t(9) = 3.25
upper limit =
16.Confidence intervalsHard
The population standard deviation is 31. 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 5?
σ = 31E = 590%
n =
17.Hypothesis tests & confidence intervalsMedium
9 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 11.95 and the standard deviation of the differences is 11.2. Find the paired test statistic t for H₀: μd = 0.
n = 9d̄ = 11.95sd = 11.2
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 10. Find the smallest sample size n.
σ = 10E = 199%
n =
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 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 = 5?
n = 6ŷ = 5
x =
3Solutions
1.Time value of money
Answer: FV = 25425 €
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 = 0.06·4 = 0.24
e0.24 = 1.271
Future value
FV = 20 000·1.271 = €25 424.98
Answer
FV = €25 424.98
2.Time value of money
Answer: N = 69 payments
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.
Rearrange the annuity formula
i = 0.0045
(1 + i)−N = 1 − 25 000·0.0045/425 = 0.7353
Solve with logarithms
N = −ln(0.7353)/ln(1.004) = 68.48
Round up to whole payments
68 full payments leave a small balance, which the last, smaller payment clears.
N = 69
Answer
N = 69 payments
3.NPV & IRR
Answer: NPV = −983.2 €
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 = −60 000
CF1/(1 + 0.06)1 = 25 000/1.06 = 23 584.91
CF2/(1 + 0.06)2 = 20 000/1.124 = 17 799.93
CF3/(1 + 0.06)3 = 21 000/1.191 = 17 632.00
Add them up
NPV = €−983.16
Answer
NPV = €−983.16
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 5.472 %
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)−6)/i = 25 000/5 000 = 5
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 5%: 5.076
i = 6%: 4.917
IRR = 5.472%
Answer
IRR = 5.472%
5.Bond pricing & yield
Answer: P = 1408 €
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 = €40.00, i = 0.0175, N = 22
Present value of the coupons
€40.00·(1 − 0.6827)/0.0175 = €725.21
Present value of the face value
€1 000·0.6827 = €682.72
Price
P = €1 407.93
Answer
P = €1 407.93
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 4.993 %
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.
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).
Market values
E = 5 000 000·€8.00 = €40 000 000
D = 40 000·€980 = €39 200 000
Cost of equity (CAPM)
re = 3% + 0.6·7% = 7.2%
Weights
V = €79 200 000, E/V = 0.5051, D/V = 0.4949
WACC
WACC = 0.5051·7.2% + 0.4949·4%·(1 − 0.3) = 5.022%
Answer
WACC = 5.022%
9.Descriptive statistics
Answer: s = 9.534
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
Standard normal table
Φ(1) = 0.8413
P = 0.8413
Answer
P = 0.8413
11.Probability
Answer: σ = 0.995
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 = 1·0.35 + 2·0.3 + 3·0.25 + 4·0.1 = 2.1
Mean of the squares
E(X²) = 1·0.35 + 4·0.3 + 9·0.25 + 16·0.1 = 5.4
Variance
Var(X) = 5.4 − 2.1² = 0.99
σ = √0.99 = 0.995
Answer
σ = 0.995
12.Probability
Answer: E(G) = 5 €
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: 15 (p = 0.25), 245 (p = 0.02), −5 (p = 0.73)
Expected value
E(G) = 15·0.25 + 245·0.02 − 5·0.73 = 5 €
Answer
E(G) = 5 €
A positive value means you win money on average.
13.Normal distribution
Answer: z = 1.95
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 − 0.0256 = 0.9744
Read the z-table backwards
Find 0.9744 inside the table and read off z.
z = 1.95
Answer
z = 1.95
14.Normal distribution
Answer: x = 59.8 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) = 0.3
z from the table
z ≈ −0.52
Back to the original units
x = 65 − 0.52·10 = 59.8 points
Answer
x = 59.8 points
15.Confidence intervals
Answer: upper limit = 43.76
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 = 9, t = 3.25
Margin of error
E = 3.25·9.5/√10 = 9.764
Interval
34 ± 9.764 → [24.24, 43.76]
Answer
upper limit = 43.76
16.Confidence intervals
Answer: n = 105
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 = (1.645·31/5)² = 104
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 105
Answer
n = 105
17.Hypothesis tests & confidence intervals
Answer: t = 3.201
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 of the mean difference
SE = 11.2/√9 = 3.733
Test statistic
t = 11.95/3.733 = 3.201
Answer
t = 3.201
With df = 8, the two-sided 5% critical value is 2.306; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: n = 664
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 = (2.576·10/1)² = 663.6
Round up
n = 664
Answer
n = 664
19.Linear regression & correlation
Answer: r = 0.9298
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 = 1397 − 27·300/6 = 47
Sxx = 139 − 27²/6 = 17.5
Syy = 15146 − 300²/6 = 146
Correlation coefficient
r = 47/√(17.5·146) = 0.9298
Answer
r = 0.9298
20.Linear regression & correlation
Answer: x = 2.905
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 = 83/6 = 13.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 151
Regression line
b = 151/70 = 2.157
a = 13.83 − 2.157·7 = −1.267
ŷ = −1.267 + 2.157·x
Solve for x
x = (5 − (−1.267))/2.157 = 2.905
Answer
x = 2.905
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