You invest €5 000 at 8% per year with continuous compounding. Find the value FV after 19 years.
PV = €5 000r = 8%t = 19
FV = €
2.Time value of moneyHard
A loan of €8 000 is repaid with equal monthly payments over 6 years. The interest rate is 5.4% per year, compounded monthly. Find the monthly payment A.
P = €8 0005.4% / 12N = 72
A = €
3.NPV & IRRMedium
A machine costs €30 000 today. It generates €4 500 at the end of each year for 7 years and can then be sold for €3 000. The cost of capital is 12%. Find the net present value NPV.
C0 = €30 000CF = €4 500S = €3 000n = 7r = 12%
NPV = €
4.NPV & IRRHard
A project costs €20 000 and returns €13 500, €12 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 13 500CF2 = 12 000
IRR = %
5.Bond pricing & yieldMedium
You buy a bond with 11 years to maturity (face value €1 000, annual coupon 2%) when its yield to maturity is 5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 6.5%. Find your holding period return HPR in percent.
F = €1 000coupon = 2%n = 11y0 = 5%y1 = 6.5%
HPR = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 6% matures in 8 years and sells for €910. Find its yield to maturity YTM in percent.
F = €1 000coupon = 6%n = 8P = €910
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €4 000 000, preferred shares worth €1 500 000 and debt worth €5 000 000. The cost of equity is 13%, the cost of preferred stock 6% and the pre-tax cost of debt 3%. The tax rate is 30%. Find the WACC.
A company's shares have a beta of 1 at its current debt-to-equity ratio of 0.4. It plans to raise its debt-to-equity ratio to 1. The tax rate is 30%, the risk-free rate 4% and the market risk premium 4.5%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 48, 25, 48, 37, 32, 34.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is above 68?
μ = 50σ = 12x = 68
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3, 4p(x): 0.25, 0.05, 0.3, 0.1, 0.3
σ =
12.ProbabilityHard
9 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?
n = 9p = 0.7
P =
13.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 490 and 582 h? Give P as a decimal.
μ = 500σ = 40x1 = 490x2 = 582
P =
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 9 items has mean 58 and standard deviation s = 12.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 9x̄ = 58s = 12.599%t(8) = 3.355
E =
16.Confidence intervalsHard
The population standard deviation is 25. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 2.5?
σ = 25E = 2.599%
n =
17.Hypothesis tests & confidence intervalsMedium
12 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 −1.82 and the standard deviation of the differences is 4.5. Find the paired test statistic t for H₀: μd = 0.
n = 12d̄ = −1.82sd = 4.5
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 49 measurements has mean 103.8. The population standard deviation is 12. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 103.8σ = 12n = 4990%
L =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. 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. A new observation (5, 13) is made. Find its residual e = y − ŷ.
n = 6(5, 13)
e =
3Solutions
1.Time value of money
Answer: FV = 22861 €
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 = 0.08·19 = 1.52
e1.52 = 4.572
Future value
FV = 5 000·4.572 = €22 861.13
Answer
FV = €22 861.13
2.Time value of money
Answer: A = 130.3 €
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.
Monthly rate and number of payments
i = 0.054000000000000006/12 = 0.0045
N = 12·6 = 72
Payment
The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.
(1 + i)−N = 0.7238
A = 8 000·0.0045/(1 − 0.7238) = €130.33
Answer
A = €130.33
3.NPV & IRR
Answer: NPV = −8106 €
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 the yearly cash flows
4 500·4.564 = 20 536.90
Present value of the salvage value
3 000/2.211 = 1 357.05
NPV
NPV = −30 000 + 20 536.90 + 1 357.05 = −€8 106.05
Answer
NPV = −€8 106.05
4.NPV & IRR
Answer: IRR = 18.24 %
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(15%) = 812.85
NPV(20%) = −416.67
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 15 + 5·812.85/(812.85 − (−416.67)) = 18.31%
exact: NPV(18.24%) = 0 → IRR = 18.24%
Answer
IRR = 18.24%
5.Bond pricing & yield
Answer: HPR = −7.233 %
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.
Equation for the yield
€910 = €60·(1 − (1 + y)⁻⁸)/y + €1 000/(1 + y)⁸
Solve numerically
P(7%) = €940.29, P(8%) = €885.07
YTM = 7.539%
Answer
YTM = 7.539%
7.CAPM & WACC
Answer: WACC = 6.81 %
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).
Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.
Unlever the beta
βU = 1/(1 + 0.7·0.4) = 0.7813
Relever at the new D/E
βL = 0.7813·(1 + 0.7·1) = 1.328
Cost of equity (CAPM)
re = 4% + 1.328·4.5% = 9.977%
Answer
re = 9.977%
9.Descriptive statistics
Answer: s = 9.158
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: σ = 1.526
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.
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.
Zero and one failure
P(X = 0) = 0.39 = 1.968 × 10⁻⁵
P(X = 1) = 9·0.7·0.38 = 0.0004133
Add them
P = 1.968 × 10⁻⁵ + 0.0004133 = 0.000433
Answer
P = 0.000433
13.Normal distribution
Answer: P = 0.5785
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z-scores
z₁ = (490 − 500)/40 = −0.25
z₂ = (582 − 500)/40 = 2.05
Area between
P = Φ(2.05) − Φ(−0.25) = 0.9798 − 0.4013 = 0.5785
Answer
P = 0.5785
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: E = 13.98
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 = 8, t = 3.355
Margin of error
E = 3.355·12.5/√9 = 13.98
Answer
E = 13.98
16.Confidence intervals
Answer: n = 664
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 = (2.576·25/2.5)² = 663.6
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 664
Answer
n = 664
17.Hypothesis tests & confidence intervals
Answer: t = −1.401
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 = 4.5/√12 = 1.299
Test statistic
t = −1.82/1.299 = −1.401
Answer
t = −1.401
With df = 11, the two-sided 5% critical value is 2.201; |t| is smaller, so H₀ is not rejected.
18.Hypothesis tests & confidence intervals
Answer: L = 101
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 = 1.645·12/√49 = 2.82
Interval limit
L = 103.8 − 2.82 = 101
Answer
L = 101
19.Linear regression & correlation
Answer: r = 0.8845
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 = 25/5 = 5
ȳ = 162/5 = 32.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 33
Syy = Σ(y − ȳ)² = 139.2
Correlation coefficient
r = 33/√(10·139.2) = 0.8845
Answer
r = 0.8845
20.Linear regression & correlation
Answer: e = −4.595
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 = 139/6 = 23.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 195
Regression line
b = 195/70 = 2.786
a = 23.17 − 2.786·7 = 3.667
ŷ = 3.667 + 2.786·x
Predicted value
ŷ = 3.667 + 2.786·5 = 17.6
Residual
e = 13 − 17.6 = −4.595
Answer
e = −4.595
The point lies below the line.
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