An annuity pays €1 500 at the end of each year for 5 years. The interest rate is 6% per year. Find the present value PV of the annuity.
A = €1 500i = 6%n = 5
PV = €
2.Time value of moneyHard
A loan of €200 000 is repaid with equal monthly payments over 15 years at 6% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €200 0006% / 12N = 180
I = €
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 4: €19 000, €20 000, €20 000, €23 000. The cost of capital is 9%. Find the net present value NPV.
A project costs €10 000 and returns €4 000, €3 500, €5 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −10 000CF1 = 4 000CF2 = 3 500CF3 = 5 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 3% and matures in 12 years. Its yield to maturity changes from 5.5% to 7%. By how much ΔP does the bond price change?
F = €1 000coupon = 3%n = 12y1 = 5.5%y2 = 7%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 6% matures in 5 years and sells for €1 065. Find its yield to maturity YTM in percent.
F = €1 000coupon = 6%n = 5P = €1 065
YTM = %
7.CAPM & WACCMedium
A company has equity worth €6 000 000 and debt worth €1 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 7% and the tax rate is 30%. Find the WACC.
E = €6 000 000D = €1 000 000re = 8%rd = 7%T = 30%
WACC = %
8.CAPM & WACCHard
A company's shares have a beta of 1.1 at its current debt-to-equity ratio of 0.25. It plans to raise its debt-to-equity ratio to 0.75. The tax rate is 25%, the risk-free rate 3% and the market risk premium 6%. Find the cost of equity re after the change.
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9147. Find z to two decimals.
A = 0.9147
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 top 95% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points95%
x = points
15.Confidence intervalsMedium
The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 8.8 and standard deviation 4.9. Find the margin of error E of the 95% confidence interval for the mean difference.
n = 8d̄ = 8.8sd = 4.995%t(7) = 2.365
E = mmHg
16.Confidence intervalsHard
The population standard deviation is 23. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 0.5?
σ = 23E = 0.595%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 50; the population standard deviation is known to be 20. A random sample of 100 observations has a mean of 46.08. Test H₀: μ = 50 against H₁: μ < 50. Find the p-value p.
μ₀ = 50σ = 20n = 100x̄ = 46.08H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A sample of 49 measurements has mean 203.4. The population standard deviation is 20. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 203.4σ = 20n = 4990%
U =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 42Σy = 181Σx² = 364Σy² = 6611Σxy = 1550
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 17.
n = 6x = 17
ŷ =
3Solutions
1.Time value of money
Answer: PV = 6319 €
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.
Annuity factor
(1 − (1 + i)−n)/i = (1 − 0.7473)/0.06 = 4.212
Present value
PV = 1 500·4.212 = €6 318.55
Answer
PV = €6 318.55
2.Time value of money
Answer: I = 103788 €
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.
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 + 0.09)1 = 19 000/1.09 = 17 431.19
CF2/(1 + 0.09)2 = 20 000/1.188 = 16 833.60
CF3/(1 + 0.09)3 = 20 000/1.295 = 15 443.67
CF4/(1 + 0.09)4 = 23 000/1.412 = 16 293.78
Add them up
NPV = €6 002.24
Answer
NPV = €6 002.24
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 11.55 %
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(10%) = 285.50
NPV(15%) = −587.66
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 10 + 5·285.50/(285.50 − (−587.66)) = 11.63%
exact: NPV(11.55%) = 0 → IRR = 11.55%
Answer
IRR = 11.55%
5.Bond pricing & yield
Answer: ΔP = −102.2 €
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
€1 065 = €60·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵
Solve numerically
P(4%) = €1 089.04, P(5%) = €1 043.29
YTM = 4.519%
Answer
YTM = 4.519%
7.CAPM & WACC
Answer: WACC = 7.557 %
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 = €7 000 000, E/V = 0.8571, D/V = 0.1429
WACC
WACC = 0.8571·8% + 0.1429·7%·(1 − 0.3) = 7.557%
Answer
WACC = 7.557%
8.CAPM & WACC
Answer: re = 11.68 %
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/(1 + 0.75·0.25) = 0.9263
Relever at the new D/E
βL = 0.9263·(1 + 0.75·0.75) = 1.447
Cost of equity (CAPM)
re = 3% + 1.447·6% = 11.68%
Answer
re = 11.68%
9.Descriptive statistics
Answer: s = 17.33
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743
Answer
P = 0.2743
11.Probability
Answer: P = 0.1515
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.
First and second draw
P = 5/12 · 4/11 = 20/132 = 0.1515
Answer
P = 0.1515
12.Probability
Answer: σY = 4.118
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.
Spread of X
μ = 2.4, Var(X) = 4.24, σX = 2.059
Linear change
σY = |−2|·2.059 = 4.118
Answer
σY = 4.118
The mean does move: E(Y) = −2·2.4 + 5 = 0.2.
13.Normal distribution
Answer: z = −1.37
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 − 0.9147 = 0.0853
Read the z-table backwards
Find 0.0853 inside the table and read off z.
z = −1.37
Answer
z = −1.37
14.Normal distribution
Answer: x = 48.6 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) = 0.05
z from the table
z ≈ −1.64
Back to the original units
x = 65 − 1.64·10 = 48.6 points
Answer
x = 48.6 points
15.Confidence intervals
Answer: E = 4.097 mmHg
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Critical value
df = 7, t = 2.365
Margin of error
E = 2.365·4.9/√8 = 4.097
Answer
E = 4.097 mmHg
16.Confidence intervals
Answer: n = 8129
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 = (1.96·23/0.5)² = 8129
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 8129
Answer
n = 8129
17.Hypothesis tests & confidence intervals
Answer: p = 0.025
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.
Test statistic
SE = σ/√n = 20/√100 = 2
z = (46.08 − 50)/2 = −1.96
Tail area
Φ(1.96) = 0.9750
1 − Φ(1.96) = 0.0250
p-value
p = 0.025
Answer
p = 0.025
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: U = 208.1
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 = 1.645·20/√49 = 4.7
Interval limit
U = 203.4 + 4.7 = 208.1
Answer
U = 208.1
19.Linear regression & correlation
Answer: r = 0.9971
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 = 1550 − 42·181/6 = 283
Sxx = 364 − 42²/6 = 70
Syy = 6611 − 181²/6 = 1151
Correlation coefficient
r = 283/√(70·1151) = 0.9971
Answer
r = 0.9971
20.Linear regression & correlation
Answer: ŷ = 92.29
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.