An annuity pays €1 500 at the end of each year for 8 years. The interest rate is 7% per year. Find the present value PV of the annuity.
A = €1 500i = 7%n = 8
PV = €
2.Time value of moneyHard
You deposit €2 000 at the start of every year for 19 years, the first deposit today. The account pays 3% per year. How much is in the account at the end of year 19? Find FV.
A = €2 000i = 3%n = 19
FV = €
3.NPV & IRRMedium
A machine costs €20 000 today, lasts 10 years and costs €1 500 per year to run (paid at the end of each year). The cost of capital is 5%. Find its equivalent annual cost EAC.
C0 = €20 000M = €1 500n = 10r = 5%
EAC = €
4.NPV & IRRHard
A project costs €50 000 and returns €36 500, €18 000, €33 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −50 000CF1 = 36 500CF2 = 18 000CF3 = 33 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 8% per year. Find the price P of the bond.
F = €1 000coupon = 6%YTM = 8%n = 12
P = €
6.Bond pricing & yieldHard
A bond trades at €1 010 with a yield to maturity of 6% (annual compounding) and a Macaulay duration of 6.6 years. Use the modified duration to estimate the new price P if the yield falls by 100 basis points.
P0 = €1 010y = 6%D = 6.6Δy = −100 bp
P = €
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €10 000 000, preferred shares worth €2 000 000 and debt worth €2 000 000. The cost of equity is 14%, the cost of preferred stock 7.5% and the pre-tax cost of debt 4.5%. The tax rate is 25%. Find the WACC.
A company's shares have a beta of 1.2 at its current debt-to-equity ratio of 0.4. It plans to raise its debt-to-equity ratio to 0.75. The tax rate is 28%, the risk-free rate 2.5% 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: 59, 36, 40, 40, 29.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is below 67.5?
μ = 60σ = 5x = 67.5
P =
11.ProbabilityMedium
A fair die is rolled 4 times. What is the probability P of getting at least one six?
n = 4
P =
12.ProbabilityHard
6 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?
n = 6p = 0.2
P =
13.Normal distributionMedium
The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 9 of them is below 365.4 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 9< 365.4 g
P =
14.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 75% of them lie (the 75th percentile).
μ = 178σ = 7P = 75%
x = cm
15.Confidence intervalsMedium
A sample of 11 items has mean 91.5 and standard deviation s = 19. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.
n = 11x̄ = 91.5s = 1995%t(10) = 2.228
E =
16.Confidence intervalsHard
The population standard deviation is 8. 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?
σ = 8E = 299%
n =
17.Hypothesis tests & confidence intervalsMedium
7 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 −8.03 and the standard deviation of the differences is 5.9. Find the paired test statistic t for H₀: μd = 0.
n = 7d̄ = −8.03sd = 5.9
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 2.5? The standard deviation is 8. Find the smallest sample size n.
σ = 8E = 2.599%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.5.
n = 5x = 3.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 8957 €
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 − 0.582)/0.07 = 5.971
Present value
PV = 1 500·5.971 = €8 956.95
Answer
PV = €8 956.95
2.Time value of money
Answer: FV = 51741 €
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 + 0.03)19 − 1)/0.03 = 25.12
Annuity due
Each deposit is invested one year longer.
FV = 2 000·25.12·1.03 = €51 740.75
Answer
FV = €51 740.75
3.NPV & IRR
Answer: EAC = 4090 €
Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.
Annuity factor
(1 − 1.05−10)/0.05 = 7.722
Annualised purchase price
20 000/7.722 = 2 590.09
EAC
EAC = 2 590.09 + 1 500 = €4 090.09
Answer
EAC = €4 090.09
4.NPV & IRR
Answer: IRR = 35.82 %
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%) = 529.39
NPV(40%) = −2 536.44
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
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 = €30.00, i = 0.04, N = 24
Present value of the coupons
€30.00·(1 − 0.3901)/0.04 = €457.41
Present value of the face value
€1 000·0.3901 = €390.12
Price
P = €847.53
Answer
P = €847.53
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: P = 1073 €
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.
Modified duration
Dmod = 6.6/1.06 = 6.226
Estimated price change
%ΔP ≈ −6.226·(−0.01) = 6.226%
P ≈ 1010·(1 + 0.06226) = €1 072.89
Answer
P = €1 072.89
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: WACC = 11.55 %
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 = 1.2/(1 + 0.72·0.4) = 0.9317
Relever at the new D/E
βL = 0.9317·(1 + 0.72·0.75) = 1.435
Cost of equity (CAPM)
re = 2.5% + 1.435·4.5% = 8.957%
Answer
re = 8.957%
9.Descriptive statistics
Answer: s = 11.12
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 = 0.9332
Answer
P = 0.9332
11.Probability
Answer: P = 0.5177
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)4 = 0.4823
P = 1 − 0.4823 = 0.5177
Answer
P = 0.5177
12.Probability
Answer: P = 0.6554
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) = 0.86 = 0.2621
P(X = 1) = 6·0.2·0.85 = 0.3932
Add them
P = 0.2621 + 0.3932 = 0.6554
Answer
P = 0.6554
13.Normal distribution
Answer: P = 0.9332
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 total
μT = 360 g, σT = 1.2·√9 = 3.6 g
Standardise
z = (365.4 − 360)/3.6 = 1.5
Probability
P = Φ(1.5) = 0.9332
Answer
P = 0.9332
14.Normal distribution
Answer: x = 182.7 cm
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 for the percentile
Φ(z) = 0.75 → z = 0.6745
Back to x
x = 178 + 0.6745·7 = 182.72 cm
Answer
x = 182.72 cm
15.Confidence intervals
Answer: E = 12.76
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 = 10, t = 2.228
Margin of error
E = 2.228·19/√11 = 12.76
Answer
E = 12.76
16.Confidence intervals
Answer: n = 107
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 = (2.576·8/2)² = 106.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 107
Answer
n = 107
17.Hypothesis tests & confidence intervals
Answer: t = −3.601
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 = 5.9/√7 = 2.23
Test statistic
t = −8.03/2.23 = −3.601
Answer
t = −3.601
With df = 6, the two-sided 5% critical value is 2.447; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: n = 68
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 = (2.576·8/2.5)² = 67.95
Round up
n = 68
Answer
n = 68
19.Linear regression & correlation
Answer: r = 0.977
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 = 11028 − 105·504/5 = 444
Sxx = 2295 − 105²/5 = 90
Syy = 53098 − 504²/5 = 2295
Correlation coefficient
r = 444/√(90·2295) = 0.977
Answer
r = 0.977
20.Linear regression & correlation
Answer: ŷ = 16.45
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.