An annuity pays €3 000 at the end of each year for 18 years. The interest rate is 8% per year. Find the present value PV of the annuity.
A = €3 000i = 8%n = 18
PV = €
2.Time value of moneyHard
You borrow €5 000 at 4.8% per year, compounded monthly, and repay €125 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 = €5 000A = €1254.8% / 12
N = payments
3.NPV & IRRMedium
A machine costs €60 000 today. It generates €11 000 at the end of each year for 8 years and can then be sold for €9 000. The cost of capital is 12%. Find the net present value NPV.
C0 = €60 000CF = €11 000S = €9 000n = 8r = 12%
NPV = €
4.NPV & IRRHard
A project costs €40 000 and returns €16 000, €27 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −40 000CF1 = 16 000CF2 = 27 000
IRR = %
5.Bond pricing & yieldMedium
You buy a bond with 7 years to maturity (face value €1 000, annual coupon 2%) when its yield to maturity is 8%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 7%. Find your holding period return HPR in percent.
F = €1 000coupon = 2%n = 7y0 = 8%y1 = 7%
HPR = %
6.Bond pricing & yieldHard
A bond with 15 years to maturity (face value €1 000, annual coupon 5%) sells for €1 239. The issuer can call it in 3 years at a call price of €1 050. Find the yield to call YTC in percent.
F = €1 000coupon = 5%P = €1 239call price = €1 050call in 3 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €4 000 000 and debt worth €4 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 3% and the tax rate is 30%. Find the WACC.
A company has 2 000 000 shares trading at €38.00; the book value of its equity is €45 600 000. It just paid a dividend of €1.50 per share, and dividends grow by 3% per year. Its debt has a market value of €22 800 000 and a pre-tax cost of 4%. The tax rate is 25%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 37, 22, 42, 10, 12, 40, 54.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is above 62.5?
μ = 70σ = 15x = 62.5
P =
11.ProbabilityMedium
A bag holds 5 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
5 / 6
P =
12.ProbabilityHard
A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 3 correct answers?
n = 12p = 0.4k = 3
P =
13.Normal distributionMedium
X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.3 mm. Round z to two decimals.
μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.3 mm
P =
14.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.
μ = 178 cmσ = 7 cm90%
x = cm
15.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 8.7. What is the margin of error E₂ of a 99% interval from the same data?
90%n₁ = 25E₁ = 8.799%
E₂ =
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 0.5?
σ = 31E = 0.590%
n =
17.Hypothesis tests & confidence intervalsMedium
6 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 11.48 and the standard deviation of the differences is 7.6. Find the paired test statistic t for H₀: μd = 0.
n = 6d̄ = 11.48sd = 7.6
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 1.5? The standard deviation is 20. Find the smallest sample size n.
σ = 20E = 1.599%
n =
19.Linear regression & correlationMedium
The table gives 5 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.
n = 5Σx = 20Σy = 168Σx² = 90Σy² = 5726Σxy = 695
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 18?
n = 6ŷ = 18
x =
3Solutions
1.Time value of money
Answer: PV = 28116 €
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 − 0.2502)/0.08 = 9.372
Present value
PV = 3 000·9.372 = €28 115.66
Answer
PV = €28 115.66
2.Time value of money
Answer: N = 44 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 = 0.004
(1 + i)−N = 1 − 5 000·0.004/125 = 0.84
Solve with logarithms
N = −ln(0.84)/ln(1.004) = 43.68
Round up to whole payments
43 full payments leave a small balance, which the last, smaller payment clears.
N = 44
Answer
N = 44 payments
3.NPV & IRR
Answer: NPV = −1721 €
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
11 000·4.968 = 54 644.04
Present value of the salvage value
9 000/2.476 = 3 634.95
NPV
NPV = −60 000 + 54 644.04 + 3 634.95 = −€1 721.01
Answer
NPV = −€1 721.01
4.NPV & IRR
Answer: IRR = 4.558 %
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(0%) = 3 000.00
NPV(5%) = −272.11
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.
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
1239 = 50·(1 − (1 + y)−3)/y + 1050/(1 + y)3
Solve numerically
y = 50%: €381.48
y = 50%: €381.48
YTC = 50%
Answer
YTC = 50%
The yield to maturity is 2.998%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.
7.CAPM & WACC
Answer: WACC = 6.55 %
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).
Weights
V = €8 000 000, E/V = 0.5, D/V = 0.5
WACC
WACC = 0.5·11% + 0.5·3%·(1 − 0.3) = 6.55%
Answer
WACC = 6.55%
8.CAPM & WACC
Answer: WACC = 6.128 %
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915
Answer
P = 0.6915
11.Probability
Answer: P = 0.1818
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 = 5/11 · 4/10 = 20/110 = 0.1818
Answer
P = 0.1818
12.Probability
Answer: P = 0.1419
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.
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 sum
μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm
Standardise
z = (199.3 − 200)/0.5 = −1.4
Upper tail
P = 1 − Φ(−1.4) = 1 − 0.08076 = 0.9192
Answer
P = 0.9192
14.Normal distribution
Answer: x = 187 cm
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) = 0.9
z from the table
z ≈ 1.28
Back to the original units
x = 178 + 1.28·7 = 187 cm
Answer
x = 187 cm
15.Confidence intervals
Answer: E₂ = 13.62
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.
Ratio of margins
E₂/E₁ = (2.576/1.645)·√(25/25) = 1.566
New margin
E₂ = 8.7·1.566 = 13.62
Answer
E₂ = 13.62
16.Confidence intervals
Answer: n = 10402
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 = (1.645·31/0.5)² = 10402
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 10402
Answer
n = 10402
17.Hypothesis tests & confidence intervals
Answer: t = 3.7
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 = 7.6/√6 = 3.103
Test statistic
t = 11.48/3.103 = 3.7
Answer
t = 3.7
With df = 5, the two-sided 5% critical value is 2.571; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: n = 1180
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 = (2.576·20/1.5)² = 1180
Round up
n = 1180
Answer
n = 1180
19.Linear regression & correlation
Answer: r = 0.8071
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 = 695 − 20·168/5 = 23
Sxx = 90 − 20²/5 = 10
Syy = 5726 − 168²/5 = 81.2
Correlation coefficient
r = 23/√(10·81.2) = 0.8071
Answer
r = 0.8071
20.Linear regression & correlation
Answer: x = 4.81
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 = 21/6 = 3.5
ȳ = 119/6 = 19.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −24.5
Regression line
b = −24.5/17.5 = −1.4
a = 19.83 − (−1.4)·3.5 = 24.73
ŷ = 24.73 − 1.4·x
Solve for x
x = (18 − 24.73)/(−1.4) = 4.81
Answer
x = 4.81
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