You invest €20 000 at a nominal rate of 6% per year, compounded monthly. Find the value FV of the investment after 10 years.
PV = €20 0006% / 12n = 10
FV = €
2.Time value of moneyHard
A loan of €15 000 is repaid with equal monthly payments over 5 years. The interest rate is 5.4% per year, compounded monthly. Find the monthly payment A.
P = €15 0005.4% / 12N = 60
A = €
3.NPV & IRRMedium
A project costs €20 000 today and returns €5 500 at the end of each year for 6 years. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €25 000 and returns €17 500, €15 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −25 000CF1 = 17 500CF2 = 15 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 3% per year. Find the price P of the bond.
F = €1 000coupon = 2%YTM = 3%n = 12
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 8% matures in 5 years and sells for €1 060. Find its yield to maturity YTM in percent.
F = €1 000coupon = 8%n = 5P = €1 060
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €6 000 000, preferred shares worth €1 000 000 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 6%. The tax rate is 25.8%. Find the WACC.
A company's shares have a beta of 1.4 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 1.5. The tax rate is 25%, the risk-free rate 1% and the market risk premium 6%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 44, 37, 46, 60, 48.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is above 155?
μ = 170σ = 15x = 155
P =
11.ProbabilityMedium
A fair die is rolled 2 times. What is the probability P of getting at least one six?
n = 2
P =
12.ProbabilityHard
12 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?
n = 12p = 0.2
P =
13.Normal distributionMedium
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 110.1 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 110.1 s
P =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 20% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g20%
x = g
15.Confidence intervalsMedium
A sample of 12 items has mean 68 and standard deviation s = 10. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.
n = 12x̄ = 68s = 1095%t(11) = 2.201
E =
16.Confidence intervalsHard
The population standard deviation is 26. 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 1?
σ = 26E = 195%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 120; the population standard deviation is known to be 24. A random sample of 36 observations has a mean of 124.4. Test H₀: μ = 120 against H₁: μ > 120. Find the p-value p.
μ₀ = 120σ = 24n = 36x̄ = 124.4H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 5. Find the smallest sample size n.
σ = 5E = 0.590%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (5, 58) is made. Find its residual e = y − ŷ.
n = 5(5, 58)
e =
3Solutions
1.Time value of money
Answer: FV = 36388 €
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.
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 = 12·5 = 60
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 = 0.7638
A = 15 000·0.0045/(1 − 0.7638) = €285.83
Answer
A = €285.83
3.NPV & IRR
Answer: DPB = 5.062 years
Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.
Discounted and cumulative cash flows
year 1: 5 500/1.12 = 4 910.71 → Σ = 4 910.71
year 2: 5 500/1.254 = 4 384.57 → Σ = 9 295.28
year 3: 5 500/1.405 = 3 914.79 → Σ = 13 210.07
year 4: 5 500/1.574 = 3 495.35 → Σ = 16 705.42
year 5: 5 500/1.762 = 3 120.85 → Σ = 19 826.27
year 6: 5 500/1.974 = 2 786.47 → Σ = 22 612.74
Fraction of the last year
DPB = 5 + 173.73/2 786.47 = 5.062
Answer
DPB = 5.062 years
The discounted payback is always longer than the simple payback (3.636 years).
4.NPV & IRR
Answer: IRR = 21.17 %
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(20%) = 347.22
NPV(25%) = −1 080.00
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 = €10.00, i = 0.015, N = 24
Present value of the coupons
€10.00·(1 − 0.6995)/0.015 = €200.30
Present value of the face value
€1 000·0.6995 = €699.54
Price
P = €899.85
Answer
P = €899.85
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 6.554 %
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
€1 060 = €80·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵
Solve numerically
P(6%) = €1 084.25, P(7%) = €1 041.00
YTM = 6.554%
Answer
YTM = 6.554%
7.CAPM & WACC
Answer: WACC = 11.16 %
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 = 1.4/(1 + 0.75·0.2) = 1.217
Relever at the new D/E
βL = 1.217·(1 + 0.75·1.5) = 2.587
Cost of equity (CAPM)
re = 1% + 2.587·6% = 16.52%
Answer
re = 16.52%
9.Descriptive statistics
Answer: s = 8.367
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) = 0.1587
P = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
11.Probability
Answer: P = 0.3056
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)2 = 0.6944
P = 1 − 0.6944 = 0.3056
Answer
P = 0.3056
12.Probability
Answer: P = 0.2749
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) = 0.812 = 0.06872
P(X = 1) = 12·0.2·0.811 = 0.2062
Add them
P = 0.06872 + 0.2062 = 0.2749
Answer
P = 0.2749
13.Normal distribution
Answer: P = 0.2119
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 = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (110.1 − 105)/6.403 = 0.8
Upper tail
P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119
Answer
P = 0.2119
14.Normal distribution
Answer: x = 254.2 g
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) = 0.8
z from the table
z ≈ 0.84
Back to the original units
x = 250 + 0.84·5 = 254.2 g
Answer
x = 254.2 g
15.Confidence intervals
Answer: E = 6.354
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 = 11, t = 2.201
Margin of error
E = 2.201·10/√12 = 6.354
Answer
E = 6.354
16.Confidence intervals
Answer: n = 2597
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 = (1.96·26/1)² = 2597
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 2597
Answer
n = 2597
17.Hypothesis tests & confidence intervals
Answer: p = 0.1357
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 = 24/√36 = 4
z = (124.4 − 120)/4 = 1.1
Tail area
Φ(1.1) = 0.8643
1 − Φ(1.1) = 0.1357
p-value
p = 0.1357
Answer
p = 0.1357
p > 0.05, so H₀ is not rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: n = 271
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 = (1.645·5/0.5)² = 270.6
Round up
n = 271
Answer
n = 271
19.Linear regression & correlation
Answer: r² = 0.8893
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 = 100/5 = 20
ȳ = 300/5 = 60
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 291
Syy = Σ(y − ȳ)² = 1058
Coefficient of determination
r² = 291²/(90·1058) = 0.8893
Answer
r² = 0.8893
About 88.93% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = −5.15
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 = 30/5 = 6
ȳ = 340/5 = 68
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 194
Regression line
b = 194/40 = 4.85
a = 68 − 4.85·6 = 38.9
ŷ = 38.9 + 4.85·x
Predicted value
ŷ = 38.9 + 4.85·5 = 63.15
Residual
e = 58 − 63.15 = −5.15
Answer
e = −5.15
The point lies below the line.
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