You invest €8 000 at a nominal rate of 3% per year, compounded quarterly. Find the value FV of the investment after 9 years.
PV = €8 0003% / 4n = 9
FV = €
2.Time value of moneyHard
A loan of €200 000 is repaid with equal monthly payments over 25 years at 3% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €200 0003% / 12N = 300
I = €
3.NPV & IRRMedium
A machine costs €60 000 today. It generates €12 000 at the end of each year for 5 years and can then be sold for €6 000. The cost of capital is 12%. Find the net present value NPV.
C0 = €60 000CF = €12 000S = €6 000n = 5r = 12%
NPV = €
4.NPV & IRRHard
An investment of €40 000 returns €7 500 at the end of each year for 6 years. Find its internal rate of return IRR in percent.
C0 = €40 000CF = €7 500n = 6
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 3 years. The yield to maturity is 3% per year. Find the price P of the bond.
F = €1 000coupon = 6%YTM = 3%n = 3
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 5% matures in 5 years and sells for €960. Find its yield to maturity YTM in percent.
F = €1 000coupon = 5%n = 5P = €960
YTM = %
7.CAPM & WACCMedium
A company has equity worth €2 000 000 and debt worth €3 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 4.5% and the tax rate is 21%. Find the WACC.
A company's shares have a beta of 1.1 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 0.6. The tax rate is 25%, the risk-free rate 3.5% and the market risk premium 4%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 35, 43, 26, 38, 29, 42, 58.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 10. Find the z-score z of a score of 165.
μ = 170σ = 10x = 165
z =
11.ProbabilityMedium
A fair die is rolled 8 times. What is the probability P of getting at least one six?
n = 8
P =
12.ProbabilityHard
7 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?
n = 7p = 0.3
P =
13.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 29.3 and 35.3 min? Give P as a decimal.
μ = 32σ = 6x1 = 29.3x2 = 35.3
P =
14.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 20% of them lie (the 20th percentile).
μ = 32σ = 6P = 20%
x = min
15.Confidence intervalsMedium
A sample of 12 items has mean 100 and standard deviation s = 5.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.
n = 12x̄ = 100s = 5.595%t(11) = 2.201
upper limit =
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 4?
σ = 23E = 495%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 203.4 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 203.4s = 10n = 49
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 74.7. The population standard deviation is 4. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 74.7σ = 4n = 6495%
U =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
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 (3, 13) is made. Find its residual e = y − ŷ.
n = 6(3, 13)
e =
3Solutions
1.Time value of money
Answer: FV = 10469 €
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 payment
i = 0.0025, N = 300
A = 200 000·0.0025/(1 − (1 + 0.0025)−300) = €948.42
Total interest
I = N·A − P = 300·948.4 − 200 000 = €84 526.79
Answer
I = €84 526.79
3.NPV & IRR
Answer: NPV = −13338 €
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.
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.
Required annuity factor
(1 − (1 + i)−6)/i = 40 000/7 500 = 5.333
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 3%: 5.417
i = 4%: 5.242
IRR = 3.473%
Answer
IRR = 3.473%
5.Bond pricing & yield
Answer: P = 1085 €
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 = 0.015, N = 6
Present value of the coupons
€30.00·(1 − 0.9145)/0.015 = €170.92
Present value of the face value
€1 000·0.9145 = €914.54
Price
P = €1 085.46
Answer
P = €1 085.46
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 5.948 %
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
€960 = €50·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵
Solve numerically
P(5%) = €1 000.00, P(6%) = €957.88
YTM = 5.948%
Answer
YTM = 5.948%
7.CAPM & WACC
Answer: WACC = 7.333 %
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 = €5 000 000, E/V = 0.4, D/V = 0.6
WACC
WACC = 0.4·13% + 0.6·4.5%·(1 − 0.21) = 7.333%
Answer
WACC = 7.333%
8.CAPM & WACC
Answer: re = 9.048 %
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 + 0.75·0.2) = 0.9565
Relever at the new D/E
βL = 0.9565·(1 + 0.75·0.6) = 1.387
Cost of equity (CAPM)
re = 3.5% + 1.387·4% = 9.048%
Answer
re = 9.048%
9.Descriptive statistics
Answer: s = 10.58
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 = (165 − 170)/10 = −0.5
Answer
z = −0.5
11.Probability
Answer: P = 0.7674
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)8 = 0.2326
P = 1 − 0.2326 = 0.7674
Answer
P = 0.7674
12.Probability
Answer: P = 0.3294
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) = 0.77 = 0.08235
P(X = 1) = 7·0.3·0.76 = 0.2471
Add them
P = 0.08235 + 0.2471 = 0.3294
Answer
P = 0.3294
13.Normal distribution
Answer: P = 0.3825
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 = (29.3 − 32)/6 = −0.45
z₂ = (35.3 − 32)/6 = 0.55
Area between
P = Φ(0.55) − Φ(−0.45) = 0.7088 − 0.3264 = 0.3825
Answer
P = 0.3825
14.Normal distribution
Answer: x = 26.95 min
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) = 0.2 → z = −0.8416
Back to x
x = 32 + −0.8416·6 = 26.95 min
Answer
x = 26.95 min
15.Confidence intervals
Answer: upper limit = 103.5
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 = 2.201·5.5/√12 = 3.495
Interval
100 ± 3.495 → [96.51, 103.5]
Answer
upper limit = 103.5
16.Confidence intervals
Answer: n = 128
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 = (1.96·23/4)² = 127
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 128
Answer
n = 128
17.Hypothesis tests & confidence intervals
Answer: t = 2.38
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
SE = 10/√49 = 10/7 = 1.429
Test statistic
t = (203.4 − 200)/1.429 = 2.38
Answer
t = 2.38
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: U = 75.68
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).
Margin of error
E = 1.96·4/√64 = 0.98
Interval limit
U = 74.7 + 0.98 = 75.68
Answer
U = 75.68
19.Linear regression & correlation
Answer: r² = 0.6385
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 = 255/6 = 42.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 15.5
Syy = Σ(y − ȳ)² = 21.5
Coefficient of determination
r² = 15.5²/(17.5·21.5) = 0.6385
Answer
r² = 0.6385
About 63.85% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = 4.71
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 = 83/6 = 13.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 97
Regression line
b = 97/70 = 1.386
a = 13.83 − 1.386·7 = 4.133
ŷ = 4.133 + 1.386·x
Predicted value
ŷ = 4.133 + 1.386·3 = 8.29
Residual
e = 13 − 8.29 = 4.71
Answer
e = 4.71
The point lies above the line.
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