You invest €1 000 at 3% per year with continuous compounding. Find the value FV after 14 years.
PV = €1 000r = 3%t = 14
FV = €
2.Time value of moneyHard
You want €100 000 in 19 years. You save the same amount at the end of every month in an account paying 5.4% per year, compounded monthly. Find the monthly deposit A.
FV = €100 0005.4% / 12N = 228
A = €
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 3: €22 000, €14 000, €20 000. The cost of capital is 15%. Find the net present value NPV.
A project costs €50 000 today. Its cash flow is €11 000 at the end of year 1 and grows by 5% per year until the last cash flow at the end of year 5. The cost of capital is 10%. Find the net present value NPV.
C0 = €50 000CF1 = €11 000g = 5%n = 5r = 10%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 4% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 5% per year. Find the price P of the bond.
F = €1 000coupon = 4%YTM = 5%n = 12
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 5% matures in 6 years and sells for €1 050. Find its yield to maturity YTM in percent.
F = €1 000coupon = 5%n = 6P = €1 050
YTM = %
7.CAPM & WACCMedium
A company has equity worth €5 000 000 and debt worth €5 000 000. Its WACC is 8.6%, its pre-tax cost of debt is 6% and the tax rate is 30%. Find the cost of equity re.
A company has 2 500 000 shares at €33.50 each and 40 000 bonds trading at €1 020 each. The pre-tax cost of debt is 7% and the tax rate is 21%. The share’s beta is 0.9, the risk-free rate 3.5% and the market return 9.5%. Find the WACC.
β = 0.9rf = 3.5%rm = 9.5%rd = 7%T = 21%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 23, 17, 45, 11, 41, 22.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 10. What proportion P of values is above 80?
μ = 100σ = 10x = 80
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.15, 0.3, 0.3, 0.25
Var(X) =
12.ProbabilityHard
X has the distribution shown, and Y = −2X + 1. Find the standard deviation σY of Y.
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0170. Find z to two decimals.
A = 0.0170
z =
14.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 75% of them lie (the 75th percentile).
μ = 32σ = 6P = 75%
x = min
15.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 4.4. What is the margin of error E₂ of a 95% interval from the same data?
99%n₁ = 50E₁ = 4.495%
E₂ =
16.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.01?
p = 0.5E = 0.0190%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 100; the population standard deviation is known to be 12. A random sample of 64 observations has a mean of 103.39. Test H₀: μ = 100 against H₁: μ ≠ 100. Find the p-value p.
μ₀ = 100σ = 12n = 64x̄ = 103.39H₁: μ ≠ μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A sample of 49 measurements has mean 501. The population standard deviation is 4. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 501σ = 4n = 4999%
U =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. 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 = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 20) is made. Find its residual e = y − ŷ.
n = 6(2.5, 20)
e =
3Solutions
1.Time value of money
Answer: FV = 1522 €
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.
Exponent
r·t = 0.03·14 = 0.42
e0.42 = 1.522
Future value
FV = 1 000·1.522 = €1 521.96
Answer
FV = €1 521.96
2.Time value of money
Answer: A = 252.3 €
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 + 0.15)1 = 22 000/1.15 = 19 130.43
CF2/(1 + 0.15)2 = 14 000/1.322 = 10 586.01
CF3/(1 + 0.15)3 = 20 000/1.521 = 13 150.32
Add them up
NPV = €−17 133.23
Answer
NPV = €−17 133.23
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: NPV = −4343 €
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 growing annuity
((1.05)/(1.1))5 = 0.7925
PV = 11 000/(0.1 − 0.05)·(1 − 0.7925) = 45 656.50
NPV
NPV = 45 656.50 − 50 000 = −€4 343.50
Answer
NPV = −€4 343.50
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: P = 910.6 €
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 = €20.00, i = 0.025, N = 24
Present value of the coupons
€20.00·(1 − 0.5529)/0.025 = €357.70
Present value of the face value
€1 000·0.5529 = €552.88
Price
P = €910.58
Answer
P = €910.58
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 4.045 %
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
€1 050 = €50·(1 − (1 + y)⁻⁶)/y + €1 000/(1 + y)⁶
Solve numerically
P(4%) = €1 052.42, P(5%) = €1 000.00
YTM = 4.045%
Answer
YTM = 4.045%
7.CAPM & WACC
Answer: re = 13 %
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 = €10 000 000, E/V = 0.5, D/V = 0.5
Debt part of the WACC
0.5·6%·(1 − 0.3) = 2.1%
Cost of equity
re = (8.6% − 2.1%)/0.5 = 13%
Answer
re = 13%
8.CAPM & WACC
Answer: WACC = 7.796 %
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772
Answer
P = 0.9772
11.Probability
Answer: Var(X) = 1.028
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.
Mean
μ = 0·0.15 + 1·0.3 + 2·0.3 + 3·0.25 = 1.65
Mean of the squares
E(X²) = 0·0.15 + 1·0.3 + 4·0.3 + 9·0.25 = 3.75
Variance
Var(X) = 3.75 − 1.65² = 1.028
Answer
Var(X) = 1.028
12.Probability
Answer: σY = 2.154
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 = 2.2, Var(X) = 1.16, σX = 1.077
Linear change
σY = |−2|·1.077 = 2.154
Answer
σY = 2.154
The mean does move: E(Y) = −2·2.2 + 1 = −3.4.
13.Normal distribution
Answer: z = 2.12
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 − 0.0170 = 0.9830
Read the z-table backwards
Find 0.9830 inside the table and read off z.
z = 2.12
Answer
z = 2.12
14.Normal distribution
Answer: x = 36.05 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) = 0.75 → z = 0.6745
Back to x
x = 32 + 0.6745·6 = 36.05 min
Answer
x = 36.05 min
15.Confidence intervals
Answer: E₂ = 3.348
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 = (1.96/2.576)·√(50/50) = 0.7609
New margin
E₂ = 4.4·0.7609 = 3.348
Answer
E₂ = 3.348
16.Confidence intervals
Answer: n = 6766
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 1.645²·0.5·0.5/0.01² = 6765
Round up
n = 6766
Answer
n = 6766
17.Hypothesis tests & confidence intervals
Answer: p = 0.02382
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 = 12/√64 = 1.5
z = (103.39 − 100)/1.5 = 2.26
Tail area
Φ(2.26) = 0.9881
1 − Φ(2.26) = 0.0119
p-value
p = 2·0.0119 = 0.02382
Answer
p = 0.02382
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: U = 502.5
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 = 2.576·4/√49 = 1.472
Interval limit
U = 501 + 1.472 = 502.5
Answer
U = 502.5
19.Linear regression & correlation
Answer: r² = 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.
Means
x̄ = 153/6 = 25.5
ȳ = 672/6 = 112
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 903
Syy = Σ(y − ȳ)² = 5192
Coefficient of determination
r² = 903²/(157.5·5192) = 0.9971
Answer
r² = 0.9971
About 99.71% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = −2.338
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 = 127/6 = 21.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −20.5
Regression line
b = −20.5/17.5 = −1.171
a = 21.17 − (−1.171)·3.5 = 25.27
ŷ = 25.27 − 1.171·x
Predicted value
ŷ = 25.27 + (−1.171)·2.5 = 22.34
Residual
e = 20 − 22.34 = −2.338
Answer
e = −2.338
The point lies below the line.
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