You invest €5 000 at 7% per year with continuous compounding. Find the value FV after 16 years.
PV = €5 000r = 7%t = 16
FV = €
2.Time value of moneyHard
You deposit €3 000 at the start of every year for 15 years, the first deposit today. The account pays 5% per year. How much is in the account at the end of year 15? Find FV.
A = €3 000i = 5%n = 15
FV = €
3.NPV & IRRMedium
A project costs €200 000 today. It generates €19 000 at the end of the first year, and the cash flow then grows by 1% per year forever. The cost of capital is 9%. Find the net present value NPV.
C0 = €200 000CF1 = €19 000g = 1%r = 9%
NPV = €
4.NPV & IRRHard
A project costs €20 000 and returns €15 000, €13 000, €13 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 15 000CF2 = 13 000CF3 = 13 500
IRR = %
5.Bond pricing & yieldMedium
You buy a bond with 12 years to maturity (face value €1 000, annual coupon 2%) when its yield to maturity is 3.5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 5%. Find your holding period return HPR in percent.
F = €1 000coupon = 2%n = 12y0 = 3.5%y1 = 5%
HPR = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 6% matures in 8 years and sells for €1 100. Find its yield to maturity YTM in percent.
F = €1 000coupon = 6%n = 8P = €1 100
YTM = %
7.CAPM & WACCMedium
A company has equity worth €5 000 000 and debt worth €1 000 000. Its WACC is 10.54%, its pre-tax cost of debt is 4.5% and the tax rate is 28%. Find the cost of equity re.
A company has 2 500 000 shares at €43.50 each and 5 000 bonds trading at €980 each. The pre-tax cost of debt is 4% and the tax rate is 30%. The share’s beta is 0.6, the risk-free rate 3% and the market return 7%. Find the WACC.
β = 0.6rf = 3%rm = 7%rd = 4%T = 30%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 13, 13, 13, 42, 40, 42.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 82?
μ = 70σ = 8x = 82
P =
11.ProbabilityMedium
In how many ways N can a committee of 5 people be chosen from a group of 9? The order does not matter.
n = 9k = 5
N =
12.ProbabilityHard
10 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?
n = 10p = 0.3
P =
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9736. Find z to two decimals.
A = 0.9736
z =
14.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.03 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
A sample of 7 items has mean 71 and standard deviation s = 13. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 7x̄ = 71s = 1390%t(6) = 1.943
upper limit =
16.Confidence intervalsHard
The population standard deviation is 38. 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 3?
σ = 38E = 395%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 100; the population standard deviation is known to be 5. A random sample of 100 observations has a mean of 98.8. Test H₀: μ = 100 against H₁: μ ≠ 100. Find the p-value p.
μ₀ = 100σ = 5n = 100x̄ = 98.8H₁: μ ≠ μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 500 against H₁: μ > 500 is done at α = 0.10 with a sample of 64; the standard deviation is known to be 12. In reality the mean is 501.95. Find the power 1 − β of the test in percent.
μ₀ = 500μ = 501.95σ = 12n = 64α = 0.10
1 − β = %
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 83?
n = 6ŷ = 83
x =
3Solutions
1.Time value of money
Answer: FV = 15324 €
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 = 0.07·16 = 1.12
e1.12 = 3.065
Future value
FV = 5 000·3.065 = €15 324.27
Answer
FV = €15 324.27
2.Time value of money
Answer: FV = 67972 €
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 + 0.05)15 − 1)/0.05 = 21.58
Annuity due
Each deposit is invested one year longer.
FV = 3 000·21.58·1.05 = €67 972.48
Answer
FV = €67 972.48
3.NPV & IRR
Answer: NPV = 37500 €
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.
Value of the cash flows
PV = 19 000/(0.09 − 0.01) = 237 500.00
NPV
NPV = 237 500.00 − 200 000 = €37 500.00
Answer
NPV = €37 500.00
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 49.02 %
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(45%) = 956.17
NPV(50%) = −222.22
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 45 + 5·956.17/(956.17 − (−222.22)) = 49.06%
exact: NPV(49.02%) = 0 → IRR = 49.02%
Answer
IRR = 49.02%
5.Bond pricing & yield
Answer: HPR = −9.852 %
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
€1 100 = €60·(1 − (1 + y)⁻⁸)/y + €1 000/(1 + y)⁸
Solve numerically
P(4%) = €1 134.65, P(5%) = €1 064.63
YTM = 4.485%
Answer
YTM = 4.485%
7.CAPM & WACC
Answer: re = 12 %
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 = €6 000 000, E/V = 0.8333, D/V = 0.1667
Debt part of the WACC
0.1667·4.5%·(1 − 0.28) = 0.54%
Cost of equity
re = (10.54% − 0.54%)/0.8333 = 12%
Answer
re = 12%
8.CAPM & WACC
Answer: WACC = 5.288 %
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 0.9332
Answer
P = 0.9332
11.Probability
Answer: N = 126
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.
Combinations
C(9, 5) = (9·8·7·6·5)/(5·4·3·2·1) = 126
Answer
N = 126
12.Probability
Answer: P = 0.1493
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) = 0.710 = 0.02825
P(X = 1) = 10·0.3·0.79 = 0.1211
Add them
P = 0.02825 + 0.1211 = 0.1493
Answer
P = 0.1493
13.Normal distribution
Answer: z = 2.22
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 + 0.9736)/2 = 0.9868
Read the z-table backwards
Find 0.9868 inside the table and read off z.
z = 2.22
Answer
z = 2.22
14.Normal distribution
Answer: P = 0.8849
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 difference
μ = 25.03 − 25 = 0.03 mm, σ = 0.025 mm
Standardise
z = (0 − 0.03)/0.025 = −1.2
Probability
P = 1 − Φ(−1.2) = 0.8849
Answer
P = 0.8849
15.Confidence intervals
Answer: upper limit = 80.55
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 = 6, t = 1.943
Margin of error
E = 1.943·13/√7 = 9.547
Interval
71 ± 9.547 → [61.45, 80.55]
Answer
upper limit = 80.55
16.Confidence intervals
Answer: n = 617
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 = (1.96·38/3)² = 616.4
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 617
Answer
n = 617
17.Hypothesis tests & confidence intervals
Answer: p = 0.0164
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 = 5/√100 = 0.5
z = (98.8 − 100)/0.5 = −2.4
Tail area
Φ(2.4) = 0.9918
1 − Φ(2.4) = 0.0082
p-value
p = 2·0.0082 = 0.0164
Answer
p = 0.0164
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: 1 − β = 50.72 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 12/√64 = 1.5
x̄c = 500 + 1.282·1.5 = 501.9
z under the true mean
z = (501.9 − 501.95)/1.5 = −0.018
Power
power = P(Z > −0.018) = 50.72%
1 − β = 50.72%
Answer
1 − β = 50.72%
19.Linear regression & correlation
Answer: r = 0.9774
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 = 47/5 = 9.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 64
Syy = Σ(y − ȳ)² = 107.2
Correlation coefficient
r = 64/√(40·107.2) = 0.9774
Answer
r = 0.9774
20.Linear regression & correlation
Answer: x = 20.92
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 = 114/6 = 19
ȳ = 443/6 = 73.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 335
Regression line
b = 335/70 = 4.786
a = 73.83 − 4.786·19 = −17.1
ŷ = −17.1 + 4.786·x
Solve for x
x = (83 − (−17.1))/4.786 = 20.92
Answer
x = 20.92
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