You invest €20 000 at 8% per year with continuous compounding. Find the value FV after 7 years.
PV = €20 000r = 8%t = 7
FV = €
2.Time value of moneyHard
You borrow €5 000 at 3% per year, compounded monthly, and repay €100 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 = €1003% / 12
N = payments
3.NPV & IRRMedium
An investment of €30 000 gives these cash flows at the end of years 1 to 4: €12 000, €13 000, €12 000, €11 000. The cost of capital is 5%. Find the net present value NPV.
A project costs €20 000 and returns €12 500, €14 500, €12 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 12 500CF2 = 14 500CF3 = 12 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 6 years. The yield to maturity is 7% per year. Find the price P of the bond.
F = €1 000coupon = 8%YTM = 7%n = 6
P = €
6.Bond pricing & yieldHard
A bond with 20 years to maturity (face value €1 000, annual coupon 9%) sells for €1 098. The issuer can call it in 4 years at a call price of €1 090. Find the yield to call YTC in percent.
F = €1 000coupon = 9%P = €1 098call price = €1 090call in 4 years
YTC = %
7.CAPM & WACCMedium
A share with a beta of 0.6 has a required return of 6.4%. The risk-free rate is 2.5%. What expected market return rm does CAPM imply?
β = 0.6re = 6.4%rf = 2.5%
rm = %
8.CAPM & WACCHard
A company's business has an unlevered (asset) beta of 0.6. It plans a debt-to-equity ratio of 0.75. The tax rate is 30%, the risk-free rate is 2% and the market risk premium is 6%. Find the cost of equity re at the new capital structure.
βU = 0.6D/E = 0.75T = 30%rf = 2%rm − rf = 6%
re = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 29, 35, 29, 27, 52.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is below 545?
μ = 500σ = 75x = 545
P =
11.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 7? The order does not matter.
n = 7k = 2
N =
12.ProbabilityHard
8 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?
n = 8p = 0.3
P =
13.Normal distributionMedium
X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 764.8 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 764.8 g
P =
14.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points80%
x = points
15.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 3.7. What is the margin of error E₂ of a 99% interval based on 144 observations?
95%n₁ = 36E₁ = 3.799%n₂ = 144
E₂ =
16.Confidence intervalsHard
The diameters of 30 machined shafts have a sample standard deviation of 10.6 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.
n = 30s = 10.6 µmχ² = 16.047, 45.722
σ = µm
17.Hypothesis tests & confidence intervalsMedium
16 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 −4.4 and the standard deviation of the differences is 8. Find the paired test statistic t for H₀: μd = 0.
n = 16d̄ = −4.4sd = 8
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 88 of 250 patients on treatment A recover, and 36 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 88n₁ = 250x₂ = 36n₂ = 120
z =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 30Σy = 134Σx² = 220Σy² = 4388Σxy = 982
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 20?
n = 5ŷ = 20
x =
3Solutions
1.Time value of money
Answer: FV = 35013 €
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 = 0.08·7 = 0.56
e0.56 = 1.751
Future value
FV = 20 000·1.751 = €35 013.45
Answer
FV = €35 013.45
2.Time value of money
Answer: N = 54 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 = 0.0025
(1 + i)−N = 1 − 5 000·0.0025/100 = 0.875
Solve with logarithms
N = −ln(0.875)/ln(1.002) = 53.48
Round up to whole payments
53 full payments leave a small balance, which the last, smaller payment clears.
N = 54
Answer
N = 54 payments
3.NPV & IRR
Answer: NPV = 12636 €
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 = −30 000
CF1/(1 + 0.05)1 = 12 000/1.05 = 11 428.57
CF2/(1 + 0.05)2 = 13 000/1.103 = 11 791.38
CF3/(1 + 0.05)3 = 12 000/1.158 = 10 366.05
CF4/(1 + 0.05)4 = 11 000/1.216 = 9 049.73
Add them up
NPV = €12 635.73
Answer
NPV = €12 635.73
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 42.74 %
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(40%) = 699.71
NPV(45%) = −546.56
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 40 + 5·699.71/(699.71 − (−546.56)) = 42.81%
exact: NPV(42.74%) = 0 → IRR = 42.74%
Answer
IRR = 42.74%
5.Bond pricing & yield
Answer: P = 1048 €
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 = €40.00, i = 0.035, N = 12
Present value of the coupons
€40.00·(1 − 0.6618)/0.035 = €386.53
Present value of the face value
€1 000·0.6618 = €661.78
Price
P = €1 048.32
Answer
P = €1 048.32
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTC = 8.035 %
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
1098 = 90·(1 − (1 + y)−4)/y + 1090/(1 + y)4
Solve numerically
y = 8%: €1 099.27
y = 9%: €1 063.76
YTC = 8.035%
Answer
YTC = 8.035%
The yield to maturity is 8.002%. 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: rm = 9 %
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).
Market risk premium
rm − rf = (6.4% − 2.5%)/0.6 = 6.5%
Market return
rm = 2.5% + 6.5% = 9%
Answer
rm = 9%
8.CAPM & WACC
Answer: re = 7.49 %
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.
Levered beta
βL = 0.6·(1 + 0.7·0.75) = 0.915
Cost of equity (CAPM)
re = 2% + 0.915·6% = 7.49%
Answer
re = 7.49%
9.Descriptive statistics
Answer: s = 10.29
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 0.7257
Answer
P = 0.7257
11.Probability
Answer: N = 21
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(7, 2) = (7·6)/(2·1) = 21
Answer
N = 21
12.Probability
Answer: P = 0.2553
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) = 0.78 = 0.05765
P(X = 1) = 8·0.3·0.77 = 0.1977
Add them
P = 0.05765 + 0.1977 = 0.2553
Answer
P = 0.2553
13.Normal distribution
Answer: P = 0.06681
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 = 750 g, σ = √(4² + 9²) = 9.849 g
Standardise
z = (764.8 − 750)/9.849 = 1.5
Upper tail
P = 1 − Φ(1.5) = 1 − 0.9332 = 0.06681
Answer
P = 0.06681
14.Normal distribution
Answer: x = 87.4 points
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) = 0.2
z from the table
z ≈ −0.84
Back to the original units
x = 100 − 0.84·15 = 87.4 points
Answer
x = 87.4 points
15.Confidence intervals
Answer: E₂ = 2.431
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 = (2.576/1.96)·√(36/144) = 0.6571
New margin
E₂ = 3.7·0.6571 = 2.431
Answer
E₂ = 2.431
16.Confidence intervals
Answer: σ = 8.442 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 29·10.6² = 3258
Limits for σ²
3258/45.722 = 71.27
3258/16.047 = 203.1
Limit for σ
σ = √71.27 = 8.442 µm
Answer
σ = 8.442 µm
95% interval for σ: [8.442, 14.25] µm.
17.Hypothesis tests & confidence intervals
Answer: t = −2.2
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 = 8/√16 = 2
Test statistic
t = −4.4/2 = −2.2
Answer
t = −2.2
With df = 15, the two-sided 5% critical value is 2.131; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: z = 0.9919
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.
Sample proportions
p̂₁ = 88/250 = 0.352
p̂₂ = 36/120 = 0.3
p̂ = 124/370 = 0.3351
Standard error under H₀
SE = √(0.3351·0.6649·(1/250 + 1/120)) = 0.05242
Test statistic
z = (0.352 − 0.3)/0.05242 = 0.9919
Answer
z = 0.9919
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.997
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 = 982 − 30·134/5 = 178
Sxx = 220 − 30²/5 = 40
Syy = 4388 − 134²/5 = 796.8
Correlation coefficient
r = 178/√(40·796.8) = 0.997
Answer
r = 0.997
20.Linear regression & correlation
Answer: x = 4.866
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 = 119/5 = 23.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 134
Regression line
b = 134/40 = 3.35
a = 23.8 − 3.35·6 = 3.7
ŷ = 3.7 + 3.35·x
Solve for x
x = (20 − 3.7)/3.35 = 4.866
Answer
x = 4.866
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