You invest €2 000 at 2% per year with continuous compounding. Find the value FV after 17 years.
PV = €2 000r = 2%t = 17
FV = €
2.Time value of moneyHard
A loan of €150 000 is repaid with equal monthly payments over 30 years. The interest rate is 2.4% per year, compounded monthly. Find the monthly payment A.
P = €150 0002.4% / 12N = 360
A = €
3.NPV & IRRMedium
A project costs €40 000 today and returns €14 000, €10 000, €16 000, €14 000 at the end of years 1 to 4. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
Project A costs €10 000 and returns €5 500, €4 000, €2 500 at the end of years 1 to 3. Project B costs €15 000 and returns €4 000, €6 000, €8 000. Find the crossover rate r* at which both projects have the same NPV, in percent.
A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 8 years. The yield to maturity is 4% per year. Find the price P of the bond.
F = €1 000coupon = 4.5%YTM = 4%n = 8
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 7% matures in 9 years and sells for €855. Find its yield to maturity YTM in percent.
F = €1 000coupon = 7%n = 9P = €855
YTM = %
7.CAPM & WACCMedium
A company has equity worth €2 000 000 and debt worth €1 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 4% and the tax rate is 21%. Find the WACC.
A company has 1 000 000 shares at €48.50 each and 5 000 bonds trading at €1 050 each. The pre-tax cost of debt is 3.5% and the tax rate is 21%. The share’s beta is 0.8, the risk-free rate 2% and the market return 8%. Find the WACC.
β = 0.8rf = 2%rm = 8%rd = 3.5%T = 21%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 12, 59, 47, 19, 21, 12.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is below 180?
μ = 170σ = 8x = 180
P =
11.ProbabilityMedium
A bag holds 5 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
5 / 6
P =
12.ProbabilityHard
9 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?
n = 9p = 0.25
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 111.4 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 111.4 s
P =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 90% is to the right of it. Find the cut-off x.
μ = 500 hσ = 40 h90%
x = h
15.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 9.4. What is the margin of error E₂ if the same study uses 400 observations instead?
90%n₁ = 100E₁ = 9.4n₂ = 400
E₂ =
16.Confidence intervalsHard
A pilot study suggests that about 0.2 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.025?
p ≈ 0.2E = 0.02599%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 119.5 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 119.5s = 5n = 100
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 1 units with a power of 90%. The standard deviation is 7. Find the minimum sample size n.
α = 0.05power = 90%δ = 1σ = 7zα = 1.645zβ = 1.282
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.5.
n = 5x = 3.5
ŷ =
3Solutions
1.Time value of money
Answer: FV = 2810 €
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 = 0.02·17 = 0.34
e0.34 = 1.405
Future value
FV = 2 000·1.405 = €2 809.90
Answer
FV = €2 809.90
2.Time value of money
Answer: A = 584.9 €
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 = 0.024/12 = 0.002
N = 12·30 = 360
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 = 0.4871
A = 150 000·0.002/(1 − 0.4871) = €584.91
Answer
A = €584.91
3.NPV & IRR
Answer: DPB = 3.915 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: 14 000/1.12 = 12 500.00 → Σ = 12 500.00
year 2: 10 000/1.254 = 7 971.94 → Σ = 20 471.94
year 3: 16 000/1.405 = 11 388.48 → Σ = 31 860.42
year 4: 14 000/1.574 = 8 897.25 → Σ = 40 757.68
Fraction of the last year
DPB = 3 + 8 139.58/8 897.25 = 3.915
Answer
DPB = 3.915 years
The discounted payback is always longer than the simple payback (3 years).
4.NPV & IRR
Answer: r* = 5.886 %
Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.
Difference in cash flows
ΔCF0 = −15 000 − (−10 000) = −5 000
ΔCF1 = 4 000 − 5 500 = −1 500
ΔCF2 = 6 000 − 4 000 = 2 000
ΔCF3 = 8 000 − 2 500 = 5 500
IRR of the difference
NPVΔ(5%) = 136.59
NPVΔ(6%) = −17.20
r* = 5.886%
Answer
r* = 5.886%
Check: at 5.886% both NPVs are €867.70.
5.Bond pricing & yield
Answer: P = 1034 €
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 = €22.50, i = 0.02, N = 16
Present value of the coupons
€22.50·(1 − 0.7284)/0.02 = €305.50
Present value of the face value
€1 000·0.7284 = €728.45
Price
P = €1 033.94
Answer
P = €1 033.94
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 9.464 %
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
€855 = €70·(1 − (1 + y)⁻⁹)/y + €1 000/(1 + y)⁹
Solve numerically
P(9%) = €880.10, P(10%) = €827.23
YTM = 9.464%
Answer
YTM = 9.464%
7.CAPM & WACC
Answer: WACC = 9.053 %
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 = €3 000 000, E/V = 0.6667, D/V = 0.3333
WACC
WACC = 0.6667·12% + 0.3333·4%·(1 − 0.21) = 9.053%
Answer
WACC = 9.053%
8.CAPM & WACC
Answer: WACC = 6.406 %
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 = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 0.8944
Answer
P = 0.8944
11.Probability
Answer: P = 0.1818
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.
First and second draw
P = 5/11 · 4/10 = 20/110 = 0.1818
Answer
P = 0.1818
12.Probability
Answer: P = 0.3003
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) = 0.759 = 0.07508
P(X = 1) = 9·0.25·0.758 = 0.2253
Add them
P = 0.07508 + 0.2253 = 0.3003
Answer
P = 0.3003
13.Normal distribution
Answer: P = 0.1587
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 = (111.4 − 105)/6.403 = 1
Upper tail
P = 1 − Φ(1) = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
14.Normal distribution
Answer: x = 449 h
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) = 0.1
z from the table
z ≈ −1.28
Back to the original units
x = 500 − 1.28·40 = 449 h
Answer
x = 449 h
15.Confidence intervals
Answer: E₂ = 4.7
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 = (1.645/1.645)·√(100/400) = 0.5
New margin
E₂ = 9.4·0.5 = 4.7
Answer
E₂ = 4.7
16.Confidence intervals
Answer: n = 1699
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 = 2.576²·0.2·0.8/0.025² = 1699
Round up
n = 1699
Answer
n = 1699
17.Hypothesis tests & confidence intervals
Answer: t = −1
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 = 5/√100 = 5/10 = 0.5
Test statistic
t = (119.5 − 120)/0.5 = −1
Answer
t = −1
t0.025; df = 99 = 1.984
|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 420
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((1.645 + 1.282)·7/1)² = 419.8
Round up
n = 420
Answer
n = 420
19.Linear regression & correlation
Answer: r = 0.9533
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 = 7386 − 90·399/5 = 204
Sxx = 1660 − 90²/5 = 40
Syy = 32985 − 399²/5 = 1145
Correlation coefficient
r = 204/√(40·1145) = 0.9533
Answer
r = 0.9533
20.Linear regression & correlation
Answer: ŷ = 13.7
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 = 15/5 = 3
ȳ = 76/5 = 15.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −30
Regression line
b = −30/10 = −3
a = 15.2 − (−3)·3 = 24.2
ŷ = 24.2 − 3·x
Prediction
ŷ = 24.2 + (−3)·3.5 = 13.7
Answer
ŷ = 13.7
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