You invest €5 000 at a nominal rate of 6% per year, compounded monthly. Find the value FV of the investment after 10 years.
PV = €5 0006% / 12n = 10
FV = €
2.Time value of moneyHard
You save €50 at the end of every month for 27 years. The account pays 3% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €503% / 12N = 324
FV = €
3.NPV & IRRMedium
A project costs €30 000 today and returns €9 500 at the end of each year for 4 years. The cost of capital is 6%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
Project A costs €30 000 and returns €16 000, €12 000, €6 000 at the end of years 1 to 3. Project B costs €35 000 and returns €9 000, €13 500, €18 000. 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 2% annual coupon in two half-yearly payments and matures in 10 years. The yield to maturity is 9% per year. Find the price P of the bond.
F = €1 000coupon = 2%YTM = 9%n = 10
P = €
6.Bond pricing & yieldHard
A bond with 17 years to maturity (face value €1 000, annual coupon 8%) sells for €1 098. The issuer can call it in 6 years at a call price of €1 080. Find the yield to call YTC in percent.
F = €1 000coupon = 8%P = €1 098call price = €1 080call in 6 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €4 000 000 and debt worth €6 000 000. Its cost of equity is 9%, its pre-tax cost of debt is 5% and the tax rate is 28%. Find the WACC.
E = €4 000 000D = €6 000 000re = 9%rd = 5%T = 28%
WACC = %
8.CAPM & WACCHard
A company has 2 000 000 shares at €30.00 each and 20 000 bonds trading at €1 050 each. The pre-tax cost of debt is 7% and the tax rate is 20%. The share’s beta is 1.2, the risk-free rate 1.5% and the market return 8.5%. Find the WACC.
β = 1.2rf = 1.5%rm = 8.5%rd = 7%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 38, 46, 24, 45, 59.
n = 5
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 155.
μ = 170σ = 10x = 155
z =
11.ProbabilityMedium
A bag holds 8 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
8 / 4
P =
12.ProbabilityHard
12 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?
n = 12p = 0.1
P =
13.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 500 and 560 h? Give P as a decimal.
μ = 500σ = 40x1 = 500x2 = 560
P =
14.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 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 = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
15.Confidence intervalsMedium
A sample of 25 items has mean 23.5 and standard deviation s = 15. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 25x̄ = 23.5s = 1595%t(24) = 2.064
lower limit =
16.Confidence intervalsHard
The population standard deviation is 11. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 4?
σ = 11E = 499%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 200.8 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 200.8s = 4n = 81
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 51.2. The population standard deviation is 8. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 51.2σ = 8n = 8190%
L =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (5, 70) is made. Find its residual e = y − ŷ.
n = 5(5, 70)
e =
3Solutions
1.Time value of money
Answer: FV = 9097 €
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.
You paid in €16 200 yourself; the rest is interest.
3.NPV & IRR
Answer: DPB = 3.612 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: 9 500/1.06 = 8 962.26 → Σ = 8 962.26
year 2: 9 500/1.124 = 8 454.97 → Σ = 17 417.23
year 3: 9 500/1.191 = 7 976.38 → Σ = 25 393.61
year 4: 9 500/1.262 = 7 524.89 → Σ = 32 918.50
Fraction of the last year
DPB = 3 + 4 606.39/7 524.89 = 3.612
Answer
DPB = 3.612 years
The discounted payback is always longer than the simple payback (3.158 years).
4.NPV & IRR
Answer: r* = 5.233 %
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 = −35 000 − (−30 000) = −5 000
ΔCF1 = 9 000 − 16 000 = −7 000
ΔCF2 = 13 500 − 12 000 = 1 500
ΔCF3 = 18 000 − 6 000 = 12 000
IRR of the difference
NPVΔ(5%) = 59.93
NPVΔ(6%) = −193.35
r* = 5.233%
Answer
r* = 5.233%
Check: at 5.233% both NPVs are €1 189.30.
5.Bond pricing & yield
Answer: P = 544.7 €
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 = €10.00, i = 0.045, N = 20
Present value of the coupons
€10.00·(1 − 0.4146)/0.045 = €130.08
Present value of the face value
€1 000·0.4146 = €414.64
Price
P = €544.72
Answer
P = €544.72
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTC = 7.057 %
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 = 80·(1 − (1 + y)−6)/y + 1080/(1 + y)6
Solve numerically
y = 7%: €1 100.97
y = 8%: €1 050.41
YTC = 7.057%
Answer
YTC = 7.057%
The yield to maturity is 6.996%. 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: WACC = 5.76 %
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 = 0.4, D/V = 0.6
WACC
WACC = 0.4·9% + 0.6·5%·(1 − 0.28) = 5.76%
Answer
WACC = 5.76%
8.CAPM & WACC
Answer: WACC = 8.785 %
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 values
E = 2 000 000·€30.00 = €60 000 000
D = 20 000·€1 050 = €21 000 000
Cost of equity (CAPM)
re = 1.5% + 1.2·7% = 9.9%
Weights
V = €81 000 000, E/V = 0.7407, D/V = 0.2593
WACC
WACC = 0.7407·9.9% + 0.2593·7%·(1 − 0.2) = 8.785%
Answer
WACC = 8.785%
9.Descriptive statistics
Answer: s = 12.78
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 = (155 − 170)/10 = −1.5
Answer
z = −1.5
11.Probability
Answer: P = 0.4242
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 = 8/12 · 7/11 = 56/132 = 0.4242
Answer
P = 0.4242
12.Probability
Answer: P = 0.659
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) = 0.912 = 0.2824
P(X = 1) = 12·0.1·0.911 = 0.3766
Add them
P = 0.2824 + 0.3766 = 0.659
Answer
P = 0.659
13.Normal distribution
Answer: P = 0.4332
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 = (500 − 500)/40 = 0
z₂ = (560 − 500)/40 = 1.5
Area between
P = Φ(1.5) − Φ(0) = 0.9332 − 0.5 = 0.4332
Answer
P = 0.4332
14.Normal distribution
Answer: P = 0.9099
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 = 0.02236 mm
Standardise
z = (0 − 0.03)/0.02236 = −1.34
Probability
P = 1 − Φ(−1.34) = 0.9099
Answer
P = 0.9099
15.Confidence intervals
Answer: lower limit = 17.31
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 = 24, t = 2.064
Margin of error
E = 2.064·15/√25 = 6.192
Interval
23.5 ± 6.192 → [17.31, 29.69]
Answer
lower limit = 17.31
16.Confidence intervals
Answer: n = 51
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 = (2.576·11/4)² = 50.18
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 51
Answer
n = 51
17.Hypothesis tests & confidence intervals
Answer: t = 1.8
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 = 4/√81 = 4/9 = 0.4444
Test statistic
t = (200.8 − 200)/0.4444 = 1.8
Answer
t = 1.8
t0.025; df = 80 = 1.99
|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: L = 49.74
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 = 1.645·8/√81 = 1.462
Interval limit
L = 51.2 − 1.462 = 49.74
Answer
L = 49.74
19.Linear regression & correlation
Answer: r = 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 = 42/6 = 7
ȳ = 185/6 = 30.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 283
Syy = Σ(y − ȳ)² = 1151
Correlation coefficient
r = 283/√(70·1151) = 0.9971
Answer
r = 0.9971
20.Linear regression & correlation
Answer: e = 2.65
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 = 367/5 = 73.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 242
Regression line
b = 242/40 = 6.05
a = 73.4 − 6.05·6 = 37.1
ŷ = 37.1 + 6.05·x
Predicted value
ŷ = 37.1 + 6.05·5 = 67.35
Residual
e = 70 − 67.35 = 2.65
Answer
e = 2.65
The point lies above the line.
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