You need €20 000 in 5 years. An account pays 4% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €20 0004% / 12n = 5
PV = €
2.Time value of moneyHard
You save €500 at the end of every month for 6 years. The account pays 4.8% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €5004.8% / 12N = 72
FV = €
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 3: €26 000, €18 000, €23 000. The cost of capital is 8%. Find the net present value NPV.
A project costs €10 000 and returns €3 000, €3 000, €5 500, €3 500 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 8%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 14 years. Its yield to maturity changes from 7% to 7.5%. By how much ΔP does the bond price change?
F = €1 000coupon = 5%n = 14y1 = 7%y2 = 7.5%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 4% and matures in 5 years. The yield to maturity is 7%. Find the Macaulay duration D in years.
F = €1 000coupon = 4%n = 5YTM = 7%
D = years
7.CAPM & WACCMedium
A company has equity worth €8 000 000 and debt worth €5 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 4% and the tax rate is 30%. Find the WACC.
E = €8 000 000D = €5 000 000re = 8%rd = 4%T = 30%
WACC = %
8.CAPM & WACCHard
A company's business has an unlevered (asset) beta of 0.8. It plans a debt-to-equity ratio of 0.5. The tax rate is 20%, the risk-free rate is 3.5% and the market risk premium is 6%. Find the cost of equity re at the new capital structure.
βU = 0.8D/E = 0.5T = 20%rf = 3.5%rm − rf = 6%
re = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 31, 28, 34, 52, 26.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 90.
μ = 70σ = 20x = 90
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3, 4p(x): 0.1, 0.2, 0.3, 0.35, 0.05
Var(X) =
12.ProbabilityHard
A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 4 correct answers?
n = 5p = 0.7k = 4
P =
13.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 4 of them exceeds 102 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 4> 102 kg
P =
14.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.02 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 40 phones, mean 84.5 h, known standard deviation 8 h. Model 2: 40 phones, mean 74 h, known standard deviation 7 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
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 0.02?
p = 0.5E = 0.0290%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 96.7 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 96.7s = 15n = 100
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 5 units with a power of 80%. The standard deviation is 21. Find the minimum sample size n.
The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 15Σy = 100Σx² = 55Σy² = 2034Σxy = 283
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 = 18?
n = 5ŷ = 18
x =
3Solutions
1.Time value of money
Answer: PV = 16380 €
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.
Rate and number of periods
i = 0.04/12 = 0.003333
N = 12·5 = 60
Present value
PV = FV/(1 + i)N = 20 000/1.221 = €16 380.06
Answer
PV = €16 380.06
2.Time value of money
Answer: FV = 41624 €
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 €36 000 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = −17236 €
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 = −75 000
CF1/(1 + 0.08)1 = 26 000/1.08 = 24 074.07
CF2/(1 + 0.08)2 = 18 000/1.166 = 15 432.10
CF3/(1 + 0.08)3 = 23 000/1.26 = 18 258.14
Add them up
NPV = €−17 235.69
Answer
NPV = €−17 235.69
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: MIRR = 13.71 %
Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.
Terminal value of the inflows
3 000·1.083 = 3 779.14
3 000·1.082 = 3 499.20
5 500·1.081 = 5 940.00
3 500·1.080 = 3 500.00
TV = 16 718.34
MIRR
MIRR = (16 718.34/10 000)1/4 − 1 = 13.71%
Answer
MIRR = 13.71%
5.Bond pricing & yield
Answer: ΔP = −37.32 €
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.
Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.
Present value of each cash flow
t = 1: 40/1.07 = 37.38, t·PV = 37.38
t = 2: 40/1.145 = 34.94, t·PV = 69.88
t = 3: 40/1.225 = 32.65, t·PV = 97.96
t = 4: 40/1.311 = 30.52, t·PV = 122.06
t = 5: 1040/1.403 = 741.51, t·PV = 3 707.53
Duration
P = 876.99
D = 4 034.81/876.99 = 4.601
Answer
D = 4.601 years
7.CAPM & WACC
Answer: WACC = 6 %
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 = €13 000 000, E/V = 0.6154, D/V = 0.3846
WACC
WACC = 0.6154·8% + 0.3846·4%·(1 − 0.3) = 6%
Answer
WACC = 6%
8.CAPM & WACC
Answer: re = 10.22 %
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 = 0.8·(1 + 0.8·0.5) = 1.12
Cost of equity (CAPM)
re = 3.5% + 1.12·6% = 10.22%
Answer
re = 10.22%
9.Descriptive statistics
Answer: s = 10.4
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 = (90 − 70)/20 = 1
Answer
z = 1
11.Probability
Answer: Var(X) = 1.147
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.
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.
Binomial formula
P(X = 4) = 5·0.74·0.31 = 5·0.2401·0.3 = 0.3601
Answer
P = 0.3601
13.Normal distribution
Answer: P = 0.02275
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 total
μT = 100 kg, σT = 0.5·√4 = 1 kg
Standardise
z = (102 − 100)/1 = 2
Probability
P = 1 − Φ(2) = 0.02275
Answer
P = 0.02275
14.Normal distribution
Answer: P = 0.8133
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 = 25.02 − 25 = 0.02 mm, σ = 0.02236 mm
Standardise
z = (0 − 0.02)/0.02236 = −0.89
Probability
P = 1 − Φ(−0.89) = 0.8133
Answer
P = 0.8133
15.Confidence intervals
Answer: lower limit = 7.735 h
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Point estimate
x̄₁ − x̄₂ = 84.5 − 74 = 10.5
Standard error
SE = √(8²/40 + 7²/40) = 1.681
Margin of error
E = 1.645·1.681 = 2.765
Interval
10.5 ± 2.765 → [7.735, 13.26]
Answer
lower limit = 7.735 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: n = 1692
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 = 1.645²·0.5·0.5/0.02² = 1691
Round up
n = 1692
Answer
n = 1692
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
SE = 15/√100 = 15/10 = 1.5
Test statistic
t = (96.7 − 100)/1.5 = −2.2
Answer
t = −2.2
t0.025; df = 99 = 1.984
|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 110
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 + 0.842)·21/5)² = 109.1
Round up
n = 110
Answer
n = 110
19.Linear regression & correlation
Answer: r = −0.922
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 = 283 − 15·100/5 = −17
Sxx = 55 − 15²/5 = 10
Syy = 2034 − 100²/5 = 34
Correlation coefficient
r = −17/√(10·34) = −0.922
Answer
r = −0.922
20.Linear regression & correlation
Answer: x = 6.898
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 = 79/5 = 15.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 98
Regression line
b = 98/40 = 2.45
a = 15.8 − 2.45·6 = 1.1
ŷ = 1.1 + 2.45·x
Solve for x
x = (18 − 1.1)/2.45 = 6.898
Answer
x = 6.898
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