An annuity pays €3 000 at the end of each year for 10 years. The interest rate is 4% per year. Find the present value PV of the annuity.
A = €3 000i = 4%n = 10
PV = €
2.Time value of moneyHard
A loan of €250 000 is repaid with equal monthly payments over 30 years. The interest rate is 5.4% per year, compounded monthly. Find the monthly payment A.
P = €250 0005.4% / 12N = 360
A = €
3.NPV & IRRMedium
An investment of €30 000 gives these cash flows at the end of years 1 to 3: €9 000, €13 000, €11 000. The cost of capital is 5%. Find the net present value NPV.
A project costs €50 000 and returns €21 500, €29 000, €18 000, €17 000 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 6%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 pays an annual coupon of 2% and matures in 6 years. Its yield to maturity changes from 4% to 3.5%. By how much ΔP does the bond price change?
F = €1 000coupon = 2%n = 6y1 = 4%y2 = 3.5%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 7% and matures in 5 years. The yield to maturity is 2.5%. Find the Macaulay duration D in years.
F = €1 000coupon = 7%n = 5YTM = 2.5%
D = years
7.CAPM & WACCMedium
A company has equity worth €2 000 000 and debt worth €1 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 5% and the tax rate is 28%. Find the WACC.
A company's business has an unlevered (asset) beta of 0.8. It plans a debt-to-equity ratio of 1.5. The tax rate is 28%, 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 = 0.8D/E = 1.5T = 28%rf = 2%rm − rf = 6%
re = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 33, 54, 35, 14, 55.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 40. Find the z-score z of a score of 420.
μ = 500σ = 40x = 420
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35
Var(X) =
12.ProbabilityHard
A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 1 correct answers?
n = 10p = 0.25k = 1
P =
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0582. Find z to two decimals.
A = 0.0582
z =
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 bottom 10% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points10%
x = points
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 60 phones, mean 81.5 h, known standard deviation 6 h. Model 2: 50 phones, mean 73.5 h, known standard deviation 9 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
The population standard deviation is 16. 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 2.5?
σ = 16E = 2.599%
n =
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 50 parts has mean length 111.5 mm, and the line's known standard deviation is 20 mm. Line 2: a sample of 80 parts has mean 114.2 mm, known standard deviation 9 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 95%. The standard deviation is 12. Find the minimum sample size n.
The table gives 6 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 = 6Σx = 42Σy = 131Σx² = 364Σy² = 3433Σxy = 1116
r =
20.Linear regression & correlationHard
The table gives 6 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 = 4.5.
n = 6x = 4.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 24333 €
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.
Annuity factor
(1 − (1 + i)−n)/i = (1 − 0.6756)/0.04 = 8.111
Present value
PV = 3 000·8.111 = €24 332.69
Answer
PV = €24 332.69
2.Time value of money
Answer: A = 1404 €
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 = 0.054000000000000006/12 = 0.0045
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 = 0.1986
A = 250 000·0.0045/(1 − 0.1986) = €1 403.83
Answer
A = €1 403.83
3.NPV & IRR
Answer: NPV = −135 €
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 = 9 000/1.05 = 8 571.43
CF2/(1 + 0.05)2 = 13 000/1.103 = 11 791.38
CF3/(1 + 0.05)3 = 11 000/1.158 = 9 502.21
Add them up
NPV = €−134.97
Answer
NPV = €−134.97
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: MIRR = 17.18 %
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
21 500·1.063 = 25 606.84
29 000·1.062 = 32 584.40
18 000·1.061 = 19 080.00
17 000·1.060 = 17 000.00
TV = 94 271.24
MIRR
MIRR = (94 271.24/50 000)1/4 − 1 = 17.18%
Answer
MIRR = 17.18%
5.Bond pricing & yield
Answer: ΔP = 24.91 €
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: 70/1.025 = 68.29, t·PV = 68.29
t = 2: 70/1.051 = 66.63, t·PV = 133.25
t = 3: 70/1.077 = 65.00, t·PV = 195.01
t = 4: 70/1.104 = 63.42, t·PV = 253.67
t = 5: 1070/1.131 = 945.72, t·PV = 4 728.62
Duration
P = 1 209.06
D = 5 378.84/1 209.06 = 4.449
Answer
D = 4.449 years
7.CAPM & WACC
Answer: WACC = 10.53 %
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 = 0.6667·14% + 0.3333·5%·(1 − 0.28) = 10.53%
Answer
WACC = 10.53%
8.CAPM & WACC
Answer: re = 11.98 %
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 + 0.72·1.5) = 1.664
Cost of equity (CAPM)
re = 2% + 1.664·6% = 11.98%
Answer
re = 11.98%
9.Descriptive statistics
Answer: s = 16.99
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 = (420 − 500)/40 = −2
Answer
z = −2
11.Probability
Answer: Var(X) = 0.7
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.
Mean
μ = 0·0.35 + 1·0.3 + 2·0.35 = 1
Mean of the squares
E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7
Variance
Var(X) = 1.7 − 1² = 0.7
Answer
Var(X) = 0.7
12.Probability
Answer: P = 0.1877
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.
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 − 0.0582 = 0.9418
Read the z-table backwards
Find 0.9418 inside the table and read off z.
z = 1.57
Answer
z = 1.57
14.Normal distribution
Answer: x = 80.8 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) = 0.1
z from the table
z ≈ −1.28
Back to the original units
x = 100 − 1.28·15 = 80.8 points
Answer
x = 80.8 points
15.Confidence intervals
Answer: lower limit = 5.549 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 = 81.5 − 73.5 = 8
Standard error
SE = √(6²/60 + 9²/50) = 1.49
Margin of error
E = 1.645·1.49 = 2.451
Interval
8 ± 2.451 → [5.549, 10.45]
Answer
lower limit = 5.549 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: n = 272
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 = (2.576·16/2.5)² = 271.8
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 272
Answer
n = 272
17.Hypothesis tests & confidence intervals
Answer: z = −0.8994
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.
Standard error of the difference
SE = √(20²/50 + 9²/80) = √9.013 = 3.002
Test statistic
z = (111.5 − 114.2)/3.002 = −0.8994
Answer
z = −0.8994
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 98
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 + 1.645)·12/4)² = 97.42
Round up
n = 98
Answer
n = 98
19.Linear regression & correlation
Answer: r = 0.9938
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 = 1116 − 42·131/6 = 199
Sxx = 364 − 42²/6 = 70
Syy = 3433 − 131²/6 = 572.8
Correlation coefficient
r = 199/√(70·572.8) = 0.9938
Answer
r = 0.9938
20.Linear regression & correlation
Answer: ŷ = 15.96
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 = 21/6 = 3.5
ȳ = 111/6 = 18.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −44.5
Regression line
b = −44.5/17.5 = −2.543
a = 18.5 − (−2.543)·3.5 = 27.4
ŷ = 27.4 − 2.543·x
Prediction
ŷ = 27.4 + (−2.543)·4.5 = 15.96
Answer
ŷ = 15.96
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