You need €50 000 in 2 years. An account pays 5% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €50 0005% / 4n = 2
PV = €
2.Time value of moneyHard
You save €250 at the end of every month for 28 years. The account pays 2.4% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €2502.4% / 12N = 336
FV = €
3.NPV & IRRMedium
A machine costs €30 000 today, lasts 6 years and costs €3 000 per year to run (paid at the end of each year). The cost of capital is 12%. Find its equivalent annual cost EAC.
C0 = €30 000M = €3 000n = 6r = 12%
EAC = €
4.NPV & IRRHard
A project costs €25 000 and returns €18 500, €10 000, €18 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −25 000CF1 = 18 500CF2 = 10 000CF3 = 18 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 9 years. The yield to maturity is 4% per year. Find the price P of the bond.
F = €1 000coupon = 7%YTM = 4%n = 9
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 6% and matures in 6 years. The yield to maturity is 5.5%. Find the Macaulay duration D in years.
F = €1 000coupon = 6%n = 6YTM = 5.5%
D = years
7.CAPM & WACCMedium
A company has equity worth €10 000 000 and debt worth €6 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 5% and the tax rate is 25%. Find the WACC.
A company's business has an unlevered (asset) beta of 0.9. It plans a debt-to-equity ratio of 0.6. The tax rate is 30%, the risk-free rate is 4% and the market risk premium is 4%. Find the cost of equity re at the new capital structure.
βU = 0.9D/E = 0.6T = 30%rf = 4%rm − rf = 4%
re = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 37, 16, 13, 38, 56, 37.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is below 45?
μ = 50σ = 10x = 45
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.05, 0.35, 0.3, 0.2, 0.1
σ =
12.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 2 correct answers?
n = 7p = 0.1k = 2
P =
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0150. Find z to two decimals.
A = 0.0150
z =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 95% of them lie (the 95th percentile).
μ = 500σ = 40P = 95%
x = h
15.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 1. What is the margin of error E₂ of a 99% interval from the same data?
90%n₁ = 64E₁ = 199%
E₂ =
16.Confidence intervalsHard
The population standard deviation is 18. 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 = 18E = 499%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 49.1 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 49.1s = 5n = 100
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 119. The population standard deviation is 4. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 119σ = 4n = 6499%
U =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.
n = 5
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. For which value x does the line predict ŷ = 45?
n = 5ŷ = 45
x =
3Solutions
1.Time value of money
Answer: PV = 45270 €
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 = 0.05/4 = 0.0125
N = 4·2 = 8
Present value
PV = FV/(1 + i)N = 50 000/1.104 = €45 269.92
Answer
PV = €45 269.92
2.Time value of money
Answer: FV = 119605 €
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 €84 000 yourself; the rest is interest.
3.NPV & IRR
Answer: EAC = 10297 €
Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.
Annuity factor
(1 − 1.12−6)/0.12 = 4.111
Annualised purchase price
30 000/4.111 = 7 296.77
EAC
EAC = 7 296.77 + 3 000 = €10 296.77
Answer
EAC = €10 296.77
4.NPV & IRR
Answer: IRR = 40.19 %
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%) = 58.31
NPV(45%) = −1 416.83
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 40 + 5·58.31/(58.31 − (−1 416.83)) = 40.2%
exact: NPV(40.19%) = 0 → IRR = 40.19%
Answer
IRR = 40.19%
5.Bond pricing & yield
Answer: P = 1225 €
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 = €35.00, i = 0.02, N = 18
Present value of the coupons
€35.00·(1 − 0.7002)/0.02 = €524.72
Present value of the face value
€1 000·0.7002 = €700.16
Price
P = €1 224.88
Answer
P = €1 224.88
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: D = 5.223 years
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: 60/1.055 = 56.87, t·PV = 56.87
t = 2: 60/1.113 = 53.91, t·PV = 107.81
t = 3: 60/1.174 = 51.10, t·PV = 153.29
t = 4: 60/1.239 = 48.43, t·PV = 193.73
t = 5: 60/1.307 = 45.91, t·PV = 229.54
t = 6: 1060/1.379 = 768.76, t·PV = 4 612.56
Duration
P = 1 024.98
D = 5 353.81/1 024.98 = 5.223
Answer
D = 5.223 years
7.CAPM & WACC
Answer: WACC = 7.656 %
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 = €16 000 000, E/V = 0.625, D/V = 0.375
WACC
WACC = 0.625·10% + 0.375·5%·(1 − 0.25) = 7.656%
Answer
WACC = 7.656%
8.CAPM & WACC
Answer: re = 9.112 %
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 = 0.9·(1 + 0.7·0.6) = 1.278
Cost of equity (CAPM)
re = 4% + 1.278·4% = 9.112%
Answer
re = 9.112%
9.Descriptive statistics
Answer: s = 15.97
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 0.3085
Answer
P = 0.3085
11.Probability
Answer: σ = 1.071
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 = 2) = 21·0.12·0.95 = 21·0.01·0.5905 = 0.124
Answer
P = 0.124
13.Normal distribution
Answer: z = 2.17
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 − 0.0150 = 0.9850
Read the z-table backwards
Find 0.9850 inside the table and read off z.
z = 2.17
Answer
z = 2.17
14.Normal distribution
Answer: x = 565.8 h
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 for the percentile
Φ(z) = 0.95 → z = 1.645
Back to x
x = 500 + 1.645·40 = 565.8 h
Answer
x = 565.8 h
15.Confidence intervals
Answer: E₂ = 1.566
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 = (2.576/1.645)·√(64/64) = 1.566
New margin
E₂ = 1·1.566 = 1.566
Answer
E₂ = 1.566
16.Confidence intervals
Answer: n = 135
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 = (2.576·18/4)² = 134.4
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 135
Answer
n = 135
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 = 5/√100 = 5/10 = 0.5
Test statistic
t = (49.1 − 50)/0.5 = −1.8
Answer
t = −1.8
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: U = 120.3
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 = 2.576·4/√64 = 1.288
Interval limit
U = 119 + 1.288 = 120.3
Answer
U = 120.3
19.Linear regression & correlation
Answer: r = 0.94
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 = 265/5 = 53
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 52
Syy = Σ(y − ȳ)² = 306
Correlation coefficient
r = 52/√(10·306) = 0.94
Answer
r = 0.94
20.Linear regression & correlation
Answer: x = 2.4
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 = 20/5 = 4
ȳ = 257/5 = 51.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 40
Regression line
b = 40/10 = 4
a = 51.4 − 4·4 = 35.4
ŷ = 35.4 + 4·x
Solve for x
x = (45 − 35.4)/4 = 2.4
Answer
x = 2.4
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