You need €50 000 in 3 years. An account pays 4% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €50 0004% / 4n = 3
PV = €
2.Time value of moneyHard
You save €150 at the end of every month for 27 years. The account pays 5.4% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €1505.4% / 12N = 324
FV = €
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 4: €30 000, €30 000, €23 000, €37 000. The cost of capital is 10%. Find the net present value NPV.
A project costs €25 000 and returns €15 000, €11 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −25 000CF1 = 15 000CF2 = 11 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 4 years and pays annual coupons. It sells for €935 when the yield to maturity is 9%. Find the coupon rate c in percent.
F = €1 000n = 4P = €935YTM = 9%
c = %
6.Bond pricing & yieldHard
A bond trades at €1 005 with a yield to maturity of 5% (annual compounding) and a Macaulay duration of 6.4 years. Use the modified duration to estimate the new price P if the yield rises by 25 basis points.
P0 = €1 005y = 5%D = 6.4Δy = +25 bp
P = €
7.CAPM & WACCMedium
A company has a debt-to-equity ratio (at market values) of 1. Its cost of equity is 13%, its pre-tax cost of debt is 6% and the tax rate is 30%. Find the WACC.
D/E = 1re = 13%rd = 6%T = 30%
WACC = %
8.CAPM & WACCHard
A company's shares have a beta of 1.3 at its current debt-to-equity ratio of 0.5. It plans to raise its debt-to-equity ratio to 0.75. The tax rate is 25%, the risk-free rate 4% and the market risk premium 5%. Find the cost of equity re after the change.
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 248.75 and 256 g? Give P as a decimal.
μ = 250σ = 5x1 = 248.75x2 = 256
P =
14.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 75% of them lie (the 75th percentile).
μ = 178σ = 7P = 75%
x = cm
15.Confidence intervalsMedium
The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 4 and standard deviation 3.9. Find the lower limit of the 90% confidence interval for the mean difference.
n = 8d̄ = 4sd = 3.990%t(7) = 1.895
lower limit = mmHg
16.Confidence intervalsHard
The diameters of 10 machined shafts have a sample standard deviation of 5.5 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.
n = 10s = 5.5 µmχ² = 2.7, 19.023
σ = µm
17.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 499 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 499s = 5n = 81
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 16 measurements has mean 97.3. The population standard deviation is 10. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 97.3σ = 10n = 1690%
U =
19.Linear regression & correlationMedium
The table gives 6 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 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (21.5, 112) is made. Find its residual e = y − ŷ.
n = 6(21.5, 112)
e =
3Solutions
1.Time value of money
Answer: PV = 44372 €
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 = 0.04/4 = 0.01
N = 4·3 = 12
Present value
PV = FV/(1 + i)N = 50 000/1.127 = €44 372.46
Answer
PV = €44 372.46
2.Time value of money
Answer: FV = 109444 €
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 €48 600 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = 19618 €
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 + 0.1)1 = 30 000/1.1 = 27 272.73
CF2/(1 + 0.1)2 = 30 000/1.21 = 24 793.39
CF3/(1 + 0.1)3 = 23 000/1.331 = 17 280.24
CF4/(1 + 0.1)4 = 37 000/1.464 = 25 271.50
Add them up
NPV = €19 617.85
Answer
NPV = €19 617.85
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 4.162 %
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(0%) = 1 500.00
NPV(5%) = −283.45
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
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.
Present value of the face value
1000·(1 + 0.09)−4 = €708.43
Coupon
C = (935 − 708.43)/3.24 = €69.94
Coupon rate
c = 69.94/1000 = 6.994%
Answer
c = 6.994%
6.Bond pricing & yield
Answer: P = 989.7 €
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.
Modified duration
Dmod = 6.4/1.05 = 6.095
Estimated price change
%ΔP ≈ −6.095·(0.0025) = −1.524%
P ≈ 1005·(1 − 0.01524) = €989.69
Answer
P = €989.69
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: WACC = 8.6 %
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 from D/E
E/V = 1/(1 + 1) = 0.5
D/V = 1/(1 + 1) = 0.5
WACC
WACC = 0.5·13% + 0.5·6%·(1 − 0.3) = 8.6%
Answer
WACC = 8.6%
8.CAPM & WACC
Answer: re = 11.39 %
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.
Unlever the beta
βU = 1.3/(1 + 0.75·0.5) = 0.9455
Relever at the new D/E
βL = 0.9455·(1 + 0.75·0.75) = 1.477
Cost of equity (CAPM)
re = 4% + 1.477·5% = 11.39%
Answer
re = 11.39%
9.Descriptive statistics
Answer: s = 20.73
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 = (167.5 − 170)/5 = −0.5
Answer
z = −0.5
11.Probability
Answer: P = 0.5981
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.
Complement
P(no six) = (5/6)5 = 0.4019
P = 1 − 0.4019 = 0.5981
Answer
P = 0.5981
12.Probability
Answer: σY = 2.088
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.
Spread of X
μ = 2.1, Var(X) = 1.09, σX = 1.044
Linear change
σY = |2|·1.044 = 2.088
Answer
σY = 2.088
The mean does move: E(Y) = 2·2.1 + 5 = 9.2.
13.Normal distribution
Answer: P = 0.4836
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 = (248.75 − 250)/5 = −0.25
z₂ = (256 − 250)/5 = 1.2
Area between
P = Φ(1.2) − Φ(−0.25) = 0.8849 − 0.4013 = 0.4836
Answer
P = 0.4836
14.Normal distribution
Answer: x = 182.7 cm
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) = 0.75 → z = 0.6745
Back to x
x = 178 + 0.6745·7 = 182.72 cm
Answer
x = 182.72 cm
15.Confidence intervals
Answer: lower limit = 1.387 mmHg
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.
Critical value
df = 7, t = 1.895
Margin of error
E = 1.895·3.9/√8 = 2.613
Interval
4 ± 2.613 → [1.387, 6.613]
Answer
lower limit = 1.387 mmHg
16.Confidence intervals
Answer: σ = 10.04 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 9·5.5² = 272.3
Limits for σ²
272.3/19.023 = 14.31
272.3/2.7 = 100.8
Limit for σ
σ = √100.8 = 10.04 µm
Answer
σ = 10.04 µm
95% interval for σ: [3.783, 10.04] µm.
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/√81 = 5/9 = 0.5556
Test statistic
t = (499 − 500)/0.5556 = −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: U = 101.4
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 = 1.645·10/√16 = 4.112
Interval limit
U = 97.3 + 4.112 = 101.4
Answer
U = 101.4
19.Linear regression & correlation
Answer: r = 0.9526
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 = 8274 − 120·402/6 = 234
Sxx = 2470 − 120²/6 = 70
Syy = 27796 − 402²/6 = 862
Correlation coefficient
r = 234/√(70·862) = 0.9526
Answer
r = 0.9526
20.Linear regression & correlation
Answer: e = −1.5
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 = 129/6 = 21.5
ȳ = 681/6 = 113.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 820.5
Regression line
b = 820.5/157.5 = 5.21
a = 113.5 − 5.21·21.5 = 1.495
ŷ = 1.495 + 5.21·x
Predicted value
ŷ = 1.495 + 5.21·21.5 = 113.5
Residual
e = 112 − 113.5 = −1.5
Answer
e = −1.5
The point lies below the line.
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