You invest €10 000 at 3% per year with continuous compounding. Find the value FV after 8 years.
PV = €10 000r = 3%t = 8
FV = €
2.Time value of moneyHard
You want €10 000 in 3 years. You save the same amount at the end of every month in an account paying 3.6% per year, compounded monthly. Find the monthly deposit A.
FV = €10 0003.6% / 12N = 36
A = €
3.NPV & IRRMedium
A machine costs €80 000 today. It generates €15 000 at the end of each year for 8 years and can then be sold for €20 000. The cost of capital is 8%. Find the net present value NPV.
C0 = €80 000CF = €15 000S = €20 000n = 8r = 8%
NPV = €
4.NPV & IRRHard
A project costs €40 000 and returns €24 000, €25 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −40 000CF1 = 24 000CF2 = 25 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 7% and matures in 11 years. Its yield to maturity changes from 3.5% to 2.5%. By how much ΔP does the bond price change?
F = €1 000coupon = 7%n = 11y1 = 3.5%y2 = 2.5%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 4 years. It sells for €892. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 2% (2×)n = 4P = €892
YTM = %
7.CAPM & WACCMedium
A company has a debt-to-equity ratio (at market values) of 1.5. Its cost of equity is 11%, its pre-tax cost of debt is 4% and the tax rate is 25%. Find the WACC.
D/E = 1.5re = 11%rd = 4%T = 25%
WACC = %
8.CAPM & WACCHard
A company's shares have a beta of 1.5 at its current debt-to-equity ratio of 0.5. It plans to raise its debt-to-equity ratio to 1. The tax rate is 25%, the risk-free rate 2.5% and the market risk premium 7%. Find the cost of equity re after the change.
Drag z so that the shaded area between −z and z under the standard normal curve is 0.8664. Find z to two decimals.
A = 0.8664
z =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 75% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g75%
x = g
15.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 4.2. What is the margin of error E₂ of a 99% interval based on 200 observations?
95%n₁ = 100E₁ = 4.299%n₂ = 200
E₂ =
16.Confidence intervalsHard
The population standard deviation is 34. 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 0.5?
σ = 34E = 0.599%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 50; the population standard deviation is known to be 12. A random sample of 144 observations has a mean of 49.04. Test H₀: μ = 50 against H₁: μ < 50. Find the p-value p.
μ₀ = 50σ = 12n = 144x̄ = 49.04H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A sample of 100 measurements has mean 51.4. The population standard deviation is 12. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 51.4σ = 12n = 10095%
U =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
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. Predict ŷ for x = 3.
n = 5x = 3
ŷ =
3Solutions
1.Time value of money
Answer: FV = 12712 €
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.
Exponent
r·t = 0.03·8 = 0.24
e0.24 = 1.271
Future value
FV = 10 000·1.271 = €12 712.49
Answer
FV = €12 712.49
2.Time value of money
Answer: A = 263.5 €
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.
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.
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(10%) = 2 479.34
NPV(15%) = −226.84
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.
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, N = 8
Solve numerically
i = 2.507% per half-year
YTM
YTM = 2·2.507% = 5.013%
Answer
YTM = 5.013%
7.CAPM & WACC
Answer: WACC = 6.2 %
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 + 1.5) = 0.4
D/V = 1.5/(1 + 1.5) = 0.6
WACC
WACC = 0.4·11% + 0.6·4%·(1 − 0.25) = 6.2%
Answer
WACC = 6.2%
8.CAPM & WACC
Answer: re = 15.86 %
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 = 1.5/(1 + 0.75·0.5) = 1.091
Relever at the new D/E
βL = 1.091·(1 + 0.75·1) = 1.909
Cost of equity (CAPM)
re = 2.5% + 1.909·7% = 15.86%
Answer
re = 15.86%
9.Descriptive statistics
Answer: s = 16.84
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 = (90.4 − 100)/12 = −0.8
Answer
z = −0.8
11.Probability
Answer: σ = 11.08
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.
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 = 1.45, Var(X) = 1.847, σX = 1.359
Linear change
σY = |5|·1.359 = 6.796
Answer
σY = 6.796
The mean does move: E(Y) = 5·1.45 + 4 = 11.25.
13.Normal distribution
Answer: z = 1.5
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 + 0.8664)/2 = 0.9332
Read the z-table backwards
Find 0.9332 inside the table and read off z.
z = 1.50
Answer
z = 1.50
14.Normal distribution
Answer: x = 253.4 g
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) = 0.75
z from the table
z ≈ 0.67
Back to the original units
x = 250 + 0.67·5 = 253.4 g
Answer
x = 253.4 g
15.Confidence intervals
Answer: E₂ = 3.903
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 = (2.576/1.96)·√(100/200) = 0.9293
New margin
E₂ = 4.2·0.9293 = 3.903
Answer
E₂ = 3.903
16.Confidence intervals
Answer: n = 30684
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 = (2.576·34/0.5)² = 30684
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 30684
Answer
n = 30684
17.Hypothesis tests & confidence intervals
Answer: p = 0.1685
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.
Test statistic
SE = σ/√n = 12/√144 = 1
z = (49.04 − 50)/1 = −0.96
Tail area
Φ(0.96) = 0.8315
1 − Φ(0.96) = 0.1685
p-value
p = 0.1685
Answer
p = 0.1685
p > 0.05, so H₀ is not rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: U = 53.75
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 = 1.96·12/√100 = 2.352
Interval limit
U = 51.4 + 2.352 = 53.75
Answer
U = 53.75
19.Linear regression & correlation
Answer: r² = 0.894
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 = 95/5 = 19
ȳ = 497/5 = 99.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 224
Syy = Σ(y − ȳ)² = 1403
Coefficient of determination
r² = 224²/(40·1403) = 0.894
Answer
r² = 0.894
About 89.4% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 4.45
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 = 47/5 = 9.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 66
Regression line
b = 66/40 = 1.65
a = 9.4 − 1.65·6 = −0.5
ŷ = −0.5 + 1.65·x
Prediction
ŷ = −0.5 + 1.65·3 = 4.45
Answer
ŷ = 4.45
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