You need €20 000 in 3 years. An account pays 4.8% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €20 0004.8% / 12n = 3
PV = €
2.Time value of moneyHard
A loan of €30 000 is repaid with equal monthly payments over 7 years at 3% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €30 0003% / 12N = 84
I = €
3.NPV & IRRMedium
An investment of €30 000 gives these cash flows at the end of years 1 to 4: €14 000, €9 000, €11 000, €13 000. The cost of capital is 10%. Find the net present value NPV.
A project costs €25 000 and returns €15 000, €13 000, €15 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −25 000CF1 = 15 000CF2 = 13 000CF3 = 15 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 3% per year. Find the price P of the bond.
F = €1 000coupon = 6%YTM = 3%n = 12
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 8% matures in 7 years and sells for €1 110. Find its yield to maturity YTM in percent.
F = €1 000coupon = 8%n = 7P = €1 110
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €6 000 000, preferred shares worth €500 000 and debt worth €2 000 000. The cost of equity is 11%, the cost of preferred stock 6.5% and the pre-tax cost of debt 5%. The tax rate is 30%. Find the WACC.
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 0.6. The tax rate is 20%, the risk-free rate 3% and the market risk premium 6%. 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 0.8969. Find z to two decimals.
A = 0.8969
z =
14.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.04 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
15.Confidence intervalsMedium
A sample of 6 items has mean 12 and standard deviation s = 14. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 6x̄ = 12s = 1490%t(5) = 2.015
E =
16.Confidence intervalsHard
The population standard deviation is 28. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 3?
σ = 28E = 395%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 195.6 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 195.6s = 20n = 100
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 500 against H₁: μ > 500 is done at α = 0.05 with a sample of 64; the standard deviation is known to be 8. In reality the mean is 502.2. Find the power 1 − β of the test in percent.
μ₀ = 500μ = 502.2σ = 8n = 64α = 0.05
1 − β = %
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 = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 45?
n = 6ŷ = 45
x =
3Solutions
1.Time value of money
Answer: PV = 17323 €
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 = 0.048/12 = 0.004
N = 12·3 = 36
Present value
PV = FV/(1 + i)N = 20 000/1.155 = €17 322.73
Answer
PV = €17 322.73
2.Time value of money
Answer: I = 3298 €
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 payment
i = 0.0025, N = 84
A = 30 000·0.0025/(1 − (1 + 0.0025)−84) = €396.40
Total interest
I = N·A − P = 84·396.4 − 30 000 = €3 297.52
Answer
I = €3 297.52
3.NPV & IRR
Answer: NPV = 7309 €
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 + 0.1)1 = 14 000/1.1 = 12 727.27
CF2/(1 + 0.1)2 = 9 000/1.21 = 7 438.02
CF3/(1 + 0.1)3 = 11 000/1.331 = 8 264.46
CF4/(1 + 0.1)4 = 13 000/1.464 = 8 879.17
Add them up
NPV = €7 308.93
Answer
NPV = €7 308.93
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 33.01 %
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(30%) = 1 058.26
NPV(35%) = −659.20
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.
Per period
C = €30.00, i = 0.015, N = 24
Present value of the coupons
€30.00·(1 − 0.6995)/0.015 = €600.91
Present value of the face value
€1 000·0.6995 = €699.54
Price
P = €1 300.46
Answer
P = €1 300.46
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 6.028 %
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.
Equation for the yield
€1 110 = €80·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷
Solve numerically
P(6%) = €1 111.65, P(7%) = €1 053.89
YTM = 6.028%
Answer
YTM = 6.028%
7.CAPM & WACC
Answer: WACC = 8.971 %
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).
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 + 0.8·0.5) = 1.071
Relever at the new D/E
βL = 1.071·(1 + 0.8·0.6) = 1.586
Cost of equity (CAPM)
re = 3% + 1.586·6% = 12.51%
Answer
re = 12.51%
9.Descriptive statistics
Answer: s = 13.84
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 0.9332
Answer
P = 0.9332
11.Probability
Answer: P = 0.2
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.
First and second draw
P = 7/15 · 6/14 = 42/210 = 0.2
Answer
P = 0.2
12.Probability
Answer: σY = 2.917
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 = 3.15, Var(X) = 2.128, σX = 1.459
Linear change
σY = |−2|·1.459 = 2.917
Answer
σY = 2.917
The mean does move: E(Y) = −2·3.15 + 4 = −2.3.
13.Normal distribution
Answer: z = 1.63
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 + 0.8969)/2 = 0.9484
Read the z-table backwards
Find 0.9484 inside the table and read off z.
z = 1.63
Answer
z = 1.63
14.Normal distribution
Answer: P = 0.9633
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 = 25.04 − 25 = 0.04 mm, σ = 0.02236 mm
Standardise
z = (0 − 0.04)/0.02236 = −1.79
Probability
P = 1 − Φ(−1.79) = 0.9633
Answer
P = 0.9633
15.Confidence intervals
Answer: E = 11.52
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.
Critical value
df = n − 1 = 5, t = 2.015
Margin of error
E = 2.015·14/√6 = 11.52
Answer
E = 11.52
16.Confidence intervals
Answer: n = 335
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 = (1.96·28/3)² = 334.6
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 335
Answer
n = 335
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 = 20/√100 = 20/10 = 2
Test statistic
t = (195.6 − 200)/2 = −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: 1 − β = 71.06 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 8/√64 = 1
x̄c = 500 + 1.645·1 = 501.6
z under the true mean
z = (501.6 − 502.2)/1 = −0.555
Power
power = P(Z > −0.555) = 71.06%
1 − β = 71.06%
Answer
1 − β = 71.06%
19.Linear regression & correlation
Answer: r = 0.988
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 = 16023 − 147·624/6 = 735
Sxx = 3759 − 147²/6 = 157.5
Syy = 68410 − 624²/6 = 3514
Correlation coefficient
r = 735/√(157.5·3514) = 0.988
Answer
r = 0.988
20.Linear regression & correlation
Answer: x = 3.453
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 = 33/6 = 5.5
ȳ = 350/6 = 58.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 114
Regression line
b = 114/17.5 = 6.514
a = 58.33 − 6.514·5.5 = 22.5
ŷ = 22.5 + 6.514·x
Solve for x
x = (45 − 22.5)/6.514 = 3.453
Answer
x = 3.453
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