An annuity pays €1 000 at the end of each year for 6 years. The interest rate is 4% per year. Find the present value PV of the annuity.
A = €1 000i = 4%n = 6
PV = €
2.Time value of moneyHard
A loan of €25 000 is repaid with equal monthly payments over 7 years at 3.6% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €25 0003.6% / 12N = 84
I = €
3.NPV & IRRMedium
A project costs €40 000 today and returns €16 000, €12 500, €14 000, €13 000 at the end of years 1 to 4. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €40 000 and returns €27 000, €15 500, €19 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −40 000CF1 = 27 000CF2 = 15 500CF3 = 19 500
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 10 years. The yield to maturity is 8% per year. Find the price P of the bond.
F = €1 000coupon = 7%YTM = 8%n = 10
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 3% annual coupon in two half-yearly payments and matures in 3 years. It sells for €869. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 3% (2×)n = 3P = €869
YTM = %
7.CAPM & WACCMedium
A company has equity worth €4 000 000 and debt worth €2 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 6% and the tax rate is 30%. Find the WACC.
A company's business has an unlevered (asset) beta of 1.1. It plans a debt-to-equity ratio of 0.6. The tax rate is 25%, the risk-free rate is 3% and the market risk premium is 5.5%. Find the cost of equity re at the new capital structure.
βU = 1.1D/E = 0.6T = 25%rf = 3%rm − rf = 5.5%
re = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 35, 11, 23, 25, 37.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 20. What proportion P of values is below 20?
μ = 60σ = 20x = 20
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3, 4p(x): 0.25, 0.2, 0.35, 0.1, 0.1
σ =
12.ProbabilityHard
X has the distribution shown, and Y = 0.5X + 4. Find the standard deviation σY of Y.
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 23.3 and 36.8 min? Give P as a decimal.
μ = 32σ = 6x1 = 23.3x2 = 36.8
P =
14.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 10% of them lie (the 10th percentile).
μ = 65σ = 10P = 10%
x = points
15.Confidence intervalsMedium
A sample of 30 items has mean 18.5 and standard deviation s = 18. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 30x̄ = 18.5s = 1890%t(29) = 1.699
lower limit =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 7 plots, mean yield 33 kg, standard deviation 2.5 kg. Plot group 2: 5 plots, mean 42 kg, standard deviation 5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 99% confidence interval for μ₁ − μ₂.
Two production lines make the same part. Line 1: a sample of 60 parts has mean length 68.5 mm, and the line's known standard deviation is 9 mm. Line 2: a sample of 60 parts has mean 70 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).
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 12. Find the smallest sample size n.
σ = 12E = 399%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (11, 33) is made. Find its residual e = y − ŷ.
n = 6(11, 33)
e =
3Solutions
1.Time value of money
Answer: PV = 5242 €
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 − 0.7903)/0.04 = 5.242
Present value
PV = 1 000·5.242 = €5 242.14
Answer
PV = €5 242.14
2.Time value of money
Answer: I = 3319 €
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 = 0.003, N = 84
A = 25 000·0.003/(1 − (1 + 0.003)−84) = €337.14
Total interest
I = N·A − P = 84·337.1 − 25 000 = €3 319.44
Answer
I = €3 319.44
3.NPV & IRR
Answer: DPB = 3.7 years
Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.
Discounted and cumulative cash flows
year 1: 16 000/1.12 = 14 285.71 → Σ = 14 285.71
year 2: 12 500/1.254 = 9 964.92 → Σ = 24 250.64
year 3: 14 000/1.405 = 9 964.92 → Σ = 34 215.56
year 4: 13 000/1.574 = 8 261.74 → Σ = 42 477.30
Fraction of the last year
DPB = 3 + 5 784.44/8 261.74 = 3.7
Answer
DPB = 3.7 years
The discounted payback is always longer than the simple payback (2.821 years).
4.NPV & IRR
Answer: IRR = 27.72 %
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(25%) = 1 504.00
NPV(30%) = −1 183.43
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 = €35.00, i = 0.04, N = 20
Present value of the coupons
€35.00·(1 − 0.4564)/0.04 = €475.66
Present value of the face value
€1 000·0.4564 = €456.39
Price
P = €932.05
Answer
P = €932.05
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 7.998 %
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 = €15.00, N = 6
Solve numerically
i = 3.999% per half-year
YTM
YTM = 2·3.999% = 7.998%
Answer
YTM = 7.998%
7.CAPM & WACC
Answer: WACC = 10.07 %
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 = €6 000 000, E/V = 0.6667, D/V = 0.3333
WACC
WACC = 0.6667·13% + 0.3333·6%·(1 − 0.3) = 10.07%
Answer
WACC = 10.07%
8.CAPM & WACC
Answer: re = 11.77 %
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 = 1.1·(1 + 0.75·0.6) = 1.595
Cost of equity (CAPM)
re = 3% + 1.595·5.5% = 11.77%
Answer
re = 11.77%
9.Descriptive statistics
Answer: s = 10.45
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 0.02275
Answer
P = 0.02275
11.Probability
Answer: σ = 1.241
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 = 3.2, Var(X) = 5.76, σX = 2.4
Linear change
σY = |0.5|·2.4 = 1.2
Answer
σY = 1.2
The mean does move: E(Y) = 0.5·3.2 + 4 = 5.6.
13.Normal distribution
Answer: P = 0.7146
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 = (23.3 − 32)/6 = −1.45
z₂ = (36.8 − 32)/6 = 0.8
Area between
P = Φ(0.8) − Φ(−1.45) = 0.7881 − 0.07353 = 0.7146
Answer
P = 0.7146
14.Normal distribution
Answer: x = 52.18 points
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) = 0.1 → z = −1.282
Back to x
x = 65 + −1.282·10 = 52.18 points
Answer
x = 52.18 points
15.Confidence intervals
Answer: lower limit = 12.92
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 = 29, t = 1.699
Margin of error
E = 1.699·18/√30 = 5.583
Interval
18.5 ± 5.583 → [12.92, 24.08]
Answer
lower limit = 12.92
16.Confidence intervals
Answer: lower limit = −15.88 kg
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.
Pooled standard deviation
sp² = (6·2.5² + 4·5²)/10 = 13.75
sp = 3.708
Margin of error
E = 3.169·3.708·√(1/7 + 1/5) = 6.881
Interval
−9 ± 6.881 → [−15.88, −2.119]
Answer
lower limit = −15.88 kg
17.Hypothesis tests & confidence intervals
Answer: z = −0.9129
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 = √(9²/60 + 9²/60) = √2.7 = 1.643
Test statistic
z = (68.5 − 70)/1.643 = −0.9129
Answer
z = −0.9129
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 107
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).
Solve for n
n = (2.576·12/3)² = 106.2
Round up
n = 107
Answer
n = 107
19.Linear regression & correlation
Answer: r = 0.9955
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 = 130/5 = 26
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 168
Syy = Σ(y − ȳ)² = 712
Correlation coefficient
r = 168/√(40·712) = 0.9955
Answer
r = 0.9955
20.Linear regression & correlation
Answer: e = 2.39
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 = 42/6 = 7
ȳ = 124/6 = 20.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 174
Regression line
b = 174/70 = 2.486
a = 20.67 − 2.486·7 = 3.267
ŷ = 3.267 + 2.486·x
Predicted value
ŷ = 3.267 + 2.486·11 = 30.61
Residual
e = 33 − 30.61 = 2.39
Answer
e = 2.39
The point lies above the line.
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