You need €15 000 in 6 years. An account pays 3.6% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €15 0003.6% / 12n = 6
PV = €
2.Time value of moneyHard
A loan of €200 000 is repaid with equal monthly payments over 25 years. The interest rate is 3% per year, compounded monthly. Find the monthly payment A.
P = €200 0003% / 12N = 300
A = €
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 4: €29 000, €25 000, €26 000, €29 000. The cost of capital is 10%. Find the net present value NPV.
A project costs €40 000 and returns €24 500, €29 000, €27 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −40 000CF1 = 24 500CF2 = 29 000CF3 = 27 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 11 years. The yield to maturity is 3.5% per year. Find the price P of the bond.
F = €1 000coupon = 5%YTM = 3.5%n = 11
P = €
6.Bond pricing & yieldHard
A bond with 13 years to maturity (face value €1 000, annual coupon 8%) sells for €1 177. The issuer can call it in 6 years at a call price of €1 080. Find the yield to call YTC in percent.
F = €1 000coupon = 8%P = €1 177call price = €1 080call in 6 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €12 000 000 and debt worth €3 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 3% and the tax rate is 28%. Find the WACC.
A company's business has an unlevered (asset) beta of 1. It plans a debt-to-equity ratio of 1.5. The tax rate is 30%, the risk-free rate is 4% and the market risk premium is 6.5%. Find the cost of equity re at the new capital structure.
βU = 1D/E = 1.5T = 30%rf = 4%rm − rf = 6.5%
re = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 29, 36, 36, 51, 44, 38, 23.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 12. Find the z-score z of a score of 62.
μ = 50σ = 12x = 62
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3, 4p(x): 0.25, 0.2, 0.25, 0.25, 0.05
Var(X) =
12.ProbabilityHard
12 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?
n = 12p = 0.1
P =
13.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 9 of them exceeds 224.3 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 9> 224.3 kg
P =
14.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
15.Confidence intervalsMedium
A sample of 8 items has mean 41.5 and standard deviation s = 12.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 8x̄ = 41.5s = 12.599%t(7) = 3.499
E =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 7 plots, mean yield 83 kg, standard deviation 4.5 kg. Plot group 2: 5 plots, mean 85 kg, standard deviation 5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
A sample of 100 students has a mean score of 497.5 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 497.5s = 8n = 100
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 10. Find the smallest sample size n.
σ = 10E = 199%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
20.Linear regression & correlationHard
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. Predict ŷ for x = 23.5.
n = 6x = 23.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 12090 €
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 = 0.036000000000000004/12 = 0.003
N = 12·6 = 72
Present value
PV = FV/(1 + i)N = 15 000/1.241 = €12 089.94
Answer
PV = €12 089.94
2.Time value of money
Answer: A = 948.4 €
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 rate and number of payments
i = 0.03/12 = 0.0025
N = 12·25 = 300
Payment
The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.
(1 + i)−N = 0.4728
A = 200 000·0.0025/(1 − 0.4728) = €948.42
Answer
A = €948.42
3.NPV & IRR
Answer: NPV = 26366 €
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 = −60 000
CF1/(1 + 0.1)1 = 29 000/1.1 = 26 363.64
CF2/(1 + 0.1)2 = 25 000/1.21 = 20 661.16
CF3/(1 + 0.1)3 = 26 000/1.331 = 19 534.18
CF4/(1 + 0.1)4 = 29 000/1.464 = 19 807.39
Add them up
NPV = €26 366.37
Answer
NPV = €26 366.37
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 44.42 %
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%) = 2 317.78
NPV(45%) = −289.88
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 = €25.00, i = 0.0175, N = 22
Present value of the coupons
€25.00·(1 − 0.6827)/0.0175 = €453.26
Present value of the face value
€1 000·0.6827 = €682.72
Price
P = €1 135.98
Answer
P = €1 135.98
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTC = 5.604 %
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
1177 = 80·(1 − (1 + y)−6)/y + 1080/(1 + y)6
Solve numerically
y = 5%: €1 211.97
y = 6%: €1 154.74
YTC = 5.604%
Answer
YTC = 5.604%
The yield to maturity is 6.001%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.
7.CAPM & WACC
Answer: WACC = 6.832 %
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 = €15 000 000, E/V = 0.8, D/V = 0.2
WACC
WACC = 0.8·8% + 0.2·3%·(1 − 0.28) = 6.832%
Answer
WACC = 6.832%
8.CAPM & WACC
Answer: re = 17.32 %
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 + 0.7·1.5) = 2.05
Cost of equity (CAPM)
re = 4% + 2.05·6.5% = 17.32%
Answer
re = 17.32%
9.Descriptive statistics
Answer: s = 9.196
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 = (62 − 50)/12 = 1
Answer
z = 1
11.Probability
Answer: Var(X) = 1.528
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.
Zero and one failure
P(X = 0) = 0.912 = 0.2824
P(X = 1) = 12·0.1·0.911 = 0.3766
Add them
P = 0.2824 + 0.3766 = 0.659
Answer
P = 0.659
13.Normal distribution
Answer: P = 0.6808
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 total
μT = 225 kg, σT = 0.5·√9 = 1.5 kg
Standardise
z = (224.3 − 225)/1.5 = −0.47
Probability
P = 1 − Φ(−0.47) = 0.6808
Answer
P = 0.6808
14.Normal distribution
Answer: P = 0.8413
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 = 500 − 490 = 10 g, σ = 10 g
Standardise
z = (0 − 10)/10 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
15.Confidence intervals
Answer: E = 15.46
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 = 7, t = 3.499
Margin of error
E = 3.499·12.5/√8 = 15.46
Answer
E = 15.46
16.Confidence intervals
Answer: upper limit = 2.993 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 = (6·4.5² + 4·5²)/10 = 22.15
sp = 4.706
Margin of error
E = 1.812·4.706·√(1/7 + 1/5) = 4.993
Interval
−2 ± 4.993 → [−6.993, 2.993]
Answer
upper limit = 2.993 kg
17.Hypothesis tests & confidence intervals
Answer: t = −3.125
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 = 8/√100 = 8/10 = 0.8
Test statistic
t = (497.5 − 500)/0.8 = −3.125
Answer
t = −3.125
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: n = 664
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 = (2.576·10/1)² = 663.6
Round up
n = 664
Answer
n = 664
19.Linear regression & correlation
Answer: r² = 0.9944
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 = 132/5 = 26.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 174
Syy = Σ(y − ȳ)² = 761.2
Coefficient of determination
r² = 174²/(40·761.2) = 0.9944
Answer
r² = 0.9944
About 99.44% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 79.33
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 = 141/6 = 23.5
ȳ = 476/6 = 79.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 573
Regression line
b = 573/157.5 = 3.638
a = 79.33 − 3.638·23.5 = −6.162
ŷ = −6.162 + 3.638·x
Prediction
ŷ = −6.162 + 3.638·23.5 = 79.33
Answer
ŷ = 79.33
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