You need €25 000 in 3 years. An account pays 4.8% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €25 0004.8% / 12n = 3
PV = €
2.Time value of moneyHard
A loan of €12 000 is repaid with equal monthly payments over 3 years. The interest rate is 4.2% per year, compounded monthly. Find the monthly payment A.
P = €12 0004.2% / 12N = 36
A = €
3.NPV & IRRMedium
A machine costs €80 000 today, lasts 7 years and costs €4 000 per year to run (paid at the end of each year). The cost of capital is 8%. Find its equivalent annual cost EAC.
C0 = €80 000M = €4 000n = 7r = 8%
EAC = €
4.NPV & IRRHard
A project costs €10 000 and returns €3 500, €3 000, €6 000, €4 500 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 12%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 5% per year. Find the price P of the bond.
F = €1 000coupon = 8%YTM = 5%n = 7
P = €
6.Bond pricing & yieldHard
A bond with 19 years to maturity (face value €1 000, annual coupon 7%) sells for €1 112. The issuer can call it in 6 years at a call price of €1 070. Find the yield to call YTC in percent.
F = €1 000coupon = 7%P = €1 112call price = €1 070call in 6 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €3 000 000 and debt worth €6 000 000. Its cost of equity is 9%, its pre-tax cost of debt is 4.5% and the tax rate is 30%. Find the WACC.
A company has 4 000 000 shares at €21.50 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 3% and the tax rate is 25.8%. The share’s beta is 1.1, the risk-free rate 3% and the market return 10%. Find the WACC.
β = 1.1rf = 3%rm = 10%rd = 3%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 41, 37, 37, 25, 37, 26, 45.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 5. What proportion P of values is below 62.5?
μ = 70σ = 5x = 62.5
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.15, 0.2, 0.3, 0.15, 0.2
σ =
12.ProbabilityHard
A game costs €3 to play. You win €12 with probability 0.15 and €150 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€12: p = 0.15€150: p = 0.05
E(G) = €
13.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 57 and 73 points? Give P as a decimal.
μ = 65σ = 10x1 = 57x2 = 73
P =
14.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 60 phones, mean 53.5 h, known standard deviation 13 h. Model 2: 80 phones, mean 57 h, known standard deviation 8 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.
The population standard deviation is 22. 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 = 22E = 0.599%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 101.6 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 101.6s = 12n = 81
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 53 of 150 patients on treatment A recover, and 41 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 53n₁ = 150x₂ = 41n₂ = 100
z =
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 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.5.
n = 5x = 3.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 21653 €
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 = 25 000/1.155 = €21 653.41
Answer
PV = €21 653.41
2.Time value of money
Answer: A = 355.4 €
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 = 0.042/12 = 0.0035
N = 12·3 = 36
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 = 0.8818
A = 12 000·0.0035/(1 − 0.8818) = €355.36
Answer
A = €355.36
3.NPV & IRR
Answer: EAC = 19366 €
Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.
Annuity factor
(1 − 1.08−7)/0.08 = 5.206
Annualised purchase price
80 000/5.206 = 15 365.79
EAC
EAC = 15 365.79 + 4 000 = €19 365.79
Answer
EAC = €19 365.79
4.NPV & IRR
Answer: MIRR = 18.77 %
Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.
Terminal value of the inflows
3 500·1.123 = 4 917.25
3 000·1.122 = 3 763.20
6 000·1.121 = 6 720.00
4 500·1.120 = 4 500.00
TV = 19 900.45
MIRR
MIRR = (19 900.45/10 000)1/4 − 1 = 18.77%
Answer
MIRR = 18.77%
5.Bond pricing & yield
Answer: P = 1175 €
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 = €40.00, i = 0.025, N = 14
Present value of the coupons
€40.00·(1 − 0.7077)/0.025 = €467.64
Present value of the face value
€1 000·0.7077 = €707.73
Price
P = €1 175.36
Answer
P = €1 175.36
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTC = 5.75 %
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
1112 = 70·(1 − (1 + y)−6)/y + 1070/(1 + y)6
Solve numerically
y = 5%: €1 153.75
y = 6%: €1 098.52
YTC = 5.75%
Answer
YTC = 5.75%
The yield to maturity is 5.997%. 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 = 5.1 %
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 = €9 000 000, E/V = 0.3333, D/V = 0.6667
WACC
WACC = 0.3333·9% + 0.6667·4.5%·(1 − 0.3) = 5.1%
Answer
WACC = 5.1%
8.CAPM & WACC
Answer: WACC = 8.084 %
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).
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) = 0.06681
P = 0.06681
Answer
P = 0.06681
11.Probability
Answer: σ = 1.322
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.
Net outcomes
G: 9 (p = 0.15), 147 (p = 0.05), −3 (p = 0.8)
Expected value
E(G) = 9·0.15 + 147·0.05 − 3·0.8 = 6.3 €
Answer
E(G) = 6.3 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.5763
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 = (57 − 65)/10 = −0.8
z₂ = (73 − 65)/10 = 0.8
Area between
P = Φ(0.8) − Φ(−0.8) = 0.7881 − 0.2119 = 0.5763
Answer
P = 0.5763
14.Normal distribution
Answer: P = 0.7389
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 = 180 − 170 = 10 cm, σ = 15.62 cm
Standardise
z = (0 − 10)/15.62 = −0.64
Probability
P = 1 − Φ(−0.64) = 0.7389
Answer
P = 0.7389
15.Confidence intervals
Answer: upper limit = 1.399 h
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.
Point estimate
x̄₁ − x̄₂ = 53.5 − 57 = −3.5
Standard error
SE = √(13²/60 + 8²/80) = 1.902
Margin of error
E = 2.576·1.902 = 4.899
Interval
−3.5 ± 4.899 → [−8.399, 1.399]
Answer
upper limit = 1.399 h
The interval contains 0, so the data do not show a significant difference.
16.Confidence intervals
Answer: n = 12847
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 = (2.576·22/0.5)² = 12847
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 12847
Answer
n = 12847
17.Hypothesis tests & confidence intervals
Answer: t = 1.2
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 = 12/√81 = 12/9 = 1.333
Test statistic
t = (101.6 − 100)/1.333 = 1.2
Answer
t = 1.2
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: z = −0.9062
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.
Sample proportions
p̂₁ = 53/150 = 0.3533
p̂₂ = 41/100 = 0.41
p̂ = 94/250 = 0.376
Standard error under H₀
SE = √(0.376·0.624·(1/150 + 1/100)) = 0.06253
Test statistic
z = (0.3533 − 0.41)/0.06253 = −0.9062
Answer
z = −0.9062
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9939
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 = 72/5 = 14.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 72
Syy = Σ(y − ȳ)² = 131.2
Correlation coefficient
r = 72/√(40·131.2) = 0.9939
Answer
r = 0.9939
20.Linear regression & correlation
Answer: ŷ = 13.7
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 = 15/5 = 3
ȳ = 76/5 = 15.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −30
Regression line
b = −30/10 = −3
a = 15.2 − (−3)·3 = 24.2
ŷ = 24.2 − 3·x
Prediction
ŷ = 24.2 + (−3)·3.5 = 13.7
Answer
ŷ = 13.7
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