You need €40 000 in 3 years. An account pays 5% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €40 0005% / 4n = 3
PV = €
2.Time value of moneyHard
You borrow €10 000 at 7.2% per year, compounded monthly, and repay €175 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.
P = €10 000A = €1757.2% / 12
N = payments
3.NPV & IRRMedium
A machine costs €80 000 today, lasts 5 years and costs €1 000 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 = €1 000n = 5r = 8%
EAC = €
4.NPV & IRRHard
A project costs €25 000 and returns €9 500, €12 500, €7 500 at the end of years 1 to 3. Cash flows can be reinvested at the cost of capital of 6%. Find the modified internal rate of return MIRR in percent.
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 5% per year. Find the price P of the bond.
F = €1 000coupon = 6%YTM = 5%n = 12
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 7% matures in 10 years and sells for €900. Find its yield to maturity YTM in percent.
F = €1 000coupon = 7%n = 10P = €900
YTM = %
7.CAPM & WACCMedium
A company has equity worth €10 000 000 and debt worth €6 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 3% and the tax rate is 25.8%. Find the WACC.
A company has 2 000 000 shares at €31.00 each and 20 000 bonds trading at €1 000 each. The pre-tax cost of debt is 6% and the tax rate is 25%. The share’s beta is 1.3, the risk-free rate 3.5% and the market return 10%. Find the WACC.
β = 1.3rf = 3.5%rm = 10%rd = 6%T = 25%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 26, 28, 38, 32, 29, 38, 39.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 52.
μ = 60σ = 10x = 52
z =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3p(x): 0.35, 0.2, 0.3, 0.15
σ =
12.ProbabilityHard
A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 1 correct answers?
n = 8p = 0.5k = 1
P =
13.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 85.75 and 115 points? Give P as a decimal.
μ = 100σ = 15x1 = 85.75x2 = 115
P =
14.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 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 = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
15.Confidence intervalsMedium
A sample of 11 items has mean 25.5 and standard deviation s = 17. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 11x̄ = 25.5s = 1795%t(10) = 2.228
lower limit =
16.Confidence intervalsHard
The diameters of 10 machined shafts have a sample standard deviation of 3.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.
n = 10s = 3.8 µmχ² = 2.7, 19.023
σ = µm
17.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 73.88 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 73.88s = 8n = 100
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 49.2. The population standard deviation is 8. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 49.2σ = 8n = 8195%
U =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 6.5.
n = 6x = 6.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 34460 €
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 = 0.05/4 = 0.0125
N = 4·3 = 12
Present value
PV = FV/(1 + i)N = 40 000/1.161 = €34 460.34
Answer
PV = €34 460.34
2.Time value of money
Answer: N = 71 payments
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.
Rearrange the annuity formula
i = 0.006
(1 + i)−N = 1 − 10 000·0.006/175 = 0.6571
Solve with logarithms
N = −ln(0.6571)/ln(1.006) = 70.19
Round up to whole payments
70 full payments leave a small balance, which the last, smaller payment clears.
N = 71
Answer
N = 71 payments
3.NPV & IRR
Answer: EAC = 21037 €
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 − 1.08−5)/0.08 = 3.993
Annualised purchase price
80 000/3.993 = 20 036.52
EAC
EAC = 20 036.52 + 1 000 = €21 036.52
Answer
EAC = €21 036.52
4.NPV & IRR
Answer: MIRR = 7.922 %
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
9 500·1.062 = 10 674.20
12 500·1.061 = 13 250.00
7 500·1.060 = 7 500.00
TV = 31 424.20
MIRR
MIRR = (31 424.20/25 000)1/3 − 1 = 7.922%
Answer
MIRR = 7.922%
5.Bond pricing & yield
Answer: P = 1089 €
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 = 0.025, N = 24
Present value of the coupons
€30.00·(1 − 0.5529)/0.025 = €536.55
Present value of the face value
€1 000·0.5529 = €552.88
Price
P = €1 089.42
Answer
P = €1 089.42
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 8.526 %
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
€900 = €70·(1 − (1 + y)⁻¹⁰)/y + €1 000/(1 + y)¹⁰
Solve numerically
P(8%) = €932.90, P(9%) = €871.65
YTM = 8.526%
Answer
YTM = 8.526%
7.CAPM & WACC
Answer: WACC = 8.335 %
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 = €16 000 000, E/V = 0.625, D/V = 0.375
WACC
WACC = 0.625·12% + 0.375·3%·(1 − 0.258) = 8.335%
Answer
WACC = 8.335%
8.CAPM & WACC
Answer: WACC = 10.13 %
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 = (52 − 60)/10 = −0.8
Answer
z = −0.8
11.Probability
Answer: σ = 1.09
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.
Mean
μ = 0·0.35 + 1·0.2 + 2·0.3 + 3·0.15 = 1.25
Mean of the squares
E(X²) = 0·0.35 + 1·0.2 + 4·0.3 + 9·0.15 = 2.75
Variance
Var(X) = 2.75 − 1.25² = 1.188
σ = √1.188 = 1.09
Answer
σ = 1.09
12.Probability
Answer: P = 0.03125
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.
Binomial formula
P(X = 1) = 8·0.51·0.57 = 8·0.5·0.007813 = 0.03125
Answer
P = 0.03125
13.Normal distribution
Answer: P = 0.6703
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 = (85.75 − 100)/15 = −0.95
z₂ = (115 − 100)/15 = 1
Area between
P = Φ(1) − Φ(−0.95) = 0.8413 − 0.1711 = 0.6703
Answer
P = 0.6703
14.Normal distribution
Answer: P = 0.8849
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 = 25.03 − 25 = 0.03 mm, σ = 0.025 mm
Standardise
z = (0 − 0.03)/0.025 = −1.2
Probability
P = 1 − Φ(−1.2) = 0.8849
Answer
P = 0.8849
15.Confidence intervals
Answer: lower limit = 14.08
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 = 10, t = 2.228
Margin of error
E = 2.228·17/√11 = 11.42
Interval
25.5 ± 11.42 → [14.08, 36.92]
Answer
lower limit = 14.08
16.Confidence intervals
Answer: σ = 2.614 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 9·3.8² = 130
Limits for σ²
130/19.023 = 6.832
130/2.7 = 48.13
Limit for σ
σ = √6.832 = 2.614 µm
Answer
σ = 2.614 µm
95% interval for σ: [2.614, 6.938] µm.
17.Hypothesis tests & confidence intervals
Answer: t = −1.4
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 = (73.88 − 75)/0.8 = −1.4
Answer
t = −1.4
t0.025; df = 99 = 1.984
|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: U = 50.94
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 = 1.96·8/√81 = 1.742
Interval limit
U = 49.2 + 1.742 = 50.94
Answer
U = 50.94
19.Linear regression & correlation
Answer: r = 0.989
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 = 533/6 = 88.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 385
Syy = Σ(y − ȳ)² = 2165
Correlation coefficient
r = 385/√(70·2165) = 0.989
Answer
r = 0.989
20.Linear regression & correlation
Answer: ŷ = 59.6
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.