You need €10 000 in 3 years. An account pays 3% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €10 0003% / 4n = 3
PV = €
2.Time value of moneyHard
A mortgage of €300 000 is repaid with equal monthly payments over 20 years at 6% per year, compounded monthly. How much is still owed just after the payment at the end of year 4? Find the outstanding balance B.
P = €300 0006% / 12N = 240k = 48
B = €
3.NPV & IRRMedium
A project costs €50 000 today and returns €18 500, €20 000, €18 000, €16 000, €21 000 at the end of years 1 to 5. The cost of capital is 8%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €50 000 and returns €36 000, €33 500, €31 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −50 000CF1 = 36 000CF2 = 33 500CF3 = 31 500
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 11 years. The yield to maturity is 2.5% per year. Find the price P of the bond.
F = €1 000coupon = 6%YTM = 2.5%n = 11
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 4% matures in 6 years and sells for €880. Find its yield to maturity YTM in percent.
F = €1 000coupon = 4%n = 6P = €880
YTM = %
7.CAPM & WACCMedium
A company has equity worth €5 000 000 and debt worth €3 000 000. Its cost of equity is 9%, its pre-tax cost of debt is 5% and the tax rate is 25%. Find the WACC.
E = €5 000 000D = €3 000 000re = 9%rd = 5%T = 25%
WACC = %
8.CAPM & WACCHard
A company has 1 000 000 shares at €18.50 each and 5 000 bonds trading at €1 000 each. The pre-tax cost of debt is 7% and the tax rate is 25.8%. The share’s beta is 1.8, the risk-free rate 1% and the market return 8%. Find the WACC.
β = 1.8rf = 1%rm = 8%rd = 7%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 45, 47, 41, 45, 24, 45.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 5. What proportion P of values is below 97.5?
μ = 100σ = 5x = 97.5
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.25, 0.25, 0.35, 0.15
Var(X) =
12.ProbabilityHard
A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 1 correct answers?
n = 10p = 0.1k = 1
P =
13.Normal distributionMedium
X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.6 mm. Round z to two decimals.
μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.6 mm
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 11 items has mean 52.5 and standard deviation s = 4.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.
n = 11x̄ = 52.5s = 4.595%t(10) = 2.228
E =
16.Confidence intervalsHard
The population standard deviation is 9. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 2.5?
σ = 9E = 2.590%
n =
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 30 parts has mean length 83 mm, and the line's known standard deviation is 18 mm. Line 2: a sample of 30 parts has mean 76.3 mm, known standard deviation 19 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
In a trial, 30 of 120 patients on treatment A recover, and 36 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 30n₁ = 120x₂ = 36n₂ = 120
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 = 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 = 42?
n = 5ŷ = 42
x =
3Solutions
1.Time value of money
Answer: PV = 9142 €
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 = 0.03/4 = 0.0075
N = 4·3 = 12
Present value
PV = FV/(1 + i)N = 10 000/1.094 = €9 142.38
Answer
PV = €9 142.38
2.Time value of money
Answer: B = 264875 €
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.
Whatever is still owed must be exactly paid off by the remaining payments.
N − k = 240 − 48 = 192
B = 2149·(1 − (1 + 0.005)−192)/0.005 = €264 874.65
Answer
B = €264 874.65
3.NPV & IRR
Answer: DPB = 3.122 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: 18 500/1.08 = 17 129.63 → Σ = 17 129.63
year 2: 20 000/1.166 = 17 146.78 → Σ = 34 276.41
year 3: 18 000/1.26 = 14 288.98 → Σ = 48 565.39
year 4: 16 000/1.36 = 11 760.48 → Σ = 60 325.86
Fraction of the last year
DPB = 3 + 1 434.61/11 760.48 = 3.122
Answer
DPB = 3.122 years
The discounted payback is always longer than the simple payback (2.639 years).
4.NPV & IRR
Answer: IRR = 46.84 %
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(45%) = 1 093.53
NPV(50%) = −1 777.78
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 = 0.0125, N = 22
Present value of the coupons
€30.00·(1 − 0.7609)/0.0125 = €573.92
Present value of the face value
€1 000·0.7609 = €760.87
Price
P = €1 334.78
Answer
P = €1 334.78
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 6.477 %
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
€880 = €40·(1 − (1 + y)⁻⁶)/y + €1 000/(1 + y)⁶
Solve numerically
P(6%) = €901.65, P(7%) = €857.00
YTM = 6.477%
Answer
YTM = 6.477%
7.CAPM & WACC
Answer: WACC = 7.031 %
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 = €8 000 000, E/V = 0.625, D/V = 0.375
WACC
WACC = 0.625·9% + 0.375·5%·(1 − 0.25) = 7.031%
Answer
WACC = 7.031%
8.CAPM & WACC
Answer: WACC = 11.81 %
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 0.3085
Answer
P = 0.3085
11.Probability
Answer: Var(X) = 1.04
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 = 1·0.25 + 2·0.25 + 3·0.35 + 4·0.15 = 2.4
Mean of the squares
E(X²) = 1·0.25 + 4·0.25 + 9·0.35 + 16·0.15 = 6.8
Variance
Var(X) = 6.8 − 2.4² = 1.04
Answer
Var(X) = 1.04
12.Probability
Answer: P = 0.3874
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) = 10·0.11·0.99 = 10·0.1·0.3874 = 0.3874
Answer
P = 0.3874
13.Normal distribution
Answer: P = 0.7881
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 sum
μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm
Standardise
z = (199.6 − 200)/0.5 = −0.8
Upper tail
P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881
Answer
P = 0.7881
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 = 3.023
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 = 2.228·4.5/√11 = 3.023
Answer
E = 3.023
16.Confidence intervals
Answer: n = 36
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 = (1.645·9/2.5)² = 35.07
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 36
Answer
n = 36
17.Hypothesis tests & confidence intervals
Answer: z = 1.402
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 = √(18²/30 + 19²/30) = √22.83 = 4.778
Test statistic
z = (83 − 76.3)/4.778 = 1.402
Answer
z = 1.402
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = −0.8674
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 = 30/120 = 0.25
p̂₂ = 36/120 = 0.3
p̂ = 66/240 = 0.275
Standard error under H₀
SE = √(0.275·0.725·(1/120 + 1/120)) = 0.05764
Test statistic
z = (0.25 − 0.3)/0.05764 = −0.8674
Answer
z = −0.8674
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9923
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 = 123/5 = 24.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 146
Syy = Σ(y − ȳ)² = 541.2
Correlation coefficient
r = 146/√(40·541.2) = 0.9923
Answer
r = 0.9923
20.Linear regression & correlation
Answer: x = 2.009
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.