You need €40 000 in 10 years. An account pays 6% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €40 0006% / 4n = 10
PV = €
2.Time value of moneyHard
A mortgage of €200 000 is repaid with equal monthly payments over 25 years at 4.2% 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 = €200 0004.2% / 12N = 300k = 48
B = €
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 3: €16 000, €26 000, €14 000. The cost of capital is 9%. Find the net present value NPV.
Project A costs €30 000 and returns €17 000, €10 500, €7 000 at the end of years 1 to 3. Project B costs €35 000 and returns €9 000, €14 000, €18 000. Find the crossover rate r* at which both projects have the same NPV, in percent.
A bond with a face value of €1 000 pays an annual coupon of 2% and matures in 6 years. Its yield to maturity changes from 4% to 5%. By how much ΔP does the bond price change?
F = €1 000coupon = 2%n = 6y1 = 4%y2 = 5%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 6% matures in 5 years and sells for €960. Find its yield to maturity YTM in percent.
F = €1 000coupon = 6%n = 5P = €960
YTM = %
7.CAPM & WACCMedium
A company has equity worth €3 000 000 and debt worth €2 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 4.5% and the tax rate is 25.8%. Find the WACC.
A company has 2 500 000 shares at €38.00 each and 20 000 bonds trading at €960 each. The pre-tax cost of debt is 3% and the tax rate is 25%. The share’s beta is 1.1, the risk-free rate 3.5% and the market return 7.5%. Find the WACC.
β = 1.1rf = 3.5%rm = 7.5%rd = 3%T = 25%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 60, 30, 43, 21, 17, 24, 39.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is below 452?
μ = 500σ = 40x = 452
P =
11.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 8? The order does not matter.
n = 8k = 2
N =
12.ProbabilityHard
X has the distribution shown, and Y = 2X + 20. Find the standard deviation σY of Y.
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
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −7.5 and standard deviation 1.3. Find the lower limit of the 99% confidence interval for the mean difference.
n = 8d̄ = −7.5sd = 1.399%t(7) = 3.499
lower limit = mmHg
16.Confidence intervalsHard
The diameters of 21 machined shafts have a sample standard deviation of 1 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.
n = 21s = 1 µmχ² = 9.591, 34.17
σ = µm
17.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 12 students, mean score 26, standard deviation 11. Group 2: 14 students, mean score 35.8, standard deviation 3. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 12x̄₁ = 26s₁ = 11n₂ = 14x̄₂ = 35.8s₂ = 3
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 199.7. The population standard deviation is 15. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 199.7σ = 15n = 2595%
L =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. 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 = 49?
n = 5ŷ = 49
x =
3Solutions
1.Time value of money
Answer: PV = 22050 €
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 = 0.06/4 = 0.015
N = 4·10 = 40
Present value
PV = FV/(1 + i)N = 40 000/1.814 = €22 050.49
Answer
PV = €22 050.49
2.Time value of money
Answer: B = 180286 €
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 = 300 − 48 = 252
B = 1078·(1 − (1 + 0.0035)−252)/0.0035 = €180 286.33
Answer
B = €180 286.33
3.NPV & IRR
Answer: NPV = −12627 €
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 + 0.09)1 = 16 000/1.09 = 14 678.90
CF2/(1 + 0.09)2 = 26 000/1.188 = 21 883.68
CF3/(1 + 0.09)3 = 14 000/1.295 = 10 810.57
Add them up
NPV = €−12 626.85
Answer
NPV = €−12 626.85
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: r* = 5.224 %
Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.
Difference in cash flows
ΔCF0 = −35 000 − (−30 000) = −5 000
ΔCF1 = 9 000 − 17 000 = −8 000
ΔCF2 = 14 000 − 10 500 = 3 500
ΔCF3 = 18 000 − 7 000 = 11 000
IRR of the difference
NPVΔ(5%) = 57.77
NPVΔ(6%) = −196.37
r* = 5.224%
Answer
r* = 5.224%
Check: at 5.224% both NPVs are €1 647.78.
5.Bond pricing & yield
Answer: ΔP = −47.43 €
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.
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
€960 = €60·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵
Solve numerically
P(6%) = €1 000.00, P(7%) = €959.00
YTM = 6.975%
Answer
YTM = 6.975%
7.CAPM & WACC
Answer: WACC = 6.136 %
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 = €5 000 000, E/V = 0.6, D/V = 0.4
WACC
WACC = 0.6·8% + 0.4·4.5%·(1 − 0.258) = 6.136%
Answer
WACC = 6.136%
8.CAPM & WACC
Answer: WACC = 6.95 %
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).
Market values
E = 2 500 000·€38.00 = €95 000 000
D = 20 000·€960 = €19 200 000
Cost of equity (CAPM)
re = 3.5% + 1.1·4% = 7.9%
Weights
V = €114 200 000, E/V = 0.8319, D/V = 0.1681
WACC
WACC = 0.8319·7.9% + 0.1681·3%·(1 − 0.25) = 6.95%
Answer
WACC = 6.95%
9.Descriptive statistics
Answer: s = 15.02
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 = −1.2
Standard normal table
Φ(−1.2) = 0.1151
P = 0.1151
Answer
P = 0.1151
11.Probability
Answer: N = 28
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.
Combinations
C(8, 2) = (8·7)/(2·1) = 28
Answer
N = 28
12.Probability
Answer: σY = 4.539
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 = 2.5, Var(X) = 5.15, σX = 2.269
Linear change
σY = |2|·2.269 = 4.539
Answer
σY = 4.539
The mean does move: E(Y) = 2·2.5 + 20 = 25.
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 = 0.7881
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 difference
μ = 52 − 48 = 4 s, σ = 5 s
Standardise
z = (0 − 4)/5 = −0.8
Probability
P = 1 − Φ(−0.8) = 0.7881
Answer
P = 0.7881
15.Confidence intervals
Answer: lower limit = −9.108 mmHg
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.
Critical value
df = 7, t = 3.499
Margin of error
E = 3.499·1.3/√8 = 1.608
Interval
−7.5 ± 1.608 → [−9.108, −5.892]
Answer
lower limit = −9.108 mmHg
16.Confidence intervals
Answer: σ = 0.7651 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 = 20·1² = 20
Limits for σ²
20/34.17 = 0.5853
20/9.591 = 2.085
Limit for σ
σ = √0.5853 = 0.7651 µm
Answer
σ = 0.7651 µm
95% interval for σ: [0.7651, 1.444] µm.
17.Hypothesis tests & confidence intervals
Answer: t = −3.207
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.
Pooled variance
sp² = (11·11² + 13·3²)/24 = 60.33
sp = 7.767
Standard error
SE = 7.767·√(1/12 + 1/14) = 3.056
Test statistic
t = (26 − 35.8)/3.056 = −3.207
Answer
t = −3.207
With df = 24, the two-sided 5% critical value is 2.064; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: L = 193.8
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 = 1.96·15/√25 = 5.88
Interval limit
L = 199.7 − 5.88 = 193.8
Answer
L = 193.8
19.Linear regression & correlation
Answer: r = 0.9867
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 = 25/5 = 5
ȳ = 306/5 = 61.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 50
Syy = Σ(y − ȳ)² = 256.8
Correlation coefficient
r = 50/√(10·256.8) = 0.9867
Answer
r = 0.9867
20.Linear regression & correlation
Answer: x = 3.4
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 = 25/5 = 5
ȳ = 281/5 = 56.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 45
Regression line
b = 45/10 = 4.5
a = 56.2 − 4.5·5 = 33.7
ŷ = 33.7 + 4.5·x
Solve for x
x = (49 − 33.7)/4.5 = 3.4
Answer
x = 3.4
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