You need €5 000 in 10 years. An account pays 4% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €5 0004% / 4n = 10
PV = €
2.Time value of moneyHard
A loan of €12 000 is repaid with equal monthly payments over 5 years. The interest rate is 4.8% per year, compounded monthly. Find the monthly payment A.
P = €12 0004.8% / 12N = 60
A = €
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 4: €17 000, €13 000, €13 000, €29 000. The cost of capital is 5%. Find the net present value NPV.
A project costs €20 000 and returns €10 500, €10 500, €8 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 10 500CF2 = 10 500CF3 = 8 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 4% annual coupon in two half-yearly payments and matures in 10 years. The yield to maturity is 5% per year. Find the price P of the bond.
F = €1 000coupon = 4%YTM = 5%n = 10
P = €
6.Bond pricing & yieldHard
A bond with 17 years to maturity (face value €1 000, annual coupon 7%) sells for €1 163. The issuer can call it in 5 years at a call price of €1 070. Find the yield to call YTC in percent.
F = €1 000coupon = 7%P = €1 163call price = €1 070call in 5 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €8 000 000 and debt worth €3 000 000. Its WACC is 10.97%, its pre-tax cost of debt is 4% and the tax rate is 28%. Find the cost of equity re.
A company's shares have a beta of 1.2 at its current debt-to-equity ratio of 0.4. It plans to raise its debt-to-equity ratio to 0.6. The tax rate is 25.8%, the risk-free rate 2.5% and the market risk premium 4.5%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 43, 56, 22, 33, 22, 41, 32.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 50. Find the z-score z of a score of 450.
μ = 500σ = 50x = 450
z =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.15, 0.3, 0.35, 0.2
σ =
12.ProbabilityHard
X has the distribution shown, and Y = 3X − 3. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 3X − 3
σY =
13.Normal distributionMedium
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 101.8 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 101.8 s
P =
14.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 5% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points5%
x = points
15.Confidence intervalsMedium
The blood pressure of 16 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 8 and standard deviation 1. Find the lower limit of the 99% confidence interval for the mean difference.
n = 16d̄ = 8sd = 199%t(15) = 2.947
lower limit = mmHg
16.Confidence intervalsHard
The population standard deviation is 36. 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?
σ = 36E = 290%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 202.9 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 202.9s = 12n = 49
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 15. Find the smallest sample size n.
σ = 15E = 195%
n =
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 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.
n = 6x = 2.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 3358 €
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 = 0.04/4 = 0.01
N = 4·10 = 40
Present value
PV = FV/(1 + i)N = 5 000/1.489 = €3 358.27
Answer
PV = €3 358.27
2.Time value of money
Answer: A = 225.4 €
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 = 0.048/12 = 0.004
N = 12·5 = 60
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 = 0.787
A = 12 000·0.004/(1 − 0.787) = €225.36
Answer
A = €225.36
3.NPV & IRR
Answer: NPV = 3070 €
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 + 0.05)1 = 17 000/1.05 = 16 190.48
CF2/(1 + 0.05)2 = 13 000/1.103 = 11 791.38
CF3/(1 + 0.05)3 = 13 000/1.158 = 11 229.89
CF4/(1 + 0.05)4 = 29 000/1.216 = 23 858.37
Add them up
NPV = €3 070.12
Answer
NPV = €3 070.12
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 22.23 %
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(20%) = 671.30
NPV(25%) = −784.00
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 20 + 5·671.30/(671.30 − (−784.00)) = 22.31%
exact: NPV(22.23%) = 0 → IRR = 22.23%
Answer
IRR = 22.23%
5.Bond pricing & yield
Answer: P = 922.1 €
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 = €20.00, i = 0.025, N = 20
Present value of the coupons
€20.00·(1 − 0.6103)/0.025 = €311.78
Present value of the face value
€1 000·0.6103 = €610.27
Price
P = €922.05
Answer
P = €922.05
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTC = 4.559 %
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
1163 = 70·(1 − (1 + y)−5)/y + 1070/(1 + y)5
Solve numerically
y = 4%: €1 191.09
y = 5%: €1 141.44
YTC = 4.559%
Answer
YTC = 4.559%
The yield to maturity is 5.5%. 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: re = 14 %
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 = €11 000 000, E/V = 0.7273, D/V = 0.2727
Debt part of the WACC
0.2727·4%·(1 − 0.28) = 0.7855%
Cost of equity
re = (10.97% − 0.7855%)/0.7273 = 14%
Answer
re = 14%
8.CAPM & WACC
Answer: re = 8.518 %
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.
Unlever the beta
βU = 1.2/(1 + 0.742·0.4) = 0.9254
Relever at the new D/E
βL = 0.9254·(1 + 0.742·0.6) = 1.337
Cost of equity (CAPM)
re = 2.5% + 1.337·4.5% = 8.518%
Answer
re = 8.518%
9.Descriptive statistics
Answer: s = 12.18
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 = (450 − 500)/50 = −1
Answer
z = −1
11.Probability
Answer: σ = 9.695
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.
Spread of X
μ = 1.05, Var(X) = 0.6475, σX = 0.8047
Linear change
σY = |3|·0.8047 = 2.414
Answer
σY = 2.414
The mean does move: E(Y) = 3·1.05 − 3 = 0.15.
13.Normal distribution
Answer: P = 0.6915
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 = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (101.8 − 105)/6.403 = −0.5
Upper tail
P = 1 − Φ(−0.5) = 1 − 0.3085 = 0.6915
Answer
P = 0.6915
14.Normal distribution
Answer: x = 75.4 points
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left
Φ(z) = 0.05
z from the table
z ≈ −1.64
Back to the original units
x = 100 − 1.64·15 = 75.4 points
Answer
x = 75.4 points
15.Confidence intervals
Answer: lower limit = 7.263 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 = 15, t = 2.947
Margin of error
E = 2.947·1/√16 = 0.7368
Interval
8 ± 0.7368 → [7.263, 8.737]
Answer
lower limit = 7.263 mmHg
16.Confidence intervals
Answer: n = 877
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 = (1.645·36/2)² = 876.8
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 877
Answer
n = 877
17.Hypothesis tests & confidence intervals
Answer: t = 1.692
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/√49 = 12/7 = 1.714
Test statistic
t = (202.9 − 200)/1.714 = 1.692
Answer
t = 1.692
t0.025; df = 48 = 2.011
|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 865
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 = (1.96·15/1)² = 864.4
Round up
n = 865
Answer
n = 865
19.Linear regression & correlation
Answer: r = 0.9617
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 = 47/5 = 9.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 46
Syy = Σ(y − ȳ)² = 57.2
Correlation coefficient
r = 46/√(40·57.2) = 0.9617
Answer
r = 0.9617
20.Linear regression & correlation
Answer: ŷ = 51.97
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 = 27/6 = 4.5
ȳ = 366/6 = 61
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 79
Regression line
b = 79/17.5 = 4.514
a = 61 − 4.514·4.5 = 40.69
ŷ = 40.69 + 4.514·x
Prediction
ŷ = 40.69 + 4.514·2.5 = 51.97
Answer
ŷ = 51.97
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