You need €5 000 in 8 years. An account pays 6% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €5 0006% / 12n = 8
PV = €
2.Time value of moneyHard
A loan of €20 000 is repaid with equal monthly payments over 6 years. The interest rate is 5.4% per year, compounded monthly. Find the monthly payment A.
P = €20 0005.4% / 12N = 72
A = €
3.NPV & IRRMedium
A project costs €20 000 today and returns €6 000 at the end of each year for 5 years. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €150 000 today. Its cash flow is €16 500 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 9. The cost of capital is 8%. Find the net present value NPV.
C0 = €150 000CF1 = €16 500g = 3%n = 9r = 8%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 8 years. The yield to maturity is 7% per year. Find the price P of the bond.
F = €1 000coupon = 8%YTM = 7%n = 8
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 7% matures in 4 years and sells for €1 070. Find its yield to maturity YTM in percent.
F = €1 000coupon = 7%n = 4P = €1 070
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €10 000 000, preferred shares worth €2 000 000 and debt worth €4 000 000. The cost of equity is 9%, the cost of preferred stock 6.5% and the pre-tax cost of debt 6%. The tax rate is 25%. Find the WACC.
A company's business has an unlevered (asset) beta of 1.1. It plans a debt-to-equity ratio of 0.6. The tax rate is 25.8%, the risk-free rate is 1% and the market risk premium is 4%. Find the cost of equity re at the new capital structure.
βU = 1.1D/E = 0.6T = 25.8%rf = 1%rm − rf = 4%
re = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 51, 50, 51, 13, 25.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is below 70?
μ = 50σ = 10x = 70
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 10, 20, 30, 40p(x): 0.05, 0.35, 0.35, 0.25
Var(X) =
12.ProbabilityHard
X has the distribution shown, and Y = 0.5X − 3. Find the standard deviation σY of Y.
The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 8 of them is below 325.1 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 8< 325.1 g
P =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 99% of them lie (the 99th percentile).
μ = 500σ = 40P = 99%
x = h
15.Confidence intervalsMedium
A sample of 12 items has mean 29 and standard deviation s = 13.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 12x̄ = 29s = 13.590%t(11) = 1.796
lower limit =
16.Confidence intervalsHard
The population standard deviation is 10. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1?
σ = 10E = 195%
n =
17.Hypothesis tests & confidence intervalsMedium
10 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 0.76 and the standard deviation of the differences is 1. Find the paired test statistic t for H₀: μd = 0.
n = 10d̄ = 0.76sd = 1
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 16 measurements has mean 494.2. The population standard deviation is 8. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 494.2σ = 8n = 1690%
L =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 5
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. For which value x does the line predict ŷ = 83?
n = 6ŷ = 83
x =
3Solutions
1.Time value of money
Answer: PV = 3098 €
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 = 0.06/12 = 0.005
N = 12·8 = 96
Present value
PV = FV/(1 + i)N = 5 000/1.614 = €3 097.62
Answer
PV = €3 097.62
2.Time value of money
Answer: A = 325.8 €
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 = 0.054000000000000006/12 = 0.0045
N = 12·6 = 72
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 = 0.7238
A = 20 000·0.0045/(1 − 0.7238) = €325.82
Answer
A = €325.82
3.NPV & IRR
Answer: DPB = 4.522 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: 6 000/1.12 = 5 357.14 → Σ = 5 357.14
year 2: 6 000/1.254 = 4 783.16 → Σ = 10 140.31
year 3: 6 000/1.405 = 4 270.68 → Σ = 14 410.99
year 4: 6 000/1.574 = 3 813.11 → Σ = 18 224.10
year 5: 6 000/1.762 = 3 404.56 → Σ = 21 628.66
Fraction of the last year
DPB = 4 + 1 775.90/3 404.56 = 4.522
Answer
DPB = 4.522 years
The discounted payback is always longer than the simple payback (3.333 years).
4.NPV & IRR
Answer: NPV = −35395 €
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.
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: P = 1060 €
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 = 0.035, N = 16
Present value of the coupons
€40.00·(1 − 0.5767)/0.035 = €483.76
Present value of the face value
€1 000·0.5767 = €576.71
Price
P = €1 060.47
Answer
P = €1 060.47
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 5.025 %
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
€1 070 = €70·(1 − (1 + y)⁻⁴)/y + €1 000/(1 + y)⁴
Solve numerically
P(5%) = €1 070.92, P(6%) = €1 034.65
YTM = 5.025%
Answer
YTM = 5.025%
7.CAPM & WACC
Answer: WACC = 7.563 %
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).
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.
Levered beta
βL = 1.1·(1 + 0.742·0.6) = 1.59
Cost of equity (CAPM)
re = 1% + 1.59·4% = 7.359%
Answer
re = 7.359%
9.Descriptive statistics
Answer: s = 17.86
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 0.9772
Answer
P = 0.9772
11.Probability
Answer: Var(X) = 76
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 = 3.1, Var(X) = 1.99, σX = 1.411
Linear change
σY = |0.5|·1.411 = 0.7053
Answer
σY = 0.7053
The mean does move: E(Y) = 0.5·3.1 − 3 = −1.45.
13.Normal distribution
Answer: P = 0.9332
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 = 320 g, σT = 1.2·√8 = 3.394 g
Standardise
z = (325.1 − 320)/3.394 = 1.5
Probability
P = Φ(1.5) = 0.9332
Answer
P = 0.9332
14.Normal distribution
Answer: x = 593.1 h
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 for the percentile
Φ(z) = 0.99 → z = 2.326
Back to x
x = 500 + 2.326·40 = 593.05 h
Answer
x = 593.05 h
15.Confidence intervals
Answer: lower limit = 22
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 = 11, t = 1.796
Margin of error
E = 1.796·13.5/√12 = 6.999
Interval
29 ± 6.999 → [22, 36]
Answer
lower limit = 22
16.Confidence intervals
Answer: n = 385
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 = (1.96·10/1)² = 384.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 385
Answer
n = 385
17.Hypothesis tests & confidence intervals
Answer: t = 2.403
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 of the mean difference
SE = 1/√10 = 0.3162
Test statistic
t = 0.76/0.3162 = 2.403
Answer
t = 2.403
With df = 9, the two-sided 5% critical value is 2.262; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: L = 490.9
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 = 1.645·8/√16 = 3.29
Interval limit
L = 494.2 − 3.29 = 490.9
Answer
L = 490.9
19.Linear regression & correlation
Answer: r = 0.969
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 = 35/5 = 7
ȳ = 391/5 = 78.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 162
Syy = Σ(y − ȳ)² = 698.8
Correlation coefficient
r = 162/√(40·698.8) = 0.969
Answer
r = 0.969
20.Linear regression & correlation
Answer: x = 11.04
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 = 391/6 = 65.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 309
Regression line
b = 309/70 = 4.414
a = 65.17 − 4.414·7 = 34.27
ŷ = 34.27 + 4.414·x
Solve for x
x = (83 − 34.27)/4.414 = 11.04
Answer
x = 11.04
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