You need €40 000 in 10 years. An account pays 5% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €40 0005% / 12n = 10
PV = €
2.Time value of moneyHard
You save €500 at the end of every month for 17 years. The account pays 7.2% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €5007.2% / 12N = 204
FV = €
3.NPV & IRRMedium
A project costs €200 000 today. It generates €11 000 at the end of the first year, and the cash flow then grows by 2% per year forever. The cost of capital is 8%. Find the net present value NPV.
C0 = €200 000CF1 = €11 000g = 2%r = 8%
NPV = €
4.NPV & IRRHard
A project costs €50 000 and returns €37 000, €33 000, €29 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −50 000CF1 = 37 000CF2 = 33 000CF3 = 29 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 12 years and pays annual coupons. It sells for €1 051 when the yield to maturity is 2.5%. Find the coupon rate c in percent.
F = €1 000n = 12P = €1 051YTM = 2.5%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 6% matures in 7 years and sells for €1 120. Find its yield to maturity YTM in percent.
F = €1 000coupon = 6%n = 7P = €1 120
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €10 000 000, preferred shares worth €500 000 and debt worth €5 000 000. The cost of equity is 10%, the cost of preferred stock 7.5% and the pre-tax cost of debt 3%. The tax rate is 20%. Find the WACC.
A company has 5 000 000 shares at €42.50 each and 40 000 bonds trading at €1 000 each. The pre-tax cost of debt is 3.5% and the tax rate is 28%. The share’s beta is 1.3, the risk-free rate 3.5% and the market return 9.5%. Find the WACC.
β = 1.3rf = 3.5%rm = 9.5%rd = 3.5%T = 28%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 20, 58, 44, 52, 54, 22, 52.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 20. Find the z-score z of a score of 120.
μ = 100σ = 20x = 120
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3, 4p(x): 0.15, 0.2, 0.25, 0.3, 0.1
Var(X) =
12.ProbabilityHard
9 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?
n = 9p = 0.3
P =
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.6102. Find z to two decimals.
A = 0.6102
z =
14.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 25% of them lie (the 25th percentile).
μ = 32σ = 6P = 25%
x = min
15.Confidence intervalsMedium
A sample of 8 items has mean 64 and standard deviation s = 3. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
n = 8x̄ = 64s = 399%t(7) = 3.499
lower limit =
16.Confidence intervalsHard
The population standard deviation is 39. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 1.5?
σ = 39E = 1.599%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 193.4 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 193.4s = 15n = 49
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 1.5 units with a power of 90%. The standard deviation is 6. Find the minimum sample size n.
The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.
n = 5x = 3
ŷ =
3Solutions
1.Time value of money
Answer: PV = 24286 €
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 = 0.05/12 = 0.004167
N = 12·10 = 120
Present value
PV = FV/(1 + i)N = 40 000/1.647 = €24 286.44
Answer
PV = €24 286.44
2.Time value of money
Answer: FV = 199029 €
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.
You paid in €102 000 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = −16667 €
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.
Value of the cash flows
PV = 11 000/(0.08 − 0.02) = 183 333.33
NPV
NPV = 183 333.33 − 200 000 = −€16 666.67
Answer
NPV = −€16 666.67
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 46.52 %
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%) = 889.34
NPV(50%) = −1 925.93
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.
Present value of the face value
1000·(1 + 0.025)−12 = €743.56
Coupon
C = (1051 − 743.56)/10.26 = €29.97
Coupon rate
c = 29.97/1000 = 2.997%
Answer
c = 2.997%
6.Bond pricing & yield
Answer: YTM = 4.001 %
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
€1 120 = €60·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷
Solve numerically
P(4%) = €1 120.04, P(5%) = €1 057.86
YTM = 4.001%
Answer
YTM = 4.001%
7.CAPM & WACC
Answer: WACC = 7.468 %
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).
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 = (120 − 100)/20 = 1
Answer
z = 1
11.Probability
Answer: Var(X) = 1.5
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.
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.
Zero and one failure
P(X = 0) = 0.79 = 0.04035
P(X = 1) = 9·0.3·0.78 = 0.1556
Add them
P = 0.04035 + 0.1556 = 0.196
Answer
P = 0.196
13.Normal distribution
Answer: z = 0.86
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 of z
Φ(z) = (1 + 0.6102)/2 = 0.8051
Read the z-table backwards
Find 0.8051 inside the table and read off z.
z = 0.86
Answer
z = 0.86
14.Normal distribution
Answer: x = 27.95 min
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) = 0.25 → z = −0.6745
Back to x
x = 32 + −0.6745·6 = 27.95 min
Answer
x = 27.95 min
15.Confidence intervals
Answer: lower limit = 60.29
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 = 7, t = 3.499
Margin of error
E = 3.499·3/√8 = 3.711
Interval
64 ± 3.711 → [60.29, 67.71]
Answer
lower limit = 60.29
16.Confidence intervals
Answer: n = 4486
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 = (2.576·39/1.5)² = 4486
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 4486
Answer
n = 4486
17.Hypothesis tests & confidence intervals
Answer: t = −3.08
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 = 15/√49 = 15/7 = 2.143
Test statistic
t = (193.4 − 200)/2.143 = −3.08
Answer
t = −3.08
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 209
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((2.326 + 1.282)·6/1.5)² = 208.3
Round up
n = 209
Answer
n = 209
19.Linear regression & correlation
Answer: r = 0.9973
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.
Centred sums
Sxy = 1919 − 25·373/5 = 54
Sxx = 135 − 25²/5 = 10
Syy = 28119 − 373²/5 = 293.2
Correlation coefficient
r = 54/√(10·293.2) = 0.9973
Answer
r = 0.9973
20.Linear regression & correlation
Answer: ŷ = 15.35
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 = 139/5 = 27.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 166
Regression line
b = 166/40 = 4.15
a = 27.8 − 4.15·6 = 2.9
ŷ = 2.9 + 4.15·x
Prediction
ŷ = 2.9 + 4.15·3 = 15.35
Answer
ŷ = 15.35
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