You invest €8 000 at a nominal rate of 5% per year, compounded quarterly. Find the value FV of the investment after 6 years.
PV = €8 0005% / 4n = 6
FV = €
2.Time value of moneyHard
You deposit €2 500 at the start of every year for 21 years, the first deposit today. The account pays 4% per year. How much is in the account at the end of year 21? Find FV.
A = €2 500i = 4%n = 21
FV = €
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 4: €20 000, €19 000, €22 000, €36 000. The cost of capital is 12%. Find the net present value NPV.
A project costs €40 000 and returns €24 500, €23 000, €15 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −40 000CF1 = 24 500CF2 = 23 000CF3 = 15 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 11 years. The yield to maturity is 6% per year. Find the price P of the bond.
F = €1 000coupon = 5%YTM = 6%n = 11
P = €
6.Bond pricing & yieldHard
A bond with 17 years to maturity (face value €1 000, annual coupon 6%) sells for €1 243. The issuer can call it in 5 years at a call price of €1 060. Find the yield to call YTC in percent.
F = €1 000coupon = 6%P = €1 243call price = €1 060call in 5 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €8 000 000 and debt worth €3 000 000. Its WACC is 7.53%, its pre-tax cost of debt is 5% and the tax rate is 28%. Find the cost of equity re.
A company has equity with a market value of €8 000 000 and a cost of equity of 9%. Its debt consists of 5 000 bonds (face value €1 000, annual coupon 6%, 10 years to maturity) that trade at €1 038 each. The tax rate is 21%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0192. Find z to two decimals.
A = 0.0192
z =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 95% of them lie (the 95th percentile).
μ = 500σ = 40P = 95%
x = h
15.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 4.7. What is the margin of error E₂ of a 99% interval based on 200 observations?
95%n₁ = 50E₁ = 4.799%n₂ = 200
E₂ =
16.Confidence intervalsHard
The population standard deviation is 26. 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 2?
σ = 26E = 295%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 106 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 106s = 20n = 16
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 100 measurements has mean 125.4. The population standard deviation is 20. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 125.4σ = 20n = 10099%
U =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (25.5, 100) is made. Find its residual e = y − ŷ.
n = 5(25.5, 100)
e =
3Solutions
1.Time value of money
Answer: FV = 10779 €
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.
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.
Ordinary annuity factor
((1 + 0.04)21 − 1)/0.04 = 31.97
Annuity due
Each deposit is invested one year longer.
FV = 2 500·31.97·1.04 = €83 119.92
Answer
FV = €83 119.92
3.NPV & IRR
Answer: NPV = −3458 €
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 = −75 000
CF1/(1 + 0.12)1 = 20 000/1.12 = 17 857.14
CF2/(1 + 0.12)2 = 19 000/1.254 = 15 146.68
CF3/(1 + 0.12)3 = 22 000/1.405 = 15 659.17
CF4/(1 + 0.12)4 = 36 000/1.574 = 22 878.65
Add them up
NPV = €−3 458.36
Answer
NPV = €−3 458.36
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 28.62 %
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(25%) = 2 000.00
NPV(30%) = −716.89
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 = €25.00, i = 0.03, N = 22
Present value of the coupons
€25.00·(1 − 0.5219)/0.03 = €398.42
Present value of the face value
€1 000·0.5219 = €521.89
Price
P = €920.32
Answer
P = €920.32
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTC = 1.998 %
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
1243 = 60·(1 − (1 + y)−5)/y + 1060/(1 + y)5
Solve numerically
y = 1%: €1 299.76
y = 2%: €1 242.88
YTC = 1.998%
Answer
YTC = 1.998%
The yield to maturity is 4.002%. 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 = 9.004 %
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
0.2727·5%·(1 − 0.28) = 0.9818%
Cost of equity
re = (7.53% − 0.9818%)/0.7273 = 9.004%
Answer
re = 9.004%
8.CAPM & WACC
Answer: WACC = 7.167 %
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 = 1
Standard normal table
Φ(1) = 0.8413
P = 0.8413
Answer
P = 0.8413
11.Probability
Answer: σ = 1.492
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 = 1.45, Var(X) = 0.9475, σX = 0.9734
Linear change
σY = |0.5|·0.9734 = 0.4867
Answer
σY = 0.4867
The mean does move: E(Y) = 0.5·1.45 + 10 = 10.72.
13.Normal distribution
Answer: z = 2.07
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 − 0.0192 = 0.9808
Read the z-table backwards
Find 0.9808 inside the table and read off z.
z = 2.07
Answer
z = 2.07
14.Normal distribution
Answer: x = 565.8 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) = 0.95 → z = 1.645
Back to x
x = 500 + 1.645·40 = 565.8 h
Answer
x = 565.8 h
15.Confidence intervals
Answer: E₂ = 3.089
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.
Ratio of margins
E₂/E₁ = (2.576/1.96)·√(50/200) = 0.6571
New margin
E₂ = 4.7·0.6571 = 3.089
Answer
E₂ = 3.089
16.Confidence intervals
Answer: n = 650
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 = (1.96·26/2)² = 649.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 650
Answer
n = 650
17.Hypothesis tests & confidence intervals
Answer: t = 1.2
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 = 20/√16 = 20/4 = 5
Test statistic
t = (106 − 100)/5 = 1.2
Answer
t = 1.2
t0.025; df = 15 = 2.131
|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: U = 130.6
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 = 2.576·20/√100 = 5.152
Interval limit
U = 125.4 + 5.152 = 130.6
Answer
U = 130.6
19.Linear regression & correlation
Answer: r² = 0.9966
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 = 157/6 = 26.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 239
Syy = Σ(y − ȳ)² = 818.8
Coefficient of determination
r² = 239²/(70·818.8) = 0.9966
Answer
r² = 0.9966
About 99.66% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = 3.65
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 = 105/5 = 21
ȳ = 376/5 = 75.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 423
Regression line
b = 423/90 = 4.7
a = 75.2 − 4.7·21 = −23.5
ŷ = −23.5 + 4.7·x
Predicted value
ŷ = −23.5 + 4.7·25.5 = 96.35
Residual
e = 100 − 96.35 = 3.65
Answer
e = 3.65
The point lies above the line.
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