You invest €1 000 at a nominal rate of 5% per year, compounded quarterly. Find the value FV of the investment after 4 years.
PV = €1 0005% / 4n = 4
FV = €
2.Time value of moneyHard
You want €30 000 in 3 years. You save the same amount at the end of every month in an account paying 4.8% per year, compounded monthly. Find the monthly deposit A.
FV = €30 0004.8% / 12N = 36
A = €
3.NPV & IRRMedium
A project costs €20 000 today and returns €6 500 at the end of each year for 6 years. The cost of capital is 6%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
Project A costs €10 000 and returns €5 500, €4 000, €2 500 at the end of years 1 to 3. Project B costs €15 000 and returns €4 000, €6 500, €8 000. Find the crossover rate r* at which both projects have the same NPV, in percent.
A bond with a face value of €1 000 matures in 7 years and pays annual coupons. It sells for €948 when the yield to maturity is 8%. Find the coupon rate c in percent.
F = €1 000n = 7P = €948YTM = 8%
c = %
6.Bond pricing & yieldHard
A bond trades at €1 080 with a yield to maturity of 6% (annual compounding) and a Macaulay duration of 5.1 years. Use the modified duration to estimate the new price P if the yield rises by 150 basis points.
P0 = €1 080y = 6%D = 5.1Δy = +150 bp
P = €
7.CAPM & WACCMedium
A company has equity worth €4 000 000 and debt worth €2 000 000. Its WACC is 9.39%, its pre-tax cost of debt is 3% and the tax rate is 28%. Find the cost of equity re.
A company's shares have a beta of 0.9 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 1.5% and the market risk premium 5%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 24, 57, 29, 28, 10.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is below 152?
μ = 170σ = 15x = 152
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3, 4p(x): 0.25, 0.2, 0.3, 0.1, 0.15
σ =
12.ProbabilityHard
10 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?
n = 10p = 0.1
P =
13.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 460 and 574 h? Give P as a decimal.
μ = 500σ = 40x1 = 460x2 = 574
P =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 85% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g85%
x = g
15.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 1.4. What is the margin of error E₂ of a 99% interval from the same data?
95%n₁ = 50E₁ = 1.499%
E₂ =
16.Confidence intervalsHard
The population standard deviation is 30. 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 2?
σ = 30E = 299%
n =
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 50 parts has mean length 55.5 mm, and the line's known standard deviation is 9 mm. Line 2: a sample of 40 parts has mean 57.8 mm, known standard deviation 14 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A sample of 64 measurements has mean 202.2. The population standard deviation is 8. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 202.2σ = 8n = 6499%
U =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 135?
n = 6ŷ = 135
x =
3Solutions
1.Time value of money
Answer: FV = 1220 €
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.
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: 6 500/1.06 = 6 132.08 → Σ = 6 132.08
year 2: 6 500/1.124 = 5 784.98 → Σ = 11 917.05
year 3: 6 500/1.191 = 5 457.53 → Σ = 17 374.58
year 4: 6 500/1.262 = 5 148.61 → Σ = 22 523.19
Fraction of the last year
DPB = 3 + 2 625.42/5 148.61 = 3.51
Answer
DPB = 3.51 years
The discounted payback is always longer than the simple payback (3.077 years).
4.NPV & IRR
Answer: r* = 8.826 %
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 = −15 000 − (−10 000) = −5 000
ΔCF1 = 4 000 − 5 500 = −1 500
ΔCF2 = 6 500 − 4 000 = 2 500
ΔCF3 = 8 000 − 2 500 = 5 500
IRR of the difference
NPVΔ(8%) = 120.54
NPVΔ(9%) = −24.94
r* = 8.826%
Answer
r* = 8.826%
Check: at 8.826% both NPVs are €371.17.
5.Bond pricing & yield
Answer: c = 7.001 %
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.
Present value of the face value
1000·(1 + 0.08)−7 = €583.49
Coupon
C = (948 − 583.49)/5.206 = €70.01
Coupon rate
c = 70.01/1000 = 7.001%
Answer
c = 7.001%
6.Bond pricing & yield
Answer: P = 1002 €
Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.
Modified duration
Dmod = 5.1/1.06 = 4.811
Estimated price change
%ΔP ≈ −4.811·(0.015) = −7.217%
P ≈ 1080·(1 − 0.07217) = €1 002.06
Answer
P = €1 002.06
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: re = 13.01 %
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 = €6 000 000, E/V = 0.6667, D/V = 0.3333
Debt part of the WACC
0.3333·3%·(1 − 0.28) = 0.72%
Cost of equity
re = (9.39% − 0.72%)/0.6667 = 13.01%
Answer
re = 13.01%
8.CAPM & WACC
Answer: re = 6.515 %
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 = 0.9/(1 + 0.742·0.4) = 0.694
Relever at the new D/E
βL = 0.694·(1 + 0.742·0.6) = 1.003
Cost of equity (CAPM)
re = 1.5% + 1.003·5% = 6.515%
Answer
re = 6.515%
9.Descriptive statistics
Answer: s = 17.1
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: σ = 1.345
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) = 0.910 = 0.3487
P(X = 1) = 10·0.1·0.99 = 0.3874
Add them
P = 0.3487 + 0.3874 = 0.7361
Answer
P = 0.7361
13.Normal distribution
Answer: P = 0.8092
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-scores
z₁ = (460 − 500)/40 = −1
z₂ = (574 − 500)/40 = 1.85
Area between
P = Φ(1.85) − Φ(−1) = 0.9678 − 0.1587 = 0.8092
Answer
P = 0.8092
14.Normal distribution
Answer: x = 255.2 g
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) = 0.85
z from the table
z ≈ 1.04
Back to the original units
x = 250 + 1.04·5 = 255.2 g
Answer
x = 255.2 g
15.Confidence intervals
Answer: E₂ = 1.84
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)·√(50/50) = 1.314
New margin
E₂ = 1.4·1.314 = 1.84
Answer
E₂ = 1.84
16.Confidence intervals
Answer: n = 1494
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 = (2.576·30/2)² = 1493
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1494
Answer
n = 1494
17.Hypothesis tests & confidence intervals
Answer: z = −0.9007
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.
Standard error of the difference
SE = √(9²/50 + 14²/40) = √6.52 = 2.553
Test statistic
z = (55.5 − 57.8)/2.553 = −0.9007
Answer
z = −0.9007
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: U = 204.8
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 = 2.576·8/√64 = 2.576
Interval limit
U = 202.2 + 2.576 = 204.8
Answer
U = 204.8
19.Linear regression & correlation
Answer: r = 0.9978
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 = 389/6 = 64.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 407
Syy = Σ(y − ȳ)² = 2377
Correlation coefficient
r = 407/√(70·2377) = 0.9978
Answer
r = 0.9978
20.Linear regression & correlation
Answer: x = 23.16
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 = 138/6 = 23
ȳ = 805/6 = 134.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 373
Regression line
b = 373/70 = 5.329
a = 134.2 − 5.329·23 = 11.61
ŷ = 11.61 + 5.329·x
Solve for x
x = (135 − 11.61)/5.329 = 23.16
Answer
x = 23.16
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