You need €20 000 in 4 years. An account pays 2% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €20 0002% / 4n = 4
PV = €
2.Time value of moneyHard
A loan of €25 000 is repaid with equal monthly payments over 5 years. The interest rate is 3.6% per year, compounded monthly. Find the monthly payment A.
P = €25 0003.6% / 12N = 60
A = €
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 4: €37 000, €37 000, €26 000, €33 000. The cost of capital is 12%. Find the net present value NPV.
Project A costs €10 000 and returns €6 000, €4 000, €2 000 at the end of years 1 to 3. Project B costs €12 000 and returns €3 000, €5 000, €6 500. Find the crossover rate r* at which both projects have the same NPV, in percent.
A bond with a face value of €1 000 pays a 3% annual coupon in two half-yearly payments and matures in 6 years. The yield to maturity is 8% per year. Find the price P of the bond.
F = €1 000coupon = 3%YTM = 8%n = 6
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 4 years. It sells for €1 110. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 7% (2×)n = 4P = €1 110
YTM = %
7.CAPM & WACCMedium
A company has a debt-to-equity ratio (at market values) of 0.5. Its cost of equity is 9%, its pre-tax cost of debt is 6% and the tax rate is 25%. Find the WACC.
D/E = 0.5re = 9%rd = 6%T = 25%
WACC = %
8.CAPM & WACCHard
A company has 4 000 000 shares at €47.50 each and 20 000 bonds trading at €1 000 each. The pre-tax cost of debt is 3.5% and the tax rate is 20%. The share’s beta is 0.8, the risk-free rate 3.5% and the market return 10.5%. Find the WACC.
β = 0.8rf = 3.5%rm = 10.5%rd = 3.5%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 14, 39, 51, 13, 59, 21, 36.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 10. What proportion P of values is above 120?
μ = 100σ = 10x = 120
P =
11.ProbabilityMedium
A bag holds 7 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
7 / 9
P =
12.ProbabilityHard
A game costs €10 to play. You win €50 with probability 0.15 and €500 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€50: p = 0.15€500: p = 0.01
E(G) = €
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9803. Find z to two decimals.
A = 0.9803
z =
14.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
A sample of 20 items has mean 87 and standard deviation s = 2. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 20x̄ = 87s = 290%t(19) = 1.729
E =
16.Confidence intervalsHard
The population standard deviation is 28. 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?
σ = 28E = 199%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 50; the population standard deviation is known to be 6. A random sample of 36 observations has a mean of 51.74. Test H₀: μ = 50 against H₁: μ > 50. Find the p-value p.
μ₀ = 50σ = 6n = 36x̄ = 51.74H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A sample of 36 measurements has mean 118.8. The population standard deviation is 8. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 118.8σ = 8n = 3690%
L =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 5
r =
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 (21.5, 94) is made. Find its residual e = y − ŷ.
n = 5(21.5, 94)
e =
3Solutions
1.Time value of money
Answer: PV = 18466 €
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 = 0.02/4 = 0.005
N = 4·4 = 16
Present value
PV = FV/(1 + i)N = 20 000/1.083 = €18 466.01
Answer
PV = €18 466.01
2.Time value of money
Answer: A = 455.9 €
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 = 0.036000000000000004/12 = 0.003
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 = 0.8355
A = 25 000·0.003/(1 − 0.8355) = €455.91
Answer
A = €455.91
3.NPV & IRR
Answer: NPV = 27010 €
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 = 37 000/1.12 = 33 035.71
CF2/(1 + 0.12)2 = 37 000/1.254 = 29 496.17
CF3/(1 + 0.12)3 = 26 000/1.405 = 18 506.29
CF4/(1 + 0.12)4 = 33 000/1.574 = 20 972.10
Add them up
NPV = €27 010.27
Answer
NPV = €27 010.27
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: r* = 4.386 %
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 = −12 000 − (−10 000) = −2 000
ΔCF1 = 3 000 − 6 000 = −3 000
ΔCF2 = 5 000 − 4 000 = 1 000
ΔCF3 = 6 500 − 2 000 = 4 500
IRR of the difference
NPVΔ(4%) = 40.42
NPVΔ(5%) = −62.84
r* = 4.386%
Answer
r* = 4.386%
Check: at 4.386% both NPVs are €1 177.08.
5.Bond pricing & yield
Answer: P = 765.4 €
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 = €15.00, i = 0.04, N = 12
Present value of the coupons
€15.00·(1 − 0.6246)/0.04 = €140.78
Present value of the face value
€1 000·0.6246 = €624.60
Price
P = €765.37
Answer
P = €765.37
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 3.997 %
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 = €35.00, N = 8
Solve numerically
i = 1.998% per half-year
YTM
YTM = 2·1.998% = 3.997%
Answer
YTM = 3.997%
7.CAPM & WACC
Answer: WACC = 7.5 %
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 from D/E
E/V = 1/(1 + 0.5) = 0.6667
D/V = 0.5/(1 + 0.5) = 0.3333
WACC
WACC = 0.6667·9% + 0.3333·6%·(1 − 0.25) = 7.5%
Answer
WACC = 7.5%
8.CAPM & WACC
Answer: WACC = 8.5 %
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 1 − 0.9772 = 0.02275
Answer
P = 0.02275
11.Probability
Answer: P = 0.175
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.
First and second draw
P = 7/16 · 6/15 = 42/240 = 0.175
Answer
P = 0.175
12.Probability
Answer: E(G) = 2.5 €
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.
Net outcomes
G: 40 (p = 0.15), 490 (p = 0.01), −10 (p = 0.84)
Expected value
E(G) = 40·0.15 + 490·0.01 − 10·0.84 = 2.5 €
Answer
E(G) = 2.5 €
A positive value means you win money on average.
13.Normal distribution
Answer: z = −2.06
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 − 0.9803 = 0.0197
Read the z-table backwards
Find 0.0197 inside the table and read off z.
z = −2.06
Answer
z = −2.06
14.Normal distribution
Answer: P = 0.7881
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 difference
μ = 52 − 48 = 4 s, σ = 5 s
Standardise
z = (0 − 4)/5 = −0.8
Probability
P = 1 − Φ(−0.8) = 0.7881
Answer
P = 0.7881
15.Confidence intervals
Answer: E = 0.7732
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 = 19, t = 1.729
Margin of error
E = 1.729·2/√20 = 0.7732
Answer
E = 0.7732
16.Confidence intervals
Answer: n = 5203
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 = (2.576·28/1)² = 5202
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 5203
Answer
n = 5203
17.Hypothesis tests & confidence intervals
Answer: p = 0.04093
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.
Test statistic
SE = σ/√n = 6/√36 = 1
z = (51.74 − 50)/1 = 1.74
Tail area
Φ(1.74) = 0.9591
1 − Φ(1.74) = 0.0409
p-value
p = 0.04093
Answer
p = 0.04093
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: L = 116.6
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 = 1.645·8/√36 = 2.193
Interval limit
L = 118.8 − 2.193 = 116.6
Answer
L = 116.6
19.Linear regression & correlation
Answer: r = 0.9957
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 = 115/5 = 23
ȳ = 503/5 = 100.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 366
Syy = Σ(y − ȳ)² = 1501
Correlation coefficient
r = 366/√(90·1501) = 0.9957
Answer
r = 0.9957
20.Linear regression & correlation
Answer: e = 4.7
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 = 100/5 = 20
ȳ = 413/5 = 82.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 402
Regression line
b = 402/90 = 4.467
a = 82.6 − 4.467·20 = −6.733
ŷ = −6.733 + 4.467·x
Predicted value
ŷ = −6.733 + 4.467·21.5 = 89.3
Residual
e = 94 − 89.3 = 4.7
Answer
e = 4.7
The point lies above the line.
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