An annuity pays €500 at the end of each year for 16 years. The interest rate is 8% per year. Find the present value PV of the annuity.
A = €500i = 8%n = 16
PV = €
2.Time value of moneyHard
You borrow €8 000 at 7.2% per year, compounded monthly, and repay €375 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.
P = €8 000A = €3757.2% / 12
N = payments
3.NPV & IRRMedium
A project costs €50 000 today and returns €16 500, €12 000, €10 500, €14 000, €14 000 at the end of years 1 to 5. The cost of capital is 6%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €40 000 and returns €28 500, €21 500, €26 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −40 000CF1 = 28 500CF2 = 21 500CF3 = 26 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 4 years. The yield to maturity is 2.5% per year. Find the price P of the bond.
F = €1 000coupon = 6%YTM = 2.5%n = 4
P = €
6.Bond pricing & yieldHard
A bond trades at €985 with a yield to maturity of 7% (annual compounding) and a Macaulay duration of 10.8 years. Use the modified duration to estimate the new price P if the yield rises by 25 basis points.
P0 = €985y = 7%D = 10.8Δy = +25 bp
P = €
7.CAPM & WACCMedium
A company has equity worth €5 000 000 and debt worth €4 000 000. Its WACC is 5.78%, its pre-tax cost of debt is 4% and the tax rate is 25%. Find the cost of equity re.
A company has 1 000 000 shares at €39.50 each and 5 000 bonds trading at €1 000 each. The pre-tax cost of debt is 7% and the tax rate is 25.8%. The share’s beta is 0.9, the risk-free rate 3% and the market return 7.5%. Find the WACC.
β = 0.9rf = 3%rm = 7.5%rd = 7%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 24, 33, 52, 19, 37, 17, 11.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 25. What proportion P of values is above 470?
μ = 500σ = 25x = 470
P =
11.ProbabilityMedium
A fair die is rolled 3 times. What is the probability P of getting at least one six?
n = 3
P =
12.ProbabilityHard
X has the distribution shown, and Y = 0.5X + 1. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 0.5X + 1
σY =
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.3616. Find z to two decimals.
A = 0.3616
z =
14.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points40%
x = points
15.Confidence intervalsMedium
A sample of 8 items has mean 55.5 and standard deviation s = 12.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 8x̄ = 55.5s = 12.590%t(7) = 1.895
upper limit =
16.Confidence intervalsHard
The population standard deviation is 32. 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 4?
σ = 32E = 495%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 50; the population standard deviation is known to be 40. A random sample of 100 observations has a mean of 42.64. Test H₀: μ = 50 against H₁: μ < 50. Find the p-value p.
μ₀ = 50σ = 40n = 100x̄ = 42.64H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 15. Find the smallest sample size n.
σ = 15E = 390%
n =
19.Linear regression & correlationMedium
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 = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 69?
n = 5ŷ = 69
x =
3Solutions
1.Time value of money
Answer: PV = 4426 €
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.
Annuity factor
(1 − (1 + i)−n)/i = (1 − 0.2919)/0.08 = 8.851
Present value
PV = 500·8.851 = €4 425.68
Answer
PV = €4 425.68
2.Time value of money
Answer: N = 23 payments
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.
Rearrange the annuity formula
i = 0.006
(1 + i)−N = 1 − 8 000·0.006/375 = 0.872
Solve with logarithms
N = −ln(0.872)/ln(1.006) = 22.9
Round up to whole payments
22 full payments leave a small balance, which the last, smaller payment clears.
N = 23
Answer
N = 23 payments
3.NPV & IRR
Answer: DPB = 4.368 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: 16 500/1.06 = 15 566.04 → Σ = 15 566.04
year 2: 12 000/1.124 = 10 679.96 → Σ = 26 246.00
year 3: 10 500/1.191 = 8 816.00 → Σ = 35 062.00
year 4: 14 000/1.262 = 11 089.31 → Σ = 46 151.31
year 5: 14 000/1.338 = 10 461.61 → Σ = 56 612.92
Fraction of the last year
DPB = 4 + 3 848.69/10 461.61 = 4.368
Answer
DPB = 4.368 years
The discounted payback is always longer than the simple payback (3.786 years).
4.NPV & IRR
Answer: IRR = 41.62 %
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(40%) = 801.75
NPV(45%) = −1 590.47
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 = €30.00, i = 0.0125, N = 8
Present value of the coupons
€30.00·(1 − 0.9054)/0.0125 = €227.04
Present value of the face value
€1 000·0.9054 = €905.40
Price
P = €1 132.44
Answer
P = €1 132.44
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: P = 960.1 €
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 = 10.8/1.07 = 10.09
Estimated price change
%ΔP ≈ −10.09·(0.0025) = −2.523%
P ≈ 985·(1 − 0.02523) = €960.14
Answer
P = €960.14
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: re = 8.004 %
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 = €9 000 000, E/V = 0.5556, D/V = 0.4444
Debt part of the WACC
0.4444·4%·(1 − 0.25) = 1.333%
Cost of equity
re = (5.78% − 1.333%)/0.5556 = 8.004%
Answer
re = 8.004%
8.CAPM & WACC
Answer: WACC = 6.841 %
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 = −1.2
Standard normal table
Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849
Answer
P = 0.8849
11.Probability
Answer: P = 0.4213
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.
Complement
P(no six) = (5/6)3 = 0.5787
P = 1 − 0.5787 = 0.4213
Answer
P = 0.4213
12.Probability
Answer: σY = 0.4023
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 = 1.05, Var(X) = 0.6475, σX = 0.8047
Linear change
σY = |0.5|·0.8047 = 0.4023
Answer
σY = 0.4023
The mean does move: E(Y) = 0.5·1.05 + 1 = 1.525.
13.Normal distribution
Answer: z = 0.47
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 + 0.3616)/2 = 0.6808
Read the z-table backwards
Find 0.6808 inside the table and read off z.
z = 0.47
Answer
z = 0.47
14.Normal distribution
Answer: x = 103.8 points
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) = 0.6
z from the table
z ≈ 0.25
Back to the original units
x = 100 + 0.25·15 = 103.8 points
Answer
x = 103.8 points
15.Confidence intervals
Answer: upper limit = 63.87
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 = 1.895
Margin of error
E = 1.895·12.5/√8 = 8.375
Interval
55.5 ± 8.375 → [47.13, 63.87]
Answer
upper limit = 63.87
16.Confidence intervals
Answer: n = 246
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 = (1.96·32/4)² = 245.9
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 246
Answer
n = 246
17.Hypothesis tests & confidence intervals
Answer: p = 0.03288
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 = 40/√100 = 4
z = (42.64 − 50)/4 = −1.84
Tail area
Φ(1.84) = 0.9671
1 − Φ(1.84) = 0.0329
p-value
p = 0.03288
Answer
p = 0.03288
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: n = 68
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).
Solve for n
n = (1.645·15/3)² = 67.65
Round up
n = 68
Answer
n = 68
19.Linear regression & correlation
Answer: r = 0.793
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 = 1426 − 25·279/5 = 31
Sxx = 135 − 25²/5 = 10
Syy = 15721 − 279²/5 = 152.8
Correlation coefficient
r = 31/√(10·152.8) = 0.793
Answer
r = 0.793
20.Linear regression & correlation
Answer: x = 9.111
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.