You need €25 000 in 7 years. An account pays 2% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €25 0002% / 4n = 7
PV = €
2.Time value of moneyHard
A loan of €150 000 is repaid with equal monthly payments over 25 years. The interest rate is 5.4% per year, compounded monthly. Find the monthly payment A.
P = €150 0005.4% / 12N = 300
A = €
3.NPV & IRRMedium
A project costs €50 000 today and returns €17 500, €21 000, €15 500, €15 000, €18 500, €21 500 at the end of years 1 to 6. The cost of capital is 10%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €25 000 and returns €11 000, €10 000, €18 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −25 000CF1 = 11 000CF2 = 10 000CF3 = 18 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 2% and matures in 6 years. Its yield to maturity changes from 2.5% to 1%. By how much ΔP does the bond price change?
F = €1 000coupon = 2%n = 6y1 = 2.5%y2 = 1%
ΔP = €
6.Bond pricing & yieldHard
A bond trades at €935 with a yield to maturity of 3% (annual compounding) and a Macaulay duration of 10.9 years. Use the modified duration to estimate the new price P if the yield falls by 50 basis points.
P0 = €935y = 3%D = 10.9Δy = −50 bp
P = €
7.CAPM & WACCMedium
A company has equity worth €10 000 000 and debt worth €2 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 3% and the tax rate is 25%. Find the WACC.
A company has 4 000 000 shares at €58.50 each and 10 000 bonds trading at €1 020 each. The pre-tax cost of debt is 5% and the tax rate is 30%. The share’s beta is 1.8, the risk-free rate 1% and the market return 6.5%. Find the WACC.
β = 1.8rf = 1%rm = 6.5%rd = 5%T = 30%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 40, 52, 44, 22, 21, 22.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is below 650?
μ = 500σ = 75x = 650
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.3, 0.2, 0.15, 0.15, 0.2
σ =
12.ProbabilityHard
A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 1 correct answers?
n = 11p = 0.4k = 1
P =
13.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 167.5 and 185.35 cm? Give P as a decimal.
μ = 178σ = 7x1 = 167.5x2 = 185.35
P =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 85% is to the right of it. Find the cut-off x.
μ = 500 hσ = 40 h85%
x = h
15.Confidence intervalsMedium
A sample of 6 items has mean 16.5 and standard deviation s = 9. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 6x̄ = 16.5s = 999%t(5) = 4.032
upper limit =
16.Confidence intervalsHard
The population standard deviation is 16. 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 = 16E = 199%
n =
17.Hypothesis tests & confidence intervalsMedium
9 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 4.68 and the standard deviation of the differences is 11.7. Find the paired test statistic t for H₀: μd = 0.
n = 9d̄ = 4.68sd = 11.7
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 2.5 units with a power of 90%. The standard deviation is 17. Find the minimum sample size n.
The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.
n = 6
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. For which value x does the line predict ŷ = 62?
n = 5ŷ = 62
x =
3Solutions
1.Time value of money
Answer: PV = 21742 €
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 = 4·7 = 28
Present value
PV = FV/(1 + i)N = 25 000/1.15 = €21 741.54
Answer
PV = €21 741.54
2.Time value of money
Answer: A = 912.2 €
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 = 0.054000000000000006/12 = 0.0045
N = 12·25 = 300
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 = 0.26
A = 150 000·0.0045/(1 − 0.26) = €912.19
Answer
A = €912.19
3.NPV & IRR
Answer: DPB = 3.497 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: 17 500/1.1 = 15 909.09 → Σ = 15 909.09
year 2: 21 000/1.21 = 17 355.37 → Σ = 33 264.46
year 3: 15 500/1.331 = 11 645.38 → Σ = 44 909.84
year 4: 15 000/1.464 = 10 245.20 → Σ = 55 155.04
Fraction of the last year
DPB = 3 + 5 090.16/10 245.20 = 3.497
Answer
DPB = 3.497 years
The discounted payback is always longer than the simple payback (2.742 years).
4.NPV & IRR
Answer: IRR = 24.19 %
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(20%) = 1 817.13
NPV(25%) = −328.00
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.
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 = 10.9/1.03 = 10.58
Estimated price change
%ΔP ≈ −10.58·(−0.005) = 5.291%
P ≈ 935·(1 + 0.05291) = €984.47
Answer
P = €984.47
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: WACC = 12.04 %
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 = €12 000 000, E/V = 0.8333, D/V = 0.1667
WACC
WACC = 0.8333·14% + 0.1667·3%·(1 − 0.25) = 12.04%
Answer
WACC = 12.04%
8.CAPM & WACC
Answer: WACC = 10.59 %
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 = 0.9772
Answer
P = 0.9772
11.Probability
Answer: σ = 1.512
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.
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 = (167.5 − 178)/7 = −1.5
z₂ = (185.35 − 178)/7 = 1.05
Area between
P = Φ(1.05) − Φ(−1.5) = 0.8531 − 0.06681 = 0.7863
Answer
P = 0.7863
14.Normal distribution
Answer: x = 458 h
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) = 0.15
z from the table
z ≈ −1.04
Back to the original units
x = 500 − 1.04·40 = 458 h
Answer
x = 458 h
15.Confidence intervals
Answer: upper limit = 31.31
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 = 5, t = 4.032
Margin of error
E = 4.032·9/√6 = 14.81
Interval
16.5 ± 14.81 → [1.685, 31.31]
Answer
upper limit = 31.31
16.Confidence intervals
Answer: n = 1699
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 = (2.576·16/1)² = 1699
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1699
Answer
n = 1699
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 of the mean difference
SE = 11.7/√9 = 3.9
Test statistic
t = 4.68/3.9 = 1.2
Answer
t = 1.2
With df = 8, the two-sided 5% critical value is 2.306; |t| is smaller, so H₀ is not rejected.
18.Hypothesis tests & confidence intervals
Answer: n = 397
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 = ((1.645 + 1.282)·17/2.5)² = 396.2
Round up
n = 397
Answer
n = 397
19.Linear regression & correlation
Answer: r = −0.9766
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 = 21/6 = 3.5
ȳ = 99/6 = 16.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −35.5
Syy = Σ(y − ȳ)² = 75.5
Correlation coefficient
r = −35.5/√(17.5·75.5) = −0.9766
Answer
r = −0.9766
20.Linear regression & correlation
Answer: x = 20.99
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 = 287/5 = 57.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 186
Regression line
b = 186/40 = 4.65
a = 57.4 − 4.65·20 = −35.6
ŷ = −35.6 + 4.65·x
Solve for x
x = (62 − (−35.6))/4.65 = 20.99
Answer
x = 20.99
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