You need €40 000 in 10 years. An account pays 4.8% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €40 0004.8% / 4n = 10
PV = €
2.Time value of moneyHard
You save €200 at the end of every month for 6 years. The account pays 7.2% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €2007.2% / 12N = 72
FV = €
3.NPV & IRRMedium
A project costs €500 000 today. It generates €44 000 at the end of the first year, and the cash flow then grows by 3% per year forever. The cost of capital is 11%. Find the net present value NPV.
C0 = €500 000CF1 = €44 000g = 3%r = 11%
NPV = €
4.NPV & IRRHard
A project costs €20 000 and returns €15 000, €14 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 15 000CF2 = 14 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 14 years. Its yield to maturity changes from 2.5% to 1%. By how much ΔP does the bond price change?
F = €1 000coupon = 5%n = 14y1 = 2.5%y2 = 1%
ΔP = €
6.Bond pricing & yieldHard
A bond with 11 years to maturity (face value €1 000, annual coupon 6%) sells for €1 128. The issuer can call it in 4 years at a call price of €1 060. Find the yield to call YTC in percent.
F = €1 000coupon = 6%P = €1 128call price = €1 060call in 4 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €3 000 000 and debt worth €2 000 000. Its WACC is 7.55%, its pre-tax cost of debt is 3% and the tax rate is 21%. Find the cost of equity re.
A company has equity with a market value of €12 000 000 and a cost of equity of 11%. Its debt consists of 8 000 bonds (face value €1 000, annual coupon 5%, 11 years to maturity) that trade at €1 088 each. The tax rate is 28%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.
Find the sample standard deviation s of the data: 38, 22, 59, 37, 48, 18.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is above 45?
μ = 60σ = 15x = 45
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.2, 0.1, 0.2, 0.35, 0.15
σ =
12.ProbabilityHard
X has the distribution shown, and Y = −2X + 20. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = −2X + 20
σY =
13.Normal distributionMedium
The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 5 of them exceeds 196 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 5> 196 g
P =
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 20% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points20%
x = points
15.Confidence intervalsMedium
A sample of 12 items has mean 107 and standard deviation s = 18.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 12x̄ = 107s = 18.590%t(11) = 1.796
upper limit =
16.Confidence intervalsHard
The population standard deviation is 19. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?
σ = 19E = 1.590%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 50; the population standard deviation is known to be 5. A random sample of 25 observations has a mean of 48.88. Test H₀: μ = 50 against H₁: μ ≠ 50. Find the p-value p.
μ₀ = 50σ = 5n = 25x̄ = 48.88H₁: μ ≠ μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 90%. The standard deviation is 13. Find the minimum sample size n.
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
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 = 71?
n = 5ŷ = 71
x =
3Solutions
1.Time value of money
Answer: PV = 24822 €
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 = 0.048/4 = 0.012
N = 4·10 = 40
Present value
PV = FV/(1 + i)N = 40 000/1.611 = €24 822.16
Answer
PV = €24 822.16
2.Time value of money
Answer: FV = 17945 €
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.
You paid in €14 400 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = 50000 €
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.
Value of the cash flows
PV = 44 000/(0.11 − 0.03) = 550 000.00
NPV
NPV = 550 000.00 − 500 000 = €50 000.00
Answer
NPV = €50 000.00
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 29.19 %
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(25%) = 960.00
NPV(30%) = −177.51
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 25 + 5·960.00/(960.00 − (−177.51)) = 29.22%
exact: NPV(29.19%) = 0 → IRR = 29.19%
Answer
IRR = 29.19%
5.Bond pricing & yield
Answer: ΔP = 227.9 €
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.
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
1128 = 60·(1 − (1 + y)−4)/y + 1060/(1 + y)4
Solve numerically
y = 3%: €1 164.82
y = 4%: €1 123.89
YTC = 3.897%
Answer
YTC = 3.897%
The yield to maturity is 4.499%. 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 = 11 %
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 = €5 000 000, E/V = 0.6, D/V = 0.4
Debt part of the WACC
0.4·3%·(1 − 0.21) = 0.948%
Cost of equity
re = (7.55% − 0.948%)/0.6 = 11%
Answer
re = 11%
8.CAPM & WACC
Answer: WACC = 7.585 %
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) = 0.1587
P = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
11.Probability
Answer: σ = 1.352
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 = 1.05, Var(X) = 0.6475, σX = 0.8047
Linear change
σY = |−2|·0.8047 = 1.609
Answer
σY = 1.609
The mean does move: E(Y) = −2·1.05 + 20 = 17.9.
13.Normal distribution
Answer: P = 0.9319
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 total
μT = 200 g, σT = 1.2·√5 = 2.683 g
Standardise
z = (196 − 200)/2.683 = −1.49
Probability
P = 1 − Φ(−1.49) = 0.9319
Answer
P = 0.9319
14.Normal distribution
Answer: x = 112.6 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) = 0.8
z from the table
z ≈ 0.84
Back to the original units
x = 100 + 0.84·15 = 112.6 points
Answer
x = 112.6 points
15.Confidence intervals
Answer: upper limit = 116.6
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 = 11, t = 1.796
Margin of error
E = 1.796·18.5/√12 = 9.592
Interval
107 ± 9.592 → [97.41, 116.6]
Answer
upper limit = 116.6
16.Confidence intervals
Answer: n = 435
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 = (1.645·19/1.5)² = 434.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 435
Answer
n = 435
17.Hypothesis tests & confidence intervals
Answer: p = 0.2627
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 = 5/√25 = 1
z = (48.88 − 50)/1 = −1.12
Tail area
Φ(1.12) = 0.8686
1 − Φ(1.12) = 0.1314
p-value
p = 2·0.1314 = 0.2627
Answer
p = 0.2627
p > 0.05, so H₀ is not rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: n = 138
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 = ((2.326 + 1.282)·13/4)² = 137.5
Round up
n = 138
Answer
n = 138
19.Linear regression & correlation
Answer: r = 0.9953
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 = 2352 − 36·326/6 = 396
Sxx = 286 − 36²/6 = 70
Syy = 19974 − 326²/6 = 2261
Correlation coefficient
r = 396/√(70·2261) = 0.9953
Answer
r = 0.9953
20.Linear regression & correlation
Answer: x = 22.1
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 = 336/5 = 67.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 138
Regression line
b = 138/40 = 3.45
a = 67.2 − 3.45·21 = −5.25
ŷ = −5.25 + 3.45·x
Solve for x
x = (71 − (−5.25))/3.45 = 22.1
Answer
x = 22.1
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