You need €50 000 in 4 years. An account pays 3.6% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €50 0003.6% / 12n = 4
PV = €
2.Time value of moneyHard
A mortgage of €300 000 is repaid with equal monthly payments over 30 years at 7.2% per year, compounded monthly. How much is still owed just after the payment at the end of year 4? Find the outstanding balance B.
P = €300 0007.2% / 12N = 360k = 48
B = €
3.NPV & IRRMedium
An investment of €45 000 gives these cash flows at the end of years 1 to 4: €11 000, €18 000, €11 000, €19 000. The cost of capital is 15%. Find the net present value NPV.
A project costs €20 000 and returns €7 000, €6 500, €11 500, €11 500 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 10%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 4 years. The yield to maturity is 3.5% per year. Find the price P of the bond.
F = €1 000coupon = 8%YTM = 3.5%n = 4
P = €
6.Bond pricing & yieldHard
A bond with 17 years to maturity (face value €1 000, annual coupon 5%) sells for €1 190. The issuer can call it in 5 years at a call price of €1 050. Find the yield to call YTC in percent.
F = €1 000coupon = 5%P = €1 190call price = €1 050call in 5 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €6 000 000 and debt worth €3 000 000. Its WACC is 8.79%, its pre-tax cost of debt is 3% and the tax rate is 21%. Find the cost of equity re.
A company has 2 500 000 shares at €28.50 each and 20 000 bonds trading at €1 000 each. The pre-tax cost of debt is 5% and the tax rate is 20%. The share’s beta is 1.1, the risk-free rate 3.5% and the market return 10.5%. Find the WACC.
β = 1.1rf = 3.5%rm = 10.5%rd = 5%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 21, 19, 51, 53, 42.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 60.
μ = 70σ = 20x = 60
z =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.25, 0.2, 0.2, 0.35
σ =
12.ProbabilityHard
8 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?
n = 8p = 0.2
P =
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.6255. Find z to two decimals.
A = 0.6255
z =
14.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 50 phones, mean 123 h, known standard deviation 18 h. Model 2: 60 phones, mean 133 h, known standard deviation 18 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.
The population standard deviation is 35. 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 3?
σ = 35E = 399%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 76.05 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 76.05s = 4n = 64
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 49 measurements has mean 53.4. The population standard deviation is 10. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 53.4σ = 10n = 4999%
L =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.
n = 6x = 3
ŷ =
3Solutions
1.Time value of money
Answer: PV = 43304 €
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 = 0.036000000000000004/12 = 0.003
N = 12·4 = 48
Present value
PV = FV/(1 + i)N = 50 000/1.155 = €43 303.72
Answer
PV = €43 303.72
2.Time value of money
Answer: B = 286897 €
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.
Whatever is still owed must be exactly paid off by the remaining payments.
N − k = 360 − 48 = 312
B = 2036·(1 − (1 + 0.006)−312)/0.006 = €286 897.13
Answer
B = €286 897.13
3.NPV & IRR
Answer: NPV = −3728 €
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 = −45 000
CF1/(1 + 0.15)1 = 11 000/1.15 = 9 565.22
CF2/(1 + 0.15)2 = 18 000/1.322 = 13 610.59
CF3/(1 + 0.15)3 = 11 000/1.521 = 7 232.68
CF4/(1 + 0.15)4 = 19 000/1.749 = 10 863.31
Add them up
NPV = €−3 728.21
Answer
NPV = €−3 728.21
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: MIRR = 19.9 %
Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.
Terminal value of the inflows
7 000·1.13 = 9 317.00
6 500·1.12 = 7 865.00
11 500·1.11 = 12 650.00
11 500·1.10 = 11 500.00
TV = 41 332.00
MIRR
MIRR = (41 332.00/20 000)1/4 − 1 = 19.9%
Answer
MIRR = 19.9%
5.Bond pricing & yield
Answer: P = 1167 €
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 = €40.00, i = 0.0175, N = 8
Present value of the coupons
€40.00·(1 − 0.8704)/0.0175 = €296.20
Present value of the face value
€1 000·0.8704 = €870.41
Price
P = €1 166.61
Answer
P = €1 166.61
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTC = 1.938 %
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
1190 = 50·(1 − (1 + y)−5)/y + 1050/(1 + y)5
Solve numerically
y = 1%: €1 241.71
y = 2%: €1 186.69
YTC = 1.938%
Answer
YTC = 1.938%
The yield to maturity is 3.498%. 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 = 12 %
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 = 0.6667, D/V = 0.3333
Debt part of the WACC
0.3333·3%·(1 − 0.21) = 0.79%
Cost of equity
re = (8.79% − 0.79%)/0.6667 = 12%
Answer
re = 12%
8.CAPM & WACC
Answer: WACC = 9.622 %
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 = (60 − 70)/20 = −0.5
Answer
z = −0.5
11.Probability
Answer: σ = 11.95
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) = 0.88 = 0.1678
P(X = 1) = 8·0.2·0.87 = 0.3355
Add them
P = 0.1678 + 0.3355 = 0.5033
Answer
P = 0.5033
13.Normal distribution
Answer: z = −0.32
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 − 0.6255 = 0.3745
Read the z-table backwards
Find 0.3745 inside the table and read off z.
z = −0.32
Answer
z = −0.32
14.Normal distribution
Answer: P = 0.9515
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 = 25.03 − 25 = 0.03 mm, σ = 0.01803 mm
Standardise
z = (0 − 0.03)/0.01803 = −1.66
Probability
P = 1 − Φ(−1.66) = 0.9515
Answer
P = 0.9515
15.Confidence intervals
Answer: E = 8.879 h
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Point estimate
x̄₁ − x̄₂ = 123 − 133 = −10
Standard error
SE = √(18²/50 + 18²/60) = 3.447
Margin of error
E = 2.576·3.447 = 8.879
Answer
E = 8.879 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: n = 904
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 = (2.576·35/3)² = 903.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 904
Answer
n = 904
17.Hypothesis tests & confidence intervals
Answer: t = 2.1
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
SE = 4/√64 = 4/8 = 0.5
Test statistic
t = (76.05 − 75)/0.5 = 2.1
Answer
t = 2.1
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: L = 49.72
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 = 2.576·10/√49 = 3.68
Interval limit
L = 53.4 − 3.68 = 49.72
Answer
L = 49.72
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.
Means
x̄ = 135/6 = 22.5
ȳ = 706/6 = 117.7
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 987
Syy = Σ(y − ȳ)² = 6243
Correlation coefficient
r = 987/√(157.5·6243) = 0.9953
Answer
r = 0.9953
20.Linear regression & correlation
Answer: ŷ = 4.262
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 = 65/6 = 10.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 115
Regression line
b = 115/70 = 1.643
a = 10.83 − 1.643·7 = −0.6667
ŷ = −0.6667 + 1.643·x
Prediction
ŷ = −0.6667 + 1.643·3 = 4.262
Answer
ŷ = 4.262
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