An annuity pays €5 000 at the end of each year for 4 years. The interest rate is 7% per year. Find the present value PV of the annuity.
A = €5 000i = 7%n = 4
PV = €
2.Time value of moneyHard
A loan of €200 000 is repaid with equal monthly payments over 25 years at 3.6% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €200 0003.6% / 12N = 300
I = €
3.NPV & IRRMedium
A project costs €50 000 today and returns €18 000 at the end of each year for 6 years. The cost of capital is 8%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €20 000 and returns €7 500, €7 500, €7 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 7 500CF2 = 7 500CF3 = 7 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 9 years and pays annual coupons. It sells for €1 114 when the yield to maturity is 3.5%. Find the coupon rate c in percent.
F = €1 000n = 9P = €1 114YTM = 3.5%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 7% matures in 7 years and sells for €1 055. Find its yield to maturity YTM in percent.
F = €1 000coupon = 7%n = 7P = €1 055
YTM = %
7.CAPM & WACCMedium
A company has equity worth €8 000 000 and debt worth €5 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 4% and the tax rate is 28%. Find the WACC.
A company has 1 000 000 shares at €50.50 each and 40 000 bonds trading at €980 each. The pre-tax cost of debt is 4.5% and the tax rate is 25%. The share’s beta is 0.6, the risk-free rate 3% and the market return 9.5%. Find the WACC.
β = 0.6rf = 3%rm = 9.5%rd = 4.5%T = 25%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 44, 21, 43, 50, 22, 21, 18.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 20. What proportion P of values is above 44?
μ = 60σ = 20x = 44
P =
11.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 10? The order does not matter.
n = 10k = 2
N =
12.ProbabilityHard
A test has 9 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 = 9p = 0.4k = 1
P =
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.4778. Find z to two decimals.
A = 0.4778
z =
14.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 90% of them lie (the 90th percentile).
μ = 32σ = 6P = 90%
x = min
15.Confidence intervalsMedium
A sample of 30 items has mean 99.5 and standard deviation s = 14.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
n = 30x̄ = 99.5s = 14.599%t(29) = 2.756
lower limit =
16.Confidence intervalsHard
The population standard deviation is 26. 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 4?
σ = 26E = 499%
n =
17.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 13 students, mean score 33, standard deviation 6.5. Group 2: 8 students, mean score 38.4, standard deviation 6. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 13x̄₁ = 33s₁ = 6.5n₂ = 8x̄₂ = 38.4s₂ = 6
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 50 against H₁: μ < 50 is done at α = 0.01 with a sample of 36; the standard deviation is known to be 12. In reality the mean is 43.6. Find the power 1 − β of the test in percent.
μ₀ = 50μ = 43.6σ = 12n = 36α = 0.01
1 − β = %
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.
n = 5x = 4.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 16936 €
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 − 0.7629)/0.07 = 3.387
Present value
PV = 5 000·3.387 = €16 936.06
Answer
PV = €16 936.06
2.Time value of money
Answer: I = 103602 €
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.
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: 18 000/1.08 = 16 666.67 → Σ = 16 666.67
year 2: 18 000/1.166 = 15 432.10 → Σ = 32 098.77
year 3: 18 000/1.26 = 14 288.98 → Σ = 46 387.75
year 4: 18 000/1.36 = 13 230.54 → Σ = 59 618.28
Fraction of the last year
DPB = 3 + 3 612.25/13 230.54 = 3.273
Answer
DPB = 3.273 years
The discounted payback is always longer than the simple payback (2.778 years).
4.NPV & IRR
Answer: IRR = 4.98 %
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(0%) = 2 000.00
NPV(5%) = −7.56
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.
Present value of the face value
1000·(1 + 0.035)−9 = €733.73
Coupon
C = (1114 − 733.73)/7.608 = €49.98
Coupon rate
c = 49.98/1000 = 4.998%
Answer
c = 4.998%
6.Bond pricing & yield
Answer: YTM = 6.014 %
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
€1 055 = €70·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷
Solve numerically
P(6%) = €1 055.82, P(7%) = €1 000.00
YTM = 6.014%
Answer
YTM = 6.014%
7.CAPM & WACC
Answer: WACC = 7.877 %
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 = €13 000 000, E/V = 0.6154, D/V = 0.3846
WACC
WACC = 0.6154·11% + 0.3846·4%·(1 − 0.28) = 7.877%
Answer
WACC = 7.877%
8.CAPM & WACC
Answer: WACC = 5.36 %
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).
Market values
E = 1 000 000·€50.50 = €50 500 000
D = 40 000·€980 = €39 200 000
Cost of equity (CAPM)
re = 3% + 0.6·6.5% = 6.9%
Weights
V = €89 700 000, E/V = 0.563, D/V = 0.437
WACC
WACC = 0.563·6.9% + 0.437·4.5%·(1 − 0.25) = 5.36%
Answer
WACC = 5.36%
9.Descriptive statistics
Answer: s = 13.68
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 = −0.8
Standard normal table
Φ(−0.8) = 0.2119
P = 1 − 0.2119 = 0.7881
Answer
P = 0.7881
11.Probability
Answer: N = 45
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.
Combinations
C(10, 2) = (10·9)/(2·1) = 45
Answer
N = 45
12.Probability
Answer: P = 0.06047
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.
Binomial formula
P(X = 1) = 9·0.41·0.68 = 9·0.4·0.0168 = 0.06047
Answer
P = 0.06047
13.Normal distribution
Answer: z = 0.64
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 + 0.4778)/2 = 0.7389
Read the z-table backwards
Find 0.7389 inside the table and read off z.
z = 0.64
Answer
z = 0.64
14.Normal distribution
Answer: x = 39.69 min
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 for the percentile
Φ(z) = 0.9 → z = 1.282
Back to x
x = 32 + 1.282·6 = 39.69 min
Answer
x = 39.69 min
15.Confidence intervals
Answer: lower limit = 92.2
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 = 29, t = 2.756
Margin of error
E = 2.756·14.5/√30 = 7.296
Interval
99.5 ± 7.296 → [92.2, 106.8]
Answer
lower limit = 92.2
16.Confidence intervals
Answer: n = 281
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 = (2.576·26/4)² = 280.4
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 281
Answer
n = 281
17.Hypothesis tests & confidence intervals
Answer: t = −1.901
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.
Pooled variance
sp² = (12·6.5² + 7·6²)/19 = 39.95
sp = 6.32
Standard error
SE = 6.32·√(1/13 + 1/8) = 2.84
Test statistic
t = (33 − 38.4)/2.84 = −1.901
Answer
t = −1.901
With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.
18.Hypothesis tests & confidence intervals
Answer: 1 − β = 80.89 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 12/√36 = 2
x̄c = 50 − 2.326·2 = 45.35
z under the true mean
z = (45.35 − 43.6)/2 = 0.874
Power
power = P(Z < 0.874) = 80.89%
1 − β = 80.89%
Answer
1 − β = 80.89%
19.Linear regression & correlation
Answer: r² = 0.9882
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 = 182/6 = 30.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 280
Syy = Σ(y − ȳ)² = 1133
Coefficient of determination
r² = 280²/(70·1133) = 0.9882
Answer
r² = 0.9882
About 98.82% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 54.15
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 = 15/5 = 3
ȳ = 237/5 = 47.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 45
Regression line
b = 45/10 = 4.5
a = 47.4 − 4.5·3 = 33.9
ŷ = 33.9 + 4.5·x
Prediction
ŷ = 33.9 + 4.5·4.5 = 54.15
Answer
ŷ = 54.15
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