You need €25 000 in 5 years. An account pays 5% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €25 0005% / 4n = 5
PV = €
2.Time value of moneyHard
A loan of €8 000 is repaid with equal monthly payments over 6 years. The interest rate is 4.8% per year, compounded monthly. Find the monthly payment A.
P = €8 0004.8% / 12N = 72
A = €
3.NPV & IRRMedium
A project costs €20 000 today and returns €7 500, €8 000, €8 000, €6 000, €4 000 at the end of years 1 to 5. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €20 000 and returns €14 500, €7 000, €12 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 14 500CF2 = 7 000CF3 = 12 500
IRR = %
5.Bond pricing & yieldMedium
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% per year. Find the price P of the bond.
F = €1 000coupon = 8%YTM = 3%n = 4
P = €
6.Bond pricing & yieldHard
A bond with 16 years to maturity (face value €1 000, annual coupon 6%) sells for €1 108. The issuer can call it in 5 years at a call price of €1 060. Find the yield to call YTC in percent.
F = €1 000coupon = 6%P = €1 108call price = €1 060call in 5 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €5 000 000 and debt worth €1 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 4% and the tax rate is 25.8%. Find the WACC.
A company has 2 000 000 shares trading at €42.50; the book value of its equity is €51 000 000. It just paid a dividend of €0.80 per share, and dividends grow by 5% per year. Its debt has a market value of €51 000 000 and a pre-tax cost of 6%. The tax rate is 25%. Find the WACC, using the dividend growth model for the cost of equity.
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 27.5 and 35 min? Give P as a decimal.
μ = 32σ = 6x1 = 27.5x2 = 35
P =
14.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 5% is to the right of it. Find the cut-off x.
μ = 178 cmσ = 7 cm5%
x = cm
15.Confidence intervalsMedium
The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 8.3 and standard deviation 6.9. Find the lower limit of the 99% confidence interval for the mean difference.
n = 10d̄ = 8.3sd = 6.999%t(9) = 3.25
lower limit = mmHg
16.Confidence intervalsHard
The diameters of 15 machined shafts have a sample standard deviation of 2.4 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.
n = 15s = 2.4 µmχ² = 5.629, 26.119
σ = µm
17.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 122.3 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 122.3s = 12n = 81
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 16 measurements has mean 122. The population standard deviation is 8. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 122σ = 8n = 1699%
U =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 54) is made. Find its residual e = y − ŷ.
n = 6(2.5, 54)
e =
3Solutions
1.Time value of money
Answer: PV = 19500 €
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 = 0.05/4 = 0.0125
N = 4·5 = 20
Present value
PV = FV/(1 + i)N = 25 000/1.282 = €19 500.21
Answer
PV = €19 500.21
2.Time value of money
Answer: A = 128.1 €
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 = 0.048/12 = 0.004
N = 12·6 = 72
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 = 0.7502
A = 8 000·0.004/(1 − 0.7502) = €128.10
Answer
A = €128.10
3.NPV & IRR
Answer: DPB = 3.323 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: 7 500/1.12 = 6 696.43 → Σ = 6 696.43
year 2: 8 000/1.254 = 6 377.55 → Σ = 13 073.98
year 3: 8 000/1.405 = 5 694.24 → Σ = 18 768.22
year 4: 6 000/1.574 = 3 813.11 → Σ = 22 581.33
Fraction of the last year
DPB = 3 + 1 231.78/3 813.11 = 3.323
Answer
DPB = 3.323 years
The discounted payback is always longer than the simple payback (2.563 years).
4.NPV & IRR
Answer: IRR = 33.67 %
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(30%) = 985.43
NPV(35%) = −337.86
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 30 + 5·985.43/(985.43 − (−337.86)) = 33.72%
exact: NPV(33.67%) = 0 → IRR = 33.67%
Answer
IRR = 33.67%
5.Bond pricing & yield
Answer: P = 1187 €
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 = 0.015, N = 8
Present value of the coupons
€40.00·(1 − 0.8877)/0.015 = €299.44
Present value of the face value
€1 000·0.8877 = €887.71
Price
P = €1 187.15
Answer
P = €1 187.15
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTC = 4.625 %
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
1108 = 60·(1 − (1 + y)−5)/y + 1060/(1 + y)5
Solve numerically
y = 4%: €1 138.35
y = 5%: €1 090.31
YTC = 4.625%
Answer
YTC = 4.625%
The yield to maturity is 5.003%. 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: WACC = 7.161 %
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 = €6 000 000, E/V = 0.8333, D/V = 0.1667
WACC
WACC = 0.8333·8% + 0.1667·4%·(1 − 0.258) = 7.161%
Answer
WACC = 7.161%
8.CAPM & WACC
Answer: WACC = 6.048 %
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: Var(X) = 94.75
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 = 26, Var(X) = 164, σX = 12.81
Linear change
σY = |3|·12.81 = 38.42
Answer
σY = 38.42
The mean does move: E(Y) = 3·26 + 10 = 88.
13.Normal distribution
Answer: P = 0.4648
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 = (27.5 − 32)/6 = −0.75
z₂ = (35 − 32)/6 = 0.5
Area between
P = Φ(0.5) − Φ(−0.75) = 0.6915 − 0.2266 = 0.4648
Answer
P = 0.4648
14.Normal distribution
Answer: x = 189.5 cm
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) = 0.95
z from the table
z ≈ 1.64
Back to the original units
x = 178 + 1.64·7 = 189.5 cm
Answer
x = 189.5 cm
15.Confidence intervals
Answer: lower limit = 1.209 mmHg
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.
Critical value
df = 9, t = 3.25
Margin of error
E = 3.25·6.9/√10 = 7.091
Interval
8.3 ± 7.091 → [1.209, 15.39]
Answer
lower limit = 1.209 mmHg
16.Confidence intervals
Answer: σ = 3.785 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 14·2.4² = 80.64
Limits for σ²
80.64/26.119 = 3.087
80.64/5.629 = 14.33
Limit for σ
σ = √14.33 = 3.785 µm
Answer
σ = 3.785 µm
95% interval for σ: [1.757, 3.785] µm.
17.Hypothesis tests & confidence intervals
Answer: t = 1.725
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 = 12/√81 = 12/9 = 1.333
Test statistic
t = (122.3 − 120)/1.333 = 1.725
Answer
t = 1.725
t0.025; df = 80 = 1.99
|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: U = 127.2
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).
Margin of error
E = 2.576·8/√16 = 5.152
Interval limit
U = 122 + 5.152 = 127.2
Answer
U = 127.2
19.Linear regression & correlation
Answer: r = 0.9785
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 = 48/6 = 8
ȳ = 344/6 = 57.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 222
Syy = Σ(y − ȳ)² = 735.3
Correlation coefficient
r = 222/√(70·735.3) = 0.9785
Answer
r = 0.9785
20.Linear regression & correlation
Answer: e = 4.433
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 = 27/6 = 4.5
ȳ = 367/6 = 61.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 101.5
Regression line
b = 101.5/17.5 = 5.8
a = 61.17 − 5.8·4.5 = 35.07
ŷ = 35.07 + 5.8·x
Predicted value
ŷ = 35.07 + 5.8·2.5 = 49.57
Residual
e = 54 − 49.57 = 4.433
Answer
e = 4.433
The point lies above the line.
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