CFA Level I quantitative methods and fixed income practice: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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1Formulas and methods

Time value of money

Future value
FV = PV·(1 + i)n
Present value
PV = FV/(1 + i)n
Several periods per year
i = r/m, N = m·n
Annuity, present value
PV = PMT·[1 − (1 + i)−n]/i
Annuity, future value
FV = PMT·[(1 + i)n − 1]/i

Method

  1. Draw a timeline with every cash flow and its timing.
  2. Match the rate to the period: a monthly rate for monthly payments.
  3. Use FV = PV·(1 + i)ⁿ for single amounts and the annuity formulas for equal payments.
  4. Check the sign and the order of magnitude of the answer.

NPV & IRR

Net present value
NPV = −C0 + Σ CFt/(1 + r)t
Equal cash flows
NPV = −C0 + CF·[1 − (1 + r)−n]/r
Internal rate of return
the r at which NPV = 0
Decision
accept when NPV > 0 (or IRR > required return)

Method

  1. List all cash flows with their year; the investment at year 0 is negative.
  2. Discount each cash flow: CFt/(1 + r)t.
  3. Add them up to get the NPV. Positive means the project earns more than r.
  4. For the IRR, find the rate that makes NPV zero by trial or interpolation.

Bond pricing & yield

Bond price
P = C·(1 − (1 + y)−n)/y + F/(1 + y)n
Semi-annual
C/2, y/2 and 2n periods
Current yield
C/P
Price and yield
yield up → price down

Method

  1. Find the coupon per period and the yield per period.
  2. Price = present value of the coupons (an annuity) + present value of the face value.
  3. For semi-annual coupons use half the coupon, half the yield and twice the periods.
  4. Check: if the yield is above the coupon rate, the price is below face value.

CAPM & WACC

CAPM
re = rf + β·(rm − rf)
WACC
E/V·re + D/V·rd·(1 − T)
Firm value
V = E + D

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

Confidence intervals

σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up

Method

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. Interval: x̄ ± E.

Hypothesis tests & confidence intervals

z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n

Method

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.

Linear regression & correlation

Sums of squares
Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
Slope
b = Sxy/Sxx
Intercept
a = ȳ − b·x̄
Correlation
r = Sxy/√(SxxSyy)
Residual
e = y − ŷ

2Practice problems

1.Time value of moneyMedium
years012345€8 000FV = ?

You invest €8 000 at a nominal rate of 6% per year, compounded quarterly. Find the value FV of the investment after 5 years.

PV = €8 0006% / 4n = 5
FV = €
2.Time value of moneyHard
months01…180…300€150 000AB = ?A2.4% / 12

A mortgage of €150 000 is repaid with equal monthly payments over 25 years at 2.4% per year, compounded monthly. How much is still owed just after the payment at the end of year 15? Find the outstanding balance B.

P = €150 0002.4% / 12N = 300k = 180
B = €
3.NPV & IRRMedium
years0123456€40 000€1 000€1 000€1 000€1 000€1 000€1 000r = 12%, EAC = ?

A machine costs €40 000 today, lasts 6 years and costs €1 000 per year to run (paid at the end of each year). The cost of capital is 12%. Find its equivalent annual cost EAC.

C0 = €40 000M = €1 000n = 6r = 12%
EAC = €
4.NPV & IRRHard
years0123−20 00010 50013 00011 000

A project costs €20 000 and returns €10 500, €13 000, €11 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 10 500CF2 = 13 000CF3 = 11 000
IRR = %
5.Bond pricing & yieldMedium
half-years0123…18€22.50€22.50€22.50 + €1 000P = ?

A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 9 years. The yield to maturity is 8% per year. Find the price P of the bond.

F = €1 000coupon = 4.5%YTM = 8%n = 9
P = €
6.Bond pricing & yieldHard
years0123€975€40€40 + €1 000

A bond with a face value of €1 000 and an annual coupon of 4% matures in 3 years and sells for €975. Find its yield to maturity YTM in percent.

F = €1 000coupon = 4%n = 3P = €975
YTM = %
7.CAPM & WACCMedium
WACC = ?equityEre​ = 12%debtD = 0.75·Erd​ = 4.5%T = 21%

A company has a debt-to-equity ratio (at market values) of 0.75. Its cost of equity is 12%, its pre-tax cost of debt is 4.5% and the tax rate is 21%. Find the WACC.

D/E = 0.75re = 12%rd = 4.5%T = 21%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €20.00D0​ = €3.00g = 2%debt€24 000 000rd​ = 4.5%T = 20%

A company has 2 000 000 shares trading at €20.00; the book value of its equity is €24 000 000. It just paid a dividend of €3.00 per share, and dividends grow by 2% per year. Its debt has a market value of €24 000 000 and a pre-tax cost of 4.5%. The tax rate is 20%. Find the WACC, using the dividend growth model for the cost of equity.

N = 2 000 000P0 = €20.00book equity = €24 000 000D0 = €3.00g = 2%D = €24 000 000rd = 4.5%T = 20%
WACC = %
9.Descriptive statisticsMedium
1520253035

Find the sample standard deviation s of the data: 14, 18, 23, 31, 39.

n = 5
s =
10.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 20. Find the z-score z of a score of 190.

μ = 170σ = 20x = 190
z =
11.ProbabilityMedium
red: 7blue: 3

A bag holds 7 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 3
P =
12.ProbabilityHard
xp(x)100.35200.3300.1400.25μ

X has the distribution shown, and Y = 5X + 4. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.35, 0.3, 0.1, 0.25Y = 5X + 4
σY =
13.Normal distributionMedium
765 kg801 kgT = X₁ + … + X₉?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 9 of them is below 801 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 9< 801 kg
P =
14.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
15.Confidence intervalsMedium
n = 9, s = 14LUx̄ = 111E = ?90%

A sample of 9 items has mean 111 and standard deviation s = 14. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 9x̄ = 111s = 1490%t(8) = 1.86
E =
16.Confidence intervalsHard
μ₁ − μ₂, t(19) = 1.7290L = ?Ux̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 12 plots, mean yield 29.5 kg, standard deviation 8.5 kg. Plot group 2: 9 plots, mean 31.5 kg, standard deviation 5.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 12x̄₁ = 29.5s₁ = 8.5n₂ = 9x̄₂ = 31.5s₂ = 5.590%t(19) = 1.729
lower limit = kg
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 40 parts has mean length 140 mm, and the line's known standard deviation is 20 mm. Line 2: a sample of 80 parts has mean 143.6 mm, known standard deviation 15 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 40x̄₁ = 140 mmσ₁ = 20 mmn₂ = 80x̄₂ = 143.6 mmσ₂ = 15 mm
z =
18.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 81 measurements has mean 50. The population standard deviation is 4. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 50σ = 4n = 8195%
L =
19.Linear regression & correlationMedium
xy144254356458565672xy246506070

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
xy12322031441158xy24101520ŷ = 13

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 13?

n = 5ŷ = 13
x =

3Solutions

1.Time value of money

Answer: FV = 10775 €

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 = r/m = 0.06/4 = 0.015
N = m·n = 4·5 = 20

Future value

FV = 8 000·(1 + 0.015)20 = 8 000·1.347 = €10 774.84

Answer

FV = €10 774.84
2.Time value of money

Answer: B = 70926 €

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 payment

i = 0.002, N = 300
A = 150 000·0.002/(1 − (1 + 0.002)−300) = €665.40

Present value of the remaining payments

Whatever is still owed must be exactly paid off by the remaining payments.

N − k = 300 − 180 = 120
B = 665.4·(1 − (1 + 0.002)−120)/0.002 = €70 925.74

Answer

B = €70 925.74
3.NPV & IRR

Answer: EAC = 10729 €

Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.

Annuity factor

(1 − 1.12−6)/0.12 = 4.111

Annualised purchase price

40 000/4.111 = 9 729.03

EAC

EAC = 9 729.03 + 1 000 = €10 729.03

Answer

EAC = €10 729.03
4.NPV & IRR

Answer: IRR = 32.71 %

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%) = 776.06
NPV(35%) = −618.30

Narrow down

Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.

IRR ≈ 30 + 5·776.06/(776.06 − (−618.30)) = 32.78%
exact: NPV(32.71%) = 0 → IRR = 32.71%

Answer

IRR = 32.71%
5.Bond pricing & yield

Answer: P = 778.5 €

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 = €22.50, i = 0.04, N = 18

Present value of the coupons

€22.50·(1 − 0.4936)/0.04 = €284.83

Present value of the face value

€1 000·0.4936 = €493.63

Price

P = €778.46

Answer

P = €778.46

The yield is above the coupon rate, so the bond sells at a discount.

6.Bond pricing & yield

Answer: YTM = 4.917 %

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

€975 = €40·(1 − (1 + y)⁻³)/y + €1 000/(1 + y)³

Solve numerically

P(4%) = €1 000.00, P(5%) = €972.77
YTM = 4.917%

Answer

YTM = 4.917%
7.CAPM & WACC

Answer: WACC = 8.381 %

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 from D/E

E/V = 1/(1 + 0.75) = 0.5714
D/V = 0.75/(1 + 0.75) = 0.4286

WACC

WACC = 0.5714·12% + 0.4286·4.5%·(1 − 0.21) = 8.381%

Answer

WACC = 8.381%
8.CAPM & WACC

Answer: WACC = 12.16 %

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).

Cost of equity

D1 = 3·1.02 = €3.06
re = 3.06/20 + 0.02 = 17.3%

Market values

E = 2 000 000·€20.00 = €40 000 000
V = €64 000 000, E/V = 0.625, D/V = 0.375

WACC

WACC = 0.625·17.3% + 0.375·4.5%·(1 − 0.2) = 12.16%

Answer

WACC = 12.16%
9.Descriptive statistics

Answer: s = 10.07

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.

Mean

x̄ = 125/5 = 25

Squared deviations

Σ(x − x̄)² = (−11)² + (−7)² + (−2)² + 6² + 14² = 406

Standard deviation

s = √(406/4) = 10.07

Answer

s = 10.07
10.Descriptive statistics

Answer: z = 1

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 = (190 − 170)/20 = 1

Answer

z = 1
11.Probability

Answer: P = 0.4667

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.

First and second draw

P = 7/10 · 6/9 = 42/90 = 0.4667

Answer

P = 0.4667
12.Probability

Answer: σY = 58.9

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 = 22.5, Var(X) = 138.8, σX = 11.78

Linear change

σY = |5|·11.78 = 58.9

Answer

σY = 58.9

The mean does move: E(Y) = 5·22.5 + 4 = 116.5.

13.Normal distribution

Answer: P = 0.8413

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 = 765 kg, σT = 12·√9 = 36 kg

Standardise

z = (801 − 765)/36 = 1

Probability

P = Φ(1) = 0.8413

Answer

P = 0.8413
14.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
15.Confidence intervals

Answer: E = 8.68

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 = 8, t = 1.86

Margin of error

E = 1.86·14/√9 = 8.68

Answer

E = 8.68
16.Confidence intervals

Answer: lower limit = −7.632 kg

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.

Pooled standard deviation

sp² = (11·8.5² + 8·5.5²)/19 = 54.57
sp = 7.387

Margin of error

E = 1.729·7.387·√(1/12 + 1/9) = 5.632

Interval

−2 ± 5.632 → [−7.632, 3.632]

Answer

lower limit = −7.632 kg
17.Hypothesis tests & confidence intervals

Answer: z = −1.006

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.

Standard error of the difference

SE = √(20²/40 + 15²/80) = √12.81 = 3.579

Test statistic

z = (140 − 143.6)/3.579 = −1.006

Answer

z = −1.006

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: L = 49.13

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 = 1.96·4/√81 = 0.8711

Interval limit

L = 50 − 0.8711 = 49.13

Answer

L = 49.13
19.Linear regression & correlation

Answer: r = 0.9744

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 = 349/6 = 58.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 87.5
Syy = Σ(y − ȳ)² = 460.8

Correlation coefficient

r = 87.5/√(17.5·460.8) = 0.9744

Answer

r = 0.9744
20.Linear regression & correlation

Answer: x = 3.564

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 = 76/5 = 15.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −39

Regression line

b = −39/10 = −3.9
a = 15.2 − (−3.9)·3 = 26.9
ŷ = 26.9 − 3.9·x

Solve for x

x = (13 − 26.9)/(−3.9) = 3.564

Answer

x = 3.564

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