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
years0123…18€2 000FV = ?

You invest €2 000 at 4% per year with continuous compounding. Find the value FV after 18 years.

PV = €2 000r = 4%t = 18
FV = €
2.Time value of moneyHard
months0123…360€150 000A?AAA

A loan of €150 000 is repaid with equal monthly payments over 30 years. The interest rate is 3% per year, compounded monthly. Find the monthly payment A.

P = €150 0003% / 12N = 360
A = €
3.NPV & IRRMedium
years0123…9−500 00036 000+3%+3%…r = 9%, NPV = ?

A project costs €500 000 today. It generates €36 000 at the end of the first year, and the cash flow then grows by 3% per year forever. The cost of capital is 9%. Find the net present value NPV.

C0 = €500 000CF1 = €36 000g = 3%r = 9%
NPV = €
4.NPV & IRRHard
years012−25 00015 50017 000

A project costs €25 000 and returns €15 500, €17 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −25 000CF1 = 15 500CF2 = 17 000
IRR = %
5.Bond pricing & yieldMedium

You buy a bond with 8 years to maturity (face value €1 000, annual coupon 2%) when its yield to maturity is 6%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 5%. Find your holding period return HPR in percent.

F = €1 000coupon = 2%n = 8y0 = 6%y1 = 5%
HPR = %
6.Bond pricing & yieldHard
years01234€920€60€60 + €1 000

A bond with a face value of €1 000 and an annual coupon of 6% matures in 4 years and sells for €920. Find its yield to maturity YTM in percent.

F = €1 000coupon = 6%n = 4P = €920
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€12 000 000re​ = 10%debt€6 000 000rd​ = 4.5%T = 25%

A company has equity worth €12 000 000 and debt worth €6 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 4.5% and the tax rate is 25%. Find the WACC.

E = €12 000 000D = €6 000 000re = 10%rd = 4.5%T = 25%
WACC = %
8.CAPM & WACCHard
rf​ = 1%, rm​ − rf​ = 7%D/E = 0.2equityβL​ = 1.5debtD/E = 0.75equityre​ = ?debtT = 25%

A company's shares have a beta of 1.5 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 0.75. The tax rate is 25%, the risk-free rate 1% and the market risk premium 7%. Find the cost of equity re after the change.

βL = 1.5D/E1 = 0.2D/E2 = 0.75T = 25%rf = 1%rm − rf = 7%
re = %
9.Descriptive statisticsMedium
25262728293031323334

Find the sample standard deviation s of the data: 33, 25, 26, 34, 32.

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

Exam scores are normally distributed with mean 100 and standard deviation 5. Find the z-score z of a score of 92.5.

μ = 100σ = 5x = 92.5
z =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6P(six) = 1/6 on each roll

A fair die is rolled 6 times. What is the probability P of getting at least one six?

n = 6
P =
12.ProbabilityHard
€p(x)−10.7410.2590.01

A game costs €1 to play. You win €2 with probability 0.25 and €10 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€2: p = 0.25€10: p = 0.01
E(G) = €
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.1093

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.1093. Find z to two decimals.

A = 0.1093
z =
14.Normal distributionHard
μ = 100x = ?20%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 20% of them lie (the 20th percentile).

μ = 100σ = 15P = 20%
x = points
15.Confidence intervalsMedium
n = 13, s = 1LU = ?x̄ = 65E90%

A sample of 13 items has mean 65 and standard deviation s = 1. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 13x̄ = 65s = 190%t(12) = 1.782
upper limit =
16.Confidence intervalsHard
μ₁ − μ₂, t(29) = 2.0450LUx̄₁ − x̄₂E = ?95%

Two fertilisers are tested. Plot group 1: 7 plots, mean yield 92 kg, standard deviation 6.5 kg. Plot group 2: 24 plots, mean 96 kg, standard deviation 1.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 7x̄₁ = 92s₁ = 6.5n₂ = 24x̄₂ = 96s₂ = 1.595%t(29) = 2.045
E = kg
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 80 parts has mean length 126 mm, and the line's known standard deviation is 12 mm. Line 2: a sample of 30 parts has mean 131.9 mm, known standard deviation 12 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 80x̄₁ = 126 mmσ₁ = 12 mmn₂ = 30x̄₂ = 131.9 mmσ₂ = 12 mm
z =
18.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 16 measurements has mean 204.5. The population standard deviation is 15. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 204.5σ = 15n = 1695%
U =
19.Linear regression & correlationMedium
xy2541462383010361243xy5102040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 151Σx² = 364Σy² = 4795Σxy = 1320
r =
20.Linear regression & correlationHard
xy15661877211012410927121xy15202580100120x = 19.5

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. Predict ŷ for x = 19.5.

n = 5x = 19.5
ŷ =

3Solutions

1.Time value of money

Answer: FV = 4109 €

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.

Exponent

r·t = 0.04·18 = 0.72
e0.72 = 2.054

Future value

FV = 2 000·2.054 = €4 108.87

Answer

FV = €4 108.87
2.Time value of money

Answer: A = 632.4 €

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 = 0.03/12 = 0.0025
N = 12·30 = 360

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 = 0.407
A = 150 000·0.0025/(1 − 0.407) = €632.41

Answer

A = €632.41
3.NPV & IRR

Answer: NPV = 100000 €

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 = 36 000/(0.09 − 0.03) = 600 000.00

NPV

NPV = 600 000.00 − 500 000 = €100 000.00

Answer

NPV = €100 000.00

NPV is positive, so the project earns more than the required return.

4.NPV & IRR

Answer: IRR = 19.1 %

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(15%) = 1 332.70
NPV(20%) = −277.78

Narrow down

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

IRR ≈ 15 + 5·1 332.70/(1 332.70 − (−277.78)) = 19.14%
exact: NPV(19.1%) = 0 → IRR = 19.1%

Answer

IRR = 19.1%
5.Bond pricing & yield

Answer: HPR = 12.61 %

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.

Purchase price

C = €20.00, i = 0.06, N = 8
€20.00·(1 − 0.6274)/0.06 + €1 000·0.6274 = €124.20 + €627.41 = €751.61

Selling price

C = €20.00, i = 0.05, N = 7
€20.00·(1 − 0.7107)/0.05 + €1 000·0.7107 = €115.73 + €710.68 = €826.41

Holding period return

HPR = (20 + 826.41 − 751.61)/751.61 = 12.61%

Answer

HPR = 12.61%
6.Bond pricing & yield

Answer: YTM = 8.439 %

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

€920 = €60·(1 − (1 + y)⁻⁴)/y + €1 000/(1 + y)⁴

Solve numerically

P(8%) = €933.76, P(9%) = €902.81
YTM = 8.439%

Answer

YTM = 8.439%
7.CAPM & WACC

Answer: WACC = 7.792 %

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 = €18 000 000, E/V = 0.6667, D/V = 0.3333

WACC

WACC = 0.6667·10% + 0.3333·4.5%·(1 − 0.25) = 7.792%

Answer

WACC = 7.792%
8.CAPM & WACC

Answer: re = 15.27 %

Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.

Unlever the beta

βU = 1.5/(1 + 0.75·0.2) = 1.304

Relever at the new D/E

βL = 1.304·(1 + 0.75·0.75) = 2.038

Cost of equity (CAPM)

re = 1% + 2.038·7% = 15.27%

Answer

re = 15.27%
9.Descriptive statistics

Answer: s = 4.183

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 = 150/5 = 30

Squared deviations

Σ(x − x̄)² = 3² + (−5)² + (−4)² + 4² + 2² = 70

Standard deviation

s = √(70/4) = 4.183

Answer

s = 4.183
10.Descriptive statistics

Answer: z = −1.5

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 = (92.5 − 100)/5 = −1.5

Answer

z = −1.5
11.Probability

Answer: P = 0.6651

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.

Complement

P(no six) = (5/6)6 = 0.3349
P = 1 − 0.3349 = 0.6651

Answer

P = 0.6651
12.Probability

Answer: E(G) = −0.4 €

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.

Net outcomes

G: 1 (p = 0.25), 9 (p = 0.01), −1 (p = 0.74)

Expected value

E(G) = 1·0.25 + 9·0.01 − 1·0.74 = −0.4 €

Answer

E(G) = −0.4 €

A negative value means you lose money on average.

13.Normal distribution

Answer: z = 1.23

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 − 0.1093 = 0.8907

Read the z-table backwards

Find 0.8907 inside the table and read off z.

z = 1.23

Answer

z = 1.23
14.Normal distribution

Answer: x = 87.38 points

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) = 0.2 → z = −0.8416

Back to x

x = 100 + −0.8416·15 = 87.38 points

Answer

x = 87.38 points
15.Confidence intervals

Answer: upper limit = 65.49

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 = 12, t = 1.782

Margin of error

E = 1.782·1/√13 = 0.4942

Interval

65 ± 0.4942 → [64.51, 65.49]

Answer

upper limit = 65.49
16.Confidence intervals

Answer: E = 2.85 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 = (6·6.5² + 23·1.5²)/29 = 10.53
sp = 3.244

Margin of error

E = 2.045·3.244·√(1/7 + 1/24) = 2.85

Answer

E = 2.85 kg
17.Hypothesis tests & confidence intervals

Answer: z = −2.297

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 = √(12²/80 + 12²/30) = √6.6 = 2.569

Test statistic

z = (126 − 131.9)/2.569 = −2.297

Answer

z = −2.297

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: U = 211.8

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 = 1.96·15/√16 = 7.35

Interval limit

U = 204.5 + 7.35 = 211.8

Answer

U = 211.8
19.Linear regression & correlation

Answer: r = 0.9966

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 = 1320 − 42·151/6 = 263
Sxx = 364 − 42²/6 = 70
Syy = 4795 − 151²/6 = 994.8

Correlation coefficient

r = 263/√(70·994.8) = 0.9966

Answer

r = 0.9966
20.Linear regression & correlation

Answer: ŷ = 87.7

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 = 474/5 = 94.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 426

Regression line

b = 426/90 = 4.733
a = 94.8 − 4.733·21 = −4.6
ŷ = −4.6 + 4.733·x

Prediction

ŷ = −4.6 + 4.733·19.5 = 87.7

Answer

ŷ = 87.7

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