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
years012PV = ?€50 000

You need €50 000 in 2 years. An account pays 5% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €50 0005% / 4n = 2
PV = €
2.Time value of moneyHard
months0123…336€250€250€250FV = ?

You save €250 at the end of every month for 28 years. The account pays 2.4% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €2502.4% / 12N = 336
FV = €
3.NPV & IRRMedium
years0123456€30 000€3 000€3 000€3 000€3 000€3 000€3 000r = 12%, EAC = ?

A machine costs €30 000 today, lasts 6 years and costs €3 000 per year to run (paid at the end of each year). The cost of capital is 12%. Find its equivalent annual cost EAC.

C0 = €30 000M = €3 000n = 6r = 12%
EAC = €
4.NPV & IRRHard
years0123−25 00018 50010 00018 500

A project costs €25 000 and returns €18 500, €10 000, €18 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −25 000CF1 = 18 500CF2 = 10 000CF3 = 18 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…18€35€35€35 + €1 000P = ?

A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 9 years. The yield to maturity is 4% per year. Find the price P of the bond.

F = €1 000coupon = 7%YTM = 4%n = 9
P = €
6.Bond pricing & yieldHard
years0123456P€60€60€60 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 6% and matures in 6 years. The yield to maturity is 5.5%. Find the Macaulay duration D in years.

F = €1 000coupon = 6%n = 6YTM = 5.5%
D = years
7.CAPM & WACCMedium
WACC = ?equity€10 000 000re​ = 10%debt€6 000 000rd​ = 5%T = 25%

A company has equity worth €10 000 000 and debt worth €6 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 5% and the tax rate is 25%. Find the WACC.

E = €10 000 000D = €6 000 000re = 10%rd = 5%T = 25%
WACC = %
8.CAPM & WACCHard
rf​ = 4%, rm​ − rf​ = 4%no debtequityβU​ = 0.9D/E = 0.6equityre​ = ?debtT = 30%

A company's business has an unlevered (asset) beta of 0.9. It plans a debt-to-equity ratio of 0.6. The tax rate is 30%, the risk-free rate is 4% and the market risk premium is 4%. Find the cost of equity re at the new capital structure.

βU = 0.9D/E = 0.6T = 30%rf = 4%rm − rf = 4%
re = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 37, 16, 13, 38, 56, 37.

n = 6
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is below 45?

μ = 50σ = 10x = 45
P =
11.ProbabilityMedium
xp(x)10.0520.3530.340.250.1μ = 2.95

For the distribution shown, find the standard deviation σ of X.

x: 1, 2, 3, 4, 5p(x): 0.05, 0.35, 0.3, 0.2, 0.1
σ =
12.ProbabilityHard
n = 7, p = 0.1: P(X = 2) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 2 correct answers?

n = 7p = 0.1k = 2
P =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0150

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0150. Find z to two decimals.

A = 0.0150
z =
14.Normal distributionHard
μ = 500x = ?95%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 95% of them lie (the 95th percentile).

μ = 500σ = 40P = 95%
x = h
15.Confidence intervalsMedium
x̄E₁ = 190%, n = 64x̄E₂ = ?99%, n = 64

A 90% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 1. What is the margin of error E₂ of a 99% interval from the same data?

90%n₁ = 64E₁ = 199%
E₂ =
16.Confidence intervalsHard
σ = 18, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 18. 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 = 18E = 499%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 49.1 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 49.1s = 5n = 100
t =
18.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 64 measurements has mean 119. The population standard deviation is 4. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 119σ = 4n = 6499%
U =
19.Linear regression & correlationMedium
xy144249350455567xy245060

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy242349453553660xy2465060ŷ = 45

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 45?

n = 5ŷ = 45
x =

3Solutions

1.Time value of money

Answer: PV = 45270 €

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 = 4·2 = 8

Present value

PV = FV/(1 + i)N = 50 000/1.104 = €45 269.92

Answer

PV = €45 269.92
2.Time value of money

Answer: FV = 119605 €

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 deposits

i = 0.024/12 = 0.002
N = 12·28 = 336

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 250·(1.957 − 1)/0.002 = €119 604.50

Answer

FV = €119 604.50

You paid in €84 000 yourself; the rest is interest.

3.NPV & IRR

Answer: EAC = 10297 €

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

30 000/4.111 = 7 296.77

EAC

EAC = 7 296.77 + 3 000 = €10 296.77

Answer

EAC = €10 296.77
4.NPV & IRR

Answer: IRR = 40.19 %

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(40%) = 58.31
NPV(45%) = −1 416.83

Narrow down

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

IRR ≈ 40 + 5·58.31/(58.31 − (−1 416.83)) = 40.2%
exact: NPV(40.19%) = 0 → IRR = 40.19%

Answer

IRR = 40.19%
5.Bond pricing & yield

Answer: P = 1225 €

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 = €35.00, i = 0.02, N = 18

Present value of the coupons

€35.00·(1 − 0.7002)/0.02 = €524.72

Present value of the face value

€1 000·0.7002 = €700.16

Price

P = €1 224.88

Answer

P = €1 224.88

The yield is below the coupon rate, so the bond sells at a premium.

6.Bond pricing & yield

Answer: D = 5.223 years

Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.

Present value of each cash flow

t = 1: 60/1.055 = 56.87, t·PV = 56.87
t = 2: 60/1.113 = 53.91, t·PV = 107.81
t = 3: 60/1.174 = 51.10, t·PV = 153.29
t = 4: 60/1.239 = 48.43, t·PV = 193.73
t = 5: 60/1.307 = 45.91, t·PV = 229.54
t = 6: 1060/1.379 = 768.76, t·PV = 4 612.56

Duration

P = 1 024.98
D = 5 353.81/1 024.98 = 5.223

Answer

D = 5.223 years
7.CAPM & WACC

Answer: WACC = 7.656 %

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 = €16 000 000, E/V = 0.625, D/V = 0.375

WACC

WACC = 0.625·10% + 0.375·5%·(1 − 0.25) = 7.656%

Answer

WACC = 7.656%
8.CAPM & WACC

Answer: re = 9.112 %

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.

Levered beta

βL = 0.9·(1 + 0.7·0.6) = 1.278

Cost of equity (CAPM)

re = 4% + 1.278·4% = 9.112%

Answer

re = 9.112%
9.Descriptive statistics

Answer: s = 15.97

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 = 197/6 = 32.83

Squared deviations

Σ(x − x̄)² = 4.167² + (−16.83)² + (−19.83)² + 5.167² + 23.17² + 4.167² = 1275

Standard deviation

s = √(1275/5) = 15.97

Answer

s = 15.97
10.Descriptive statistics

Answer: P = 0.3085

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 0.3085

Answer

P = 0.3085
11.Probability

Answer: σ = 1.071

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.

Mean

μ = 1·0.05 + 2·0.35 + 3·0.3 + 4·0.2 + 5·0.1 = 2.95

Mean of the squares

E(X²) = 1·0.05 + 4·0.35 + 9·0.3 + 16·0.2 + 25·0.1 = 9.85

Variance

Var(X) = 9.85 − 2.95² = 1.147
σ = √1.147 = 1.071

Answer

σ = 1.071
12.Probability

Answer: P = 0.124

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 = 2) = 21·0.12·0.95 = 21·0.01·0.5905 = 0.124

Answer

P = 0.124
13.Normal distribution

Answer: z = 2.17

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 − 0.0150 = 0.9850

Read the z-table backwards

Find 0.9850 inside the table and read off z.

z = 2.17

Answer

z = 2.17
14.Normal distribution

Answer: x = 565.8 h

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) = 0.95 → z = 1.645

Back to x

x = 500 + 1.645·40 = 565.8 h

Answer

x = 565.8 h
15.Confidence intervals

Answer: E₂ = 1.566

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.

Ratio of margins

E₂/E₁ = (2.576/1.645)·√(64/64) = 1.566

New margin

E₂ = 1·1.566 = 1.566

Answer

E₂ = 1.566
16.Confidence intervals

Answer: n = 135

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 = (2.576·18/4)² = 134.4

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 135

Answer

n = 135
17.Hypothesis tests & confidence intervals

Answer: t = −1.8

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).

Standard error

SE = 5/√100 = 5/10 = 0.5

Test statistic

t = (49.1 − 50)/0.5 = −1.8

Answer

t = −1.8
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: U = 120.3

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 = 2.576·4/√64 = 1.288

Interval limit

U = 119 + 1.288 = 120.3

Answer

U = 120.3
19.Linear regression & correlation

Answer: r = 0.94

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 = 265/5 = 53

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 52
Syy = Σ(y − ȳ)² = 306

Correlation coefficient

r = 52/√(10·306) = 0.94

Answer

r = 0.94
20.Linear regression & correlation

Answer: x = 2.4

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 = 20/5 = 4
ȳ = 257/5 = 51.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 40

Regression line

b = 40/10 = 4
a = 51.4 − 4·4 = 35.4
ŷ = 35.4 + 4·x

Solve for x

x = (45 − 35.4)/4 = 2.4

Answer

x = 2.4

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