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
years01234€3 000FV = ?

You invest €3 000 at a nominal rate of 4% per year, compounded monthly. Find the value FV of the investment after 4 years.

PV = €3 0004% / 12n = 4
FV = €
2.Time value of moneyHard
months0123…336€50€50€50FV = ?

You save €50 at the end of every month for 28 years. The account pays 3% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €503% / 12N = 336
FV = €
3.NPV & IRRMedium
years012345€30 000€1 500€1 500€1 500€1 500€1 500r = 10%, EAC = ?

A machine costs €30 000 today, lasts 5 years and costs €1 500 per year to run (paid at the end of each year). The cost of capital is 10%. Find its equivalent annual cost EAC.

C0 = €30 000M = €1 500n = 5r = 10%
EAC = €
4.NPV & IRRHard
years0123−50 00021 50036 00022 500

A project costs €50 000 and returns €21 500, €36 000, €22 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 21 500CF2 = 36 000CF3 = 22 500
IRR = %
5.Bond pricing & yieldMedium
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. Its yield to maturity changes from 9% to 8%. By how much ΔP does the bond price change?

F = €1 000coupon = 6%n = 6y1 = 9%y2 = 8%
ΔP = €
6.Bond pricing & yieldHard
years012345€1 045€40€40 + €1 000

A bond with a face value of €1 000 and an annual coupon of 4% matures in 5 years and sells for €1 045. Find its yield to maturity YTM in percent.

F = €1 000coupon = 4%n = 5P = €1 045
YTM = %
7.CAPM & WACCMedium
WACC = 7.52%equity€8 000 000re​ = ?debt€1 000 000rd​ = 5%T = 25.8%

A company has equity worth €8 000 000 and debt worth €1 000 000. Its WACC is 7.52%, its pre-tax cost of debt is 5% and the tax rate is 25.8%. Find the cost of equity re.

E = €8 000 000D = €1 000 000WACC = 7.52%rd = 5%T = 25.8%
re = %
8.CAPM & WACCHard
rf​ = 4%, rm​ − rf​ = 4%D/E = 0.4equityβL​ = 1.3debtD/E = 1.5equityre​ = ?debtT = 28%

A company's shares have a beta of 1.3 at its current debt-to-equity ratio of 0.4. It plans to raise its debt-to-equity ratio to 1.5. The tax rate is 28%, the risk-free rate 4% and the market risk premium 4%. Find the cost of equity re after the change.

βL = 1.3D/E1 = 0.4D/E2 = 1.5T = 28%rf = 4%rm − rf = 4%
re = %
9.Descriptive statisticsMedium
101520253035404550

Find the sample standard deviation s of the data: 26, 42, 52, 38, 26, 28, 10.

n = 7
s =
10.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 70 and standard deviation 10. Find the z-score z of a score of 50.

μ = 70σ = 10x = 50
z =
11.ProbabilityMedium
xp(x)00.310.3520.230.140.05μ = 1.25

For the distribution shown, find the variance Var(X) of X.

x: 0, 1, 2, 3, 4p(x): 0.3, 0.35, 0.2, 0.1, 0.05
Var(X) =
12.ProbabilityHard
n = 12, p = 0.3: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?

n = 12p = 0.3
P =
13.Normal distributionMedium
200 mm201 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 201 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 201 mm
P =
14.Normal distributionHard
157164171178185192199cmbottom 90%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm90%
x = cm
15.Confidence intervalsMedium
n = 7, s = 8L = ?Ux̄ = 18.5E99%

A sample of 7 items has mean 18.5 and standard deviation s = 8. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 7x̄ = 18.5s = 899%t(6) = 3.707
lower limit =
16.Confidence intervalsHard
σ = 33, n = ?x̄ − Ex̄ + Ex̄E = 2.590%

The population standard deviation is 33. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 2.5?

σ = 33E = 2.590%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.88p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 48. A random sample of 144 observations has a mean of 242.48. Test H₀: μ = 250 against H₁: μ < 250. Find the p-value p.

μ₀ = 250σ = 48n = 144x̄ = 242.48H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 21, power 95%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 95%. The standard deviation is 21. Find the minimum sample size n.

α = 0.01power = 95%δ = 4σ = 21zα = 2.326zβ = 1.645
n =
19.Linear regression & correlationMedium
xy242349455555662xy2465060

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

n = 5Σx = 20Σy = 263Σx² = 90Σy² = 14059Σxy = 1098
r =
20.Linear regression & correlationHard
xy125225321419513xy24152025ŷ = 19

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 = 19?

n = 5ŷ = 19
x =

3Solutions

1.Time value of money

Answer: FV = 3520 €

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 = 0.04/12 = 0.003333
N = m·n = 12·4 = 48

Future value

FV = 3 000·(1 + 0.003333)48 = 3 000·1.173 = €3 519.60

Answer

FV = €3 519.60
2.Time value of money

Answer: FV = 26279 €

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 = 0.03/12 = 0.0025
N = 12·28 = 336

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 50·(2.314 − 1)/0.0025 = €26 278.80

Answer

FV = €26 278.80

You paid in €16 800 yourself; the rest is interest.

3.NPV & IRR

Answer: EAC = 9414 €

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 − 1.1−5)/0.1 = 3.791

Annualised purchase price

30 000/3.791 = 7 913.92

EAC

EAC = 7 913.92 + 1 500 = €9 413.92

Answer

EAC = €9 413.92
4.NPV & IRR

Answer: IRR = 27.31 %

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(25%) = 1 760.00
NPV(30%) = −1 918.53

Narrow down

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

IRR ≈ 25 + 5·1 760.00/(1 760.00 − (−1 918.53)) = 27.39%
exact: NPV(27.31%) = 0 → IRR = 27.31%

Answer

IRR = 27.31%
5.Bond pricing & yield

Answer: ΔP = 42.12 €

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.

Price at the old yield

C = €60.00, i = 0.09, N = 6
€60.00·(1 − 0.5963)/0.09 + €1 000·0.5963 = €269.16 + €596.27 = €865.42

Price at the new yield

C = €60.00, i = 0.08, N = 6
€60.00·(1 − 0.6302)/0.08 + €1 000·0.6302 = €277.37 + €630.17 = €907.54

Change

ΔP = 907.54 − 865.42 = €42.12

Answer

ΔP = €42.12
6.Bond pricing & yield

Answer: YTM = 3.017 %

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

€1 045 = €40·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵

Solve numerically

P(3%) = €1 045.80, P(4%) = €1 000.00
YTM = 3.017%

Answer

YTM = 3.017%
7.CAPM & WACC

Answer: re = 7.996 %

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 = €9 000 000, E/V = 0.8889, D/V = 0.1111

Debt part of the WACC

0.1111·5%·(1 − 0.258) = 0.4122%

Cost of equity

re = (7.52% − 0.4122%)/0.8889 = 7.996%

Answer

re = 7.996%
8.CAPM & WACC

Answer: re = 12.4 %

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 = 1.3/(1 + 0.72·0.4) = 1.009

Relever at the new D/E

βL = 1.009·(1 + 0.72·1.5) = 2.099

Cost of equity (CAPM)

re = 4% + 2.099·4% = 12.4%

Answer

re = 12.4%
9.Descriptive statistics

Answer: s = 13.59

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 = 222/7 = 31.71

Squared deviations

Σ(x − x̄)² = (−5.714)² + 10.29² + 20.29² + 6.286² + (−5.714)² + (−3.714)² + (−21.71)² = 1107

Standard deviation

s = √(1107/6) = 13.59

Answer

s = 13.59
10.Descriptive statistics

Answer: z = −2

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 = (50 − 70)/10 = −2

Answer

z = −2
11.Probability

Answer: Var(X) = 1.287

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 = 0·0.3 + 1·0.35 + 2·0.2 + 3·0.1 + 4·0.05 = 1.25

Mean of the squares

E(X²) = 0·0.3 + 1·0.35 + 4·0.2 + 9·0.1 + 16·0.05 = 2.85

Variance

Var(X) = 2.85 − 1.25² = 1.287

Answer

Var(X) = 1.287
12.Probability

Answer: P = 0.08503

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.

Zero and one failure

P(X = 0) = 0.712 = 0.01384
P(X = 1) = 12·0.3·0.711 = 0.07118

Add them

P = 0.01384 + 0.07118 = 0.08503

Answer

P = 0.08503
13.Normal distribution

Answer: P = 0.02275

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 sum

μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (201 − 200)/0.5 = 2

Upper tail

P = 1 − Φ(2) = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
14.Normal distribution

Answer: x = 187 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) = 0.9

z from the table

z ≈ 1.28

Back to the original units

x = 178 + 1.28·7 = 187 cm

Answer

x = 187 cm
15.Confidence intervals

Answer: lower limit = 7.291

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 = 6, t = 3.707

Margin of error

E = 3.707·8/√7 = 11.21

Interval

18.5 ± 11.21 → [7.291, 29.71]

Answer

lower limit = 7.291
16.Confidence intervals

Answer: n = 472

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 = (1.645·33/2.5)² = 471.5

Round up

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

n = 472

Answer

n = 472
17.Hypothesis tests & confidence intervals

Answer: p = 0.03005

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.

Test statistic

SE = σ/√n = 48/√144 = 4
z = (242.48 − 250)/4 = −1.88

Tail area

Φ(1.88) = 0.9699
1 − Φ(1.88) = 0.0301

p-value

p = 0.03005

Answer

p = 0.03005

p ≤ 0.05, so H₀ is rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: n = 435

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((2.326 + 1.645)·21/4)² = 434.6

Round up

n = 435

Answer

n = 435
19.Linear regression & correlation

Answer: r = 0.9693

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 = 1098 − 20·263/5 = 46
Sxx = 90 − 20²/5 = 10
Syy = 14059 − 263²/5 = 225.2

Correlation coefficient

r = 46/√(10·225.2) = 0.9693

Answer

r = 0.9693
20.Linear regression & correlation

Answer: x = 3.533

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 = 103/5 = 20.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −30

Regression line

b = −30/10 = −3
a = 20.6 − (−3)·3 = 29.6
ŷ = 29.6 − 3·x

Solve for x

x = (19 − 29.6)/(−3) = 3.533

Answer

x = 3.533

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