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€5 000FV = ?

You invest €5 000 at 3% per year with continuous compounding. Find the value FV after 3 years.

PV = €5 000r = 3%t = 3
FV = €
2.Time value of moneyHard
months0123…N = ?€25 000€475€475€475€4753.6% / 12

You borrow €25 000 at 3.6% per year, compounded monthly, and repay €475 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.

P = €25 000A = €4753.6% / 12
N = payments
3.NPV & IRRMedium
years0123456€40 000€2 000€2 000€2 000€2 000€2 000€2 000r = 10%, EAC = ?

A machine costs €40 000 today, lasts 6 years and costs €2 000 per year to run (paid at the end of each year). The cost of capital is 10%. Find its equivalent annual cost EAC.

C0 = €40 000M = €2 000n = 6r = 10%
EAC = €
4.NPV & IRRHard
years0123−80 00028 50028 50028 500

An investment of €80 000 returns €28 500 at the end of each year for 3 years. Find its internal rate of return IRR in percent.

C0 = €80 000CF = €28 500n = 3
IRR = %
5.Bond pricing & yieldMedium
half-years0123…14€30€30€30 + €1 000P = ?

A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 3.5% per year. Find the price P of the bond.

F = €1 000coupon = 6%YTM = 3.5%n = 7
P = €
6.Bond pricing & yieldHard
years01234€965€50€50 + €1 000

A bond with a face value of €1 000 and an annual coupon of 5% matures in 4 years and sells for €965. Find its yield to maturity YTM in percent.

F = €1 000coupon = 5%n = 4P = €965
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€3 000 000re​ = 14%debt€1 000 000rd​ = 7%T = 30%

A company has equity worth €3 000 000 and debt worth €1 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 7% and the tax rate is 30%. Find the WACC.

E = €3 000 000D = €1 000 000re = 14%rd = 7%T = 30%
WACC = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €58.50β = 1.2rf​ = 4%, rm​ = 8.5%debt20 000 × €960rd​ = 5%T = 28%

A company has 1 000 000 shares at €58.50 each and 20 000 bonds trading at €960 each. The pre-tax cost of debt is 5% and the tax rate is 28%. The share’s beta is 1.2, the risk-free rate 4% and the market return 8.5%. Find the WACC.

β = 1.2rf = 4%rm = 8.5%rd = 5%T = 28%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 50, 14, 29, 58, 41, 15.

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

Exam scores are normally distributed with mean 70 and standard deviation 15. Find the z-score z of a score of 100.

μ = 70σ = 15x = 100
z =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7P(six) = 1/6 on each roll

A fair die is rolled 7 times. What is the probability P of getting at least one six?

n = 7
P =
12.ProbabilityHard
n = 7, p = 0.7: P(X = 1) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 1 correct answers?

n = 7p = 0.7k = 1
P =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0143

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0143. Find z to two decimals.

A = 0.0143
z =
14.Normal distributionHard
μ = 65x = ?80%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 80% of them lie (the 80th percentile).

μ = 65σ = 10P = 80%
x = points
15.Confidence intervalsMedium
n = 16, s = 10LU = ?x̄ = 33E95%

A sample of 16 items has mean 33 and standard deviation s = 10. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 16x̄ = 33s = 1095%t(15) = 2.131
upper limit =
16.Confidence intervalsHard
p ≈ 0.8, n = ?p̂ − Ep̂ + Ep̂E = 0.0395%

A pilot study suggests that about 0.8 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.03?

p ≈ 0.8E = 0.0395%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 202.4 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 202.4s = 8n = 64
t =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2.5α = 0.05βσ = 7, power 80%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 2.5 units with a power of 80%. The standard deviation is 7. Find the minimum sample size n.

α = 0.05power = 80%δ = 2.5σ = 7zα = 1.645zβ = 0.842
n =
19.Linear regression & correlationMedium
xy122220321418517xy24182022

The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy142417312040235726682971xy152025304060ŷ = 50

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 50?

n = 6ŷ = 50
x =

3Solutions

1.Time value of money

Answer: FV = 5471 €

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 = 0.03·3 = 0.09
e0.09 = 1.094

Future value

FV = 5 000·1.094 = €5 470.87

Answer

FV = €5 470.87
2.Time value of money

Answer: N = 58 payments

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.

Rearrange the annuity formula

i = 0.003
(1 + i)−N = 1 − 25 000·0.003/475 = 0.8421

Solve with logarithms

N = −ln(0.8421)/ln(1.003) = 57.37

Round up to whole payments

57 full payments leave a small balance, which the last, smaller payment clears.

N = 58

Answer

N = 58 payments
3.NPV & IRR

Answer: EAC = 11184 €

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 − 1.1−6)/0.1 = 4.355

Annualised purchase price

40 000/4.355 = 9 184.30

EAC

EAC = 9 184.30 + 2 000 = €11 184.30

Answer

EAC = €11 184.30
4.NPV & IRR

Answer: IRR = 3.4 %

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.

Required annuity factor

(1 − (1 + i)−3)/i = 80 000/28 500 = 2.807

Solve for the rate

The annuity factor falls as the rate rises; bracket the rate and narrow down.

i = 3%: 2.829
i = 4%: 2.775
IRR = 3.4%

Answer

IRR = 3.4%
5.Bond pricing & yield

Answer: P = 1154 €

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 = €30.00, i = 0.0175, N = 14

Present value of the coupons

€30.00·(1 − 0.7844)/0.0175 = €369.66

Present value of the face value

€1 000·0.7844 = €784.36

Price

P = €1 154.03

Answer

P = €1 154.03

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

6.Bond pricing & yield

Answer: YTM = 6.01 %

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

€965 = €50·(1 − (1 + y)⁻⁴)/y + €1 000/(1 + y)⁴

Solve numerically

P(6%) = €965.35, P(7%) = €932.26
YTM = 6.01%

Answer

YTM = 6.01%
7.CAPM & WACC

Answer: WACC = 11.72 %

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 = €4 000 000, E/V = 0.75, D/V = 0.25

WACC

WACC = 0.75·14% + 0.25·7%·(1 − 0.3) = 11.72%

Answer

WACC = 11.72%
8.CAPM & WACC

Answer: WACC = 7.967 %

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).

Market values

E = 1 000 000·€58.50 = €58 500 000
D = 20 000·€960 = €19 200 000

Cost of equity (CAPM)

re = 4% + 1.2·4.5% = 9.4%

Weights

V = €77 700 000, E/V = 0.7529, D/V = 0.2471

WACC

WACC = 0.7529·9.4% + 0.2471·5%·(1 − 0.28) = 7.967%

Answer

WACC = 7.967%
9.Descriptive statistics

Answer: s = 18.25

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 = 207/6 = 34.5

Squared deviations

Σ(x − x̄)² = 15.5² + (−20.5)² + (−5.5)² + 23.5² + 6.5² + (−19.5)² = 1666

Standard deviation

s = √(1666/5) = 18.25

Answer

s = 18.25
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 = (100 − 70)/15 = 2

Answer

z = 2
11.Probability

Answer: P = 0.7209

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)7 = 0.2791
P = 1 − 0.2791 = 0.7209

Answer

P = 0.7209
12.Probability

Answer: P = 0.003572

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 = 1) = 7·0.71·0.36 = 7·0.7·0.000729 = 0.003572

Answer

P = 0.003572
13.Normal distribution

Answer: z = 2.19

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 − 0.0143 = 0.9857

Read the z-table backwards

Find 0.9857 inside the table and read off z.

z = 2.19

Answer

z = 2.19
14.Normal distribution

Answer: x = 73.42 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) = 0.8 → z = 0.8416

Back to x

x = 65 + 0.8416·10 = 73.42 points

Answer

x = 73.42 points
15.Confidence intervals

Answer: upper limit = 38.33

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 = 15, t = 2.131

Margin of error

E = 2.131·10/√16 = 5.327

Interval

33 ± 5.327 → [27.67, 38.33]

Answer

upper limit = 38.33
16.Confidence intervals

Answer: n = 683

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.96²·0.8·0.2/0.03² = 683

Round up

n = 683

Answer

n = 683
17.Hypothesis tests & confidence intervals

Answer: t = 2.4

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 = 8/√64 = 8/8 = 1

Test statistic

t = (202.4 − 200)/1 = 2.4

Answer

t = 2.4
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 49

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 = ((1.645 + 0.842)·7/2.5)² = 48.49

Round up

n = 49

Answer

n = 49
19.Linear regression & correlation

Answer: r² = 0.8372

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 = 98/5 = 19.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −12
Syy = Σ(y − ȳ)² = 17.2

Coefficient of determination

r² = −12²/(10·17.2) = 0.8372

Answer

r² = 0.8372

About 83.72% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: x = 21.93

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 = 129/6 = 21.5
ȳ = 291/6 = 48.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 544.5

Regression line

b = 544.5/157.5 = 3.457
a = 48.5 − 3.457·21.5 = −25.83
ŷ = −25.83 + 3.457·x

Solve for x

x = (50 − (−25.83))/3.457 = 21.93

Answer

x = 21.93

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