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…8€3 000FV = ?

You invest €3 000 at a nominal rate of 2% per year, compounded monthly. Find the value FV of the investment after 8 years.

PV = €3 0002% / 12n = 8
FV = €
2.Time value of moneyHard
months0123…N = ?€15 000€450€450€450€4503.6% / 12

You borrow €15 000 at 3.6% per year, compounded monthly, and repay €450 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 = €15 000A = €4503.6% / 12
N = payments
3.NPV & IRRMedium
years0123456€20 000€1 000€1 000€1 000€1 000€1 000€1 000r = 12%, EAC = ?

A machine costs €20 000 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 = €20 000M = €1 000n = 6r = 12%
EAC = €
4.NPV & IRRHard
years012−25 00016 00015 000

A project costs €25 000 and returns €16 000, €15 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −25 000CF1 = 16 000CF2 = 15 000
IRR = %
5.Bond pricing & yieldMedium
years0123…9P€50€50€50 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 9 years. Its yield to maturity changes from 8% to 7.5%. By how much ΔP does the bond price change?

F = €1 000coupon = 5%n = 9y1 = 8%y2 = 7.5%
ΔP = €
6.Bond pricing & yieldHard

A bond trades at €920 with a yield to maturity of 6% (annual compounding) and a Macaulay duration of 7.8 years. Use the modified duration to estimate the new price P if the yield falls by 75 basis points.

P0 = €920y = 6%D = 7.8Δy = −75 bp
P = €
7.CAPM & WACCMedium
WACC = ?ordinary shares€10 000 000re​ = 9%preferred shares€1 000 000rp​ = 7.5%debt€2 000 000rd​ = 4.5%T = 20%

A company is financed with ordinary shares worth €10 000 000, preferred shares worth €1 000 000 and debt worth €2 000 000. The cost of equity is 9%, the cost of preferred stock 7.5% and the pre-tax cost of debt 4.5%. The tax rate is 20%. Find the WACC.

E = €10 000 000Ps = €1 000 000D = €2 000 000re = 9%rp = 7.5%rd = 4.5%T = 20%
WACC = %
8.CAPM & WACCHard
rf​ = 3.5%, rm​ − rf​ = 5%no debtequityβU​ = 0.9D/E = 1equityre​ = ?debtT = 25.8%

A company's business has an unlevered (asset) beta of 0.9. It plans a debt-to-equity ratio of 1. The tax rate is 25.8%, the risk-free rate is 3.5% and the market risk premium is 5%. Find the cost of equity re at the new capital structure.

βU = 0.9D/E = 1T = 25.8%rf = 3.5%rm − rf = 5%
re = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 11, 46, 45, 52, 42, 22, 22.

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

A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is below 66?

μ = 60σ = 15x = 66
P =
11.ProbabilityMedium
n = 15k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 15? The order does not matter.

n = 15k = 2
N =
12.ProbabilityHard
n = 10, p = 0.2: P(X = 4) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 4 correct answers?

n = 10p = 0.2k = 4
P =
13.Normal distributionMedium
32 min34 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 34 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 34 min
P =
14.Normal distributionHard
157164171178185192199cmtop 60%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 60% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm60%
x = cm
15.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 80 phones, mean 94 h, known standard deviation 12 h. Model 2: 50 phones, mean 104 h, known standard deviation 17 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 94 hσ₁ = 12 hn₂ = 50x̄₂ = 104 hσ₂ = 17 h99%
upper limit = h
16.Confidence intervalsHard
σ = 25, n = ?x̄ − Ex̄ + Ex̄E = 390%

The population standard deviation is 25. 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 3?

σ = 25E = 390%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 78 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 78s = 15n = 36
t =
18.Hypothesis tests & confidence intervalsHard
σ = 5, n = ?x̄ − Ex̄ + Ex̄E = 2.595%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 5. Find the smallest sample size n.

σ = 5E = 2.595%
n =
19.Linear regression & correlationMedium
xy125224319418518xy242025

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

n = 5Σx = 15Σy = 104Σx² = 55Σy² = 2210Σxy = 292
r =
20.Linear regression & correlationHard
xy125218316417513611xy24610152025x = 2.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.

n = 6x = 2.5
ŷ =

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 = 0.02/12 = 0.001667
N = m·n = 12·8 = 96

Future value

FV = 3 000·(1 + 0.001667)96 = 3 000·1.173 = €3 520.06

Answer

FV = €3 520.06
2.Time value of money

Answer: N = 36 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 − 15 000·0.003/450 = 0.9

Solve with logarithms

N = −ln(0.9)/ln(1.003) = 35.17

Round up to whole payments

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

N = 36

Answer

N = 36 payments
3.NPV & IRR

Answer: EAC = 5865 €

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

20 000/4.111 = 4 864.51

EAC

EAC = 4 864.51 + 1 000 = €5 864.51

Answer

EAC = €5 864.51
4.NPV & IRR

Answer: IRR = 15.81 %

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%) = 255.20
NPV(20%) = −1 250.00

Narrow down

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

IRR ≈ 15 + 5·255.20/(255.20 − (−1 250.00)) = 15.85%
exact: NPV(15.81%) = 0 → IRR = 15.81%

Answer

IRR = 15.81%
5.Bond pricing & yield

Answer: ΔP = 27.93 €

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 = €50.00, i = 0.08, N = 9
€50.00·(1 − 0.5002)/0.08 + €1 000·0.5002 = €312.34 + €500.25 = €812.59

Price at the new yield

C = €50.00, i = 0.075, N = 9
€50.00·(1 − 0.5216)/0.075 + €1 000·0.5216 = €318.94 + €521.58 = €840.53

Change

ΔP = 840.53 − 812.59 = €27.93

Answer

ΔP = €27.93
6.Bond pricing & yield

Answer: P = 970.8 €

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.

Modified duration

Dmod = 7.8/1.06 = 7.358

Estimated price change

%ΔP ≈ −7.358·(−0.0075) = 5.519%
P ≈ 920·(1 + 0.05519) = €970.77

Answer

P = €970.77

Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.

7.CAPM & WACC

Answer: WACC = 8.054 %

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 = 10 000 000 + 1 000 000 + 2 000 000 = €13 000 000
E/V = 0.7692, Ps/V = 0.07692, D/V = 0.1538

WACC

WACC = 0.7692·9% + 0.07692·7.5% + 0.1538·4.5%·(1 − 0.2) = 8.054%

Answer

WACC = 8.054%
8.CAPM & WACC

Answer: re = 11.34 %

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 + 0.742·1) = 1.568

Cost of equity (CAPM)

re = 3.5% + 1.568·5% = 11.34%

Answer

re = 11.34%
9.Descriptive statistics

Answer: s = 15.65

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 = 240/7 = 34.29

Squared deviations

Σ(x − x̄)² = (−23.29)² + 11.71² + 10.71² + 17.71² + 7.714² + (−12.29)² + (−12.29)² = 1469

Standard deviation

s = √(1469/6) = 15.65

Answer

s = 15.65
10.Descriptive statistics

Answer: P = 0.6554

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 0.6554

Answer

P = 0.6554
11.Probability

Answer: N = 105

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.

Combinations

C(15, 2) = (15·14)/(2·1) = 105

Answer

N = 105
12.Probability

Answer: P = 0.08808

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 = 4) = 210·0.24·0.86 = 210·0.0016·0.2621 = 0.08808

Answer

P = 0.08808
13.Normal distribution

Answer: P = 0.2119

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 = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (34 − 32)/2.5 = 0.8

Upper tail

P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119

Answer

P = 0.2119
14.Normal distribution

Answer: x = 176.3 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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 178 − 0.25·7 = 176.3 cm

Answer

x = 176.3 cm
15.Confidence intervals

Answer: upper limit = −2.908 h

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.

Point estimate

x̄₁ − x̄₂ = 94 − 104 = −10

Standard error

SE = √(12²/80 + 17²/50) = 2.753

Margin of error

E = 2.576·2.753 = 7.092

Interval

−10 ± 7.092 → [−17.09, −2.908]

Answer

upper limit = −2.908 h

The interval does not contain 0, so the means differ significantly.

16.Confidence intervals

Answer: n = 188

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 = (1.645·25/3)² = 187.9

Round up

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

n = 188

Answer

n = 188
17.Hypothesis tests & confidence intervals

Answer: t = 1.2

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 = 15/√36 = 15/6 = 2.5

Test statistic

t = (78 − 75)/2.5 = 1.2

Answer

t = 1.2
t0.025; df = 35 = 2.03

|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 16

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).

Solve for n

n = (1.96·5/2.5)² = 15.37

Round up

n = 16

Answer

n = 16
19.Linear regression & correlation

Answer: r = −0.9245

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 = 292 − 15·104/5 = −20
Sxx = 55 − 15²/5 = 10
Syy = 2210 − 104²/5 = 46.8

Correlation coefficient

r = −20/√(10·46.8) = −0.9245

Answer

r = −0.9245
20.Linear regression & correlation

Answer: ŷ = 19.07

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 = 100/6 = 16.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −42

Regression line

b = −42/17.5 = −2.4
a = 16.67 − (−2.4)·3.5 = 25.07
ŷ = 25.07 − 2.4·x

Prediction

ŷ = 25.07 + (−2.4)·2.5 = 19.07

Answer

ŷ = 19.07

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