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…8PV = ?€25 000

You need €25 000 in 8 years. An account pays 3% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €25 0003% / 12n = 8
PV = €
2.Time value of moneyHard
years0123…10€1 000€1 000€1 000FV = ?

You deposit €1 000 at the start of every year for 10 years, the first deposit today. The account pays 3% per year. How much is in the account at the end of year 10? Find FV.

A = €1 000i = 3%n = 10
FV = €
3.NPV & IRRMedium
years0123−60 00020 00026 00025 000

An investment of €60 000 gives these cash flows at the end of years 1 to 3: €20 000, €26 000, €25 000. The cost of capital is 10%. Find the net present value NPV.

r = 10%CF0 = −60 000CF1 = 20 000CF2 = 26 000CF3 = 25 000
NPV = €
4.NPV & IRRHard
years0123−10 0006 5006 0007 000

A project costs €10 000 and returns €6 500, €6 000, €7 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −10 000CF1 = 6 500CF2 = 6 000CF3 = 7 000
IRR = %
5.Bond pricing & yieldMedium
years0123…8P€30€30€30 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 3% and matures in 8 years. Its yield to maturity changes from 5.5% to 4%. By how much ΔP does the bond price change?

F = €1 000coupon = 3%n = 8y1 = 5.5%y2 = 4%
ΔP = €
6.Bond pricing & yieldHard
years0123€1 025€60€60 + €1 000

A bond with a face value of €1 000 and an annual coupon of 6% matures in 3 years and sells for €1 025. Find its yield to maturity YTM in percent.

F = €1 000coupon = 6%n = 3P = €1 025
YTM = %
7.CAPM & WACCMedium
WACC = 9.56%equity€6 000 000re​ = ?debt€2 000 000rd​ = 3%T = 25.8%

A company has equity worth €6 000 000 and debt worth €2 000 000. Its WACC is 9.56%, its pre-tax cost of debt is 3% and the tax rate is 25.8%. Find the cost of equity re.

E = €6 000 000D = €2 000 000WACC = 9.56%rd = 3%T = 25.8%
re = %
8.CAPM & WACCHard
rf​ = 1%, rm​ − rf​ = 4%D/E = 0.25equityβL​ = 1.2debtD/E = 1.25equityre​ = ?debtT = 25%

A company's shares have a beta of 1.2 at its current debt-to-equity ratio of 0.25. It plans to raise its debt-to-equity ratio to 1.25. The tax rate is 25%, the risk-free rate 1% and the market risk premium 4%. Find the cost of equity re after the change.

βL = 1.2D/E1 = 0.25D/E2 = 1.25T = 25%rf = 1%rm − rf = 4%
re = %
9.Descriptive statisticsMedium
1015202530354045

Find the sample standard deviation s of the data: 47, 35, 21, 10, 25.

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

A quantity is normally distributed with mean 60 and standard deviation 12. What proportion P of values is below 84?

μ = 60σ = 12x = 84
P =
11.ProbabilityMedium
red: 8blue: 5

A bag holds 8 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 5
P =
12.ProbabilityHard
€p(x)−10.7420.2590.01

A game costs €1 to play. You win €3 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€3: p = 0.25€10: p = 0.01
E(G) = €
13.Normal distributionMedium
160 g164.8 gT = X₁ + … + X₄?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 4 of them exceeds 164.8 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 4> 164.8 g
P =
14.Normal distributionHard
μ = 178x = ?75%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 75% of them lie (the 75th percentile).

μ = 178σ = 7P = 75%
x = cm
15.Confidence intervalsMedium
n = 8, s = 18.5LUx̄ = 12.5E = ?90%

A sample of 8 items has mean 12.5 and standard deviation s = 18.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 8x̄ = 12.5s = 18.590%t(7) = 1.895
E =
16.Confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 299%

The population standard deviation is 15. 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 2?

σ = 15E = 299%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.4472.447t = ?df = 6

7 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −5.65 and the standard deviation of the differences is 8.3. Find the paired test statistic t for H₀: μd = 0.

n = 7d̄ = −5.65sd = 8.3
t =
18.Hypothesis tests & confidence intervalsHard
σ = 8, 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 8. Find the smallest sample size n.

σ = 8E = 2.595%
n =
19.Linear regression & correlationMedium
xy126224320420516xy242025

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 = 106Σx² = 55Σy² = 2308Σxy = 294
r =
20.Linear regression & correlationHard
xy21142062583410431252xy5102040ŷ = 15

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 15?

n = 6ŷ = 15
x =

3Solutions

1.Time value of money

Answer: PV = 19672 €

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 = 0.03/12 = 0.0025
N = 12·8 = 96

Present value

PV = FV/(1 + i)N = 25 000/1.271 = €19 671.59

Answer

PV = €19 671.59
2.Time value of money

Answer: FV = 11808 €

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.

Ordinary annuity factor

((1 + 0.03)10 − 1)/0.03 = 11.46

Annuity due

Each deposit is invested one year longer.

FV = 1 000·11.46·1.03 = €11 807.80

Answer

FV = €11 807.80
3.NPV & IRR

Answer: NPV = −1548 €

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.

Present value of each cash flow

CF0 = −60 000
CF1/(1 + 0.1)1 = 20 000/1.1 = 18 181.82
CF2/(1 + 0.1)2 = 26 000/1.21 = 21 487.60
CF3/(1 + 0.1)3 = 25 000/1.331 = 18 782.87

Add them up

NPV = €−1 547.71

Answer

NPV = €−1 547.71

NPV is negative, so the project earns less than the required return.

4.NPV & IRR

Answer: IRR = 41.98 %

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%) = 255.10
NPV(45%) = −367.38

Narrow down

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

IRR ≈ 40 + 5·255.10/(255.10 − (−367.38)) = 42.05%
exact: NPV(41.98%) = 0 → IRR = 41.98%

Answer

IRR = 41.98%
5.Bond pricing & yield

Answer: ΔP = 91.04 €

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 = €30.00, i = 0.055, N = 8
€30.00·(1 − 0.6516)/0.055 + €1 000·0.6516 = €190.04 + €651.60 = €841.64

Price at the new yield

C = €30.00, i = 0.04, N = 8
€30.00·(1 − 0.7307)/0.04 + €1 000·0.7307 = €201.98 + €730.69 = €932.67

Change

ΔP = 932.67 − 841.64 = €91.04

Answer

ΔP = €91.04
6.Bond pricing & yield

Answer: YTM = 5.081 %

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

€1 025 = €60·(1 − (1 + y)⁻³)/y + €1 000/(1 + y)³

Solve numerically

P(5%) = €1 027.23, P(6%) = €1 000.00
YTM = 5.081%

Answer

YTM = 5.081%
7.CAPM & WACC

Answer: re = 12 %

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 = €8 000 000, E/V = 0.75, D/V = 0.25

Debt part of the WACC

0.25·3%·(1 − 0.258) = 0.5565%

Cost of equity

re = (9.56% − 0.5565%)/0.75 = 12%

Answer

re = 12%
8.CAPM & WACC

Answer: re = 8.832 %

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 = 1.2/(1 + 0.75·0.25) = 1.011

Relever at the new D/E

βL = 1.011·(1 + 0.75·1.25) = 1.958

Cost of equity (CAPM)

re = 1% + 1.958·4% = 8.832%

Answer

re = 8.832%
9.Descriptive statistics

Answer: s = 14.06

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 = 138/5 = 27.6

Squared deviations

Σ(x − x̄)² = 19.4² + 7.4² + (−6.6)² + (−17.6)² + (−2.6)² = 791.2

Standard deviation

s = √(791.2/4) = 14.06

Answer

s = 14.06
10.Descriptive statistics

Answer: P = 0.9772

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 0.9772

Answer

P = 0.9772
11.Probability

Answer: P = 0.359

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.

First and second draw

P = 8/13 · 7/12 = 56/156 = 0.359

Answer

P = 0.359
12.Probability

Answer: E(G) = −0.15 €

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

Expected value

E(G) = 2·0.25 + 9·0.01 − 1·0.74 = −0.15 €

Answer

E(G) = −0.15 €

A negative value means you lose money on average.

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 total

μT = 160 g, σT = 1.2·√4 = 2.4 g

Standardise

z = (164.8 − 160)/2.4 = 2

Probability

P = 1 − Φ(2) = 0.02275

Answer

P = 0.02275
14.Normal distribution

Answer: x = 182.7 cm

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) = 0.75 → z = 0.6745

Back to x

x = 178 + 0.6745·7 = 182.72 cm

Answer

x = 182.72 cm
15.Confidence intervals

Answer: E = 12.39

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 = 7, t = 1.895

Margin of error

E = 1.895·18.5/√8 = 12.39

Answer

E = 12.39
16.Confidence intervals

Answer: n = 374

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 = (2.576·15/2)² = 373.3

Round up

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

n = 374

Answer

n = 374
17.Hypothesis tests & confidence intervals

Answer: t = −1.801

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 of the mean difference

SE = 8.3/√7 = 3.137

Test statistic

t = −5.65/3.137 = −1.801

Answer

t = −1.801

With df = 6, the two-sided 5% critical value is 2.447; |t| is smaller, so H₀ is not rejected.

18.Hypothesis tests & confidence intervals

Answer: n = 40

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 = (1.96·8/2.5)² = 39.34

Round up

n = 40

Answer

n = 40
19.Linear regression & correlation

Answer: r = −0.9733

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 = 294 − 15·106/5 = −24
Sxx = 55 − 15²/5 = 10
Syy = 2308 − 106²/5 = 60.8

Correlation coefficient

r = −24/√(10·60.8) = −0.9733

Answer

r = −0.9733
20.Linear regression & correlation

Answer: x = 3.084

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 = 42/6 = 7
ȳ = 185/6 = 30.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 283

Regression line

b = 283/70 = 4.043
a = 30.83 − 4.043·7 = 2.533
ŷ = 2.533 + 4.043·x

Solve for x

x = (15 − 2.533)/4.043 = 3.084

Answer

x = 3.084

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