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 = ?€1 500€1 500€1 500€1 500

An annuity pays €1 500 at the end of each year for 8 years. The interest rate is 7% per year. Find the present value PV of the annuity.

A = €1 500i = 7%n = 8
PV = €
2.Time value of moneyHard
years0123…19€2 000€2 000€2 000FV = ?

You deposit €2 000 at the start of every year for 19 years, the first deposit today. The account pays 3% per year. How much is in the account at the end of year 19? Find FV.

A = €2 000i = 3%n = 19
FV = €
3.NPV & IRRMedium
years0123…10€20 000€1 500€1 500€1 500€1 500r = 5%, EAC = ?

A machine costs €20 000 today, lasts 10 years and costs €1 500 per year to run (paid at the end of each year). The cost of capital is 5%. Find its equivalent annual cost EAC.

C0 = €20 000M = €1 500n = 10r = 5%
EAC = €
4.NPV & IRRHard
years0123−50 00036 50018 00033 500

A project costs €50 000 and returns €36 500, €18 000, €33 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 36 500CF2 = 18 000CF3 = 33 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…24€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 12 years. The yield to maturity is 8% per year. Find the price P of the bond.

F = €1 000coupon = 6%YTM = 8%n = 12
P = €
6.Bond pricing & yieldHard

A bond trades at €1 010 with a yield to maturity of 6% (annual compounding) and a Macaulay duration of 6.6 years. Use the modified duration to estimate the new price P if the yield falls by 100 basis points.

P0 = €1 010y = 6%D = 6.6Δy = −100 bp
P = €
7.CAPM & WACCMedium
WACC = ?ordinary shares€10 000 000re​ = 14%preferred shares€2 000 000rp​ = 7.5%debt€2 000 000rd​ = 4.5%T = 25%

A company is financed with ordinary shares worth €10 000 000, preferred shares worth €2 000 000 and debt worth €2 000 000. The cost of equity is 14%, the cost of preferred stock 7.5% and the pre-tax cost of debt 4.5%. The tax rate is 25%. Find the WACC.

E = €10 000 000Ps = €2 000 000D = €2 000 000re = 14%rp = 7.5%rd = 4.5%T = 25%
WACC = %
8.CAPM & WACCHard
rf​ = 2.5%, rm​ − rf​ = 4.5%D/E = 0.4equityβL​ = 1.2debtD/E = 0.75equityre​ = ?debtT = 28%

A company's shares have a beta of 1.2 at its current debt-to-equity ratio of 0.4. It plans to raise its debt-to-equity ratio to 0.75. The tax rate is 28%, the risk-free rate 2.5% and the market risk premium 4.5%. Find the cost of equity re after the change.

βL = 1.2D/E1 = 0.4D/E2 = 0.75T = 28%rf = 2.5%rm − rf = 4.5%
re = %
9.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 59, 36, 40, 40, 29.

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

A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is below 67.5?

μ = 60σ = 5x = 67.5
P =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4P(six) = 1/6 on each roll

A fair die is rolled 4 times. What is the probability P of getting at least one six?

n = 4
P =
12.ProbabilityHard
n = 6, p = 0.2: P(X ≤ 1) = ?k0123456

6 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 6p = 0.2
P =
13.Normal distributionMedium
360 g365.4 gT = X₁ + … + X₉?

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 9 of them is below 365.4 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 9< 365.4 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 = 11, s = 19LUx̄ = 91.5E = ?95%

A sample of 11 items has mean 91.5 and standard deviation s = 19. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 11x̄ = 91.5s = 1995%t(10) = 2.228
E =
16.Confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 299%

The population standard deviation is 8. 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 = 8E = 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 −8.03 and the standard deviation of the differences is 5.9. Find the paired test statistic t for H₀: μd = 0.

n = 7d̄ = −8.03sd = 5.9
t =
18.Hypothesis tests & confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 2.599%

How large must a sample be so that the 99% 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 = 2.599%
n =
19.Linear regression & correlationMedium
xy1573188821922412027131xy15202580100120

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 105Σy = 504Σx² = 2295Σy² = 53098Σxy = 11028
r =
20.Linear regression & correlationHard
xy120220319415513xy241520x = 3.5

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. Predict ŷ for x = 3.5.

n = 5x = 3.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 8957 €

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.

Annuity factor

(1 − (1 + i)−n)/i = (1 − 0.582)/0.07 = 5.971

Present value

PV = 1 500·5.971 = €8 956.95

Answer

PV = €8 956.95
2.Time value of money

Answer: FV = 51741 €

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)19 − 1)/0.03 = 25.12

Annuity due

Each deposit is invested one year longer.

FV = 2 000·25.12·1.03 = €51 740.75

Answer

FV = €51 740.75
3.NPV & IRR

Answer: EAC = 4090 €

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 − 1.05−10)/0.05 = 7.722

Annualised purchase price

20 000/7.722 = 2 590.09

EAC

EAC = 2 590.09 + 1 500 = €4 090.09

Answer

EAC = €4 090.09
4.NPV & IRR

Answer: IRR = 35.82 %

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(35%) = 529.39
NPV(40%) = −2 536.44

Narrow down

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

IRR ≈ 35 + 5·529.39/(529.39 − (−2 536.44)) = 35.86%
exact: NPV(35.82%) = 0 → IRR = 35.82%

Answer

IRR = 35.82%
5.Bond pricing & yield

Answer: P = 847.5 €

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 = 0.04, N = 24

Present value of the coupons

€30.00·(1 − 0.3901)/0.04 = €457.41

Present value of the face value

€1 000·0.3901 = €390.12

Price

P = €847.53

Answer

P = €847.53

The yield is above the coupon rate, so the bond sells at a discount.

6.Bond pricing & yield

Answer: P = 1073 €

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 = 6.6/1.06 = 6.226

Estimated price change

%ΔP ≈ −6.226·(−0.01) = 6.226%
P ≈ 1010·(1 + 0.06226) = €1 072.89

Answer

P = €1 072.89

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

7.CAPM & WACC

Answer: WACC = 11.55 %

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 + 2 000 000 + 2 000 000 = €14 000 000
E/V = 0.7143, Ps/V = 0.1429, D/V = 0.1429

WACC

WACC = 0.7143·14% + 0.1429·7.5% + 0.1429·4.5%·(1 − 0.25) = 11.55%

Answer

WACC = 11.55%
8.CAPM & WACC

Answer: re = 8.957 %

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 + 0.72·0.4) = 0.9317

Relever at the new D/E

βL = 0.9317·(1 + 0.72·0.75) = 1.435

Cost of equity (CAPM)

re = 2.5% + 1.435·4.5% = 8.957%

Answer

re = 8.957%
9.Descriptive statistics

Answer: s = 11.12

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 = 204/5 = 40.8

Squared deviations

Σ(x − x̄)² = 18.2² + (−4.8)² + (−0.8)² + (−0.8)² + (−11.8)² = 494.8

Standard deviation

s = √(494.8/4) = 11.12

Answer

s = 11.12
10.Descriptive statistics

Answer: P = 0.9332

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 0.9332

Answer

P = 0.9332
11.Probability

Answer: P = 0.5177

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)4 = 0.4823
P = 1 − 0.4823 = 0.5177

Answer

P = 0.5177
12.Probability

Answer: P = 0.6554

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) = 0.86 = 0.2621
P(X = 1) = 6·0.2·0.85 = 0.3932

Add them

P = 0.2621 + 0.3932 = 0.6554

Answer

P = 0.6554
13.Normal distribution

Answer: P = 0.9332

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 = 360 g, σT = 1.2·√9 = 3.6 g

Standardise

z = (365.4 − 360)/3.6 = 1.5

Probability

P = Φ(1.5) = 0.9332

Answer

P = 0.9332
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 = 12.76

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 = 10, t = 2.228

Margin of error

E = 2.228·19/√11 = 12.76

Answer

E = 12.76
16.Confidence intervals

Answer: n = 107

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 = (2.576·8/2)² = 106.2

Round up

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

n = 107

Answer

n = 107
17.Hypothesis tests & confidence intervals

Answer: t = −3.601

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 = 5.9/√7 = 2.23

Test statistic

t = −8.03/2.23 = −3.601

Answer

t = −3.601

With df = 6, the two-sided 5% critical value is 2.447; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: n = 68

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 = (2.576·8/2.5)² = 67.95

Round up

n = 68

Answer

n = 68
19.Linear regression & correlation

Answer: r = 0.977

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 = 11028 − 105·504/5 = 444
Sxx = 2295 − 105²/5 = 90
Syy = 53098 − 504²/5 = 2295

Correlation coefficient

r = 444/√(90·2295) = 0.977

Answer

r = 0.977
20.Linear regression & correlation

Answer: ŷ = 16.45

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 = 87/5 = 17.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −19

Regression line

b = −19/10 = −1.9
a = 17.4 − (−1.9)·3 = 23.1
ŷ = 23.1 − 1.9·x

Prediction

ŷ = 23.1 + (−1.9)·3.5 = 16.45

Answer

ŷ = 16.45

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