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…9PV = ?€10 000+4%+4%…r = 10%

A property is expected to earn net rent of €10 000 next year. The rent is expected to grow by 4% per year forever. The required return is 10% per year. Find the value PV of the property today.

C1 = €10 000g = 4%r = 10%
PV = €
2.Time value of moneyHard
months0123…N = ?€8 000€200€200€200€2004.2% / 12

You borrow €8 000 at 4.2% per year, compounded monthly, and repay €200 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 = €8 000A = €2004.2% / 12
N = payments
3.NPV & IRRMedium
years01234−60 00023 00020 00023 00018 000

An investment of €60 000 gives these cash flows at the end of years 1 to 4: €23 000, €20 000, €23 000, €18 000. The cost of capital is 12%. Find the net present value NPV.

r = 12%CF0 = −60 000CF1 = 23 000CF2 = 20 000CF3 = 23 000CF4 = 18 000
NPV = €
4.NPV & IRRHard
years0123…10−80 00011 00011 00011 00011 000

An investment of €80 000 returns €11 000 at the end of each year for 10 years. Find its internal rate of return IRR in percent.

C0 = €80 000CF = €11 000n = 10
IRR = %
5.Bond pricing & yieldMedium
half-years0123…18€35€35€35 + €1 000P = ?

A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 9 years. The yield to maturity is 9% per year. Find the price P of the bond.

F = €1 000coupon = 7%YTM = 9%n = 9
P = €
6.Bond pricing & yieldHard
years01234€1 299€80€80€80 + €1 080

A bond with 20 years to maturity (face value €1 000, annual coupon 8%) sells for €1 299. The issuer can call it in 4 years at a call price of €1 080. Find the yield to call YTC in percent.

F = €1 000coupon = 8%P = €1 299call price = €1 080call in 4 years
YTC = %
7.CAPM & WACCMedium
βr (%)SMLrf​ = 1%β = 1rm​ = ?β = 1.4re​ = 7.3%

A share with a beta of 1.4 has a required return of 7.3%. The risk-free rate is 1%. What expected market return rm does CAPM imply?

β = 1.4re = 7.3%rf = 1%
rm = %
8.CAPM & WACCHard
rf​ = 3%, rm​ − rf​ = 5%D/E = 0.5equityβL​ = 1.2debtD/E = 1.25equityre​ = ?debtT = 28%

A company's shares have a beta of 1.2 at its current debt-to-equity ratio of 0.5. It plans to raise its debt-to-equity ratio to 1.25. The tax rate is 28%, the risk-free rate 3% and the market risk premium 5%. Find the cost of equity re after the change.

βL = 1.2D/E1 = 0.5D/E2 = 1.25T = 28%rf = 3%rm − rf = 5%
re = %
9.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 58, 18, 50, 46, 35, 22.

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

A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is above 40?

μ = 50σ = 5x = 40
P =
11.ProbabilityMedium
xp(x)00.2520.340.3560.1μ = 2.6

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

x: 0, 2, 4, 6p(x): 0.25, 0.3, 0.35, 0.1
Var(X) =
12.ProbabilityHard
n = 7, p = 0.7: P(X ≤ 1) = ?k01234567

7 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?

n = 7p = 0.7
P =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8789

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8789. Find z to two decimals.

A = 0.8789
z =
14.Normal distributionHard
μ = 250x = ?25%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 25% of them lie (the 25th percentile).

μ = 250σ = 5P = 25%
x = g
15.Confidence intervalsMedium
n = 25, s = 17L = ?Ux̄ = 59E99%

A sample of 25 items has mean 59 and standard deviation s = 17. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 25x̄ = 59s = 1799%t(24) = 2.797
lower limit =
16.Confidence intervalsHard
n = 30σ = ?Us = 2.595%

The diameters of 30 machined shafts have a sample standard deviation of 2.5 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 2.5 µmχ² = 16.047, 45.722
σ = µm
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 50 parts has mean length 49 mm, and the line's known standard deviation is 17 mm. Line 2: a sample of 50 parts has mean 43.4 mm, known standard deviation 18 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 50x̄₁ = 49 mmσ₁ = 17 mmn₂ = 50x̄₂ = 43.4 mmσ₂ = 18 mm
z =
18.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 36 measurements has mean 66. The population standard deviation is 20. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 66σ = 20n = 3690%
L =
19.Linear regression & correlationMedium
xy163418412044225324532665xy2025405060

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy329428534633742843xy468303540x = 6.5

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (6.5, 42) is made. Find its residual e = y − ŷ.

n = 6(6.5, 42)
e =

3Solutions

1.Time value of money

Answer: PV = 166667 €

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.

Growing perpetuity

PV = C1/(r − g) = 10 000/(0.1 − 0.04) = 10 000/0.06 = €166 666.67

Answer

PV = €166 666.67
2.Time value of money

Answer: N = 44 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 = 0.0035
(1 + i)−N = 1 − 8 000·0.0035/200 = 0.86

Solve with logarithms

N = −ln(0.86)/ln(1.004) = 43.17

Round up to whole payments

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

N = 44

Answer

N = 44 payments
3.NPV & IRR

Answer: NPV = 4290 €

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 + 0.12)1 = 23 000/1.12 = 20 535.71
CF2/(1 + 0.12)2 = 20 000/1.254 = 15 943.88
CF3/(1 + 0.12)3 = 23 000/1.405 = 16 370.95
CF4/(1 + 0.12)4 = 18 000/1.574 = 11 439.33

Add them up

NPV = €4 289.86

Answer

NPV = €4 289.86

NPV is positive, so the project earns more than the required return.

4.NPV & IRR

Answer: IRR = 6.253 %

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)−10)/i = 80 000/11 000 = 7.273

Solve for the rate

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

i = 6%: 7.36
i = 7%: 7.024
IRR = 6.253%

Answer

IRR = 6.253%
5.Bond pricing & yield

Answer: P = 878.4 €

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 = €35.00, i = 0.045, N = 18

Present value of the coupons

€35.00·(1 − 0.4528)/0.045 = €425.60

Present value of the face value

€1 000·0.4528 = €452.80

Price

P = €878.40

Answer

P = €878.40

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

6.Bond pricing & yield

Answer: YTC = 2.073 %

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

1299 = 80·(1 − (1 + y)−4)/y + 1080/(1 + y)4

Solve numerically

y = 2%: €1 302.37
y = 3%: €1 256.93
YTC = 2.073%

Answer

YTC = 2.073%

The yield to maturity is 5.498%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.

7.CAPM & WACC

Answer: rm = 5.5 %

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 risk premium

rm − rf = (7.3% − 1%)/1.4 = 4.5%

Market return

rm = 1% + 4.5% = 5.5%

Answer

rm = 5.5%
8.CAPM & WACC

Answer: re = 11.38 %

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 + 0.72·0.5) = 0.8824

Relever at the new D/E

βL = 0.8824·(1 + 0.72·1.25) = 1.676

Cost of equity (CAPM)

re = 3% + 1.676·5% = 11.38%

Answer

re = 11.38%
9.Descriptive statistics

Answer: s = 15.96

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 = 229/6 = 38.17

Squared deviations

Σ(x − x̄)² = 19.83² + (−20.17)² + 11.83² + 7.833² + (−3.167)² + (−16.17)² = 1273

Standard deviation

s = √(1273/5) = 15.96

Answer

s = 15.96
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) = 0.02275
P = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
11.Probability

Answer: Var(X) = 3.64

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 = 0·0.25 + 2·0.3 + 4·0.35 + 6·0.1 = 2.6

Mean of the squares

E(X²) = 0·0.25 + 4·0.3 + 16·0.35 + 36·0.1 = 10.4

Variance

Var(X) = 10.4 − 2.6² = 3.64

Answer

Var(X) = 3.64
12.Probability

Answer: P = 0.003791

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) = 0.37 = 0.0002187
P(X = 1) = 7·0.7·0.36 = 0.003572

Add them

P = 0.0002187 + 0.003572 = 0.003791

Answer

P = 0.003791
13.Normal distribution

Answer: z = 1.55

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 + 0.8789)/2 = 0.9394

Read the z-table backwards

Find 0.9394 inside the table and read off z.

z = 1.55

Answer

z = 1.55
14.Normal distribution

Answer: x = 246.6 g

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) = 0.25 → z = −0.6745

Back to x

x = 250 + −0.6745·5 = 246.63 g

Answer

x = 246.63 g
15.Confidence intervals

Answer: lower limit = 49.49

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 = 24, t = 2.797

Margin of error

E = 2.797·17/√25 = 9.51

Interval

59 ± 9.51 → [49.49, 68.51]

Answer

lower limit = 49.49
16.Confidence intervals

Answer: σ = 1.991 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 29·2.5² = 181.3

Limits for σ²

181.3/45.722 = 3.964
181.3/16.047 = 11.29

Limit for σ

σ = √3.964 = 1.991 µm

Answer

σ = 1.991 µm

95% interval for σ: [1.991, 3.361] µm.

17.Hypothesis tests & confidence intervals

Answer: z = 1.599

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.

Standard error of the difference

SE = √(17²/50 + 18²/50) = √12.26 = 3.501

Test statistic

z = (49 − 43.4)/3.501 = 1.599

Answer

z = 1.599

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: L = 60.52

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).

Margin of error

E = 1.645·20/√36 = 5.483

Interval limit

L = 66 − 5.483 = 60.52

Answer

L = 60.52
19.Linear regression & correlation

Answer: r = 0.9764

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 = 126/6 = 21
ȳ = 290/6 = 48.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 200
Syy = Σ(y − ȳ)² = 599.3

Correlation coefficient

r = 200/√(70·599.3) = 0.9764

Answer

r = 0.9764
20.Linear regression & correlation

Answer: e = 3.995

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 = 33/6 = 5.5
ȳ = 209/6 = 34.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 55.5

Regression line

b = 55.5/17.5 = 3.171
a = 34.83 − 3.171·5.5 = 17.39
ŷ = 17.39 + 3.171·x

Predicted value

ŷ = 17.39 + 3.171·6.5 = 38

Residual

e = 42 − 38 = 3.995

Answer

e = 3.995

The point lies above the line.

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