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 = ?€20 000+3%+3%…r = 6%

A property is expected to earn net rent of €20 000 next year. The rent is expected to grow by 3% per year forever. The required return is 6% per year. Find the value PV of the property today.

C1 = €20 000g = 3%r = 6%
PV = €
2.Time value of moneyHard
months01…108…300€300 000AB = ?A4.2% / 12

A mortgage of €300 000 is repaid with equal monthly payments over 25 years at 4.2% per year, compounded monthly. How much is still owed just after the payment at the end of year 9? Find the outstanding balance B.

P = €300 0004.2% / 12N = 300k = 108
B = €
3.NPV & IRRMedium
years0123−40 00014 00014 00022 000

A machine costs €40 000 today. It generates €14 000 at the end of each year for 3 years and can then be sold for €8 000. The cost of capital is 12%. Find the net present value NPV.

C0 = €40 000CF = €14 000S = €8 000n = 3r = 12%
NPV = €
4.NPV & IRRHard
years0123−50 00034 00028 50029 000

A project costs €50 000 and returns €34 000, €28 500, €29 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 34 000CF2 = 28 500CF3 = 29 000
IRR = %
5.Bond pricing & yieldMedium
half-years0123…24€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 12 years. The yield to maturity is 3.5% per year. Find the price P of the bond.

F = €1 000coupon = 7%YTM = 3.5%n = 12
P = €
6.Bond pricing & yieldHard
years0123…10€1 080€50€50 + €1 000

A bond with a face value of €1 000 and an annual coupon of 5% matures in 10 years and sells for €1 080. Find its yield to maturity YTM in percent.

F = €1 000coupon = 5%n = 10P = €1 080
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€2 000 000re​ = 13%debt€6 000 000rd​ = 6%T = 20%

A company has equity worth €2 000 000 and debt worth €6 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 6% and the tax rate is 20%. Find the WACC.

E = €2 000 000D = €6 000 000re = 13%rd = 6%T = 20%
WACC = %
8.CAPM & WACCHard
rf​ = 2%, rm​ − rf​ = 5%no debtequityβU​ = 0.7D/E = 1equityre​ = ?debtT = 21%

A company's business has an unlevered (asset) beta of 0.7. It plans a debt-to-equity ratio of 1. The tax rate is 21%, the risk-free rate is 2% and the market risk premium is 5%. Find the cost of equity re at the new capital structure.

βU = 0.7D/E = 1T = 21%rf = 2%rm − rf = 5%
re = %
9.Descriptive statisticsMedium
152025303540

Find the sample standard deviation s of the data: 39, 42, 23, 13, 16, 37, 40.

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

Exam scores are normally distributed with mean 60 and standard deviation 5. Find the z-score z of a score of 50.

μ = 60σ = 5x = 50
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 = 12, p = 0.1: P(X = 2) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 2 correct answers?

n = 12p = 0.1k = 2
P =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9051

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9051. Find z to two decimals.

A = 0.9051
z =
14.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
15.Confidence intervalsMedium
n = 7, s = 2.5LU = ?x̄ = 64.5E95%

A sample of 7 items has mean 64.5 and standard deviation s = 2.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 7x̄ = 64.5s = 2.595%t(6) = 2.447
upper limit =
16.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0195%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.01?

p = 0.5E = 0.0195%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 498.6 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 498.6s = 5n = 81
t =
18.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 0.590%
n =
19.Linear regression & correlationMedium
xy21041562082810351245xy5102040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy2104186258311042xy51010203040ŷ = 14

The table gives 5 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 = 14?

n = 5ŷ = 14
x =

3Solutions

1.Time value of money

Answer: PV = 666667 €

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) = 20 000/(0.06 − 0.03) = 20 000/0.03 = €666 666.67

Answer

PV = €666 666.67
2.Time value of money

Answer: B = 225762 €

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.

Monthly payment

i = 0.0035, N = 300
A = 300 000·0.0035/(1 − (1 + 0.0035)−300) = €1 616.83

Present value of the remaining payments

Whatever is still owed must be exactly paid off by the remaining payments.

N − k = 300 − 108 = 192
B = 1617·(1 − (1 + 0.0035)−192)/0.0035 = €225 761.84

Answer

B = €225 761.84
3.NPV & IRR

Answer: NPV = −680.1 €

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 the yearly cash flows

14 000·2.402 = 33 625.64

Present value of the salvage value

8 000/1.405 = 5 694.24

NPV

NPV = −40 000 + 33 625.64 + 5 694.24 = −€680.12

Answer

NPV = −€680.12
4.NPV & IRR

Answer: IRR = 39.02 %

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%) = 2 609.87
NPV(40%) = −604.96

Narrow down

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

IRR ≈ 35 + 5·2 609.87/(2 609.87 − (−604.96)) = 39.06%
exact: NPV(39.02%) = 0 → IRR = 39.02%

Answer

IRR = 39.02%
5.Bond pricing & yield

Answer: P = 1341 €

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 = 0.0175, N = 24

Present value of the coupons

€35.00·(1 − 0.6594)/0.0175 = €681.12

Present value of the face value

€1 000·0.6594 = €659.44

Price

P = €1 340.56

Answer

P = €1 340.56

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

6.Bond pricing & yield

Answer: YTM = 4.013 %

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

€1 080 = €50·(1 − (1 + y)⁻¹⁰)/y + €1 000/(1 + y)¹⁰

Solve numerically

P(4%) = €1 081.11, P(5%) = €1 000.00
YTM = 4.013%

Answer

YTM = 4.013%
7.CAPM & WACC

Answer: WACC = 6.85 %

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 = 0.25, D/V = 0.75

WACC

WACC = 0.25·13% + 0.75·6%·(1 − 0.2) = 6.85%

Answer

WACC = 6.85%
8.CAPM & WACC

Answer: re = 8.265 %

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 = 0.7·(1 + 0.79·1) = 1.253

Cost of equity (CAPM)

re = 2% + 1.253·5% = 8.265%

Answer

re = 8.265%
9.Descriptive statistics

Answer: s = 12.3

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 = 210/7 = 30

Squared deviations

Σ(x − x̄)² = 9² + 12² + (−7)² + (−17)² + (−14)² + 7² + 10² = 908

Standard deviation

s = √(908/6) = 12.3

Answer

s = 12.3
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 = (50 − 60)/5 = −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 = 0.2301

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 = 2) = 66·0.12·0.910 = 66·0.01·0.3487 = 0.2301

Answer

P = 0.2301
13.Normal distribution

Answer: z = 1.67

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 + 0.9051)/2 = 0.9525

Read the z-table backwards

Find 0.9525 inside the table and read off z.

z = 1.67

Answer

z = 1.67
14.Normal distribution

Answer: P = 0.8413

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 difference

μ = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
15.Confidence intervals

Answer: upper limit = 66.81

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 = 6, t = 2.447

Margin of error

E = 2.447·2.5/√7 = 2.312

Interval

64.5 ± 2.312 → [62.19, 66.81]

Answer

upper limit = 66.81
16.Confidence intervals

Answer: n = 9604

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 = 1.96²·0.5·0.5/0.01² = 9604

Round up

n = 9604

Answer

n = 9604
17.Hypothesis tests & confidence intervals

Answer: t = −2.52

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).

Standard error

SE = 5/√81 = 5/9 = 0.5556

Test statistic

t = (498.6 − 500)/0.5556 = −2.52

Answer

t = −2.52
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 1559

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 = (1.645·12/0.5)² = 1559

Round up

n = 1559

Answer

n = 1559
19.Linear regression & correlation

Answer: r² = 0.9837

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 = 153/6 = 25.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 243
Syy = Σ(y − ȳ)² = 857.5

Coefficient of determination

r² = 243²/(70·857.5) = 0.9837

Answer

r² = 0.9837

About 98.37% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: x = 3.091

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 = 30/5 = 6
ȳ = 126/5 = 25.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 154

Regression line

b = 154/40 = 3.85
a = 25.2 − 3.85·6 = 2.1
ŷ = 2.1 + 3.85·x

Solve for x

x = (14 − 2.1)/3.85 = 3.091

Answer

x = 3.091

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