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…19PV = ?€2 000€2 000€2 000€2 000

An annuity pays €2 000 at the end of each year for 19 years. The interest rate is 7% per year. Find the present value PV of the annuity.

A = €2 000i = 7%n = 19
PV = €
2.Time value of moneyHard
months0123…192€300€300€300FV = ?

You save €300 at the end of every month for 16 years. The account pays 4.2% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €3004.2% / 12N = 192
FV = €
3.NPV & IRRMedium
years01234€60 000€1 500€1 500€1 500€1 500r = 12%, EAC = ?

A machine costs €60 000 today, lasts 4 years and costs €1 500 per year to run (paid at the end of each year). The cost of capital is 12%. Find its equivalent annual cost EAC.

C0 = €60 000M = €1 500n = 4r = 12%
EAC = €
4.NPV & IRRHard
years012−50 00033 00032 000

A project costs €50 000 and returns €33 000, €32 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 33 000CF2 = 32 000
IRR = %
5.Bond pricing & yieldMedium
years0123…12€734C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 12 years and pays annual coupons. It sells for €734 when the yield to maturity is 5%. Find the coupon rate c in percent.

F = €1 000n = 12P = €734YTM = 5%
c = %
6.Bond pricing & yieldHard
years012345€960€60€60 + €1 000

A bond with a face value of €1 000 and an annual coupon of 6% matures in 5 years and sells for €960. Find its yield to maturity YTM in percent.

F = €1 000coupon = 6%n = 5P = €960
YTM = %
7.CAPM & WACCMedium
WACC = 8.41%equity€4 000 000re​ = ?debt€2 000 000rd​ = 4.5%T = 28%

A company has equity worth €4 000 000 and debt worth €2 000 000. Its WACC is 8.41%, its pre-tax cost of debt is 4.5% and the tax rate is 28%. Find the cost of equity re.

E = €4 000 000D = €2 000 000WACC = 8.41%rd = 4.5%T = 28%
re = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €55.00β = 0.9rf​ = 4%, rm​ = 11%debt40 000 × €960rd​ = 4.5%T = 25%

A company has 1 000 000 shares at €55.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 4.5% and the tax rate is 25%. The share’s beta is 0.9, the risk-free rate 4% and the market return 11%. Find the WACC.

β = 0.9rf = 4%rm = 11%rd = 4.5%T = 25%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 35, 24, 17, 52, 13, 59, 21.

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 65.

μ = 60σ = 5x = 65
z =
11.ProbabilityMedium
xp(x)00.3510.320.35μ = 1

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2p(x): 0.35, 0.3, 0.35
σ =
12.ProbabilityHard
xp(x)100.25200.3300.15400.3μ

X has the distribution shown, and Y = 5X − 3. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.25, 0.3, 0.15, 0.3Y = 5X − 3
σY =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.4039

Drag z so that the shaded area between −z and z under the standard normal curve is 0.4039. Find z to two decimals.

A = 0.4039
z =
14.Normal distributionHard
557085100115130145pointstop 25%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 25% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points25%
x = points
15.Confidence intervalsMedium
n = 8, s = 10LU = ?x̄ = 112.5E95%

A sample of 8 items has mean 112.5 and standard deviation s = 10. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 8x̄ = 112.5s = 1095%t(7) = 2.365
upper limit =
16.Confidence intervalsHard
p ≈ 0.6, n = ?p̂ − Ep̂ + Ep̂E = 0.0590%

A pilot study suggests that about 0.6 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.05?

p ≈ 0.6E = 0.0590%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 200.8 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 200.8s = 4n = 81
t =
18.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 295%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 295%
n =
19.Linear regression & correlationMedium
xy2104136218281032xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy130228324426525623xy2462530ŷ = 24

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. For which value x does the line predict ŷ = 24?

n = 6ŷ = 24
x =

3Solutions

1.Time value of money

Answer: PV = 20671 €

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 − 0.2765)/0.07 = 10.34

Present value

PV = 2 000·10.34 = €20 671.19

Answer

PV = €20 671.19
2.Time value of money

Answer: FV = 81930 €

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 rate and number of deposits

i = 0.042/12 = 0.0035
N = 12·16 = 192

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 300·(1.956 − 1)/0.0035 = €81 930.31

Answer

FV = €81 930.31

You paid in €57 600 yourself; the rest is interest.

3.NPV & IRR

Answer: EAC = 21254 €

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 − 1.12−4)/0.12 = 3.037

Annualised purchase price

60 000/3.037 = 19 754.07

EAC

EAC = 19 754.07 + 1 500 = €21 254.07

Answer

EAC = €21 254.07
4.NPV & IRR

Answer: IRR = 19.54 %

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%) = 2 892.25
NPV(20%) = −277.78

Narrow down

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

IRR ≈ 15 + 5·2 892.25/(2 892.25 − (−277.78)) = 19.56%
exact: NPV(19.54%) = 0 → IRR = 19.54%

Answer

IRR = 19.54%
5.Bond pricing & yield

Answer: c = 1.999 %

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.

Present value of the face value

1000·(1 + 0.05)−12 = €556.84

Coupon

C = (734 − 556.84)/8.863 = €19.99

Coupon rate

c = 19.99/1000 = 1.999%

Answer

c = 1.999%
6.Bond pricing & yield

Answer: YTM = 6.975 %

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

€960 = €60·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵

Solve numerically

P(6%) = €1 000.00, P(7%) = €959.00
YTM = 6.975%

Answer

YTM = 6.975%
7.CAPM & WACC

Answer: re = 11 %

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 = €6 000 000, E/V = 0.6667, D/V = 0.3333

Debt part of the WACC

0.3333·4.5%·(1 − 0.28) = 1.08%

Cost of equity

re = (8.41% − 1.08%)/0.6667 = 11%

Answer

re = 11%
8.CAPM & WACC

Answer: WACC = 7.453 %

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 values

E = 1 000 000·€55.00 = €55 000 000
D = 40 000·€960 = €38 400 000

Cost of equity (CAPM)

re = 4% + 0.9·7% = 10.3%

Weights

V = €93 400 000, E/V = 0.5889, D/V = 0.4111

WACC

WACC = 0.5889·10.3% + 0.4111·4.5%·(1 − 0.25) = 7.453%

Answer

WACC = 7.453%
9.Descriptive statistics

Answer: s = 17.83

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 = 221/7 = 31.57

Squared deviations

Σ(x − x̄)² = 3.429² + (−7.571)² + (−14.57)² + 20.43² + (−18.57)² + 27.43² + (−10.57)² = 1908

Standard deviation

s = √(1908/6) = 17.83

Answer

s = 17.83
10.Descriptive statistics

Answer: z = 1

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 = (65 − 60)/5 = 1

Answer

z = 1
11.Probability

Answer: σ = 0.8367

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 = 0·0.35 + 1·0.3 + 2·0.35 = 1

Mean of the squares

E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7

Variance

Var(X) = 1.7 − 1² = 0.7
σ = √0.7 = 0.8367

Answer

σ = 0.8367
12.Probability

Answer: σY = 58.09

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.

Spread of X

μ = 25, Var(X) = 135, σX = 11.62

Linear change

σY = |5|·11.62 = 58.09

Answer

σY = 58.09

The mean does move: E(Y) = 5·25 − 3 = 122.

13.Normal distribution

Answer: z = 0.53

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 + 0.4039)/2 = 0.7019

Read the z-table backwards

Find 0.7019 inside the table and read off z.

z = 0.53

Answer

z = 0.53
14.Normal distribution

Answer: x = 110.1 points

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) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 100 + 0.67·15 = 110.1 points

Answer

x = 110.1 points
15.Confidence intervals

Answer: upper limit = 120.9

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 = 2.365

Margin of error

E = 2.365·10/√8 = 8.362

Interval

112.5 ± 8.362 → [104.1, 120.9]

Answer

upper limit = 120.9
16.Confidence intervals

Answer: n = 260

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 = 1.645²·0.6·0.4/0.05² = 259.8

Round up

n = 260

Answer

n = 260
17.Hypothesis tests & confidence intervals

Answer: t = 1.8

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 = 4/√81 = 4/9 = 0.4444

Test statistic

t = (200.8 − 200)/0.4444 = 1.8

Answer

t = 1.8
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 139

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 = (1.96·12/2)² = 138.3

Round up

n = 139

Answer

n = 139
19.Linear regression & correlation

Answer: r² = 0.9811

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 = 104/5 = 20.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 118
Syy = Σ(y − ȳ)² = 354.8

Coefficient of determination

r² = 118²/(40·354.8) = 0.9811

Answer

r² = 0.9811

About 98.11% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: x = 5.167

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 = 156/6 = 26

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −21

Regression line

b = −21/17.5 = −1.2
a = 26 − (−1.2)·3.5 = 30.2
ŷ = 30.2 − 1.2·x

Solve for x

x = (24 − 30.2)/(−1.2) = 5.167

Answer

x = 5.167

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