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
years01234€20 000FV = ?

You invest €20 000 at 6% per year with continuous compounding. Find the value FV after 4 years.

PV = €20 000r = 6%t = 4
FV = €
2.Time value of moneyHard
months0123…180A?AA€30 000

You want €30 000 in 15 years. You save the same amount at the end of every month in an account paying 4.8% per year, compounded monthly. Find the monthly deposit A.

FV = €30 0004.8% / 12N = 180
A = €
3.NPV & IRRMedium
years0123−60 00019 00027 00015 000

An investment of €60 000 gives these cash flows at the end of years 1 to 3: €19 000, €27 000, €15 000. The cost of capital is 9%. Find the net present value NPV.

r = 9%CF0 = −60 000CF1 = 19 000CF2 = 27 000CF3 = 15 000
NPV = €
4.NPV & IRRHard
years012345−100 00022 000+3%+3%+3%+3%r = 10%, NPV = ?

A project costs €100 000 today. Its cash flow is €22 000 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 5. The cost of capital is 10%. Find the net present value NPV.

C0 = €100 000CF1 = €22 000g = 3%n = 5r = 10%
NPV = €
5.Bond pricing & yieldMedium
half-years0123…14€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 7 years. The yield to maturity is 4% per year. Find the price P of the bond.

F = €1 000coupon = 7%YTM = 4%n = 7
P = €
6.Bond pricing & yieldHard
years0123€940€80€80 + €1 000

A bond with a face value of €1 000 and an annual coupon of 8% matures in 3 years and sells for €940. Find its yield to maturity YTM in percent.

F = €1 000coupon = 8%n = 3P = €940
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€3 000 000re​ = 9%debt€3 000 000rd​ = 4%T = 21%

A company has equity worth €3 000 000 and debt worth €3 000 000. Its cost of equity is 9%, its pre-tax cost of debt is 4% and the tax rate is 21%. Find the WACC.

E = €3 000 000D = €3 000 000re = 9%rd = 4%T = 21%
WACC = %
8.CAPM & WACCHard
WACC = ?equity4 000 000 × €52.00β = 1.3rf​ = 2.5%, rm​ = 7.5%debt40 000 × €960rd​ = 7%T = 28%

A company has 4 000 000 shares at €52.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 7% and the tax rate is 28%. The share’s beta is 1.3, the risk-free rate 2.5% and the market return 7.5%. Find the WACC.

β = 1.3rf = 2.5%rm = 7.5%rd = 7%T = 28%
WACC = %
9.Descriptive statisticsMedium
3035404550

Find the sample standard deviation s of the data: 44, 41, 27, 36, 52.

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

Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 60.4.

μ = 70σ = 8x = 60.4
z =
11.ProbabilityMedium
P(at least one six) = ??1?2?3P(six) = 1/6 on each roll

A fair die is rolled 3 times. What is the probability P of getting at least one six?

n = 3
P =
12.ProbabilityHard
€p(x)−50.79100.22450.01

A game costs €5 to play. You win €15 with probability 0.2 and €250 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€15: p = 0.2€250: p = 0.01
E(G) = €
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.3328

Drag z so that the shaded area between −z and z under the standard normal curve is 0.3328. Find z to two decimals.

A = 0.3328
z =
14.Normal distributionHard
μ = 65x = ?10%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 10% of them lie (the 10th percentile).

μ = 65σ = 10P = 10%
x = points
15.Confidence intervalsMedium
n = 21, s = 16.5LUx̄ = 115.5E = ?90%

A sample of 21 items has mean 115.5 and standard deviation s = 16.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 21x̄ = 115.5s = 16.590%t(20) = 1.725
E =
16.Confidence intervalsHard
σ = 2, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

The population standard deviation is 2. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 0.5?

σ = 2E = 0.590%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 102.5 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 102.5s = 12n = 100
t =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2.5α = 0.05βσ = 6, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 2.5 units with a power of 95%. The standard deviation is 6. Find the minimum sample size n.

α = 0.05power = 95%δ = 2.5σ = 6zα = 1.645zβ = 1.645
n =
19.Linear regression & correlationMedium
xy1887208222912410126114xy20258090100110

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy25447167688710102xy5106080100x = 9

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 9.

n = 5x = 9
ŷ =

3Solutions

1.Time value of money

Answer: FV = 25425 €

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.

Exponent

r·t = 0.06·4 = 0.24
e0.24 = 1.271

Future value

FV = 20 000·1.271 = €25 424.98

Answer

FV = €25 424.98
2.Time value of money

Answer: A = 114.1 €

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 = 0.048/12 = 0.004
N = 12·15 = 180

Deposit

A = FV·i/((1 + i)N − 1) = 30 000·0.004/(2.051 − 1) = €114.12

Answer

A = €114.12
3.NPV & IRR

Answer: NPV = −8261 €

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 + 0.09)1 = 19 000/1.09 = 17 431.19
CF2/(1 + 0.09)2 = 27 000/1.188 = 22 725.36
CF3/(1 + 0.09)3 = 15 000/1.295 = 11 582.75

Add them up

NPV = €−8 260.70

Answer

NPV = €−8 260.70

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

4.NPV & IRR

Answer: NPV = −11943 €

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 growing annuity

((1.03)/(1.1))5 = 0.7198
PV = 22 000/(0.1 − 0.03)·(1 − 0.7198) = 88 057.20

NPV

NPV = 88 057.20 − 100 000 = −€11 942.80

Answer

NPV = −€11 942.80

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

5.Bond pricing & yield

Answer: P = 1182 €

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 = 0.02, N = 14

Present value of the coupons

€35.00·(1 − 0.7579)/0.02 = €423.72

Present value of the face value

€1 000·0.7579 = €757.88

Price

P = €1 181.59

Answer

P = €1 181.59

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

6.Bond pricing & yield

Answer: YTM = 10.43 %

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

€940 = €80·(1 − (1 + y)⁻³)/y + €1 000/(1 + y)³

Solve numerically

P(10%) = €950.26, P(11%) = €926.69
YTM = 10.43%

Answer

YTM = 10.43%
7.CAPM & WACC

Answer: WACC = 6.08 %

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 = 0.5, D/V = 0.5

WACC

WACC = 0.5·9% + 0.5·4%·(1 − 0.21) = 6.08%

Answer

WACC = 6.08%
8.CAPM & WACC

Answer: WACC = 8.383 %

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 = 4 000 000·€52.00 = €208 000 000
D = 40 000·€960 = €38 400 000

Cost of equity (CAPM)

re = 2.5% + 1.3·5% = 9%

Weights

V = €246 400 000, E/V = 0.8442, D/V = 0.1558

WACC

WACC = 0.8442·9% + 0.1558·7%·(1 − 0.28) = 8.383%

Answer

WACC = 8.383%
9.Descriptive statistics

Answer: s = 9.301

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 = 200/5 = 40

Squared deviations

Σ(x − x̄)² = 4² + 1² + (−13)² + (−4)² + 12² = 346

Standard deviation

s = √(346/4) = 9.301

Answer

s = 9.301
10.Descriptive statistics

Answer: z = −1.2

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 = (60.4 − 70)/8 = −1.2

Answer

z = −1.2
11.Probability

Answer: P = 0.4213

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)3 = 0.5787
P = 1 − 0.5787 = 0.4213

Answer

P = 0.4213
12.Probability

Answer: E(G) = 0.5 €

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: 10 (p = 0.2), 245 (p = 0.01), −5 (p = 0.79)

Expected value

E(G) = 10·0.2 + 245·0.01 − 5·0.79 = 0.5 €

Answer

E(G) = 0.5 €

A positive value means you win money on average.

13.Normal distribution

Answer: z = 0.43

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 + 0.3328)/2 = 0.6664

Read the z-table backwards

Find 0.6664 inside the table and read off z.

z = 0.43

Answer

z = 0.43
14.Normal distribution

Answer: x = 52.18 points

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) = 0.1 → z = −1.282

Back to x

x = 65 + −1.282·10 = 52.18 points

Answer

x = 52.18 points
15.Confidence intervals

Answer: E = 6.211

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 = 20, t = 1.725

Margin of error

E = 1.725·16.5/√21 = 6.211

Answer

E = 6.211
16.Confidence intervals

Answer: n = 44

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 = (1.645·2/0.5)² = 43.3

Round up

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

n = 44

Answer

n = 44
17.Hypothesis tests & confidence intervals

Answer: t = 2.083

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 = 12/√100 = 12/10 = 1.2

Test statistic

t = (102.5 − 100)/1.2 = 2.083

Answer

t = 2.083
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 63

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 1.645)·6/2.5)² = 62.35

Round up

n = 63

Answer

n = 63
19.Linear regression & correlation

Answer: r² = 0.8249

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 = 110/5 = 22
ȳ = 475/5 = 95

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 146
Syy = Σ(y − ȳ)² = 646

Coefficient of determination

r² = 146²/(40·646) = 0.8249

Answer

r² = 0.8249

About 82.49% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 94.8

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 = 390/5 = 78

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 224

Regression line

b = 224/40 = 5.6
a = 78 − 5.6·6 = 44.4
ŷ = 44.4 + 5.6·x

Prediction

ŷ = 44.4 + 5.6·9 = 94.8

Answer

ŷ = 94.8

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