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…9€1 000FV = ?

You invest €1 000 at a nominal rate of 4% per year, compounded quarterly. Find the value FV of the investment after 9 years.

PV = €1 0004% / 4n = 9
FV = €
2.Time value of moneyHard
years0123…18€6 000€6 000€6 000FV = ?

You deposit €6 000 at the start of every year for 18 years, the first deposit today. The account pays 7% per year. How much is in the account at the end of year 18? Find FV.

A = €6 000i = 7%n = 18
FV = €
3.NPV & IRRMedium
years0123−75 00023 00037 00033 000

An investment of €75 000 gives these cash flows at the end of years 1 to 3: €23 000, €37 000, €33 000. The cost of capital is 10%. Find the net present value NPV.

r = 10%CF0 = −75 000CF1 = 23 000CF2 = 37 000CF3 = 33 000
NPV = €
4.NPV & IRRHard
years0123…10−50 0005 500+2%+2%+2%r = 12%, NPV = ?

A project costs €50 000 today. Its cash flow is €5 500 at the end of year 1 and grows by 2% per year until the last cash flow at the end of year 10. The cost of capital is 12%. Find the net present value NPV.

C0 = €50 000CF1 = €5 500g = 2%n = 10r = 12%
NPV = €
5.Bond pricing & yieldMedium
half-years0123…16€15€15€15 + €1 000P = ?

A bond with a face value of €1 000 pays a 3% annual coupon in two half-yearly payments and matures in 8 years. The yield to maturity is 6% per year. Find the price P of the bond.

F = €1 000coupon = 3%YTM = 6%n = 8
P = €
6.Bond pricing & yieldHard
years012345P€50€50€50 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 5 years. The yield to maturity is 7%. Find the Macaulay duration D in years.

F = €1 000coupon = 5%n = 5YTM = 7%
D = years
7.CAPM & WACCMedium
WACC = ?equity€4 000 000re​ = 10%debt€1 000 000rd​ = 3%T = 25.8%

A company has equity worth €4 000 000 and debt worth €1 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 3% and the tax rate is 25.8%. Find the WACC.

E = €4 000 000D = €1 000 000re = 10%rd = 3%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €53.50β = 1.3rf​ = 4%, rm​ = 9.5%debt40 000 × €960rd​ = 3%T = 25%

A company has 1 000 000 shares at €53.50 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 3% and the tax rate is 25%. The share’s beta is 1.3, the risk-free rate 4% and the market return 9.5%. Find the WACC.

β = 1.3rf = 4%rm = 9.5%rd = 3%T = 25%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 50, 17, 58, 18, 12, 26.

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

A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is above 190?

μ = 170σ = 10x = 190
P =
11.ProbabilityMedium
xp(x)00.310.3520.35μ = 1.05

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

x: 0, 1, 2p(x): 0.3, 0.35, 0.35
σ =
12.ProbabilityHard
n = 5, p = 0.1: P(X = 1) = ?k012345

A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 1 correct answers?

n = 5p = 0.1k = 1
P =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9127

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9127. Find z to two decimals.

A = 0.9127
z =
14.Normal distributionHard
557085100115130145pointstop 10%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 10% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points10%
x = points
15.Confidence intervalsMedium
n = 10, s = 17LUx̄ = 107.5E = ?90%

A sample of 10 items has mean 107.5 and standard deviation s = 17. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 10x̄ = 107.5s = 1790%t(9) = 1.833
E =
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
−2.2282.228t = ?df = 10

11 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 1.77 and the standard deviation of the differences is 4.2. Find the paired test statistic t for H₀: μd = 0.

n = 11d̄ = 1.77sd = 4.2
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 75 of 150 patients on treatment A recover, and 50 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 75n₁ = 150x₂ = 50n₂ = 80
z =
19.Linear regression & correlationMedium
xy254166238281035xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 107Σx² = 220Σy² = 2819Σxy = 786
r =
20.Linear regression & correlationHard
xy1654187120752283249626100xy20256080100x = 25

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 25.

n = 6x = 25
ŷ =

3Solutions

1.Time value of money

Answer: FV = 1431 €

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.

Rate and number of periods

i = r/m = 0.04/4 = 0.01
N = m·n = 4·9 = 36

Future value

FV = 1 000·(1 + 0.01)36 = 1 000·1.431 = €1 430.77

Answer

FV = €1 430.77
2.Time value of money

Answer: FV = 218274 €

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 + 0.07)18 − 1)/0.07 = 34

Annuity due

Each deposit is invested one year longer.

FV = 6 000·34·1.07 = €218 273.79

Answer

FV = €218 273.79
3.NPV & IRR

Answer: NPV = 1281 €

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 = −75 000
CF1/(1 + 0.1)1 = 23 000/1.1 = 20 909.09
CF2/(1 + 0.1)2 = 37 000/1.21 = 30 578.51
CF3/(1 + 0.1)3 = 33 000/1.331 = 24 793.39

Add them up

NPV = €1 280.99

Answer

NPV = €1 280.99

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

4.NPV & IRR

Answer: NPV = −16587 €

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

((1.02)/(1.12))10 = 0.3925
PV = 5 500/(0.12 − 0.02)·(1 − 0.3925) = 33 413.40

NPV

NPV = 33 413.40 − 50 000 = −€16 586.60

Answer

NPV = −€16 586.60

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

5.Bond pricing & yield

Answer: P = 811.6 €

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 = €15.00, i = 0.03, N = 16

Present value of the coupons

€15.00·(1 − 0.6232)/0.03 = €188.42

Present value of the face value

€1 000·0.6232 = €623.17

Price

P = €811.58

Answer

P = €811.58

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

6.Bond pricing & yield

Answer: D = 4.523 years

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.

Present value of each cash flow

t = 1: 50/1.07 = 46.73, t·PV = 46.73
t = 2: 50/1.145 = 43.67, t·PV = 87.34
t = 3: 50/1.225 = 40.81, t·PV = 122.44
t = 4: 50/1.311 = 38.14, t·PV = 152.58
t = 5: 1050/1.403 = 748.64, t·PV = 3 743.18

Duration

P = 918.00
D = 4 152.27/918.00 = 4.523

Answer

D = 4.523 years
7.CAPM & WACC

Answer: WACC = 8.445 %

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 = €5 000 000, E/V = 0.8, D/V = 0.2

WACC

WACC = 0.8·10% + 0.2·3%·(1 − 0.258) = 8.445%

Answer

WACC = 8.445%
8.CAPM & WACC

Answer: WACC = 7.431 %

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·€53.50 = €53 500 000
D = 40 000·€960 = €38 400 000

Cost of equity (CAPM)

re = 4% + 1.3·5.5% = 11.15%

Weights

V = €91 900 000, E/V = 0.5822, D/V = 0.4178

WACC

WACC = 0.5822·11.15% + 0.4178·3%·(1 − 0.25) = 7.431%

Answer

WACC = 7.431%
9.Descriptive statistics

Answer: s = 19.17

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 = 181/6 = 30.17

Squared deviations

Σ(x − x̄)² = 19.83² + (−13.17)² + 27.83² + (−12.17)² + (−18.17)² + (−4.167)² = 1837

Standard deviation

s = √(1837/5) = 19.17

Answer

s = 19.17
10.Descriptive statistics

Answer: P = 0.02275

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) = 0.9772
P = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
11.Probability

Answer: σ = 0.8047

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 = 0·0.3 + 1·0.35 + 2·0.35 = 1.05

Mean of the squares

E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75

Variance

Var(X) = 1.75 − 1.05² = 0.6475
σ = √0.6475 = 0.8047

Answer

σ = 0.8047
12.Probability

Answer: P = 0.3281

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 = 1) = 5·0.11·0.94 = 5·0.1·0.6561 = 0.3281

Answer

P = 0.3281
13.Normal distribution

Answer: z = 1.71

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 + 0.9127)/2 = 0.9564

Read the z-table backwards

Find 0.9564 inside the table and read off z.

z = 1.71

Answer

z = 1.71
14.Normal distribution

Answer: x = 119.2 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) = 0.9

z from the table

z ≈ 1.28

Back to the original units

x = 100 + 1.28·15 = 119.2 points

Answer

x = 119.2 points
15.Confidence intervals

Answer: E = 9.854

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 = 9, t = 1.833

Margin of error

E = 1.833·17/√10 = 9.854

Answer

E = 9.854
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 = 1.398

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 = 4.2/√11 = 1.266

Test statistic

t = 1.77/1.266 = 1.398

Answer

t = 1.398

With df = 10, the two-sided 5% critical value is 2.228; |t| is smaller, so H₀ is not rejected.

18.Hypothesis tests & confidence intervals

Answer: z = −1.813

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.

Sample proportions

p̂₁ = 75/150 = 0.5
p̂₂ = 50/80 = 0.625
p̂ = 125/230 = 0.5435

Standard error under H₀

SE = √(0.5435·0.4565·(1/150 + 1/80)) = 0.06896

Test statistic

z = (0.5 − 0.625)/0.06896 = −1.813

Answer

z = −1.813

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

19.Linear regression & correlation

Answer: r = 0.9897

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 = 786 − 30·107/5 = 144
Sxx = 220 − 30²/5 = 40
Syy = 2819 − 107²/5 = 529.2

Correlation coefficient

r = 144/√(40·529.2) = 0.9897

Answer

r = 0.9897
20.Linear regression & correlation

Answer: ŷ = 97.72

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 = 479/6 = 79.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 313

Regression line

b = 313/70 = 4.471
a = 79.83 − 4.471·21 = −14.07
ŷ = −14.07 + 4.471·x

Prediction

ŷ = −14.07 + 4.471·25 = 97.72

Answer

ŷ = 97.72

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