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 = ?€12 000+2.5%+2.5%…r = 7.5%

A property is expected to earn net rent of €12 000 next year. The rent is expected to grow by 2.5% per year forever. The required return is 7.5% per year. Find the value PV of the property today.

C1 = €12 000g = 2.5%r = 7.5%
PV = €
2.Time value of moneyHard
months0123…N = ?€20 000€475€475€475€4755.4% / 12

You borrow €20 000 at 5.4% per year, compounded monthly, and repay €475 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 = €20 000A = €4755.4% / 12
N = payments
3.NPV & IRRMedium
years01234€20 000€3 000€3 000€3 000€3 000r = 10%, EAC = ?

A machine costs €20 000 today, lasts 4 years and costs €3 000 per year to run (paid at the end of each year). The cost of capital is 10%. Find its equivalent annual cost EAC.

C0 = €20 000M = €3 000n = 4r = 10%
EAC = €
4.NPV & IRRHard
years0123…8−10 0001 5001 5001 5001 500

An investment of €10 000 returns €1 500 at the end of each year for 8 years. Find its internal rate of return IRR in percent.

C0 = €10 000CF = €1 500n = 8
IRR = %
5.Bond pricing & yieldMedium
half-years0123…12€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 6 years. The yield to maturity is 5.5% per year. Find the price P of the bond.

F = €1 000coupon = 7%YTM = 5.5%n = 6
P = €
6.Bond pricing & yieldHard
years012345€1 359€50€50€50 + €1 050

A bond with 18 years to maturity (face value €1 000, annual coupon 5%) sells for €1 359. The issuer can call it in 5 years at a call price of €1 050. Find the yield to call YTC in percent.

F = €1 000coupon = 5%P = €1 359call price = €1 050call in 5 years
YTC = %
7.CAPM & WACCMedium
WACC = ?equity€2 000 000re​ = 12%debt€6 000 000rd​ = 5%T = 25%

A company has equity worth €2 000 000 and debt worth €6 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 5% and the tax rate is 25%. Find the WACC.

E = €2 000 000D = €6 000 000re = 12%rd = 5%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity5 000 000 × €58.00D0​ = €2.30g = 4%debt€232 000 000rd​ = 3%T = 25.8%

A company has 5 000 000 shares trading at €58.00; the book value of its equity is €87 000 000. It just paid a dividend of €2.30 per share, and dividends grow by 4% per year. Its debt has a market value of €232 000 000 and a pre-tax cost of 3%. The tax rate is 25.8%. Find the WACC, using the dividend growth model for the cost of equity.

N = 5 000 000P0 = €58.00book equity = €87 000 000D0 = €2.30g = 4%D = €232 000 000rd = 3%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
36384042444648505254

Find the sample standard deviation s of the data: 43, 55, 47, 37, 35, 45.

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

A quantity is normally distributed with mean 500 and standard deviation 25. What proportion P of values is above 550?

μ = 500σ = 25x = 550
P =
11.ProbabilityMedium
xp(x)00.2510.3520.2530.0540.1μ = 1.4

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

x: 0, 1, 2, 3, 4p(x): 0.25, 0.35, 0.25, 0.05, 0.1
Var(X) =
12.ProbabilityHard
€p(x)−50.84150.152450.01

A game costs €5 to play. You win €20 with probability 0.15 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€20: p = 0.15€250: p = 0.01
E(G) = €
13.Normal distributionMedium
200 mm200.6 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 200.6 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.6 mm
P =
14.Normal distributionHard
μ = 178x = ?95%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 95% of them lie (the 95th percentile).

μ = 178σ = 7P = 95%
x = cm
15.Confidence intervalsMedium
n = 8, s = 13LU = ?x̄ = 15E99%

A sample of 8 items has mean 15 and standard deviation s = 13. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 8x̄ = 15s = 1399%t(7) = 3.499
upper limit =
16.Confidence intervalsHard
μ₁ − μ₂, t(19) = 2.8610LUx̄₁ − x̄₂E = ?99%

Two fertilisers are tested. Plot group 1: 13 plots, mean yield 42 kg, standard deviation 10 kg. Plot group 2: 8 plots, mean 35.5 kg, standard deviation 9 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 13x̄₁ = 42s₁ = 10n₂ = 8x̄₂ = 35.5s₂ = 999%t(19) = 2.861
E = kg
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 30 parts has mean length 115 mm, and the line's known standard deviation is 18 mm. Line 2: a sample of 30 parts has mean 107.7 mm, known standard deviation 11 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 30x̄₁ = 115 mmσ₁ = 18 mmn₂ = 30x̄₂ = 107.7 mmσ₂ = 11 mm
z =
18.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 16 measurements has mean 78.5. The population standard deviation is 10. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 78.5σ = 10n = 1690%
U =
19.Linear regression & correlationMedium
xy350450560658758xy46505560

The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy339440547650747852xy468404550ŷ = 51

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. For which value x does the line predict ŷ = 51?

n = 6ŷ = 51
x =

3Solutions

1.Time value of money

Answer: PV = 240000 €

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) = 12 000/(0.075 − 0.025) = 12 000/0.05 = €240 000.00

Answer

PV = €240 000.00
2.Time value of money

Answer: N = 47 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 = 0.0045
(1 + i)−N = 1 − 20 000·0.0045/475 = 0.8105

Solve with logarithms

N = −ln(0.8105)/ln(1.004) = 46.79

Round up to whole payments

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

N = 47

Answer

N = 47 payments
3.NPV & IRR

Answer: EAC = 9309 €

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 − 1.1−4)/0.1 = 3.17

Annualised purchase price

20 000/3.17 = 6 309.42

EAC

EAC = 6 309.42 + 3 000 = €9 309.42

Answer

EAC = €9 309.42
4.NPV & IRR

Answer: IRR = 4.239 %

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)−8)/i = 10 000/1 500 = 6.667

Solve for the rate

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

i = 4%: 6.733
i = 5%: 6.463
IRR = 4.239%

Answer

IRR = 4.239%
5.Bond pricing & yield

Answer: P = 1076 €

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 = 0.0275, N = 12

Present value of the coupons

€35.00·(1 − 0.7221)/0.0275 = €353.65

Present value of the face value

€1 000·0.7221 = €722.13

Price

P = €1 075.78

Answer

P = €1 075.78

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

6.Bond pricing & yield

Answer: YTC = 50 %

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

1359 = 50·(1 − (1 + y)−5)/y + 1050/(1 + y)5

Solve numerically

y = 50%: €225.10
y = 50%: €225.10
YTC = 50%

Answer

YTC = 50%

The yield to maturity is 2.499%. 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: WACC = 5.813 %

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 = 0.25·12% + 0.75·5%·(1 − 0.25) = 5.813%

Answer

WACC = 5.813%
8.CAPM & WACC

Answer: WACC = 5.503 %

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).

Cost of equity

D1 = 2.3·1.04 = €2.39
re = 2.392/58 + 0.04 = 8.124%

Market values

E = 5 000 000·€58.00 = €290 000 000
V = €522 000 000, E/V = 0.5556, D/V = 0.4444

WACC

WACC = 0.5556·8.124% + 0.4444·3%·(1 − 0.258) = 5.503%

Answer

WACC = 5.503%
9.Descriptive statistics

Answer: s = 7.23

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 = 262/6 = 43.67

Squared deviations

Σ(x − x̄)² = (−0.6667)² + 11.33² + 3.333² + (−6.667)² + (−8.667)² + 1.333² = 261.3

Standard deviation

s = √(261.3/5) = 7.23

Answer

s = 7.23
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: Var(X) = 1.44

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 + 1·0.35 + 2·0.25 + 3·0.05 + 4·0.1 = 1.4

Mean of the squares

E(X²) = 0·0.25 + 1·0.35 + 4·0.25 + 9·0.05 + 16·0.1 = 3.4

Variance

Var(X) = 3.4 − 1.4² = 1.44

Answer

Var(X) = 1.44
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: 15 (p = 0.15), 245 (p = 0.01), −5 (p = 0.84)

Expected value

E(G) = 15·0.15 + 245·0.01 − 5·0.84 = 0.5 €

Answer

E(G) = 0.5 €

A positive value means you win money on average.

13.Normal distribution

Answer: P = 0.1151

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 sum

μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (200.6 − 200)/0.5 = 1.2

Upper tail

P = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151

Answer

P = 0.1151
14.Normal distribution

Answer: x = 189.5 cm

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) = 0.95 → z = 1.645

Back to x

x = 178 + 1.645·7 = 189.51 cm

Answer

x = 189.51 cm
15.Confidence intervals

Answer: upper limit = 31.08

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 = 3.499

Margin of error

E = 3.499·13/√8 = 16.08

Interval

15 ± 16.08 → [−1.082, 31.08]

Answer

upper limit = 31.08
16.Confidence intervals

Answer: E = 12.4 kg

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Pooled standard deviation

sp² = (12·10² + 7·9²)/19 = 93
sp = 9.644

Margin of error

E = 2.861·9.644·√(1/13 + 1/8) = 12.4

Answer

E = 12.4 kg
17.Hypothesis tests & confidence intervals

Answer: z = 1.895

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 = √(18²/30 + 11²/30) = √14.83 = 3.851

Test statistic

z = (115 − 107.7)/3.851 = 1.895

Answer

z = 1.895

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

18.Hypothesis tests & confidence intervals

Answer: U = 82.61

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 = 1.645·10/√16 = 4.112

Interval limit

U = 78.5 + 4.112 = 82.61

Answer

U = 82.61
19.Linear regression & correlation

Answer: r² = 0.6207

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 = 25/5 = 5
ȳ = 276/5 = 55.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 24
Syy = Σ(y − ȳ)² = 92.8

Coefficient of determination

r² = 24²/(10·92.8) = 0.6207

Answer

r² = 0.6207

About 62.07% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: x = 7.532

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 = 275/6 = 45.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 44.5

Regression line

b = 44.5/17.5 = 2.543
a = 45.83 − 2.543·5.5 = 31.85
ŷ = 31.85 + 2.543·x

Solve for x

x = (51 − 31.85)/2.543 = 7.532

Answer

x = 7.532

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