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
periods0123…1211 + EAR = ?16% / 12

A credit card charges a nominal interest rate of 16% per year, compounded monthly. Find the effective annual rate EAR in percent.

r = 16%m = 12
EAR = %
2.Time value of moneyHard
months0123…N = ?€10 000€450€450€450€4506% / 12

You borrow €10 000 at 6% per year, compounded monthly, and repay €450 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 = €10 000A = €4506% / 12
N = payments
3.NPV & IRRMedium
years0123456€40 000€2 000€2 000€2 000€2 000€2 000€2 000r = 8%, EAC = ?

A machine costs €40 000 today, lasts 6 years and costs €2 000 per year to run (paid at the end of each year). The cost of capital is 8%. Find its equivalent annual cost EAC.

C0 = €40 000M = €2 000n = 6r = 8%
EAC = €
4.NPV & IRRHard
years0123−10 0007 0004 5007 500

A project costs €10 000 and returns €7 000, €4 500, €7 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −10 000CF1 = 7 000CF2 = 4 500CF3 = 7 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123456€20€20€20 + €1 000P = ?

A bond with a face value of €1 000 pays a 4% annual coupon in two half-yearly payments and matures in 3 years. The yield to maturity is 2.5% per year. Find the price P of the bond.

F = €1 000coupon = 4%YTM = 2.5%n = 3
P = €
6.Bond pricing & yieldHard

A bond trades at €945 with a yield to maturity of 5% (annual compounding) and a Macaulay duration of 5.6 years. Use the modified duration to estimate the new price P if the yield rises by 50 basis points.

P0 = €945y = 5%D = 5.6Δy = +50 bp
P = €
7.CAPM & WACCMedium
WACC = ?equity€5 000 000re​ = 12%debt€1 000 000rd​ = 5%T = 21%

A company has equity worth €5 000 000 and debt worth €1 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 5% and the tax rate is 21%. Find the WACC.

E = €5 000 000D = €1 000 000re = 12%rd = 5%T = 21%
WACC = %
8.CAPM & WACCHard
WACC = ?equity4 000 000 × €34.50D0​ = €2.00g = 2%debt€69 000 000rd​ = 4.5%T = 28%

A company has 4 000 000 shares trading at €34.50; the book value of its equity is €82 800 000. It just paid a dividend of €2.00 per share, and dividends grow by 2% per year. Its debt has a market value of €69 000 000 and a pre-tax cost of 4.5%. The tax rate is 28%. Find the WACC, using the dividend growth model for the cost of equity.

N = 4 000 000P0 = €34.50book equity = €82 800 000D0 = €2.00g = 2%D = €69 000 000rd = 4.5%T = 28%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 39, 11, 20, 57, 33.

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

A quantity is normally distributed with mean 60 and standard deviation 10. What proportion P of values is above 75?

μ = 60σ = 10x = 75
P =
11.ProbabilityMedium
xp(x)00.1510.2520.2530.35μ = 1.8

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

x: 0, 1, 2, 3p(x): 0.15, 0.25, 0.25, 0.35
σ =
12.ProbabilityHard
n = 12, p = 0.4: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?

n = 12p = 0.4
P =
13.Normal distributionMedium
300 kg298.3 kgT = X₁ + … + X₁₂?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 12 of them exceeds 298.3 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 12> 298.3 kg
P =
14.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
x̄E₁ = 11.499%, n = 36x̄E₂ = ?90%, n = 36

A 99% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 11.4. What is the margin of error E₂ of a 90% interval from the same data?

99%n₁ = 36E₁ = 11.490%
E₂ =
16.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0499%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.04?

p = 0.5E = 0.0499%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.86p = ?

The mean of a population is claimed to be 120; the population standard deviation is known to be 10. A random sample of 25 observations has a mean of 116.28. Test H₀: μ = 120 against H₁: μ < 120. Find the p-value p.

μ₀ = 120σ = 10n = 25x̄ = 116.28H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2.5α = 0.05βσ = 22, power 80%, 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 80%. The standard deviation is 22. Find the minimum sample size n.

α = 0.05power = 80%δ = 2.5σ = 22zα = 1.645zβ = 0.842
n =
19.Linear regression & correlationMedium
xy1548185421652476278830107xy152025306080100

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy124221319418513613xy246152025x = 4.5

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. A new observation (4.5, 18) is made. Find its residual e = y − ŷ.

n = 6(4.5, 18)
e =

3Solutions

1.Time value of money

Answer: EAR = 17.23 %

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 per period

r/m = 0.16/12 = 0.01333

Compound over one year

EAR = (1 + 0.01333)12 − 1 = 1.172 − 1 = 17.23%

Answer

EAR = 17.23%

The EAR is always above the nominal rate when interest compounds more than once a year.

2.Time value of money

Answer: N = 24 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 = 0.005
(1 + i)−N = 1 − 10 000·0.005/450 = 0.8889

Solve with logarithms

N = −ln(0.8889)/ln(1.005) = 23.62

Round up to whole payments

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

N = 24

Answer

N = 24 payments
3.NPV & IRR

Answer: EAC = 10653 €

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 − 1.08−6)/0.08 = 4.623

Annualised purchase price

40 000/4.623 = 8 652.62

EAC

EAC = 8 652.62 + 2 000 = €10 652.62

Answer

EAC = €10 652.62
4.NPV & IRR

Answer: IRR = 40.23 %

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(40%) = 29.15
NPV(45%) = −571.98

Narrow down

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

IRR ≈ 40 + 5·29.15/(29.15 − (−571.98)) = 40.24%
exact: NPV(40.23%) = 0 → IRR = 40.23%

Answer

IRR = 40.23%
5.Bond pricing & yield

Answer: P = 1043 €

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 = €20.00, i = 0.0125, N = 6

Present value of the coupons

€20.00·(1 − 0.9282)/0.0125 = €114.92

Present value of the face value

€1 000·0.9282 = €928.17

Price

P = €1 043.10

Answer

P = €1 043.10

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

6.Bond pricing & yield

Answer: P = 919.8 €

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.

Modified duration

Dmod = 5.6/1.05 = 5.333

Estimated price change

%ΔP ≈ −5.333·(0.005) = −2.667%
P ≈ 945·(1 − 0.02667) = €919.80

Answer

P = €919.80

Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.

7.CAPM & WACC

Answer: WACC = 10.66 %

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 = 0.8333, D/V = 0.1667

WACC

WACC = 0.8333·12% + 0.1667·5%·(1 − 0.21) = 10.66%

Answer

WACC = 10.66%
8.CAPM & WACC

Answer: WACC = 6.355 %

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 = 2·1.02 = €2.04
re = 2.04/34.5 + 0.02 = 7.913%

Market values

E = 4 000 000·€34.50 = €138 000 000
V = €207 000 000, E/V = 0.6667, D/V = 0.3333

WACC

WACC = 0.6667·7.913% + 0.3333·4.5%·(1 − 0.28) = 6.355%

Answer

WACC = 6.355%
9.Descriptive statistics

Answer: s = 17.75

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 = 160/5 = 32

Squared deviations

Σ(x − x̄)² = 7² + (−21)² + (−12)² + 25² + 1² = 1260

Standard deviation

s = √(1260/4) = 17.75

Answer

s = 17.75
10.Descriptive statistics

Answer: P = 0.06681

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
11.Probability

Answer: σ = 1.077

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 = 0·0.15 + 1·0.25 + 2·0.25 + 3·0.35 = 1.8

Mean of the squares

E(X²) = 0·0.15 + 1·0.25 + 4·0.25 + 9·0.35 = 4.4

Variance

Var(X) = 4.4 − 1.8² = 1.16
σ = √1.16 = 1.077

Answer

σ = 1.077
12.Probability

Answer: P = 0.01959

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.

Zero and one failure

P(X = 0) = 0.612 = 0.002177
P(X = 1) = 12·0.4·0.611 = 0.01741

Add them

P = 0.002177 + 0.01741 = 0.01959

Answer

P = 0.01959
13.Normal distribution

Answer: P = 0.8365

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 total

μT = 300 kg, σT = 0.5·√12 = 1.732 kg

Standardise

z = (298.3 − 300)/1.732 = −0.98

Probability

P = 1 − Φ(−0.98) = 0.8365

Answer

P = 0.8365
14.Normal distribution

Answer: P = 0.9452

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 = 25.04 − 25 = 0.04 mm, σ = 0.025 mm

Standardise

z = (0 − 0.04)/0.025 = −1.6

Probability

P = 1 − Φ(−1.6) = 0.9452

Answer

P = 0.9452
15.Confidence intervals

Answer: E₂ = 7.28

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.

Ratio of margins

E₂/E₁ = (1.645/2.576)·√(36/36) = 0.6386

New margin

E₂ = 11.4·0.6386 = 7.28

Answer

E₂ = 7.28
16.Confidence intervals

Answer: n = 1037

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 = 2.576²·0.5·0.5/0.04² = 1037

Round up

n = 1037

Answer

n = 1037
17.Hypothesis tests & confidence intervals

Answer: p = 0.03144

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.

Test statistic

SE = σ/√n = 10/√25 = 2
z = (116.28 − 120)/2 = −1.86

Tail area

Φ(1.86) = 0.9686
1 − Φ(1.86) = 0.0314

p-value

p = 0.03144

Answer

p = 0.03144

p ≤ 0.05, so H₀ is rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: n = 479

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 + 0.842)·22/2.5)² = 479

Round up

n = 479

Answer

n = 479
19.Linear regression & correlation

Answer: r² = 0.9746

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 = 135/6 = 22.5
ȳ = 438/6 = 73

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 612
Syy = Σ(y − ȳ)² = 2440

Coefficient of determination

r² = 612²/(157.5·2440) = 0.9746

Answer

r² = 0.9746

About 97.46% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = 2.286

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 = 108/6 = 18

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −40

Regression line

b = −40/17.5 = −2.286
a = 18 − (−2.286)·3.5 = 26
ŷ = 26 − 2.286·x

Predicted value

ŷ = 26 + (−2.286)·4.5 = 15.71

Residual

e = 18 − 15.71 = 2.286

Answer

e = 2.286

The point lies above the line.

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