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 = ?20% / 12

A credit card charges a nominal interest rate of 20% per year, compounded monthly. Find the effective annual rate EAR in percent.

r = 20%m = 12
EAR = %
2.Time value of moneyHard
months0123…N = ?€8 000€350€350€350€3503.6% / 12

You borrow €8 000 at 3.6% per year, compounded monthly, and repay €350 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 = €8 000A = €3503.6% / 12
N = payments
3.NPV & IRRMedium
years0123−20 0004 0009 0004 000

An investment of €20 000 gives these cash flows at the end of years 1 to 3: €4 000, €9 000, €4 000. The cost of capital is 10%. Find the net present value NPV.

r = 10%CF0 = −20 000CF1 = 4 000CF2 = 9 000CF3 = 4 000
NPV = €
4.NPV & IRRHard
years012−50 00030 00031 500

A project costs €50 000 and returns €30 000, €31 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 30 000CF2 = 31 500
IRR = %
5.Bond pricing & yieldMedium
years0123…14€907C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 14 years and pays annual coupons. It sells for €907 when the yield to maturity is 6%. Find the coupon rate c in percent.

F = €1 000n = 14P = €907YTM = 6%
c = %
6.Bond pricing & yieldHard
half-years0123…20€1 327€40€40€40 + €1 000

A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 10 years. It sells for €1 327. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 8% (2×)n = 10P = €1 327
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€10 000 000re​ = 10%debt€2 000 000rd​ = 4.5%T = 20%

A company has equity worth €10 000 000 and debt worth €2 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 4.5% and the tax rate is 20%. Find the WACC.

E = €10 000 000D = €2 000 000re = 10%rd = 4.5%T = 20%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €52.00β = 1.1rf​ = 2%, rm​ = 6%debt40 000 × €960rd​ = 5%T = 25%

A company has 2 000 000 shares at €52.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 5% and the tax rate is 25%. The share’s beta is 1.1, the risk-free rate 2% and the market return 6%. Find the WACC.

β = 1.1rf = 2%rm = 6%rd = 5%T = 25%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 36, 14, 56, 23, 26, 25, 44.

n = 7
s =
10.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 70 and standard deviation 12. Find the z-score z of a score of 77.2.

μ = 70σ = 12x = 77.2
z =
11.ProbabilityMedium
n = 10k = 5N = ?

In how many ways N can a committee of 5 people be chosen from a group of 10? The order does not matter.

n = 10k = 5
N =
12.ProbabilityHard
€p(x)−30.890.15270.05

A game costs €3 to play. You win €12 with probability 0.15 and €30 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€12: p = 0.15€30: p = 0.05
E(G) = €
13.Normal distributionMedium
48 kg47.2 kgT = X₁ + … + X₄?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 4 of them exceeds 47.2 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 4> 47.2 kg
P =
14.Normal distributionHard
μ = 100x = ?95%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 95% of them lie (the 95th percentile).

μ = 100σ = 15P = 95%
x = points
15.Confidence intervalsMedium
n = 12, sd​ = 1.50LU = ?d̄ = 8.1E90%

The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 8.1 and standard deviation 1.5. Find the upper limit of the 90% confidence interval for the mean difference.

n = 12d̄ = 8.1sd = 1.590%t(11) = 1.796
upper limit = mmHg
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
−2.142.14p = ?

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 124.28. Test H₀: μ = 120 against H₁: μ ≠ 120. Find the p-value p.

μ₀ = 120σ = 10n = 25x̄ = 124.28H₁: μ ≠ μ₀
p =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.01 with a sample of 36; the standard deviation is known to be 9. In reality the mean is 55.85. Find the probability of a Type II error β in percent.

μ₀ = 50μ = 55.85σ = 9n = 36α = 0.01
β = %
19.Linear regression & correlationMedium
xy22456108151017xy51051015

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
xy124220321418517xy2420x = 3.5

The table gives 5 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 (3.5, 15) is made. Find its residual e = y − ŷ.

n = 5(3.5, 15)
e =

3Solutions

1.Time value of money

Answer: EAR = 21.94 %

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 = 0.2/12 = 0.01667

Compound over one year

EAR = (1 + 0.01667)12 − 1 = 1.219 − 1 = 21.94%

Answer

EAR = 21.94%

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 = 0.003
(1 + i)−N = 1 − 8 000·0.003/350 = 0.9314

Solve with logarithms

N = −ln(0.9314)/ln(1.003) = 23.71

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: NPV = −5920 €

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 = −20 000
CF1/(1 + 0.1)1 = 4 000/1.1 = 3 636.36
CF2/(1 + 0.1)2 = 9 000/1.21 = 7 438.02
CF3/(1 + 0.1)3 = 4 000/1.331 = 3 005.26

Add them up

NPV = €−5 920.36

Answer

NPV = €−5 920.36

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

4.NPV & IRR

Answer: IRR = 14.85 %

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(10%) = 3 305.79
NPV(15%) = −94.52

Narrow down

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

IRR ≈ 10 + 5·3 305.79/(3 305.79 − (−94.52)) = 14.86%
exact: NPV(14.85%) = 0 → IRR = 14.85%

Answer

IRR = 14.85%
5.Bond pricing & yield

Answer: c = 4.999 %

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 + 0.06)−14 = €442.30

Coupon

C = (907 − 442.30)/9.295 = €49.99

Coupon rate

c = 49.99/1000 = 4.999%

Answer

c = 4.999%
6.Bond pricing & yield

Answer: YTM = 4 %

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 = €40.00, N = 20

Solve numerically

i = 2% per half-year

YTM

YTM = 2·2% = 4%

Answer

YTM = 4%
7.CAPM & WACC

Answer: WACC = 8.933 %

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 = €12 000 000, E/V = 0.8333, D/V = 0.1667

WACC

WACC = 0.8333·10% + 0.1667·4.5%·(1 − 0.2) = 8.933%

Answer

WACC = 8.933%
8.CAPM & WACC

Answer: WACC = 5.685 %

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 = 2 000 000·€52.00 = €104 000 000
D = 40 000·€960 = €38 400 000

Cost of equity (CAPM)

re = 2% + 1.1·4% = 6.4%

Weights

V = €142 400 000, E/V = 0.7303, D/V = 0.2697

WACC

WACC = 0.7303·6.4% + 0.2697·5%·(1 − 0.25) = 5.685%

Answer

WACC = 5.685%
9.Descriptive statistics

Answer: s = 14.29

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 = 224/7 = 32

Squared deviations

Σ(x − x̄)² = 4² + (−18)² + 24² + (−9)² + (−6)² + (−7)² + 12² = 1226

Standard deviation

s = √(1226/6) = 14.29

Answer

s = 14.29
10.Descriptive statistics

Answer: z = 0.6

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 = (77.2 − 70)/12 = 0.6

Answer

z = 0.6
11.Probability

Answer: N = 252

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.

Combinations

C(10, 5) = (10·9·8·7·6)/(5·4·3·2·1) = 252

Answer

N = 252
12.Probability

Answer: E(G) = 0.3 €

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: 9 (p = 0.15), 27 (p = 0.05), −3 (p = 0.8)

Expected value

E(G) = 9·0.15 + 27·0.05 − 3·0.8 = 0.3 €

Answer

E(G) = 0.3 €

A positive value means you win money on average.

13.Normal distribution

Answer: P = 0.6915

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 = 48 kg, σT = 0.8·√4 = 1.6 kg

Standardise

z = (47.2 − 48)/1.6 = −0.5

Probability

P = 1 − Φ(−0.5) = 0.6915

Answer

P = 0.6915
14.Normal distribution

Answer: x = 124.7 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) = 0.95 → z = 1.645

Back to x

x = 100 + 1.645·15 = 124.67 points

Answer

x = 124.67 points
15.Confidence intervals

Answer: upper limit = 8.878 mmHg

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.

Critical value

df = 11, t = 1.796

Margin of error

E = 1.796·1.5/√12 = 0.7777

Interval

8.1 ± 0.7777 → [7.322, 8.878]

Answer

upper limit = 8.878 mmHg
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 = 0.03235

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 = (124.28 − 120)/2 = 2.14

Tail area

Φ(2.14) = 0.9838
1 − Φ(2.14) = 0.0162

p-value

p = 2·0.0162 = 0.03235

Answer

p = 0.03235

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

18.Hypothesis tests & confidence intervals

Answer: β = 5.774 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 9/√36 = 1.5
x̄c = 50 + 2.326·1.5 = 53.49

z under the true mean

z = (53.49 − 55.85)/1.5 = −1.574

Type II error

power = P(Z > −1.574) = 94.23%
β = 100% − 94.23% = 5.774%

Answer

β = 5.774%
19.Linear regression & correlation

Answer: r² = 0.9828

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 = 49/5 = 9.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 80
Syy = Σ(y − ȳ)² = 162.8

Coefficient of determination

r² = 80²/(40·162.8) = 0.9828

Answer

r² = 0.9828

About 98.28% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = −4.2

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 = 15/5 = 3
ȳ = 100/5 = 20

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −16

Regression line

b = −16/10 = −1.6
a = 20 − (−1.6)·3 = 24.8
ŷ = 24.8 − 1.6·x

Predicted value

ŷ = 24.8 + (−1.6)·3.5 = 19.2

Residual

e = 15 − 19.2 = −4.2

Answer

e = −4.2

The point lies below the line.

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