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
years012€20 000FV = ?

You invest €20 000 at a nominal rate of 3% per year, compounded monthly. Find the value FV of the investment after 2 years.

PV = €20 0003% / 12n = 2
FV = €
2.Time value of moneyHard
months0123…36€25 000A?AAA

A loan of €25 000 is repaid with equal monthly payments over 3 years. The interest rate is 2.4% per year, compounded monthly. Find the monthly payment A.

P = €25 0002.4% / 12N = 36
A = €
3.NPV & IRRMedium
years012345−60 00015 50015 50015 50015 50015 500

A project costs €60 000 today and returns €15 500 at the end of each year for 5 years. The cost of capital is 6%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 6%CF0 = −60 000CF1 = 15 500CF2 = 15 500CF3 = 15 500CF4 = 15 500CF5 = 15 500
DPB = years
4.NPV & IRRHard
years0123…10−40 0007 0007 0007 0007 000

An investment of €40 000 returns €7 000 at the end of each year for 10 years. Find its internal rate of return IRR in percent.

C0 = €40 000CF = €7 000n = 10
IRR = %
5.Bond pricing & yieldMedium
half-years0123…14€10€10€10 + €1 000P = ?

A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 9% per year. Find the price P of the bond.

F = €1 000coupon = 2%YTM = 9%n = 7
P = €
6.Bond pricing & yieldHard
half-years0123…16€1 131€35€35€35 + €1 000

A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 8 years. It sells for €1 131. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 7% (2×)n = 8P = €1 131
YTM = %
7.CAPM & WACCMedium
WACC = ?equityEre​ = 10%debtD = 0.5·Erd​ = 7%T = 20%

A company has a debt-to-equity ratio (at market values) of 0.5. Its cost of equity is 10%, its pre-tax cost of debt is 7% and the tax rate is 20%. Find the WACC.

D/E = 0.5re = 10%rd = 7%T = 20%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €38.00β = 1.2rf​ = 3.5%, rm​ = 9.5%debt10 000 × €1 020rd​ = 3.5%T = 25%

A company has 2 000 000 shares at €38.00 each and 10 000 bonds trading at €1 020 each. The pre-tax cost of debt is 3.5% and the tax rate is 25%. The share’s beta is 1.2, the risk-free rate 3.5% and the market return 9.5%. Find the WACC.

β = 1.2rf = 3.5%rm = 9.5%rd = 3.5%T = 25%
WACC = %
9.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 53, 57, 40, 25, 35, 41.

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

A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is above 92.5?

μ = 100σ = 15x = 92.5
P =
11.ProbabilityMedium
red: 8blue: 5

A bag holds 8 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 5
P =
12.ProbabilityHard
€p(x)−50.7200.25950.05

A game costs €5 to play. You win €25 with probability 0.25 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€25: p = 0.25€100: p = 0.05
E(G) = €
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0606

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0606. Find z to two decimals.

A = 0.0606
z =
14.Normal distributionHard
μ = 500x = ?20%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 20% of them lie (the 20th percentile).

μ = 500σ = 40P = 20%
x = h
15.Confidence intervalsMedium
n = 30, s = 15.5LUx̄ = 83E = ?95%

A sample of 30 items has mean 83 and standard deviation s = 15.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 30x̄ = 83s = 15.595%t(29) = 2.045
E =
16.Confidence intervalsHard
σ = 31, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 31. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 4?

σ = 31E = 499%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.84p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 10. A random sample of 100 observations has a mean of 248.16. Test H₀: μ = 250 against H₁: μ < 250. Find the p-value p.

μ₀ = 250σ = 10n = 100x̄ = 248.16H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 200 against H₁: μ > 200 is done at α = 0.01 with a sample of 100; the standard deviation is known to be 10. In reality the mean is 201.9. Find the probability of a Type II error β in percent.

μ₀ = 200μ = 201.9σ = 10n = 100α = 0.01
β = %
19.Linear regression & correlationMedium
xy2641562282910331244xy5102040

The table gives 6 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 = 6Σx = 42Σy = 149Σx² = 364Σy² = 4611Σxy = 1294
r =
20.Linear regression & correlationHard
xy3684755856857908101xy46880100x = 5.5

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.5.

n = 6x = 5.5
ŷ =

3Solutions

1.Time value of money

Answer: FV = 21235 €

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 = 0.03/12 = 0.0025
N = m·n = 12·2 = 24

Future value

FV = 20 000·(1 + 0.0025)24 = 20 000·1.062 = €21 235.14

Answer

FV = €21 235.14
2.Time value of money

Answer: A = 720.4 €

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 payments

i = 0.024/12 = 0.002
N = 12·3 = 36

Payment

The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.

(1 + i)−N = 0.9306
A = 25 000·0.002/(1 − 0.9306) = €720.44

Answer

A = €720.44
3.NPV & IRR

Answer: DPB = 4.543 years

Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.

Discounted and cumulative cash flows

year 1: 15 500/1.06 = 14 622.64 → Σ = 14 622.64
year 2: 15 500/1.124 = 13 794.94 → Σ = 28 417.59
year 3: 15 500/1.191 = 13 014.10 → Σ = 41 431.69
year 4: 15 500/1.262 = 12 277.45 → Σ = 53 709.14
year 5: 15 500/1.338 = 11 582.50 → Σ = 65 291.64

Fraction of the last year

DPB = 4 + 6 290.86/11 582.50 = 4.543

Answer

DPB = 4.543 years

The discounted payback is always longer than the simple payback (3.871 years).

4.NPV & IRR

Answer: IRR = 11.73 %

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)−10)/i = 40 000/7 000 = 5.714

Solve for the rate

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

i = 11%: 5.889
i = 12%: 5.65
IRR = 11.73%

Answer

IRR = 11.73%
5.Bond pricing & yield

Answer: P = 642.2 €

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 = €10.00, i = 0.045, N = 14

Present value of the coupons

€10.00·(1 − 0.54)/0.045 = €102.23

Present value of the face value

€1 000·0.54 = €539.97

Price

P = €642.20

Answer

P = €642.20

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

6.Bond pricing & yield

Answer: YTM = 4.994 %

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, N = 16

Solve numerically

i = 2.497% per half-year

YTM

YTM = 2·2.497% = 4.994%

Answer

YTM = 4.994%
7.CAPM & WACC

Answer: WACC = 8.533 %

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 from D/E

E/V = 1/(1 + 0.5) = 0.6667
D/V = 0.5/(1 + 0.5) = 0.3333

WACC

WACC = 0.6667·10% + 0.3333·7%·(1 − 0.2) = 8.533%

Answer

WACC = 8.533%
8.CAPM & WACC

Answer: WACC = 9.744 %

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·€38.00 = €76 000 000
D = 10 000·€1 020 = €10 200 000

Cost of equity (CAPM)

re = 3.5% + 1.2·6% = 10.7%

Weights

V = €86 200 000, E/V = 0.8817, D/V = 0.1183

WACC

WACC = 0.8817·10.7% + 0.1183·3.5%·(1 − 0.25) = 9.744%

Answer

WACC = 9.744%
9.Descriptive statistics

Answer: s = 11.74

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 = 251/6 = 41.83

Squared deviations

Σ(x − x̄)² = 11.17² + 15.17² + (−1.833)² + (−16.83)² + (−6.833)² + (−0.8333)² = 688.8

Standard deviation

s = √(688.8/5) = 11.74

Answer

s = 11.74
10.Descriptive statistics

Answer: P = 0.6915

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
11.Probability

Answer: P = 0.359

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.

First and second draw

P = 8/13 · 7/12 = 56/156 = 0.359

Answer

P = 0.359
12.Probability

Answer: E(G) = 6.25 €

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: 20 (p = 0.25), 95 (p = 0.05), −5 (p = 0.7)

Expected value

E(G) = 20·0.25 + 95·0.05 − 5·0.7 = 6.25 €

Answer

E(G) = 6.25 €

A positive value means you win money on average.

13.Normal distribution

Answer: z = 1.55

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 − 0.0606 = 0.9394

Read the z-table backwards

Find 0.9394 inside the table and read off z.

z = 1.55

Answer

z = 1.55
14.Normal distribution

Answer: x = 466.3 h

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) = 0.2 → z = −0.8416

Back to x

x = 500 + −0.8416·40 = 466.34 h

Answer

x = 466.34 h
15.Confidence intervals

Answer: E = 5.787

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 = 29, t = 2.045

Margin of error

E = 2.045·15.5/√30 = 5.787

Answer

E = 5.787
16.Confidence intervals

Answer: n = 399

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 = (2.576·31/4)² = 398.6

Round up

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

n = 399

Answer

n = 399
17.Hypothesis tests & confidence intervals

Answer: p = 0.03288

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/√100 = 1
z = (248.16 − 250)/1 = −1.84

Tail area

Φ(1.84) = 0.9671
1 − Φ(1.84) = 0.0329

p-value

p = 0.03288

Answer

p = 0.03288

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

18.Hypothesis tests & confidence intervals

Answer: β = 66.49 %

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 = 10/√100 = 1
x̄c = 200 + 2.326·1 = 202.3

z under the true mean

z = (202.3 − 201.9)/1 = 0.426

Type II error

power = P(Z > 0.426) = 33.51%
β = 100% − 33.51% = 66.49%

Answer

β = 66.49%
19.Linear regression & correlation

Answer: r = 0.994

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 = 1294 − 42·149/6 = 251
Sxx = 364 − 42²/6 = 70
Syy = 4611 − 149²/6 = 910.8

Correlation coefficient

r = 251/√(70·910.8) = 0.994

Answer

r = 0.994
20.Linear regression & correlation

Answer: ŷ = 84

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 = 504/6 = 84

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 105

Regression line

b = 105/17.5 = 6
a = 84 − 6·5.5 = 51
ŷ = 51 + 6·x

Prediction

ŷ = 51 + 6·5.5 = 84

Answer

ŷ = 84

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