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…8€2 000FV = ?

You invest €2 000 at a nominal rate of 5% per year, compounded quarterly. Find the value FV of the investment after 8 years.

PV = €2 0005% / 4n = 8
FV = €
2.Time value of moneyHard
months0123…48€25 000A?AAA

A loan of €25 000 is repaid with equal monthly payments over 4 years. The interest rate is 6% per year, compounded monthly. Find the monthly payment A.

P = €25 0006% / 12N = 48
A = €
3.NPV & IRRMedium
years0123−20 0007 0004 0008 000

An investment of €20 000 gives these cash flows at the end of years 1 to 3: €7 000, €4 000, €8 000. The cost of capital is 8%. Find the net present value NPV.

r = 8%CF0 = −20 000CF1 = 7 000CF2 = 4 000CF3 = 8 000
NPV = €
4.NPV & IRRHard
r (%)NPVABr* = ?0

Project A costs €10 000 and returns €5 500, €4 000, €2 500 at the end of years 1 to 3. Project B costs €12 000 and returns €3 000, €5 000, €7 000. Find the crossover rate r* at which both projects have the same NPV, in percent.

A: −10 000, 5 500, 4 000, 2 500B: −12 000, 3 000, 5 000, 7 000
r* = %
5.Bond pricing & yieldMedium

You buy a bond with 11 years to maturity (face value €1 000, annual coupon 2%) when its yield to maturity is 3.5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 4%. Find your holding period return HPR in percent.

F = €1 000coupon = 2%n = 11y0 = 3.5%y1 = 4%
HPR = %
6.Bond pricing & yieldHard
half-years0123…12€1 134€30€30€30 + €1 000

A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 6 years. It sells for €1 134. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 6% (2×)n = 6P = €1 134
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€12 000 000re​ = 14%preferred shares€500 000rp​ = 6.5%debt€2 000 000rd​ = 5%T = 21%

A company is financed with ordinary shares worth €12 000 000, preferred shares worth €500 000 and debt worth €2 000 000. The cost of equity is 14%, the cost of preferred stock 6.5% and the pre-tax cost of debt 5%. The tax rate is 21%. Find the WACC.

E = €12 000 000Ps = €500 000D = €2 000 000re = 14%rp = 6.5%rd = 5%T = 21%
WACC = %
8.CAPM & WACCHard
rf​ = 3%, rm​ − rf​ = 4.5%no debtequityβU​ = 1.1D/E = 0.75equityre​ = ?debtT = 21%

A company's business has an unlevered (asset) beta of 1.1. It plans a debt-to-equity ratio of 0.75. The tax rate is 21%, the risk-free rate is 3% and the market risk premium is 4.5%. Find the cost of equity re at the new capital structure.

βU = 1.1D/E = 0.75T = 21%rf = 3%rm − rf = 4.5%
re = %
9.Descriptive statisticsMedium
15202530354045505560

Find the sample standard deviation s of the data: 50, 53, 15, 60, 22, 24.

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

Exam scores are normally distributed with mean 50 and standard deviation 8. Find the z-score z of a score of 60.

μ = 50σ = 8x = 60
z =
11.ProbabilityMedium
n = 9k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 9? The order does not matter.

n = 9k = 4
N =
12.ProbabilityHard
€p(x)−10.8920.190.01

A game costs €1 to play. You win €3 with probability 0.1 and €10 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€3: p = 0.1€10: p = 0.01
E(G) = €
13.Normal distributionMedium
6573P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 65 and 73 points? Give P as a decimal.

μ = 65σ = 10x1 = 65x2 = 73
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
n = 15, s = 12LU = ?x̄ = 58.5E99%

A sample of 15 items has mean 58.5 and standard deviation s = 12. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 15x̄ = 58.5s = 1299%t(14) = 2.977
upper limit =
16.Confidence intervalsHard
σ = 31, n = ?x̄ − Ex̄ + Ex̄E = 1.590%

The population standard deviation is 31. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?

σ = 31E = 1.590%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 78 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 78s = 5n = 16
t =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ < 50 is done at α = 0.05 with a sample of 64; the standard deviation is known to be 12. In reality the mean is 46.85. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 46.85σ = 12n = 64α = 0.05
1 − β = %
19.Linear regression & correlationMedium
xy230328433541640xy246303540

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy274156208241031xy510102030x = 5

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.

n = 5x = 5
ŷ =

3Solutions

1.Time value of money

Answer: FV = 2976 €

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 = 0.05/4 = 0.0125
N = m·n = 4·8 = 32

Future value

FV = 2 000·(1 + 0.0125)32 = 2 000·1.488 = €2 976.26

Answer

FV = €2 976.26
2.Time value of money

Answer: A = 587.1 €

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 = 0.06/12 = 0.005
N = 12·4 = 48

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 = 0.7871
A = 25 000·0.005/(1 − 0.7871) = €587.13

Answer

A = €587.13
3.NPV & IRR

Answer: NPV = −3739 €

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 + 0.08)1 = 7 000/1.08 = 6 481.48
CF2/(1 + 0.08)2 = 4 000/1.166 = 3 429.36
CF3/(1 + 0.08)3 = 8 000/1.26 = 6 350.66

Add them up

NPV = €−3 738.51

Answer

NPV = €−3 738.51

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

4.NPV & IRR

Answer: r* = 9.256 %

Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.

Difference in cash flows

ΔCF0 = −12 000 − (−10 000) = −2 000
ΔCF1 = 3 000 − 5 500 = −2 500
ΔCF2 = 5 000 − 4 000 = 1 000
ΔCF3 = 7 000 − 2 500 = 4 500

IRR of the difference

NPVΔ(9%) = 22.93
NPVΔ(10%) = −65.36
r* = 9.256%

Answer

r* = 9.256%

Check: at 9.256% both NPVs are €301.94.

5.Bond pricing & yield

Answer: HPR = −0.8318 %

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.

Purchase price

C = €20.00, i = 0.035, N = 11
€20.00·(1 − 0.6849)/0.035 + €1 000·0.6849 = €180.03 + €684.95 = €864.98

Selling price

C = €20.00, i = 0.04, N = 10
€20.00·(1 − 0.6756)/0.04 + €1 000·0.6756 = €162.22 + €675.56 = €837.78

Holding period return

HPR = (20 + 837.78 − 864.98)/864.98 = −0.8318%

Answer

HPR = −0.8318%
6.Bond pricing & yield

Answer: YTM = 3.504 %

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 = €30.00, N = 12

Solve numerically

i = 1.752% per half-year

YTM

YTM = 2·1.752% = 3.504%

Answer

YTM = 3.504%
7.CAPM & WACC

Answer: WACC = 12.36 %

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 + 500 000 + 2 000 000 = €14 500 000
E/V = 0.8276, Ps/V = 0.03448, D/V = 0.1379

WACC

WACC = 0.8276·14% + 0.03448·6.5% + 0.1379·5%·(1 − 0.21) = 12.36%

Answer

WACC = 12.36%
8.CAPM & WACC

Answer: re = 10.88 %

Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.

Levered beta

βL = 1.1·(1 + 0.79·0.75) = 1.752

Cost of equity (CAPM)

re = 3% + 1.752·4.5% = 10.88%

Answer

re = 10.88%
9.Descriptive statistics

Answer: s = 19.14

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 = 224/6 = 37.33

Squared deviations

Σ(x − x̄)² = 12.67² + 15.67² + (−22.33)² + 22.67² + (−15.33)² + (−13.33)² = 1831

Standard deviation

s = √(1831/5) = 19.14

Answer

s = 19.14
10.Descriptive statistics

Answer: z = 1.25

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 = (60 − 50)/8 = 1.25

Answer

z = 1.25
11.Probability

Answer: N = 126

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(9, 4) = (9·8·7·6)/(4·3·2·1) = 126

Answer

N = 126
12.Probability

Answer: E(G) = −0.6 €

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: 2 (p = 0.1), 9 (p = 0.01), −1 (p = 0.89)

Expected value

E(G) = 2·0.1 + 9·0.01 − 1·0.89 = −0.6 €

Answer

E(G) = −0.6 €

A negative value means you lose money on average.

13.Normal distribution

Answer: P = 0.2881

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-scores

z₁ = (65 − 65)/10 = 0
z₂ = (73 − 65)/10 = 0.8

Area between

P = Φ(0.8) − Φ(0) = 0.7881 − 0.5 = 0.2881

Answer

P = 0.2881
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: upper limit = 67.72

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 = 14, t = 2.977

Margin of error

E = 2.977·12/√15 = 9.224

Interval

58.5 ± 9.224 → [49.28, 67.72]

Answer

upper limit = 67.72
16.Confidence intervals

Answer: n = 1156

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 = (1.645·31/1.5)² = 1156

Round up

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

n = 1156

Answer

n = 1156
17.Hypothesis tests & confidence intervals

Answer: t = 2.4

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

SE = 5/√16 = 5/4 = 1.25

Test statistic

t = (78 − 75)/1.25 = 2.4

Answer

t = 2.4
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: 1 − β = 67.54 %

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 = 12/√64 = 1.5
x̄c = 50 − 1.645·1.5 = 47.53

z under the true mean

z = (47.53 − 46.85)/1.5 = 0.455

Power

power = P(Z < 0.455) = 67.54%
1 − β = 67.54%

Answer

1 − β = 67.54%
19.Linear regression & correlation

Answer: r² = 0.7937

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 = 20/5 = 4
ȳ = 172/5 = 34.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 33
Syy = Σ(y − ȳ)² = 137.2

Coefficient of determination

r² = 33²/(10·137.2) = 0.7937

Answer

r² = 0.7937

About 79.37% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 16.55

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 = 97/5 = 19.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 114

Regression line

b = 114/40 = 2.85
a = 19.4 − 2.85·6 = 2.3
ŷ = 2.3 + 2.85·x

Prediction

ŷ = 2.3 + 2.85·5 = 16.55

Answer

ŷ = 16.55

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