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
periods01211 + EAR = ?16% / 2

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

r = 16%m = 2
EAR = %
2.Time value of moneyHard
months0123…360€200 000A?AAA

A loan of €200 000 is repaid with equal monthly payments over 30 years. The interest rate is 2.4% per year, compounded monthly. Find the monthly payment A.

P = €200 0002.4% / 12N = 360
A = €
3.NPV & IRRMedium
years01234−75 00034 00020 00018 00026 000

An investment of €75 000 gives these cash flows at the end of years 1 to 4: €34 000, €20 000, €18 000, €26 000. The cost of capital is 5%. Find the net present value NPV.

r = 5%CF0 = −75 000CF1 = 34 000CF2 = 20 000CF3 = 18 000CF4 = 26 000
NPV = €
4.NPV & IRRHard
years0123456−200 00036 500+3%+3%+3%+3%+3%r = 8%, NPV = ?

A project costs €200 000 today. Its cash flow is €36 500 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 6. The cost of capital is 8%. Find the net present value NPV.

C0 = €200 000CF1 = €36 500g = 3%n = 6r = 8%
NPV = €
5.Bond pricing & yieldMedium
half-years0123…10€25€25€25 + €1 000P = ?

A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 5 years. The yield to maturity is 7% per year. Find the price P of the bond.

F = €1 000coupon = 5%YTM = 7%n = 5
P = €
6.Bond pricing & yieldHard
years0123…10€935€80€80 + €1 000

A bond with a face value of €1 000 and an annual coupon of 8% matures in 10 years and sells for €935. Find its yield to maturity YTM in percent.

F = €1 000coupon = 8%n = 10P = €935
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€10 000 000re​ = 11%debt€2 000 000rd​ = 3.5%T = 25.8%

A company has equity worth €10 000 000 and debt worth €2 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 3.5% and the tax rate is 25.8%. Find the WACC.

E = €10 000 000D = €2 000 000re = 11%rd = 3.5%T = 25.8%
WACC = %
8.CAPM & WACCHard
rf​ = 2.5%, rm​ − rf​ = 4.5%no debtequityβU​ = 1.2D/E = 1.5equityre​ = ?debtT = 20%

A company's business has an unlevered (asset) beta of 1.2. It plans a debt-to-equity ratio of 1.5. The tax rate is 20%, the risk-free rate is 2.5% and the market risk premium is 4.5%. Find the cost of equity re at the new capital structure.

βU = 1.2D/E = 1.5T = 20%rf = 2.5%rm − rf = 4.5%
re = %
9.Descriptive statisticsMedium
15202530354045505560

Find the sample standard deviation s of the data: 60, 60, 28, 41, 13, 52.

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

A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is below 38?

μ = 50σ = 12x = 38
P =
11.ProbabilityMedium
xp(x)00.2510.220.3530.1540.05μ = 1.55

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

x: 0, 1, 2, 3, 4p(x): 0.25, 0.2, 0.35, 0.15, 0.05
Var(X) =
12.ProbabilityHard
n = 5, p = 0.1: P(X ≤ 1) = ?k012345

5 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?

n = 5p = 0.1
P =
13.Normal distributionMedium
105 s113 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 113 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 113 s
P =
14.Normal distributionHard
μ = 32x = ?20%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 20% of them lie (the 20th percentile).

μ = 32σ = 6P = 20%
x = min
15.Confidence intervalsMedium
x̄E₁ = 3.295%, n = 25x̄E₂ = ?95%, n = 50

A 95% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 3.2. What is the margin of error E₂ if the same study uses 50 observations instead?

95%n₁ = 25E₁ = 3.2n₂ = 50
E₂ =
16.Confidence intervalsHard
σ = 7, n = ?x̄ − Ex̄ + Ex̄E = 1.590%

The population standard deviation is 7. 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 = 7E = 1.590%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 101.7 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 101.7s = 10n = 100
t =
18.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 64 measurements has mean 199.8. The population standard deviation is 5. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 199.8σ = 5n = 6499%
L =
19.Linear regression & correlationMedium
xy294116138201023xy510101520

The table gives 5 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 = 5Σx = 30Σy = 76Σx² = 220Σy² = 1300Σxy = 530
r =
20.Linear regression & correlationHard
xy231333437532636xy246323436x = 4.5

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.

n = 5x = 4.5
ŷ =

3Solutions

1.Time value of money

Answer: EAR = 16.64 %

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 = 0.16/2 = 0.08

Compound over one year

EAR = (1 + 0.08)2 − 1 = 1.166 − 1 = 16.64%

Answer

EAR = 16.64%

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

2.Time value of money

Answer: A = 779.9 €

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 = 12·30 = 360

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 = 0.4871
A = 200 000·0.002/(1 − 0.4871) = €779.88

Answer

A = €779.88
3.NPV & IRR

Answer: NPV = 12461 €

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 = −75 000
CF1/(1 + 0.05)1 = 34 000/1.05 = 32 380.95
CF2/(1 + 0.05)2 = 20 000/1.103 = 18 140.59
CF3/(1 + 0.05)3 = 18 000/1.158 = 15 549.08
CF4/(1 + 0.05)4 = 26 000/1.216 = 21 390.26

Add them up

NPV = €12 460.88

Answer

NPV = €12 460.88

NPV is positive, so the project earns more than the required return.

4.NPV & IRR

Answer: NPV = −19293 €

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 the growing annuity

((1.03)/(1.08))6 = 0.7525
PV = 36 500/(0.08 − 0.03)·(1 − 0.7525) = 180 707.49

NPV

NPV = 180 707.49 − 200 000 = −€19 292.51

Answer

NPV = −€19 292.51

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

5.Bond pricing & yield

Answer: P = 916.8 €

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 = €25.00, i = 0.035, N = 10

Present value of the coupons

€25.00·(1 − 0.7089)/0.035 = €207.92

Present value of the face value

€1 000·0.7089 = €708.92

Price

P = €916.83

Answer

P = €916.83

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

6.Bond pricing & yield

Answer: YTM = 9.013 %

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

€935 = €80·(1 − (1 + y)⁻¹⁰)/y + €1 000/(1 + y)¹⁰

Solve numerically

P(9%) = €935.82, P(10%) = €877.11
YTM = 9.013%

Answer

YTM = 9.013%
7.CAPM & WACC

Answer: WACC = 9.6 %

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 = 0.8333·11% + 0.1667·3.5%·(1 − 0.258) = 9.6%

Answer

WACC = 9.6%
8.CAPM & WACC

Answer: re = 14.38 %

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 = 1.2·(1 + 0.8·1.5) = 2.64

Cost of equity (CAPM)

re = 2.5% + 2.64·4.5% = 14.38%

Answer

re = 14.38%
9.Descriptive statistics

Answer: s = 18.9

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 = 254/6 = 42.33

Squared deviations

Σ(x − x̄)² = 17.67² + 17.67² + (−14.33)² + (−1.333)² + (−29.33)² + 9.667² = 1785

Standard deviation

s = √(1785/5) = 18.9

Answer

s = 18.9
10.Descriptive statistics

Answer: P = 0.1587

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

Standard normal table

Φ(−1) = 0.1587
P = 0.1587

Answer

P = 0.1587
11.Probability

Answer: Var(X) = 1.348

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 + 1·0.2 + 2·0.35 + 3·0.15 + 4·0.05 = 1.55

Mean of the squares

E(X²) = 0·0.25 + 1·0.2 + 4·0.35 + 9·0.15 + 16·0.05 = 3.75

Variance

Var(X) = 3.75 − 1.55² = 1.348

Answer

Var(X) = 1.348
12.Probability

Answer: P = 0.9185

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) = 0.95 = 0.5905
P(X = 1) = 5·0.1·0.94 = 0.3281

Add them

P = 0.5905 + 0.3281 = 0.9185

Answer

P = 0.9185
13.Normal distribution

Answer: P = 0.1056

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 = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (113 − 105)/6.403 = 1.25

Upper tail

P = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
14.Normal distribution

Answer: x = 26.95 min

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 = 32 + −0.8416·6 = 26.95 min

Answer

x = 26.95 min
15.Confidence intervals

Answer: E₂ = 2.263

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 = (1.96/1.96)·√(25/50) = 0.7071

New margin

E₂ = 3.2·0.7071 = 2.263

Answer

E₂ = 2.263
16.Confidence intervals

Answer: n = 59

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 = (1.645·7/1.5)² = 58.93

Round up

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

n = 59

Answer

n = 59
17.Hypothesis tests & confidence intervals

Answer: t = 1.7

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 = 10/√100 = 10/10 = 1

Test statistic

t = (101.7 − 100)/1 = 1.7

Answer

t = 1.7
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: L = 198.2

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 = 2.576·5/√64 = 1.61

Interval limit

L = 199.8 − 1.61 = 198.2

Answer

L = 198.2
19.Linear regression & correlation

Answer: r = 0.9723

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 = 530 − 30·76/5 = 74
Sxx = 220 − 30²/5 = 40
Syy = 1300 − 76²/5 = 144.8

Correlation coefficient

r = 74/√(40·144.8) = 0.9723

Answer

r = 0.9723
20.Linear regression & correlation

Answer: ŷ = 34.25

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 = 169/5 = 33.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 9

Regression line

b = 9/10 = 0.9
a = 33.8 − 0.9·4 = 30.2
ŷ = 30.2 + 0.9·x

Prediction

ŷ = 30.2 + 0.9·4.5 = 34.25

Answer

ŷ = 34.25

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