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
years01234€20 000FV = ?

You invest €20 000 at 6% per year with continuous compounding. Find the value FV after 4 years.

PV = €20 000r = 6%t = 4
FV = €
2.Time value of moneyHard
months0123…N = ?€25 000€425€425€425€4255.4% / 12

You borrow €25 000 at 5.4% per year, compounded monthly, and repay €425 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 = €25 000A = €4255.4% / 12
N = payments
3.NPV & IRRMedium
years0123−60 00025 00020 00021 000

An investment of €60 000 gives these cash flows at the end of years 1 to 3: €25 000, €20 000, €21 000. The cost of capital is 6%. Find the net present value NPV.

r = 6%CF0 = −60 000CF1 = 25 000CF2 = 20 000CF3 = 21 000
NPV = €
4.NPV & IRRHard
years0123456−25 0005 0005 0005 0005 0005 0005 000

An investment of €25 000 returns €5 000 at the end of each year for 6 years. Find its internal rate of return IRR in percent.

C0 = €25 000CF = €5 000n = 6
IRR = %
5.Bond pricing & yieldMedium
half-years0123…22€40€40€40 + €1 000P = ?

A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 11 years. The yield to maturity is 3.5% per year. Find the price P of the bond.

F = €1 000coupon = 8%YTM = 3.5%n = 11
P = €
6.Bond pricing & yieldHard
years0123…10€1 155€70€70 + €1 000

A bond with a face value of €1 000 and an annual coupon of 7% matures in 10 years and sells for €1 155. Find its yield to maturity YTM in percent.

F = €1 000coupon = 7%n = 10P = €1 155
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€10 000 000re​ = 12%preferred shares€2 000 000rp​ = 6%debt€6 000 000rd​ = 3%T = 20%

A company is financed with ordinary shares worth €10 000 000, preferred shares worth €2 000 000 and debt worth €6 000 000. The cost of equity is 12%, the cost of preferred stock 6% and the pre-tax cost of debt 3%. The tax rate is 20%. Find the WACC.

E = €10 000 000Ps = €2 000 000D = €6 000 000re = 12%rp = 6%rd = 3%T = 20%
WACC = %
8.CAPM & WACCHard
WACC = ?equity5 000 000 × €8.00β = 0.6rf​ = 3%, rm​ = 10%debt40 000 × €980rd​ = 4%T = 30%

A company has 5 000 000 shares at €8.00 each and 40 000 bonds trading at €980 each. The pre-tax cost of debt is 4% and the tax rate is 30%. The share’s beta is 0.6, the risk-free rate 3% and the market return 10%. Find the WACC.

β = 0.6rf = 3%rm = 10%rd = 4%T = 30%
WACC = %
9.Descriptive statisticsMedium
101520253035

Find the sample standard deviation s of the data: 16, 18, 10, 32, 32, 23, 35.

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

A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is below 55?

μ = 50σ = 5x = 55
P =
11.ProbabilityMedium
xp(x)10.3520.330.2540.1μ = 2.1

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

x: 1, 2, 3, 4p(x): 0.35, 0.3, 0.25, 0.1
σ =
12.ProbabilityHard
€p(x)−50.73150.252450.02

A game costs €5 to play. You win €20 with probability 0.25 and €250 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€20: p = 0.25€250: p = 0.02
E(G) = €
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0256

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0256. Find z to two decimals.

A = 0.0256
z =
14.Normal distributionHard
35455565758595pointsbottom 30%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 30% is to the left of it. Find the cut-off x.

μ = 65 pointsσ = 10 points30%
x = points
15.Confidence intervalsMedium
n = 10, s = 9.5LU = ?x̄ = 34E99%

A sample of 10 items has mean 34 and standard deviation s = 9.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 10x̄ = 34s = 9.599%t(9) = 3.25
upper limit =
16.Confidence intervalsHard
σ = 31, n = ?x̄ − Ex̄ + Ex̄E = 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 5?

σ = 31E = 590%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.3062.306t = ?df = 8

9 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 11.95 and the standard deviation of the differences is 11.2. Find the paired test statistic t for H₀: μd = 0.

n = 9d̄ = 11.95sd = 11.2
t =
18.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 199%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 199%
n =
19.Linear regression & correlationMedium
xy242348447554652757xy246455055

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 27Σy = 300Σx² = 139Σy² = 15146Σxy = 1397
r =
20.Linear regression & correlationHard
xy234861181510221224xy5101020ŷ = 5

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 5?

n = 6ŷ = 5
x =

3Solutions

1.Time value of money

Answer: FV = 25425 €

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.

Exponent

r·t = 0.06·4 = 0.24
e0.24 = 1.271

Future value

FV = 20 000·1.271 = €25 424.98

Answer

FV = €25 424.98
2.Time value of money

Answer: N = 69 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 = 0.0045
(1 + i)−N = 1 − 25 000·0.0045/425 = 0.7353

Solve with logarithms

N = −ln(0.7353)/ln(1.004) = 68.48

Round up to whole payments

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

N = 69

Answer

N = 69 payments
3.NPV & IRR

Answer: NPV = −983.2 €

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 = −60 000
CF1/(1 + 0.06)1 = 25 000/1.06 = 23 584.91
CF2/(1 + 0.06)2 = 20 000/1.124 = 17 799.93
CF3/(1 + 0.06)3 = 21 000/1.191 = 17 632.00

Add them up

NPV = €−983.16

Answer

NPV = €−983.16

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

4.NPV & IRR

Answer: IRR = 5.472 %

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)−6)/i = 25 000/5 000 = 5

Solve for the rate

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

i = 5%: 5.076
i = 6%: 4.917
IRR = 5.472%

Answer

IRR = 5.472%
5.Bond pricing & yield

Answer: P = 1408 €

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, i = 0.0175, N = 22

Present value of the coupons

€40.00·(1 − 0.6827)/0.0175 = €725.21

Present value of the face value

€1 000·0.6827 = €682.72

Price

P = €1 407.93

Answer

P = €1 407.93

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

6.Bond pricing & yield

Answer: YTM = 4.993 %

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

€1 155 = €70·(1 − (1 + y)⁻¹⁰)/y + €1 000/(1 + y)¹⁰

Solve numerically

P(4%) = €1 243.33, P(5%) = €1 154.43
YTM = 4.993%

Answer

YTM = 4.993%
7.CAPM & WACC

Answer: WACC = 8.133 %

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 = 10 000 000 + 2 000 000 + 6 000 000 = €18 000 000
E/V = 0.5556, Ps/V = 0.1111, D/V = 0.3333

WACC

WACC = 0.5556·12% + 0.1111·6% + 0.3333·3%·(1 − 0.2) = 8.133%

Answer

WACC = 8.133%
8.CAPM & WACC

Answer: WACC = 5.022 %

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 = 5 000 000·€8.00 = €40 000 000
D = 40 000·€980 = €39 200 000

Cost of equity (CAPM)

re = 3% + 0.6·7% = 7.2%

Weights

V = €79 200 000, E/V = 0.5051, D/V = 0.4949

WACC

WACC = 0.5051·7.2% + 0.4949·4%·(1 − 0.3) = 5.022%

Answer

WACC = 5.022%
9.Descriptive statistics

Answer: s = 9.534

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 = 166/7 = 23.71

Squared deviations

Σ(x − x̄)² = (−7.714)² + (−5.714)² + (−13.71)² + 8.286² + 8.286² + (−0.7143)² + 11.29² = 545.4

Standard deviation

s = √(545.4/6) = 9.534

Answer

s = 9.534
10.Descriptive statistics

Answer: P = 0.8413

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) = 0.8413
P = 0.8413

Answer

P = 0.8413
11.Probability

Answer: σ = 0.995

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 = 1·0.35 + 2·0.3 + 3·0.25 + 4·0.1 = 2.1

Mean of the squares

E(X²) = 1·0.35 + 4·0.3 + 9·0.25 + 16·0.1 = 5.4

Variance

Var(X) = 5.4 − 2.1² = 0.99
σ = √0.99 = 0.995

Answer

σ = 0.995
12.Probability

Answer: E(G) = 5 €

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: 15 (p = 0.25), 245 (p = 0.02), −5 (p = 0.73)

Expected value

E(G) = 15·0.25 + 245·0.02 − 5·0.73 = 5 €

Answer

E(G) = 5 €

A positive value means you win money on average.

13.Normal distribution

Answer: z = 1.95

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 − 0.0256 = 0.9744

Read the z-table backwards

Find 0.9744 inside the table and read off z.

z = 1.95

Answer

z = 1.95
14.Normal distribution

Answer: x = 59.8 points

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

Φ(z) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 65 − 0.52·10 = 59.8 points

Answer

x = 59.8 points
15.Confidence intervals

Answer: upper limit = 43.76

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 = 9, t = 3.25

Margin of error

E = 3.25·9.5/√10 = 9.764

Interval

34 ± 9.764 → [24.24, 43.76]

Answer

upper limit = 43.76
16.Confidence intervals

Answer: n = 105

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 = (1.645·31/5)² = 104

Round up

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

n = 105

Answer

n = 105
17.Hypothesis tests & confidence intervals

Answer: t = 3.201

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 of the mean difference

SE = 11.2/√9 = 3.733

Test statistic

t = 11.95/3.733 = 3.201

Answer

t = 3.201

With df = 8, the two-sided 5% critical value is 2.306; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: n = 664

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).

Solve for n

n = (2.576·10/1)² = 663.6

Round up

n = 664

Answer

n = 664
19.Linear regression & correlation

Answer: r = 0.9298

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 = 1397 − 27·300/6 = 47
Sxx = 139 − 27²/6 = 17.5
Syy = 15146 − 300²/6 = 146

Correlation coefficient

r = 47/√(17.5·146) = 0.9298

Answer

r = 0.9298
20.Linear regression & correlation

Answer: x = 2.905

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 = 42/6 = 7
ȳ = 83/6 = 13.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 151

Regression line

b = 151/70 = 2.157
a = 13.83 − 2.157·7 = −1.267
ŷ = −1.267 + 2.157·x

Solve for x

x = (5 − (−1.267))/2.157 = 2.905

Answer

x = 2.905

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