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
years0123456€8 000FV = ?

You invest €8 000 at a nominal rate of 4.8% per year, compounded quarterly. Find the value FV of the investment after 6 years.

PV = €8 0004.8% / 4n = 6
FV = €
2.Time value of moneyHard
months0123…228€100€100€100FV = ?

You save €100 at the end of every month for 19 years. The account pays 3.6% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €1003.6% / 12N = 228
FV = €
3.NPV & IRRMedium
years012345€80 000€3 000€3 000€3 000€3 000€3 000r = 10%, EAC = ?

A machine costs €80 000 today, lasts 5 years and costs €3 000 per year to run (paid at the end of each year). The cost of capital is 10%. Find its equivalent annual cost EAC.

C0 = €80 000M = €3 000n = 5r = 10%
EAC = €
4.NPV & IRRHard
years012−40 00028 50019 500

A project costs €40 000 and returns €28 500, €19 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −40 000CF1 = 28 500CF2 = 19 500
IRR = %
5.Bond pricing & yieldMedium

You buy a bond with 3 years to maturity (face value €1 000, annual coupon 2%) when its yield to maturity is 3%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 4.5%. Find your holding period return HPR in percent.

F = €1 000coupon = 2%n = 3y0 = 3%y1 = 4.5%
HPR = %
6.Bond pricing & yieldHard
years0123€975€40€40 + €1 000

A bond with a face value of €1 000 and an annual coupon of 4% matures in 3 years and sells for €975. Find its yield to maturity YTM in percent.

F = €1 000coupon = 4%n = 3P = €975
YTM = %
7.CAPM & WACCMedium
βr (%)SMLrf​ = 4%β = 1rm​ = ?β = 1.6re​ = 14.4%

A share with a beta of 1.6 has a required return of 14.4%. The risk-free rate is 4%. What expected market return rm does CAPM imply?

β = 1.6re = 14.4%rf = 4%
rm = %
8.CAPM & WACCHard
WACC = ?equity€8 000 000re​ = 10%bonds4 000 × €1 128coupon 6%, n = 11T = 20%

A company has equity with a market value of €8 000 000 and a cost of equity of 10%. Its debt consists of 4 000 bonds (face value €1 000, annual coupon 6%, 11 years to maturity) that trade at €1 128 each. The tax rate is 20%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.

E = €8 000 000re = 10%bonds: 4 000 × €1 128coupon = 6%n = 11T = 20%
WACC = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 12, 53, 39, 24, 24.

n = 5
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 500 and standard deviation 60. What proportion P of values is above 428?

μ = 500σ = 60x = 428
P =
11.ProbabilityMedium
n = 6k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 6? The order does not matter.

n = 6k = 3
N =
12.ProbabilityHard
n = 7, p = 0.25: P(X = 1) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 1 correct answers?

n = 7p = 0.25k = 1
P =
13.Normal distributionMedium
70118.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 70 and 118.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 70x2 = 118.75
P =
14.Normal distributionHard
380420460500540580620htop 70%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.

μ = 500 hσ = 40 h70%
x = h
15.Confidence intervalsMedium
x̄E₁ = 4.890%, n = 64x̄E₂ = ?99%, n = 576

A 90% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 4.8. What is the margin of error E₂ of a 99% interval based on 576 observations?

90%n₁ = 64E₁ = 4.899%n₂ = 576
E₂ =
16.Confidence intervalsHard
σ = 25, n = ?x̄ − Ex̄ + Ex̄E = 1.590%

The population standard deviation is 25. 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 = 25E = 1.590%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 51.2 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 51.2s = 4n = 64
t =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.05βσ = 10, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 3 units with a power of 95%. The standard deviation is 10. Find the minimum sample size n.

α = 0.05power = 95%δ = 3σ = 10zα = 1.645zβ = 1.645
n =
19.Linear regression & correlationMedium
xy125221320416515xy24152025

The table gives 5 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy252466673889109212108xy5106080100x = 3

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (3, 56) is made. Find its residual e = y − ŷ.

n = 6(3, 56)
e =

3Solutions

1.Time value of money

Answer: FV = 10652 €

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 = 0.048/4 = 0.012
N = m·n = 4·6 = 24

Future value

FV = 8 000·(1 + 0.012)24 = 8 000·1.331 = €10 651.78

Answer

FV = €10 651.78
2.Time value of money

Answer: FV = 32659 €

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 deposits

i = 0.036000000000000004/12 = 0.003
N = 12·19 = 228

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 100·(1.98 − 1)/0.003 = €32 658.69

Answer

FV = €32 658.69

You paid in €22 800 yourself; the rest is interest.

3.NPV & IRR

Answer: EAC = 24104 €

Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.

Annuity factor

(1 − 1.1−5)/0.1 = 3.791

Annualised purchase price

80 000/3.791 = 21 103.80

EAC

EAC = 21 103.80 + 3 000 = €24 103.80

Answer

EAC = €24 103.80
4.NPV & IRR

Answer: IRR = 14.01 %

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%) = 2 024.79
NPV(15%) = −472.59

Narrow down

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

IRR ≈ 10 + 5·2 024.79/(2 024.79 − (−472.59)) = 14.05%
exact: NPV(14.01%) = 0 → IRR = 14.01%

Answer

IRR = 14.01%
5.Bond pricing & yield

Answer: HPR = 0.1512 %

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 = 0.03, N = 3
€20.00·(1 − 0.9151)/0.03 + €1 000·0.9151 = €56.57 + €915.14 = €971.71

Selling price

C = €20.00, i = 0.045, N = 2
€20.00·(1 − 0.9157)/0.045 + €1 000·0.9157 = €37.45 + €915.73 = €953.18

Holding period return

HPR = (20 + 953.18 − 971.71)/971.71 = 0.1512%

Answer

HPR = 0.1512%
6.Bond pricing & yield

Answer: YTM = 4.917 %

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

€975 = €40·(1 − (1 + y)⁻³)/y + €1 000/(1 + y)³

Solve numerically

P(4%) = €1 000.00, P(5%) = €972.77
YTM = 4.917%

Answer

YTM = 4.917%
7.CAPM & WACC

Answer: rm = 10.5 %

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 risk premium

rm − rf = (14.4% − 4%)/1.6 = 6.5%

Market return

rm = 4% + 6.5% = 10.5%

Answer

rm = 10.5%
8.CAPM & WACC

Answer: WACC = 7.692 %

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).

Cost of debt (YTM)

1128 = 60·(1 − (1 + y)−11)/y + 1000/(1 + y)11
y = 4.499%

Market values

D = 4 000·€1 128 = €4 512 000
V = €12 512 000, E/V = 0.6394, D/V = 0.3606

WACC

WACC = 0.6394·10% + 0.3606·4.499%·(1 − 0.2) = 7.692%

Answer

WACC = 7.692%
9.Descriptive statistics

Answer: s = 15.85

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 = 152/5 = 30.4

Squared deviations

Σ(x − x̄)² = (−18.4)² + 22.6² + 8.6² + (−6.4)² + (−6.4)² = 1005

Standard deviation

s = √(1005/4) = 15.85

Answer

s = 15.85
10.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
11.Probability

Answer: N = 20

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(6, 3) = (6·5·4)/(3·2·1) = 20

Answer

N = 20
12.Probability

Answer: P = 0.3115

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.

Binomial formula

P(X = 1) = 7·0.251·0.756 = 7·0.25·0.178 = 0.3115

Answer

P = 0.3115
13.Normal distribution

Answer: P = 0.8716

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 = (70 − 100)/15 = −2
z₂ = (118.75 − 100)/15 = 1.25

Area between

P = Φ(1.25) − Φ(−2) = 0.8944 − 0.02275 = 0.8716

Answer

P = 0.8716
14.Normal distribution

Answer: x = 479 h

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 = 500 − 0.52·40 = 479 h

Answer

x = 479 h
15.Confidence intervals

Answer: E₂ = 2.506

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 = (2.576/1.645)·√(64/576) = 0.522

New margin

E₂ = 4.8·0.522 = 2.506

Answer

E₂ = 2.506
16.Confidence intervals

Answer: n = 752

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 = (1.645·25/1.5)² = 751.7

Round up

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

n = 752

Answer

n = 752
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 = 4/√64 = 4/8 = 0.5

Test statistic

t = (51.2 − 50)/0.5 = 2.4

Answer

t = 2.4
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 121

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 1.645)·10/3)² = 120.3

Round up

n = 121

Answer

n = 121
19.Linear regression & correlation

Answer: r = −0.9791

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 = 97/5 = 19.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −25
Syy = Σ(y − ȳ)² = 65.2

Correlation coefficient

r = −25/√(10·65.2) = −0.9791

Answer

r = −0.9791
20.Linear regression & correlation

Answer: e = −2.629

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 = 480/6 = 80

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 374

Regression line

b = 374/70 = 5.343
a = 80 − 5.343·7 = 42.6
ŷ = 42.6 + 5.343·x

Predicted value

ŷ = 42.6 + 5.343·3 = 58.63

Residual

e = 56 − 58.63 = −2.629

Answer

e = −2.629

The point lies below the line.

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