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…9€20 000FV = ?

You invest €20 000 at 6% per year with continuous compounding. Find the value FV after 9 years.

PV = €20 000r = 6%t = 9
FV = €
2.Time value of moneyHard
months0123…324€250€250€250FV = ?

You save €250 at the end of every month for 27 years. The account pays 5.4% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €2505.4% / 12N = 324
FV = €
3.NPV & IRRMedium
years0123…9−500 00065 000+1%+1%…r = 11%, NPV = ?

A project costs €500 000 today. It generates €65 000 at the end of the first year, and the cash flow then grows by 1% per year forever. The cost of capital is 11%. Find the net present value NPV.

C0 = €500 000CF1 = €65 000g = 1%r = 11%
NPV = €
4.NPV & IRRHard
years0123−40 00014 00020 50022 000

A project costs €40 000 and returns €14 000, €20 500, €22 000 at the end of years 1 to 3. Cash flows can be reinvested at the cost of capital of 8%. Find the modified internal rate of return MIRR in percent.

r = 8%CF0 = −40 000CF1 = 14 000CF2 = 20 500CF3 = 22 000
MIRR = %
5.Bond pricing & yieldMedium
years012345€1 229C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 5 years and pays annual coupons. It sells for €1 229 when the yield to maturity is 3%. Find the coupon rate c in percent.

F = €1 000n = 5P = €1 229YTM = 3%
c = %
6.Bond pricing & yieldHard
half-years0123…20€961€22.50€22.50€22.50 + €1 000

A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 10 years. It sells for €961. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 4.5% (2×)n = 10P = €961
YTM = %
7.CAPM & WACCMedium
βr (%)SMLrf​ = 2%β = 1rm​ = ?β = 0.9re​ = 8.3%

A share with a beta of 0.9 has a required return of 8.3%. The risk-free rate is 2%. What expected market return rm does CAPM imply?

β = 0.9re = 8.3%rf = 2%
rm = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €37.00D0​ = €2.90g = 4%debt€18 500 000rd​ = 5%T = 25.8%

A company has 1 000 000 shares trading at €37.00; the book value of its equity is €22 200 000. It just paid a dividend of €2.90 per share, and dividends grow by 4% per year. Its debt has a market value of €18 500 000 and a pre-tax cost of 5%. The tax rate is 25.8%. Find the WACC, using the dividend growth model for the cost of equity.

N = 1 000 000P0 = €37.00book equity = €22 200 000D0 = €2.90g = 4%D = €18 500 000rd = 5%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 40, 31, 57, 11, 21.

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

Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 510.

μ = 500σ = 25x = 510
z =
11.ProbabilityMedium
red: 4blue: 6

A bag holds 4 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

4 / 6
P =
12.ProbabilityHard
n = 5, p = 0.5: P(X = 2) = ?k012345

A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 2 correct answers?

n = 5p = 0.5k = 2
P =
13.Normal distributionMedium
26.333.5P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 26.3 and 33.5 min? Give P as a decimal.

μ = 32σ = 6x1 = 26.3x2 = 33.5
P =
14.Normal distributionHard
μ = 500x = ?25%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 25% of them lie (the 25th percentile).

μ = 500σ = 40P = 25%
x = h
15.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 80 phones, mean 86 h, known standard deviation 18 h. Model 2: 80 phones, mean 97 h, known standard deviation 9 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 86 hσ₁ = 18 hn₂ = 80x̄₂ = 97 hσ₂ = 9 h90%
lower limit = h
16.Confidence intervalsHard
μ₁ − μ₂, t(19) = 2.0930LU = ?x̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 7 plots, mean yield 19.5 kg, standard deviation 5.5 kg. Plot group 2: 14 plots, mean 13 kg, standard deviation 3 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 7x̄₁ = 19.5s₁ = 5.5n₂ = 14x̄₂ = 13s₂ = 395%t(19) = 2.093
upper limit = kg
17.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 200.5 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 200.5s = 4n = 81
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 75 of 250 patients on treatment A recover, and 34 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 75n₁ = 250x₂ = 34n₂ = 80
z =
19.Linear regression & correlationMedium
xy2941862783610451250xy5102040

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 = 185Σx² = 364Σy² = 6955Σxy = 1590
r =
20.Linear regression & correlationHard
xy23486148151023xy5101020x = 3

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 = 3.

n = 5x = 3
ŷ =

3Solutions

1.Time value of money

Answer: FV = 34320 €

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 = 0.06·9 = 0.54
e0.54 = 1.716

Future value

FV = 20 000·1.716 = €34 320.14

Answer

FV = €34 320.14
2.Time value of money

Answer: FV = 182407 €

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 = 0.054000000000000006/12 = 0.0045
N = 12·27 = 324

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 250·(4.283 − 1)/0.0045 = €182 406.88

Answer

FV = €182 406.88

You paid in €81 000 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = 150000 €

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.

Value of the cash flows

PV = 65 000/(0.11 − 0.01) = 650 000.00

NPV

NPV = 650 000.00 − 500 000 = €150 000.00

Answer

NPV = €150 000.00

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

4.NPV & IRR

Answer: MIRR = 14.77 %

Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.

Terminal value of the inflows

14 000·1.082 = 16 329.60
20 500·1.081 = 22 140.00
22 000·1.080 = 22 000.00
TV = 60 469.60

MIRR

MIRR = (60 469.60/40 000)1/3 − 1 = 14.77%

Answer

MIRR = 14.77%
5.Bond pricing & yield

Answer: c = 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.

Present value of the face value

1000·(1 + 0.03)−5 = €862.61

Coupon

C = (1229 − 862.61)/4.58 = €80.00

Coupon rate

c = 80/1000 = 8%

Answer

c = 8%
6.Bond pricing & yield

Answer: YTM = 5 %

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 = €22.50, N = 20

Solve numerically

i = 2.5% per half-year

YTM

YTM = 2·2.5% = 5%

Answer

YTM = 5%
7.CAPM & WACC

Answer: rm = 9 %

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 = (8.3% − 2%)/0.9 = 7%

Market return

rm = 2% + 7% = 9%

Answer

rm = 9%
8.CAPM & WACC

Answer: WACC = 9.338 %

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 equity

D1 = 2.9·1.04 = €3.02
re = 3.016/37 + 0.04 = 12.15%

Market values

E = 1 000 000·€37.00 = €37 000 000
V = €55 500 000, E/V = 0.6667, D/V = 0.3333

WACC

WACC = 0.6667·12.15% + 0.3333·5%·(1 − 0.258) = 9.338%

Answer

WACC = 9.338%
9.Descriptive statistics

Answer: s = 17.69

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 = 160/5 = 32

Squared deviations

Σ(x − x̄)² = 8² + (−1)² + 25² + (−21)² + (−11)² = 1252

Standard deviation

s = √(1252/4) = 17.69

Answer

s = 17.69
10.Descriptive statistics

Answer: z = 0.4

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 = (510 − 500)/25 = 0.4

Answer

z = 0.4
11.Probability

Answer: P = 0.1333

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 = 4/10 · 3/9 = 12/90 = 0.1333

Answer

P = 0.1333
12.Probability

Answer: P = 0.3125

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 = 2) = 10·0.52·0.53 = 10·0.25·0.125 = 0.3125

Answer

P = 0.3125
13.Normal distribution

Answer: P = 0.4277

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 = (26.3 − 32)/6 = −0.95
z₂ = (33.5 − 32)/6 = 0.25

Area between

P = Φ(0.25) − Φ(−0.95) = 0.5987 − 0.1711 = 0.4277

Answer

P = 0.4277
14.Normal distribution

Answer: x = 473 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) = 0.25 → z = −0.6745

Back to x

x = 500 + −0.6745·40 = 473.02 h

Answer

x = 473.02 h
15.Confidence intervals

Answer: lower limit = −14.7 h

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Point estimate

x̄₁ − x̄₂ = 86 − 97 = −11

Standard error

SE = √(18²/80 + 9²/80) = 2.25

Margin of error

E = 1.645·2.25 = 3.701

Interval

−11 ± 3.701 → [−14.7, −7.299]

Answer

lower limit = −14.7 h

The interval does not contain 0, so the means differ significantly.

16.Confidence intervals

Answer: upper limit = 10.34 kg

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Pooled standard deviation

sp² = (6·5.5² + 13·3²)/19 = 15.71
sp = 3.964

Margin of error

E = 2.093·3.964·√(1/7 + 1/14) = 3.84

Interval

6.5 ± 3.84 → [2.66, 10.34]

Answer

upper limit = 10.34 kg
17.Hypothesis tests & confidence intervals

Answer: t = 1.125

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/√81 = 4/9 = 0.4444

Test statistic

t = (200.5 − 200)/0.4444 = 1.125

Answer

t = 1.125
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = −2.069

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.

Sample proportions

p̂₁ = 75/250 = 0.3
p̂₂ = 34/80 = 0.425
p̂ = 109/330 = 0.3303

Standard error under H₀

SE = √(0.3303·0.6697·(1/250 + 1/80)) = 0.06041

Test statistic

z = (0.3 − 0.425)/0.06041 = −2.069

Answer

z = −2.069

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

19.Linear regression & correlation

Answer: r = 0.9969

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 = 1590 − 42·185/6 = 295
Sxx = 364 − 42²/6 = 70
Syy = 6955 − 185²/6 = 1251

Correlation coefficient

r = 295/√(70·1251) = 0.9969

Answer

r = 0.9969
20.Linear regression & correlation

Answer: ŷ = 5.55

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 = 63/5 = 12.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 94

Regression line

b = 94/40 = 2.35
a = 12.6 − 2.35·6 = −1.5
ŷ = −1.5 + 2.35·x

Prediction

ŷ = −1.5 + 2.35·3 = 5.55

Answer

ŷ = 5.55

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