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…16€5 000FV = ?

You invest €5 000 at 7% per year with continuous compounding. Find the value FV after 16 years.

PV = €5 000r = 7%t = 16
FV = €
2.Time value of moneyHard
years0123…15€3 000€3 000€3 000FV = ?

You deposit €3 000 at the start of every year for 15 years, the first deposit today. The account pays 5% per year. How much is in the account at the end of year 15? Find FV.

A = €3 000i = 5%n = 15
FV = €
3.NPV & IRRMedium
years0123…9−200 00019 000+1%+1%…r = 9%, NPV = ?

A project costs €200 000 today. It generates €19 000 at the end of the first year, and the cash flow then grows by 1% per year forever. The cost of capital is 9%. Find the net present value NPV.

C0 = €200 000CF1 = €19 000g = 1%r = 9%
NPV = €
4.NPV & IRRHard
years0123−20 00015 00013 00013 500

A project costs €20 000 and returns €15 000, €13 000, €13 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 15 000CF2 = 13 000CF3 = 13 500
IRR = %
5.Bond pricing & yieldMedium

You buy a bond with 12 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 5%. Find your holding period return HPR in percent.

F = €1 000coupon = 2%n = 12y0 = 3.5%y1 = 5%
HPR = %
6.Bond pricing & yieldHard
years0123…8€1 100€60€60 + €1 000

A bond with a face value of €1 000 and an annual coupon of 6% matures in 8 years and sells for €1 100. Find its yield to maturity YTM in percent.

F = €1 000coupon = 6%n = 8P = €1 100
YTM = %
7.CAPM & WACCMedium
WACC = 10.54%equity€5 000 000re​ = ?debt€1 000 000rd​ = 4.5%T = 28%

A company has equity worth €5 000 000 and debt worth €1 000 000. Its WACC is 10.54%, its pre-tax cost of debt is 4.5% and the tax rate is 28%. Find the cost of equity re.

E = €5 000 000D = €1 000 000WACC = 10.54%rd = 4.5%T = 28%
re = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €43.50β = 0.6rf​ = 3%, rm​ = 7%debt5 000 × €980rd​ = 4%T = 30%

A company has 2 500 000 shares at €43.50 each and 5 000 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 7%. Find the WACC.

β = 0.6rf = 3%rm = 7%rd = 4%T = 30%
WACC = %
9.Descriptive statisticsMedium
152025303540

Find the sample standard deviation s of the data: 13, 13, 13, 42, 40, 42.

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

A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 82?

μ = 70σ = 8x = 82
P =
11.ProbabilityMedium
n = 9k = 5N = ?

In how many ways N can a committee of 5 people be chosen from a group of 9? The order does not matter.

n = 9k = 5
N =
12.ProbabilityHard
n = 10, p = 0.3: P(X ≤ 1) = ?k012345678910

10 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?

n = 10p = 0.3
P =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9736

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9736. Find z to two decimals.

A = 0.9736
z =
14.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 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 = 25.03 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
n = 7, s = 13LU = ?x̄ = 71E90%

A sample of 7 items has mean 71 and standard deviation s = 13. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 7x̄ = 71s = 1390%t(6) = 1.943
upper limit =
16.Confidence intervalsHard
σ = 38, n = ?x̄ − Ex̄ + Ex̄E = 395%

The population standard deviation is 38. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 3?

σ = 38E = 395%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.42.4p = ?

The mean of a population is claimed to be 100; the population standard deviation is known to be 5. A random sample of 100 observations has a mean of 98.8. Test H₀: μ = 100 against H₁: μ ≠ 100. Find the p-value p.

μ₀ = 100σ = 5n = 100x̄ = 98.8H₁: μ ≠ μ₀
p =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 500 against H₁: μ > 500 is done at α = 0.10 with a sample of 64; the standard deviation is known to be 12. In reality the mean is 501.95. Find the power 1 − β of the test in percent.

μ₀ = 500μ = 501.95σ = 12n = 64α = 0.10
1 − β = %
19.Linear regression & correlationMedium
xy2248698121016xy51051015

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy145016581874207322902498xy1520256080100ŷ = 83

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 83?

n = 6ŷ = 83
x =

3Solutions

1.Time value of money

Answer: FV = 15324 €

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 = 0.07·16 = 1.12
e1.12 = 3.065

Future value

FV = 5 000·3.065 = €15 324.27

Answer

FV = €15 324.27
2.Time value of money

Answer: FV = 67972 €

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.

Ordinary annuity factor

((1 + 0.05)15 − 1)/0.05 = 21.58

Annuity due

Each deposit is invested one year longer.

FV = 3 000·21.58·1.05 = €67 972.48

Answer

FV = €67 972.48
3.NPV & IRR

Answer: NPV = 37500 €

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 = 19 000/(0.09 − 0.01) = 237 500.00

NPV

NPV = 237 500.00 − 200 000 = €37 500.00

Answer

NPV = €37 500.00

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

4.NPV & IRR

Answer: IRR = 49.02 %

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(45%) = 956.17
NPV(50%) = −222.22

Narrow down

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

IRR ≈ 45 + 5·956.17/(956.17 − (−222.22)) = 49.06%
exact: NPV(49.02%) = 0 → IRR = 49.02%

Answer

IRR = 49.02%
5.Bond pricing & yield

Answer: HPR = −9.852 %

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 = 12
€20.00·(1 − 0.6618)/0.035 + €1 000·0.6618 = €193.27 + €661.78 = €855.05

Selling price

C = €20.00, i = 0.05, N = 11
€20.00·(1 − 0.5847)/0.05 + €1 000·0.5847 = €166.13 + €584.68 = €750.81

Holding period return

HPR = (20 + 750.81 − 855.05)/855.05 = −9.852%

Answer

HPR = −9.852%
6.Bond pricing & yield

Answer: YTM = 4.485 %

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

€1 100 = €60·(1 − (1 + y)⁻⁸)/y + €1 000/(1 + y)⁸

Solve numerically

P(4%) = €1 134.65, P(5%) = €1 064.63
YTM = 4.485%

Answer

YTM = 4.485%
7.CAPM & WACC

Answer: re = 12 %

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 = €6 000 000, E/V = 0.8333, D/V = 0.1667

Debt part of the WACC

0.1667·4.5%·(1 − 0.28) = 0.54%

Cost of equity

re = (10.54% − 0.54%)/0.8333 = 12%

Answer

re = 12%
8.CAPM & WACC

Answer: WACC = 5.288 %

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 = 2 500 000·€43.50 = €108 750 000
D = 5 000·€980 = €4 900 000

Cost of equity (CAPM)

re = 3% + 0.6·4% = 5.4%

Weights

V = €113 650 000, E/V = 0.9569, D/V = 0.04311

WACC

WACC = 0.9569·5.4% + 0.04311·4%·(1 − 0.3) = 5.288%

Answer

WACC = 5.288%
9.Descriptive statistics

Answer: s = 15.54

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 = 163/6 = 27.17

Squared deviations

Σ(x − x̄)² = (−14.17)² + (−14.17)² + (−14.17)² + 14.83² + 12.83² + 14.83² = 1207

Standard deviation

s = √(1207/5) = 15.54

Answer

s = 15.54
10.Descriptive statistics

Answer: P = 0.9332

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 0.9332

Answer

P = 0.9332
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, 5) = (9·8·7·6·5)/(5·4·3·2·1) = 126

Answer

N = 126
12.Probability

Answer: P = 0.1493

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) = 0.710 = 0.02825
P(X = 1) = 10·0.3·0.79 = 0.1211

Add them

P = 0.02825 + 0.1211 = 0.1493

Answer

P = 0.1493
13.Normal distribution

Answer: z = 2.22

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 + 0.9736)/2 = 0.9868

Read the z-table backwards

Find 0.9868 inside the table and read off z.

z = 2.22

Answer

z = 2.22
14.Normal distribution

Answer: P = 0.8849

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 = 25.03 − 25 = 0.03 mm, σ = 0.025 mm

Standardise

z = (0 − 0.03)/0.025 = −1.2

Probability

P = 1 − Φ(−1.2) = 0.8849

Answer

P = 0.8849
15.Confidence intervals

Answer: upper limit = 80.55

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 = 6, t = 1.943

Margin of error

E = 1.943·13/√7 = 9.547

Interval

71 ± 9.547 → [61.45, 80.55]

Answer

upper limit = 80.55
16.Confidence intervals

Answer: n = 617

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 = (1.96·38/3)² = 616.4

Round up

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

n = 617

Answer

n = 617
17.Hypothesis tests & confidence intervals

Answer: p = 0.0164

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.

Test statistic

SE = σ/√n = 5/√100 = 0.5
z = (98.8 − 100)/0.5 = −2.4

Tail area

Φ(2.4) = 0.9918
1 − Φ(2.4) = 0.0082

p-value

p = 2·0.0082 = 0.0164

Answer

p = 0.0164

p ≤ 0.05, so H₀ is rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: 1 − β = 50.72 %

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 = 500 + 1.282·1.5 = 501.9

z under the true mean

z = (501.9 − 501.95)/1.5 = −0.018

Power

power = P(Z > −0.018) = 50.72%
1 − β = 50.72%

Answer

1 − β = 50.72%
19.Linear regression & correlation

Answer: r = 0.9774

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 = 47/5 = 9.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 64
Syy = Σ(y − ȳ)² = 107.2

Correlation coefficient

r = 64/√(40·107.2) = 0.9774

Answer

r = 0.9774
20.Linear regression & correlation

Answer: x = 20.92

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 = 114/6 = 19
ȳ = 443/6 = 73.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 335

Regression line

b = 335/70 = 4.786
a = 73.83 − 4.786·19 = −17.1
ŷ = −17.1 + 4.786·x

Solve for x

x = (83 − (−17.1))/4.786 = 20.92

Answer

x = 20.92

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