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
years0123PV = ?€10 000

You need €10 000 in 3 years. An account pays 3% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €10 0003% / 4n = 3
PV = €
2.Time value of moneyHard
months01…48…240€300 000AB = ?A6% / 12

A mortgage of €300 000 is repaid with equal monthly payments over 20 years at 6% per year, compounded monthly. How much is still owed just after the payment at the end of year 4? Find the outstanding balance B.

P = €300 0006% / 12N = 240k = 48
B = €
3.NPV & IRRMedium
years012345−50 00018 50020 00018 00016 00021 000

A project costs €50 000 today and returns €18 500, €20 000, €18 000, €16 000, €21 000 at the end of years 1 to 5. The cost of capital is 8%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 8%CF0 = −50 000CF1 = 18 500CF2 = 20 000CF3 = 18 000CF4 = 16 000CF5 = 21 000
DPB = years
4.NPV & IRRHard
years0123−50 00036 00033 50031 500

A project costs €50 000 and returns €36 000, €33 500, €31 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 36 000CF2 = 33 500CF3 = 31 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…22€30€30€30 + €1 000P = ?

A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 11 years. The yield to maturity is 2.5% per year. Find the price P of the bond.

F = €1 000coupon = 6%YTM = 2.5%n = 11
P = €
6.Bond pricing & yieldHard
years0123456€880€40€40 + €1 000

A bond with a face value of €1 000 and an annual coupon of 4% matures in 6 years and sells for €880. Find its yield to maturity YTM in percent.

F = €1 000coupon = 4%n = 6P = €880
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€5 000 000re​ = 9%debt€3 000 000rd​ = 5%T = 25%

A company has equity worth €5 000 000 and debt worth €3 000 000. Its cost of equity is 9%, its pre-tax cost of debt is 5% and the tax rate is 25%. Find the WACC.

E = €5 000 000D = €3 000 000re = 9%rd = 5%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €18.50β = 1.8rf​ = 1%, rm​ = 8%debt5 000 × €1 000rd​ = 7%T = 25.8%

A company has 1 000 000 shares at €18.50 each and 5 000 bonds trading at €1 000 each. The pre-tax cost of debt is 7% and the tax rate is 25.8%. The share’s beta is 1.8, the risk-free rate 1% and the market return 8%. Find the WACC.

β = 1.8rf = 1%rm = 8%rd = 7%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
2530354045

Find the sample standard deviation s of the data: 45, 47, 41, 45, 24, 45.

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

A quantity is normally distributed with mean 100 and standard deviation 5. What proportion P of values is below 97.5?

μ = 100σ = 5x = 97.5
P =
11.ProbabilityMedium
xp(x)10.2520.2530.3540.15μ = 2.4

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

x: 1, 2, 3, 4p(x): 0.25, 0.25, 0.35, 0.15
Var(X) =
12.ProbabilityHard
n = 10, p = 0.1: P(X = 1) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 1 correct answers?

n = 10p = 0.1k = 1
P =
13.Normal distributionMedium
200 mm199.6 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.6 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.6 mm
P =
14.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
15.Confidence intervalsMedium
n = 11, s = 4.5LUx̄ = 52.5E = ?95%

A sample of 11 items has mean 52.5 and standard deviation s = 4.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 11x̄ = 52.5s = 4.595%t(10) = 2.228
E =
16.Confidence intervalsHard
σ = 9, n = ?x̄ − Ex̄ + Ex̄E = 2.590%

The population standard deviation is 9. 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 2.5?

σ = 9E = 2.590%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 30 parts has mean length 83 mm, and the line's known standard deviation is 18 mm. Line 2: a sample of 30 parts has mean 76.3 mm, known standard deviation 19 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 30x̄₁ = 83 mmσ₁ = 18 mmn₂ = 30x̄₂ = 76.3 mmσ₂ = 19 mm
z =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 30 of 120 patients on treatment A recover, and 36 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 30n₁ = 120x₂ = 36n₂ = 120
z =
19.Linear regression & correlationMedium
xy2114156268321039xy51010203040

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
xy133351559774979xy5406080ŷ = 42

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. For which value x does the line predict ŷ = 42?

n = 5ŷ = 42
x =

3Solutions

1.Time value of money

Answer: PV = 9142 €

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 = 0.03/4 = 0.0075
N = 4·3 = 12

Present value

PV = FV/(1 + i)N = 10 000/1.094 = €9 142.38

Answer

PV = €9 142.38
2.Time value of money

Answer: B = 264875 €

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 payment

i = 0.005, N = 240
A = 300 000·0.005/(1 − (1 + 0.005)−240) = €2 149.29

Present value of the remaining payments

Whatever is still owed must be exactly paid off by the remaining payments.

N − k = 240 − 48 = 192
B = 2149·(1 − (1 + 0.005)−192)/0.005 = €264 874.65

Answer

B = €264 874.65
3.NPV & IRR

Answer: DPB = 3.122 years

Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.

Discounted and cumulative cash flows

year 1: 18 500/1.08 = 17 129.63 → Σ = 17 129.63
year 2: 20 000/1.166 = 17 146.78 → Σ = 34 276.41
year 3: 18 000/1.26 = 14 288.98 → Σ = 48 565.39
year 4: 16 000/1.36 = 11 760.48 → Σ = 60 325.86

Fraction of the last year

DPB = 3 + 1 434.61/11 760.48 = 3.122

Answer

DPB = 3.122 years

The discounted payback is always longer than the simple payback (2.639 years).

4.NPV & IRR

Answer: IRR = 46.84 %

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%) = 1 093.53
NPV(50%) = −1 777.78

Narrow down

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

IRR ≈ 45 + 5·1 093.53/(1 093.53 − (−1 777.78)) = 46.9%
exact: NPV(46.84%) = 0 → IRR = 46.84%

Answer

IRR = 46.84%
5.Bond pricing & yield

Answer: P = 1335 €

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 = €30.00, i = 0.0125, N = 22

Present value of the coupons

€30.00·(1 − 0.7609)/0.0125 = €573.92

Present value of the face value

€1 000·0.7609 = €760.87

Price

P = €1 334.78

Answer

P = €1 334.78

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

6.Bond pricing & yield

Answer: YTM = 6.477 %

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

€880 = €40·(1 − (1 + y)⁻⁶)/y + €1 000/(1 + y)⁶

Solve numerically

P(6%) = €901.65, P(7%) = €857.00
YTM = 6.477%

Answer

YTM = 6.477%
7.CAPM & WACC

Answer: WACC = 7.031 %

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 = €8 000 000, E/V = 0.625, D/V = 0.375

WACC

WACC = 0.625·9% + 0.375·5%·(1 − 0.25) = 7.031%

Answer

WACC = 7.031%
8.CAPM & WACC

Answer: WACC = 11.81 %

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 = 1 000 000·€18.50 = €18 500 000
D = 5 000·€1 000 = €5 000 000

Cost of equity (CAPM)

re = 1% + 1.8·7% = 13.6%

Weights

V = €23 500 000, E/V = 0.7872, D/V = 0.2128

WACC

WACC = 0.7872·13.6% + 0.2128·7%·(1 − 0.258) = 11.81%

Answer

WACC = 11.81%
9.Descriptive statistics

Answer: s = 8.635

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 = 247/6 = 41.17

Squared deviations

Σ(x − x̄)² = 3.833² + 5.833² + (−0.1667)² + 3.833² + (−17.17)² + 3.833² = 372.8

Standard deviation

s = √(372.8/5) = 8.635

Answer

s = 8.635
10.Descriptive statistics

Answer: P = 0.3085

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 0.3085

Answer

P = 0.3085
11.Probability

Answer: Var(X) = 1.04

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 = 1·0.25 + 2·0.25 + 3·0.35 + 4·0.15 = 2.4

Mean of the squares

E(X²) = 1·0.25 + 4·0.25 + 9·0.35 + 16·0.15 = 6.8

Variance

Var(X) = 6.8 − 2.4² = 1.04

Answer

Var(X) = 1.04
12.Probability

Answer: P = 0.3874

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) = 10·0.11·0.99 = 10·0.1·0.3874 = 0.3874

Answer

P = 0.3874
13.Normal distribution

Answer: P = 0.7881

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 = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (199.6 − 200)/0.5 = −0.8

Upper tail

P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
14.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
15.Confidence intervals

Answer: E = 3.023

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 = 10, t = 2.228

Margin of error

E = 2.228·4.5/√11 = 3.023

Answer

E = 3.023
16.Confidence intervals

Answer: n = 36

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 = (1.645·9/2.5)² = 35.07

Round up

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

n = 36

Answer

n = 36
17.Hypothesis tests & confidence intervals

Answer: z = 1.402

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.

Standard error of the difference

SE = √(18²/30 + 19²/30) = √22.83 = 4.778

Test statistic

z = (83 − 76.3)/4.778 = 1.402

Answer

z = 1.402

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = −0.8674

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 = 30/120 = 0.25
p̂₂ = 36/120 = 0.3
p̂ = 66/240 = 0.275

Standard error under H₀

SE = √(0.275·0.725·(1/120 + 1/120)) = 0.05764

Test statistic

z = (0.25 − 0.3)/0.05764 = −0.8674

Answer

z = −0.8674

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

19.Linear regression & correlation

Answer: r = 0.9923

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 = 123/5 = 24.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 146
Syy = Σ(y − ȳ)² = 541.2

Correlation coefficient

r = 146/√(40·541.2) = 0.9923

Answer

r = 0.9923
20.Linear regression & correlation

Answer: x = 2.009

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 = 25/5 = 5
ȳ = 296/5 = 59.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 230

Regression line

b = 230/40 = 5.75
a = 59.2 − 5.75·5 = 30.45
ŷ = 30.45 + 5.75·x

Solve for x

x = (42 − 30.45)/5.75 = 2.009

Answer

x = 2.009

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