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
periods0123411 + EAR = ?17.5% / 4

A credit card charges a nominal interest rate of 17.5% per year, compounded quarterly. Find the effective annual rate EAR in percent.

r = 17.5%m = 4
EAR = %
2.Time value of moneyHard
months01…60…360€250 000AB = ?A6% / 12

A mortgage of €250 000 is repaid with equal monthly payments over 30 years at 6% per year, compounded monthly. How much is still owed just after the payment at the end of year 5? Find the outstanding balance B.

P = €250 0006% / 12N = 360k = 60
B = €
3.NPV & IRRMedium
years01234−75 00028 00021 00022 00037 000

An investment of €75 000 gives these cash flows at the end of years 1 to 4: €28 000, €21 000, €22 000, €37 000. The cost of capital is 8%. Find the net present value NPV.

r = 8%CF0 = −75 000CF1 = 28 000CF2 = 21 000CF3 = 22 000CF4 = 37 000
NPV = €
4.NPV & IRRHard
years012−20 0008 00014 500

A project costs €20 000 and returns €8 000, €14 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 8 000CF2 = 14 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…16€35€35€35 + €1 000P = ?

A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 8 years. The yield to maturity is 3.5% per year. Find the price P of the bond.

F = €1 000coupon = 7%YTM = 3.5%n = 8
P = €
6.Bond pricing & yieldHard
years012345€1 090€50€50 + €1 000

A bond with a face value of €1 000 and an annual coupon of 5% matures in 5 years and sells for €1 090. Find its yield to maturity YTM in percent.

F = €1 000coupon = 5%n = 5P = €1 090
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€12 000 000re​ = 12%debt€2 000 000rd​ = 3%T = 25%

A company has equity worth €12 000 000 and debt worth €2 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 3% and the tax rate is 25%. Find the WACC.

E = €12 000 000D = €2 000 000re = 12%rd = 3%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €36.00β = 1.1rf​ = 1.5%, rm​ = 5.5%debt40 000 × €960rd​ = 6%T = 28%

A company has 1 000 000 shares at €36.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 6% and the tax rate is 28%. The share’s beta is 1.1, the risk-free rate 1.5% and the market return 5.5%. Find the WACC.

β = 1.1rf = 1.5%rm = 5.5%rd = 6%T = 28%
WACC = %
9.Descriptive statisticsMedium
4045505560

Find the sample standard deviation s of the data: 54, 36, 49, 51, 60.

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

Exam scores are normally distributed with mean 170 and standard deviation 15. Find the z-score z of a score of 152.

μ = 170σ = 15x = 152
z =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7?8P(six) = 1/6 on each roll

A fair die is rolled 8 times. What is the probability P of getting at least one six?

n = 8
P =
12.ProbabilityHard
n = 7, p = 0.2: P(X = 2) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 2 correct answers?

n = 7p = 0.2k = 2
P =
13.Normal distributionMedium
32 min33.3 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 33.3 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 33.3 min
P =
14.Normal distributionHard
380420460500540580620hbottom 75%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 75% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h75%
x = h
15.Confidence intervalsMedium
n = 6, s = 11L = ?Ux̄ = 110.5E90%

A sample of 6 items has mean 110.5 and standard deviation s = 11. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 6x̄ = 110.5s = 1190%t(5) = 2.015
lower limit =
16.Confidence intervalsHard
n = 21Lσ = ?s = 7.495%

The diameters of 21 machined shafts have a sample standard deviation of 7.4 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.

n = 21s = 7.4 µmχ² = 9.591, 34.17
σ = µm
17.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 82.2 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 82.2s = 15n = 25
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 28 of 80 patients on treatment A recover, and 100 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 28n₁ = 80x₂ = 100n₂ = 200
z =
19.Linear regression & correlationMedium
xy16461956225725712884xy20256080

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 110Σy = 314Σx² = 2510Σy² = 20598Σxy = 7181
r =
20.Linear regression & correlationHard
xy12422131741351169xy24610152025x = 4.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.

n = 6x = 4.5
ŷ =

3Solutions

1.Time value of money

Answer: EAR = 18.68 %

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 per period

r/m = 0.175/4 = 0.04375

Compound over one year

EAR = (1 + 0.04375)4 − 1 = 1.187 − 1 = 18.68%

Answer

EAR = 18.68%

The EAR is always above the nominal rate when interest compounds more than once a year.

2.Time value of money

Answer: B = 232636 €

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 = 360
A = 250 000·0.005/(1 − (1 + 0.005)−360) = €1 498.88

Present value of the remaining payments

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

N − k = 360 − 60 = 300
B = 1499·(1 − (1 + 0.005)−300)/0.005 = €232 635.89

Answer

B = €232 635.89
3.NPV & IRR

Answer: NPV = 13590 €

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 = −75 000
CF1/(1 + 0.08)1 = 28 000/1.08 = 25 925.93
CF2/(1 + 0.08)2 = 21 000/1.166 = 18 004.12
CF3/(1 + 0.08)3 = 22 000/1.26 = 17 464.31
CF4/(1 + 0.08)4 = 37 000/1.36 = 27 196.10

Add them up

NPV = €13 590.46

Answer

NPV = €13 590.46

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

4.NPV & IRR

Answer: IRR = 7.464 %

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(5%) = 770.98
NPV(10%) = −743.80

Narrow down

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

IRR ≈ 5 + 5·770.98/(770.98 − (−743.80)) = 7.545%
exact: NPV(7.464%) = 0 → IRR = 7.464%

Answer

IRR = 7.464%
5.Bond pricing & yield

Answer: P = 1242 €

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 = €35.00, i = 0.0175, N = 16

Present value of the coupons

€35.00·(1 − 0.7576)/0.0175 = €484.77

Present value of the face value

€1 000·0.7576 = €757.62

Price

P = €1 242.38

Answer

P = €1 242.38

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

6.Bond pricing & yield

Answer: YTM = 3.033 %

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

€1 090 = €50·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵

Solve numerically

P(3%) = €1 091.59, P(4%) = €1 044.52
YTM = 3.033%

Answer

YTM = 3.033%
7.CAPM & WACC

Answer: WACC = 10.61 %

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 = €14 000 000, E/V = 0.8571, D/V = 0.1429

WACC

WACC = 0.8571·12% + 0.1429·3%·(1 − 0.25) = 10.61%

Answer

WACC = 10.61%
8.CAPM & WACC

Answer: WACC = 5.085 %

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·€36.00 = €36 000 000
D = 40 000·€960 = €38 400 000

Cost of equity (CAPM)

re = 1.5% + 1.1·4% = 5.9%

Weights

V = €74 400 000, E/V = 0.4839, D/V = 0.5161

WACC

WACC = 0.4839·5.9% + 0.5161·6%·(1 − 0.28) = 5.085%

Answer

WACC = 5.085%
9.Descriptive statistics

Answer: s = 8.86

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 = 250/5 = 50

Squared deviations

Σ(x − x̄)² = 4² + (−14)² + (−1)² + 1² + 10² = 314

Standard deviation

s = √(314/4) = 8.86

Answer

s = 8.86
10.Descriptive statistics

Answer: z = −1.2

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 = (152 − 170)/15 = −1.2

Answer

z = −1.2
11.Probability

Answer: P = 0.7674

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.

Complement

P(no six) = (5/6)8 = 0.2326
P = 1 − 0.2326 = 0.7674

Answer

P = 0.7674
12.Probability

Answer: P = 0.2753

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) = 21·0.22·0.85 = 21·0.04·0.3277 = 0.2753

Answer

P = 0.2753
13.Normal distribution

Answer: P = 0.3015

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 = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (33.3 − 32)/2.5 = 0.52

Upper tail

P = 1 − Φ(0.52) = 1 − 0.6985 = 0.3015

Answer

P = 0.3015
14.Normal distribution

Answer: x = 527 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) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 500 + 0.67·40 = 527 h

Answer

x = 527 h
15.Confidence intervals

Answer: lower limit = 101.5

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 = 5, t = 2.015

Margin of error

E = 2.015·11/√6 = 9.049

Interval

110.5 ± 9.049 → [101.5, 119.5]

Answer

lower limit = 101.5
16.Confidence intervals

Answer: σ = 10.69 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 20·7.4² = 1095

Limits for σ²

1095/34.17 = 32.05
1095/9.591 = 114.2

Limit for σ

σ = √114.2 = 10.69 µm

Answer

σ = 10.69 µm

95% interval for σ: [5.661, 10.69] µm.

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 = 15/√25 = 15/5 = 3

Test statistic

t = (82.2 − 75)/3 = 2.4

Answer

t = 2.4
t0.025; df = 24 = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = −2.276

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 = 28/80 = 0.35
p̂₂ = 100/200 = 0.5
p̂ = 128/280 = 0.4571

Standard error under H₀

SE = √(0.4571·0.5429·(1/80 + 1/200)) = 0.0659

Test statistic

z = (0.35 − 0.5)/0.0659 = −2.276

Answer

z = −2.276

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

19.Linear regression & correlation

Answer: r = 0.9707

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 = 7181 − 110·314/5 = 273
Sxx = 2510 − 110²/5 = 90
Syy = 20598 − 314²/5 = 878.8

Correlation coefficient

r = 273/√(90·878.8) = 0.9707

Answer

r = 0.9707
20.Linear regression & correlation

Answer: ŷ = 12.72

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 = 21/6 = 3.5
ȳ = 95/6 = 15.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −54.5

Regression line

b = −54.5/17.5 = −3.114
a = 15.83 − (−3.114)·3.5 = 26.73
ŷ = 26.73 − 3.114·x

Prediction

ŷ = 26.73 + (−3.114)·4.5 = 12.72

Answer

ŷ = 12.72

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