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…8€20 000FV = ?

You invest €20 000 at a nominal rate of 4.8% per year, compounded monthly. Find the value FV of the investment after 8 years.

PV = €20 0004.8% / 12n = 8
FV = €
2.Time value of moneyHard
months0123…72€20 000A?AAA

A loan of €20 000 is repaid with equal monthly payments over 6 years. The interest rate is 3% per year, compounded monthly. Find the monthly payment A.

P = €20 0003% / 12N = 72
A = €
3.NPV & IRRMedium
years0123−45 00014 00013 00018 000

An investment of €45 000 gives these cash flows at the end of years 1 to 3: €14 000, €13 000, €18 000. The cost of capital is 6%. Find the net present value NPV.

r = 6%CF0 = −45 000CF1 = 14 000CF2 = 13 000CF3 = 18 000
NPV = €
4.NPV & IRRHard
years0123…10−50 0005 000+2%+2%+2%r = 10%, NPV = ?

A project costs €50 000 today. Its cash flow is €5 000 at the end of year 1 and grows by 2% per year until the last cash flow at the end of year 10. The cost of capital is 10%. Find the net present value NPV.

C0 = €50 000CF1 = €5 000g = 2%n = 10r = 10%
NPV = €
5.Bond pricing & yieldMedium
years0123…14P€60€60€60 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 6% and matures in 14 years. Its yield to maturity changes from 8% to 6.5%. By how much ΔP does the bond price change?

F = €1 000coupon = 6%n = 14y1 = 8%y2 = 6.5%
ΔP = €
6.Bond pricing & yieldHard
half-years0123…8€769€10€10€10 + €1 000

A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 4 years. It sells for €769. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 2% (2×)n = 4P = €769
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€4 000 000re​ = 13%debt€3 000 000rd​ = 6%T = 30%

A company has equity worth €4 000 000 and debt worth €3 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 6% and the tax rate is 30%. Find the WACC.

E = €4 000 000D = €3 000 000re = 13%rd = 6%T = 30%
WACC = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €23.00D0​ = €2.60g = 4%debt€13 800 000rd​ = 4%T = 30%

A company has 1 000 000 shares trading at €23.00; the book value of its equity is €9 200 000. It just paid a dividend of €2.60 per share, and dividends grow by 4% per year. Its debt has a market value of €13 800 000 and a pre-tax cost of 4%. The tax rate is 30%. Find the WACC, using the dividend growth model for the cost of equity.

N = 1 000 000P0 = €23.00book equity = €9 200 000D0 = €2.60g = 4%D = €13 800 000rd = 4%T = 30%
WACC = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 51, 40, 15, 45, 24.

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

Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 46.

μ = 70σ = 20x = 46
z =
11.ProbabilityMedium
n = 15k = 5N = ?

In how many ways N can a committee of 5 people be chosen from a group of 15? The order does not matter.

n = 15k = 5
N =
12.ProbabilityHard
n = 11, p = 0.2: P(X = 4) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 4 correct answers?

n = 11p = 0.2k = 4
P =
13.Normal distributionMedium
240.25254P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 240.25 and 254 g? Give P as a decimal.

μ = 250σ = 5x1 = 240.25x2 = 254
P =
14.Normal distributionHard
μ = 32x = ?97.5%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 32σ = 6P = 97.5%
x = min
15.Confidence intervalsMedium
n = 21, s = 4.5LUx̄ = 91E = ?95%

A sample of 21 items has mean 91 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 = 21x̄ = 91s = 4.595%t(20) = 2.086
E =
16.Confidence intervalsHard
p ≈ 0.15, n = ?p̂ − Ep̂ + Ep̂E = 0.0599%

A pilot study suggests that about 0.15 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.05?

p ≈ 0.15E = 0.0599%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 197.1 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 197.1s = 8n = 36
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 50 of 250 patients on treatment A recover, and 24 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 50n₁ = 250x₂ = 24n₂ = 200
z =
19.Linear regression & correlationMedium
xy23496981210141219xy5105101520

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy249358458563663xy246505560x = 4.5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.

n = 5x = 4.5
ŷ =

3Solutions

1.Time value of money

Answer: FV = 29340 €

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 = 0.048/12 = 0.004
N = m·n = 12·8 = 96

Future value

FV = 20 000·(1 + 0.004)96 = 20 000·1.467 = €29 340.43

Answer

FV = €29 340.43
2.Time value of money

Answer: A = 303.9 €

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 payments

i = 0.03/12 = 0.0025
N = 12·6 = 72

Payment

The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.

(1 + i)−N = 0.8355
A = 20 000·0.0025/(1 − 0.8355) = €303.87

Answer

A = €303.87
3.NPV & IRR

Answer: NPV = −5109 €

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 = −45 000
CF1/(1 + 0.06)1 = 14 000/1.06 = 13 207.55
CF2/(1 + 0.06)2 = 13 000/1.124 = 11 569.95
CF3/(1 + 0.06)3 = 18 000/1.191 = 15 113.15

Add them up

NPV = €−5 109.35

Answer

NPV = €−5 109.35

NPV is negative, so the project earns less than the required return.

4.NPV & IRR

Answer: NPV = −16873 €

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 the growing annuity

((1.02)/(1.1))10 = 0.47
PV = 5 000/(0.1 − 0.02)·(1 − 0.47) = 33 126.56

NPV

NPV = 33 126.56 − 50 000 = −€16 873.44

Answer

NPV = −€16 873.44

NPV is negative, so the project earns less than the required return.

5.Bond pricing & yield

Answer: ΔP = 119.8 €

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.

Price at the old yield

C = €60.00, i = 0.08, N = 14
€60.00·(1 − 0.3405)/0.08 + €1 000·0.3405 = €494.65 + €340.46 = €835.12

Price at the new yield

C = €60.00, i = 0.065, N = 14
€60.00·(1 − 0.4141)/0.065 + €1 000·0.4141 = €540.83 + €414.10 = €954.93

Change

ΔP = 954.93 − 835.12 = €119.82

Answer

ΔP = €119.82
6.Bond pricing & yield

Answer: YTM = 9.005 %

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 = €10.00, N = 8

Solve numerically

i = 4.503% per half-year

YTM

YTM = 2·4.503% = 9.005%

Answer

YTM = 9.005%
7.CAPM & WACC

Answer: WACC = 9.229 %

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 = €7 000 000, E/V = 0.5714, D/V = 0.4286

WACC

WACC = 0.5714·13% + 0.4286·6%·(1 − 0.3) = 9.229%

Answer

WACC = 9.229%
8.CAPM & WACC

Answer: WACC = 10.9 %

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).

Cost of equity

D1 = 2.6·1.04 = €2.70
re = 2.704/23 + 0.04 = 15.76%

Market values

E = 1 000 000·€23.00 = €23 000 000
V = €36 800 000, E/V = 0.625, D/V = 0.375

WACC

WACC = 0.625·15.76% + 0.375·4%·(1 − 0.3) = 10.9%

Answer

WACC = 10.9%
9.Descriptive statistics

Answer: s = 15.02

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 = 175/5 = 35

Squared deviations

Σ(x − x̄)² = 16² + 5² + (−20)² + 10² + (−11)² = 902

Standard deviation

s = √(902/4) = 15.02

Answer

s = 15.02
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 = (46 − 70)/20 = −1.2

Answer

z = −1.2
11.Probability

Answer: N = 3003

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(15, 5) = (15·14·13·12·11)/(5·4·3·2·1) = 3003

Answer

N = 3003
12.Probability

Answer: P = 0.1107

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 = 4) = 330·0.24·0.87 = 330·0.0016·0.2097 = 0.1107

Answer

P = 0.1107
13.Normal distribution

Answer: P = 0.7626

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 = (240.25 − 250)/5 = −1.95
z₂ = (254 − 250)/5 = 0.8

Area between

P = Φ(0.8) − Φ(−1.95) = 0.7881 − 0.02559 = 0.7626

Answer

P = 0.7626
14.Normal distribution

Answer: x = 43.76 min

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) = 0.975 → z = 1.96

Back to x

x = 32 + 1.96·6 = 43.76 min

Answer

x = 43.76 min
15.Confidence intervals

Answer: E = 2.048

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 = 20, t = 2.086

Margin of error

E = 2.086·4.5/√21 = 2.048

Answer

E = 2.048
16.Confidence intervals

Answer: n = 339

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 2.576²·0.15·0.85/0.05² = 338.4

Round up

n = 339

Answer

n = 339
17.Hypothesis tests & confidence intervals

Answer: t = −2.175

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 = 8/√36 = 8/6 = 1.333

Test statistic

t = (197.1 − 200)/1.333 = −2.175

Answer

t = −2.175
t0.025; df = 35 = 2.03

|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = 2.275

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 = 50/250 = 0.2
p̂₂ = 24/200 = 0.12
p̂ = 74/450 = 0.1644

Standard error under H₀

SE = √(0.1644·0.8356·(1/250 + 1/200)) = 0.03517

Test statistic

z = (0.2 − 0.12)/0.03517 = 2.275

Answer

z = 2.275

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

19.Linear regression & correlation

Answer: r = 0.9694

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 = 66/6 = 11

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 98
Syy = Σ(y − ȳ)² = 146

Correlation coefficient

r = 98/√(70·146) = 0.9694

Answer

r = 0.9694
20.Linear regression & correlation

Answer: ŷ = 59.85

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 = 20/5 = 4
ȳ = 291/5 = 58.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 33

Regression line

b = 33/10 = 3.3
a = 58.2 − 3.3·4 = 45
ŷ = 45 + 3.3·x

Prediction

ŷ = 45 + 3.3·4.5 = 59.85

Answer

ŷ = 59.85

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