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€8 000FV = ?

You invest €8 000 at a nominal rate of 3% per year, compounded quarterly. Find the value FV of the investment after 9 years.

PV = €8 0003% / 4n = 9
FV = €
2.Time value of moneyHard
months0123…300€200 000AAAA3% / 12, I = ?

A loan of €200 000 is repaid with equal monthly payments over 25 years at 3% per year, compounded monthly. How much interest I is paid in total over the life of the loan?

P = €200 0003% / 12N = 300
I = €
3.NPV & IRRMedium
years012345−60 00012 00012 00012 00012 00018 000

A machine costs €60 000 today. It generates €12 000 at the end of each year for 5 years and can then be sold for €6 000. The cost of capital is 12%. Find the net present value NPV.

C0 = €60 000CF = €12 000S = €6 000n = 5r = 12%
NPV = €
4.NPV & IRRHard
years0123456−40 0007 5007 5007 5007 5007 5007 500

An investment of €40 000 returns €7 500 at the end of each year for 6 years. Find its internal rate of return IRR in percent.

C0 = €40 000CF = €7 500n = 6
IRR = %
5.Bond pricing & yieldMedium
half-years0123456€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 3 years. The yield to maturity is 3% per year. Find the price P of the bond.

F = €1 000coupon = 6%YTM = 3%n = 3
P = €
6.Bond pricing & yieldHard
years012345€960€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 €960. Find its yield to maturity YTM in percent.

F = €1 000coupon = 5%n = 5P = €960
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€2 000 000re​ = 13%debt€3 000 000rd​ = 4.5%T = 21%

A company has equity worth €2 000 000 and debt worth €3 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 4.5% and the tax rate is 21%. Find the WACC.

E = €2 000 000D = €3 000 000re = 13%rd = 4.5%T = 21%
WACC = %
8.CAPM & WACCHard
rf​ = 3.5%, rm​ − rf​ = 4%D/E = 0.2equityβL​ = 1.1debtD/E = 0.6equityre​ = ?debtT = 25%

A company's shares have a beta of 1.1 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 0.6. The tax rate is 25%, the risk-free rate 3.5% and the market risk premium 4%. Find the cost of equity re after the change.

βL = 1.1D/E1 = 0.2D/E2 = 0.6T = 25%rf = 3.5%rm − rf = 4%
re = %
9.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 35, 43, 26, 38, 29, 42, 58.

n = 7
s =
10.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 10. Find the z-score z of a score of 165.

μ = 170σ = 10x = 165
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 = 0.3: P(X ≤ 1) = ?k01234567

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

n = 7p = 0.3
P =
13.Normal distributionMedium
29.335.3P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 29.3 and 35.3 min? Give P as a decimal.

μ = 32σ = 6x1 = 29.3x2 = 35.3
P =
14.Normal distributionHard
μ = 32x = ?20%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 20% of them lie (the 20th percentile).

μ = 32σ = 6P = 20%
x = min
15.Confidence intervalsMedium
n = 12, s = 5.5LU = ?x̄ = 100E95%

A sample of 12 items has mean 100 and standard deviation s = 5.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 12x̄ = 100s = 5.595%t(11) = 2.201
upper limit =
16.Confidence intervalsHard
σ = 23, n = ?x̄ − Ex̄ + Ex̄E = 495%

The population standard deviation is 23. 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 4?

σ = 23E = 495%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 203.4 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 203.4s = 10n = 49
t =
18.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 64 measurements has mean 74.7. The population standard deviation is 4. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 74.7σ = 4n = 6495%
U =
19.Linear regression & correlationMedium
xy140243340443544645xy246404244

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy284961381310181222xy510101520x = 3

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (3, 13) is made. Find its residual e = y − ŷ.

n = 6(3, 13)
e =

3Solutions

1.Time value of money

Answer: FV = 10469 €

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 = 0.03/4 = 0.0075
N = m·n = 4·9 = 36

Future value

FV = 8 000·(1 + 0.0075)36 = 8 000·1.309 = €10 469.16

Answer

FV = €10 469.16
2.Time value of money

Answer: I = 84527 €

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 = 0.0025, N = 300
A = 200 000·0.0025/(1 − (1 + 0.0025)−300) = €948.42

Total interest

I = N·A − P = 300·948.4 − 200 000 = €84 526.79

Answer

I = €84 526.79
3.NPV & IRR

Answer: NPV = −13338 €

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 yearly cash flows

12 000·3.605 = 43 257.31

Present value of the salvage value

6 000/1.762 = 3 404.56

NPV

NPV = −60 000 + 43 257.31 + 3 404.56 = −€13 338.12

Answer

NPV = −€13 338.12
4.NPV & IRR

Answer: IRR = 3.473 %

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.

Required annuity factor

(1 − (1 + i)−6)/i = 40 000/7 500 = 5.333

Solve for the rate

The annuity factor falls as the rate rises; bracket the rate and narrow down.

i = 3%: 5.417
i = 4%: 5.242
IRR = 3.473%

Answer

IRR = 3.473%
5.Bond pricing & yield

Answer: P = 1085 €

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 = 0.015, N = 6

Present value of the coupons

€30.00·(1 − 0.9145)/0.015 = €170.92

Present value of the face value

€1 000·0.9145 = €914.54

Price

P = €1 085.46

Answer

P = €1 085.46

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

6.Bond pricing & yield

Answer: YTM = 5.948 %

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

€960 = €50·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵

Solve numerically

P(5%) = €1 000.00, P(6%) = €957.88
YTM = 5.948%

Answer

YTM = 5.948%
7.CAPM & WACC

Answer: WACC = 7.333 %

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 = €5 000 000, E/V = 0.4, D/V = 0.6

WACC

WACC = 0.4·13% + 0.6·4.5%·(1 − 0.21) = 7.333%

Answer

WACC = 7.333%
8.CAPM & WACC

Answer: re = 9.048 %

Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.

Unlever the beta

βU = 1.1/(1 + 0.75·0.2) = 0.9565

Relever at the new D/E

βL = 0.9565·(1 + 0.75·0.6) = 1.387

Cost of equity (CAPM)

re = 3.5% + 1.387·4% = 9.048%

Answer

re = 9.048%
9.Descriptive statistics

Answer: s = 10.58

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 = 271/7 = 38.71

Squared deviations

Σ(x − x̄)² = (−3.714)² + 4.286² + (−12.71)² + (−0.7143)² + (−9.714)² + 3.286² + 19.29² = 671.4

Standard deviation

s = √(671.4/6) = 10.58

Answer

s = 10.58
10.Descriptive statistics

Answer: z = −0.5

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 = (165 − 170)/10 = −0.5

Answer

z = −0.5
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 = 0.3294

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) = 0.77 = 0.08235
P(X = 1) = 7·0.3·0.76 = 0.2471

Add them

P = 0.08235 + 0.2471 = 0.3294

Answer

P = 0.3294
13.Normal distribution

Answer: P = 0.3825

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 = (29.3 − 32)/6 = −0.45
z₂ = (35.3 − 32)/6 = 0.55

Area between

P = Φ(0.55) − Φ(−0.45) = 0.7088 − 0.3264 = 0.3825

Answer

P = 0.3825
14.Normal distribution

Answer: x = 26.95 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) = 0.2 → z = −0.8416

Back to x

x = 32 + −0.8416·6 = 26.95 min

Answer

x = 26.95 min
15.Confidence intervals

Answer: upper limit = 103.5

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 = 11, t = 2.201

Margin of error

E = 2.201·5.5/√12 = 3.495

Interval

100 ± 3.495 → [96.51, 103.5]

Answer

upper limit = 103.5
16.Confidence intervals

Answer: n = 128

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 = (1.96·23/4)² = 127

Round up

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

n = 128

Answer

n = 128
17.Hypothesis tests & confidence intervals

Answer: t = 2.38

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 = 10/√49 = 10/7 = 1.429

Test statistic

t = (203.4 − 200)/1.429 = 2.38

Answer

t = 2.38
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: U = 75.68

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).

Margin of error

E = 1.96·4/√64 = 0.98

Interval limit

U = 74.7 + 0.98 = 75.68

Answer

U = 75.68
19.Linear regression & correlation

Answer: r² = 0.6385

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 = 255/6 = 42.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 15.5
Syy = Σ(y − ȳ)² = 21.5

Coefficient of determination

r² = 15.5²/(17.5·21.5) = 0.6385

Answer

r² = 0.6385

About 63.85% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = 4.71

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 = 83/6 = 13.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 97

Regression line

b = 97/70 = 1.386
a = 13.83 − 1.386·7 = 4.133
ŷ = 4.133 + 1.386·x

Predicted value

ŷ = 4.133 + 1.386·3 = 8.29

Residual

e = 13 − 8.29 = 4.71

Answer

e = 4.71

The point lies above the line.

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