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…10€5 000FV = ?

You invest €5 000 at a nominal rate of 6% per year, compounded monthly. Find the value FV of the investment after 10 years.

PV = €5 0006% / 12n = 10
FV = €
2.Time value of moneyHard
months0123…324€50€50€50FV = ?

You save €50 at the end of every month for 27 years. The account pays 3% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €503% / 12N = 324
FV = €
3.NPV & IRRMedium
years01234−30 0009 5009 5009 5009 500

A project costs €30 000 today and returns €9 500 at the end of each year for 4 years. The cost of capital is 6%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 6%CF0 = −30 000CF1 = 9 500CF2 = 9 500CF3 = 9 500CF4 = 9 500
DPB = years
4.NPV & IRRHard
r (%)NPVABr* = ?0

Project A costs €30 000 and returns €16 000, €12 000, €6 000 at the end of years 1 to 3. Project B costs €35 000 and returns €9 000, €13 500, €18 000. Find the crossover rate r* at which both projects have the same NPV, in percent.

A: −30 000, 16 000, 12 000, 6 000B: −35 000, 9 000, 13 500, 18 000
r* = %
5.Bond pricing & yieldMedium
half-years0123…20€10€10€10 + €1 000P = ?

A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 10 years. The yield to maturity is 9% per year. Find the price P of the bond.

F = €1 000coupon = 2%YTM = 9%n = 10
P = €
6.Bond pricing & yieldHard
years0123456€1 098€80€80€80 + €1 080

A bond with 17 years to maturity (face value €1 000, annual coupon 8%) sells for €1 098. The issuer can call it in 6 years at a call price of €1 080. Find the yield to call YTC in percent.

F = €1 000coupon = 8%P = €1 098call price = €1 080call in 6 years
YTC = %
7.CAPM & WACCMedium
WACC = ?equity€4 000 000re​ = 9%debt€6 000 000rd​ = 5%T = 28%

A company has equity worth €4 000 000 and debt worth €6 000 000. Its cost of equity is 9%, its pre-tax cost of debt is 5% and the tax rate is 28%. Find the WACC.

E = €4 000 000D = €6 000 000re = 9%rd = 5%T = 28%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €30.00β = 1.2rf​ = 1.5%, rm​ = 8.5%debt20 000 × €1 050rd​ = 7%T = 20%

A company has 2 000 000 shares at €30.00 each and 20 000 bonds trading at €1 050 each. The pre-tax cost of debt is 7% and the tax rate is 20%. The share’s beta is 1.2, the risk-free rate 1.5% and the market return 8.5%. Find the WACC.

β = 1.2rf = 1.5%rm = 8.5%rd = 7%T = 20%
WACC = %
9.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 38, 46, 24, 45, 59.

n = 5
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 155.

μ = 170σ = 10x = 155
z =
11.ProbabilityMedium
red: 8blue: 4

A bag holds 8 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 4
P =
12.ProbabilityHard
n = 12, p = 0.1: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?

n = 12p = 0.1
P =
13.Normal distributionMedium
500560P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 500 and 560 h? Give P as a decimal.

μ = 500σ = 40x1 = 500x2 = 560
P =
14.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 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 = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
15.Confidence intervalsMedium
n = 25, s = 15L = ?Ux̄ = 23.5E95%

A sample of 25 items has mean 23.5 and standard deviation s = 15. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 25x̄ = 23.5s = 1595%t(24) = 2.064
lower limit =
16.Confidence intervalsHard
σ = 11, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 11. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 4?

σ = 11E = 499%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 200.8 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 200.8s = 4n = 81
t =
18.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 81 measurements has mean 51.2. The population standard deviation is 8. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 51.2σ = 8n = 8190%
L =
19.Linear regression & correlationMedium
xy21041962883510411252xy5102040

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
xy2474626768871095xy5106080100x = 5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (5, 70) is made. Find its residual e = y − ŷ.

n = 5(5, 70)
e =

3Solutions

1.Time value of money

Answer: FV = 9097 €

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 = 0.06/12 = 0.005
N = m·n = 12·10 = 120

Future value

FV = 5 000·(1 + 0.005)120 = 5 000·1.819 = €9 096.98

Answer

FV = €9 096.98
2.Time value of money

Answer: FV = 24913 €

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 deposits

i = 0.03/12 = 0.0025
N = 12·27 = 324

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 50·(2.246 − 1)/0.0025 = €24 912.74

Answer

FV = €24 912.74

You paid in €16 200 yourself; the rest is interest.

3.NPV & IRR

Answer: DPB = 3.612 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: 9 500/1.06 = 8 962.26 → Σ = 8 962.26
year 2: 9 500/1.124 = 8 454.97 → Σ = 17 417.23
year 3: 9 500/1.191 = 7 976.38 → Σ = 25 393.61
year 4: 9 500/1.262 = 7 524.89 → Σ = 32 918.50

Fraction of the last year

DPB = 3 + 4 606.39/7 524.89 = 3.612

Answer

DPB = 3.612 years

The discounted payback is always longer than the simple payback (3.158 years).

4.NPV & IRR

Answer: r* = 5.233 %

Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.

Difference in cash flows

ΔCF0 = −35 000 − (−30 000) = −5 000
ΔCF1 = 9 000 − 16 000 = −7 000
ΔCF2 = 13 500 − 12 000 = 1 500
ΔCF3 = 18 000 − 6 000 = 12 000

IRR of the difference

NPVΔ(5%) = 59.93
NPVΔ(6%) = −193.35
r* = 5.233%

Answer

r* = 5.233%

Check: at 5.233% both NPVs are €1 189.30.

5.Bond pricing & yield

Answer: P = 544.7 €

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, i = 0.045, N = 20

Present value of the coupons

€10.00·(1 − 0.4146)/0.045 = €130.08

Present value of the face value

€1 000·0.4146 = €414.64

Price

P = €544.72

Answer

P = €544.72

The yield is above the coupon rate, so the bond sells at a discount.

6.Bond pricing & yield

Answer: YTC = 7.057 %

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

1098 = 80·(1 − (1 + y)−6)/y + 1080/(1 + y)6

Solve numerically

y = 7%: €1 100.97
y = 8%: €1 050.41
YTC = 7.057%

Answer

YTC = 7.057%

The yield to maturity is 6.996%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.

7.CAPM & WACC

Answer: WACC = 5.76 %

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 = €10 000 000, E/V = 0.4, D/V = 0.6

WACC

WACC = 0.4·9% + 0.6·5%·(1 − 0.28) = 5.76%

Answer

WACC = 5.76%
8.CAPM & WACC

Answer: WACC = 8.785 %

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 = 2 000 000·€30.00 = €60 000 000
D = 20 000·€1 050 = €21 000 000

Cost of equity (CAPM)

re = 1.5% + 1.2·7% = 9.9%

Weights

V = €81 000 000, E/V = 0.7407, D/V = 0.2593

WACC

WACC = 0.7407·9.9% + 0.2593·7%·(1 − 0.2) = 8.785%

Answer

WACC = 8.785%
9.Descriptive statistics

Answer: s = 12.78

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 = 212/5 = 42.4

Squared deviations

Σ(x − x̄)² = (−4.4)² + 3.6² + (−18.4)² + 2.6² + 16.6² = 653.2

Standard deviation

s = √(653.2/4) = 12.78

Answer

s = 12.78
10.Descriptive statistics

Answer: z = −1.5

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 = (155 − 170)/10 = −1.5

Answer

z = −1.5
11.Probability

Answer: P = 0.4242

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.

First and second draw

P = 8/12 · 7/11 = 56/132 = 0.4242

Answer

P = 0.4242
12.Probability

Answer: P = 0.659

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) = 0.912 = 0.2824
P(X = 1) = 12·0.1·0.911 = 0.3766

Add them

P = 0.2824 + 0.3766 = 0.659

Answer

P = 0.659
13.Normal distribution

Answer: P = 0.4332

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 = (500 − 500)/40 = 0
z₂ = (560 − 500)/40 = 1.5

Area between

P = Φ(1.5) − Φ(0) = 0.9332 − 0.5 = 0.4332

Answer

P = 0.4332
14.Normal distribution

Answer: P = 0.9099

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 = 0.02236 mm

Standardise

z = (0 − 0.03)/0.02236 = −1.34

Probability

P = 1 − Φ(−1.34) = 0.9099

Answer

P = 0.9099
15.Confidence intervals

Answer: lower limit = 17.31

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 = 24, t = 2.064

Margin of error

E = 2.064·15/√25 = 6.192

Interval

23.5 ± 6.192 → [17.31, 29.69]

Answer

lower limit = 17.31
16.Confidence intervals

Answer: n = 51

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 = (2.576·11/4)² = 50.18

Round up

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

n = 51

Answer

n = 51
17.Hypothesis tests & confidence intervals

Answer: t = 1.8

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 = 4/√81 = 4/9 = 0.4444

Test statistic

t = (200.8 − 200)/0.4444 = 1.8

Answer

t = 1.8
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: L = 49.74

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 = 1.645·8/√81 = 1.462

Interval limit

L = 51.2 − 1.462 = 49.74

Answer

L = 49.74
19.Linear regression & correlation

Answer: r = 0.9971

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 = 185/6 = 30.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 283
Syy = Σ(y − ȳ)² = 1151

Correlation coefficient

r = 283/√(70·1151) = 0.9971

Answer

r = 0.9971
20.Linear regression & correlation

Answer: e = 2.65

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 = 367/5 = 73.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 242

Regression line

b = 242/40 = 6.05
a = 73.4 − 6.05·6 = 37.1
ŷ = 37.1 + 6.05·x

Predicted value

ŷ = 37.1 + 6.05·5 = 67.35

Residual

e = 70 − 67.35 = 2.65

Answer

e = 2.65

The point lies above the line.

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