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
years01234€8 000FV = ?

You invest €8 000 at a nominal rate of 3.6% per year, compounded quarterly. Find the value FV of the investment after 4 years.

PV = €8 0003.6% / 4n = 4
FV = €
2.Time value of moneyHard
months0123…36A?AA€75 000

You want €75 000 in 3 years. You save the same amount at the end of every month in an account paying 4.2% per year, compounded monthly. Find the monthly deposit A.

FV = €75 0004.2% / 12N = 36
A = €
3.NPV & IRRMedium
years012345−30 0008 0008 0008 0008 0008 000

A project costs €30 000 today and returns €8 000 at the end of each year for 5 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 = 8 000CF2 = 8 000CF3 = 8 000CF4 = 8 000CF5 = 8 000
DPB = years
4.NPV & IRRHard
r (%)NPVABr* = ?0

Project A costs €40 000 and returns €22 500, €16 000, €9 000 at the end of years 1 to 3. Project B costs €45 000 and returns €11 500, €17 000, €26 000. Find the crossover rate r* at which both projects have the same NPV, in percent.

A: −40 000, 22 500, 16 000, 9 000B: −45 000, 11 500, 17 000, 26 000
r* = %
5.Bond pricing & yieldMedium
years01234P€80€80€80 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 8% and matures in 4 years. Its yield to maturity changes from 2.5% to 1%. By how much ΔP does the bond price change?

F = €1 000coupon = 8%n = 4y1 = 2.5%y2 = 1%
ΔP = €
6.Bond pricing & yieldHard
years0123€960€30€30 + €1 000

A bond with a face value of €1 000 and an annual coupon of 3% matures in 3 years and sells for €960. Find its yield to maturity YTM in percent.

F = €1 000coupon = 3%n = 3P = €960
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€3 000 000re​ = 14%debt€2 000 000rd​ = 7%T = 21%

A company has equity worth €3 000 000 and debt worth €2 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 7% and the tax rate is 21%. Find the WACC.

E = €3 000 000D = €2 000 000re = 14%rd = 7%T = 21%
WACC = %
8.CAPM & WACCHard
rf​ = 3%, rm​ − rf​ = 4%D/E = 0.2equityβL​ = 1.1debtD/E = 1equityre​ = ?debtT = 25.8%

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 1. The tax rate is 25.8%, the risk-free rate 3% and the market risk premium 4%. Find the cost of equity re after the change.

βL = 1.1D/E1 = 0.2D/E2 = 1T = 25.8%rf = 3%rm − rf = 4%
re = %
9.Descriptive statisticsMedium
10152025303540455055

Find the sample standard deviation s of the data: 50, 10, 20, 58, 31, 20, 46.

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

A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is above 34?

μ = 50σ = 8x = 34
P =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7P(six) = 1/6 on each roll

A fair die is rolled 7 times. What is the probability P of getting at least one six?

n = 7
P =
12.ProbabilityHard
n = 11, p = 0.2: P(X ≤ 1) = ?k01234567891011

11 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 11p = 0.2
P =
13.Normal distributionMedium
105 s98.6 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 98.6 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 98.6 s
P =
14.Normal distributionHard
μ = 500x = ?90%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 90% of them lie (the 90th percentile).

μ = 500σ = 40P = 90%
x = h
15.Confidence intervalsMedium
n = 21, s = 13LUx̄ = 23E = ?99%

A sample of 21 items has mean 23 and standard deviation s = 13. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 21x̄ = 23s = 1399%t(20) = 2.845
E =
16.Confidence intervalsHard
μ₁ − μ₂, t(12) = 1.7820LUx̄₁ − x̄₂E = ?90%

Two fertilisers are tested. Plot group 1: 7 plots, mean yield 90 kg, standard deviation 1 kg. Plot group 2: 7 plots, mean 93 kg, standard deviation 1.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 7x̄₁ = 90s₁ = 1n₂ = 7x̄₂ = 93s₂ = 1.590%t(12) = 1.782
E = kg
17.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 124.8 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 124.8s = 10n = 36
t =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.01βσ = 7, power 95%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 3 units with a power of 95%. The standard deviation is 7. Find the minimum sample size n.

α = 0.01power = 95%δ = 3σ = 7zα = 2.326zβ = 1.645
n =
19.Linear regression & correlationMedium
xy17361939214123602556xy2025405060

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy159117101191062112223132xy1520100120x = 20

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (20, 119) is made. Find its residual e = y − ŷ.

n = 5(20, 119)
e =

3Solutions

1.Time value of money

Answer: FV = 9233 €

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 = 0.036000000000000004/4 = 0.009
N = m·n = 4·4 = 16

Future value

FV = 8 000·(1 + 0.009)16 = 8 000·1.154 = €9 233.12

Answer

FV = €9 233.12
2.Time value of money

Answer: A = 1958 €

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 = 0.042/12 = 0.0035
N = 12·3 = 36

Deposit

A = FV·i/((1 + i)N − 1) = 75 000·0.0035/(1.134 − 1) = €1 958.48

Answer

A = €1 958.48
3.NPV & IRR

Answer: DPB = 4.381 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: 8 000/1.06 = 7 547.17 → Σ = 7 547.17
year 2: 8 000/1.124 = 7 119.97 → Σ = 14 667.14
year 3: 8 000/1.191 = 6 716.95 → Σ = 21 384.10
year 4: 8 000/1.262 = 6 336.75 → Σ = 27 720.84
year 5: 8 000/1.338 = 5 978.07 → Σ = 33 698.91

Fraction of the last year

DPB = 4 + 2 279.16/5 978.07 = 4.381

Answer

DPB = 4.381 years

The discounted payback is always longer than the simple payback (3.75 years).

4.NPV & IRR

Answer: r* = 5.347 %

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 = −45 000 − (−40 000) = −5 000
ΔCF1 = 11 500 − 22 500 = −11 000
ΔCF2 = 17 000 − 16 000 = 1 000
ΔCF3 = 26 000 − 9 000 = 17 000

IRR of the difference

NPVΔ(5%) = 116.08
NPVΔ(6%) = −213.83
r* = 5.347%

Answer

r* = 5.347%

Check: at 5.347% both NPVs are €3 473.01.

5.Bond pricing & yield

Answer: ΔP = 66.23 €

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 = €80.00, i = 0.025, N = 4
€80.00·(1 − 0.906)/0.025 + €1 000·0.906 = €300.96 + €905.95 = €1 206.91

Price at the new yield

C = €80.00, i = 0.01, N = 4
€80.00·(1 − 0.961)/0.01 + €1 000·0.961 = €312.16 + €960.98 = €1 273.14

Change

ΔP = 1 273.14 − 1 206.91 = €66.23

Answer

ΔP = €66.23
6.Bond pricing & yield

Answer: YTM = 4.454 %

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 = €30·(1 − (1 + y)⁻³)/y + €1 000/(1 + y)³

Solve numerically

P(4%) = €972.25, P(5%) = €945.54
YTM = 4.454%

Answer

YTM = 4.454%
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 = €5 000 000, E/V = 0.6, D/V = 0.4

WACC

WACC = 0.6·14% + 0.4·7%·(1 − 0.21) = 10.61%

Answer

WACC = 10.61%
8.CAPM & WACC

Answer: re = 9.674 %

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 + 0.742·0.2) = 0.9579

Relever at the new D/E

βL = 0.9579·(1 + 0.742·1) = 1.669

Cost of equity (CAPM)

re = 3% + 1.669·4% = 9.674%

Answer

re = 9.674%
9.Descriptive statistics

Answer: s = 18.04

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 = 235/7 = 33.57

Squared deviations

Σ(x − x̄)² = 16.43² + (−23.57)² + (−13.57)² + 24.43² + (−2.571)² + (−13.57)² + 12.43² = 1952

Standard deviation

s = √(1952/6) = 18.04

Answer

s = 18.04
10.Descriptive statistics

Answer: P = 0.9772

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
11.Probability

Answer: P = 0.7209

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)7 = 0.2791
P = 1 − 0.2791 = 0.7209

Answer

P = 0.7209
12.Probability

Answer: P = 0.3221

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) = 0.811 = 0.0859
P(X = 1) = 11·0.2·0.810 = 0.2362

Add them

P = 0.0859 + 0.2362 = 0.3221

Answer

P = 0.3221
13.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 sum

μ = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (98.6 − 105)/6.403 = −1

Upper tail

P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
14.Normal distribution

Answer: x = 551.3 h

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) = 0.9 → z = 1.282

Back to x

x = 500 + 1.282·40 = 551.26 h

Answer

x = 551.26 h
15.Confidence intervals

Answer: E = 8.071

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 = 2.845

Margin of error

E = 2.845·13/√21 = 8.071

Answer

E = 8.071
16.Confidence intervals

Answer: E = 1.214 kg

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Pooled standard deviation

sp² = (6·1² + 6·1.5²)/12 = 1.625
sp = 1.275

Margin of error

E = 1.782·1.275·√(1/7 + 1/7) = 1.214

Answer

E = 1.214 kg
17.Hypothesis tests & confidence intervals

Answer: t = 2.88

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/√36 = 10/6 = 1.667

Test statistic

t = (124.8 − 120)/1.667 = 2.88

Answer

t = 2.88
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: n = 86

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((2.326 + 1.645)·7/3)² = 85.85

Round up

n = 86

Answer

n = 86
19.Linear regression & correlation

Answer: r² = 0.7931

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 = 105/5 = 21
ȳ = 232/5 = 46.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 122
Syy = Σ(y − ȳ)² = 469.2

Coefficient of determination

r² = 122²/(40·469.2) = 0.7931

Answer

r² = 0.7931

About 79.31% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = 3.45

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 = 95/5 = 19
ȳ = 552/5 = 110.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 206

Regression line

b = 206/40 = 5.15
a = 110.4 − 5.15·19 = 12.55
ŷ = 12.55 + 5.15·x

Predicted value

ŷ = 12.55 + 5.15·20 = 115.5

Residual

e = 119 − 115.5 = 3.45

Answer

e = 3.45

The point lies above the line.

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