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…17€2 000FV = ?

You invest €2 000 at 2% per year with continuous compounding. Find the value FV after 17 years.

PV = €2 000r = 2%t = 17
FV = €
2.Time value of moneyHard
months0123…360€150 000A?AAA

A loan of €150 000 is repaid with equal monthly payments over 30 years. The interest rate is 2.4% per year, compounded monthly. Find the monthly payment A.

P = €150 0002.4% / 12N = 360
A = €
3.NPV & IRRMedium
years01234−40 00014 00010 00016 00014 000

A project costs €40 000 today and returns €14 000, €10 000, €16 000, €14 000 at the end of years 1 to 4. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 12%CF0 = −40 000CF1 = 14 000CF2 = 10 000CF3 = 16 000CF4 = 14 000
DPB = years
4.NPV & IRRHard
r (%)NPVABr* = ?0

Project A costs €10 000 and returns €5 500, €4 000, €2 500 at the end of years 1 to 3. Project B costs €15 000 and returns €4 000, €6 000, €8 000. Find the crossover rate r* at which both projects have the same NPV, in percent.

A: −10 000, 5 500, 4 000, 2 500B: −15 000, 4 000, 6 000, 8 000
r* = %
5.Bond pricing & yieldMedium
half-years0123…16€22.50€22.50€22.50 + €1 000P = ?

A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 8 years. The yield to maturity is 4% per year. Find the price P of the bond.

F = €1 000coupon = 4.5%YTM = 4%n = 8
P = €
6.Bond pricing & yieldHard
years0123…9€855€70€70 + €1 000

A bond with a face value of €1 000 and an annual coupon of 7% matures in 9 years and sells for €855. Find its yield to maturity YTM in percent.

F = €1 000coupon = 7%n = 9P = €855
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€2 000 000re​ = 12%debt€1 000 000rd​ = 4%T = 21%

A company has equity worth €2 000 000 and debt worth €1 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 4% and the tax rate is 21%. Find the WACC.

E = €2 000 000D = €1 000 000re = 12%rd = 4%T = 21%
WACC = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €48.50β = 0.8rf​ = 2%, rm​ = 8%debt5 000 × €1 050rd​ = 3.5%T = 21%

A company has 1 000 000 shares at €48.50 each and 5 000 bonds trading at €1 050 each. The pre-tax cost of debt is 3.5% and the tax rate is 21%. The share’s beta is 0.8, the risk-free rate 2% and the market return 8%. Find the WACC.

β = 0.8rf = 2%rm = 8%rd = 3.5%T = 21%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 12, 59, 47, 19, 21, 12.

n = 6
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is below 180?

μ = 170σ = 8x = 180
P =
11.ProbabilityMedium
red: 5blue: 6

A bag holds 5 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 6
P =
12.ProbabilityHard
n = 9, p = 0.25: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?

n = 9p = 0.25
P =
13.Normal distributionMedium
105 s111.4 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 111.4 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 111.4 s
P =
14.Normal distributionHard
380420460500540580620htop 90%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 90% is to the right of it. Find the cut-off x.

μ = 500 hσ = 40 h90%
x = h
15.Confidence intervalsMedium
x̄E₁ = 9.490%, n = 100x̄E₂ = ?90%, n = 400

A 90% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 9.4. What is the margin of error E₂ if the same study uses 400 observations instead?

90%n₁ = 100E₁ = 9.4n₂ = 400
E₂ =
16.Confidence intervalsHard
p ≈ 0.2, n = ?p̂ − Ep̂ + Ep̂E = 0.02599%

A pilot study suggests that about 0.2 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 0.025?

p ≈ 0.2E = 0.02599%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 119.5 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 119.5s = 5n = 100
t =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 1α = 0.05βσ = 7, power 90%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 1 units with a power of 90%. The standard deviation is 7. Find the minimum sample size n.

α = 0.05power = 90%δ = 1σ = 7zα = 1.645zβ = 1.282
n =
19.Linear regression & correlationMedium
xy14621662188620902299xy15206080100

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 = 90Σy = 399Σx² = 1660Σy² = 32985Σxy = 7386
r =
20.Linear regression & correlationHard
xy121219315411510xy24101520x = 3.5

The table gives 5 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 = 3.5.

n = 5x = 3.5
ŷ =

3Solutions

1.Time value of money

Answer: FV = 2810 €

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.

Exponent

r·t = 0.02·17 = 0.34
e0.34 = 1.405

Future value

FV = 2 000·1.405 = €2 809.90

Answer

FV = €2 809.90
2.Time value of money

Answer: A = 584.9 €

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 = 0.024/12 = 0.002
N = 12·30 = 360

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 = 0.4871
A = 150 000·0.002/(1 − 0.4871) = €584.91

Answer

A = €584.91
3.NPV & IRR

Answer: DPB = 3.915 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: 14 000/1.12 = 12 500.00 → Σ = 12 500.00
year 2: 10 000/1.254 = 7 971.94 → Σ = 20 471.94
year 3: 16 000/1.405 = 11 388.48 → Σ = 31 860.42
year 4: 14 000/1.574 = 8 897.25 → Σ = 40 757.68

Fraction of the last year

DPB = 3 + 8 139.58/8 897.25 = 3.915

Answer

DPB = 3.915 years

The discounted payback is always longer than the simple payback (3 years).

4.NPV & IRR

Answer: r* = 5.886 %

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 = −15 000 − (−10 000) = −5 000
ΔCF1 = 4 000 − 5 500 = −1 500
ΔCF2 = 6 000 − 4 000 = 2 000
ΔCF3 = 8 000 − 2 500 = 5 500

IRR of the difference

NPVΔ(5%) = 136.59
NPVΔ(6%) = −17.20
r* = 5.886%

Answer

r* = 5.886%

Check: at 5.886% both NPVs are €867.70.

5.Bond pricing & yield

Answer: P = 1034 €

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 = €22.50, i = 0.02, N = 16

Present value of the coupons

€22.50·(1 − 0.7284)/0.02 = €305.50

Present value of the face value

€1 000·0.7284 = €728.45

Price

P = €1 033.94

Answer

P = €1 033.94

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

6.Bond pricing & yield

Answer: YTM = 9.464 %

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

€855 = €70·(1 − (1 + y)⁻⁹)/y + €1 000/(1 + y)⁹

Solve numerically

P(9%) = €880.10, P(10%) = €827.23
YTM = 9.464%

Answer

YTM = 9.464%
7.CAPM & WACC

Answer: WACC = 9.053 %

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 = €3 000 000, E/V = 0.6667, D/V = 0.3333

WACC

WACC = 0.6667·12% + 0.3333·4%·(1 − 0.21) = 9.053%

Answer

WACC = 9.053%
8.CAPM & WACC

Answer: WACC = 6.406 %

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·€48.50 = €48 500 000
D = 5 000·€1 050 = €5 250 000

Cost of equity (CAPM)

re = 2% + 0.8·6% = 6.8%

Weights

V = €53 750 000, E/V = 0.9023, D/V = 0.09767

WACC

WACC = 0.9023·6.8% + 0.09767·3.5%·(1 − 0.21) = 6.406%

Answer

WACC = 6.406%
9.Descriptive statistics

Answer: s = 19.82

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 = 170/6 = 28.33

Squared deviations

Σ(x − x̄)² = (−16.33)² + 30.67² + 18.67² + (−9.333)² + (−7.333)² + (−16.33)² = 1963

Standard deviation

s = √(1963/5) = 19.82

Answer

s = 19.82
10.Descriptive statistics

Answer: P = 0.8944

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 0.8944

Answer

P = 0.8944
11.Probability

Answer: P = 0.1818

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 = 5/11 · 4/10 = 20/110 = 0.1818

Answer

P = 0.1818
12.Probability

Answer: P = 0.3003

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.

Zero and one failure

P(X = 0) = 0.759 = 0.07508
P(X = 1) = 9·0.25·0.758 = 0.2253

Add them

P = 0.07508 + 0.2253 = 0.3003

Answer

P = 0.3003
13.Normal distribution

Answer: P = 0.1587

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 = (111.4 − 105)/6.403 = 1

Upper tail

P = 1 − Φ(1) = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
14.Normal distribution

Answer: x = 449 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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 500 − 1.28·40 = 449 h

Answer

x = 449 h
15.Confidence intervals

Answer: E₂ = 4.7

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.

Ratio of margins

E₂/E₁ = (1.645/1.645)·√(100/400) = 0.5

New margin

E₂ = 9.4·0.5 = 4.7

Answer

E₂ = 4.7
16.Confidence intervals

Answer: n = 1699

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 = 2.576²·0.2·0.8/0.025² = 1699

Round up

n = 1699

Answer

n = 1699
17.Hypothesis tests & confidence intervals

Answer: t = −1

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 = 5/√100 = 5/10 = 0.5

Test statistic

t = (119.5 − 120)/0.5 = −1

Answer

t = −1
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 420

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 = ((1.645 + 1.282)·7/1)² = 419.8

Round up

n = 420

Answer

n = 420
19.Linear regression & correlation

Answer: r = 0.9533

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 = 7386 − 90·399/5 = 204
Sxx = 1660 − 90²/5 = 40
Syy = 32985 − 399²/5 = 1145

Correlation coefficient

r = 204/√(40·1145) = 0.9533

Answer

r = 0.9533
20.Linear regression & correlation

Answer: ŷ = 13.7

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 = 15/5 = 3
ȳ = 76/5 = 15.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −30

Regression line

b = −30/10 = −3
a = 15.2 − (−3)·3 = 24.2
ŷ = 24.2 − 3·x

Prediction

ŷ = 24.2 + (−3)·3.5 = 13.7

Answer

ŷ = 13.7

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