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€1 000FV = ?

You invest €1 000 at 5% per year with continuous compounding. Find the value FV after 4 years.

PV = €1 000r = 5%t = 4
FV = €
2.Time value of moneyHard
months0123…48€20 000A?AAA

A loan of €20 000 is repaid with equal monthly payments over 4 years. The interest rate is 4.8% per year, compounded monthly. Find the monthly payment A.

P = €20 0004.8% / 12N = 48
A = €
3.NPV & IRRMedium
years0123−45 00021 0009 00011 000

An investment of €45 000 gives these cash flows at the end of years 1 to 3: €21 000, €9 000, €11 000. The cost of capital is 6%. Find the net present value NPV.

r = 6%CF0 = −45 000CF1 = 21 000CF2 = 9 000CF3 = 11 000
NPV = €
4.NPV & IRRHard
r (%)NPVABr* = ?0

Project A costs €40 000 and returns €22 500, €16 500, €9 000 at the end of years 1 to 3. Project B costs €45 000 and returns €14 000, €18 000, €23 500. Find the crossover rate r* at which both projects have the same NPV, in percent.

A: −40 000, 22 500, 16 500, 9 000B: −45 000, 14 000, 18 000, 23 500
r* = %
5.Bond pricing & yieldMedium
years0123…15€436C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 15 years and pays annual coupons. It sells for €436 when the yield to maturity is 9%. Find the coupon rate c in percent.

F = €1 000n = 15P = €436YTM = 9%
c = %
6.Bond pricing & yieldHard
years012345P€80€80€80 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 8% and matures in 5 years. The yield to maturity is 5%. Find the Macaulay duration D in years.

F = €1 000coupon = 8%n = 5YTM = 5%
D = years
7.CAPM & WACCMedium
WACC = ?equity€10 000 000re​ = 13%debt€1 000 000rd​ = 3.5%T = 25.8%

A company has equity worth €10 000 000 and debt worth €1 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 3.5% and the tax rate is 25.8%. Find the WACC.

E = €10 000 000D = €1 000 000re = 13%rd = 3.5%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity4 000 000 × €15.00β = 1.8rf​ = 2.5%, rm​ = 8.5%debt40 000 × €1 050rd​ = 6%T = 25.8%

A company has 4 000 000 shares at €15.00 each and 40 000 bonds trading at €1 050 each. The pre-tax cost of debt is 6% and the tax rate is 25.8%. The share’s beta is 1.8, the risk-free rate 2.5% and the market return 8.5%. Find the WACC.

β = 1.8rf = 2.5%rm = 8.5%rd = 6%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
2628303234363840424446

Find the sample standard deviation s of the data: 35, 34, 40, 41, 46, 34, 26.

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

A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is above 109?

μ = 100σ = 15x = 109
P =
11.ProbabilityMedium
red: 4blue: 7

A bag holds 4 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

4 / 7
P =
12.ProbabilityHard
n = 9, p = 0.7: P(X = 3) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 3 correct answers?

n = 9p = 0.7k = 3
P =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.3594

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.3594. Find z to two decimals.

A = 0.3594
z =
14.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
n = 15, s = 16.5LUx̄ = 113.5E = ?95%

A sample of 15 items has mean 113.5 and standard deviation s = 16.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 15x̄ = 113.5s = 16.595%t(14) = 2.145
E =
16.Confidence intervalsHard
σ = 30, n = ?x̄ − Ex̄ + Ex̄E = 2.595%

The population standard deviation is 30. 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 2.5?

σ = 30E = 2.595%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.2282.228t = ?df = 10

11 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 1.93 and the standard deviation of the differences is 3.2. Find the paired test statistic t for H₀: μd = 0.

n = 11d̄ = 1.93sd = 3.2
t =
18.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 36 measurements has mean 199.4. The population standard deviation is 12. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 199.4σ = 12n = 3695%
L =
19.Linear regression & correlationMedium
xy146017682073239226100xy1520256080100

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
xy140242343443555xy2440455055x = 2.5

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 = 2.5.

n = 5x = 2.5
ŷ =

3Solutions

1.Time value of money

Answer: FV = 1221 €

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 = 0.05·4 = 0.2
e0.2 = 1.221

Future value

FV = 1 000·1.221 = €1 221.40

Answer

FV = €1 221.40
2.Time value of money

Answer: A = 458.8 €

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 = 0.048/12 = 0.004
N = 12·4 = 48

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 = 0.8256
A = 20 000·0.004/(1 − 0.8256) = €458.78

Answer

A = €458.78
3.NPV & IRR

Answer: NPV = −7943 €

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 = 21 000/1.06 = 19 811.32
CF2/(1 + 0.06)2 = 9 000/1.124 = 8 009.97
CF3/(1 + 0.06)3 = 11 000/1.191 = 9 235.81

Add them up

NPV = €−7 942.90

Answer

NPV = €−7 942.90

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

4.NPV & IRR

Answer: r* = 7.729 %

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 = 14 000 − 22 500 = −8 500
ΔCF2 = 18 000 − 16 500 = 1 500
ΔCF3 = 23 500 − 9 000 = 14 500

IRR of the difference

NPVΔ(7%) = 202.55
NPVΔ(8%) = −73.79
r* = 7.729%

Answer

r* = 7.729%

Check: at 7.729% both NPVs are €2 301.71.

5.Bond pricing & yield

Answer: c = 2.003 %

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.

Present value of the face value

1000·(1 + 0.09)−15 = €274.54

Coupon

C = (436 − 274.54)/8.061 = €20.03

Coupon rate

c = 20.03/1000 = 2.003%

Answer

c = 2.003%
6.Bond pricing & yield

Answer: D = 4.357 years

Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.

Present value of each cash flow

t = 1: 80/1.05 = 76.19, t·PV = 76.19
t = 2: 80/1.103 = 72.56, t·PV = 145.12
t = 3: 80/1.158 = 69.11, t·PV = 207.32
t = 4: 80/1.216 = 65.82, t·PV = 263.26
t = 5: 1080/1.276 = 846.21, t·PV = 4 231.04

Duration

P = 1 129.88
D = 4 922.94/1 129.88 = 4.357

Answer

D = 4.357 years
7.CAPM & WACC

Answer: WACC = 12.05 %

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 = €11 000 000, E/V = 0.9091, D/V = 0.09091

WACC

WACC = 0.9091·13% + 0.09091·3.5%·(1 − 0.258) = 12.05%

Answer

WACC = 12.05%
8.CAPM & WACC

Answer: WACC = 9.657 %

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 = 4 000 000·€15.00 = €60 000 000
D = 40 000·€1 050 = €42 000 000

Cost of equity (CAPM)

re = 2.5% + 1.8·6% = 13.3%

Weights

V = €102 000 000, E/V = 0.5882, D/V = 0.4118

WACC

WACC = 0.5882·13.3% + 0.4118·6%·(1 − 0.258) = 9.657%

Answer

WACC = 9.657%
9.Descriptive statistics

Answer: s = 6.425

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 = 256/7 = 36.57

Squared deviations

Σ(x − x̄)² = (−1.571)² + (−2.571)² + 3.429² + 4.429² + 9.429² + (−2.571)² + (−10.57)² = 247.7

Standard deviation

s = √(247.7/6) = 6.425

Answer

s = 6.425
10.Descriptive statistics

Answer: P = 0.2743

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
11.Probability

Answer: P = 0.1091

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 = 4/11 · 3/10 = 12/110 = 0.1091

Answer

P = 0.1091
12.Probability

Answer: P = 0.021

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 = 3) = 84·0.73·0.36 = 84·0.343·0.000729 = 0.021

Answer

P = 0.021
13.Normal distribution

Answer: z = 0.36

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 of z

Φ(z) = 1 − 0.3594 = 0.6406

Read the z-table backwards

Find 0.6406 inside the table and read off z.

z = 0.36

Answer

z = 0.36
14.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
15.Confidence intervals

Answer: E = 9.138

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 = 14, t = 2.145

Margin of error

E = 2.145·16.5/√15 = 9.138

Answer

E = 9.138
16.Confidence intervals

Answer: n = 554

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 = (1.96·30/2.5)² = 553.2

Round up

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

n = 554

Answer

n = 554
17.Hypothesis tests & confidence intervals

Answer: t = 2

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 of the mean difference

SE = 3.2/√11 = 0.9648

Test statistic

t = 1.93/0.9648 = 2

Answer

t = 2

With df = 10, the two-sided 5% critical value is 2.228; |t| is smaller, so H₀ is not rejected.

18.Hypothesis tests & confidence intervals

Answer: L = 195.5

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 = 1.96·12/√36 = 3.92

Interval limit

L = 199.4 − 3.92 = 195.5

Answer

L = 195.5
19.Linear regression & correlation

Answer: r² = 0.9595

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 = 100/5 = 20
ȳ = 393/5 = 78.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 312
Syy = Σ(y − ȳ)² = 1127

Coefficient of determination

r² = 312²/(90·1127) = 0.9595

Answer

r² = 0.9595

About 95.95% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 43.05

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 = 223/5 = 44.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 31

Regression line

b = 31/10 = 3.1
a = 44.6 − 3.1·3 = 35.3
ŷ = 35.3 + 3.1·x

Prediction

ŷ = 35.3 + 3.1·2.5 = 43.05

Answer

ŷ = 43.05

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