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
years01234567€5 000FV = ?

You invest €5 000 at 6% per year with continuous compounding. Find the value FV after 7 years.

PV = €5 000r = 6%t = 7
FV = €
2.Time value of moneyHard
months0123…N = ?€20 000€250€250€250€2503% / 12

You borrow €20 000 at 3% per year, compounded monthly, and repay €250 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.

P = €20 000A = €2503% / 12
N = payments
3.NPV & IRRMedium
years01234−20 0009 0006 0008 0006 000

An investment of €20 000 gives these cash flows at the end of years 1 to 4: €9 000, €6 000, €8 000, €6 000. The cost of capital is 7.000000000000001%. Find the net present value NPV.

r = 7.000000000000001%CF0 = −20 000CF1 = 9 000CF2 = 6 000CF3 = 8 000CF4 = 6 000
NPV = €
4.NPV & IRRHard
years01234−200 00055 000+3%+3%+3%r = 12%, NPV = ?

A project costs €200 000 today. Its cash flow is €55 000 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 4. The cost of capital is 12%. Find the net present value NPV.

C0 = €200 000CF1 = €55 000g = 3%n = 4r = 12%
NPV = €
5.Bond pricing & yieldMedium
half-years0123…12€20€20€20 + €1 000P = ?

A bond with a face value of €1 000 pays a 4% annual coupon in two half-yearly payments and matures in 6 years. The yield to maturity is 5% per year. Find the price P of the bond.

F = €1 000coupon = 4%YTM = 5%n = 6
P = €
6.Bond pricing & yieldHard
years01234567€1 060€60€60 + €1 000

A bond with a face value of €1 000 and an annual coupon of 6% matures in 7 years and sells for €1 060. Find its yield to maturity YTM in percent.

F = €1 000coupon = 6%n = 7P = €1 060
YTM = %
7.CAPM & WACCMedium
βr (%)SMLrf​ = 1%β = 1rm​ = ?β = 1.6re​ = 10.6%

A share with a beta of 1.6 has a required return of 10.6%. The risk-free rate is 1%. What expected market return rm does CAPM imply?

β = 1.6re = 10.6%rf = 1%
rm = %
8.CAPM & WACCHard
WACC = ?equity€6 000 000re​ = 11%bonds2 000 × €1 042coupon 4%, n = 10T = 30%

A company has equity with a market value of €6 000 000 and a cost of equity of 11%. Its debt consists of 2 000 bonds (face value €1 000, annual coupon 4%, 10 years to maturity) that trade at €1 042 each. The tax rate is 30%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.

E = €6 000 000re = 11%bonds: 2 000 × €1 042coupon = 4%n = 10T = 30%
WACC = %
9.Descriptive statisticsMedium
253035404550

Find the sample standard deviation s of the data: 54, 24, 42, 53, 51.

n = 5
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is below 182?

μ = 170σ = 12x = 182
P =
11.ProbabilityMedium
red: 3blue: 8

A bag holds 3 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

3 / 8
P =
12.ProbabilityHard
n = 10, p = 0.7: P(X ≤ 1) = ?k012345678910

10 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?

n = 10p = 0.7
P =
13.Normal distributionMedium
65.567P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 65.5 and 67 points? Give P as a decimal.

μ = 65σ = 10x1 = 65.5x2 = 67
P =
14.Normal distributionHard
557085100115130145pointstop 95%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points95%
x = points
15.Confidence intervalsMedium
x̄E₁ = 7.699%, n = 100x̄E₂ = ?95%, n = 200

A 99% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 7.6. What is the margin of error E₂ of a 95% interval based on 200 observations?

99%n₁ = 100E₁ = 7.695%n₂ = 200
E₂ =
16.Confidence intervalsHard
σ = 21, n = ?x̄ − Ex̄ + Ex̄E = 590%

The population standard deviation is 21. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 5?

σ = 21E = 590%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 96.88 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 96.88s = 12n = 100
t =
18.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 25 measurements has mean 50.2. The population standard deviation is 12. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 50.2σ = 12n = 2590%
U =
19.Linear regression & correlationMedium
xy3355367409471144xy510354045

The table gives 5 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 = 5
r² =
20.Linear regression & correlationHard
xy231333430534641xy246303540x = 3.5

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (3.5, 31) is made. Find its residual e = y − ŷ.

n = 5(3.5, 31)
e =

3Solutions

1.Time value of money

Answer: FV = 7610 €

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 = 0.06·7 = 0.42
e0.42 = 1.522

Future value

FV = 5 000·1.522 = €7 609.81

Answer

FV = €7 609.81
2.Time value of money

Answer: N = 90 payments

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.

Rearrange the annuity formula

i = 0.0025
(1 + i)−N = 1 − 20 000·0.0025/250 = 0.8

Solve with logarithms

N = −ln(0.8)/ln(1.002) = 89.37

Round up to whole payments

89 full payments leave a small balance, which the last, smaller payment clears.

N = 90

Answer

N = 90 payments
3.NPV & IRR

Answer: NPV = 4760 €

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 = −20 000
CF1/(1 + 0.07)1 = 9 000/1.07 = 8 411.21
CF2/(1 + 0.07)2 = 6 000/1.145 = 5 240.63
CF3/(1 + 0.07)3 = 8 000/1.225 = 6 530.38
CF4/(1 + 0.07)4 = 6 000/1.311 = 4 577.37

Add them up

NPV = €4 759.60

Answer

NPV = €4 759.60

NPV is positive, so the project earns more than the required return.

4.NPV & IRR

Answer: NPV = −26005 €

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 growing annuity

((1.03)/(1.12))4 = 0.7153
PV = 55 000/(0.12 − 0.03)·(1 − 0.7153) = 173 994.82

NPV

NPV = 173 994.82 − 200 000 = −€26 005.18

Answer

NPV = −€26 005.18

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

5.Bond pricing & yield

Answer: P = 948.7 €

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 = €20.00, i = 0.025, N = 12

Present value of the coupons

€20.00·(1 − 0.7436)/0.025 = €205.16

Present value of the face value

€1 000·0.7436 = €743.56

Price

P = €948.71

Answer

P = €948.71

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

6.Bond pricing & yield

Answer: YTM = 4.964 %

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

€1 060 = €60·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷

Solve numerically

P(4%) = €1 120.04, P(5%) = €1 057.86
YTM = 4.964%

Answer

YTM = 4.964%
7.CAPM & WACC

Answer: rm = 7 %

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 risk premium

rm − rf = (10.6% − 1%)/1.6 = 6%

Market return

rm = 1% + 6% = 7%

Answer

rm = 7%
8.CAPM & WACC

Answer: WACC = 8.795 %

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).

Cost of debt (YTM)

1042 = 40·(1 − (1 + y)−10)/y + 1000/(1 + y)10
y = 3.495%

Market values

D = 2 000·€1 042 = €2 084 000
V = €8 084 000, E/V = 0.7422, D/V = 0.2578

WACC

WACC = 0.7422·11% + 0.2578·3.495%·(1 − 0.3) = 8.795%

Answer

WACC = 8.795%
9.Descriptive statistics

Answer: s = 12.56

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 = 224/5 = 44.8

Squared deviations

Σ(x − x̄)² = 9.2² + (−20.8)² + (−2.8)² + 8.2² + 6.2² = 630.8

Standard deviation

s = √(630.8/4) = 12.56

Answer

s = 12.56
10.Descriptive statistics

Answer: P = 0.8413

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

Standard normal table

Φ(1) = 0.8413
P = 0.8413

Answer

P = 0.8413
11.Probability

Answer: P = 0.05455

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 = 3/11 · 2/10 = 6/110 = 0.05455

Answer

P = 0.05455
12.Probability

Answer: P = 0.0001437

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) = 0.310 = 5.905 × 10⁻⁶
P(X = 1) = 10·0.7·0.39 = 0.0001378

Add them

P = 5.905 × 10⁻⁶ + 0.0001378 = 0.0001437

Answer

P = 0.0001437
13.Normal distribution

Answer: P = 0.05932

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 = (65.5 − 65)/10 = 0.05
z₂ = (67 − 65)/10 = 0.2

Area between

P = Φ(0.2) − Φ(0.05) = 0.5793 − 0.5199 = 0.05932

Answer

P = 0.05932
14.Normal distribution

Answer: x = 75.4 points

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) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 100 − 1.64·15 = 75.4 points

Answer

x = 75.4 points
15.Confidence intervals

Answer: E₂ = 4.089

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 = (1.96/2.576)·√(100/200) = 0.538

New margin

E₂ = 7.6·0.538 = 4.089

Answer

E₂ = 4.089
16.Confidence intervals

Answer: n = 48

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 = (1.645·21/5)² = 47.73

Round up

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

n = 48

Answer

n = 48
17.Hypothesis tests & confidence intervals

Answer: t = −2.6

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 = 12/√100 = 12/10 = 1.2

Test statistic

t = (96.88 − 100)/1.2 = −2.6

Answer

t = −2.6
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: U = 54.15

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 = 1.645·12/√25 = 3.948

Interval limit

U = 50.2 + 3.948 = 54.15

Answer

U = 54.15
19.Linear regression & correlation

Answer: r² = 0.7994

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 = 35/5 = 7
ȳ = 202/5 = 40.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 58
Syy = Σ(y − ȳ)² = 105.2

Coefficient of determination

r² = 58²/(40·105.2) = 0.7994

Answer

r² = 0.7994

About 79.94% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = −1.75

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 = 20/5 = 4
ȳ = 169/5 = 33.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 21

Regression line

b = 21/10 = 2.1
a = 33.8 − 2.1·4 = 25.4
ŷ = 25.4 + 2.1·x

Predicted value

ŷ = 25.4 + 2.1·3.5 = 32.75

Residual

e = 31 − 32.75 = −1.75

Answer

e = −1.75

The point lies below the line.

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