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
years0123456€8 000FV = ?

You invest €8 000 at a nominal rate of 5% per year, compounded quarterly. Find the value FV of the investment after 6 years.

PV = €8 0005% / 4n = 6
FV = €
2.Time value of moneyHard
years0123…21€2 500€2 500€2 500FV = ?

You deposit €2 500 at the start of every year for 21 years, the first deposit today. The account pays 4% per year. How much is in the account at the end of year 21? Find FV.

A = €2 500i = 4%n = 21
FV = €
3.NPV & IRRMedium
years01234−75 00020 00019 00022 00036 000

An investment of €75 000 gives these cash flows at the end of years 1 to 4: €20 000, €19 000, €22 000, €36 000. The cost of capital is 12%. Find the net present value NPV.

r = 12%CF0 = −75 000CF1 = 20 000CF2 = 19 000CF3 = 22 000CF4 = 36 000
NPV = €
4.NPV & IRRHard
years0123−40 00024 50023 00015 000

A project costs €40 000 and returns €24 500, €23 000, €15 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −40 000CF1 = 24 500CF2 = 23 000CF3 = 15 000
IRR = %
5.Bond pricing & yieldMedium
half-years0123…22€25€25€25 + €1 000P = ?

A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 11 years. The yield to maturity is 6% per year. Find the price P of the bond.

F = €1 000coupon = 5%YTM = 6%n = 11
P = €
6.Bond pricing & yieldHard
years012345€1 243€60€60€60 + €1 060

A bond with 17 years to maturity (face value €1 000, annual coupon 6%) sells for €1 243. The issuer can call it in 5 years at a call price of €1 060. Find the yield to call YTC in percent.

F = €1 000coupon = 6%P = €1 243call price = €1 060call in 5 years
YTC = %
7.CAPM & WACCMedium
WACC = 7.53%equity€8 000 000re​ = ?debt€3 000 000rd​ = 5%T = 28%

A company has equity worth €8 000 000 and debt worth €3 000 000. Its WACC is 7.53%, its pre-tax cost of debt is 5% and the tax rate is 28%. Find the cost of equity re.

E = €8 000 000D = €3 000 000WACC = 7.53%rd = 5%T = 28%
re = %
8.CAPM & WACCHard
WACC = ?equity€8 000 000re​ = 9%bonds5 000 × €1 038coupon 6%, n = 10T = 21%

A company has equity with a market value of €8 000 000 and a cost of equity of 9%. Its debt consists of 5 000 bonds (face value €1 000, annual coupon 6%, 10 years to maturity) that trade at €1 038 each. The tax rate is 21%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.

E = €8 000 000re = 9%bonds: 5 000 × €1 038coupon = 6%n = 10T = 21%
WACC = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 22, 11, 52, 35, 15.

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

A quantity is normally distributed with mean 70 and standard deviation 5. What proportion P of values is below 75?

μ = 70σ = 5x = 75
P =
11.ProbabilityMedium
xp(x)10.320.2530.1540.150.2μ = 2.65

For the distribution shown, find the standard deviation σ of X.

x: 1, 2, 3, 4, 5p(x): 0.3, 0.25, 0.15, 0.1, 0.2
σ =
12.ProbabilityHard
xp(x)00.210.320.3530.15μ

X has the distribution shown, and Y = 0.5X + 10. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.2, 0.3, 0.35, 0.15Y = 0.5X + 10
σY =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0192

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0192. Find z to two decimals.

A = 0.0192
z =
14.Normal distributionHard
μ = 500x = ?95%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 95% of them lie (the 95th percentile).

μ = 500σ = 40P = 95%
x = h
15.Confidence intervalsMedium
x̄E₁ = 4.795%, n = 50x̄E₂ = ?99%, n = 200

A 95% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 4.7. What is the margin of error E₂ of a 99% interval based on 200 observations?

95%n₁ = 50E₁ = 4.799%n₂ = 200
E₂ =
16.Confidence intervalsHard
σ = 26, n = ?x̄ − Ex̄ + Ex̄E = 295%

The population standard deviation is 26. 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?

σ = 26E = 295%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 106 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 106s = 20n = 16
t =
18.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 100 measurements has mean 125.4. The population standard deviation is 20. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 125.4σ = 20n = 10099%
U =
19.Linear regression & correlationMedium
xy2941562483010361243xy5102040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy155118592173248427109xy1520256080100x = 25.5

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 (25.5, 100) is made. Find its residual e = y − ŷ.

n = 5(25.5, 100)
e =

3Solutions

1.Time value of money

Answer: FV = 10779 €

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 = 0.05/4 = 0.0125
N = m·n = 4·6 = 24

Future value

FV = 8 000·(1 + 0.0125)24 = 8 000·1.347 = €10 778.81

Answer

FV = €10 778.81
2.Time value of money

Answer: FV = 83120 €

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.

Ordinary annuity factor

((1 + 0.04)21 − 1)/0.04 = 31.97

Annuity due

Each deposit is invested one year longer.

FV = 2 500·31.97·1.04 = €83 119.92

Answer

FV = €83 119.92
3.NPV & IRR

Answer: NPV = −3458 €

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 = −75 000
CF1/(1 + 0.12)1 = 20 000/1.12 = 17 857.14
CF2/(1 + 0.12)2 = 19 000/1.254 = 15 146.68
CF3/(1 + 0.12)3 = 22 000/1.405 = 15 659.17
CF4/(1 + 0.12)4 = 36 000/1.574 = 22 878.65

Add them up

NPV = €−3 458.36

Answer

NPV = €−3 458.36

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

4.NPV & IRR

Answer: IRR = 28.62 %

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.

Bracket the IRR

NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.

NPV(25%) = 2 000.00
NPV(30%) = −716.89

Narrow down

Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.

IRR ≈ 25 + 5·2 000.00/(2 000.00 − (−716.89)) = 28.68%
exact: NPV(28.62%) = 0 → IRR = 28.62%

Answer

IRR = 28.62%
5.Bond pricing & yield

Answer: P = 920.3 €

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 = €25.00, i = 0.03, N = 22

Present value of the coupons

€25.00·(1 − 0.5219)/0.03 = €398.42

Present value of the face value

€1 000·0.5219 = €521.89

Price

P = €920.32

Answer

P = €920.32

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

6.Bond pricing & yield

Answer: YTC = 1.998 %

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

1243 = 60·(1 − (1 + y)−5)/y + 1060/(1 + y)5

Solve numerically

y = 1%: €1 299.76
y = 2%: €1 242.88
YTC = 1.998%

Answer

YTC = 1.998%

The yield to maturity is 4.002%. 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: re = 9.004 %

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 = 0.7273, D/V = 0.2727

Debt part of the WACC

0.2727·5%·(1 − 0.28) = 0.9818%

Cost of equity

re = (7.53% − 0.9818%)/0.7273 = 9.004%

Answer

re = 9.004%
8.CAPM & WACC

Answer: WACC = 7.167 %

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)

1038 = 60·(1 − (1 + y)−10)/y + 1000/(1 + y)10
y = 5.496%

Market values

D = 5 000·€1 038 = €5 190 000
V = €13 190 000, E/V = 0.6065, D/V = 0.3935

WACC

WACC = 0.6065·9% + 0.3935·5.496%·(1 − 0.21) = 7.167%

Answer

WACC = 7.167%
9.Descriptive statistics

Answer: s = 16.69

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 = 135/5 = 27

Squared deviations

Σ(x − x̄)² = (−5)² + (−16)² + 25² + 8² + (−12)² = 1114

Standard deviation

s = √(1114/4) = 16.69

Answer

s = 16.69
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: σ = 1.492

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Mean

μ = 1·0.3 + 2·0.25 + 3·0.15 + 4·0.1 + 5·0.2 = 2.65

Mean of the squares

E(X²) = 1·0.3 + 4·0.25 + 9·0.15 + 16·0.1 + 25·0.2 = 9.25

Variance

Var(X) = 9.25 − 2.65² = 2.228
σ = √2.228 = 1.492

Answer

σ = 1.492
12.Probability

Answer: σY = 0.4867

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Spread of X

μ = 1.45, Var(X) = 0.9475, σX = 0.9734

Linear change

σY = |0.5|·0.9734 = 0.4867

Answer

σY = 0.4867

The mean does move: E(Y) = 0.5·1.45 + 10 = 10.72.

13.Normal distribution

Answer: z = 2.07

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 − 0.0192 = 0.9808

Read the z-table backwards

Find 0.9808 inside the table and read off z.

z = 2.07

Answer

z = 2.07
14.Normal distribution

Answer: x = 565.8 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) = 0.95 → z = 1.645

Back to x

x = 500 + 1.645·40 = 565.8 h

Answer

x = 565.8 h
15.Confidence intervals

Answer: E₂ = 3.089

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 = (2.576/1.96)·√(50/200) = 0.6571

New margin

E₂ = 4.7·0.6571 = 3.089

Answer

E₂ = 3.089
16.Confidence intervals

Answer: n = 650

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 = (1.96·26/2)² = 649.2

Round up

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

n = 650

Answer

n = 650
17.Hypothesis tests & confidence intervals

Answer: t = 1.2

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

SE = 20/√16 = 20/4 = 5

Test statistic

t = (106 − 100)/5 = 1.2

Answer

t = 1.2
t0.025; df = 15 = 2.131

|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: U = 130.6

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).

Margin of error

E = 2.576·20/√100 = 5.152

Interval limit

U = 125.4 + 5.152 = 130.6

Answer

U = 130.6
19.Linear regression & correlation

Answer: r² = 0.9966

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 = 157/6 = 26.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 239
Syy = Σ(y − ȳ)² = 818.8

Coefficient of determination

r² = 239²/(70·818.8) = 0.9966

Answer

r² = 0.9966

About 99.66% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = 3.65

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 = 105/5 = 21
ȳ = 376/5 = 75.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 423

Regression line

b = 423/90 = 4.7
a = 75.2 − 4.7·21 = −23.5
ŷ = −23.5 + 4.7·x

Predicted value

ŷ = −23.5 + 4.7·25.5 = 96.35

Residual

e = 100 − 96.35 = 3.65

Answer

e = 3.65

The point lies above the line.

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