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…9PV = ?€8 000+3%+3%…r = 7%

A property is expected to earn net rent of €8 000 next year. The rent is expected to grow by 3% per year forever. The required return is 7% per year. Find the value PV of the property today.

C1 = €8 000g = 3%r = 7%
PV = €
2.Time value of moneyHard
months0123…240€250 000A?AAA

A loan of €250 000 is repaid with equal monthly payments over 20 years. The interest rate is 3% per year, compounded monthly. Find the monthly payment A.

P = €250 0003% / 12N = 240
A = €
3.NPV & IRRMedium
years0123456€30 000€2 000€2 000€2 000€2 000€2 000€2 000r = 12%, EAC = ?

A machine costs €30 000 today, lasts 6 years and costs €2 000 per year to run (paid at the end of each year). The cost of capital is 12%. Find its equivalent annual cost EAC.

C0 = €30 000M = €2 000n = 6r = 12%
EAC = €
4.NPV & IRRHard
years0123−25 00016 50011 00013 500

A project costs €25 000 and returns €16 500, €11 000, €13 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −25 000CF1 = 16 500CF2 = 11 000CF3 = 13 500
IRR = %
5.Bond pricing & yieldMedium

You buy a bond with 3 years to maturity (face value €1 000, annual coupon 8%) when its yield to maturity is 2.5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 3%. Find your holding period return HPR in percent.

F = €1 000coupon = 8%n = 3y0 = 2.5%y1 = 3%
HPR = %
6.Bond pricing & yieldHard
years01234€1 055€50€50 + €1 000

A bond with a face value of €1 000 and an annual coupon of 5% matures in 4 years and sells for €1 055. Find its yield to maturity YTM in percent.

F = €1 000coupon = 5%n = 4P = €1 055
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€8 000 000re​ = 11%preferred shares€2 000 000rp​ = 8%debt€2 000 000rd​ = 4%T = 20%

A company is financed with ordinary shares worth €8 000 000, preferred shares worth €2 000 000 and debt worth €2 000 000. The cost of equity is 11%, the cost of preferred stock 8% and the pre-tax cost of debt 4%. The tax rate is 20%. Find the WACC.

E = €8 000 000Ps = €2 000 000D = €2 000 000re = 11%rp = 8%rd = 4%T = 20%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €44.50β = 1.3rf​ = 2%, rm​ = 7%debt10 000 × €980rd​ = 5%T = 28%

A company has 2 000 000 shares at €44.50 each and 10 000 bonds trading at €980 each. The pre-tax cost of debt is 5% and the tax rate is 28%. The share’s beta is 1.3, the risk-free rate 2% and the market return 7%. Find the WACC.

β = 1.3rf = 2%rm = 7%rd = 5%T = 28%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 48, 13, 59, 16, 44, 55, 12.

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

Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 62.

μ = 70σ = 8x = 62
z =
11.ProbabilityMedium
xp(x)10.2520.2530.340.0550.15μ = 2.6

For the distribution shown, find the variance Var(X) of X.

x: 1, 2, 3, 4, 5p(x): 0.25, 0.25, 0.3, 0.05, 0.15
Var(X) =
12.ProbabilityHard
xp(x)00.3510.320.35μ

X has the distribution shown, and Y = 0.5X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 0.5X + 1
σY =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8029

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8029. Find z to two decimals.

A = 0.8029
z =
14.Normal distributionHard
0.05 mm0D − d?

Hole diameters D are normal with mean 25.05 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.05 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
x̄E₁ = 2.695%, n = 25x̄E₂ = ?99%, n = 25

A 95% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 2.6. What is the margin of error E₂ of a 99% interval from the same data?

95%n₁ = 25E₁ = 2.699%
E₂ =
16.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0595%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.05?

p = 0.5E = 0.0595%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 60 parts has mean length 140 mm, and the line's known standard deviation is 18 mm. Line 2: a sample of 50 parts has mean 145.3 mm, known standard deviation 4 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 60x̄₁ = 140 mmσ₁ = 18 mmn₂ = 50x̄₂ = 145.3 mmσ₂ = 4 mm
z =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 98 of 150 patients on treatment A recover, and 85 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 98n₁ = 150x₂ = 85n₂ = 120
z =
19.Linear regression & correlationMedium
xy294126178241029xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy251355464564673777xy24650607080x = 4.5

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.

n = 6x = 4.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 200000 €

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.

Growing perpetuity

PV = C1/(r − g) = 8 000/(0.07 − 0.03) = 8 000/0.04 = €200 000.00

Answer

PV = €200 000.00
2.Time value of money

Answer: A = 1386 €

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 = 0.03/12 = 0.0025
N = 12·20 = 240

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 = 0.5492
A = 250 000·0.0025/(1 − 0.5492) = €1 386.49

Answer

A = €1 386.49
3.NPV & IRR

Answer: EAC = 9297 €

Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.

Annuity factor

(1 − 1.12−6)/0.12 = 4.111

Annualised purchase price

30 000/4.111 = 7 296.77

EAC

EAC = 7 296.77 + 2 000 = €9 296.77

Answer

EAC = €9 296.77
4.NPV & IRR

Answer: IRR = 31.03 %

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(30%) = 345.93
NPV(35%) = −1 255.14

Narrow down

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

IRR ≈ 30 + 5·345.93/(345.93 − (−1 255.14)) = 31.08%
exact: NPV(31.03%) = 0 → IRR = 31.03%

Answer

IRR = 31.03%
5.Bond pricing & yield

Answer: HPR = 1.607 %

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.

Purchase price

C = €80.00, i = 0.025, N = 3
€80.00·(1 − 0.9286)/0.025 + €1 000·0.9286 = €228.48 + €928.60 = €1 157.08

Selling price

C = €80.00, i = 0.03, N = 2
€80.00·(1 − 0.9426)/0.03 + €1 000·0.9426 = €153.08 + €942.60 = €1 095.67

Holding period return

HPR = (80 + 1 095.67 − 1 157.08)/1 157.08 = 1.607%

Answer

HPR = 1.607%
6.Bond pricing & yield

Answer: YTM = 3.503 %

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

€1 055 = €50·(1 − (1 + y)⁻⁴)/y + €1 000/(1 + y)⁴

Solve numerically

P(3%) = €1 074.34, P(4%) = €1 036.30
YTM = 3.503%

Answer

YTM = 3.503%
7.CAPM & WACC

Answer: WACC = 9.2 %

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 = 8 000 000 + 2 000 000 + 2 000 000 = €12 000 000
E/V = 0.6667, Ps/V = 0.1667, D/V = 0.1667

WACC

WACC = 0.6667·11% + 0.1667·8% + 0.1667·4%·(1 − 0.2) = 9.2%

Answer

WACC = 9.2%
8.CAPM & WACC

Answer: WACC = 8.014 %

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 = 2 000 000·€44.50 = €89 000 000
D = 10 000·€980 = €9 800 000

Cost of equity (CAPM)

re = 2% + 1.3·5% = 8.5%

Weights

V = €98 800 000, E/V = 0.9008, D/V = 0.09919

WACC

WACC = 0.9008·8.5% + 0.09919·5%·(1 − 0.28) = 8.014%

Answer

WACC = 8.014%
9.Descriptive statistics

Answer: s = 20.81

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 = 247/7 = 35.29

Squared deviations

Σ(x − x̄)² = 12.71² + (−22.29)² + 23.71² + (−19.29)² + 8.714² + 19.71² + (−23.29)² = 2599

Standard deviation

s = √(2599/6) = 20.81

Answer

s = 20.81
10.Descriptive statistics

Answer: z = −1

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 = (62 − 70)/8 = −1

Answer

z = −1
11.Probability

Answer: Var(X) = 1.74

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 = 1·0.25 + 2·0.25 + 3·0.3 + 4·0.05 + 5·0.15 = 2.6

Mean of the squares

E(X²) = 1·0.25 + 4·0.25 + 9·0.3 + 16·0.05 + 25·0.15 = 8.5

Variance

Var(X) = 8.5 − 2.6² = 1.74

Answer

Var(X) = 1.74
12.Probability

Answer: σY = 0.4183

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, Var(X) = 0.7, σX = 0.8367

Linear change

σY = |0.5|·0.8367 = 0.4183

Answer

σY = 0.4183

The mean does move: E(Y) = 0.5·1 + 1 = 1.5.

13.Normal distribution

Answer: z = 1.29

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 + 0.8029)/2 = 0.9015

Read the z-table backwards

Find 0.9015 inside the table and read off z.

z = 1.29

Answer

z = 1.29
14.Normal distribution

Answer: P = 0.9772

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 = 25.05 − 25 = 0.05 mm, σ = 0.025 mm

Standardise

z = (0 − 0.05)/0.025 = −2

Probability

P = 1 − Φ(−2) = 0.9772

Answer

P = 0.9772
15.Confidence intervals

Answer: E₂ = 3.417

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)·√(25/25) = 1.314

New margin

E₂ = 2.6·1.314 = 3.417

Answer

E₂ = 3.417
16.Confidence intervals

Answer: n = 385

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 = 1.96²·0.5·0.5/0.05² = 384.2

Round up

n = 385

Answer

n = 385
17.Hypothesis tests & confidence intervals

Answer: z = −2.216

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Standard error of the difference

SE = √(18²/60 + 4²/50) = √5.72 = 2.392

Test statistic

z = (140 − 145.3)/2.392 = −2.216

Answer

z = −2.216

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = −0.9609

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Sample proportions

p̂₁ = 98/150 = 0.6533
p̂₂ = 85/120 = 0.7083
p̂ = 183/270 = 0.6778

Standard error under H₀

SE = √(0.6778·0.3222·(1/150 + 1/120)) = 0.05724

Test statistic

z = (0.6533 − 0.7083)/0.05724 = −0.9609

Answer

z = −0.9609

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

19.Linear regression & correlation

Answer: r = 0.992

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 = 30/5 = 6
ȳ = 91/5 = 18.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 104
Syy = Σ(y − ȳ)² = 274.8

Correlation coefficient

r = 104/√(40·274.8) = 0.992

Answer

r = 0.992
20.Linear regression & correlation

Answer: ŷ = 64

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 = 27/6 = 4.5
ȳ = 384/6 = 64

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 92

Regression line

b = 92/17.5 = 5.257
a = 64 − 5.257·4.5 = 40.34
ŷ = 40.34 + 5.257·x

Prediction

ŷ = 40.34 + 5.257·4.5 = 64

Answer

ŷ = 64

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