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
periods0123…1211 + EAR = ?18% / 12

A credit card charges a nominal interest rate of 18% per year, compounded monthly. Find the effective annual rate EAR in percent.

r = 18%m = 12
EAR = %
2.Time value of moneyHard
months01…156…300€200 000AB = ?A2.4% / 12

A mortgage of €200 000 is repaid with equal monthly payments over 25 years at 2.4% per year, compounded monthly. How much is still owed just after the payment at the end of year 13? Find the outstanding balance B.

P = €200 0002.4% / 12N = 300k = 156
B = €
3.NPV & IRRMedium
years01234−30 0006 5008 50013 0008 000

A project costs €30 000 today and returns €6 500, €8 500, €13 000, €8 000 at the end of years 1 to 4. The cost of capital is 6%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 6%CF0 = −30 000CF1 = 6 500CF2 = 8 500CF3 = 13 000CF4 = 8 000
DPB = years
4.NPV & IRRHard
years0123456−10 0002 5002 5002 5002 5002 5002 500

An investment of €10 000 returns €2 500 at the end of each year for 6 years. Find its internal rate of return IRR in percent.

C0 = €10 000CF = €2 500n = 6
IRR = %
5.Bond pricing & yieldMedium
half-years0123…12€40€40€40 + €1 000P = ?

A bond with a face value of €1 000 pays an 8% 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 = 8%YTM = 5%n = 6
P = €
6.Bond pricing & yieldHard

A bond trades at €935 with a yield to maturity of 7% (annual compounding) and a Macaulay duration of 7 years. Use the modified duration to estimate the new price P if the yield rises by 25 basis points.

P0 = €935y = 7%D = 7Δy = +25 bp
P = €
7.CAPM & WACCMedium
βr (%)SMLrf​ = 1%β = 1rm​ = ?β = 1.5re​ = 10%

A share with a beta of 1.5 has a required return of 10%. The risk-free rate is 1%. What expected market return rm does CAPM imply?

β = 1.5re = 10%rf = 1%
rm = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €28.50D0​ = €1.50g = 3%debt€14 300 000rd​ = 3%T = 30%

A company has 1 000 000 shares trading at €28.50; the book value of its equity is €17 100 000. It just paid a dividend of €1.50 per share, and dividends grow by 3% per year. Its debt has a market value of €14 300 000 and a pre-tax cost of 3%. The tax rate is 30%. Find the WACC, using the dividend growth model for the cost of equity.

N = 1 000 000P0 = €28.50book equity = €17 100 000D0 = €1.50g = 3%D = €14 300 000rd = 3%T = 30%
WACC = %
9.Descriptive statisticsMedium
1015202530354045505560

Find the sample standard deviation s of the data: 24, 38, 60, 48, 10.

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

A quantity is normally distributed with mean 500 and standard deviation 60. What proportion P of values is above 536?

μ = 500σ = 60x = 536
P =
11.ProbabilityMedium
xp(x)00.310.220.1530.35μ = 1.55

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

x: 0, 1, 2, 3p(x): 0.3, 0.2, 0.15, 0.35
σ =
12.ProbabilityHard
xp(x)10.120.330.2540.250.15μ

X has the distribution shown, and Y = 0.5X − 3. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.1, 0.3, 0.25, 0.2, 0.15Y = 0.5X − 3
σY =
13.Normal distributionMedium
45.585P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 45.5 and 85 points? Give P as a decimal.

μ = 65σ = 10x1 = 45.5x2 = 85
P =
14.Normal distributionHard
557085100115130145pointsbottom 70%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 70% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points70%
x = points
15.Confidence intervalsMedium
n = 15, s = 16LUx̄ = 98E = ?99%

A sample of 15 items has mean 98 and standard deviation s = 16. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 15x̄ = 98s = 1699%t(14) = 2.977
E =
16.Confidence intervalsHard
p ≈ 0.3, n = ?p̂ − Ep̂ + Ep̂E = 0.0499%

A pilot study suggests that about 0.3 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 0.04?

p ≈ 0.3E = 0.0499%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 100 parts has mean length 65 mm, and the line's known standard deviation is 17 mm. Line 2: a sample of 40 parts has mean 69.5 mm, known standard deviation 7 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 100x̄₁ = 65 mmσ₁ = 17 mmn₂ = 40x̄₂ = 69.5 mmσ₂ = 7 mm
z =
18.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 25 measurements has mean 116.4. The population standard deviation is 12. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 116.4σ = 12n = 2590%
U =
19.Linear regression & correlationMedium
xy346447552655765xy465060

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy2324396548611073xy5104060ŷ = 47

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. For which value x does the line predict ŷ = 47?

n = 5ŷ = 47
x =

3Solutions

1.Time value of money

Answer: EAR = 19.56 %

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 per period

r/m = 0.18/12 = 0.015

Compound over one year

EAR = (1 + 0.015)12 − 1 = 1.196 − 1 = 19.56%

Answer

EAR = 19.56%

The EAR is always above the nominal rate when interest compounds more than once a year.

2.Time value of money

Answer: B = 110909 €

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 payment

i = 0.002, N = 300
A = 200 000·0.002/(1 − (1 + 0.002)−300) = €887.19

Present value of the remaining payments

Whatever is still owed must be exactly paid off by the remaining payments.

N − k = 300 − 156 = 144
B = 887.2·(1 − (1 + 0.002)−144)/0.002 = €110 909.40

Answer

B = €110 909.40
3.NPV & IRR

Answer: DPB = 3.85 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: 6 500/1.06 = 6 132.08 → Σ = 6 132.08
year 2: 8 500/1.124 = 7 564.97 → Σ = 13 697.05
year 3: 13 000/1.191 = 10 915.05 → Σ = 24 612.10
year 4: 8 000/1.262 = 6 336.75 → Σ = 30 948.85

Fraction of the last year

DPB = 3 + 5 387.90/6 336.75 = 3.85

Answer

DPB = 3.85 years

The discounted payback is always longer than the simple payback (3.25 years).

4.NPV & IRR

Answer: IRR = 12.98 %

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.

Required annuity factor

(1 − (1 + i)−6)/i = 10 000/2 500 = 4

Solve for the rate

The annuity factor falls as the rate rises; bracket the rate and narrow down.

i = 12%: 4.111
i = 13%: 3.998
IRR = 12.98%

Answer

IRR = 12.98%
5.Bond pricing & yield

Answer: P = 1154 €

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 = €40.00, i = 0.025, N = 12

Present value of the coupons

€40.00·(1 − 0.7436)/0.025 = €410.31

Present value of the face value

€1 000·0.7436 = €743.56

Price

P = €1 153.87

Answer

P = €1 153.87

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

6.Bond pricing & yield

Answer: P = 919.7 €

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.

Modified duration

Dmod = 7/1.07 = 6.542

Estimated price change

%ΔP ≈ −6.542·(0.0025) = −1.636%
P ≈ 935·(1 − 0.01636) = €919.71

Answer

P = €919.71

Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.

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% − 1%)/1.5 = 6%

Market return

rm = 1% + 6% = 7%

Answer

rm = 7%
8.CAPM & WACC

Answer: WACC = 6.309 %

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 equity

D1 = 1.5·1.03 = €1.54
re = 1.545/28.5 + 0.03 = 8.421%

Market values

E = 1 000 000·€28.50 = €28 500 000
V = €42 800 000, E/V = 0.6659, D/V = 0.3341

WACC

WACC = 0.6659·8.421% + 0.3341·3%·(1 − 0.3) = 6.309%

Answer

WACC = 6.309%
9.Descriptive statistics

Answer: s = 19.65

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 = 180/5 = 36

Squared deviations

Σ(x − x̄)² = (−12)² + 2² + 24² + 12² + (−26)² = 1544

Standard deviation

s = √(1544/4) = 19.65

Answer

s = 19.65
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: σ = 1.244

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 = 0·0.3 + 1·0.2 + 2·0.15 + 3·0.35 = 1.55

Mean of the squares

E(X²) = 0·0.3 + 1·0.2 + 4·0.15 + 9·0.35 = 3.95

Variance

Var(X) = 3.95 − 1.55² = 1.548
σ = √1.548 = 1.244

Answer

σ = 1.244
12.Probability

Answer: σY = 0.6124

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 = 3, Var(X) = 1.5, σX = 1.225

Linear change

σY = |0.5|·1.225 = 0.6124

Answer

σY = 0.6124

The mean does move: E(Y) = 0.5·3 − 3 = −1.5.

13.Normal distribution

Answer: P = 0.9517

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 = (45.5 − 65)/10 = −1.95
z₂ = (85 − 65)/10 = 2

Area between

P = Φ(2) − Φ(−1.95) = 0.9772 − 0.02559 = 0.9517

Answer

P = 0.9517
14.Normal distribution

Answer: x = 107.8 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) = 0.7

z from the table

z ≈ 0.52

Back to the original units

x = 100 + 0.52·15 = 107.8 points

Answer

x = 107.8 points
15.Confidence intervals

Answer: E = 12.3

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 = 2.977

Margin of error

E = 2.977·16/√15 = 12.3

Answer

E = 12.3
16.Confidence intervals

Answer: n = 871

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 = 2.576²·0.3·0.7/0.04² = 870.9

Round up

n = 871

Answer

n = 871
17.Hypothesis tests & confidence intervals

Answer: z = −2.218

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 = √(17²/100 + 7²/40) = √4.115 = 2.029

Test statistic

z = (65 − 69.5)/2.029 = −2.218

Answer

z = −2.218

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

18.Hypothesis tests & confidence intervals

Answer: U = 120.3

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 = 116.4 + 3.948 = 120.3

Answer

U = 120.3
19.Linear regression & correlation

Answer: r = 0.9509

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 = 25/5 = 5
ȳ = 265/5 = 53

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 46
Syy = Σ(y − ȳ)² = 234

Correlation coefficient

r = 46/√(10·234) = 0.9509

Answer

r = 0.9509
20.Linear regression & correlation

Answer: x = 5.077

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 = 259/5 = 51.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 208

Regression line

b = 208/40 = 5.2
a = 51.8 − 5.2·6 = 20.6
ŷ = 20.6 + 5.2·x

Solve for x

x = (47 − 20.6)/5.2 = 5.077

Answer

x = 5.077

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