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…8€8 000FV = ?

You invest €8 000 at a nominal rate of 2% per year, compounded quarterly. Find the value FV of the investment after 8 years.

PV = €8 0002% / 4n = 8
FV = €
2.Time value of moneyHard
months0123…300€200 000AAAA5.4% / 12, I = ?

A loan of €200 000 is repaid with equal monthly payments over 25 years at 5.4% per year, compounded monthly. How much interest I is paid in total over the life of the loan?

P = €200 0005.4% / 12N = 300
I = €
3.NPV & IRRMedium
years01234−30 00013 00015 00011 00015 000

An investment of €30 000 gives these cash flows at the end of years 1 to 4: €13 000, €15 000, €11 000, €15 000. The cost of capital is 6%. Find the net present value NPV.

r = 6%CF0 = −30 000CF1 = 13 000CF2 = 15 000CF3 = 11 000CF4 = 15 000
NPV = €
4.NPV & IRRHard
years0123…8−50 00010 50010 50010 50010 500

An investment of €50 000 returns €10 500 at the end of each year for 8 years. Find its internal rate of return IRR in percent.

C0 = €50 000CF = €10 500n = 8
IRR = %
5.Bond pricing & yieldMedium
half-years0123…8€35€35€35 + €1 000P = ?

A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 4 years. The yield to maturity is 5.5% per year. Find the price P of the bond.

F = €1 000coupon = 7%YTM = 5.5%n = 4
P = €
6.Bond pricing & yieldHard
years0123…9€930€40€40 + €1 000

A bond with a face value of €1 000 and an annual coupon of 4% matures in 9 years and sells for €930. Find its yield to maturity YTM in percent.

F = €1 000coupon = 4%n = 9P = €930
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€8 000 000re​ = 8%debt€5 000 000rd​ = 7%T = 25%

A company has equity worth €8 000 000 and debt worth €5 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 7% and the tax rate is 25%. Find the WACC.

E = €8 000 000D = €5 000 000re = 8%rd = 7%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity4 000 000 × €28.50D0​ = €1.60g = 4%debt€91 200 000rd​ = 4.5%T = 28%

A company has 4 000 000 shares trading at €28.50; the book value of its equity is €34 200 000. It just paid a dividend of €1.60 per share, and dividends grow by 4% per year. Its debt has a market value of €91 200 000 and a pre-tax cost of 4.5%. The tax rate is 28%. Find the WACC, using the dividend growth model for the cost of equity.

N = 4 000 000P0 = €28.50book equity = €34 200 000D0 = €1.60g = 4%D = €91 200 000rd = 4.5%T = 28%
WACC = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 15, 37, 26, 12, 54.

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

A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is below 140?

μ = 100σ = 20x = 140
P =
11.ProbabilityMedium
xp(x)00.220.2540.360.25μ = 3.2

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

x: 0, 2, 4, 6p(x): 0.2, 0.25, 0.3, 0.25
Var(X) =
12.ProbabilityHard
n = 9, p = 0.25: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?

n = 9p = 0.25
P =
13.Normal distributionMedium
750 g759.8 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 759.8 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 759.8 g
P =
14.Normal distributionHard
μ = 100x = ?90%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 90% of them lie (the 90th percentile).

μ = 100σ = 15P = 90%
x = points
15.Confidence intervalsMedium
x̄E₁ = 11.299%, n = 25x̄E₂ = ?90%, n = 25

A 99% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 11.2. What is the margin of error E₂ of a 90% interval from the same data?

99%n₁ = 25E₁ = 11.290%
E₂ =
16.Confidence intervalsHard
σ = 37, n = ?x̄ − Ex̄ + Ex̄E = 195%

The population standard deviation is 37. 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 1?

σ = 37E = 195%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

20 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 −7.31 and the standard deviation of the differences is 9.9. Find the paired test statistic t for H₀: μd = 0.

n = 20d̄ = −7.31sd = 9.9
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 90 of 150 patients on treatment A recover, and 96 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 90n₁ = 150x₂ = 96n₂ = 150
z =
19.Linear regression & correlationMedium
xy33754474995811641369xy51040506070

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 48Σy = 321Σx² = 454Σy² = 17927Σxy = 2797
r =
20.Linear regression & correlationHard
xy124221316415512xy24152025ŷ = 19

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 19?

n = 5ŷ = 19
x =

3Solutions

1.Time value of money

Answer: FV = 9384 €

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 = 0.02/4 = 0.005
N = m·n = 4·8 = 32

Future value

FV = 8 000·(1 + 0.005)32 = 8 000·1.173 = €9 384.34

Answer

FV = €9 384.34
2.Time value of money

Answer: I = 164878 €

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 = 0.0045, N = 300
A = 200 000·0.0045/(1 − (1 + 0.0045)−300) = €1 216.26

Total interest

I = N·A − P = 300·1216 − 200 000 = €164 877.95

Answer

I = €164 877.95
3.NPV & IRR

Answer: NPV = 16731 €

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 = −30 000
CF1/(1 + 0.06)1 = 13 000/1.06 = 12 264.15
CF2/(1 + 0.06)2 = 15 000/1.124 = 13 349.95
CF3/(1 + 0.06)3 = 11 000/1.191 = 9 235.81
CF4/(1 + 0.06)4 = 15 000/1.262 = 11 881.40

Add them up

NPV = €16 731.31

Answer

NPV = €16 731.31

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

4.NPV & IRR

Answer: IRR = 13.23 %

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)−8)/i = 50 000/10 500 = 4.762

Solve for the rate

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

i = 13%: 4.799
i = 14%: 4.639
IRR = 13.23%

Answer

IRR = 13.23%
5.Bond pricing & yield

Answer: P = 1053 €

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 = €35.00, i = 0.0275, N = 8

Present value of the coupons

€35.00·(1 − 0.8049)/0.0275 = €248.30

Present value of the face value

€1 000·0.8049 = €804.91

Price

P = €1 053.21

Answer

P = €1 053.21

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

6.Bond pricing & yield

Answer: YTM = 4.984 %

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

€930 = €40·(1 − (1 + y)⁻⁹)/y + €1 000/(1 + y)⁹

Solve numerically

P(4%) = €1 000.00, P(5%) = €928.92
YTM = 4.984%

Answer

YTM = 4.984%
7.CAPM & WACC

Answer: WACC = 6.942 %

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 = €13 000 000, E/V = 0.6154, D/V = 0.3846

WACC

WACC = 0.6154·8% + 0.3846·7%·(1 − 0.25) = 6.942%

Answer

WACC = 6.942%
8.CAPM & WACC

Answer: WACC = 6.906 %

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 = 1.6·1.04 = €1.66
re = 1.664/28.5 + 0.04 = 9.839%

Market values

E = 4 000 000·€28.50 = €114 000 000
V = €205 200 000, E/V = 0.5556, D/V = 0.4444

WACC

WACC = 0.5556·9.839% + 0.4444·4.5%·(1 − 0.28) = 6.906%

Answer

WACC = 6.906%
9.Descriptive statistics

Answer: s = 17.2

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 = 144/5 = 28.8

Squared deviations

Σ(x − x̄)² = (−13.8)² + 8.2² + (−2.8)² + (−16.8)² + 25.2² = 1183

Standard deviation

s = √(1183/4) = 17.2

Answer

s = 17.2
10.Descriptive statistics

Answer: P = 0.9772

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 0.9772

Answer

P = 0.9772
11.Probability

Answer: Var(X) = 4.56

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 = 0·0.2 + 2·0.25 + 4·0.3 + 6·0.25 = 3.2

Mean of the squares

E(X²) = 0·0.2 + 4·0.25 + 16·0.3 + 36·0.25 = 14.8

Variance

Var(X) = 14.8 − 3.2² = 4.56

Answer

Var(X) = 4.56
12.Probability

Answer: P = 0.3003

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) = 0.759 = 0.07508
P(X = 1) = 9·0.25·0.758 = 0.2253

Add them

P = 0.07508 + 0.2253 = 0.3003

Answer

P = 0.3003
13.Normal distribution

Answer: P = 0.1587

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 sum

μ = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (759.8 − 750)/9.849 = 1

Upper tail

P = 1 − Φ(1) = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
14.Normal distribution

Answer: x = 119.2 points

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) = 0.9 → z = 1.282

Back to x

x = 100 + 1.282·15 = 119.22 points

Answer

x = 119.22 points
15.Confidence intervals

Answer: E₂ = 7.152

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 = (1.645/2.576)·√(25/25) = 0.6386

New margin

E₂ = 11.2·0.6386 = 7.152

Answer

E₂ = 7.152
16.Confidence intervals

Answer: n = 5260

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 = (1.96·37/1)² = 5259

Round up

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

n = 5260

Answer

n = 5260
17.Hypothesis tests & confidence intervals

Answer: t = −3.302

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 = 9.9/√20 = 2.214

Test statistic

t = −7.31/2.214 = −3.302

Answer

t = −3.302

With df = 19, the two-sided 5% critical value is 2.093; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: z = −0.7137

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 = 90/150 = 0.6
p̂₂ = 96/150 = 0.64
p̂ = 186/300 = 0.62

Standard error under H₀

SE = √(0.62·0.38·(1/150 + 1/150)) = 0.05605

Test statistic

z = (0.6 − 0.64)/0.05605 = −0.7137

Answer

z = −0.7137

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

19.Linear regression & correlation

Answer: r = 0.9971

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.

Centred sums

Sxy = 2797 − 48·321/6 = 229
Sxx = 454 − 48²/6 = 70
Syy = 17927 − 321²/6 = 753.5

Correlation coefficient

r = 229/√(70·753.5) = 0.9971

Answer

r = 0.9971
20.Linear regression & correlation

Answer: x = 2.533

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 = 88/5 = 17.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −30

Regression line

b = −30/10 = −3
a = 17.6 − (−3)·3 = 26.6
ŷ = 26.6 − 3·x

Solve for x

x = (19 − 26.6)/(−3) = 2.533

Answer

x = 2.533

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