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+2.5%+2.5%…r = 6.5%

A property is expected to earn net rent of €8 000 next year. The rent is expected to grow by 2.5% per year forever. The required return is 6.5% per year. Find the value PV of the property today.

C1 = €8 000g = 2.5%r = 6.5%
PV = €
2.Time value of moneyHard
months0123…N = ?€20 000€275€275€275€2753% / 12

You borrow €20 000 at 3% per year, compounded monthly, and repay €275 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 = €2753% / 12
N = payments
3.NPV & IRRMedium
years01234−60 00025 00028 00014 00027 000

An investment of €60 000 gives these cash flows at the end of years 1 to 4: €25 000, €28 000, €14 000, €27 000. The cost of capital is 10%. Find the net present value NPV.

r = 10%CF0 = −60 000CF1 = 25 000CF2 = 28 000CF3 = 14 000CF4 = 27 000
NPV = €
4.NPV & IRRHard
years0123−20 00011 5008 50010 000

A project costs €20 000 and returns €11 500, €8 500, €10 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 11 500CF2 = 8 500CF3 = 10 000
IRR = %
5.Bond pricing & yieldMedium
half-years0123…22€10€10€10 + €1 000P = ?

A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 11 years. The yield to maturity is 3% per year. Find the price P of the bond.

F = €1 000coupon = 2%YTM = 3%n = 11
P = €
6.Bond pricing & yieldHard
years0123…9€1 075€50€50 + €1 000

A bond with a face value of €1 000 and an annual coupon of 5% matures in 9 years and sells for €1 075. Find its yield to maturity YTM in percent.

F = €1 000coupon = 5%n = 9P = €1 075
YTM = %
7.CAPM & WACCMedium
WACC = ?equityEre​ = 9%debtD = 1·Erd​ = 4%T = 30%

A company has a debt-to-equity ratio (at market values) of 1. Its cost of equity is 9%, its pre-tax cost of debt is 4% and the tax rate is 30%. Find the WACC.

D/E = 1re = 9%rd = 4%T = 30%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €49.50β = 0.8rf​ = 1.5%, rm​ = 8%debt20 000 × €980rd​ = 4.5%T = 20%

A company has 2 500 000 shares at €49.50 each and 20 000 bonds trading at €980 each. The pre-tax cost of debt is 4.5% and the tax rate is 20%. The share’s beta is 0.8, the risk-free rate 1.5% and the market return 8%. Find the WACC.

β = 0.8rf = 1.5%rm = 8%rd = 4.5%T = 20%
WACC = %
9.Descriptive statisticsMedium
3035404550

Find the sample standard deviation s of the data: 35, 34, 51, 30, 39.

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

Exam scores are normally distributed with mean 170 and standard deviation 12. Find the z-score z of a score of 194.

μ = 170σ = 12x = 194
z =
11.ProbabilityMedium
P(at least one six) = ??1?2?3P(six) = 1/6 on each roll

A fair die is rolled 3 times. What is the probability P of getting at least one six?

n = 3
P =
12.ProbabilityHard
n = 9, p = 0.1: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?

n = 9p = 0.1
P =
13.Normal distributionMedium
32 min28.3 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 28.3 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 28.3 min
P =
14.Normal distributionHard
35455565758595pointstop 30%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 30% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points30%
x = points
15.Confidence intervalsMedium
n = 21, s = 14LUx̄ = 78E = ?95%

A sample of 21 items has mean 78 and standard deviation s = 14. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 21x̄ = 78s = 1495%t(20) = 2.086
E =
16.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.02590%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.025?

p = 0.5E = 0.02590%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 48.8 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 48.8s = 12n = 81
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 38 of 150 patients on treatment A recover, and 98 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 38n₁ = 150x₂ = 98n₂ = 250
z =
19.Linear regression & correlationMedium
xy127227325422518619xy2462025

The table gives 6 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy166919812294251122812431136xy15202530100x = 17.5

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (17.5, 73) is made. Find its residual e = y − ŷ.

n = 6(17.5, 73)
e =

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/(0.065 − 0.025) = 8 000/0.04 = €200 000.00

Answer

PV = €200 000.00
2.Time value of money

Answer: N = 81 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 − 20 000·0.0025/275 = 0.8182

Solve with logarithms

N = −ln(0.8182)/ln(1.002) = 80.37

Round up to whole payments

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

N = 81

Answer

N = 81 payments
3.NPV & IRR

Answer: NPV = 14828 €

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 = −60 000
CF1/(1 + 0.1)1 = 25 000/1.1 = 22 727.27
CF2/(1 + 0.1)2 = 28 000/1.21 = 23 140.50
CF3/(1 + 0.1)3 = 14 000/1.331 = 10 518.41
CF4/(1 + 0.1)4 = 27 000/1.464 = 18 441.36

Add them up

NPV = €14 827.54

Answer

NPV = €14 827.54

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

4.NPV & IRR

Answer: IRR = 24.16 %

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(20%) = 1 273.15
NPV(25%) = −240.00

Narrow down

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

IRR ≈ 20 + 5·1 273.15/(1 273.15 − (−240.00)) = 24.21%
exact: NPV(24.16%) = 0 → IRR = 24.16%

Answer

IRR = 24.16%
5.Bond pricing & yield

Answer: P = 906.9 €

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 = €10.00, i = 0.015, N = 22

Present value of the coupons

€10.00·(1 − 0.7207)/0.015 = €186.21

Present value of the face value

€1 000·0.7207 = €720.69

Price

P = €906.90

Answer

P = €906.90

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

6.Bond pricing & yield

Answer: YTM = 3.992 %

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

€1 075 = €50·(1 − (1 + y)⁻⁹)/y + €1 000/(1 + y)⁹

Solve numerically

P(3%) = €1 155.72, P(4%) = €1 074.35
YTM = 3.992%

Answer

YTM = 3.992%
7.CAPM & WACC

Answer: WACC = 5.9 %

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 from D/E

E/V = 1/(1 + 1) = 0.5
D/V = 1/(1 + 1) = 0.5

WACC

WACC = 0.5·9% + 0.5·4%·(1 − 0.3) = 5.9%

Answer

WACC = 5.9%
8.CAPM & WACC

Answer: WACC = 6.276 %

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 = 2 500 000·€49.50 = €123 750 000
D = 20 000·€980 = €19 600 000

Cost of equity (CAPM)

re = 1.5% + 0.8·6.5% = 6.7%

Weights

V = €143 350 000, E/V = 0.8633, D/V = 0.1367

WACC

WACC = 0.8633·6.7% + 0.1367·4.5%·(1 − 0.2) = 6.276%

Answer

WACC = 6.276%
9.Descriptive statistics

Answer: s = 8.044

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 = 189/5 = 37.8

Squared deviations

Σ(x − x̄)² = (−2.8)² + (−3.8)² + 13.2² + (−7.8)² + 1.2² = 258.8

Standard deviation

s = √(258.8/4) = 8.044

Answer

s = 8.044
10.Descriptive statistics

Answer: z = 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.

Standardise

z = (194 − 170)/12 = 2

Answer

z = 2
11.Probability

Answer: P = 0.4213

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.

Complement

P(no six) = (5/6)3 = 0.5787
P = 1 − 0.5787 = 0.4213

Answer

P = 0.4213
12.Probability

Answer: P = 0.7748

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) = 0.99 = 0.3874
P(X = 1) = 9·0.1·0.98 = 0.3874

Add them

P = 0.3874 + 0.3874 = 0.7748

Answer

P = 0.7748
13.Normal distribution

Answer: P = 0.9306

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 = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (28.3 − 32)/2.5 = −1.48

Upper tail

P = 1 − Φ(−1.48) = 1 − 0.06944 = 0.9306

Answer

P = 0.9306
14.Normal distribution

Answer: x = 70.2 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 = 65 + 0.52·10 = 70.2 points

Answer

x = 70.2 points
15.Confidence intervals

Answer: E = 6.373

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 = 20, t = 2.086

Margin of error

E = 2.086·14/√21 = 6.373

Answer

E = 6.373
16.Confidence intervals

Answer: n = 1083

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 = 1.645²·0.5·0.5/0.025² = 1082

Round up

n = 1083

Answer

n = 1083
17.Hypothesis tests & confidence intervals

Answer: t = −0.9

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/√81 = 12/9 = 1.333

Test statistic

t = (48.8 − 50)/1.333 = −0.9

Answer

t = −0.9
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = −2.834

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 = 38/150 = 0.2533
p̂₂ = 98/250 = 0.392
p̂ = 136/400 = 0.34

Standard error under H₀

SE = √(0.34·0.66·(1/150 + 1/250)) = 0.04892

Test statistic

z = (0.2533 − 0.392)/0.04892 = −2.834

Answer

z = −2.834

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

19.Linear regression & correlation

Answer: r² = 0.8974

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 = 21/6 = 3.5
ȳ = 138/6 = 23

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −35
Syy = Σ(y − ȳ)² = 78

Coefficient of determination

r² = −35²/(17.5·78) = 0.8974

Answer

r² = 0.8974

About 89.74% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = −2.124

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 = 141/6 = 23.5
ȳ = 616/6 = 102.7

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 723

Regression line

b = 723/157.5 = 4.59
a = 102.7 − 4.59·23.5 = −5.21
ŷ = −5.21 + 4.59·x

Predicted value

ŷ = −5.21 + 4.59·17.5 = 75.12

Residual

e = 73 − 75.12 = −2.124

Answer

e = −2.124

The point lies below the line.

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