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€5 000FV = ?

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

PV = €5 0005% / 4n = 8
FV = €
2.Time value of moneyHard
months0123…240€200 000A?AAA

A loan of €200 000 is repaid with equal monthly payments over 20 years. The interest rate is 4.2% per year, compounded monthly. Find the monthly payment A.

P = €200 0004.2% / 12N = 240
A = €
3.NPV & IRRMedium
years012345−100 00020 00020 00020 00020 00035 000

A machine costs €100 000 today. It generates €20 000 at the end of each year for 5 years and can then be sold for €15 000. The cost of capital is 10%. Find the net present value NPV.

C0 = €100 000CF = €20 000S = €15 000n = 5r = 10%
NPV = €
4.NPV & IRRHard
years012−10 0004 5007 000

A project costs €10 000 and returns €4 500, €7 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −10 000CF1 = 4 500CF2 = 7 000
IRR = %
5.Bond pricing & yieldMedium
years0123…12€1 435C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 12 years and pays annual coupons. It sells for €1 435 when the yield to maturity is 3.5%. Find the coupon rate c in percent.

F = €1 000n = 12P = €1 435YTM = 3.5%
c = %
6.Bond pricing & yieldHard
years012345€1 215€80€80€80 + €1 080

A bond with 12 years to maturity (face value €1 000, annual coupon 8%) sells for €1 215. The issuer can call it in 5 years at a call price of €1 080. Find the yield to call YTC in percent.

F = €1 000coupon = 8%P = €1 215call price = €1 080call in 5 years
YTC = %
7.CAPM & WACCMedium
WACC = ?equity€3 000 000re​ = 11%debt€5 000 000rd​ = 4%T = 25%

A company has equity worth €3 000 000 and debt worth €5 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 4% and the tax rate is 25%. Find the WACC.

E = €3 000 000D = €5 000 000re = 11%rd = 4%T = 25%
WACC = %
8.CAPM & WACCHard
rf​ = 3%, rm​ − rf​ = 6%D/E = 0.5equityβL​ = 1.4debtD/E = 1.5equityre​ = ?debtT = 21%

A company's shares have a beta of 1.4 at its current debt-to-equity ratio of 0.5. It plans to raise its debt-to-equity ratio to 1.5. The tax rate is 21%, the risk-free rate 3% and the market risk premium 6%. Find the cost of equity re after the change.

βL = 1.4D/E1 = 0.5D/E2 = 1.5T = 21%rf = 3%rm − rf = 6%
re = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 23, 44, 36, 54, 34, 14.

n = 6
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 60 and standard deviation 10. What proportion P of values is above 48?

μ = 60σ = 10x = 48
P =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5P(six) = 1/6 on each roll

A fair die is rolled 5 times. What is the probability P of getting at least one six?

n = 5
P =
12.ProbabilityHard
n = 9, p = 0.6: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 9p = 0.6
P =
13.Normal distributionMedium
250 kg248.4 kgT = X₁ + … + X₁₀?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 10 of them exceeds 248.4 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 10> 248.4 kg
P =
14.Normal distributionHard
157164171178185192199cmbottom 90%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm90%
x = cm
15.Confidence intervalsMedium
x̄E₁ = 10.890%, n = 36x̄E₂ = ?95%, n = 324

A 90% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 10.8. What is the margin of error E₂ of a 95% interval based on 324 observations?

90%n₁ = 36E₁ = 10.895%n₂ = 324
E₂ =
16.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.02599%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.025?

p = 0.5E = 0.02599%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 123.6 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 123.6s = 12n = 49
t =
18.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 290%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 290%
n =
19.Linear regression & correlationMedium
xy2641261882110291232xy510102030

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 118Σx² = 364Σy² = 2810Σxy = 1010
r =
20.Linear regression & correlationHard
xy14431750206623792679xy1520256080ŷ = 79

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. For which value x does the line predict ŷ = 79?

n = 5ŷ = 79
x =

3Solutions

1.Time value of money

Answer: FV = 7441 €

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 = 4·8 = 32

Future value

FV = 5 000·(1 + 0.0125)32 = 5 000·1.488 = €7 440.65

Answer

FV = €7 440.65
2.Time value of money

Answer: A = 1233 €

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 = 0.042/12 = 0.0035
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 = 0.4323
A = 200 000·0.0035/(1 − 0.4323) = €1 233.14

Answer

A = €1 233.14
3.NPV & IRR

Answer: NPV = −14870 €

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 the yearly cash flows

20 000·3.791 = 75 815.74

Present value of the salvage value

15 000/1.611 = 9 313.82

NPV

NPV = −100 000 + 75 815.74 + 9 313.82 = −€14 870.44

Answer

NPV = −€14 870.44
4.NPV & IRR

Answer: IRR = 9.139 %

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(5%) = 634.92
NPV(10%) = −123.97

Narrow down

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

IRR ≈ 5 + 5·634.92/(634.92 − (−123.97)) = 9.183%
exact: NPV(9.139%) = 0 → IRR = 9.139%

Answer

IRR = 9.139%
5.Bond pricing & yield

Answer: c = 8.002 %

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.

Present value of the face value

1000·(1 + 0.035)−12 = €661.78

Coupon

C = (1435 − 661.78)/9.663 = €80.02

Coupon rate

c = 80.02/1000 = 8.002%

Answer

c = 8.002%
6.Bond pricing & yield

Answer: YTC = 4.556 %

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

1215 = 80·(1 − (1 + y)−5)/y + 1080/(1 + y)5

Solve numerically

y = 4%: €1 243.83
y = 5%: €1 192.57
YTC = 4.556%

Answer

YTC = 4.556%

The yield to maturity is 5.505%. 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: WACC = 6 %

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, E/V = 0.375, D/V = 0.625

WACC

WACC = 0.375·11% + 0.625·4%·(1 − 0.25) = 6%

Answer

WACC = 6%
8.CAPM & WACC

Answer: re = 16.16 %

Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.

Unlever the beta

βU = 1.4/(1 + 0.79·0.5) = 1.004

Relever at the new D/E

βL = 1.004·(1 + 0.79·1.5) = 2.193

Cost of equity (CAPM)

re = 3% + 2.193·6% = 16.16%

Answer

re = 16.16%
9.Descriptive statistics

Answer: s = 14.32

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 = 205/6 = 34.17

Squared deviations

Σ(x − x̄)² = (−11.17)² + 9.833² + 1.833² + 19.83² + (−0.1667)² + (−20.17)² = 1025

Standard deviation

s = √(1025/5) = 14.32

Answer

s = 14.32
10.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
11.Probability

Answer: P = 0.5981

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)5 = 0.4019
P = 1 − 0.4019 = 0.5981

Answer

P = 0.5981
12.Probability

Answer: P = 0.003801

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) = 0.49 = 0.0002621
P(X = 1) = 9·0.6·0.48 = 0.003539

Add them

P = 0.0002621 + 0.003539 = 0.003801

Answer

P = 0.003801
13.Normal distribution

Answer: P = 0.8438

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 total

μT = 250 kg, σT = 0.5·√10 = 1.581 kg

Standardise

z = (248.4 − 250)/1.581 = −1.01

Probability

P = 1 − Φ(−1.01) = 0.8438

Answer

P = 0.8438
14.Normal distribution

Answer: x = 187 cm

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) = 0.9

z from the table

z ≈ 1.28

Back to the original units

x = 178 + 1.28·7 = 187 cm

Answer

x = 187 cm
15.Confidence intervals

Answer: E₂ = 4.289

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 = (1.96/1.645)·√(36/324) = 0.3972

New margin

E₂ = 10.8·0.3972 = 4.289

Answer

E₂ = 4.289
16.Confidence intervals

Answer: n = 2655

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 = 2.576²·0.5·0.5/0.025² = 2654

Round up

n = 2655

Answer

n = 2655
17.Hypothesis tests & confidence intervals

Answer: t = 2.1

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/√49 = 12/7 = 1.714

Test statistic

t = (123.6 − 120)/1.714 = 2.1

Answer

t = 2.1
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 98

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).

Solve for n

n = (1.645·12/2)² = 97.42

Round up

n = 98

Answer

n = 98
19.Linear regression & correlation

Answer: r = 0.9942

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 = 1010 − 42·118/6 = 184
Sxx = 364 − 42²/6 = 70
Syy = 2810 − 118²/6 = 489.3

Correlation coefficient

r = 184/√(70·489.3) = 0.9942

Answer

r = 0.9942
20.Linear regression & correlation

Answer: x = 24.63

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 = 100/5 = 20
ȳ = 317/5 = 63.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 303

Regression line

b = 303/90 = 3.367
a = 63.4 − 3.367·20 = −3.933
ŷ = −3.933 + 3.367·x

Solve for x

x = (79 − (−3.933))/3.367 = 24.63

Answer

x = 24.63

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