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
years0123456PV = ?€15 000

You need €15 000 in 6 years. An account pays 3.6% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €15 0003.6% / 12n = 6
PV = €
2.Time value of moneyHard
months0123…300€200 000A?AAA

A loan of €200 000 is repaid with equal monthly payments over 25 years. The interest rate is 3% per year, compounded monthly. Find the monthly payment A.

P = €200 0003% / 12N = 300
A = €
3.NPV & IRRMedium
years01234−60 00029 00025 00026 00029 000

An investment of €60 000 gives these cash flows at the end of years 1 to 4: €29 000, €25 000, €26 000, €29 000. The cost of capital is 10%. Find the net present value NPV.

r = 10%CF0 = −60 000CF1 = 29 000CF2 = 25 000CF3 = 26 000CF4 = 29 000
NPV = €
4.NPV & IRRHard
years0123−40 00024 50029 00027 500

A project costs €40 000 and returns €24 500, €29 000, €27 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −40 000CF1 = 24 500CF2 = 29 000CF3 = 27 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…22€25€25€25 + €1 000P = ?

A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 11 years. The yield to maturity is 3.5% per year. Find the price P of the bond.

F = €1 000coupon = 5%YTM = 3.5%n = 11
P = €
6.Bond pricing & yieldHard
years0123456€1 177€80€80€80 + €1 080

A bond with 13 years to maturity (face value €1 000, annual coupon 8%) sells for €1 177. The issuer can call it in 6 years at a call price of €1 080. Find the yield to call YTC in percent.

F = €1 000coupon = 8%P = €1 177call price = €1 080call in 6 years
YTC = %
7.CAPM & WACCMedium
WACC = ?equity€12 000 000re​ = 8%debt€3 000 000rd​ = 3%T = 28%

A company has equity worth €12 000 000 and debt worth €3 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 3% and the tax rate is 28%. Find the WACC.

E = €12 000 000D = €3 000 000re = 8%rd = 3%T = 28%
WACC = %
8.CAPM & WACCHard
rf​ = 4%, rm​ − rf​ = 6.5%no debtequityβU​ = 1D/E = 1.5equityre​ = ?debtT = 30%

A company's business has an unlevered (asset) beta of 1. It plans a debt-to-equity ratio of 1.5. The tax rate is 30%, the risk-free rate is 4% and the market risk premium is 6.5%. Find the cost of equity re at the new capital structure.

βU = 1D/E = 1.5T = 30%rf = 4%rm − rf = 6.5%
re = %
9.Descriptive statisticsMedium
253035404550

Find the sample standard deviation s of the data: 29, 36, 36, 51, 44, 38, 23.

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

Exam scores are normally distributed with mean 50 and standard deviation 12. Find the z-score z of a score of 62.

μ = 50σ = 12x = 62
z =
11.ProbabilityMedium
xp(x)00.2510.220.2530.2540.05μ = 1.65

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

x: 0, 1, 2, 3, 4p(x): 0.25, 0.2, 0.25, 0.25, 0.05
Var(X) =
12.ProbabilityHard
n = 12, p = 0.1: P(X ≤ 1) = ?k0123456789101112

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

n = 12p = 0.1
P =
13.Normal distributionMedium
225 kg224.3 kgT = X₁ + … + X₉?

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 9 of them exceeds 224.3 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 9> 224.3 kg
P =
14.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
15.Confidence intervalsMedium
n = 8, s = 12.5LUx̄ = 41.5E = ?99%

A sample of 8 items has mean 41.5 and standard deviation s = 12.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 8x̄ = 41.5s = 12.599%t(7) = 3.499
E =
16.Confidence intervalsHard
μ₁ − μ₂, t(10) = 1.8120LU = ?x̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 7 plots, mean yield 83 kg, standard deviation 4.5 kg. Plot group 2: 5 plots, mean 85 kg, standard deviation 5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 7x̄₁ = 83s₁ = 4.5n₂ = 5x̄₂ = 85s₂ = 590%t(10) = 1.812
upper limit = kg
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 497.5 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 497.5s = 8n = 100
t =
18.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 199%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 199%
n =
19.Linear regression & correlationMedium
xy2104176258361044xy5102040

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy1651196022762589289831102xy152025306080100x = 23.5

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. Predict ŷ for x = 23.5.

n = 6x = 23.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 12090 €

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 = 0.036000000000000004/12 = 0.003
N = 12·6 = 72

Present value

PV = FV/(1 + i)N = 15 000/1.241 = €12 089.94

Answer

PV = €12 089.94
2.Time value of money

Answer: A = 948.4 €

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 = 12·25 = 300

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 = 0.4728
A = 200 000·0.0025/(1 − 0.4728) = €948.42

Answer

A = €948.42
3.NPV & IRR

Answer: NPV = 26366 €

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 = 29 000/1.1 = 26 363.64
CF2/(1 + 0.1)2 = 25 000/1.21 = 20 661.16
CF3/(1 + 0.1)3 = 26 000/1.331 = 19 534.18
CF4/(1 + 0.1)4 = 29 000/1.464 = 19 807.39

Add them up

NPV = €26 366.37

Answer

NPV = €26 366.37

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

4.NPV & IRR

Answer: IRR = 44.42 %

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(40%) = 2 317.78
NPV(45%) = −289.88

Narrow down

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

IRR ≈ 40 + 5·2 317.78/(2 317.78 − (−289.88)) = 44.44%
exact: NPV(44.42%) = 0 → IRR = 44.42%

Answer

IRR = 44.42%
5.Bond pricing & yield

Answer: P = 1136 €

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 = €25.00, i = 0.0175, N = 22

Present value of the coupons

€25.00·(1 − 0.6827)/0.0175 = €453.26

Present value of the face value

€1 000·0.6827 = €682.72

Price

P = €1 135.98

Answer

P = €1 135.98

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

6.Bond pricing & yield

Answer: YTC = 5.604 %

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

1177 = 80·(1 − (1 + y)−6)/y + 1080/(1 + y)6

Solve numerically

y = 5%: €1 211.97
y = 6%: €1 154.74
YTC = 5.604%

Answer

YTC = 5.604%

The yield to maturity is 6.001%. 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 = 6.832 %

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 = €15 000 000, E/V = 0.8, D/V = 0.2

WACC

WACC = 0.8·8% + 0.2·3%·(1 − 0.28) = 6.832%

Answer

WACC = 6.832%
8.CAPM & WACC

Answer: re = 17.32 %

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.

Levered beta

βL = 1·(1 + 0.7·1.5) = 2.05

Cost of equity (CAPM)

re = 4% + 2.05·6.5% = 17.32%

Answer

re = 17.32%
9.Descriptive statistics

Answer: s = 9.196

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 = 257/7 = 36.71

Squared deviations

Σ(x − x̄)² = (−7.714)² + (−0.7143)² + (−0.7143)² + 14.29² + 7.286² + 1.286² + (−13.71)² = 507.4

Standard deviation

s = √(507.4/6) = 9.196

Answer

s = 9.196
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 − 50)/12 = 1

Answer

z = 1
11.Probability

Answer: Var(X) = 1.528

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 = 0·0.25 + 1·0.2 + 2·0.25 + 3·0.25 + 4·0.05 = 1.65

Mean of the squares

E(X²) = 0·0.25 + 1·0.2 + 4·0.25 + 9·0.25 + 16·0.05 = 4.25

Variance

Var(X) = 4.25 − 1.65² = 1.528

Answer

Var(X) = 1.528
12.Probability

Answer: P = 0.659

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) = 0.912 = 0.2824
P(X = 1) = 12·0.1·0.911 = 0.3766

Add them

P = 0.2824 + 0.3766 = 0.659

Answer

P = 0.659
13.Normal distribution

Answer: P = 0.6808

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 = 225 kg, σT = 0.5·√9 = 1.5 kg

Standardise

z = (224.3 − 225)/1.5 = −0.47

Probability

P = 1 − Φ(−0.47) = 0.6808

Answer

P = 0.6808
14.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
15.Confidence intervals

Answer: E = 15.46

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 = 7, t = 3.499

Margin of error

E = 3.499·12.5/√8 = 15.46

Answer

E = 15.46
16.Confidence intervals

Answer: upper limit = 2.993 kg

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Pooled standard deviation

sp² = (6·4.5² + 4·5²)/10 = 22.15
sp = 4.706

Margin of error

E = 1.812·4.706·√(1/7 + 1/5) = 4.993

Interval

−2 ± 4.993 → [−6.993, 2.993]

Answer

upper limit = 2.993 kg
17.Hypothesis tests & confidence intervals

Answer: t = −3.125

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 = 8/√100 = 8/10 = 0.8

Test statistic

t = (497.5 − 500)/0.8 = −3.125

Answer

t = −3.125
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 664

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 = (2.576·10/1)² = 663.6

Round up

n = 664

Answer

n = 664
19.Linear regression & correlation

Answer: r² = 0.9944

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 = 132/5 = 26.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 174
Syy = Σ(y − ȳ)² = 761.2

Coefficient of determination

r² = 174²/(40·761.2) = 0.9944

Answer

r² = 0.9944

About 99.44% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 79.33

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 = 476/6 = 79.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 573

Regression line

b = 573/157.5 = 3.638
a = 79.33 − 3.638·23.5 = −6.162
ŷ = −6.162 + 3.638·x

Prediction

ŷ = −6.162 + 3.638·23.5 = 79.33

Answer

ŷ = 79.33

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