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…36511 + EAR = ?22% / 365

A credit card charges a nominal interest rate of 22% per year, compounded daily. Find the effective annual rate EAR in percent.

r = 22%m = 365
EAR = %
2.Time value of moneyHard
months0123…228€100€100€100FV = ?

You save €100 at the end of every month for 19 years. The account pays 5.4% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €1005.4% / 12N = 228
FV = €
3.NPV & IRRMedium
years01234−40 00013 00013 00013 00013 000

A project costs €40 000 today and returns €13 000 at the end of each year for 4 years. 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 = −40 000CF1 = 13 000CF2 = 13 000CF3 = 13 000CF4 = 13 000
DPB = years
4.NPV & IRRHard
r (%)NPVABr* = ?0

Project A costs €30 000 and returns €16 000, €11 500, €7 000 at the end of years 1 to 3. Project B costs €35 000 and returns €9 500, €13 500, €18 000. Find the crossover rate r* at which both projects have the same NPV, in percent.

A: −30 000, 16 000, 11 500, 7 000B: −35 000, 9 500, 13 500, 18 000
r* = %
5.Bond pricing & yieldMedium
half-years0123…10€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 5 years. The yield to maturity is 7% per year. Find the price P of the bond.

F = €1 000coupon = 8%YTM = 7%n = 5
P = €
6.Bond pricing & yieldHard
years012345€1 115€70€70€70 + €1 070

A bond with 20 years to maturity (face value €1 000, annual coupon 7%) sells for €1 115. The issuer can call it in 5 years at a call price of €1 070. Find the yield to call YTC in percent.

F = €1 000coupon = 7%P = €1 115call price = €1 070call in 5 years
YTC = %
7.CAPM & WACCMedium
WACC = ?equity€12 000 000re​ = 10%debt€6 000 000rd​ = 3%T = 30%

A company has equity worth €12 000 000 and debt worth €6 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 3% and the tax rate is 30%. Find the WACC.

E = €12 000 000D = €6 000 000re = 10%rd = 3%T = 30%
WACC = %
8.CAPM & WACCHard
WACC = ?equity4 000 000 × €56.50β = 0.6rf​ = 2%, rm​ = 8%debt40 000 × €1 020rd​ = 4%T = 20%

A company has 4 000 000 shares at €56.50 each and 40 000 bonds trading at €1 020 each. The pre-tax cost of debt is 4% and the tax rate is 20%. The share’s beta is 0.6, the risk-free rate 2% and the market return 8%. Find the WACC.

β = 0.6rf = 2%rm = 8%rd = 4%T = 20%
WACC = %
9.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 55, 51, 54, 19, 51.

n = 5
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?5?6?7?8P(six) = 1/6 on each roll

A fair die is rolled 8 times. What is the probability P of getting at least one six?

n = 8
P =
12.ProbabilityHard
€p(x)−20.740.25380.05

A game costs €2 to play. You win €6 with probability 0.25 and €40 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€6: p = 0.25€40: p = 0.05
E(G) = €
13.Normal distributionMedium
750 g735.2 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 735.2 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 735.2 g
P =
14.Normal distributionHard
557085100115130145pointsbottom 10%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 10% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points10%
x = points
15.Confidence intervalsMedium
n = 13, s = 9LUx̄ = 115E = ?95%

A sample of 13 items has mean 115 and standard deviation s = 9. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 13x̄ = 115s = 995%t(12) = 2.179
E =
16.Confidence intervalsHard
σ = 27, n = ?x̄ − Ex̄ + Ex̄E = 590%

The population standard deviation is 27. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 5?

σ = 27E = 590%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 30 parts has mean length 82.5 mm, and the line's known standard deviation is 11 mm. Line 2: a sample of 30 parts has mean 92.6 mm, known standard deviation 14 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 30x̄₁ = 82.5 mmσ₁ = 11 mmn₂ = 30x̄₂ = 92.6 mmσ₂ = 14 mm
z =
18.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 390%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 390%
n =
19.Linear regression & correlationMedium
xy184721542467277930903389xy2025306080

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy127222321414512612xy246152025ŷ = 21

The table gives 6 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 = 21?

n = 6ŷ = 21
x =

3Solutions

1.Time value of money

Answer: EAR = 24.6 %

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 = 0.22/365 = 0.0006027

Compound over one year

EAR = (1 + 0.0006027)365 − 1 = 1.246 − 1 = 24.6%

Answer

EAR = 24.6%

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

2.Time value of money

Answer: FV = 39633 €

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 deposits

i = 0.054000000000000006/12 = 0.0045
N = 12·19 = 228

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 100·(2.783 − 1)/0.0045 = €39 632.67

Answer

FV = €39 632.67

You paid in €22 800 yourself; the rest is interest.

3.NPV & IRR

Answer: DPB = 3.51 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: 13 000/1.06 = 12 264.15 → Σ = 12 264.15
year 2: 13 000/1.124 = 11 569.95 → Σ = 23 834.10
year 3: 13 000/1.191 = 10 915.05 → Σ = 34 749.16
year 4: 13 000/1.262 = 10 297.22 → Σ = 45 046.37

Fraction of the last year

DPB = 3 + 5 250.84/10 297.22 = 3.51

Answer

DPB = 3.51 years

The discounted payback is always longer than the simple payback (3.077 years).

4.NPV & IRR

Answer: r* = 5.515 %

Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.

Difference in cash flows

ΔCF0 = −35 000 − (−30 000) = −5 000
ΔCF1 = 9 500 − 16 000 = −6 500
ΔCF2 = 13 500 − 11 500 = 2 000
ΔCF3 = 18 000 − 7 000 = 11 000

IRR of the difference

NPVΔ(5%) = 125.80
NPVΔ(6%) = −116.27
r* = 5.515%

Answer

r* = 5.515%

Check: at 5.515% both NPVs are €1 451.98.

5.Bond pricing & yield

Answer: P = 1042 €

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 = 0.035, N = 10

Present value of the coupons

€40.00·(1 − 0.7089)/0.035 = €332.66

Present value of the face value

€1 000·0.7089 = €708.92

Price

P = €1 041.58

Answer

P = €1 041.58

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

6.Bond pricing & yield

Answer: YTC = 5.556 %

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

1115 = 70·(1 − (1 + y)−5)/y + 1070/(1 + y)5

Solve numerically

y = 5%: €1 141.44
y = 6%: €1 094.43
YTC = 5.556%

Answer

YTC = 5.556%

The yield to maturity is 5.998%. 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 = 7.367 %

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 = €18 000 000, E/V = 0.6667, D/V = 0.3333

WACC

WACC = 0.6667·10% + 0.3333·3%·(1 − 0.3) = 7.367%

Answer

WACC = 7.367%
8.CAPM & WACC

Answer: WACC = 5.233 %

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 = 4 000 000·€56.50 = €226 000 000
D = 40 000·€1 020 = €40 800 000

Cost of equity (CAPM)

re = 2% + 0.6·6% = 5.6%

Weights

V = €266 800 000, E/V = 0.8471, D/V = 0.1529

WACC

WACC = 0.8471·5.6% + 0.1529·4%·(1 − 0.2) = 5.233%

Answer

WACC = 5.233%
9.Descriptive statistics

Answer: s = 15.2

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 = 230/5 = 46

Squared deviations

Σ(x − x̄)² = 9² + 5² + 8² + (−27)² + 5² = 924

Standard deviation

s = √(924/4) = 15.2

Answer

s = 15.2
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 = 0.7674

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)8 = 0.2326
P = 1 − 0.2326 = 0.7674

Answer

P = 0.7674
12.Probability

Answer: E(G) = 1.5 €

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.

Net outcomes

G: 4 (p = 0.25), 38 (p = 0.05), −2 (p = 0.7)

Expected value

E(G) = 4·0.25 + 38·0.05 − 2·0.7 = 1.5 €

Answer

E(G) = 1.5 €

A positive value means you win money on average.

13.Normal distribution

Answer: P = 0.9332

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 = (735.2 − 750)/9.849 = −1.5

Upper tail

P = 1 − Φ(−1.5) = 1 − 0.06681 = 0.9332

Answer

P = 0.9332
14.Normal distribution

Answer: x = 80.8 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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 100 − 1.28·15 = 80.8 points

Answer

x = 80.8 points
15.Confidence intervals

Answer: E = 5.439

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 = 12, t = 2.179

Margin of error

E = 2.179·9/√13 = 5.439

Answer

E = 5.439
16.Confidence intervals

Answer: n = 79

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 = (1.645·27/5)² = 78.91

Round up

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

n = 79

Answer

n = 79
17.Hypothesis tests & confidence intervals

Answer: z = −3.107

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 = √(11²/30 + 14²/30) = √10.57 = 3.251

Test statistic

z = (82.5 − 92.6)/3.251 = −3.107

Answer

z = −3.107

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

18.Hypothesis tests & confidence intervals

Answer: n = 31

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 = (1.645·10/3)² = 30.07

Round up

n = 31

Answer

n = 31
19.Linear regression & correlation

Answer: r² = 0.9544

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 = 153/6 = 25.5
ȳ = 426/6 = 71

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 495
Syy = Σ(y − ȳ)² = 1630

Coefficient of determination

r² = 495²/(157.5·1630) = 0.9544

Answer

r² = 0.9544

About 95.44% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: x = 2.563

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 = 108/6 = 18

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −56

Regression line

b = −56/17.5 = −3.2
a = 18 − (−3.2)·3.5 = 29.2
ŷ = 29.2 − 3.2·x

Solve for x

x = (21 − 29.2)/(−3.2) = 2.563

Answer

x = 2.563

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