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
years01234567PV = ?€10 000

You need €10 000 in 7 years. An account pays 3% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €10 0003% / 12n = 7
PV = €
2.Time value of moneyHard
months01…144…240€200 000AB = ?A7.2% / 12

A mortgage of €200 000 is repaid with equal monthly payments over 20 years at 7.2% per year, compounded monthly. How much is still owed just after the payment at the end of year 12? Find the outstanding balance B.

P = €200 0007.2% / 12N = 240k = 144
B = €
3.NPV & IRRMedium
years0123456−50 00021 50015 50020 50020 00013 00011 500

A project costs €50 000 today and returns €21 500, €15 500, €20 500, €20 000, €13 000, €11 500 at the end of years 1 to 6. 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 = −50 000CF1 = 21 500CF2 = 15 500CF3 = 20 500CF4 = 20 000CF5 = 13 000CF6 = 11 500
DPB = years
4.NPV & IRRHard
years0123−50 00019 50027 00021 500

A project costs €50 000 and returns €19 500, €27 000, €21 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 19 500CF2 = 27 000CF3 = 21 500
IRR = %
5.Bond pricing & yieldMedium

You buy a bond with 5 years to maturity (face value €1 000, annual coupon 3%) when its yield to maturity is 7%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 8%. Find your holding period return HPR in percent.

F = €1 000coupon = 3%n = 5y0 = 7%y1 = 8%
HPR = %
6.Bond pricing & yieldHard
years01234€1 129€70€70€70 + €1 070

A bond with 12 years to maturity (face value €1 000, annual coupon 7%) sells for €1 129. The issuer can call it in 4 years at a call price of €1 070. Find the yield to call YTC in percent.

F = €1 000coupon = 7%P = €1 129call price = €1 070call in 4 years
YTC = %
7.CAPM & WACCMedium
WACC = ?equity€5 000 000re​ = 9%debt€4 000 000rd​ = 4%T = 25%

A company has equity worth €5 000 000 and debt worth €4 000 000. Its cost of equity is 9%, its pre-tax cost of debt is 4% and the tax rate is 25%. Find the WACC.

E = €5 000 000D = €4 000 000re = 9%rd = 4%T = 25%
WACC = %
8.CAPM & WACCHard
rf​ = 2%, rm​ − rf​ = 6.5%no debtequityβU​ = 0.6D/E = 0.25equityre​ = ?debtT = 20%

A company's business has an unlevered (asset) beta of 0.6. It plans a debt-to-equity ratio of 0.25. The tax rate is 20%, the risk-free rate is 2% and the market risk premium is 6.5%. Find the cost of equity re at the new capital structure.

βU = 0.6D/E = 0.25T = 20%rf = 2%rm − rf = 6.5%
re = %
9.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 41, 17, 18, 49, 59.

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

Exam scores are normally distributed with mean 170 and standard deviation 10. Find the z-score z of a score of 180.

μ = 170σ = 10x = 180
z =
11.ProbabilityMedium
red: 7blue: 6

A bag holds 7 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 6
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
174.5192.35P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 174.5 and 192.35 cm? Give P as a decimal.

μ = 178σ = 7x1 = 174.5x2 = 192.35
P =
14.Normal distributionHard
μ = 32x = ?99%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 99% of them lie (the 99th percentile).

μ = 32σ = 6P = 99%
x = min
15.Confidence intervalsMedium
n = 11, s = 18L = ?Ux̄ = 40E99%

A sample of 11 items has mean 40 and standard deviation s = 18. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 11x̄ = 40s = 1899%t(10) = 3.169
lower limit =
16.Confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 20. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 4?

σ = 20E = 499%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.2282.228t = ?df = 10

11 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 8.23 and the standard deviation of the differences is 7. Find the paired test statistic t for H₀: μd = 0.

n = 11d̄ = 8.23sd = 7
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 40 of 100 patients on treatment A recover, and 58 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 40n₁ = 100x₂ = 58n₂ = 120
z =
19.Linear regression & correlationMedium
xy232342450553658xy24630405060

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy2841962383310401246xy5102040x = 9

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 9.

n = 6x = 9
ŷ =

3Solutions

1.Time value of money

Answer: PV = 8108 €

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 = 0.03/12 = 0.0025
N = 12·7 = 84

Present value

PV = FV/(1 + i)N = 10 000/1.233 = €8 107.97

Answer

PV = €8 107.97
2.Time value of money

Answer: B = 114661 €

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 = 0.006, N = 240
A = 200 000·0.006/(1 − (1 + 0.006)−240) = €1 574.70

Present value of the remaining payments

Whatever is still owed must be exactly paid off by the remaining payments.

N − k = 240 − 144 = 96
B = 1575·(1 − (1 + 0.006)−96)/0.006 = €114 661.47

Answer

B = €114 661.47
3.NPV & IRR

Answer: DPB = 2.925 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: 21 500/1.06 = 20 283.02 → Σ = 20 283.02
year 2: 15 500/1.124 = 13 794.94 → Σ = 34 077.96
year 3: 20 500/1.191 = 17 212.20 → Σ = 51 290.16

Fraction of the last year

DPB = 2 + 15 922.04/17 212.20 = 2.925

Answer

DPB = 2.925 years

The discounted payback is always longer than the simple payback (2.634 years).

4.NPV & IRR

Answer: IRR = 16.78 %

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(15%) = 1 509.00
NPV(20%) = −2 557.87

Narrow down

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

IRR ≈ 15 + 5·1 509.00/(1 509.00 − (−2 557.87)) = 16.86%
exact: NPV(16.78%) = 0 → IRR = 16.78%

Answer

IRR = 16.78%
5.Bond pricing & yield

Answer: HPR = 3.397 %

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.

Purchase price

C = €30.00, i = 0.07, N = 5
€30.00·(1 − 0.713)/0.07 + €1 000·0.713 = €123.01 + €712.99 = €835.99

Selling price

C = €30.00, i = 0.08, N = 4
€30.00·(1 − 0.735)/0.08 + €1 000·0.735 = €99.36 + €735.03 = €834.39

Holding period return

HPR = (30 + 834.39 − 835.99)/835.99 = 3.397%

Answer

HPR = 3.397%
6.Bond pricing & yield

Answer: YTC = 4.987 %

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

1129 = 70·(1 − (1 + y)−4)/y + 1070/(1 + y)4

Solve numerically

y = 4%: €1 168.73
y = 5%: €1 128.51
YTC = 4.987%

Answer

YTC = 4.987%

The yield to maturity is 5.503%. 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 = 6.333 %

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 = €9 000 000, E/V = 0.5556, D/V = 0.4444

WACC

WACC = 0.5556·9% + 0.4444·4%·(1 − 0.25) = 6.333%

Answer

WACC = 6.333%
8.CAPM & WACC

Answer: re = 6.68 %

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 = 0.6·(1 + 0.8·0.25) = 0.72

Cost of equity (CAPM)

re = 2% + 0.72·6.5% = 6.68%

Answer

re = 6.68%
9.Descriptive statistics

Answer: s = 18.74

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 = 184/5 = 36.8

Squared deviations

Σ(x − x̄)² = 4.2² + (−19.8)² + (−18.8)² + 12.2² + 22.2² = 1405

Standard deviation

s = √(1405/4) = 18.74

Answer

s = 18.74
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 = (180 − 170)/10 = 1

Answer

z = 1
11.Probability

Answer: P = 0.2692

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.

First and second draw

P = 7/13 · 6/12 = 42/156 = 0.2692

Answer

P = 0.2692
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 = 0.6713

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-scores

z₁ = (174.5 − 178)/7 = −0.5
z₂ = (192.35 − 178)/7 = 2.05

Area between

P = Φ(2.05) − Φ(−0.5) = 0.9798 − 0.3085 = 0.6713

Answer

P = 0.6713
14.Normal distribution

Answer: x = 45.96 min

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) = 0.99 → z = 2.326

Back to x

x = 32 + 2.326·6 = 45.96 min

Answer

x = 45.96 min
15.Confidence intervals

Answer: lower limit = 22.8

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 = 10, t = 3.169

Margin of error

E = 3.169·18/√11 = 17.2

Interval

40 ± 17.2 → [22.8, 57.2]

Answer

lower limit = 22.8
16.Confidence intervals

Answer: n = 166

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 = (2.576·20/4)² = 165.9

Round up

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

n = 166

Answer

n = 166
17.Hypothesis tests & confidence intervals

Answer: t = 3.899

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 = 7/√11 = 2.111

Test statistic

t = 8.23/2.111 = 3.899

Answer

t = 3.899

With df = 10, the two-sided 5% critical value is 2.228; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: z = −1.238

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 = 40/100 = 0.4
p̂₂ = 58/120 = 0.4833
p̂ = 98/220 = 0.4455

Standard error under H₀

SE = √(0.4455·0.5545·(1/100 + 1/120)) = 0.0673

Test statistic

z = (0.4 − 0.4833)/0.0673 = −1.238

Answer

z = −1.238

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

19.Linear regression & correlation

Answer: r = 0.9768

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 = 20/5 = 4
ȳ = 235/5 = 47

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 63
Syy = Σ(y − ȳ)² = 416

Correlation coefficient

r = 63/√(10·416) = 0.9768

Answer

r = 0.9768
20.Linear regression & correlation

Answer: ŷ = 35.68

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 = 42/6 = 7
ȳ = 169/6 = 28.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 263

Regression line

b = 263/70 = 3.757
a = 28.17 − 3.757·7 = 1.867
ŷ = 1.867 + 3.757·x

Prediction

ŷ = 1.867 + 3.757·9 = 35.68

Answer

ŷ = 35.68

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