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…19€1 000FV = ?

You invest €1 000 at 6% per year with continuous compounding. Find the value FV after 19 years.

PV = €1 000r = 6%t = 19
FV = €
2.Time value of moneyHard
years012345€2 500€2 500€2 500FV = ?

You deposit €2 500 at the start of every year for 5 years, the first deposit today. The account pays 7% per year. How much is in the account at the end of year 5? Find FV.

A = €2 500i = 7%n = 5
FV = €
3.NPV & IRRMedium
years012345−50 00015 00015 00015 00015 00015 000

A project costs €50 000 today and returns €15 000 at the end of each year for 5 years. The cost of capital is 10%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 10%CF0 = −50 000CF1 = 15 000CF2 = 15 000CF3 = 15 000CF4 = 15 000CF5 = 15 000
DPB = years
4.NPV & IRRHard
years012−20 0009 00013 500

A project costs €20 000 and returns €9 000, €13 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 9 000CF2 = 13 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…16€22.50€22.50€22.50 + €1 000P = ?

A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 8 years. The yield to maturity is 3% per year. Find the price P of the bond.

F = €1 000coupon = 4.5%YTM = 3%n = 8
P = €
6.Bond pricing & yieldHard
half-years0123…16€1 173€30€30€30 + €1 000

A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 8 years. It sells for €1 173. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 6% (2×)n = 8P = €1 173
YTM = %
7.CAPM & WACCMedium
WACC = ?equityEre​ = 14%debtD = 0.4·Erd​ = 4%T = 20%

A company has a debt-to-equity ratio (at market values) of 0.4. Its cost of equity is 14%, its pre-tax cost of debt is 4% and the tax rate is 20%. Find the WACC.

D/E = 0.4re = 14%rd = 4%T = 20%
WACC = %
8.CAPM & WACCHard
WACC = ?equity€6 000 000re​ = 10%bonds5 000 × €1 054coupon 4%, n = 6T = 21%

A company has equity with a market value of €6 000 000 and a cost of equity of 10%. Its debt consists of 5 000 bonds (face value €1 000, annual coupon 4%, 6 years to maturity) that trade at €1 054 each. The tax rate is 21%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.

E = €6 000 000re = 10%bonds: 5 000 × €1 054coupon = 4%n = 6T = 21%
WACC = %
9.Descriptive statisticsMedium
202530354045

Find the sample standard deviation s of the data: 19, 31, 18, 18, 20, 47.

n = 6
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 44.

μ = 50σ = 12x = 44
z =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7P(six) = 1/6 on each roll

A fair die is rolled 7 times. What is the probability P of getting at least one six?

n = 7
P =
12.ProbabilityHard
n = 6, p = 0.6: P(X = 3) = ?k0123456

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 3 correct answers?

n = 6p = 0.6k = 3
P =
13.Normal distributionMedium
200 mm199.8 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.8 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.8 mm
P =
14.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
n = 6, s = 5LUx̄ = 88.5E = ?90%

A sample of 6 items has mean 88.5 and standard deviation s = 5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 6x̄ = 88.5s = 590%t(5) = 2.015
E =
16.Confidence intervalsHard
μ₁ − μ₂, t(10) = 2.2280L = ?Ux̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 6 plots, mean yield 14.5 kg, standard deviation 4 kg. Plot group 2: 6 plots, mean 4 kg, standard deviation 10 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 6x̄₁ = 14.5s₁ = 4n₂ = 6x̄₂ = 4s₂ = 1095%t(10) = 2.228
lower limit = kg
17.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 501.5 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 501.5s = 15n = 49
t =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 80 against H₁: μ < 80 is done at α = 0.05 with a sample of 25; the standard deviation is known to be 2.5. In reality the mean is 78.9. Find the power 1 − β of the test in percent.

μ₀ = 80μ = 78.9σ = 2.5n = 25α = 0.05
1 − β = %
19.Linear regression & correlationMedium
xy125221320421519xy24202224

The table gives 5 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 = 5
r² =
20.Linear regression & correlationHard
xy14481652186920692279xy152050607080x = 17

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. Predict ŷ for x = 17.

n = 5x = 17
ŷ =

3Solutions

1.Time value of money

Answer: FV = 3127 €

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.

Exponent

r·t = 0.06·19 = 1.14
e1.14 = 3.127

Future value

FV = 1 000·3.127 = €3 126.77

Answer

FV = €3 126.77
2.Time value of money

Answer: FV = 15383 €

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.

Ordinary annuity factor

((1 + 0.07)5 − 1)/0.07 = 5.751

Annuity due

Each deposit is invested one year longer.

FV = 2 500·5.751·1.07 = €15 383.23

Answer

FV = €15 383.23
3.NPV & IRR

Answer: DPB = 4.263 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: 15 000/1.1 = 13 636.36 → Σ = 13 636.36
year 2: 15 000/1.21 = 12 396.69 → Σ = 26 033.06
year 3: 15 000/1.331 = 11 269.72 → Σ = 37 302.78
year 4: 15 000/1.464 = 10 245.20 → Σ = 47 547.98
year 5: 15 000/1.611 = 9 313.82 → Σ = 56 861.80

Fraction of the last year

DPB = 4 + 2 452.02/9 313.82 = 4.263

Answer

DPB = 4.263 years

The discounted payback is always longer than the simple payback (3.333 years).

4.NPV & IRR

Answer: IRR = 7.684 %

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%) = 816.33
NPV(10%) = −661.16

Narrow down

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

IRR ≈ 5 + 5·816.33/(816.33 − (−661.16)) = 7.763%
exact: NPV(7.684%) = 0 → IRR = 7.684%

Answer

IRR = 7.684%
5.Bond pricing & yield

Answer: P = 1106 €

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 = €22.50, i = 0.015, N = 16

Present value of the coupons

€22.50·(1 − 0.788)/0.015 = €317.95

Present value of the face value

€1 000·0.788 = €788.03

Price

P = €1 105.98

Answer

P = €1 105.98

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

6.Bond pricing & yield

Answer: YTM = 3.502 %

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 = €30.00, N = 16

Solve numerically

i = 1.751% per half-year

YTM

YTM = 2·1.751% = 3.502%

Answer

YTM = 3.502%
7.CAPM & WACC

Answer: WACC = 10.91 %

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 + 0.4) = 0.7143
D/V = 0.4/(1 + 0.4) = 0.2857

WACC

WACC = 0.7143·14% + 0.2857·4%·(1 − 0.2) = 10.91%

Answer

WACC = 10.91%
8.CAPM & WACC

Answer: WACC = 6.433 %

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).

Cost of debt (YTM)

1054 = 40·(1 − (1 + y)−6)/y + 1000/(1 + y)6
y = 3.003%

Market values

D = 5 000·€1 054 = €5 270 000
V = €11 270 000, E/V = 0.5324, D/V = 0.4676

WACC

WACC = 0.5324·10% + 0.4676·3.003%·(1 − 0.21) = 6.433%

Answer

WACC = 6.433%
9.Descriptive statistics

Answer: s = 11.64

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 = 153/6 = 25.5

Squared deviations

Σ(x − x̄)² = (−6.5)² + 5.5² + (−7.5)² + (−7.5)² + (−5.5)² + 21.5² = 677.5

Standard deviation

s = √(677.5/5) = 11.64

Answer

s = 11.64
10.Descriptive statistics

Answer: z = −0.5

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 = (44 − 50)/12 = −0.5

Answer

z = −0.5
11.Probability

Answer: P = 0.7209

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)7 = 0.2791
P = 1 − 0.2791 = 0.7209

Answer

P = 0.7209
12.Probability

Answer: P = 0.2765

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.

Binomial formula

P(X = 3) = 20·0.63·0.43 = 20·0.216·0.064 = 0.2765

Answer

P = 0.2765
13.Normal distribution

Answer: P = 0.6554

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 = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (199.8 − 200)/0.5 = −0.4

Upper tail

P = 1 − Φ(−0.4) = 1 − 0.3446 = 0.6554

Answer

P = 0.6554
14.Normal distribution

Answer: P = 0.9207

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 = 25.03 − 25 = 0.03 mm, σ = 0.02121 mm

Standardise

z = (0 − 0.03)/0.02121 = −1.41

Probability

P = 1 − Φ(−1.41) = 0.9207

Answer

P = 0.9207
15.Confidence intervals

Answer: E = 4.113

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 = 5, t = 2.015

Margin of error

E = 2.015·5/√6 = 4.113

Answer

E = 4.113
16.Confidence intervals

Answer: lower limit = 0.7036 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 = (5·4² + 5·10²)/10 = 58
sp = 7.616

Margin of error

E = 2.228·7.616·√(1/6 + 1/6) = 9.796

Interval

10.5 ± 9.796 → [0.7036, 20.3]

Answer

lower limit = 0.7036 kg
17.Hypothesis tests & confidence intervals

Answer: t = 0.7

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 = 15/√49 = 15/7 = 2.143

Test statistic

t = (501.5 − 500)/2.143 = 0.7

Answer

t = 0.7
t0.025; df = 48 = 2.011

|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: 1 − β = 71.06 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 2.5/√25 = 0.5
x̄c = 80 − 1.645·0.5 = 79.18

z under the true mean

z = (79.18 − 78.9)/0.5 = 0.555

Power

power = P(Z < 0.555) = 71.06%
1 − β = 71.06%

Answer

1 − β = 71.06%
19.Linear regression & correlation

Answer: r² = 0.6923

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 = 15/5 = 3
ȳ = 106/5 = 21.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −12
Syy = Σ(y − ȳ)² = 20.8

Coefficient of determination

r² = −12²/(10·20.8) = 0.6923

Answer

r² = 0.6923

About 69.23% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 59.45

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 = 90/5 = 18
ȳ = 317/5 = 63.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 158

Regression line

b = 158/40 = 3.95
a = 63.4 − 3.95·18 = −7.7
ŷ = −7.7 + 3.95·x

Prediction

ŷ = −7.7 + 3.95·17 = 59.45

Answer

ŷ = 59.45

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