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
years0123PV = ?€40 000

You need €40 000 in 3 years. An account pays 4% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €40 0004% / 4n = 3
PV = €
2.Time value of moneyHard
months0123…96A?AA€75 000

You want €75 000 in 8 years. You save the same amount at the end of every month in an account paying 2.4% per year, compounded monthly. Find the monthly deposit A.

FV = €75 0002.4% / 12N = 96
A = €
3.NPV & IRRMedium
years01234−30 0008 5008 5008 50013 000

A machine costs €30 000 today. It generates €8 500 at the end of each year for 4 years and can then be sold for €4 500. The cost of capital is 10%. Find the net present value NPV.

C0 = €30 000CF = €8 500S = €4 500n = 4r = 10%
NPV = €
4.NPV & IRRHard
years0123−20 00010 00012 0008 500

A project costs €20 000 and returns €10 000, €12 000, €8 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 10 000CF2 = 12 000CF3 = 8 500
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 7.5%. Find your holding period return HPR in percent.

F = €1 000coupon = 3%n = 5y0 = 7%y1 = 7.5%
HPR = %
6.Bond pricing & yieldHard
years01234567€1 060€60€60 + €1 000

A bond with a face value of €1 000 and an annual coupon of 6% matures in 7 years and sells for €1 060. Find its yield to maturity YTM in percent.

F = €1 000coupon = 6%n = 7P = €1 060
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€3 000 000re​ = 11%debt€3 000 000rd​ = 4.5%T = 20%

A company has equity worth €3 000 000 and debt worth €3 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 4.5% and the tax rate is 20%. Find the WACC.

E = €3 000 000D = €3 000 000re = 11%rd = 4.5%T = 20%
WACC = %
8.CAPM & WACCHard
WACC = ?equity€6 000 000re​ = 11%bonds8 000 × €1 035coupon 6%, n = 4T = 25.8%

A company has equity with a market value of €6 000 000 and a cost of equity of 11%. Its debt consists of 8 000 bonds (face value €1 000, annual coupon 6%, 4 years to maturity) that trade at €1 035 each. The tax rate is 25.8%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.

E = €6 000 000re = 11%bonds: 8 000 × €1 035coupon = 6%n = 4T = 25.8%
WACC = %
9.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 47, 18, 54, 38, 49.

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 150.

μ = 170σ = 10x = 150
z =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6P(six) = 1/6 on each roll

A fair die is rolled 6 times. What is the probability P of getting at least one six?

n = 6
P =
12.ProbabilityHard
n = 5, p = 0.3: P(X = 4) = ?k012345

A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 4 correct answers?

n = 5p = 0.3k = 4
P =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.7357

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7357. Find z to two decimals.

A = 0.7357
z =
14.Normal distributionHard
557085100115130145pointstop 40%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points40%
x = points
15.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?99%

The battery life of two phone models is compared. Model 1: 50 phones, mean 119 h, known standard deviation 12 h. Model 2: 60 phones, mean 119.5 h, known standard deviation 17 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 119 hσ₁ = 12 hn₂ = 60x̄₂ = 119.5 hσ₂ = 17 h99%
E = h
16.Confidence intervalsHard
σ = 31, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

The population standard deviation is 31. 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 1.5?

σ = 31E = 1.599%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.52p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 4. A random sample of 64 observations has a mean of 49.24. Test H₀: μ = 50 against H₁: μ < 50. Find the p-value p.

μ₀ = 50σ = 4n = 64x̄ = 49.24H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 113 of 250 patients on treatment A recover, and 42 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 113n₁ = 250x₂ = 42n₂ = 120
z =
19.Linear regression & correlationMedium
xy149255353464558xy2450556065

The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy2641161582010271233xy510102030x = 3

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 = 3.

n = 6x = 3
ŷ =

3Solutions

1.Time value of money

Answer: PV = 35498 €

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 = 0.04/4 = 0.01
N = 4·3 = 12

Present value

PV = FV/(1 + i)N = 40 000/1.127 = €35 497.97

Answer

PV = €35 497.97
2.Time value of money

Answer: A = 709.4 €

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 deposits

i = 0.024/12 = 0.002
N = 12·8 = 96

Deposit

A = FV·i/((1 + i)N − 1) = 75 000·0.002/(1.211 − 1) = €709.43

Answer

A = €709.43
3.NPV & IRR

Answer: NPV = 17.42 €

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.

Present value of the yearly cash flows

8 500·3.17 = 26 943.86

Present value of the salvage value

4 500/1.464 = 3 073.56

NPV

NPV = −30 000 + 26 943.86 + 3 073.56 = €17.42

Answer

NPV = €17.42
4.NPV & IRR

Answer: IRR = 25.11 %

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(25%) = 32.00
NPV(30%) = −1 338.19

Narrow down

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

IRR ≈ 25 + 5·32.00/(32.00 − (−1 338.19)) = 25.12%
exact: NPV(25.11%) = 0 → IRR = 25.11%

Answer

IRR = 25.11%
5.Bond pricing & yield

Answer: HPR = 5.178 %

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 = 0.075, N = 4
€30.00·(1 − 0.7488)/0.075 + €1 000·0.7488 = €100.48 + €748.80 = €849.28

Holding period return

HPR = (30 + 849.28 − 835.99)/835.99 = 5.178%

Answer

HPR = 5.178%
6.Bond pricing & yield

Answer: YTM = 4.964 %

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

€1 060 = €60·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷

Solve numerically

P(4%) = €1 120.04, P(5%) = €1 057.86
YTM = 4.964%

Answer

YTM = 4.964%
7.CAPM & WACC

Answer: WACC = 7.3 %

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 = €6 000 000, E/V = 0.5, D/V = 0.5

WACC

WACC = 0.5·11% + 0.5·4.5%·(1 − 0.2) = 7.3%

Answer

WACC = 7.3%
8.CAPM & WACC

Answer: WACC = 6.778 %

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)

1035 = 60·(1 − (1 + y)−4)/y + 1000/(1 + y)4
y = 5.013%

Market values

D = 8 000·€1 035 = €8 280 000
V = €14 280 000, E/V = 0.4202, D/V = 0.5798

WACC

WACC = 0.4202·11% + 0.5798·5.013%·(1 − 0.258) = 6.778%

Answer

WACC = 6.778%
9.Descriptive statistics

Answer: s = 14.2

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 = 206/5 = 41.2

Squared deviations

Σ(x − x̄)² = 5.8² + (−23.2)² + 12.8² + (−3.2)² + 7.8² = 806.8

Standard deviation

s = √(806.8/4) = 14.2

Answer

s = 14.2
10.Descriptive statistics

Answer: z = −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.

Standardise

z = (150 − 170)/10 = −2

Answer

z = −2
11.Probability

Answer: P = 0.6651

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)6 = 0.3349
P = 1 − 0.3349 = 0.6651

Answer

P = 0.6651
12.Probability

Answer: P = 0.02835

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 = 4) = 5·0.34·0.71 = 5·0.0081·0.7 = 0.02835

Answer

P = 0.02835
13.Normal distribution

Answer: z = −0.63

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 of z

Φ(z) = 1 − 0.7357 = 0.2643

Read the z-table backwards

Find 0.2643 inside the table and read off z.

z = −0.63

Answer

z = −0.63
14.Normal distribution

Answer: x = 103.8 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) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 100 + 0.25·15 = 103.8 points

Answer

x = 103.8 points
15.Confidence intervals

Answer: E = 7.147 h

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.

Point estimate

x̄₁ − x̄₂ = 119 − 119.5 = −0.5

Standard error

SE = √(12²/50 + 17²/60) = 2.774

Margin of error

E = 2.576·2.774 = 7.147

Answer

E = 7.147 h

The interval contains 0, so the data do not show a significant difference.

16.Confidence intervals

Answer: n = 2835

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 = (2.576·31/1.5)² = 2834

Round up

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

n = 2835

Answer

n = 2835
17.Hypothesis tests & confidence intervals

Answer: p = 0.06426

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.

Test statistic

SE = σ/√n = 4/√64 = 0.5
z = (49.24 − 50)/0.5 = −1.52

Tail area

Φ(1.52) = 0.9357
1 − Φ(1.52) = 0.0643

p-value

p = 0.06426

Answer

p = 0.06426

p > 0.05, so H₀ is not rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: z = 1.862

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 = 113/250 = 0.452
p̂₂ = 42/120 = 0.35
p̂ = 155/370 = 0.4189

Standard error under H₀

SE = √(0.4189·0.5811·(1/250 + 1/120)) = 0.05479

Test statistic

z = (0.452 − 0.35)/0.05479 = 1.862

Answer

z = 1.862

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

19.Linear regression & correlation

Answer: r² = 0.5749

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 = 279/5 = 55.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 27
Syy = Σ(y − ȳ)² = 126.8

Coefficient of determination

r² = 27²/(10·126.8) = 0.5749

Answer

r² = 0.5749

About 57.49% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 7.924

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 = 112/6 = 18.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 188

Regression line

b = 188/70 = 2.686
a = 18.67 − 2.686·7 = −0.1333
ŷ = −0.1333 + 2.686·x

Prediction

ŷ = −0.1333 + 2.686·3 = 7.924

Answer

ŷ = 7.924

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