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€5 000FV = ?

You invest €5 000 at 8% per year with continuous compounding. Find the value FV after 19 years.

PV = €5 000r = 8%t = 19
FV = €
2.Time value of moneyHard
months0123…72€8 000A?AAA

A loan of €8 000 is repaid with equal monthly payments over 6 years. The interest rate is 5.4% per year, compounded monthly. Find the monthly payment A.

P = €8 0005.4% / 12N = 72
A = €
3.NPV & IRRMedium
years01234567−30 0004 5004 5004 5004 5004 5004 5007 500

A machine costs €30 000 today. It generates €4 500 at the end of each year for 7 years and can then be sold for €3 000. The cost of capital is 12%. Find the net present value NPV.

C0 = €30 000CF = €4 500S = €3 000n = 7r = 12%
NPV = €
4.NPV & IRRHard
years012−20 00013 50012 000

A project costs €20 000 and returns €13 500, €12 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 13 500CF2 = 12 000
IRR = %
5.Bond pricing & yieldMedium

You buy a bond with 11 years to maturity (face value €1 000, annual coupon 2%) when its yield to maturity is 5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 6.5%. Find your holding period return HPR in percent.

F = €1 000coupon = 2%n = 11y0 = 5%y1 = 6.5%
HPR = %
6.Bond pricing & yieldHard
years0123…8€910€60€60 + €1 000

A bond with a face value of €1 000 and an annual coupon of 6% matures in 8 years and sells for €910. Find its yield to maturity YTM in percent.

F = €1 000coupon = 6%n = 8P = €910
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€4 000 000re​ = 13%preferred shares€1 500 000rp​ = 6%debt€5 000 000rd​ = 3%T = 30%

A company is financed with ordinary shares worth €4 000 000, preferred shares worth €1 500 000 and debt worth €5 000 000. The cost of equity is 13%, the cost of preferred stock 6% and the pre-tax cost of debt 3%. The tax rate is 30%. Find the WACC.

E = €4 000 000Ps = €1 500 000D = €5 000 000re = 13%rp = 6%rd = 3%T = 30%
WACC = %
8.CAPM & WACCHard
rf​ = 4%, rm​ − rf​ = 4.5%D/E = 0.4equityβL​ = 1debtD/E = 1equityre​ = ?debtT = 30%

A company's shares have a beta of 1 at its current debt-to-equity ratio of 0.4. It plans to raise its debt-to-equity ratio to 1. The tax rate is 30%, the risk-free rate 4% and the market risk premium 4.5%. Find the cost of equity re after the change.

βL = 1D/E1 = 0.4D/E2 = 1T = 30%rf = 4%rm − rf = 4.5%
re = %
9.Descriptive statisticsMedium
2530354045

Find the sample standard deviation s of the data: 48, 25, 48, 37, 32, 34.

n = 6
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is above 68?

μ = 50σ = 12x = 68
P =
11.ProbabilityMedium
xp(x)00.2510.0520.330.140.3μ = 2.15

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2, 3, 4p(x): 0.25, 0.05, 0.3, 0.1, 0.3
σ =
12.ProbabilityHard
n = 9, p = 0.7: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?

n = 9p = 0.7
P =
13.Normal distributionMedium
490582P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 490 and 582 h? Give P as a decimal.

μ = 500σ = 40x1 = 490x2 = 582
P =
14.Normal distributionHard
35455565758595pointsbottom 30%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 30% is to the left of it. Find the cut-off x.

μ = 65 pointsσ = 10 points30%
x = points
15.Confidence intervalsMedium
n = 9, s = 12.5LUx̄ = 58E = ?99%

A sample of 9 items has mean 58 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 = 9x̄ = 58s = 12.599%t(8) = 3.355
E =
16.Confidence intervalsHard
σ = 25, n = ?x̄ − Ex̄ + Ex̄E = 2.599%

The population standard deviation is 25. 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 2.5?

σ = 25E = 2.599%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.2012.201t = ?df = 11

12 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 −1.82 and the standard deviation of the differences is 4.5. Find the paired test statistic t for H₀: μd = 0.

n = 12d̄ = −1.82sd = 4.5
t =
18.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 49 measurements has mean 103.8. The population standard deviation is 12. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 103.8σ = 12n = 4990%
L =
19.Linear regression & correlationMedium
xy325427537636737xy46253035

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
xy2841662182610311237xy51010203040x = 5

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. A new observation (5, 13) is made. Find its residual e = y − ŷ.

n = 6(5, 13)
e =

3Solutions

1.Time value of money

Answer: FV = 22861 €

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 = 0.08·19 = 1.52
e1.52 = 4.572

Future value

FV = 5 000·4.572 = €22 861.13

Answer

FV = €22 861.13
2.Time value of money

Answer: A = 130.3 €

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 = 0.054000000000000006/12 = 0.0045
N = 12·6 = 72

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 = 0.7238
A = 8 000·0.0045/(1 − 0.7238) = €130.33

Answer

A = €130.33
3.NPV & IRR

Answer: NPV = −8106 €

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

4 500·4.564 = 20 536.90

Present value of the salvage value

3 000/2.211 = 1 357.05

NPV

NPV = −30 000 + 20 536.90 + 1 357.05 = −€8 106.05

Answer

NPV = −€8 106.05
4.NPV & IRR

Answer: IRR = 18.24 %

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%) = 812.85
NPV(20%) = −416.67

Narrow down

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

IRR ≈ 15 + 5·812.85/(812.85 − (−416.67)) = 18.31%
exact: NPV(18.24%) = 0 → IRR = 18.24%

Answer

IRR = 18.24%
5.Bond pricing & yield

Answer: HPR = −7.233 %

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 = €20.00, i = 0.05, N = 11
€20.00·(1 − 0.5847)/0.05 + €1 000·0.5847 = €166.13 + €584.68 = €750.81

Selling price

C = €20.00, i = 0.065, N = 10
€20.00·(1 − 0.5327)/0.065 + €1 000·0.5327 = €143.78 + €532.73 = €676.50

Holding period return

HPR = (20 + 676.50 − 750.81)/750.81 = −7.233%

Answer

HPR = −7.233%
6.Bond pricing & yield

Answer: YTM = 7.539 %

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

€910 = €60·(1 − (1 + y)⁻⁸)/y + €1 000/(1 + y)⁸

Solve numerically

P(7%) = €940.29, P(8%) = €885.07
YTM = 7.539%

Answer

YTM = 7.539%
7.CAPM & WACC

Answer: WACC = 6.81 %

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 = 4 000 000 + 1 500 000 + 5 000 000 = €10 500 000
E/V = 0.381, Ps/V = 0.1429, D/V = 0.4762

WACC

WACC = 0.381·13% + 0.1429·6% + 0.4762·3%·(1 − 0.3) = 6.81%

Answer

WACC = 6.81%
8.CAPM & WACC

Answer: re = 9.977 %

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.

Unlever the beta

βU = 1/(1 + 0.7·0.4) = 0.7813

Relever at the new D/E

βL = 0.7813·(1 + 0.7·1) = 1.328

Cost of equity (CAPM)

re = 4% + 1.328·4.5% = 9.977%

Answer

re = 9.977%
9.Descriptive statistics

Answer: s = 9.158

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 = 224/6 = 37.33

Squared deviations

Σ(x − x̄)² = 10.67² + (−12.33)² + 10.67² + (−0.3333)² + (−5.333)² + (−3.333)² = 419.3

Standard deviation

s = √(419.3/5) = 9.158

Answer

s = 9.158
10.Descriptive statistics

Answer: P = 0.06681

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 = 1.5

Standard normal table

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

Answer

P = 0.06681
11.Probability

Answer: σ = 1.526

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 + 1·0.05 + 2·0.3 + 3·0.1 + 4·0.3 = 2.15

Mean of the squares

E(X²) = 0·0.25 + 1·0.05 + 4·0.3 + 9·0.1 + 16·0.3 = 6.95

Variance

Var(X) = 6.95 − 2.15² = 2.327
σ = √2.327 = 1.526

Answer

σ = 1.526
12.Probability

Answer: P = 0.000433

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) = 0.39 = 1.968 × 10⁻⁵
P(X = 1) = 9·0.7·0.38 = 0.0004133

Add them

P = 1.968 × 10⁻⁵ + 0.0004133 = 0.000433

Answer

P = 0.000433
13.Normal distribution

Answer: P = 0.5785

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 = (490 − 500)/40 = −0.25
z₂ = (582 − 500)/40 = 2.05

Area between

P = Φ(2.05) − Φ(−0.25) = 0.9798 − 0.4013 = 0.5785

Answer

P = 0.5785
14.Normal distribution

Answer: x = 59.8 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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 65 − 0.52·10 = 59.8 points

Answer

x = 59.8 points
15.Confidence intervals

Answer: E = 13.98

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 = 8, t = 3.355

Margin of error

E = 3.355·12.5/√9 = 13.98

Answer

E = 13.98
16.Confidence intervals

Answer: n = 664

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 = (2.576·25/2.5)² = 663.6

Round up

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

n = 664

Answer

n = 664
17.Hypothesis tests & confidence intervals

Answer: t = −1.401

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 = 4.5/√12 = 1.299

Test statistic

t = −1.82/1.299 = −1.401

Answer

t = −1.401

With df = 11, the two-sided 5% critical value is 2.201; |t| is smaller, so H₀ is not rejected.

18.Hypothesis tests & confidence intervals

Answer: L = 101

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).

Margin of error

E = 1.645·12/√49 = 2.82

Interval limit

L = 103.8 − 2.82 = 101

Answer

L = 101
19.Linear regression & correlation

Answer: r = 0.8845

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 = 25/5 = 5
ȳ = 162/5 = 32.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 33
Syy = Σ(y − ȳ)² = 139.2

Correlation coefficient

r = 33/√(10·139.2) = 0.8845

Answer

r = 0.8845
20.Linear regression & correlation

Answer: e = −4.595

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 = 139/6 = 23.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 195

Regression line

b = 195/70 = 2.786
a = 23.17 − 2.786·7 = 3.667
ŷ = 3.667 + 2.786·x

Predicted value

ŷ = 3.667 + 2.786·5 = 17.6

Residual

e = 13 − 17.6 = −4.595

Answer

e = −4.595

The point lies below the line.

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