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 5% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €40 0005% / 4n = 3
PV = €
2.Time value of moneyHard
months0123…N = ?€10 000€175€175€175€1757.2% / 12

You borrow €10 000 at 7.2% per year, compounded monthly, and repay €175 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.

P = €10 000A = €1757.2% / 12
N = payments
3.NPV & IRRMedium
years012345€80 000€1 000€1 000€1 000€1 000€1 000r = 8%, EAC = ?

A machine costs €80 000 today, lasts 5 years and costs €1 000 per year to run (paid at the end of each year). The cost of capital is 8%. Find its equivalent annual cost EAC.

C0 = €80 000M = €1 000n = 5r = 8%
EAC = €
4.NPV & IRRHard
years0123−25 0009 50012 5007 500

A project costs €25 000 and returns €9 500, €12 500, €7 500 at the end of years 1 to 3. Cash flows can be reinvested at the cost of capital of 6%. Find the modified internal rate of return MIRR in percent.

r = 6%CF0 = −25 000CF1 = 9 500CF2 = 12 500CF3 = 7 500
MIRR = %
5.Bond pricing & yieldMedium
half-years0123…24€30€30€30 + €1 000P = ?

A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 5% per year. Find the price P of the bond.

F = €1 000coupon = 6%YTM = 5%n = 12
P = €
6.Bond pricing & yieldHard
years0123…10€900€70€70 + €1 000

A bond with a face value of €1 000 and an annual coupon of 7% matures in 10 years and sells for €900. Find its yield to maturity YTM in percent.

F = €1 000coupon = 7%n = 10P = €900
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€10 000 000re​ = 12%debt€6 000 000rd​ = 3%T = 25.8%

A company has equity worth €10 000 000 and debt worth €6 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 3% and the tax rate is 25.8%. Find the WACC.

E = €10 000 000D = €6 000 000re = 12%rd = 3%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €31.00β = 1.3rf​ = 3.5%, rm​ = 10%debt20 000 × €1 000rd​ = 6%T = 25%

A company has 2 000 000 shares at €31.00 each and 20 000 bonds trading at €1 000 each. The pre-tax cost of debt is 6% and the tax rate is 25%. The share’s beta is 1.3, the risk-free rate 3.5% and the market return 10%. Find the WACC.

β = 1.3rf = 3.5%rm = 10%rd = 6%T = 25%
WACC = %
9.Descriptive statisticsMedium
26283032343638

Find the sample standard deviation s of the data: 26, 28, 38, 32, 29, 38, 39.

n = 7
s =
10.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 52.

μ = 60σ = 10x = 52
z =
11.ProbabilityMedium
xp(x)00.3510.220.330.15μ = 1.25

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

x: 0, 1, 2, 3p(x): 0.35, 0.2, 0.3, 0.15
σ =
12.ProbabilityHard
n = 8, p = 0.5: P(X = 1) = ?k012345678

A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 1 correct answers?

n = 8p = 0.5k = 1
P =
13.Normal distributionMedium
85.75115P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 85.75 and 115 points? Give P as a decimal.

μ = 100σ = 15x1 = 85.75x2 = 115
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 0.02 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 = 0.02 mm
P =
15.Confidence intervalsMedium
n = 11, s = 17L = ?Ux̄ = 25.5E95%

A sample of 11 items has mean 25.5 and standard deviation s = 17. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 11x̄ = 25.5s = 1795%t(10) = 2.228
lower limit =
16.Confidence intervalsHard
n = 10σ = ?Us = 3.895%

The diameters of 10 machined shafts have a sample standard deviation of 3.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.

n = 10s = 3.8 µmχ² = 2.7, 19.023
σ = µm
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 73.88 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 73.88s = 8n = 100
t =
18.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 81 measurements has mean 49.2. The population standard deviation is 8. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 49.2σ = 8n = 8195%
U =
19.Linear regression & correlationMedium
xy2584786828941010412117xy5106080100120

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy246352453557654763xy2465060x = 6.5

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 6.5.

n = 6x = 6.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 34460 €

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 = 0.05/4 = 0.0125
N = 4·3 = 12

Present value

PV = FV/(1 + i)N = 40 000/1.161 = €34 460.34

Answer

PV = €34 460.34
2.Time value of money

Answer: N = 71 payments

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.

Rearrange the annuity formula

i = 0.006
(1 + i)−N = 1 − 10 000·0.006/175 = 0.6571

Solve with logarithms

N = −ln(0.6571)/ln(1.006) = 70.19

Round up to whole payments

70 full payments leave a small balance, which the last, smaller payment clears.

N = 71

Answer

N = 71 payments
3.NPV & IRR

Answer: EAC = 21037 €

Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.

Annuity factor

(1 − 1.08−5)/0.08 = 3.993

Annualised purchase price

80 000/3.993 = 20 036.52

EAC

EAC = 20 036.52 + 1 000 = €21 036.52

Answer

EAC = €21 036.52
4.NPV & IRR

Answer: MIRR = 7.922 %

Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.

Terminal value of the inflows

9 500·1.062 = 10 674.20
12 500·1.061 = 13 250.00
7 500·1.060 = 7 500.00
TV = 31 424.20

MIRR

MIRR = (31 424.20/25 000)1/3 − 1 = 7.922%

Answer

MIRR = 7.922%
5.Bond pricing & yield

Answer: P = 1089 €

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, i = 0.025, N = 24

Present value of the coupons

€30.00·(1 − 0.5529)/0.025 = €536.55

Present value of the face value

€1 000·0.5529 = €552.88

Price

P = €1 089.42

Answer

P = €1 089.42

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

6.Bond pricing & yield

Answer: YTM = 8.526 %

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

€900 = €70·(1 − (1 + y)⁻¹⁰)/y + €1 000/(1 + y)¹⁰

Solve numerically

P(8%) = €932.90, P(9%) = €871.65
YTM = 8.526%

Answer

YTM = 8.526%
7.CAPM & WACC

Answer: WACC = 8.335 %

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 = €16 000 000, E/V = 0.625, D/V = 0.375

WACC

WACC = 0.625·12% + 0.375·3%·(1 − 0.258) = 8.335%

Answer

WACC = 8.335%
8.CAPM & WACC

Answer: WACC = 10.13 %

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 = 2 000 000·€31.00 = €62 000 000
D = 20 000·€1 000 = €20 000 000

Cost of equity (CAPM)

re = 3.5% + 1.3·6.5% = 11.95%

Weights

V = €82 000 000, E/V = 0.7561, D/V = 0.2439

WACC

WACC = 0.7561·11.95% + 0.2439·6%·(1 − 0.25) = 10.13%

Answer

WACC = 10.13%
9.Descriptive statistics

Answer: s = 5.429

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 = 230/7 = 32.86

Squared deviations

Σ(x − x̄)² = (−6.857)² + (−4.857)² + 5.143² + (−0.8571)² + (−3.857)² + 5.143² + 6.143² = 176.9

Standard deviation

s = √(176.9/6) = 5.429

Answer

s = 5.429
10.Descriptive statistics

Answer: z = −0.8

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 = (52 − 60)/10 = −0.8

Answer

z = −0.8
11.Probability

Answer: σ = 1.09

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 = 0·0.35 + 1·0.2 + 2·0.3 + 3·0.15 = 1.25

Mean of the squares

E(X²) = 0·0.35 + 1·0.2 + 4·0.3 + 9·0.15 = 2.75

Variance

Var(X) = 2.75 − 1.25² = 1.188
σ = √1.188 = 1.09

Answer

σ = 1.09
12.Probability

Answer: P = 0.03125

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 = 1) = 8·0.51·0.57 = 8·0.5·0.007813 = 0.03125

Answer

P = 0.03125
13.Normal distribution

Answer: P = 0.6703

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 = (85.75 − 100)/15 = −0.95
z₂ = (115 − 100)/15 = 1

Area between

P = Φ(1) − Φ(−0.95) = 0.8413 − 0.1711 = 0.6703

Answer

P = 0.6703
14.Normal distribution

Answer: P = 0.8849

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 = 0.025 mm

Standardise

z = (0 − 0.03)/0.025 = −1.2

Probability

P = 1 − Φ(−1.2) = 0.8849

Answer

P = 0.8849
15.Confidence intervals

Answer: lower limit = 14.08

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 = 2.228

Margin of error

E = 2.228·17/√11 = 11.42

Interval

25.5 ± 11.42 → [14.08, 36.92]

Answer

lower limit = 14.08
16.Confidence intervals

Answer: σ = 2.614 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 9·3.8² = 130

Limits for σ²

130/19.023 = 6.832
130/2.7 = 48.13

Limit for σ

σ = √6.832 = 2.614 µm

Answer

σ = 2.614 µm

95% interval for σ: [2.614, 6.938] µm.

17.Hypothesis tests & confidence intervals

Answer: t = −1.4

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 = 8/√100 = 8/10 = 0.8

Test statistic

t = (73.88 − 75)/0.8 = −1.4

Answer

t = −1.4
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: U = 50.94

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 = 1.96·8/√81 = 1.742

Interval limit

U = 49.2 + 1.742 = 50.94

Answer

U = 50.94
19.Linear regression & correlation

Answer: r = 0.989

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 = 533/6 = 88.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 385
Syy = Σ(y − ȳ)² = 2165

Correlation coefficient

r = 385/√(70·2165) = 0.989

Answer

r = 0.989
20.Linear regression & correlation

Answer: ŷ = 59.6

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 = 27/6 = 4.5
ȳ = 325/6 = 54.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 47.5

Regression line

b = 47.5/17.5 = 2.714
a = 54.17 − 2.714·4.5 = 41.95
ŷ = 41.95 + 2.714·x

Prediction

ŷ = 41.95 + 2.714·6.5 = 59.6

Answer

ŷ = 59.6

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