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…10PV = ?€3 000€3 000€3 000€3 000

An annuity pays €3 000 at the end of each year for 10 years. The interest rate is 4% per year. Find the present value PV of the annuity.

A = €3 000i = 4%n = 10
PV = €
2.Time value of moneyHard
months0123…360€250 000A?AAA

A loan of €250 000 is repaid with equal monthly payments over 30 years. The interest rate is 5.4% per year, compounded monthly. Find the monthly payment A.

P = €250 0005.4% / 12N = 360
A = €
3.NPV & IRRMedium
years0123−30 0009 00013 00011 000

An investment of €30 000 gives these cash flows at the end of years 1 to 3: €9 000, €13 000, €11 000. The cost of capital is 5%. Find the net present value NPV.

r = 5%CF0 = −30 000CF1 = 9 000CF2 = 13 000CF3 = 11 000
NPV = €
4.NPV & IRRHard
years01234−50 00021 50029 00018 00017 000

A project costs €50 000 and returns €21 500, €29 000, €18 000, €17 000 at the end of years 1 to 4. 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 = −50 000CF1 = 21 500CF2 = 29 000CF3 = 18 000CF4 = 17 000
MIRR = %
5.Bond pricing & yieldMedium
years0123456P€20€20€20 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 2% and matures in 6 years. Its yield to maturity changes from 4% to 3.5%. By how much ΔP does the bond price change?

F = €1 000coupon = 2%n = 6y1 = 4%y2 = 3.5%
ΔP = €
6.Bond pricing & yieldHard
years012345P€70€70€70 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 7% and matures in 5 years. The yield to maturity is 2.5%. Find the Macaulay duration D in years.

F = €1 000coupon = 7%n = 5YTM = 2.5%
D = years
7.CAPM & WACCMedium
WACC = ?equity€2 000 000re​ = 14%debt€1 000 000rd​ = 5%T = 28%

A company has equity worth €2 000 000 and debt worth €1 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 5% and the tax rate is 28%. Find the WACC.

E = €2 000 000D = €1 000 000re = 14%rd = 5%T = 28%
WACC = %
8.CAPM & WACCHard
rf​ = 2%, rm​ − rf​ = 6%no debtequityβU​ = 0.8D/E = 1.5equityre​ = ?debtT = 28%

A company's business has an unlevered (asset) beta of 0.8. It plans a debt-to-equity ratio of 1.5. The tax rate is 28%, the risk-free rate is 2% and the market risk premium is 6%. Find the cost of equity re at the new capital structure.

βU = 0.8D/E = 1.5T = 28%rf = 2%rm − rf = 6%
re = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 33, 54, 35, 14, 55.

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

Exam scores are normally distributed with mean 500 and standard deviation 40. Find the z-score z of a score of 420.

μ = 500σ = 40x = 420
z =
11.ProbabilityMedium
xp(x)00.3510.320.35μ = 1

For the distribution shown, find the variance Var(X) of X.

x: 0, 1, 2p(x): 0.35, 0.3, 0.35
Var(X) =
12.ProbabilityHard
n = 10, p = 0.25: P(X = 1) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 1 correct answers?

n = 10p = 0.25k = 1
P =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0582

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0582. Find z to two decimals.

A = 0.0582
z =
14.Normal distributionHard
557085100115130145pointsbottom 10%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 10% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points10%
x = points
15.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 60 phones, mean 81.5 h, known standard deviation 6 h. Model 2: 50 phones, mean 73.5 h, known standard deviation 9 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 81.5 hσ₁ = 6 hn₂ = 50x̄₂ = 73.5 hσ₂ = 9 h90%
lower limit = h
16.Confidence intervalsHard
σ = 16, n = ?x̄ − Ex̄ + Ex̄E = 2.599%

The population standard deviation is 16. 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 = 16E = 2.599%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 50 parts has mean length 111.5 mm, and the line's known standard deviation is 20 mm. Line 2: a sample of 80 parts has mean 114.2 mm, known standard deviation 9 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 50x̄₁ = 111.5 mmσ₁ = 20 mmn₂ = 80x̄₂ = 114.2 mmσ₂ = 9 mm
z =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.05βσ = 12, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 95%. The standard deviation is 12. Find the minimum sample size n.

α = 0.05power = 95%δ = 4σ = 12zα = 1.645zβ = 1.645
n =
19.Linear regression & correlationMedium
xy2741561882510291237xy51010203040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 131Σx² = 364Σy² = 3433Σxy = 1116
r =
20.Linear regression & correlationHard
xy125220321419516610xy24610152025x = 4.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.

n = 6x = 4.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 24333 €

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.

Annuity factor

(1 − (1 + i)−n)/i = (1 − 0.6756)/0.04 = 8.111

Present value

PV = 3 000·8.111 = €24 332.69

Answer

PV = €24 332.69
2.Time value of money

Answer: A = 1404 €

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 = 12·30 = 360

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 = 0.1986
A = 250 000·0.0045/(1 − 0.1986) = €1 403.83

Answer

A = €1 403.83
3.NPV & IRR

Answer: NPV = −135 €

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 each cash flow

CF0 = −30 000
CF1/(1 + 0.05)1 = 9 000/1.05 = 8 571.43
CF2/(1 + 0.05)2 = 13 000/1.103 = 11 791.38
CF3/(1 + 0.05)3 = 11 000/1.158 = 9 502.21

Add them up

NPV = €−134.97

Answer

NPV = €−134.97

NPV is negative, so the project earns less than the required return.

4.NPV & IRR

Answer: MIRR = 17.18 %

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

21 500·1.063 = 25 606.84
29 000·1.062 = 32 584.40
18 000·1.061 = 19 080.00
17 000·1.060 = 17 000.00
TV = 94 271.24

MIRR

MIRR = (94 271.24/50 000)1/4 − 1 = 17.18%

Answer

MIRR = 17.18%
5.Bond pricing & yield

Answer: ΔP = 24.91 €

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.

Price at the old yield

C = €20.00, i = 0.04, N = 6
€20.00·(1 − 0.7903)/0.04 + €1 000·0.7903 = €104.84 + €790.31 = €895.16

Price at the new yield

C = €20.00, i = 0.035, N = 6
€20.00·(1 − 0.8135)/0.035 + €1 000·0.8135 = €106.57 + €813.50 = €920.07

Change

ΔP = 920.07 − 895.16 = €24.91

Answer

ΔP = €24.91
6.Bond pricing & yield

Answer: D = 4.449 years

Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.

Present value of each cash flow

t = 1: 70/1.025 = 68.29, t·PV = 68.29
t = 2: 70/1.051 = 66.63, t·PV = 133.25
t = 3: 70/1.077 = 65.00, t·PV = 195.01
t = 4: 70/1.104 = 63.42, t·PV = 253.67
t = 5: 1070/1.131 = 945.72, t·PV = 4 728.62

Duration

P = 1 209.06
D = 5 378.84/1 209.06 = 4.449

Answer

D = 4.449 years
7.CAPM & WACC

Answer: WACC = 10.53 %

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 = €3 000 000, E/V = 0.6667, D/V = 0.3333

WACC

WACC = 0.6667·14% + 0.3333·5%·(1 − 0.28) = 10.53%

Answer

WACC = 10.53%
8.CAPM & WACC

Answer: re = 11.98 %

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 = 0.8·(1 + 0.72·1.5) = 1.664

Cost of equity (CAPM)

re = 2% + 1.664·6% = 11.98%

Answer

re = 11.98%
9.Descriptive statistics

Answer: s = 16.99

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 = 191/5 = 38.2

Squared deviations

Σ(x − x̄)² = (−5.2)² + 15.8² + (−3.2)² + (−24.2)² + 16.8² = 1155

Standard deviation

s = √(1155/4) = 16.99

Answer

s = 16.99
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 = (420 − 500)/40 = −2

Answer

z = −2
11.Probability

Answer: Var(X) = 0.7

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 + 1·0.3 + 2·0.35 = 1

Mean of the squares

E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7

Variance

Var(X) = 1.7 − 1² = 0.7

Answer

Var(X) = 0.7
12.Probability

Answer: P = 0.1877

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) = 10·0.251·0.759 = 10·0.25·0.07508 = 0.1877

Answer

P = 0.1877
13.Normal distribution

Answer: z = 1.57

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 − 0.0582 = 0.9418

Read the z-table backwards

Find 0.9418 inside the table and read off z.

z = 1.57

Answer

z = 1.57
14.Normal distribution

Answer: x = 80.8 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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 100 − 1.28·15 = 80.8 points

Answer

x = 80.8 points
15.Confidence intervals

Answer: lower limit = 5.549 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 = 81.5 − 73.5 = 8

Standard error

SE = √(6²/60 + 9²/50) = 1.49

Margin of error

E = 1.645·1.49 = 2.451

Interval

8 ± 2.451 → [5.549, 10.45]

Answer

lower limit = 5.549 h

The interval does not contain 0, so the means differ significantly.

16.Confidence intervals

Answer: n = 272

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 = (2.576·16/2.5)² = 271.8

Round up

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

n = 272

Answer

n = 272
17.Hypothesis tests & confidence intervals

Answer: z = −0.8994

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.

Standard error of the difference

SE = √(20²/50 + 9²/80) = √9.013 = 3.002

Test statistic

z = (111.5 − 114.2)/3.002 = −0.8994

Answer

z = −0.8994

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

18.Hypothesis tests & confidence intervals

Answer: n = 98

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 1.645)·12/4)² = 97.42

Round up

n = 98

Answer

n = 98
19.Linear regression & correlation

Answer: r = 0.9938

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.

Centred sums

Sxy = 1116 − 42·131/6 = 199
Sxx = 364 − 42²/6 = 70
Syy = 3433 − 131²/6 = 572.8

Correlation coefficient

r = 199/√(70·572.8) = 0.9938

Answer

r = 0.9938
20.Linear regression & correlation

Answer: ŷ = 15.96

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 = 21/6 = 3.5
ȳ = 111/6 = 18.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −44.5

Regression line

b = −44.5/17.5 = −2.543
a = 18.5 − (−2.543)·3.5 = 27.4
ŷ = 27.4 − 2.543·x

Prediction

ŷ = 27.4 + (−2.543)·4.5 = 15.96

Answer

ŷ = 15.96

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