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
years0123456PV = ?€1 500€1 500€1 500€1 500

An annuity pays €1 500 at the end of each year for 6 years. The interest rate is 3% per year. Find the present value PV of the annuity.

A = €1 500i = 3%n = 6
PV = €
2.Time value of moneyHard
months0123…N = ?€20 000€350€350€350€3506% / 12

You borrow €20 000 at 6% per year, compounded monthly, and repay €350 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 = €20 000A = €3506% / 12
N = payments
3.NPV & IRRMedium
years01234567€40 000€2 000€2 000€2 000€2 000€2 000€2 000€2 000r = 8%, EAC = ?

A machine costs €40 000 today, lasts 7 years and costs €2 000 per year to run (paid at the end of each year). The cost of capital is 8%. Find its equivalent annual cost EAC.

C0 = €40 000M = €2 000n = 7r = 8%
EAC = €
4.NPV & IRRHard
r (%)NPVABr* = ?0

Project A costs €30 000 and returns €17 000, €12 000, €7 000 at the end of years 1 to 3. Project B costs €35 000 and returns €11 000, €14 500, €18 000. Find the crossover rate r* at which both projects have the same NPV, in percent.

A: −30 000, 17 000, 12 000, 7 000B: −35 000, 11 000, 14 500, 18 000
r* = %
5.Bond pricing & yieldMedium

You buy a bond with 11 years to maturity (face value €1 000, annual coupon 8%) when its yield to maturity is 6%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 5%. Find your holding period return HPR in percent.

F = €1 000coupon = 8%n = 11y0 = 6%y1 = 5%
HPR = %
6.Bond pricing & yieldHard

A bond trades at €1 085 with a yield to maturity of 4% (annual compounding) and a Macaulay duration of 4.1 years. Use the modified duration to estimate the new price P if the yield rises by 50 basis points.

P0 = €1 085y = 4%D = 4.1Δy = +50 bp
P = €
7.CAPM & WACCMedium
WACC = ?ordinary shares€4 000 000re​ = 13%preferred shares€500 000rp​ = 8%debt€5 000 000rd​ = 3%T = 21%

A company is financed with ordinary shares worth €4 000 000, preferred shares worth €500 000 and debt worth €5 000 000. The cost of equity is 13%, the cost of preferred stock 8% and the pre-tax cost of debt 3%. The tax rate is 21%. Find the WACC.

E = €4 000 000Ps = €500 000D = €5 000 000re = 13%rp = 8%rd = 3%T = 21%
WACC = %
8.CAPM & WACCHard
rf​ = 4%, rm​ − rf​ = 7%D/E = 0.2equityβL​ = 1.4debtD/E = 1.5equityre​ = ?debtT = 25%

A company's shares have a beta of 1.4 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 1.5. The tax rate is 25%, the risk-free rate 4% and the market risk premium 7%. Find the cost of equity re after the change.

βL = 1.4D/E1 = 0.2D/E2 = 1.5T = 25%rf = 4%rm − rf = 7%
re = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 17, 20, 14, 23, 15, 45, 52.

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

Exam scores are normally distributed with mean 60 and standard deviation 5. Find the z-score z of a score of 52.5.

μ = 60σ = 5x = 52.5
z =
11.ProbabilityMedium
n = 7k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 7? The order does not matter.

n = 7k = 3
N =
12.ProbabilityHard
n = 5, p = 0.1: P(X = 2) = ?k012345

A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 2 correct answers?

n = 5p = 0.1k = 2
P =
13.Normal distributionMedium
167.5179.75P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 167.5 and 179.75 cm? Give P as a decimal.

μ = 178σ = 7x1 = 167.5x2 = 179.75
P =
14.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
15.Confidence intervalsMedium
x̄E₁ = 11.895%, n = 100x̄E₂ = ?95%, n = 400

A 95% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 11.8. What is the margin of error E₂ if the same study uses 400 observations instead?

95%n₁ = 100E₁ = 11.8n₂ = 400
E₂ =
16.Confidence intervalsHard
p ≈ 0.7, n = ?p̂ − Ep̂ + Ep̂E = 0.0199%

A pilot study suggests that about 0.7 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.01?

p ≈ 0.7E = 0.0199%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 120.4 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 120.4s = 5n = 64
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 75 of 150 patients on treatment A recover, and 133 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 75n₁ = 150x₂ = 133n₂ = 250
z =
19.Linear regression & correlationMedium
xy253359463571673783xy246607080

The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy23944665586210631268xy51040506070x = 7

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 = 7.

n = 6x = 7
ŷ =

3Solutions

1.Time value of money

Answer: PV = 8126 €

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 − 0.8375)/0.03 = 5.417

Present value

PV = 1 500·5.417 = €8 125.79

Answer

PV = €8 125.79
2.Time value of money

Answer: N = 68 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 = 0.005
(1 + i)−N = 1 − 20 000·0.005/350 = 0.7143

Solve with logarithms

N = −ln(0.7143)/ln(1.005) = 67.46

Round up to whole payments

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

N = 68

Answer

N = 68 payments
3.NPV & IRR

Answer: EAC = 9683 €

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 − 1.08−7)/0.08 = 5.206

Annualised purchase price

40 000/5.206 = 7 682.90

EAC

EAC = 7 682.90 + 2 000 = €9 682.90

Answer

EAC = €9 682.90
4.NPV & IRR

Answer: r* = 9.426 %

Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.

Difference in cash flows

ΔCF0 = −35 000 − (−30 000) = −5 000
ΔCF1 = 11 000 − 17 000 = −6 000
ΔCF2 = 14 500 − 12 000 = 2 500
ΔCF3 = 18 000 − 7 000 = 11 000

IRR of the difference

NPVΔ(9%) = 93.63
NPVΔ(10%) = −123.97
r* = 9.426%

Answer

r* = 9.426%

Check: at 9.426% both NPVs are €899.92.

5.Bond pricing & yield

Answer: HPR = 13.29 %

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 = €80.00, i = 0.06, N = 11
€80.00·(1 − 0.5268)/0.06 + €1 000·0.5268 = €630.95 + €526.79 = €1 157.74

Selling price

C = €80.00, i = 0.05, N = 10
€80.00·(1 − 0.6139)/0.05 + €1 000·0.6139 = €617.74 + €613.91 = €1 231.65

Holding period return

HPR = (80 + 1 231.65 − 1 157.74)/1 157.74 = 13.29%

Answer

HPR = 13.29%
6.Bond pricing & yield

Answer: P = 1064 €

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.

Modified duration

Dmod = 4.1/1.04 = 3.942

Estimated price change

%ΔP ≈ −3.942·(0.005) = −1.971%
P ≈ 1085·(1 − 0.01971) = €1 063.61

Answer

P = €1 063.61

Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.

7.CAPM & WACC

Answer: WACC = 7.142 %

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 + 500 000 + 5 000 000 = €9 500 000
E/V = 0.4211, Ps/V = 0.05263, D/V = 0.5263

WACC

WACC = 0.4211·13% + 0.05263·8% + 0.5263·3%·(1 − 0.21) = 7.142%

Answer

WACC = 7.142%
8.CAPM & WACC

Answer: re = 22.11 %

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 = 1.4/(1 + 0.75·0.2) = 1.217

Relever at the new D/E

βL = 1.217·(1 + 0.75·1.5) = 2.587

Cost of equity (CAPM)

re = 4% + 2.587·7% = 22.11%

Answer

re = 22.11%
9.Descriptive statistics

Answer: s = 15.41

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 = 186/7 = 26.57

Squared deviations

Σ(x − x̄)² = (−9.571)² + (−6.571)² + (−12.57)² + (−3.571)² + (−11.57)² + 18.43² + 25.43² = 1426

Standard deviation

s = √(1426/6) = 15.41

Answer

s = 15.41
10.Descriptive statistics

Answer: z = −1.5

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 = (52.5 − 60)/5 = −1.5

Answer

z = −1.5
11.Probability

Answer: N = 35

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.

Combinations

C(7, 3) = (7·6·5)/(3·2·1) = 35

Answer

N = 35
12.Probability

Answer: P = 0.0729

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 = 2) = 10·0.12·0.93 = 10·0.01·0.729 = 0.0729

Answer

P = 0.0729
13.Normal distribution

Answer: P = 0.5319

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 = (167.5 − 178)/7 = −1.5
z₂ = (179.75 − 178)/7 = 0.25

Area between

P = Φ(0.25) − Φ(−1.5) = 0.5987 − 0.06681 = 0.5319

Answer

P = 0.5319
14.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
15.Confidence intervals

Answer: E₂ = 5.9

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.

Ratio of margins

E₂/E₁ = (1.96/1.96)·√(100/400) = 0.5

New margin

E₂ = 11.8·0.5 = 5.9

Answer

E₂ = 5.9
16.Confidence intervals

Answer: n = 13936

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 2.576²·0.7·0.3/0.01² = 13935

Round up

n = 13936

Answer

n = 13936
17.Hypothesis tests & confidence intervals

Answer: t = 0.64

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 = 5/√64 = 5/8 = 0.625

Test statistic

t = (120.4 − 120)/0.625 = 0.64

Answer

t = 0.64
t0.025; df = 63 = 1.998

|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = −0.6202

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 = 75/150 = 0.5
p̂₂ = 133/250 = 0.532
p̂ = 208/400 = 0.52

Standard error under H₀

SE = √(0.52·0.48·(1/150 + 1/250)) = 0.0516

Test statistic

z = (0.5 − 0.532)/0.0516 = −0.6202

Answer

z = −0.6202

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

19.Linear regression & correlation

Answer: r² = 0.9785

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 = 402/6 = 67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 100
Syy = Σ(y − ȳ)² = 584

Coefficient of determination

r² = 100²/(17.5·584) = 0.9785

Answer

r² = 0.9785

About 97.85% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 55.5

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 = 333/6 = 55.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 203

Regression line

b = 203/70 = 2.9
a = 55.5 − 2.9·7 = 35.2
ŷ = 35.2 + 2.9·x

Prediction

ŷ = 35.2 + 2.9·7 = 55.5

Answer

ŷ = 55.5

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