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
years01234€1 000FV = ?

You invest €1 000 at a nominal rate of 5% per year, compounded quarterly. Find the value FV of the investment after 4 years.

PV = €1 0005% / 4n = 4
FV = €
2.Time value of moneyHard
months0123…36A?AA€30 000

You want €30 000 in 3 years. You save the same amount at the end of every month in an account paying 4.8% per year, compounded monthly. Find the monthly deposit A.

FV = €30 0004.8% / 12N = 36
A = €
3.NPV & IRRMedium
years0123456−20 0006 5006 5006 5006 5006 5006 500

A project costs €20 000 today and returns €6 500 at the end of each year for 6 years. The cost of capital is 6%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 6%CF0 = −20 000CF1 = 6 500CF2 = 6 500CF3 = 6 500CF4 = 6 500CF5 = 6 500CF6 = 6 500
DPB = years
4.NPV & IRRHard
r (%)NPVABr* = ?0

Project A costs €10 000 and returns €5 500, €4 000, €2 500 at the end of years 1 to 3. Project B costs €15 000 and returns €4 000, €6 500, €8 000. Find the crossover rate r* at which both projects have the same NPV, in percent.

A: −10 000, 5 500, 4 000, 2 500B: −15 000, 4 000, 6 500, 8 000
r* = %
5.Bond pricing & yieldMedium
years01234567€948C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 7 years and pays annual coupons. It sells for €948 when the yield to maturity is 8%. Find the coupon rate c in percent.

F = €1 000n = 7P = €948YTM = 8%
c = %
6.Bond pricing & yieldHard

A bond trades at €1 080 with a yield to maturity of 6% (annual compounding) and a Macaulay duration of 5.1 years. Use the modified duration to estimate the new price P if the yield rises by 150 basis points.

P0 = €1 080y = 6%D = 5.1Δy = +150 bp
P = €
7.CAPM & WACCMedium
WACC = 9.39%equity€4 000 000re​ = ?debt€2 000 000rd​ = 3%T = 28%

A company has equity worth €4 000 000 and debt worth €2 000 000. Its WACC is 9.39%, its pre-tax cost of debt is 3% and the tax rate is 28%. Find the cost of equity re.

E = €4 000 000D = €2 000 000WACC = 9.39%rd = 3%T = 28%
re = %
8.CAPM & WACCHard
rf​ = 1.5%, rm​ − rf​ = 5%D/E = 0.4equityβL​ = 0.9debtD/E = 0.6equityre​ = ?debtT = 25.8%

A company's shares have a beta of 0.9 at its current debt-to-equity ratio of 0.4. It plans to raise its debt-to-equity ratio to 0.6. The tax rate is 25.8%, the risk-free rate 1.5% and the market risk premium 5%. Find the cost of equity re after the change.

βL = 0.9D/E1 = 0.4D/E2 = 0.6T = 25.8%rf = 1.5%rm − rf = 5%
re = %
9.Descriptive statisticsMedium
10152025303540455055

Find the sample standard deviation s of the data: 24, 57, 29, 28, 10.

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

A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is below 152?

μ = 170σ = 15x = 152
P =
11.ProbabilityMedium
xp(x)00.2510.220.330.140.15μ = 1.7

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

x: 0, 1, 2, 3, 4p(x): 0.25, 0.2, 0.3, 0.1, 0.15
σ =
12.ProbabilityHard
n = 10, p = 0.1: P(X ≤ 1) = ?k012345678910

10 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?

n = 10p = 0.1
P =
13.Normal distributionMedium
460574P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 460 and 574 h? Give P as a decimal.

μ = 500σ = 40x1 = 460x2 = 574
P =
14.Normal distributionHard
235240245250255260265gbottom 85%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 85% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g85%
x = g
15.Confidence intervalsMedium
x̄E₁ = 1.495%, n = 50x̄E₂ = ?99%, n = 50

A 95% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 1.4. What is the margin of error E₂ of a 99% interval from the same data?

95%n₁ = 50E₁ = 1.499%
E₂ =
16.Confidence intervalsHard
σ = 30, n = ?x̄ − Ex̄ + Ex̄E = 299%

The population standard deviation is 30. 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?

σ = 30E = 299%
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 55.5 mm, and the line's known standard deviation is 9 mm. Line 2: a sample of 40 parts has mean 57.8 mm, known standard deviation 14 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 50x̄₁ = 55.5 mmσ₁ = 9 mmn₂ = 40x̄₂ = 57.8 mmσ₂ = 14 mm
z =
18.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 64 measurements has mean 202.2. The population standard deviation is 8. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 202.2σ = 8n = 6499%
U =
19.Linear regression & correlationMedium
xy23544865887210841292xy510406080

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy181112011922122241412614728165xy2025120140160ŷ = 135

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 135?

n = 6ŷ = 135
x =

3Solutions

1.Time value of money

Answer: FV = 1220 €

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 = r/m = 0.05/4 = 0.0125
N = m·n = 4·4 = 16

Future value

FV = 1 000·(1 + 0.0125)16 = 1 000·1.22 = €1 219.89

Answer

FV = €1 219.89
2.Time value of money

Answer: A = 776.4 €

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 = 0.048/12 = 0.004
N = 12·3 = 36

Deposit

A = FV·i/((1 + i)N − 1) = 30 000·0.004/(1.155 − 1) = €776.44

Answer

A = €776.44
3.NPV & IRR

Answer: DPB = 3.51 years

Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.

Discounted and cumulative cash flows

year 1: 6 500/1.06 = 6 132.08 → Σ = 6 132.08
year 2: 6 500/1.124 = 5 784.98 → Σ = 11 917.05
year 3: 6 500/1.191 = 5 457.53 → Σ = 17 374.58
year 4: 6 500/1.262 = 5 148.61 → Σ = 22 523.19

Fraction of the last year

DPB = 3 + 2 625.42/5 148.61 = 3.51

Answer

DPB = 3.51 years

The discounted payback is always longer than the simple payback (3.077 years).

4.NPV & IRR

Answer: r* = 8.826 %

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 = −15 000 − (−10 000) = −5 000
ΔCF1 = 4 000 − 5 500 = −1 500
ΔCF2 = 6 500 − 4 000 = 2 500
ΔCF3 = 8 000 − 2 500 = 5 500

IRR of the difference

NPVΔ(8%) = 120.54
NPVΔ(9%) = −24.94
r* = 8.826%

Answer

r* = 8.826%

Check: at 8.826% both NPVs are €371.17.

5.Bond pricing & yield

Answer: c = 7.001 %

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.

Present value of the face value

1000·(1 + 0.08)−7 = €583.49

Coupon

C = (948 − 583.49)/5.206 = €70.01

Coupon rate

c = 70.01/1000 = 7.001%

Answer

c = 7.001%
6.Bond pricing & yield

Answer: P = 1002 €

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 = 5.1/1.06 = 4.811

Estimated price change

%ΔP ≈ −4.811·(0.015) = −7.217%
P ≈ 1080·(1 − 0.07217) = €1 002.06

Answer

P = €1 002.06

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

7.CAPM & WACC

Answer: re = 13.01 %

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 = 0.6667, D/V = 0.3333

Debt part of the WACC

0.3333·3%·(1 − 0.28) = 0.72%

Cost of equity

re = (9.39% − 0.72%)/0.6667 = 13.01%

Answer

re = 13.01%
8.CAPM & WACC

Answer: re = 6.515 %

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 = 0.9/(1 + 0.742·0.4) = 0.694

Relever at the new D/E

βL = 0.694·(1 + 0.742·0.6) = 1.003

Cost of equity (CAPM)

re = 1.5% + 1.003·5% = 6.515%

Answer

re = 6.515%
9.Descriptive statistics

Answer: s = 17.1

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 = 148/5 = 29.6

Squared deviations

Σ(x − x̄)² = (−5.6)² + 27.4² + (−0.6)² + (−1.6)² + (−19.6)² = 1169

Standard deviation

s = √(1169/4) = 17.1

Answer

s = 17.1
10.Descriptive statistics

Answer: P = 0.1151

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 0.1151

Answer

P = 0.1151
11.Probability

Answer: σ = 1.345

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 + 1·0.2 + 2·0.3 + 3·0.1 + 4·0.15 = 1.7

Mean of the squares

E(X²) = 0·0.25 + 1·0.2 + 4·0.3 + 9·0.1 + 16·0.15 = 4.7

Variance

Var(X) = 4.7 − 1.7² = 1.81
σ = √1.81 = 1.345

Answer

σ = 1.345
12.Probability

Answer: P = 0.7361

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) = 0.910 = 0.3487
P(X = 1) = 10·0.1·0.99 = 0.3874

Add them

P = 0.3487 + 0.3874 = 0.7361

Answer

P = 0.7361
13.Normal distribution

Answer: P = 0.8092

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 = (460 − 500)/40 = −1
z₂ = (574 − 500)/40 = 1.85

Area between

P = Φ(1.85) − Φ(−1) = 0.9678 − 0.1587 = 0.8092

Answer

P = 0.8092
14.Normal distribution

Answer: x = 255.2 g

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) = 0.85

z from the table

z ≈ 1.04

Back to the original units

x = 250 + 1.04·5 = 255.2 g

Answer

x = 255.2 g
15.Confidence intervals

Answer: E₂ = 1.84

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 = (2.576/1.96)·√(50/50) = 1.314

New margin

E₂ = 1.4·1.314 = 1.84

Answer

E₂ = 1.84
16.Confidence intervals

Answer: n = 1494

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 = (2.576·30/2)² = 1493

Round up

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

n = 1494

Answer

n = 1494
17.Hypothesis tests & confidence intervals

Answer: z = −0.9007

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 = √(9²/50 + 14²/40) = √6.52 = 2.553

Test statistic

z = (55.5 − 57.8)/2.553 = −0.9007

Answer

z = −0.9007

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

18.Hypothesis tests & confidence intervals

Answer: U = 204.8

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 = 2.576·8/√64 = 2.576

Interval limit

U = 202.2 + 2.576 = 204.8

Answer

U = 204.8
19.Linear regression & correlation

Answer: r = 0.9978

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 = 389/6 = 64.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 407
Syy = Σ(y − ȳ)² = 2377

Correlation coefficient

r = 407/√(70·2377) = 0.9978

Answer

r = 0.9978
20.Linear regression & correlation

Answer: x = 23.16

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 = 138/6 = 23
ȳ = 805/6 = 134.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 373

Regression line

b = 373/70 = 5.329
a = 134.2 − 5.329·23 = 11.61
ŷ = 11.61 + 5.329·x

Solve for x

x = (135 − 11.61)/5.329 = 23.16

Answer

x = 23.16

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