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
years0123456€20 000FV = ?

You invest €20 000 at a nominal rate of 5% per year, compounded monthly. Find the value FV of the investment after 6 years.

PV = €20 0005% / 12n = 6
FV = €
2.Time value of moneyHard
months0123…360€150 000A?AAA

A loan of €150 000 is repaid with equal monthly payments over 30 years. The interest rate is 3% per year, compounded monthly. Find the monthly payment A.

P = €150 0003% / 12N = 360
A = €
3.NPV & IRRMedium
years01234−20 0006 0008 0007 0009 000

An investment of €20 000 gives these cash flows at the end of years 1 to 4: €6 000, €8 000, €7 000, €9 000. The cost of capital is 10%. Find the net present value NPV.

r = 10%CF0 = −20 000CF1 = 6 000CF2 = 8 000CF3 = 7 000CF4 = 9 000
NPV = €
4.NPV & IRRHard
years0123−10 0005 0004 0005 000

A project costs €10 000 and returns €5 000, €4 000, €5 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −10 000CF1 = 5 000CF2 = 4 000CF3 = 5 000
IRR = %
5.Bond pricing & yieldMedium
years0123€895C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 3 years and pays annual coupons. It sells for €895 when the yield to maturity is 7%. Find the coupon rate c in percent.

F = €1 000n = 3P = €895YTM = 7%
c = %
6.Bond pricing & yieldHard
years0123456€1 110€40€40 + €1 000

A bond with a face value of €1 000 and an annual coupon of 4% matures in 6 years and sells for €1 110. Find its yield to maturity YTM in percent.

F = €1 000coupon = 4%n = 6P = €1 110
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€5 000 000re​ = 14%debt€2 000 000rd​ = 6%T = 28%

A company has equity worth €5 000 000 and debt worth €2 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 6% and the tax rate is 28%. Find the WACC.

E = €5 000 000D = €2 000 000re = 14%rd = 6%T = 28%
WACC = %
8.CAPM & WACCHard
rf​ = 3%, rm​ − rf​ = 4.5%D/E = 0.2equityβL​ = 1.2debtD/E = 1equityre​ = ?debtT = 30%

A company's shares have a beta of 1.2 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 1. The tax rate is 30%, the risk-free rate 3% and the market risk premium 4.5%. Find the cost of equity re after the change.

βL = 1.2D/E1 = 0.2D/E2 = 1T = 30%rf = 3%rm − rf = 4.5%
re = %
9.Descriptive statisticsMedium
101520253035404550

Find the sample standard deviation s of the data: 34, 10, 10, 19, 51, 54, 31.

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

Exam scores are normally distributed with mean 170 and standard deviation 20. Find the z-score z of a score of 190.

μ = 170σ = 20x = 190
z =
11.ProbabilityMedium
red: 4blue: 4

A bag holds 4 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

4 / 4
P =
12.ProbabilityHard
€p(x)−30.8560.1570.05

A game costs €3 to play. You win €9 with probability 0.1 and €60 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€9: p = 0.1€60: p = 0.05
E(G) = €
13.Normal distributionMedium
460520P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 460 and 520 h? Give P as a decimal.

μ = 500σ = 40x1 = 460x2 = 520
P =
14.Normal distributionHard
35455565758595pointstop 95%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points95%
x = points
15.Confidence intervalsMedium
n = 8, s = 5.5LU = ?x̄ = 71.5E90%

A sample of 8 items has mean 71.5 and standard deviation s = 5.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 8x̄ = 71.5s = 5.590%t(7) = 1.895
upper limit =
16.Confidence intervalsHard
n = 30Lσ = ?s = 6.895%

The diameters of 30 machined shafts have a sample standard deviation of 6.8 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 6.8 µmχ² = 16.047, 45.722
σ = µm
17.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 48.5 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 48.5s = 10n = 36
t =
18.Hypothesis tests & confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 2.590%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 8. Find the smallest sample size n.

σ = 8E = 2.590%
n =
19.Linear regression & correlationMedium
xy3535627809871110513117xy5106080100120

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

n = 6Σx = 48Σy = 504Σx² = 454Σy² = 45336Σxy = 4488
r =
20.Linear regression & correlationHard
xy152255357464568xy24556065x = 2.5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.

n = 5x = 2.5
ŷ =

3Solutions

1.Time value of money

Answer: FV = 26980 €

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 = 0.05/12 = 0.004167
N = m·n = 12·6 = 72

Future value

FV = 20 000·(1 + 0.004167)72 = 20 000·1.349 = €26 980.35

Answer

FV = €26 980.35
2.Time value of money

Answer: A = 632.4 €

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 payments

i = 0.03/12 = 0.0025
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 = 0.407
A = 150 000·0.0025/(1 − 0.407) = €632.41

Answer

A = €632.41
3.NPV & IRR

Answer: NPV = 3472 €

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 = −20 000
CF1/(1 + 0.1)1 = 6 000/1.1 = 5 454.55
CF2/(1 + 0.1)2 = 8 000/1.21 = 6 611.57
CF3/(1 + 0.1)3 = 7 000/1.331 = 5 259.20
CF4/(1 + 0.1)4 = 9 000/1.464 = 6 147.12

Add them up

NPV = €3 472.44

Answer

NPV = €3 472.44

NPV is positive, so the project earns more than the required return.

4.NPV & IRR

Answer: IRR = 18.96 %

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%) = 659.98
NPV(20%) = −162.04

Narrow down

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

IRR ≈ 15 + 5·659.98/(659.98 − (−162.04)) = 19.01%
exact: NPV(18.96%) = 0 → IRR = 18.96%

Answer

IRR = 18.96%
5.Bond pricing & yield

Answer: c = 2.999 %

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 + 0.07)−3 = €816.30

Coupon

C = (895 − 816.30)/2.624 = €29.99

Coupon rate

c = 29.99/1000 = 2.999%

Answer

c = 2.999%
6.Bond pricing & yield

Answer: YTM = 2.034 %

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

€1 110 = €40·(1 − (1 + y)⁻⁶)/y + €1 000/(1 + y)⁶

Solve numerically

P(2%) = €1 112.03, P(3%) = €1 054.17
YTM = 2.034%

Answer

YTM = 2.034%
7.CAPM & WACC

Answer: WACC = 11.23 %

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 = €7 000 000, E/V = 0.7143, D/V = 0.2857

WACC

WACC = 0.7143·14% + 0.2857·6%·(1 − 0.28) = 11.23%

Answer

WACC = 11.23%
8.CAPM & WACC

Answer: re = 11.05 %

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 = 1.2/(1 + 0.7·0.2) = 1.053

Relever at the new D/E

βL = 1.053·(1 + 0.7·1) = 1.789

Cost of equity (CAPM)

re = 3% + 1.789·4.5% = 11.05%

Answer

re = 11.05%
9.Descriptive statistics

Answer: s = 18.05

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 = 209/7 = 29.86

Squared deviations

Σ(x − x̄)² = 4.143² + (−19.86)² + (−19.86)² + (−10.86)² + 21.14² + 24.14² + 1.143² = 1955

Standard deviation

s = √(1955/6) = 18.05

Answer

s = 18.05
10.Descriptive statistics

Answer: z = 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.

Standardise

z = (190 − 170)/20 = 1

Answer

z = 1
11.Probability

Answer: P = 0.2143

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.

First and second draw

P = 4/8 · 3/7 = 12/56 = 0.2143

Answer

P = 0.2143
12.Probability

Answer: E(G) = 0.9 €

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.

Net outcomes

G: 6 (p = 0.1), 57 (p = 0.05), −3 (p = 0.85)

Expected value

E(G) = 6·0.1 + 57·0.05 − 3·0.85 = 0.9 €

Answer

E(G) = 0.9 €

A positive value means you win money on average.

13.Normal distribution

Answer: P = 0.5328

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 = (520 − 500)/40 = 0.5

Area between

P = Φ(0.5) − Φ(−1) = 0.6915 − 0.1587 = 0.5328

Answer

P = 0.5328
14.Normal distribution

Answer: x = 48.6 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) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 65 − 1.64·10 = 48.6 points

Answer

x = 48.6 points
15.Confidence intervals

Answer: upper limit = 75.18

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 = 7, t = 1.895

Margin of error

E = 1.895·5.5/√8 = 3.685

Interval

71.5 ± 3.685 → [67.82, 75.18]

Answer

upper limit = 75.18
16.Confidence intervals

Answer: σ = 9.141 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 = 29·6.8² = 1341

Limits for σ²

1341/45.722 = 29.33
1341/16.047 = 83.56

Limit for σ

σ = √83.56 = 9.141 µm

Answer

σ = 9.141 µm

95% interval for σ: [5.416, 9.141] µm.

17.Hypothesis tests & confidence intervals

Answer: t = −0.9

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 = 10/√36 = 10/6 = 1.667

Test statistic

t = (48.5 − 50)/1.667 = −0.9

Answer

t = −0.9
t0.025; df = 35 = 2.03

|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 28

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).

Solve for n

n = (1.645·8/2.5)² = 27.71

Round up

n = 28

Answer

n = 28
19.Linear regression & correlation

Answer: r = 0.9951

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 = 4488 − 48·504/6 = 456
Sxx = 454 − 48²/6 = 70
Syy = 45336 − 504²/6 = 3000

Correlation coefficient

r = 456/√(70·3000) = 0.9951

Answer

r = 0.9951
20.Linear regression & correlation

Answer: ŷ = 57.15

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 = 15/5 = 3
ȳ = 296/5 = 59.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 41

Regression line

b = 41/10 = 4.1
a = 59.2 − 4.1·3 = 46.9
ŷ = 46.9 + 4.1·x

Prediction

ŷ = 46.9 + 4.1·2.5 = 57.15

Answer

ŷ = 57.15

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