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€2 000FV = ?

You invest €2 000 at a nominal rate of 2% per year, compounded quarterly. Find the value FV of the investment after 6 years.

PV = €2 0002% / 4n = 6
FV = €
2.Time value of moneyHard
months0123…60€8 000A?AAA

A loan of €8 000 is repaid with equal monthly payments over 5 years. The interest rate is 3.6% per year, compounded monthly. Find the monthly payment A.

P = €8 0003.6% / 12N = 60
A = €
3.NPV & IRRMedium
years0123456−30 0009 0009 0009 0009 0009 0009 000

A project costs €30 000 today and returns €9 000 at the end of each year for 6 years. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 12%CF0 = −30 000CF1 = 9 000CF2 = 9 000CF3 = 9 000CF4 = 9 000CF5 = 9 000CF6 = 9 000
DPB = years
4.NPV & IRRHard
years0123−50 00028 00026 00018 000

A project costs €50 000 and returns €28 000, €26 000, €18 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 28 000CF2 = 26 000CF3 = 18 000
IRR = %
5.Bond pricing & yieldMedium
half-years0123…10€10€10€10 + €1 000P = ?

A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 5 years. The yield to maturity is 3.5% per year. Find the price P of the bond.

F = €1 000coupon = 2%YTM = 3.5%n = 5
P = €
6.Bond pricing & yieldHard
years01234€950€80€80 + €1 000

A bond with a face value of €1 000 and an annual coupon of 8% matures in 4 years and sells for €950. Find its yield to maturity YTM in percent.

F = €1 000coupon = 8%n = 4P = €950
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€12 000 000re​ = 14%debt€2 000 000rd​ = 3.5%T = 25%

A company has equity worth €12 000 000 and debt worth €2 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 3.5% and the tax rate is 25%. Find the WACC.

E = €12 000 000D = €2 000 000re = 14%rd = 3.5%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity5 000 000 × €56.50β = 1.5rf​ = 2.5%, rm​ = 9.5%debt20 000 × €980rd​ = 4%T = 21%

A company has 5 000 000 shares at €56.50 each and 20 000 bonds trading at €980 each. The pre-tax cost of debt is 4% and the tax rate is 21%. The share’s beta is 1.5, the risk-free rate 2.5% and the market return 9.5%. Find the WACC.

β = 1.5rf = 2.5%rm = 9.5%rd = 4%T = 21%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 28, 51, 19, 32, 46, 12, 56.

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

A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is below 440?

μ = 500σ = 50x = 440
P =
11.ProbabilityMedium
xp(x)100.15200.35300.3400.2μ = 25.5

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

x: 10, 20, 30, 40p(x): 0.15, 0.35, 0.3, 0.2
σ =
12.ProbabilityHard
€p(x)−30.6530.3270.05

A game costs €3 to play. You win €6 with probability 0.3 and €30 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€6: p = 0.3€30: p = 0.05
E(G) = €
13.Normal distributionMedium
480540P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 480 and 540 h? Give P as a decimal.

μ = 500σ = 40x1 = 480x2 = 540
P =
14.Normal distributionHard
235240245250255260265gtop 60%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 60% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g60%
x = g
15.Confidence intervalsMedium
n = 11, s = 10L = ?Ux̄ = 44.5E99%

A sample of 11 items has mean 44.5 and standard deviation s = 10. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 11x̄ = 44.5s = 1099%t(10) = 3.169
lower limit =
16.Confidence intervalsHard
p ≈ 0.15, n = ?p̂ − Ep̂ + Ep̂E = 0.0399%

A pilot study suggests that about 0.15 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 0.03?

p ≈ 0.15E = 0.0399%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 77.1 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 77.1s = 8n = 64
t =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 200 against H₁: μ > 200 is done at α = 0.10 with a sample of 64; the standard deviation is known to be 24. In reality the mean is 205.4. Find the probability of a Type II error β in percent.

μ₀ = 200μ = 205.4σ = 24n = 64α = 0.10
β = %
19.Linear regression & correlationMedium
xy129226326421521620xy246202530

The table gives 6 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy27486168181022xy510101520x = 5

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.

n = 5x = 5
ŷ =

3Solutions

1.Time value of money

Answer: FV = 2254 €

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 = 0.02/4 = 0.005
N = m·n = 4·6 = 24

Future value

FV = 2 000·(1 + 0.005)24 = 2 000·1.127 = €2 254.32

Answer

FV = €2 254.32
2.Time value of money

Answer: A = 145.9 €

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 = 0.036000000000000004/12 = 0.003
N = 12·5 = 60

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 = 0.8355
A = 8 000·0.003/(1 − 0.8355) = €145.89

Answer

A = €145.89
3.NPV & IRR

Answer: DPB = 4.522 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: 9 000/1.12 = 8 035.71 → Σ = 8 035.71
year 2: 9 000/1.254 = 7 174.74 → Σ = 15 210.46
year 3: 9 000/1.405 = 6 406.02 → Σ = 21 616.48
year 4: 9 000/1.574 = 5 719.66 → Σ = 27 336.14
year 5: 9 000/1.762 = 5 106.84 → Σ = 32 442.99

Fraction of the last year

DPB = 4 + 2 663.86/5 106.84 = 4.522

Answer

DPB = 4.522 years

The discounted payback is always longer than the simple payback (3.333 years).

4.NPV & IRR

Answer: IRR = 22.46 %

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(20%) = 1 805.56
NPV(25%) = −1 744.00

Narrow down

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

IRR ≈ 20 + 5·1 805.56/(1 805.56 − (−1 744.00)) = 22.54%
exact: NPV(22.46%) = 0 → IRR = 22.46%

Answer

IRR = 22.46%
5.Bond pricing & yield

Answer: P = 931.7 €

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 = €10.00, i = 0.0175, N = 10

Present value of the coupons

€10.00·(1 − 0.8407)/0.0175 = €91.01

Present value of the face value

€1 000·0.8407 = €840.73

Price

P = €931.74

Answer

P = €931.74

The yield is above the coupon rate, so the bond sells at a discount.

6.Bond pricing & yield

Answer: YTM = 9.562 %

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

€950 = €80·(1 − (1 + y)⁻⁴)/y + €1 000/(1 + y)⁴

Solve numerically

P(9%) = €967.60, P(10%) = €936.60
YTM = 9.562%

Answer

YTM = 9.562%
7.CAPM & WACC

Answer: WACC = 12.38 %

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 = €14 000 000, E/V = 0.8571, D/V = 0.1429

WACC

WACC = 0.8571·14% + 0.1429·3.5%·(1 − 0.25) = 12.38%

Answer

WACC = 12.38%
8.CAPM & WACC

Answer: WACC = 12.36 %

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 = 5 000 000·€56.50 = €282 500 000
D = 20 000·€980 = €19 600 000

Cost of equity (CAPM)

re = 2.5% + 1.5·7% = 13%

Weights

V = €302 100 000, E/V = 0.9351, D/V = 0.06488

WACC

WACC = 0.9351·13% + 0.06488·4%·(1 − 0.21) = 12.36%

Answer

WACC = 12.36%
9.Descriptive statistics

Answer: s = 16.64

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 = 244/7 = 34.86

Squared deviations

Σ(x − x̄)² = (−6.857)² + 16.14² + (−15.86)² + (−2.857)² + 11.14² + (−22.86)² + 21.14² = 1661

Standard deviation

s = √(1661/6) = 16.64

Answer

s = 16.64
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 = 9.734

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 = 10·0.15 + 20·0.35 + 30·0.3 + 40·0.2 = 25.5

Mean of the squares

E(X²) = 100·0.15 + 400·0.35 + 900·0.3 + 1600·0.2 = 745

Variance

Var(X) = 745 − 25.5² = 94.75
σ = √94.75 = 9.734

Answer

σ = 9.734
12.Probability

Answer: E(G) = 0.3 €

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: 3 (p = 0.3), 27 (p = 0.05), −3 (p = 0.65)

Expected value

E(G) = 3·0.3 + 27·0.05 − 3·0.65 = 0.3 €

Answer

E(G) = 0.3 €

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 = (480 − 500)/40 = −0.5
z₂ = (540 − 500)/40 = 1

Area between

P = Φ(1) − Φ(−0.5) = 0.8413 − 0.3085 = 0.5328

Answer

P = 0.5328
14.Normal distribution

Answer: x = 248.8 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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 250 − 0.25·5 = 248.8 g

Answer

x = 248.8 g
15.Confidence intervals

Answer: lower limit = 34.95

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 = 3.169

Margin of error

E = 3.169·10/√11 = 9.555

Interval

44.5 ± 9.555 → [34.95, 54.05]

Answer

lower limit = 34.95
16.Confidence intervals

Answer: n = 941

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 = 2.576²·0.15·0.85/0.03² = 940.1

Round up

n = 941

Answer

n = 941
17.Hypothesis tests & confidence intervals

Answer: t = 2.1

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/√64 = 8/8 = 1

Test statistic

t = (77.1 − 75)/1 = 2.1

Answer

t = 2.1
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: β = 30.22 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 24/√64 = 3
x̄c = 200 + 1.282·3 = 203.8

z under the true mean

z = (203.8 − 205.4)/3 = −0.518

Type II error

power = P(Z > −0.518) = 69.78%
β = 100% − 69.78% = 30.22%

Answer

β = 30.22%
19.Linear regression & correlation

Answer: r² = 0.9031

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 = 143/6 = 23.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −32.5
Syy = Σ(y − ȳ)² = 66.83

Coefficient of determination

r² = −32.5²/(17.5·66.83) = 0.9031

Answer

r² = 0.9031

About 90.31% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 12.2

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 = 30/5 = 6
ȳ = 71/5 = 14.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 80

Regression line

b = 80/40 = 2
a = 14.2 − 2·6 = 2.2
ŷ = 2.2 + 2·x

Prediction

ŷ = 2.2 + 2·5 = 12.2

Answer

ŷ = 12.2

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