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…9PV = ?€12 000+3%+3%…r = 5%

A property is expected to earn net rent of €12 000 next year. The rent is expected to grow by 3% per year forever. The required return is 5% per year. Find the value PV of the property today.

C1 = €12 000g = 3%r = 5%
PV = €
2.Time value of moneyHard
months0123…324€500€500€500FV = ?

You save €500 at the end of every month for 27 years. The account pays 2.4% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €5002.4% / 12N = 324
FV = €
3.NPV & IRRMedium
years01234−20 0004 0008 00010 0007 000

An investment of €20 000 gives these cash flows at the end of years 1 to 4: €4 000, €8 000, €10 000, €7 000. The cost of capital is 7.000000000000001%. Find the net present value NPV.

r = 7.000000000000001%CF0 = −20 000CF1 = 4 000CF2 = 8 000CF3 = 10 000CF4 = 7 000
NPV = €
4.NPV & IRRHard
years012−20 00013 50012 500

A project costs €20 000 and returns €13 500, €12 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 13 500CF2 = 12 500
IRR = %
5.Bond pricing & yieldMedium
years01234567€950C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 7 years and pays annual coupons. It sells for €950 when the yield to maturity is 9%. Find the coupon rate c in percent.

F = €1 000n = 7P = €950YTM = 9%
c = %
6.Bond pricing & yieldHard
years0123€1 045€30€30 + €1 000

A bond with a face value of €1 000 and an annual coupon of 3% matures in 3 years and sells for €1 045. Find its yield to maturity YTM in percent.

F = €1 000coupon = 3%n = 3P = €1 045
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€5 000 000re​ = 14%preferred shares€500 000rp​ = 6%debt€6 000 000rd​ = 4%T = 28%

A company is financed with ordinary shares worth €5 000 000, preferred shares worth €500 000 and debt worth €6 000 000. The cost of equity is 14%, the cost of preferred stock 6% and the pre-tax cost of debt 4%. The tax rate is 28%. Find the WACC.

E = €5 000 000Ps = €500 000D = €6 000 000re = 14%rp = 6%rd = 4%T = 28%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €33.50β = 1.3rf​ = 1.5%, rm​ = 5.5%debt5 000 × €1 000rd​ = 4%T = 25%

A company has 2 500 000 shares at €33.50 each and 5 000 bonds trading at €1 000 each. The pre-tax cost of debt is 4% and the tax rate is 25%. The share’s beta is 1.3, the risk-free rate 1.5% and the market return 5.5%. Find the WACC.

β = 1.3rf = 1.5%rm = 5.5%rd = 4%T = 25%
WACC = %
9.Descriptive statisticsMedium
2530354045

Find the sample standard deviation s of the data: 43, 22, 38, 45, 25.

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

Exam scores are normally distributed with mean 50 and standard deviation 12. Find the z-score z of a score of 26.

μ = 50σ = 12x = 26
z =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7P(six) = 1/6 on each roll

A fair die is rolled 7 times. What is the probability P of getting at least one six?

n = 7
P =
12.ProbabilityHard
xp(x)00.3510.320.35μ

X has the distribution shown, and Y = 2X + 10. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 2X + 10
σY =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9783

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9783. Find z to two decimals.

A = 0.9783
z =
14.Normal distributionHard
μ = 65x = ?97.5%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 65σ = 10P = 97.5%
x = points
15.Confidence intervalsMedium
x̄E₁ = 9.290%, n = 64x̄E₂ = ?95%, n = 64

A 90% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 9.2. What is the margin of error E₂ of a 95% interval from the same data?

90%n₁ = 64E₁ = 9.295%
E₂ =
16.Confidence intervalsHard
σ = 29, n = ?x̄ − Ex̄ + Ex̄E = 599%

The population standard deviation is 29. 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 5?

σ = 29E = 599%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.2282.228t = ?df = 10

11 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 2.75 and the standard deviation of the differences is 2.4. Find the paired test statistic t for H₀: μd = 0.

n = 11d̄ = 2.75sd = 2.4
t =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 80 against H₁: μ < 80 is done at α = 0.01 with a sample of 64; the standard deviation is known to be 20. In reality the mean is 70.5. Find the power 1 − β of the test in percent.

μ₀ = 80μ = 70.5σ = 20n = 64α = 0.01
1 − β = %
19.Linear regression & correlationMedium
xy12221931541151166xy2465101520

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

n = 6Σx = 21Σy = 84Σx² = 91Σy² = 1348Σxy = 240
r =
20.Linear regression & correlationHard
xy17611976218023932599xy20256080100x = 20

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 20.

n = 5x = 20
ŷ =

3Solutions

1.Time value of money

Answer: PV = 600000 €

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.

Growing perpetuity

PV = C1/(r − g) = 12 000/(0.05 − 0.03) = 12 000/0.02 = €600 000.00

Answer

PV = €600 000.00
2.Time value of money

Answer: FV = 227619 €

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 = 0.024/12 = 0.002
N = 12·27 = 324

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 500·(1.91 − 1)/0.002 = €227 619.21

Answer

FV = €227 619.21

You paid in €162 000 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = 4229 €

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 + 0.07)1 = 4 000/1.07 = 3 738.32
CF2/(1 + 0.07)2 = 8 000/1.145 = 6 987.51
CF3/(1 + 0.07)3 = 10 000/1.225 = 8 162.98
CF4/(1 + 0.07)4 = 7 000/1.311 = 5 340.27

Add them up

NPV = €4 229.07

Answer

NPV = €4 229.07

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

4.NPV & IRR

Answer: IRR = 19.71 %

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%) = 1 190.93
NPV(20%) = −69.44

Narrow down

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

IRR ≈ 15 + 5·1 190.93/(1 190.93 − (−69.44)) = 19.72%
exact: NPV(19.71%) = 0 → IRR = 19.71%

Answer

IRR = 19.71%
5.Bond pricing & yield

Answer: c = 8.007 %

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 + 0.09)−7 = €547.03

Coupon

C = (950 − 547.03)/5.033 = €80.07

Coupon rate

c = 80.07/1000 = 8.007%

Answer

c = 8.007%
6.Bond pricing & yield

Answer: YTM = 1.456 %

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

€1 045 = €30·(1 − (1 + y)⁻³)/y + €1 000/(1 + y)³

Solve numerically

P(1%) = €1 058.82, P(2%) = €1 028.84
YTM = 1.456%

Answer

YTM = 1.456%
7.CAPM & WACC

Answer: WACC = 7.85 %

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 = 5 000 000 + 500 000 + 6 000 000 = €11 500 000
E/V = 0.4348, Ps/V = 0.04348, D/V = 0.5217

WACC

WACC = 0.4348·14% + 0.04348·6% + 0.5217·4%·(1 − 0.28) = 7.85%

Answer

WACC = 7.85%
8.CAPM & WACC

Answer: WACC = 6.492 %

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 = 2 500 000·€33.50 = €83 750 000
D = 5 000·€1 000 = €5 000 000

Cost of equity (CAPM)

re = 1.5% + 1.3·4% = 6.7%

Weights

V = €88 750 000, E/V = 0.9437, D/V = 0.05634

WACC

WACC = 0.9437·6.7% + 0.05634·4%·(1 − 0.25) = 6.492%

Answer

WACC = 6.492%
9.Descriptive statistics

Answer: s = 10.5

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.

Mean

x̄ = 173/5 = 34.6

Squared deviations

Σ(x − x̄)² = 8.4² + (−12.6)² + 3.4² + 10.4² + (−9.6)² = 441.2

Standard deviation

s = √(441.2/4) = 10.5

Answer

s = 10.5
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 = (26 − 50)/12 = −2

Answer

z = −2
11.Probability

Answer: P = 0.7209

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.

Complement

P(no six) = (5/6)7 = 0.2791
P = 1 − 0.2791 = 0.7209

Answer

P = 0.7209
12.Probability

Answer: σY = 1.673

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.

Spread of X

μ = 1, Var(X) = 0.7, σX = 0.8367

Linear change

σY = |2|·0.8367 = 1.673

Answer

σY = 1.673

The mean does move: E(Y) = 2·1 + 10 = 12.

13.Normal distribution

Answer: z = −2.02

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 − 0.9783 = 0.0217

Read the z-table backwards

Find 0.0217 inside the table and read off z.

z = −2.02

Answer

z = −2.02
14.Normal distribution

Answer: x = 84.6 points

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 for the percentile

Φ(z) = 0.975 → z = 1.96

Back to x

x = 65 + 1.96·10 = 84.6 points

Answer

x = 84.6 points
15.Confidence intervals

Answer: E₂ = 10.96

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 = (1.96/1.645)·√(64/64) = 1.191

New margin

E₂ = 9.2·1.191 = 10.96

Answer

E₂ = 10.96
16.Confidence intervals

Answer: n = 224

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 = (2.576·29/5)² = 223.2

Round up

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

n = 224

Answer

n = 224
17.Hypothesis tests & confidence intervals

Answer: t = 3.8

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).

Standard error of the mean difference

SE = 2.4/√11 = 0.7236

Test statistic

t = 2.75/0.7236 = 3.8

Answer

t = 3.8

With df = 10, the two-sided 5% critical value is 2.228; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: 1 − β = 92.98 %

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 = 20/√64 = 2.5
x̄c = 80 − 2.326·2.5 = 74.19

z under the true mean

z = (74.19 − 70.5)/2.5 = 1.474

Power

power = P(Z < 1.474) = 92.98%
1 − β = 92.98%

Answer

1 − β = 92.98%
19.Linear regression & correlation

Answer: r = −0.9843

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 = 240 − 21·84/6 = −54
Sxx = 91 − 21²/6 = 17.5
Syy = 1348 − 84²/6 = 172

Correlation coefficient

r = −54/√(17.5·172) = −0.9843

Answer

r = −0.9843
20.Linear regression & correlation

Answer: ŷ = 77.15

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 = 105/5 = 21
ȳ = 409/5 = 81.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 186

Regression line

b = 186/40 = 4.65
a = 81.8 − 4.65·21 = −15.85
ŷ = −15.85 + 4.65·x

Prediction

ŷ = −15.85 + 4.65·20 = 77.15

Answer

ŷ = 77.15

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