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 = ?€5 000+2%+2%…r = 4%

A property is expected to earn net rent of €5 000 next year. The rent is expected to grow by 2% per year forever. The required return is 4% per year. Find the value PV of the property today.

C1 = €5 000g = 2%r = 4%
PV = €
2.Time value of moneyHard
months0123…72€25 000A?AAA

A loan of €25 000 is repaid with equal monthly payments over 6 years. The interest rate is 7.2% per year, compounded monthly. Find the monthly payment A.

P = €25 0007.2% / 12N = 72
A = €
3.NPV & IRRMedium
years0123−60 00015 00023 00019 000

An investment of €60 000 gives these cash flows at the end of years 1 to 3: €15 000, €23 000, €19 000. The cost of capital is 7.000000000000001%. Find the net present value NPV.

r = 7.000000000000001%CF0 = −60 000CF1 = 15 000CF2 = 23 000CF3 = 19 000
NPV = €
4.NPV & IRRHard
years0123−20 00011 50013 50013 000

A project costs €20 000 and returns €11 500, €13 500, €13 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 11 500CF2 = 13 500CF3 = 13 000
IRR = %
5.Bond pricing & yieldMedium
years01234567€1 349C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 7 years and pays annual coupons. It sells for €1 349 when the yield to maturity is 2.5%. Find the coupon rate c in percent.

F = €1 000n = 7P = €1 349YTM = 2.5%
c = %
6.Bond pricing & yieldHard
half-years0123…24€1 258€25€25€25 + €1 000

A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 12 years. It sells for €1 258. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 5% (2×)n = 12P = €1 258
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€6 000 000re​ = 10%preferred shares€1 500 000rp​ = 7.5%debt€6 000 000rd​ = 3%T = 25%

A company is financed with ordinary shares worth €6 000 000, preferred shares worth €1 500 000 and debt worth €6 000 000. The cost of equity is 10%, the cost of preferred stock 7.5% and the pre-tax cost of debt 3%. The tax rate is 25%. Find the WACC.

E = €6 000 000Ps = €1 500 000D = €6 000 000re = 10%rp = 7.5%rd = 3%T = 25%
WACC = %
8.CAPM & WACCHard
rf​ = 3.5%, rm​ − rf​ = 5.5%D/E = 0.5equityβL​ = 1.5debtD/E = 1equityre​ = ?debtT = 25%

A company's shares have a beta of 1.5 at its current debt-to-equity ratio of 0.5. It plans to raise its debt-to-equity ratio to 1. The tax rate is 25%, the risk-free rate 3.5% and the market risk premium 5.5%. Find the cost of equity re after the change.

βL = 1.5D/E1 = 0.5D/E2 = 1T = 25%rf = 3.5%rm − rf = 5.5%
re = %
9.Descriptive statisticsMedium
253035404550

Find the sample standard deviation s of the data: 43, 34, 25, 22, 54.

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

A quantity is normally distributed with mean 50 and standard deviation 15. What proportion P of values is above 20?

μ = 50σ = 15x = 20
P =
11.ProbabilityMedium
xp(x)10.320.230.1540.35μ = 2.55

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

x: 1, 2, 3, 4p(x): 0.3, 0.2, 0.15, 0.35
σ =
12.ProbabilityHard
n = 7, p = 0.25: P(X ≤ 1) = ?k01234567

7 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?

n = 7p = 0.25
P =
13.Normal distributionMedium
32 min30 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 30 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 30 min
P =
14.Normal distributionHard
380420460500540580620hbottom 25%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 25% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h25%
x = h
15.Confidence intervalsMedium
x̄E₁ = 7.595%, n = 50x̄E₂ = ?90%, n = 450

A 95% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 7.5. What is the margin of error E₂ of a 90% interval based on 450 observations?

95%n₁ = 50E₁ = 7.590%n₂ = 450
E₂ =
16.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0199%

Nothing is known in advance about the proportion of defective products. 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 = 0.5E = 0.0199%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 50.96 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 50.96s = 4n = 100
t =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 5α = 0.05βσ = 27, power 90%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 5 units with a power of 90%. The standard deviation is 27. Find the minimum sample size n.

α = 0.05power = 90%δ = 5σ = 27zα = 1.645zβ = 1.282
n =
19.Linear regression & correlationMedium
xy148255353456567667xy2465060

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy2741161982310321239xy51010203040ŷ = 35

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 35?

n = 6ŷ = 35
x =

3Solutions

1.Time value of money

Answer: PV = 250000 €

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) = 5 000/(0.04 − 0.02) = 5 000/0.02 = €250 000.00

Answer

PV = €250 000.00
2.Time value of money

Answer: A = 428.6 €

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 = 0.07200000000000001/12 = 0.006
N = 12·6 = 72

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 = 0.65
A = 25 000·0.006/(1 − 0.65) = €428.63

Answer

A = €428.63
3.NPV & IRR

Answer: NPV = −10383 €

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 = −60 000
CF1/(1 + 0.07)1 = 15 000/1.07 = 14 018.69
CF2/(1 + 0.07)2 = 23 000/1.145 = 20 089.09
CF3/(1 + 0.07)3 = 19 000/1.225 = 15 509.66

Add them up

NPV = €−10 382.56

Answer

NPV = €−10 382.56

NPV is negative, so the project earns less than the required return.

4.NPV & IRR

Answer: IRR = 39.38 %

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(35%) = 1 209.67
NPV(40%) = −160.35

Narrow down

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

IRR ≈ 35 + 5·1 209.67/(1 209.67 − (−160.35)) = 39.41%
exact: NPV(39.38%) = 0 → IRR = 39.38%

Answer

IRR = 39.38%
5.Bond pricing & yield

Answer: c = 7.997 %

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 + 0.025)−7 = €841.27

Coupon

C = (1349 − 841.27)/6.349 = €79.97

Coupon rate

c = 79.97/1000 = 7.997%

Answer

c = 7.997%
6.Bond pricing & yield

Answer: YTM = 2.498 %

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 = €25.00, N = 24

Solve numerically

i = 1.249% per half-year

YTM

YTM = 2·1.249% = 2.498%

Answer

YTM = 2.498%
7.CAPM & WACC

Answer: WACC = 6.278 %

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 + 1 500 000 + 6 000 000 = €13 500 000
E/V = 0.4444, Ps/V = 0.1111, D/V = 0.4444

WACC

WACC = 0.4444·10% + 0.1111·7.5% + 0.4444·3%·(1 − 0.25) = 6.278%

Answer

WACC = 6.278%
8.CAPM & WACC

Answer: re = 14 %

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 = 1.5/(1 + 0.75·0.5) = 1.091

Relever at the new D/E

βL = 1.091·(1 + 0.75·1) = 1.909

Cost of equity (CAPM)

re = 3.5% + 1.909·5.5% = 14%

Answer

re = 14%
9.Descriptive statistics

Answer: s = 13.16

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 = 178/5 = 35.6

Squared deviations

Σ(x − x̄)² = 7.4² + (−1.6)² + (−10.6)² + (−13.6)² + 18.4² = 693.2

Standard deviation

s = √(693.2/4) = 13.16

Answer

s = 13.16
10.Descriptive statistics

Answer: P = 0.9772

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
11.Probability

Answer: σ = 1.244

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 = 1·0.3 + 2·0.2 + 3·0.15 + 4·0.35 = 2.55

Mean of the squares

E(X²) = 1·0.3 + 4·0.2 + 9·0.15 + 16·0.35 = 8.05

Variance

Var(X) = 8.05 − 2.55² = 1.548
σ = √1.548 = 1.244

Answer

σ = 1.244
12.Probability

Answer: P = 0.4449

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) = 0.757 = 0.1335
P(X = 1) = 7·0.25·0.756 = 0.3115

Add them

P = 0.1335 + 0.3115 = 0.4449

Answer

P = 0.4449
13.Normal distribution

Answer: P = 0.7881

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 sum

μ = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (30 − 32)/2.5 = −0.8

Upper tail

P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
14.Normal distribution

Answer: x = 473 h

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) = 0.25

z from the table

z ≈ −0.67

Back to the original units

x = 500 − 0.67·40 = 473 h

Answer

x = 473 h
15.Confidence intervals

Answer: E₂ = 2.098

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 = (1.645/1.96)·√(50/450) = 0.2798

New margin

E₂ = 7.5·0.2798 = 2.098

Answer

E₂ = 2.098
16.Confidence intervals

Answer: n = 16590

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 = 2.576²·0.5·0.5/0.01² = 16589

Round up

n = 16590

Answer

n = 16590
17.Hypothesis tests & confidence intervals

Answer: t = 2.4

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 = 4/√100 = 4/10 = 0.4

Test statistic

t = (50.96 − 50)/0.4 = 2.4

Answer

t = 2.4
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 250

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 1.282)·27/5)² = 249.8

Round up

n = 250

Answer

n = 250
19.Linear regression & correlation

Answer: r = 0.9257

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 = 346/6 = 57.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 67
Syy = Σ(y − ȳ)² = 299.3

Correlation coefficient

r = 67/√(17.5·299.3) = 0.9257

Answer

r = 0.9257
20.Linear regression & correlation

Answer: x = 11.06

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 = 131/6 = 21.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 227

Regression line

b = 227/70 = 3.243
a = 21.83 − 3.243·7 = −0.8667
ŷ = −0.8667 + 3.243·x

Solve for x

x = (35 − (−0.8667))/3.243 = 11.06

Answer

x = 11.06

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