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 = ?€8 000+3%+3%…r = 8%

A property is expected to earn net rent of €8 000 next year. The rent is expected to grow by 3% per year forever. The required return is 8% per year. Find the value PV of the property today.

C1 = €8 000g = 3%r = 8%
PV = €
2.Time value of moneyHard
months0123…60€15 000A?AAA

A loan of €15 000 is repaid with equal monthly payments over 5 years. The interest rate is 4.2% per year, compounded monthly. Find the monthly payment A.

P = €15 0004.2% / 12N = 60
A = €
3.NPV & IRRMedium
years0123…8−80 00011 50011 50011 50019 500

A machine costs €80 000 today. It generates €11 500 at the end of each year for 8 years and can then be sold for €8 000. The cost of capital is 6%. Find the net present value NPV.

C0 = €80 000CF = €11 500S = €8 000n = 8r = 6%
NPV = €
4.NPV & IRRHard
years012−20 00012 50013 500

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

CF0 = −20 000CF1 = 12 500CF2 = 13 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…18€40€40€40 + €1 000P = ?

A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 9 years. The yield to maturity is 7% per year. Find the price P of the bond.

F = €1 000coupon = 8%YTM = 7%n = 9
P = €
6.Bond pricing & yieldHard
years0123456€1 261€90€90€90 + €1 090

A bond with 18 years to maturity (face value €1 000, annual coupon 9%) sells for €1 261. The issuer can call it in 6 years at a call price of €1 090. Find the yield to call YTC in percent.

F = €1 000coupon = 9%P = €1 261call price = €1 090call in 6 years
YTC = %
7.CAPM & WACCMedium
WACC = ?equityEre​ = 11%debtD = 0.4·Erd​ = 5%T = 25%

A company has a debt-to-equity ratio (at market values) of 0.4. Its cost of equity is 11%, its pre-tax cost of debt is 5% and the tax rate is 25%. Find the WACC.

D/E = 0.4re = 11%rd = 5%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €47.50β = 1.5rf​ = 2%, rm​ = 8.5%debt40 000 × €1 050rd​ = 6%T = 25.8%

A company has 2 000 000 shares at €47.50 each and 40 000 bonds trading at €1 050 each. The pre-tax cost of debt is 6% and the tax rate is 25.8%. The share’s beta is 1.5, the risk-free rate 2% and the market return 8.5%. Find the WACC.

β = 1.5rf = 2%rm = 8.5%rd = 6%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 54, 15, 32, 50, 33, 11.

n = 6
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 70 and standard deviation 5. What proportion P of values is below 66?

μ = 70σ = 5x = 66
P =
11.ProbabilityMedium
n = 7k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 7? The order does not matter.

n = 7k = 4
N =
12.ProbabilityHard
xp(x)00.310.3520.35μ

X has the distribution shown, and Y = 10X − 3. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 10X − 3
σY =
13.Normal distributionMedium
72 kg70 kgT = X₁ + … + X₆?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 6 of them exceeds 70 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 6> 70 kg
P =
14.Normal distributionHard
157164171178185192199cmbottom 30%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 30% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm30%
x = cm
15.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?99%

The battery life of two phone models is compared. Model 1: 30 phones, mean 94 h, known standard deviation 7 h. Model 2: 60 phones, mean 98 h, known standard deviation 9 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 94 hσ₁ = 7 hn₂ = 60x̄₂ = 98 hσ₂ = 9 h99%
E = h
16.Confidence intervalsHard
p ≈ 0.7, n = ?p̂ − Ep̂ + Ep̂E = 0.0495%

A pilot study suggests that about 0.7 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.04?

p ≈ 0.7E = 0.0495%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 201.5 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 201.5s = 5n = 49
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 30 of 150 patients on treatment A recover, and 24 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 30n₁ = 150x₂ = 24n₂ = 100
z =
19.Linear regression & correlationMedium
xy2741261782310321236xy510102030

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 127Σx² = 364Σy² = 3331Σxy = 1100
r =
20.Linear regression & correlationHard
xy22743464185310541265xy5104060ŷ = 53

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 53?

n = 6ŷ = 53
x =

3Solutions

1.Time value of money

Answer: PV = 160000 €

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) = 8 000/(0.08 − 0.03) = 8 000/0.05 = €160 000.00

Answer

PV = €160 000.00
2.Time value of money

Answer: A = 277.6 €

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 = 0.042/12 = 0.0035
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 = 0.8109
A = 15 000·0.0035/(1 − 0.8109) = €277.60

Answer

A = €277.60
3.NPV & IRR

Answer: NPV = −3568 €

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 the yearly cash flows

11 500·6.21 = 71 412.63

Present value of the salvage value

8 000/1.594 = 5 019.30

NPV

NPV = −80 000 + 71 412.63 + 5 019.30 = −€3 568.07

Answer

NPV = −€3 568.07
4.NPV & IRR

Answer: IRR = 19.15 %

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%) = 1 077.50
NPV(20%) = −208.33

Narrow down

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

IRR ≈ 15 + 5·1 077.50/(1 077.50 − (−208.33)) = 19.19%
exact: NPV(19.15%) = 0 → IRR = 19.15%

Answer

IRR = 19.15%
5.Bond pricing & yield

Answer: P = 1066 €

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 = €40.00, i = 0.035, N = 18

Present value of the coupons

€40.00·(1 − 0.5384)/0.035 = €527.59

Present value of the face value

€1 000·0.5384 = €538.36

Price

P = €1 065.95

Answer

P = €1 065.95

The yield is below the coupon rate, so the bond sells at a premium.

6.Bond pricing & yield

Answer: YTC = 5.151 %

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

1261 = 90·(1 − (1 + y)−6)/y + 1090/(1 + y)6

Solve numerically

y = 5%: €1 270.19
y = 6%: €1 210.97
YTC = 5.151%

Answer

YTC = 5.151%

The yield to maturity is 6.498%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.

7.CAPM & WACC

Answer: WACC = 8.929 %

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 from D/E

E/V = 1/(1 + 0.4) = 0.7143
D/V = 0.4/(1 + 0.4) = 0.2857

WACC

WACC = 0.7143·11% + 0.2857·5%·(1 − 0.25) = 8.929%

Answer

WACC = 8.929%
8.CAPM & WACC

Answer: WACC = 9.513 %

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 = 2 000 000·€47.50 = €95 000 000
D = 40 000·€1 050 = €42 000 000

Cost of equity (CAPM)

re = 2% + 1.5·6.5% = 11.75%

Weights

V = €137 000 000, E/V = 0.6934, D/V = 0.3066

WACC

WACC = 0.6934·11.75% + 0.3066·6%·(1 − 0.258) = 9.513%

Answer

WACC = 9.513%
9.Descriptive statistics

Answer: s = 17.54

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 = 195/6 = 32.5

Squared deviations

Σ(x − x̄)² = 21.5² + (−17.5)² + (−0.5)² + 17.5² + 0.5² + (−21.5)² = 1538

Standard deviation

s = √(1538/5) = 17.54

Answer

s = 17.54
10.Descriptive statistics

Answer: P = 0.2119

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 0.2119

Answer

P = 0.2119
11.Probability

Answer: N = 35

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.

Combinations

C(7, 4) = (7·6·5·4)/(4·3·2·1) = 35

Answer

N = 35
12.Probability

Answer: σY = 8.047

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 = 1.05, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |10|·0.8047 = 8.047

Answer

σY = 8.047

The mean does move: E(Y) = 10·1.05 − 3 = 7.5.

13.Normal distribution

Answer: P = 0.8461

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 total

μT = 72 kg, σT = 0.8·√6 = 1.96 kg

Standardise

z = (70 − 72)/1.96 = −1.02

Probability

P = 1 − Φ(−1.02) = 0.8461

Answer

P = 0.8461
14.Normal distribution

Answer: x = 174.4 cm

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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 178 − 0.52·7 = 174.4 cm

Answer

x = 174.4 cm
15.Confidence intervals

Answer: E = 4.449 h

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Point estimate

x̄₁ − x̄₂ = 94 − 98 = −4

Standard error

SE = √(7²/30 + 9²/60) = 1.727

Margin of error

E = 2.576·1.727 = 4.449

Answer

E = 4.449 h

The interval contains 0, so the data do not show a significant difference.

16.Confidence intervals

Answer: n = 505

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 = 1.96²·0.7·0.3/0.04² = 504.2

Round up

n = 505

Answer

n = 505
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 = 5/√49 = 5/7 = 0.7143

Test statistic

t = (201.5 − 200)/0.7143 = 2.1

Answer

t = 2.1
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = −0.7529

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.

Sample proportions

p̂₁ = 30/150 = 0.2
p̂₂ = 24/100 = 0.24
p̂ = 54/250 = 0.216

Standard error under H₀

SE = √(0.216·0.784·(1/150 + 1/100)) = 0.05313

Test statistic

z = (0.2 − 0.24)/0.05313 = −0.7529

Answer

z = −0.7529

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

19.Linear regression & correlation

Answer: r = 0.9947

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 = 1100 − 42·127/6 = 211
Sxx = 364 − 42²/6 = 70
Syy = 3331 − 127²/6 = 642.8

Correlation coefficient

r = 211/√(70·642.8) = 0.9947

Answer

r = 0.9947
20.Linear regression & correlation

Answer: x = 8.959

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 = 274/6 = 45.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 262

Regression line

b = 262/70 = 3.743
a = 45.67 − 3.743·7 = 19.47
ŷ = 19.47 + 3.743·x

Solve for x

x = (53 − 19.47)/3.743 = 8.959

Answer

x = 8.959

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