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…12€2 000FV = ?

You invest €2 000 at 6% per year with continuous compounding. Find the value FV after 12 years.

PV = €2 000r = 6%t = 12
FV = €
2.Time value of moneyHard
months0123…336€300€300€300FV = ?

You save €300 at the end of every month for 28 years. The account pays 5.4% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €3005.4% / 12N = 336
FV = €
3.NPV & IRRMedium
years01234−75 00030 00031 00025 00019 000

An investment of €75 000 gives these cash flows at the end of years 1 to 4: €30 000, €31 000, €25 000, €19 000. The cost of capital is 6%. Find the net present value NPV.

r = 6%CF0 = −75 000CF1 = 30 000CF2 = 31 000CF3 = 25 000CF4 = 19 000
NPV = €
4.NPV & IRRHard
years012−25 00018 50016 000

A project costs €25 000 and returns €18 500, €16 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −25 000CF1 = 18 500CF2 = 16 000
IRR = %
5.Bond pricing & yieldMedium
years0123…14P€30€30€30 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 3% and matures in 14 years. Its yield to maturity changes from 3.5% to 2%. By how much ΔP does the bond price change?

F = €1 000coupon = 3%n = 14y1 = 3.5%y2 = 2%
ΔP = €
6.Bond pricing & yieldHard
years0123…9€935€60€60 + €1 000

A bond with a face value of €1 000 and an annual coupon of 6% matures in 9 years and sells for €935. Find its yield to maturity YTM in percent.

F = €1 000coupon = 6%n = 9P = €935
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€6 000 000re​ = 14%debt€4 000 000rd​ = 5%T = 28%

A company has equity worth €6 000 000 and debt worth €4 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 5% and the tax rate is 28%. Find the WACC.

E = €6 000 000D = €4 000 000re = 14%rd = 5%T = 28%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €31.00β = 1.2rf​ = 3.5%, rm​ = 9%debt40 000 × €960rd​ = 3%T = 20%

A company has 2 000 000 shares at €31.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 3% and the tax rate is 20%. The share’s beta is 1.2, the risk-free rate 3.5% and the market return 9%. Find the WACC.

β = 1.2rf = 3.5%rm = 9%rd = 3%T = 20%
WACC = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 43, 48, 14, 52, 23, 12, 22.

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

Exam scores are normally distributed with mean 60 and standard deviation 5. Find the z-score z of a score of 63.

μ = 60σ = 5x = 63
z =
11.ProbabilityMedium
n = 10k = 5N = ?

In how many ways N can a committee of 5 people be chosen from a group of 10? The order does not matter.

n = 10k = 5
N =
12.ProbabilityHard
n = 6, p = 0.1: P(X = 2) = ?k0123456

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 2 correct answers?

n = 6p = 0.1k = 2
P =
13.Normal distributionMedium
200 kg198.6 kgT = X₁ + … + X₈?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 8 of them is below 198.6 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 8< 198.6 kg
P =
14.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
n = 6, s = 3.5L = ?Ux̄ = 55E90%

A sample of 6 items has mean 55 and standard deviation s = 3.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 6x̄ = 55s = 3.590%t(5) = 2.015
lower limit =
16.Confidence intervalsHard
μ₁ − μ₂, t(20) = 2.0860LUx̄₁ − x̄₂E = ?95%

Two fertilisers are tested. Plot group 1: 10 plots, mean yield 21.5 kg, standard deviation 4 kg. Plot group 2: 12 plots, mean 15.5 kg, standard deviation 7 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 10x̄₁ = 21.5s₁ = 4n₂ = 12x̄₂ = 15.5s₂ = 795%t(20) = 2.086
E = kg
17.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 115 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 115s = 8n = 25
t =
18.Hypothesis tests & confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 299%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 20. Find the smallest sample size n.

σ = 20E = 299%
n =
19.Linear regression & correlationMedium
xy355457565665775xy466070

The table gives 5 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 = 5Σx = 25Σy = 317Σx² = 135Σy² = 20349Σxy = 1633
r =
20.Linear regression & correlationHard
xy337442546641753xy46404550x = 4.5

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (4.5, 37) is made. Find its residual e = y − ŷ.

n = 5(4.5, 37)
e =

3Solutions

1.Time value of money

Answer: FV = 4109 €

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.

Exponent

r·t = 0.06·12 = 0.72
e0.72 = 2.054

Future value

FV = 2 000·2.054 = €4 108.87

Answer

FV = €4 108.87
2.Time value of money

Answer: FV = 234696 €

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 = 0.054000000000000006/12 = 0.0045
N = 12·28 = 336

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 300·(4.52 − 1)/0.0045 = €234 695.65

Answer

FV = €234 695.65

You paid in €100 800 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = 16932 €

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 = −75 000
CF1/(1 + 0.06)1 = 30 000/1.06 = 28 301.89
CF2/(1 + 0.06)2 = 31 000/1.124 = 27 589.89
CF3/(1 + 0.06)3 = 25 000/1.191 = 20 990.48
CF4/(1 + 0.06)4 = 19 000/1.262 = 15 049.78

Add them up

NPV = €16 932.04

Answer

NPV = €16 932.04

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

4.NPV & IRR

Answer: IRR = 25.14 %

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(25%) = 40.00
NPV(30%) = −1 301.78

Narrow down

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

IRR ≈ 25 + 5·40.00/(40.00 − (−1 301.78)) = 25.15%
exact: NPV(25.14%) = 0 → IRR = 25.14%

Answer

IRR = 25.14%
5.Bond pricing & yield

Answer: ΔP = 175.7 €

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.

Price at the old yield

C = €30.00, i = 0.035, N = 14
€30.00·(1 − 0.6178)/0.035 + €1 000·0.6178 = €327.62 + €617.78 = €945.40

Price at the new yield

C = €30.00, i = 0.02, N = 14
€30.00·(1 − 0.7579)/0.02 + €1 000·0.7579 = €363.19 + €757.88 = €1 121.06

Change

ΔP = 1 121.06 − 945.40 = €175.67

Answer

ΔP = €175.67
6.Bond pricing & yield

Answer: YTM = 6.998 %

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

€935 = €60·(1 − (1 + y)⁻⁹)/y + €1 000/(1 + y)⁹

Solve numerically

P(6%) = €1 000.00, P(7%) = €934.85
YTM = 6.998%

Answer

YTM = 6.998%
7.CAPM & WACC

Answer: WACC = 9.84 %

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 = €10 000 000, E/V = 0.6, D/V = 0.4

WACC

WACC = 0.6·14% + 0.4·5%·(1 − 0.28) = 9.84%

Answer

WACC = 9.84%
8.CAPM & WACC

Answer: WACC = 7.155 %

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·€31.00 = €62 000 000
D = 40 000·€960 = €38 400 000

Cost of equity (CAPM)

re = 3.5% + 1.2·5.5% = 10.1%

Weights

V = €100 400 000, E/V = 0.6175, D/V = 0.3825

WACC

WACC = 0.6175·10.1% + 0.3825·3%·(1 − 0.2) = 7.155%

Answer

WACC = 7.155%
9.Descriptive statistics

Answer: s = 16.67

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 = 214/7 = 30.57

Squared deviations

Σ(x − x̄)² = 12.43² + 17.43² + (−16.57)² + 21.43² + (−7.571)² + (−18.57)² + (−8.571)² = 1668

Standard deviation

s = √(1668/6) = 16.67

Answer

s = 16.67
10.Descriptive statistics

Answer: z = 0.6

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 = (63 − 60)/5 = 0.6

Answer

z = 0.6
11.Probability

Answer: N = 252

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(10, 5) = (10·9·8·7·6)/(5·4·3·2·1) = 252

Answer

N = 252
12.Probability

Answer: P = 0.09842

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.

Binomial formula

P(X = 2) = 15·0.12·0.94 = 15·0.01·0.6561 = 0.09842

Answer

P = 0.09842
13.Normal distribution

Answer: P = 0.1611

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 = 200 kg, σT = 0.5·√8 = 1.414 kg

Standardise

z = (198.6 − 200)/1.414 = −0.99

Probability

P = Φ(−0.99) = 0.1611

Answer

P = 0.1611
14.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 difference

μ = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
15.Confidence intervals

Answer: lower limit = 52.12

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 = 5, t = 2.015

Margin of error

E = 2.015·3.5/√6 = 2.879

Interval

55 ± 2.879 → [52.12, 57.88]

Answer

lower limit = 52.12
16.Confidence intervals

Answer: E = 5.22 kg

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.

Pooled standard deviation

sp² = (9·4² + 11·7²)/20 = 34.15
sp = 5.844

Margin of error

E = 2.086·5.844·√(1/10 + 1/12) = 5.22

Answer

E = 5.22 kg
17.Hypothesis tests & confidence intervals

Answer: t = −3.125

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/√25 = 8/5 = 1.6

Test statistic

t = (115 − 120)/1.6 = −3.125

Answer

t = −3.125
t0.025; df = 24 = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 664

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 = (2.576·20/2)² = 663.6

Round up

n = 664

Answer

n = 664
19.Linear regression & correlation

Answer: r = 0.9577

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 = 1633 − 25·317/5 = 48
Sxx = 135 − 25²/5 = 10
Syy = 20349 − 317²/5 = 251.2

Correlation coefficient

r = 48/√(10·251.2) = 0.9577

Answer

r = 0.9577
20.Linear regression & correlation

Answer: e = −5.25

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 = 25/5 = 5
ȳ = 219/5 = 43.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 31

Regression line

b = 31/10 = 3.1
a = 43.8 − 3.1·5 = 28.3
ŷ = 28.3 + 3.1·x

Predicted value

ŷ = 28.3 + 3.1·4.5 = 42.25

Residual

e = 37 − 42.25 = −5.25

Answer

e = −5.25

The point lies below the line.

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