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
years012€2 000FV = ?

You invest €2 000 at a nominal rate of 4% per year, compounded quarterly. Find the value FV of the investment after 2 years.

PV = €2 0004% / 4n = 2
FV = €
2.Time value of moneyHard
months0123…72€10 000AAAA7.2% / 12, I = ?

A loan of €10 000 is repaid with equal monthly payments over 6 years at 7.2% per year, compounded monthly. How much interest I is paid in total over the life of the loan?

P = €10 0007.2% / 12N = 72
I = €
3.NPV & IRRMedium
years0123…8−40 0006 0006 0006 00014 000

A machine costs €40 000 today. It generates €6 000 at the end of each year for 8 years and can then be sold for €8 000. The cost of capital is 8%. Find the net present value NPV.

C0 = €40 000CF = €6 000S = €8 000n = 8r = 8%
NPV = €
4.NPV & IRRHard
years0123−20 0009 5009 5009 500

An investment of €20 000 returns €9 500 at the end of each year for 3 years. Find its internal rate of return IRR in percent.

C0 = €20 000CF = €9 500n = 3
IRR = %
5.Bond pricing & yieldMedium
half-years0123…8€25€25€25 + €1 000P = ?

A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 4 years. The yield to maturity is 6% per year. Find the price P of the bond.

F = €1 000coupon = 5%YTM = 6%n = 4
P = €
6.Bond pricing & yieldHard
half-years0123…12€924€20€20€20 + €1 000

A bond with a face value of €1 000 pays a 4% annual coupon in two half-yearly payments and matures in 6 years. It sells for €924. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 4% (2×)n = 6P = €924
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€5 000 000re​ = 11%debt€2 000 000rd​ = 6%T = 25.8%

A company has equity worth €5 000 000 and debt worth €2 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 6% and the tax rate is 25.8%. Find the WACC.

E = €5 000 000D = €2 000 000re = 11%rd = 6%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €44.50D0​ = €2.70g = 3%debt€44 500 000rd​ = 4.5%T = 25%

A company has 2 000 000 shares trading at €44.50; the book value of its equity is €53 400 000. It just paid a dividend of €2.70 per share, and dividends grow by 3% per year. Its debt has a market value of €44 500 000 and a pre-tax cost of 4.5%. The tax rate is 25%. Find the WACC, using the dividend growth model for the cost of equity.

N = 2 000 000P0 = €44.50book equity = €53 400 000D0 = €2.70g = 3%D = €44 500 000rd = 4.5%T = 25%
WACC = %
9.Descriptive statisticsMedium
101520253035404550

Find the sample standard deviation s of the data: 10, 19, 50, 39, 19, 43, 32.

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

A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is below 180?

μ = 170σ = 10x = 180
P =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4P(six) = 1/6 on each roll

A fair die is rolled 4 times. What is the probability P of getting at least one six?

n = 4
P =
12.ProbabilityHard
€p(x)−50.65150.3450.05

A game costs €5 to play. You win €20 with probability 0.3 and €50 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€20: p = 0.3€50: p = 0.05
E(G) = €
13.Normal distributionMedium
250.25260P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 250.25 and 260 g? Give P as a decimal.

μ = 250σ = 5x1 = 250.25x2 = 260
P =
14.Normal distributionHard
μ = 32x = ?95%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 95% of them lie (the 95th percentile).

μ = 32σ = 6P = 95%
x = min
15.Confidence intervalsMedium
n = 10, sd​ = 6.80L = ?Ud̄ = 3.7E99%

The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 3.7 and standard deviation 6.8. Find the lower limit of the 99% confidence interval for the mean difference.

n = 10d̄ = 3.7sd = 6.899%t(9) = 3.25
lower limit = mmHg
16.Confidence intervalsHard
σ = 17, n = ?x̄ − Ex̄ + Ex̄E = 490%

The population standard deviation is 17. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 4?

σ = 17E = 490%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 122.9 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 122.9s = 5n = 25
t =
18.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 64 measurements has mean 120.8. The population standard deviation is 8. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 120.8σ = 8n = 6499%
L =
19.Linear regression & correlationMedium
xy16521860205822732483xy2050607080

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 100Σy = 326Σx² = 2040Σy² = 21886Σxy = 6670
r =
20.Linear regression & correlationHard
xy125221321419514610xy24610152025ŷ = 17

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 17?

n = 6ŷ = 17
x =

3Solutions

1.Time value of money

Answer: FV = 2166 €

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 = 0.04/4 = 0.01
N = m·n = 4·2 = 8

Future value

FV = 2 000·(1 + 0.01)8 = 2 000·1.083 = €2 165.71

Answer

FV = €2 165.71
2.Time value of money

Answer: I = 2345 €

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 payment

i = 0.006, N = 72
A = 10 000·0.006/(1 − (1 + 0.006)−72) = €171.45

Total interest

I = N·A − P = 72·171.5 − 10 000 = €2 344.55

Answer

I = €2 344.55
3.NPV & IRR

Answer: NPV = −1198 €

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

6 000·5.747 = 34 479.83

Present value of the salvage value

8 000/1.851 = 4 322.15

NPV

NPV = −40 000 + 34 479.83 + 4 322.15 = −€1 198.02

Answer

NPV = −€1 198.02
4.NPV & IRR

Answer: IRR = 20.04 %

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.

Required annuity factor

(1 − (1 + i)−3)/i = 20 000/9 500 = 2.105

Solve for the rate

The annuity factor falls as the rate rises; bracket the rate and narrow down.

i = 20%: 2.106
i = 21%: 2.074
IRR = 20.04%

Answer

IRR = 20.04%
5.Bond pricing & yield

Answer: P = 964.9 €

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, i = 0.03, N = 8

Present value of the coupons

€25.00·(1 − 0.7894)/0.03 = €175.49

Present value of the face value

€1 000·0.7894 = €789.41

Price

P = €964.90

Answer

P = €964.90

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

6.Bond pricing & yield

Answer: YTM = 5.505 %

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 = €20.00, N = 12

Solve numerically

i = 2.752% per half-year

YTM

YTM = 2·2.752% = 5.505%

Answer

YTM = 5.505%
7.CAPM & WACC

Answer: WACC = 9.129 %

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 = €7 000 000, E/V = 0.7143, D/V = 0.2857

WACC

WACC = 0.7143·11% + 0.2857·6%·(1 − 0.258) = 9.129%

Answer

WACC = 9.129%
8.CAPM & WACC

Answer: WACC = 7.291 %

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).

Cost of equity

D1 = 2.7·1.03 = €2.78
re = 2.781/44.5 + 0.03 = 9.249%

Market values

E = 2 000 000·€44.50 = €89 000 000
V = €133 500 000, E/V = 0.6667, D/V = 0.3333

WACC

WACC = 0.6667·9.249% + 0.3333·4.5%·(1 − 0.25) = 7.291%

Answer

WACC = 7.291%
9.Descriptive statistics

Answer: s = 14.69

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 = 212/7 = 30.29

Squared deviations

Σ(x − x̄)² = (−20.29)² + (−11.29)² + 19.71² + 8.714² + (−11.29)² + 12.71² + 1.714² = 1295

Standard deviation

s = √(1295/6) = 14.69

Answer

s = 14.69
10.Descriptive statistics

Answer: P = 0.8413

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

Standard normal table

Φ(1) = 0.8413
P = 0.8413

Answer

P = 0.8413
11.Probability

Answer: P = 0.5177

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)4 = 0.4823
P = 1 − 0.4823 = 0.5177

Answer

P = 0.5177
12.Probability

Answer: E(G) = 3.5 €

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: 15 (p = 0.3), 45 (p = 0.05), −5 (p = 0.65)

Expected value

E(G) = 15·0.3 + 45·0.05 − 5·0.65 = 3.5 €

Answer

E(G) = 3.5 €

A positive value means you win money on average.

13.Normal distribution

Answer: P = 0.4573

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 = (250.25 − 250)/5 = 0.05
z₂ = (260 − 250)/5 = 2

Area between

P = Φ(2) − Φ(0.05) = 0.9772 − 0.5199 = 0.4573

Answer

P = 0.4573
14.Normal distribution

Answer: x = 41.87 min

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) = 0.95 → z = 1.645

Back to x

x = 32 + 1.645·6 = 41.87 min

Answer

x = 41.87 min
15.Confidence intervals

Answer: lower limit = −3.289 mmHg

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.

Critical value

df = 9, t = 3.25

Margin of error

E = 3.25·6.8/√10 = 6.989

Interval

3.7 ± 6.989 → [−3.289, 10.69]

Answer

lower limit = −3.289 mmHg
16.Confidence intervals

Answer: n = 49

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 = (1.645·17/4)² = 48.88

Round up

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

n = 49

Answer

n = 49
17.Hypothesis tests & confidence intervals

Answer: t = 2.9

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/√25 = 5/5 = 1

Test statistic

t = (122.9 − 120)/1 = 2.9

Answer

t = 2.9
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: L = 118.2

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).

Margin of error

E = 2.576·8/√64 = 2.576

Interval limit

L = 120.8 − 2.576 = 118.2

Answer

L = 118.2
19.Linear regression & correlation

Answer: r = 0.9443

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 = 6670 − 100·326/5 = 150
Sxx = 2040 − 100²/5 = 40
Syy = 21886 − 326²/5 = 630.8

Correlation coefficient

r = 150/√(40·630.8) = 0.9443

Answer

r = 0.9443
20.Linear regression & correlation

Answer: x = 3.976

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 = 110/6 = 18.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −49

Regression line

b = −49/17.5 = −2.8
a = 18.33 − (−2.8)·3.5 = 28.13
ŷ = 28.13 − 2.8·x

Solve for x

x = (17 − 28.13)/(−2.8) = 3.976

Answer

x = 3.976

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