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
years01234567€12 000FV = ?

You invest €12 000 at a nominal rate of 4.8% per year, compounded quarterly. Find the value FV of the investment after 7 years.

PV = €12 0004.8% / 4n = 7
FV = €
2.Time value of moneyHard
months0123…N = ?€10 000€150€150€150€1507.2% / 12

You borrow €10 000 at 7.2% per year, compounded monthly, and repay €150 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.

P = €10 000A = €1507.2% / 12
N = payments
3.NPV & IRRMedium
years01234−75 00020 00034 00026 00016 000

An investment of €75 000 gives these cash flows at the end of years 1 to 4: €20 000, €34 000, €26 000, €16 000. The cost of capital is 10%. Find the net present value NPV.

r = 10%CF0 = −75 000CF1 = 20 000CF2 = 34 000CF3 = 26 000CF4 = 16 000
NPV = €
4.NPV & IRRHard
years0123−25 00012 00013 00014 000

A project costs €25 000 and returns €12 000, €13 000, €14 000 at the end of years 1 to 3. Cash flows can be reinvested at the cost of capital of 6%. Find the modified internal rate of return MIRR in percent.

r = 6%CF0 = −25 000CF1 = 12 000CF2 = 13 000CF3 = 14 000
MIRR = %
5.Bond pricing & yieldMedium
years0123…12P€40€40€40 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 4% and matures in 12 years. Its yield to maturity changes from 3.5% to 4%. By how much ΔP does the bond price change?

F = €1 000coupon = 4%n = 12y1 = 3.5%y2 = 4%
ΔP = €
6.Bond pricing & yieldHard
years012345€1 228€80€80€80 + €1 080

A bond with 13 years to maturity (face value €1 000, annual coupon 8%) sells for €1 228. The issuer can call it in 5 years at a call price of €1 080. Find the yield to call YTC in percent.

F = €1 000coupon = 8%P = €1 228call price = €1 080call in 5 years
YTC = %
7.CAPM & WACCMedium
WACC = 7.5%equity€6 000 000re​ = ?debt€1 000 000rd​ = 6%T = 25%

A company has equity worth €6 000 000 and debt worth €1 000 000. Its WACC is 7.5%, its pre-tax cost of debt is 6% and the tax rate is 25%. Find the cost of equity re.

E = €6 000 000D = €1 000 000WACC = 7.5%rd = 6%T = 25%
re = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €35.00β = 1.5rf​ = 2.5%, rm​ = 9.5%debt5 000 × €1 020rd​ = 3%T = 30%

A company has 2 500 000 shares at €35.00 each and 5 000 bonds trading at €1 020 each. The pre-tax cost of debt is 3% and the tax rate is 30%. The share’s beta is 1.5, the risk-free rate 2.5% and the market return 9.5%. Find the WACC.

β = 1.5rf = 2.5%rm = 9.5%rd = 3%T = 30%
WACC = %
9.Descriptive statisticsMedium
1015202530354045505560

Find the sample standard deviation s of the data: 32, 60, 28, 45, 10, 59, 59.

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

Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 462.5.

μ = 500σ = 25x = 462.5
z =
11.ProbabilityMedium
red: 3blue: 8

A bag holds 3 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

3 / 8
P =
12.ProbabilityHard
n = 7, p = 0.3: P(X = 2) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 2 correct answers?

n = 7p = 0.3k = 2
P =
13.Normal distributionMedium
29.333.2P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 29.3 and 33.2 min? Give P as a decimal.

μ = 32σ = 6x1 = 29.3x2 = 33.2
P =
14.Normal distributionHard
μ = 65x = ?10%

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

μ = 65σ = 10P = 10%
x = points
15.Confidence intervalsMedium
n = 10, sd​ = 4.20LU = ?d̄ = 3.3E99%

The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 3.3 and standard deviation 4.2. Find the upper limit of the 99% confidence interval for the mean difference.

n = 10d̄ = 3.3sd = 4.299%t(9) = 3.25
upper limit = mmHg
16.Confidence intervalsHard
σ = 18, n = ?x̄ − Ex̄ + Ex̄E = 599%

The population standard deviation is 18. 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 = 18E = 599%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.0452.045t = ?df = 29

Two teaching methods are compared. Group 1: 21 students, mean score 22.5, standard deviation 5.5. Group 2: 10 students, mean score 27.1, standard deviation 9.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 21x̄₁ = 22.5s₁ = 5.5n₂ = 10x̄₂ = 27.1s₂ = 9.5
t =
18.Hypothesis tests & confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 199%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 20. Find the smallest sample size n.

σ = 20E = 199%
n =
19.Linear regression & correlationMedium
xy23144465786410731289xy510406080

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy2941962583010391243xy5102040ŷ = 29

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 = 29?

n = 6ŷ = 29
x =

3Solutions

1.Time value of money

Answer: FV = 16759 €

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 = 0.048/4 = 0.012
N = m·n = 4·7 = 28

Future value

FV = 12 000·(1 + 0.012)28 = 12 000·1.397 = €16 758.52

Answer

FV = €16 758.52
2.Time value of money

Answer: N = 86 payments

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.

Rearrange the annuity formula

i = 0.006
(1 + i)−N = 1 − 10 000·0.006/150 = 0.6

Solve with logarithms

N = −ln(0.6)/ln(1.006) = 85.39

Round up to whole payments

85 full payments leave a small balance, which the last, smaller payment clears.

N = 86

Answer

N = 86 payments
3.NPV & IRR

Answer: NPV = 1743 €

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 + 0.1)1 = 20 000/1.1 = 18 181.82
CF2/(1 + 0.1)2 = 34 000/1.21 = 28 099.17
CF3/(1 + 0.1)3 = 26 000/1.331 = 19 534.18
CF4/(1 + 0.1)4 = 16 000/1.464 = 10 928.22

Add them up

NPV = €1 743.39

Answer

NPV = €1 743.39

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

4.NPV & IRR

Answer: MIRR = 18.18 %

Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.

Terminal value of the inflows

12 000·1.062 = 13 483.20
13 000·1.061 = 13 780.00
14 000·1.060 = 14 000.00
TV = 41 263.20

MIRR

MIRR = (41 263.20/25 000)1/3 − 1 = 18.18%

Answer

MIRR = 18.18%
5.Bond pricing & yield

Answer: ΔP = −48.32 €

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 = €40.00, i = 0.035, N = 12
€40.00·(1 − 0.6618)/0.035 + €1 000·0.6618 = €386.53 + €661.78 = €1 048.32

Price at the new yield

C = €40.00, i = 0.04, N = 12
€40.00·(1 − 0.6246)/0.04 + €1 000·0.6246 = €375.40 + €624.60 = €1 000.00

Change

ΔP = 1 000.00 − 1 048.32 = −€48.32

Answer

ΔP = −€48.32
6.Bond pricing & yield

Answer: YTC = 4.303 %

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

1228 = 80·(1 − (1 + y)−5)/y + 1080/(1 + y)5

Solve numerically

y = 4%: €1 243.83
y = 5%: €1 192.57
YTC = 4.303%

Answer

YTC = 4.303%

The yield to maturity is 5.499%. 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: re = 8 %

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 = 0.8571, D/V = 0.1429

Debt part of the WACC

0.1429·6%·(1 − 0.25) = 0.6429%

Cost of equity

re = (7.5% − 0.6429%)/0.8571 = 8%

Answer

re = 8%
8.CAPM & WACC

Answer: WACC = 12.4 %

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·€35.00 = €87 500 000
D = 5 000·€1 020 = €5 100 000

Cost of equity (CAPM)

re = 2.5% + 1.5·7% = 13%

Weights

V = €92 600 000, E/V = 0.9449, D/V = 0.05508

WACC

WACC = 0.9449·13% + 0.05508·3%·(1 − 0.3) = 12.4%

Answer

WACC = 12.4%
9.Descriptive statistics

Answer: s = 19.28

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 = 293/7 = 41.86

Squared deviations

Σ(x − x̄)² = (−9.857)² + 18.14² + (−13.86)² + 3.143² + (−31.86)² + 17.14² + 17.14² = 2231

Standard deviation

s = √(2231/6) = 19.28

Answer

s = 19.28
10.Descriptive statistics

Answer: z = −1.5

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.

Standardise

z = (462.5 − 500)/25 = −1.5

Answer

z = −1.5
11.Probability

Answer: P = 0.05455

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.

First and second draw

P = 3/11 · 2/10 = 6/110 = 0.05455

Answer

P = 0.05455
12.Probability

Answer: P = 0.3177

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) = 21·0.32·0.75 = 21·0.09·0.1681 = 0.3177

Answer

P = 0.3177
13.Normal distribution

Answer: P = 0.2529

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 = (29.3 − 32)/6 = −0.45
z₂ = (33.2 − 32)/6 = 0.2

Area between

P = Φ(0.2) − Φ(−0.45) = 0.5793 − 0.3264 = 0.2529

Answer

P = 0.2529
14.Normal distribution

Answer: x = 52.18 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) = 0.1 → z = −1.282

Back to x

x = 65 + −1.282·10 = 52.18 points

Answer

x = 52.18 points
15.Confidence intervals

Answer: upper limit = 7.617 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 = 3.25·4.2/√10 = 4.317

Interval

3.3 ± 4.317 → [−1.017, 7.617]

Answer

upper limit = 7.617 mmHg
16.Confidence intervals

Answer: n = 86

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 = (2.576·18/5)² = 86

Round up

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

n = 86

Answer

n = 86
17.Hypothesis tests & confidence intervals

Answer: t = −1.713

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.

Pooled variance

sp² = (20·5.5² + 9·9.5²)/29 = 48.87
sp = 6.991

Standard error

SE = 6.991·√(1/21 + 1/10) = 2.686

Test statistic

t = (22.5 − 27.1)/2.686 = −1.713

Answer

t = −1.713

With df = 29, the two-sided 5% critical value is 2.045; |t| is smaller, so H₀ is not rejected.

18.Hypothesis tests & confidence intervals

Answer: n = 2655

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 = (2.576·20/1)² = 2654

Round up

n = 2655

Answer

n = 2655
19.Linear regression & correlation

Answer: r² = 0.9884

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 = 358/6 = 59.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 384
Syy = Σ(y − ȳ)² = 2131

Coefficient of determination

r² = 384²/(70·2131) = 0.9884

Answer

r² = 0.9884

About 98.84% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: x = 7.447

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 = 165/6 = 27.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 235

Regression line

b = 235/70 = 3.357
a = 27.5 − 3.357·7 = 4
ŷ = 4 + 3.357·x

Solve for x

x = (29 − 4)/3.357 = 7.447

Answer

x = 7.447

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