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
years012PV = ?€20 000

You need €20 000 in 2 years. An account pays 5% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €20 0005% / 12n = 2
PV = €
2.Time value of moneyHard
months01…180…240€150 000AB = ?A4.8% / 12

A mortgage of €150 000 is repaid with equal monthly payments over 20 years at 4.8% per year, compounded monthly. How much is still owed just after the payment at the end of year 15? Find the outstanding balance B.

P = €150 0004.8% / 12N = 240k = 180
B = €
3.NPV & IRRMedium
years0123456−60 00013 50013 50013 50013 50013 50019 500

A machine costs €60 000 today. It generates €13 500 at the end of each year for 6 years and can then be sold for €6 000. The cost of capital is 12%. Find the net present value NPV.

C0 = €60 000CF = €13 500S = €6 000n = 6r = 12%
NPV = €
4.NPV & IRRHard
years01234567−80 00011 500+3%+3%+3%+3%+3%+3%r = 12%, NPV = ?

A project costs €80 000 today. Its cash flow is €11 500 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 7. The cost of capital is 12%. Find the net present value NPV.

C0 = €80 000CF1 = €11 500g = 3%n = 7r = 12%
NPV = €
5.Bond pricing & yieldMedium
half-years0123…24€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 12 years. The yield to maturity is 7% per year. Find the price P of the bond.

F = €1 000coupon = 5%YTM = 7%n = 12
P = €
6.Bond pricing & yieldHard
half-years0123…14€642€10€10€10 + €1 000

A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 7 years. It sells for €642. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 2% (2×)n = 7P = €642
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€12 000 000re​ = 10%preferred shares€1 500 000rp​ = 6.5%debt€6 000 000rd​ = 4%T = 28%

A company is financed with ordinary shares worth €12 000 000, preferred shares worth €1 500 000 and debt worth €6 000 000. The cost of equity is 10%, the cost of preferred stock 6.5% and the pre-tax cost of debt 4%. The tax rate is 28%. Find the WACC.

E = €12 000 000Ps = €1 500 000D = €6 000 000re = 10%rp = 6.5%rd = 4%T = 28%
WACC = %
8.CAPM & WACCHard
WACC = ?equity4 000 000 × €44.00β = 0.8rf​ = 2%, rm​ = 6.5%debt5 000 × €1 050rd​ = 6%T = 21%

A company has 4 000 000 shares at €44.00 each and 5 000 bonds trading at €1 050 each. The pre-tax cost of debt is 6% and the tax rate is 21%. The share’s beta is 0.8, the risk-free rate 2% and the market return 6.5%. Find the WACC.

β = 0.8rf = 2%rm = 6.5%rd = 6%T = 21%
WACC = %
9.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 49, 30, 58, 44, 30.

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

Exam scores are normally distributed with mean 50 and standard deviation 20. Find the z-score z of a score of 26.

μ = 50σ = 20x = 26
z =
11.ProbabilityMedium
red: 4blue: 6

A bag holds 4 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

4 / 6
P =
12.ProbabilityHard
n = 12, p = 0.3: P(X = 2) = ?k0123456789101112

A test has 12 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 = 12p = 0.3k = 2
P =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9692

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9692. Find z to two decimals.

A = 0.9692
z =
14.Normal distributionHard
μ = 500x = ?80%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 80% of them lie (the 80th percentile).

μ = 500σ = 40P = 80%
x = h
15.Confidence intervalsMedium
n = 10, sd​ = 5.90L = ?Ud̄ = 6.5E99%

The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.5 and standard deviation 5.9. Find the lower limit of the 99% confidence interval for the mean difference.

n = 10d̄ = 6.5sd = 5.999%t(9) = 3.25
lower limit = mmHg
16.Confidence intervalsHard
σ = 30, n = ?x̄ − Ex̄ + Ex̄E = 195%

The population standard deviation is 30. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1?

σ = 30E = 195%
n =
17.Hypothesis tests & confidence intervalsMedium
0.9p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 25. A random sample of 100 observations has a mean of 252.25. Test H₀: μ = 250 against H₁: μ > 250. Find the p-value p.

μ₀ = 250σ = 25n = 100x̄ = 252.25H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 80 against H₁: μ > 80 is done at α = 0.10 with a sample of 25; the standard deviation is known to be 15. In reality the mean is 84.5. Find the probability of a Type II error β in percent.

μ₀ = 80μ = 84.5σ = 15n = 25α = 0.10
β = %
19.Linear regression & correlationMedium
xy185420522258246926652874xy20256070

The table gives 6 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 = 6Σx = 138Σy = 372Σx² = 3244Σy² = 23446Σxy = 8706
r =
20.Linear regression & correlationHard
xy124223324418518xy2418202224ŷ = 20

The table gives 5 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 = 20?

n = 5ŷ = 20
x =

3Solutions

1.Time value of money

Answer: PV = 18101 €

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 = 0.05/12 = 0.004167
N = 12·2 = 24

Present value

PV = FV/(1 + i)N = 20 000/1.105 = €18 100.51

Answer

PV = €18 100.51
2.Time value of money

Answer: B = 51834 €

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 = 0.004, N = 240
A = 150 000·0.004/(1 − (1 + 0.004)−240) = €973.44

Present value of the remaining payments

Whatever is still owed must be exactly paid off by the remaining payments.

N − k = 240 − 180 = 60
B = 973.4·(1 − (1 + 0.004)−60)/0.004 = €51 834.38

Answer

B = €51 834.38
3.NPV & IRR

Answer: NPV = −1456 €

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

13 500·4.111 = 55 504.00

Present value of the salvage value

6 000/1.974 = 3 039.79

NPV

NPV = −60 000 + 55 504.00 + 3 039.79 = −€1 456.21

Answer

NPV = −€1 456.21
4.NPV & IRR

Answer: NPV = −23309 €

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 growing annuity

((1.03)/(1.12))7 = 0.5563
PV = 11 500/(0.12 − 0.03)·(1 − 0.5563) = 56 690.85

NPV

NPV = 56 690.85 − 80 000 = −€23 309.15

Answer

NPV = −€23 309.15

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

5.Bond pricing & yield

Answer: P = 839.4 €

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 = 0.035, N = 24

Present value of the coupons

€25.00·(1 − 0.438)/0.035 = €401.46

Present value of the face value

€1 000·0.438 = €437.96

Price

P = €839.42

Answer

P = €839.42

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

6.Bond pricing & yield

Answer: YTM = 9.005 %

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 = €10.00, N = 14

Solve numerically

i = 4.503% per half-year

YTM

YTM = 2·4.503% = 9.005%

Answer

YTM = 9.005%
7.CAPM & WACC

Answer: WACC = 7.54 %

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 = 12 000 000 + 1 500 000 + 6 000 000 = €19 500 000
E/V = 0.6154, Ps/V = 0.07692, D/V = 0.3077

WACC

WACC = 0.6154·10% + 0.07692·6.5% + 0.3077·4%·(1 − 0.28) = 7.54%

Answer

WACC = 7.54%
8.CAPM & WACC

Answer: WACC = 5.575 %

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 = 4 000 000·€44.00 = €176 000 000
D = 5 000·€1 050 = €5 250 000

Cost of equity (CAPM)

re = 2% + 0.8·4.5% = 5.6%

Weights

V = €181 250 000, E/V = 0.971, D/V = 0.02897

WACC

WACC = 0.971·5.6% + 0.02897·6%·(1 − 0.21) = 5.575%

Answer

WACC = 5.575%
9.Descriptive statistics

Answer: s = 12.21

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 = 211/5 = 42.2

Squared deviations

Σ(x − x̄)² = 6.8² + (−12.2)² + 15.8² + 1.8² + (−12.2)² = 596.8

Standard deviation

s = √(596.8/4) = 12.21

Answer

s = 12.21
10.Descriptive statistics

Answer: z = −1.2

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 = (26 − 50)/20 = −1.2

Answer

z = −1.2
11.Probability

Answer: P = 0.1333

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 = 4/10 · 3/9 = 12/90 = 0.1333

Answer

P = 0.1333
12.Probability

Answer: P = 0.1678

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) = 66·0.32·0.710 = 66·0.09·0.02825 = 0.1678

Answer

P = 0.1678
13.Normal distribution

Answer: z = 2.16

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 of z

Φ(z) = (1 + 0.9692)/2 = 0.9846

Read the z-table backwards

Find 0.9846 inside the table and read off z.

z = 2.16

Answer

z = 2.16
14.Normal distribution

Answer: x = 533.7 h

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) = 0.8 → z = 0.8416

Back to x

x = 500 + 0.8416·40 = 533.66 h

Answer

x = 533.66 h
15.Confidence intervals

Answer: lower limit = 0.4363 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 = 3.25·5.9/√10 = 6.064

Interval

6.5 ± 6.064 → [0.4363, 12.56]

Answer

lower limit = 0.4363 mmHg
16.Confidence intervals

Answer: n = 3458

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 = (1.96·30/1)² = 3457

Round up

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

n = 3458

Answer

n = 3458
17.Hypothesis tests & confidence intervals

Answer: p = 0.1841

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.

Test statistic

SE = σ/√n = 25/√100 = 2.5
z = (252.25 − 250)/2.5 = 0.9

Tail area

Φ(0.9) = 0.8159
1 − Φ(0.9) = 0.1841

p-value

p = 0.1841

Answer

p = 0.1841

p > 0.05, so H₀ is not rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: β = 41.37 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 15/√25 = 3
x̄c = 80 + 1.282·3 = 83.85

z under the true mean

z = (83.85 − 84.5)/3 = −0.218

Type II error

power = P(Z > −0.218) = 58.63%
β = 100% − 58.63% = 41.37%

Answer

β = 41.37%
19.Linear regression & correlation

Answer: r = 0.9173

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 = 8706 − 138·372/6 = 150
Sxx = 3244 − 138²/6 = 70
Syy = 23446 − 372²/6 = 382

Correlation coefficient

r = 150/√(70·382) = 0.9173

Answer

r = 0.9173
20.Linear regression & correlation

Answer: x = 3.824

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 = 15/5 = 3
ȳ = 107/5 = 21.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −17

Regression line

b = −17/10 = −1.7
a = 21.4 − (−1.7)·3 = 26.5
ŷ = 26.5 − 1.7·x

Solve for x

x = (20 − 26.5)/(−1.7) = 3.824

Answer

x = 3.824

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