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…8PV = ?€25 000

You need €25 000 in 8 years. An account pays 2% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €25 0002% / 4n = 8
PV = €
2.Time value of moneyHard
months0123…72€15 000A?AAA

A loan of €15 000 is repaid with equal monthly payments over 6 years. The interest rate is 6% per year, compounded monthly. Find the monthly payment A.

P = €15 0006% / 12N = 72
A = €
3.NPV & IRRMedium
years01234−40 00014 00014 00014 00014 000

A project costs €40 000 today and returns €14 000 at the end of each year for 4 years. The cost of capital is 8%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 8%CF0 = −40 000CF1 = 14 000CF2 = 14 000CF3 = 14 000CF4 = 14 000
DPB = years
4.NPV & IRRHard
years0123…9−100 00013 500+2%+2%+2%r = 10%, NPV = ?

A project costs €100 000 today. Its cash flow is €13 500 at the end of year 1 and grows by 2% per year until the last cash flow at the end of year 9. The cost of capital is 10%. Find the net present value NPV.

C0 = €100 000CF1 = €13 500g = 2%n = 9r = 10%
NPV = €
5.Bond pricing & yieldMedium
half-years0123…20€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 10 years. The yield to maturity is 9% per year. Find the price P of the bond.

F = €1 000coupon = 8%YTM = 9%n = 10
P = €
6.Bond pricing & yieldHard
years0123456P€80€80€80 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 8% and matures in 6 years. The yield to maturity is 6%. Find the Macaulay duration D in years.

F = €1 000coupon = 8%n = 6YTM = 6%
D = years
7.CAPM & WACCMedium
WACC = ?equity€5 000 000re​ = 8%debt€1 000 000rd​ = 5%T = 20%

A company has equity worth €5 000 000 and debt worth €1 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 5% and the tax rate is 20%. Find the WACC.

E = €5 000 000D = €1 000 000re = 8%rd = 5%T = 20%
WACC = %
8.CAPM & WACCHard
WACC = ?equity5 000 000 × €13.50β = 0.6rf​ = 4%, rm​ = 8.5%debt5 000 × €1 000rd​ = 6%T = 25%

A company has 5 000 000 shares at €13.50 each and 5 000 bonds trading at €1 000 each. The pre-tax cost of debt is 6% and the tax rate is 25%. The share’s beta is 0.6, the risk-free rate 4% and the market return 8.5%. Find the WACC.

β = 0.6rf = 4%rm = 8.5%rd = 6%T = 25%
WACC = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 19, 15, 50, 13, 50, 46.

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

A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is below 460?

μ = 500σ = 50x = 460
P =
11.ProbabilityMedium
n = 6k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 6? The order does not matter.

n = 6k = 2
N =
12.ProbabilityHard
€p(x)−20.6580.3980.05

A game costs €2 to play. You win €10 with probability 0.3 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€10: p = 0.3€100: p = 0.05
E(G) = €
13.Normal distributionMedium
242.5254.25P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 242.5 and 254.25 g? Give P as a decimal.

μ = 250σ = 5x1 = 242.5x2 = 254.25
P =
14.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
n = 9, s = 15LUx̄ = 52E = ?90%

A sample of 9 items has mean 52 and standard deviation s = 15. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 9x̄ = 52s = 1590%t(8) = 1.86
E =
16.Confidence intervalsHard
n = 20σ = ?Us = 7.495%

The diameters of 20 machined shafts have a sample standard deviation of 7.4 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 19, 2.5% of the χ² distribution lies below 8.907 and 2.5% lies above 32.852.

n = 20s = 7.4 µmχ² = 8.907, 32.852
σ = µm
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 40 parts has mean length 107.5 mm, and the line's known standard deviation is 13 mm. Line 2: a sample of 80 parts has mean 109.7 mm, known standard deviation 11 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 40x̄₁ = 107.5 mmσ₁ = 13 mmn₂ = 80x̄₂ = 109.7 mmσ₂ = 11 mm
z =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ < 50 is done at α = 0.01 with a sample of 36; the standard deviation is known to be 9. In reality the mean is 46.85. Find the probability of a Type II error β in percent.

μ₀ = 50μ = 46.85σ = 9n = 36α = 0.01
β = %
19.Linear regression & correlationMedium
xy124223318414514xy24152025

The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy124221319414512610xy24610152025x = 1.5

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. Predict ŷ for x = 1.5.

n = 6x = 1.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 21312 €

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 = 0.02/4 = 0.005
N = 4·8 = 32

Present value

PV = FV/(1 + i)N = 25 000/1.173 = €21 312.09

Answer

PV = €21 312.09
2.Time value of money

Answer: A = 248.6 €

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 = 0.06/12 = 0.005
N = 12·6 = 72

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 = 0.6983
A = 15 000·0.005/(1 − 0.6983) = €248.59

Answer

A = €248.59
3.NPV & IRR

Answer: DPB = 3.381 years

Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.

Discounted and cumulative cash flows

year 1: 14 000/1.08 = 12 962.96 → Σ = 12 962.96
year 2: 14 000/1.166 = 12 002.74 → Σ = 24 965.71
year 3: 14 000/1.26 = 11 113.65 → Σ = 36 079.36
year 4: 14 000/1.36 = 10 290.42 → Σ = 46 369.78

Fraction of the last year

DPB = 3 + 3 920.64/10 290.42 = 3.381

Answer

DPB = 3.381 years

The discounted payback is always longer than the simple payback (2.857 years).

4.NPV & IRR

Answer: NPV = −16779 €

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

((1.02)/(1.1))9 = 0.5068
PV = 13 500/(0.1 − 0.02)·(1 − 0.5068) = 83 221.44

NPV

NPV = 83 221.44 − 100 000 = −€16 778.56

Answer

NPV = −€16 778.56

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

5.Bond pricing & yield

Answer: P = 935 €

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 = 0.045, N = 20

Present value of the coupons

€40.00·(1 − 0.4146)/0.045 = €520.32

Present value of the face value

€1 000·0.4146 = €414.64

Price

P = €934.96

Answer

P = €934.96

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

6.Bond pricing & yield

Answer: D = 5.044 years

Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.

Present value of each cash flow

t = 1: 80/1.06 = 75.47, t·PV = 75.47
t = 2: 80/1.124 = 71.20, t·PV = 142.40
t = 3: 80/1.191 = 67.17, t·PV = 201.51
t = 4: 80/1.262 = 63.37, t·PV = 253.47
t = 5: 80/1.338 = 59.78, t·PV = 298.90
t = 6: 1080/1.419 = 761.36, t·PV = 4 568.14

Duration

P = 1 098.35
D = 5 539.90/1 098.35 = 5.044

Answer

D = 5.044 years
7.CAPM & WACC

Answer: WACC = 7.333 %

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 = €6 000 000, E/V = 0.8333, D/V = 0.1667

WACC

WACC = 0.8333·8% + 0.1667·5%·(1 − 0.2) = 7.333%

Answer

WACC = 7.333%
8.CAPM & WACC

Answer: WACC = 6.548 %

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 = 5 000 000·€13.50 = €67 500 000
D = 5 000·€1 000 = €5 000 000

Cost of equity (CAPM)

re = 4% + 0.6·4.5% = 6.7%

Weights

V = €72 500 000, E/V = 0.931, D/V = 0.06897

WACC

WACC = 0.931·6.7% + 0.06897·6%·(1 − 0.25) = 6.548%

Answer

WACC = 6.548%
9.Descriptive statistics

Answer: s = 18.24

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 = 193/6 = 32.17

Squared deviations

Σ(x − x̄)² = (−13.17)² + (−17.17)² + 17.83² + (−19.17)² + 17.83² + 13.83² = 1663

Standard deviation

s = √(1663/5) = 18.24

Answer

s = 18.24
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 = 15

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(6, 2) = (6·5)/(2·1) = 15

Answer

N = 15
12.Probability

Answer: E(G) = 6 €

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: 8 (p = 0.3), 98 (p = 0.05), −2 (p = 0.65)

Expected value

E(G) = 8·0.3 + 98·0.05 − 2·0.65 = 6 €

Answer

E(G) = 6 €

A positive value means you win money on average.

13.Normal distribution

Answer: P = 0.7355

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 = (242.5 − 250)/5 = −1.5
z₂ = (254.25 − 250)/5 = 0.85

Area between

P = Φ(0.85) − Φ(−1.5) = 0.8023 − 0.06681 = 0.7355

Answer

P = 0.7355
14.Normal distribution

Answer: P = 0.8849

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 = 25.03 − 25 = 0.03 mm, σ = 0.025 mm

Standardise

z = (0 − 0.03)/0.025 = −1.2

Probability

P = 1 − Φ(−1.2) = 0.8849

Answer

P = 0.8849
15.Confidence intervals

Answer: E = 9.3

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 = 8, t = 1.86

Margin of error

E = 1.86·15/√9 = 9.3

Answer

E = 9.3
16.Confidence intervals

Answer: σ = 5.628 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 19·7.4² = 1040

Limits for σ²

1040/32.852 = 31.67
1040/8.907 = 116.8

Limit for σ

σ = √31.67 = 5.628 µm

Answer

σ = 5.628 µm

95% interval for σ: [5.628, 10.81] µm.

17.Hypothesis tests & confidence intervals

Answer: z = −0.9185

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.

Standard error of the difference

SE = √(13²/40 + 11²/80) = √5.737 = 2.395

Test statistic

z = (107.5 − 109.7)/2.395 = −0.9185

Answer

z = −0.9185

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

18.Hypothesis tests & confidence intervals

Answer: β = 58.94 %

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 = 9/√36 = 1.5
x̄c = 50 − 2.326·1.5 = 46.51

z under the true mean

z = (46.51 − 46.85)/1.5 = −0.226

Type II error

power = P(Z < −0.226) = 41.06%
β = 100% − 41.06% = 58.94%

Answer

β = 58.94%
19.Linear regression & correlation

Answer: r² = 0.9221

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 = 93/5 = 18.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −29
Syy = Σ(y − ȳ)² = 91.2

Coefficient of determination

r² = −29²/(10·91.2) = 0.9221

Answer

r² = 0.9221

About 92.21% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 22.5

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 = 100/6 = 16.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −51

Regression line

b = −51/17.5 = −2.914
a = 16.67 − (−2.914)·3.5 = 26.87
ŷ = 26.87 − 2.914·x

Prediction

ŷ = 26.87 + (−2.914)·1.5 = 22.5

Answer

ŷ = 22.5

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