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€5 000FV = ?

You invest €5 000 at a nominal rate of 4.8% per year, compounded quarterly. Find the value FV of the investment after 3 years.

PV = €5 0004.8% / 4n = 3
FV = €
2.Time value of moneyHard
months0123…240€200 000AAAA5.4% / 12, I = ?

A loan of €200 000 is repaid with equal monthly payments over 20 years at 5.4% per year, compounded monthly. How much interest I is paid in total over the life of the loan?

P = €200 0005.4% / 12N = 240
I = €
3.NPV & IRRMedium
years01234567−80 00011 50011 50011 50011 50011 50011 50019 500

A machine costs €80 000 today. It generates €11 500 at the end of each year for 7 years and can then be sold for €8 000. The cost of capital is 6%. Find the net present value NPV.

C0 = €80 000CF = €11 500S = €8 000n = 7r = 6%
NPV = €
4.NPV & IRRHard
years012−10 0007 0007 000

A project costs €10 000 and returns €7 000, €7 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −10 000CF1 = 7 000CF2 = 7 000
IRR = %
5.Bond pricing & yieldMedium
years0123…13P€40€40€40 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 4% and matures in 13 years. Its yield to maturity changes from 5.5% to 3.5%. By how much ΔP does the bond price change?

F = €1 000coupon = 4%n = 13y1 = 5.5%y2 = 3.5%
ΔP = €
6.Bond pricing & yieldHard
years0123P€45€45€45 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 4.5% and matures in 3 years. The yield to maturity is 6%. Find the Macaulay duration D in years.

F = €1 000coupon = 4.5%n = 3YTM = 6%
D = years
7.CAPM & WACCMedium
βr (%)SMLrf​ = 1%β = 1rm​ = ?β = 0.9re​ = 5.95%

A share with a beta of 0.9 has a required return of 5.95%. The risk-free rate is 1%. What expected market return rm does CAPM imply?

β = 0.9re = 5.95%rf = 1%
rm = %
8.CAPM & WACCHard
WACC = ?equity€15 000 000re​ = 14%bonds8 000 × €925coupon 7%, n = 12T = 21%

A company has equity with a market value of €15 000 000 and a cost of equity of 14%. Its debt consists of 8 000 bonds (face value €1 000, annual coupon 7%, 12 years to maturity) that trade at €925 each. The tax rate is 21%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.

E = €15 000 000re = 14%bonds: 8 000 × €925coupon = 7%n = 12T = 21%
WACC = %
9.Descriptive statisticsMedium
2025303540

Find the sample standard deviation s of the data: 27, 19, 38, 44, 30.

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

A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is above 106?

μ = 100σ = 15x = 106
P =
11.ProbabilityMedium
n = 9k = 5N = ?

In how many ways N can a committee of 5 people be chosen from a group of 9? The order does not matter.

n = 9k = 5
N =
12.ProbabilityHard
€p(x)−20.8480.15380.01

A game costs €2 to play. You win €10 with probability 0.15 and €40 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€10: p = 0.15€40: p = 0.01
E(G) = €
13.Normal distributionMedium
2939.5P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 29 and 39.5 min? Give P as a decimal.

μ = 32σ = 6x1 = 29x2 = 39.5
P =
14.Normal distributionHard
μ = 500x = ?75%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 75% of them lie (the 75th percentile).

μ = 500σ = 40P = 75%
x = h
15.Confidence intervalsMedium
n = 11, s = 15.5L = ?Ux̄ = 36.5E95%

A sample of 11 items has mean 36.5 and standard deviation s = 15.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 11x̄ = 36.5s = 15.595%t(10) = 2.228
lower limit =
16.Confidence intervalsHard
n = 30Lσ = ?s = 4.695%

The diameters of 30 machined shafts have a sample standard deviation of 4.6 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 4.6 µmχ² = 16.047, 45.722
σ = µm
17.Hypothesis tests & confidence intervalsMedium
1.36p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 40. A random sample of 100 observations has a mean of 255.44. Test H₀: μ = 250 against H₁: μ > 250. Find the p-value p.

μ₀ = 250σ = 40n = 100x̄ = 255.44H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 105 of 150 patients on treatment A recover, and 160 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 105n₁ = 150x₂ = 160n₂ = 200
z =
19.Linear regression & correlationMedium
xy154252356462565673xy2466070

The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 21Σy = 362Σx² = 91Σy² = 22154Σxy = 1337
r =
20.Linear regression & correlationHard
xy238336443541652xy24635404550x = 2.5

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. Predict ŷ for x = 2.5.

n = 5x = 2.5
ŷ =

3Solutions

1.Time value of money

Answer: FV = 5769 €

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 = 4·3 = 12

Future value

FV = 5 000·(1 + 0.012)12 = 5 000·1.154 = €5 769.47

Answer

FV = €5 769.47
2.Time value of money

Answer: I = 127481 €

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 = 0.0045, N = 240
A = 200 000·0.0045/(1 − (1 + 0.0045)−240) = €1 364.50

Total interest

I = N·A − P = 240·1365 − 200 000 = €127 480.76

Answer

I = €127 480.76
3.NPV & IRR

Answer: NPV = −10482 €

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

11 500·5.582 = 64 197.39

Present value of the salvage value

8 000/1.504 = 5 320.46

NPV

NPV = −80 000 + 64 197.39 + 5 320.46 = −€10 482.16

Answer

NPV = −€10 482.16
4.NPV & IRR

Answer: IRR = 25.69 %

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%) = 80.00
NPV(30%) = −473.37

Narrow down

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

IRR ≈ 25 + 5·80.00/(80.00 − (−473.37)) = 25.72%
exact: NPV(25.69%) = 0 → IRR = 25.69%

Answer

IRR = 25.69%
5.Bond pricing & yield

Answer: ΔP = 188.3 €

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 = 0.055, N = 13
€40.00·(1 − 0.4986)/0.055 + €1 000·0.4986 = €364.68 + €498.56 = €863.24

Price at the new yield

C = €40.00, i = 0.035, N = 13
€40.00·(1 − 0.6394)/0.035 + €1 000·0.6394 = €412.11 + €639.40 = €1 051.51

Change

ΔP = 1 051.51 − 863.24 = €188.27

Answer

ΔP = €188.27
6.Bond pricing & yield

Answer: D = 2.87 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: 45/1.06 = 42.45, t·PV = 42.45
t = 2: 45/1.124 = 40.05, t·PV = 80.10
t = 3: 1045/1.191 = 877.40, t·PV = 2 632.21

Duration

P = 959.90
D = 2 754.76/959.90 = 2.87

Answer

D = 2.87 years
7.CAPM & WACC

Answer: rm = 6.5 %

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 risk premium

rm − rf = (5.95% − 1%)/0.9 = 5.5%

Market return

rm = 1% + 5.5% = 6.5%

Answer

rm = 6.5%
8.CAPM & WACC

Answer: WACC = 11.46 %

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 debt (YTM)

925 = 70·(1 − (1 + y)−12)/y + 1000/(1 + y)12
y = 7.995%

Market values

D = 8 000·€925 = €7 400 000
V = €22 400 000, E/V = 0.6696, D/V = 0.3304

WACC

WACC = 0.6696·14% + 0.3304·7.995%·(1 − 0.21) = 11.46%

Answer

WACC = 11.46%
9.Descriptive statistics

Answer: s = 9.711

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 = 158/5 = 31.6

Squared deviations

Σ(x − x̄)² = (−4.6)² + (−12.6)² + 6.4² + 12.4² + (−1.6)² = 377.2

Standard deviation

s = √(377.2/4) = 9.711

Answer

s = 9.711
10.Descriptive statistics

Answer: P = 0.3446

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446

Answer

P = 0.3446
11.Probability

Answer: N = 126

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(9, 5) = (9·8·7·6·5)/(5·4·3·2·1) = 126

Answer

N = 126
12.Probability

Answer: E(G) = −0.1 €

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 = 0.15), 38 (p = 0.01), −2 (p = 0.84)

Expected value

E(G) = 8·0.15 + 38·0.01 − 2·0.84 = −0.1 €

Answer

E(G) = −0.1 €

A negative value means you lose money on average.

13.Normal distribution

Answer: P = 0.5858

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 − 32)/6 = −0.5
z₂ = (39.5 − 32)/6 = 1.25

Area between

P = Φ(1.25) − Φ(−0.5) = 0.8944 − 0.3085 = 0.5858

Answer

P = 0.5858
14.Normal distribution

Answer: x = 527 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) = 0.75 → z = 0.6745

Back to x

x = 500 + 0.6745·40 = 526.98 h

Answer

x = 526.98 h
15.Confidence intervals

Answer: lower limit = 26.09

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 = 10, t = 2.228

Margin of error

E = 2.228·15.5/√11 = 10.41

Interval

36.5 ± 10.41 → [26.09, 46.91]

Answer

lower limit = 26.09
16.Confidence intervals

Answer: σ = 6.184 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 = 29·4.6² = 613.6

Limits for σ²

613.6/45.722 = 13.42
613.6/16.047 = 38.24

Limit for σ

σ = √38.24 = 6.184 µm

Answer

σ = 6.184 µm

95% interval for σ: [3.663, 6.184] µm.

17.Hypothesis tests & confidence intervals

Answer: p = 0.08692

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 = 40/√100 = 4
z = (255.44 − 250)/4 = 1.36

Tail area

Φ(1.36) = 0.9131
1 − Φ(1.36) = 0.0869

p-value

p = 0.08692

Answer

p = 0.08692

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

18.Hypothesis tests & confidence intervals

Answer: z = −2.159

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.

Sample proportions

p̂₁ = 105/150 = 0.7
p̂₂ = 160/200 = 0.8
p̂ = 265/350 = 0.7571

Standard error under H₀

SE = √(0.7571·0.2429·(1/150 + 1/200)) = 0.04632

Test statistic

z = (0.7 − 0.8)/0.04632 = −2.159

Answer

z = −2.159

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

19.Linear regression & correlation

Answer: r = 0.9453

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 = 1337 − 21·362/6 = 70
Sxx = 91 − 21²/6 = 17.5
Syy = 22154 − 362²/6 = 313.3

Correlation coefficient

r = 70/√(17.5·313.3) = 0.9453

Answer

r = 0.9453
20.Linear regression & correlation

Answer: ŷ = 37.05

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 = 20/5 = 4
ȳ = 210/5 = 42

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 33

Regression line

b = 33/10 = 3.3
a = 42 − 3.3·4 = 28.8
ŷ = 28.8 + 3.3·x

Prediction

ŷ = 28.8 + 3.3·2.5 = 37.05

Answer

ŷ = 37.05

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