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
years01234€1 000FV = ?

You invest €1 000 at a nominal rate of 3.6% per year, compounded monthly. Find the value FV of the investment after 4 years.

PV = €1 0003.6% / 12n = 4
FV = €
2.Time value of moneyHard
months0123…336€50€50€50FV = ?

You save €50 at the end of every month for 28 years. The account pays 4.2% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €504.2% / 12N = 336
FV = €
3.NPV & IRRMedium
years012345−80 00019 00019 00019 00019 00039 000

A machine costs €80 000 today. It generates €19 000 at the end of each year for 5 years and can then be sold for €20 000. The cost of capital is 8%. Find the net present value NPV.

C0 = €80 000CF = €19 000S = €20 000n = 5r = 8%
NPV = €
4.NPV & IRRHard
years0123−50 00029 50033 00036 500

A project costs €50 000 and returns €29 500, €33 000, €36 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 29 500CF2 = 33 000CF3 = 36 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…18€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 9 years. The yield to maturity is 9% per year. Find the price P of the bond.

F = €1 000coupon = 5%YTM = 9%n = 9
P = €
6.Bond pricing & yieldHard
years01234€950€80€80 + €1 000

A bond with a face value of €1 000 and an annual coupon of 8% matures in 4 years and sells for €950. Find its yield to maturity YTM in percent.

F = €1 000coupon = 8%n = 4P = €950
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€6 000 000re​ = 14%preferred shares€1 500 000rp​ = 6%debt€6 000 000rd​ = 4%T = 25%

A company is financed with ordinary shares worth €6 000 000, preferred shares worth €1 500 000 and debt worth €6 000 000. The cost of equity is 14%, the cost of preferred stock 6% and the pre-tax cost of debt 4%. The tax rate is 25%. Find the WACC.

E = €6 000 000Ps = €1 500 000D = €6 000 000re = 14%rp = 6%rd = 4%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €47.00β = 1.5rf​ = 3.5%, rm​ = 9.5%debt20 000 × €1 000rd​ = 6%T = 25%

A company has 2 500 000 shares at €47.00 each and 20 000 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 1.5, the risk-free rate 3.5% and the market return 9.5%. Find the WACC.

β = 1.5rf = 3.5%rm = 9.5%rd = 6%T = 25%
WACC = %
9.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 31, 27, 46, 57, 52, 47, 41.

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

A quantity is normally distributed with mean 50 and standard deviation 15. What proportion P of values is below 32?

μ = 50σ = 15x = 32
P =
11.ProbabilityMedium
n = 15k = 5N = ?

In how many ways N can a committee of 5 people be chosen from a group of 15? The order does not matter.

n = 15k = 5
N =
12.ProbabilityHard
€p(x)−10.6830.3490.02

A game costs €1 to play. You win €4 with probability 0.3 and €50 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€4: p = 0.3€50: p = 0.02
E(G) = €
13.Normal distributionMedium
150 kg151.8 kgT = X₁ + … + X₆?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 6 of them is below 151.8 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 6< 151.8 kg
P =
14.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.01 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 = 25.04 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
n = 8, sd​ = 0.60LU = ?d̄ = −5.6E90%

The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −5.6 and standard deviation 0.6. Find the upper limit of the 90% confidence interval for the mean difference.

n = 8d̄ = −5.6sd = 0.690%t(7) = 1.895
upper limit = mmHg
16.Confidence intervalsHard
n = 20σ = ?Us = 3.995%

The diameters of 20 machined shafts have a sample standard deviation of 3.9 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 = 3.9 µmχ² = 8.907, 32.852
σ = µm
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 102.4 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 102.4s = 20n = 100
t =
18.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 49 measurements has mean 43.6. The population standard deviation is 15. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 43.6σ = 15n = 4995%
L =
19.Linear regression & correlationMedium
xy364468577682793899xy46880100

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 = 33Σy = 483Σx² = 199Σy² = 39823Σxy = 2784
r =
20.Linear regression & correlationHard
xy254146228291036xy510102030x = 9

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 9.

n = 5x = 9
ŷ =

3Solutions

1.Time value of money

Answer: FV = 1155 €

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 = 0.036000000000000004/12 = 0.003
N = m·n = 12·4 = 48

Future value

FV = 1 000·(1 + 0.003)48 = 1 000·1.155 = €1 154.64

Answer

FV = €1 154.64
2.Time value of money

Answer: FV = 31925 €

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 deposits

i = 0.042/12 = 0.0035
N = 12·28 = 336

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 50·(3.235 − 1)/0.0035 = €31 924.78

Answer

FV = €31 924.78

You paid in €16 800 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = 9473 €

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

19 000·3.993 = 75 861.49

Present value of the salvage value

20 000/1.469 = 13 611.66

NPV

NPV = −80 000 + 75 861.49 + 13 611.66 = €9 473.15

Answer

NPV = €9 473.15
4.NPV & IRR

Answer: IRR = 41.83 %

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(40%) = 1 209.91
NPV(45%) = −1 986.96

Narrow down

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

IRR ≈ 40 + 5·1 209.91/(1 209.91 − (−1 986.96)) = 41.89%
exact: NPV(41.83%) = 0 → IRR = 41.83%

Answer

IRR = 41.83%
5.Bond pricing & yield

Answer: P = 756.8 €

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 = 0.045, N = 18

Present value of the coupons

€25.00·(1 − 0.4528)/0.045 = €304.00

Present value of the face value

€1 000·0.4528 = €452.80

Price

P = €756.80

Answer

P = €756.80

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

6.Bond pricing & yield

Answer: YTM = 9.562 %

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

€950 = €80·(1 − (1 + y)⁻⁴)/y + €1 000/(1 + y)⁴

Solve numerically

P(9%) = €967.60, P(10%) = €936.60
YTM = 9.562%

Answer

YTM = 9.562%
7.CAPM & WACC

Answer: WACC = 8.222 %

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 + 1 500 000 + 6 000 000 = €13 500 000
E/V = 0.4444, Ps/V = 0.1111, D/V = 0.4444

WACC

WACC = 0.4444·14% + 0.1111·6% + 0.4444·4%·(1 − 0.25) = 8.222%

Answer

WACC = 8.222%
8.CAPM & WACC

Answer: WACC = 11.34 %

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·€47.00 = €117 500 000
D = 20 000·€1 000 = €20 000 000

Cost of equity (CAPM)

re = 3.5% + 1.5·6% = 12.5%

Weights

V = €137 500 000, E/V = 0.8545, D/V = 0.1455

WACC

WACC = 0.8545·12.5% + 0.1455·6%·(1 − 0.25) = 11.34%

Answer

WACC = 11.34%
9.Descriptive statistics

Answer: s = 10.85

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 = 301/7 = 43

Squared deviations

Σ(x − x̄)² = (−12)² + (−16)² + 3² + 14² + 9² + 4² + (−2)² = 706

Standard deviation

s = √(706/6) = 10.85

Answer

s = 10.85
10.Descriptive statistics

Answer: P = 0.1151

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 0.1151

Answer

P = 0.1151
11.Probability

Answer: N = 3003

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(15, 5) = (15·14·13·12·11)/(5·4·3·2·1) = 3003

Answer

N = 3003
12.Probability

Answer: E(G) = 1.2 €

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: 3 (p = 0.3), 49 (p = 0.02), −1 (p = 0.68)

Expected value

E(G) = 3·0.3 + 49·0.02 − 1·0.68 = 1.2 €

Answer

E(G) = 1.2 €

A positive value means you win money on average.

13.Normal distribution

Answer: P = 0.9292

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 total

μT = 150 kg, σT = 0.5·√6 = 1.225 kg

Standardise

z = (151.8 − 150)/1.225 = 1.47

Probability

P = Φ(1.47) = 0.9292

Answer

P = 0.9292
14.Normal distribution

Answer: P = 0.9868

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 = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.04)/0.01803 = −2.22

Probability

P = 1 − Φ(−2.22) = 0.9868

Answer

P = 0.9868
15.Confidence intervals

Answer: upper limit = −5.198 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 = 7, t = 1.895

Margin of error

E = 1.895·0.6/√8 = 0.402

Interval

−5.6 ± 0.402 → [−6.002, −5.198]

Answer

upper limit = −5.198 mmHg
16.Confidence intervals

Answer: σ = 2.966 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 = 19·3.9² = 289

Limits for σ²

289/32.852 = 8.797
289/8.907 = 32.45

Limit for σ

σ = √8.797 = 2.966 µm

Answer

σ = 2.966 µm

95% interval for σ: [2.966, 5.696] µm.

17.Hypothesis tests & confidence intervals

Answer: t = 1.2

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).

Standard error

SE = 20/√100 = 20/10 = 2

Test statistic

t = (102.4 − 100)/2 = 1.2

Answer

t = 1.2
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: L = 39.4

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 = 1.96·15/√49 = 4.2

Interval limit

L = 43.6 − 4.2 = 39.4

Answer

L = 39.4
19.Linear regression & correlation

Answer: r = 0.9933

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 = 2784 − 33·483/6 = 127.5
Sxx = 199 − 33²/6 = 17.5
Syy = 39823 − 483²/6 = 941.5

Correlation coefficient

r = 127.5/√(17.5·941.5) = 0.9933

Answer

r = 0.9933
20.Linear regression & correlation

Answer: ŷ = 32.75

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 = 30/5 = 6
ȳ = 106/5 = 21.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 154

Regression line

b = 154/40 = 3.85
a = 21.2 − 3.85·6 = −1.9
ŷ = −1.9 + 3.85·x

Prediction

ŷ = −1.9 + 3.85·9 = 32.75

Answer

ŷ = 32.75

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