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
years0123456PV = ?€20 000

You need €20 000 in 6 years. An account pays 2% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €20 0002% / 12n = 6
PV = €
2.Time value of moneyHard
years0123…30€5 000€5 000€5 000FV = ?

You deposit €5 000 at the start of every year for 30 years, the first deposit today. The account pays 5% per year. How much is in the account at the end of year 30? Find FV.

A = €5 000i = 5%n = 30
FV = €
3.NPV & IRRMedium
years01234−20 0006 0008 0004 0009 000

An investment of €20 000 gives these cash flows at the end of years 1 to 4: €6 000, €8 000, €4 000, €9 000. The cost of capital is 12%. Find the net present value NPV.

r = 12%CF0 = −20 000CF1 = 6 000CF2 = 8 000CF3 = 4 000CF4 = 9 000
NPV = €
4.NPV & IRRHard
years0123−50 00035 50018 50027 000

A project costs €50 000 and returns €35 500, €18 500, €27 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 35 500CF2 = 18 500CF3 = 27 000
IRR = %
5.Bond pricing & yieldMedium
years0123…12€1 154C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 12 years and pays annual coupons. It sells for €1 154 when the yield to maturity is 2.5%. Find the coupon rate c in percent.

F = €1 000n = 12P = €1 154YTM = 2.5%
c = %
6.Bond pricing & yieldHard
half-years0123…12€1 028€15€15€15 + €1 000

A bond with a face value of €1 000 pays a 3% annual coupon in two half-yearly payments and matures in 6 years. It sells for €1 028. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 3% (2×)n = 6P = €1 028
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€4 000 000re​ = 14%debt€3 000 000rd​ = 4%T = 25.8%

A company has equity worth €4 000 000 and debt worth €3 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 4% and the tax rate is 25.8%. Find the WACC.

E = €4 000 000D = €3 000 000re = 14%rd = 4%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €13.00β = 0.9rf​ = 1%, rm​ = 8%debt40 000 × €960rd​ = 5%T = 28%

A company has 2 500 000 shares at €13.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 5% and the tax rate is 28%. The share’s beta is 0.9, the risk-free rate 1% and the market return 8%. Find the WACC.

β = 0.9rf = 1%rm = 8%rd = 5%T = 28%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 28, 15, 27, 58, 57, 51.

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

A quantity is normally distributed with mean 50 and standard deviation 20. What proportion P of values is above 40?

μ = 50σ = 20x = 40
P =
11.ProbabilityMedium
xp(x)10.220.130.240.3550.15μ = 3.15

For the distribution shown, find the variance Var(X) of X.

x: 1, 2, 3, 4, 5p(x): 0.2, 0.1, 0.2, 0.35, 0.15
Var(X) =
12.ProbabilityHard
n = 6, p = 0.4: P(X = 4) = ?k0123456

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 4 correct answers?

n = 6p = 0.4k = 4
P =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0749

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0749. Find z to two decimals.

A = 0.0749
z =
14.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 50 phones, mean 101 h, known standard deviation 15 h. Model 2: 100 phones, mean 97 h, known standard deviation 15 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 101 hσ₁ = 15 hn₂ = 100x̄₂ = 97 hσ₂ = 15 h90%
lower limit = h
16.Confidence intervalsHard
σ = 16, n = ?x̄ − Ex̄ + Ex̄E = 0.595%

The population standard deviation is 16. 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 0.5?

σ = 16E = 0.595%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

Two teaching methods are compared. Group 1: 12 students, mean score 86.5, standard deviation 10. Group 2: 9 students, mean score 98.8, standard deviation 9. 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 = 12x̄₁ = 86.5s₁ = 10n₂ = 9x̄₂ = 98.8s₂ = 9
t =
18.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 81 measurements has mean 501.1. The population standard deviation is 5. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 501.1σ = 5n = 8195%
U =
19.Linear regression & correlationMedium
xy175019462162236125692773xy2025506070

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy347455556657762xy46505560x = 4.5

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 = 4.5.

n = 5x = 4.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 17740 €

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 = 0.02/12 = 0.001667
N = 12·6 = 72

Present value

PV = FV/(1 + i)N = 20 000/1.127 = €17 740.18

Answer

PV = €17 740.18
2.Time value of money

Answer: FV = 348804 €

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.

Ordinary annuity factor

((1 + 0.05)30 − 1)/0.05 = 66.44

Annuity due

Each deposit is invested one year longer.

FV = 5 000·66.44·1.05 = €348 803.95

Answer

FV = €348 803.95
3.NPV & IRR

Answer: NPV = 301.5 €

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 = −20 000
CF1/(1 + 0.12)1 = 6 000/1.12 = 5 357.14
CF2/(1 + 0.12)2 = 8 000/1.254 = 6 377.55
CF3/(1 + 0.12)3 = 4 000/1.405 = 2 847.12
CF4/(1 + 0.12)4 = 9 000/1.574 = 5 719.66

Add them up

NPV = €301.48

Answer

NPV = €301.48

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

4.NPV & IRR

Answer: IRR = 30.83 %

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(30%) = 543.92
NPV(35%) = −2 578.88

Narrow down

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

IRR ≈ 30 + 5·543.92/(543.92 − (−2 578.88)) = 30.87%
exact: NPV(30.83%) = 0 → IRR = 30.83%

Answer

IRR = 30.83%
5.Bond pricing & yield

Answer: c = 4.001 %

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.

Present value of the face value

1000·(1 + 0.025)−12 = €743.56

Coupon

C = (1154 − 743.56)/10.26 = €40.01

Coupon rate

c = 40.01/1000 = 4.001%

Answer

c = 4.001%
6.Bond pricing & yield

Answer: YTM = 2.495 %

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 = €15.00, N = 12

Solve numerically

i = 1.247% per half-year

YTM

YTM = 2·1.247% = 2.495%

Answer

YTM = 2.495%
7.CAPM & WACC

Answer: WACC = 9.272 %

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 = 0.5714, D/V = 0.4286

WACC

WACC = 0.5714·14% + 0.4286·4%·(1 − 0.258) = 9.272%

Answer

WACC = 9.272%
8.CAPM & WACC

Answer: WACC = 5.296 %

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·€13.00 = €32 500 000
D = 40 000·€960 = €38 400 000

Cost of equity (CAPM)

re = 1% + 0.9·7% = 7.3%

Weights

V = €70 900 000, E/V = 0.4584, D/V = 0.5416

WACC

WACC = 0.4584·7.3% + 0.5416·5%·(1 − 0.28) = 5.296%

Answer

WACC = 5.296%
9.Descriptive statistics

Answer: s = 18.27

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 = 236/6 = 39.33

Squared deviations

Σ(x − x̄)² = (−11.33)² + (−24.33)² + (−12.33)² + 18.67² + 17.67² + 11.67² = 1669

Standard deviation

s = √(1669/5) = 18.27

Answer

s = 18.27
10.Descriptive statistics

Answer: P = 0.6915

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
11.Probability

Answer: Var(X) = 1.828

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.

Mean

μ = 1·0.2 + 2·0.1 + 3·0.2 + 4·0.35 + 5·0.15 = 3.15

Mean of the squares

E(X²) = 1·0.2 + 4·0.1 + 9·0.2 + 16·0.35 + 25·0.15 = 11.75

Variance

Var(X) = 11.75 − 3.15² = 1.828

Answer

Var(X) = 1.828
12.Probability

Answer: P = 0.1382

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 = 4) = 15·0.44·0.62 = 15·0.0256·0.36 = 0.1382

Answer

P = 0.1382
13.Normal distribution

Answer: z = 1.44

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 − 0.0749 = 0.9251

Read the z-table backwards

Find 0.9251 inside the table and read off z.

z = 1.44

Answer

z = 1.44
14.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
15.Confidence intervals

Answer: lower limit = −0.2738 h

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.

Point estimate

x̄₁ − x̄₂ = 101 − 97 = 4

Standard error

SE = √(15²/50 + 15²/100) = 2.598

Margin of error

E = 1.645·2.598 = 4.274

Interval

4 ± 4.274 → [−0.2738, 8.274]

Answer

lower limit = −0.2738 h

The interval contains 0, so the data do not show a significant difference.

16.Confidence intervals

Answer: n = 3934

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 = (1.96·16/0.5)² = 3934

Round up

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

n = 3934

Answer

n = 3934
17.Hypothesis tests & confidence intervals

Answer: t = −2.908

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 = (11·10² + 8·9²)/19 = 92
sp = 9.592

Standard error

SE = 9.592·√(1/12 + 1/9) = 4.23

Test statistic

t = (86.5 − 98.8)/4.23 = −2.908

Answer

t = −2.908

With df = 19, the two-sided 5% critical value is 2.093; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: U = 502.2

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).

Margin of error

E = 1.96·5/√81 = 1.089

Interval limit

U = 501.1 + 1.089 = 502.2

Answer

U = 502.2
19.Linear regression & correlation

Answer: r² = 0.8685

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 = 132/6 = 22
ȳ = 361/6 = 60.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 183
Syy = Σ(y − ȳ)² = 550.8

Coefficient of determination

r² = 183²/(70·550.8) = 0.8685

Answer

r² = 0.8685

About 86.85% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 53.8

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 = 25/5 = 5
ȳ = 277/5 = 55.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 32

Regression line

b = 32/10 = 3.2
a = 55.4 − 3.2·5 = 39.4
ŷ = 39.4 + 3.2·x

Prediction

ŷ = 39.4 + 3.2·4.5 = 53.8

Answer

ŷ = 53.8

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