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 = ?€8 000€8 000€8 000€8 000

An annuity pays €8 000 at the end of each year for 6 years. The interest rate is 8% per year. Find the present value PV of the annuity.

A = €8 000i = 8%n = 6
PV = €
2.Time value of moneyHard
months0123…300€150 000A?AAA

A loan of €150 000 is repaid with equal monthly payments over 25 years. The interest rate is 3.6% per year, compounded monthly. Find the monthly payment A.

P = €150 0003.6% / 12N = 300
A = €
3.NPV & IRRMedium
years0123…9−100 0007 000+2%+2%…r = 8%, NPV = ?

A project costs €100 000 today. It generates €7 000 at the end of the first year, and the cash flow then grows by 2% per year forever. The cost of capital is 8%. Find the net present value NPV.

C0 = €100 000CF1 = €7 000g = 2%r = 8%
NPV = €
4.NPV & IRRHard
years012−50 00034 50026 000

A project costs €50 000 and returns €34 500, €26 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 34 500CF2 = 26 000
IRR = %
5.Bond pricing & yieldMedium
years0123€1 028C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 3 years and pays annual coupons. It sells for €1 028 when the yield to maturity is 3.5%. Find the coupon rate c in percent.

F = €1 000n = 3P = €1 028YTM = 3.5%
c = %
6.Bond pricing & yieldHard

A bond trades at €1 005 with a yield to maturity of 3% (annual compounding) and a Macaulay duration of 11.1 years. Use the modified duration to estimate the new price P if the yield falls by 25 basis points.

P0 = €1 005y = 3%D = 11.1Δy = −25 bp
P = €
7.CAPM & WACCMedium
WACC = ?equityEre​ = 13%debtD = 0.25·Erd​ = 6%T = 28%

A company has a debt-to-equity ratio (at market values) of 0.25. Its cost of equity is 13%, its pre-tax cost of debt is 6% and the tax rate is 28%. Find the WACC.

D/E = 0.25re = 13%rd = 6%T = 28%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €55.00β = 1.2rf​ = 2%, rm​ = 6.5%debt20 000 × €1 020rd​ = 3%T = 25%

A company has 2 000 000 shares at €55.00 each and 20 000 bonds trading at €1 020 each. The pre-tax cost of debt is 3% and the tax rate is 25%. The share’s beta is 1.2, the risk-free rate 2% and the market return 6.5%. Find the WACC.

β = 1.2rf = 2%rm = 6.5%rd = 3%T = 25%
WACC = %
9.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 45, 20, 54, 47, 47.

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

A quantity is normally distributed with mean 100 and standard deviation 12. What proportion P of values is above 85.6?

μ = 100σ = 12x = 85.6
P =
11.ProbabilityMedium
xp(x)00.1520.2540.3560.25μ = 3.4

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

x: 0, 2, 4, 6p(x): 0.15, 0.25, 0.35, 0.25
Var(X) =
12.ProbabilityHard
€p(x)−10.8440.15490.01

A game costs €1 to play. You win €5 with probability 0.15 and €50 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€5: p = 0.15€50: p = 0.01
E(G) = €
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.2661

Drag z so that the shaded area between −z and z under the standard normal curve is 0.2661. Find z to two decimals.

A = 0.2661
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
n = 21, s = 17LU = ?x̄ = 118.5E99%

A sample of 21 items has mean 118.5 and standard deviation s = 17. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 21x̄ = 118.5s = 1799%t(20) = 2.845
upper limit =
16.Confidence intervalsHard
σ = 14, n = ?x̄ − Ex̄ + Ex̄E = 2.595%

The population standard deviation is 14. 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 2.5?

σ = 14E = 2.595%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.1312.131t = ?df = 15

Two teaching methods are compared. Group 1: 9 students, mean score 105.5, standard deviation 10. Group 2: 8 students, mean score 100.4, standard deviation 11. 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 = 9x̄₁ = 105.5s₁ = 10n₂ = 8x̄₂ = 100.4s₂ = 11
t =
18.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 25 measurements has mean 48.3. The population standard deviation is 4. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 48.3σ = 4n = 2595%
L =
19.Linear regression & correlationMedium
xy1523595687719791189xy5106080

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 = 36Σy = 418Σx² = 286Σy² = 30012Σxy = 2756
r =
20.Linear regression & correlationHard
xy127222322419514xy24152025ŷ = 25

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 = 25?

n = 5ŷ = 25
x =

3Solutions

1.Time value of money

Answer: PV = 36983 €

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.

Annuity factor

(1 − (1 + i)−n)/i = (1 − 0.6302)/0.08 = 4.623

Present value

PV = 8 000·4.623 = €36 983.04

Answer

PV = €36 983.04
2.Time value of money

Answer: A = 759 €

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 = 0.036000000000000004/12 = 0.003
N = 12·25 = 300

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 = 0.4071
A = 150 000·0.003/(1 − 0.4071) = €759.00

Answer

A = €759.00
3.NPV & IRR

Answer: NPV = 16667 €

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.

Value of the cash flows

PV = 7 000/(0.08 − 0.02) = 116 666.67

NPV

NPV = 116 666.67 − 100 000 = €16 666.67

Answer

NPV = €16 666.67

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

4.NPV & IRR

Answer: IRR = 14.44 %

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(10%) = 2 851.24
NPV(15%) = −340.26

Narrow down

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

IRR ≈ 10 + 5·2 851.24/(2 851.24 − (−340.26)) = 14.47%
exact: NPV(14.44%) = 0 → IRR = 14.44%

Answer

IRR = 14.44%
5.Bond pricing & yield

Answer: c = 4.499 %

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 + 0.035)−3 = €901.94

Coupon

C = (1028 − 901.94)/2.802 = €44.99

Coupon rate

c = 44.99/1000 = 4.499%

Answer

c = 4.499%
6.Bond pricing & yield

Answer: P = 1032 €

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.

Modified duration

Dmod = 11.1/1.03 = 10.78

Estimated price change

%ΔP ≈ −10.78·(−0.0025) = 2.694%
P ≈ 1005·(1 + 0.02694) = €1 032.08

Answer

P = €1 032.08

Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.

7.CAPM & WACC

Answer: WACC = 11.26 %

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 from D/E

E/V = 1/(1 + 0.25) = 0.8
D/V = 0.25/(1 + 0.25) = 0.2

WACC

WACC = 0.8·13% + 0.2·6%·(1 − 0.28) = 11.26%

Answer

WACC = 11.26%
8.CAPM & WACC

Answer: WACC = 6.594 %

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 = 2 000 000·€55.00 = €110 000 000
D = 20 000·€1 020 = €20 400 000

Cost of equity (CAPM)

re = 2% + 1.2·4.5% = 7.4%

Weights

V = €130 400 000, E/V = 0.8436, D/V = 0.1564

WACC

WACC = 0.8436·7.4% + 0.1564·3%·(1 − 0.25) = 6.594%

Answer

WACC = 6.594%
9.Descriptive statistics

Answer: s = 13.09

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 = 213/5 = 42.6

Squared deviations

Σ(x − x̄)² = 2.4² + (−22.6)² + 11.4² + 4.4² + 4.4² = 685.2

Standard deviation

s = √(685.2/4) = 13.09

Answer

s = 13.09
10.Descriptive statistics

Answer: P = 0.8849

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 = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
11.Probability

Answer: Var(X) = 4.04

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 = 0·0.15 + 2·0.25 + 4·0.35 + 6·0.25 = 3.4

Mean of the squares

E(X²) = 0·0.15 + 4·0.25 + 16·0.35 + 36·0.25 = 15.6

Variance

Var(X) = 15.6 − 3.4² = 4.04

Answer

Var(X) = 4.04
12.Probability

Answer: E(G) = 0.25 €

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: 4 (p = 0.15), 49 (p = 0.01), −1 (p = 0.84)

Expected value

E(G) = 4·0.15 + 49·0.01 − 1·0.84 = 0.25 €

Answer

E(G) = 0.25 €

A positive value means you win money on average.

13.Normal distribution

Answer: z = 0.34

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 + 0.2661)/2 = 0.6331

Read the z-table backwards

Find 0.6331 inside the table and read off z.

z = 0.34

Answer

z = 0.34
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: upper limit = 129.1

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 = 20, t = 2.845

Margin of error

E = 2.845·17/√21 = 10.55

Interval

118.5 ± 10.55 → [107.9, 129.1]

Answer

upper limit = 129.1
16.Confidence intervals

Answer: n = 121

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 = (1.96·14/2.5)² = 120.5

Round up

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

n = 121

Answer

n = 121
17.Hypothesis tests & confidence intervals

Answer: t = 1.002

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 = (8·10² + 7·11²)/15 = 109.8
sp = 10.48

Standard error

SE = 10.48·√(1/9 + 1/8) = 5.092

Test statistic

t = (105.5 − 100.4)/5.092 = 1.002

Answer

t = 1.002

With df = 15, the two-sided 5% critical value is 2.131; |t| is smaller, so H₀ is not rejected.

18.Hypothesis tests & confidence intervals

Answer: L = 46.73

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 = 1.96·4/√25 = 1.568

Interval limit

L = 48.3 − 1.568 = 46.73

Answer

L = 46.73
19.Linear regression & correlation

Answer: r = 0.9928

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 = 2756 − 36·418/6 = 248
Sxx = 286 − 36²/6 = 70
Syy = 30012 − 418²/6 = 891.3

Correlation coefficient

r = 248/√(70·891.3) = 0.9928

Answer

r = 0.9928
20.Linear regression & correlation

Answer: x = 1.552

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 = 104/5 = 20.8

Sums of squares and products

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

Regression line

b = −29/10 = −2.9
a = 20.8 − (−2.9)·3 = 29.5
ŷ = 29.5 − 2.9·x

Solve for x

x = (25 − 29.5)/(−2.9) = 1.552

Answer

x = 1.552

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