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
years012345PV = ?€1 500€1 500€1 500€1 500

An annuity pays €1 500 at the end of each year for 5 years. The interest rate is 6% per year. Find the present value PV of the annuity.

A = €1 500i = 6%n = 5
PV = €
2.Time value of moneyHard
months0123…180€200 000AAAA6% / 12, I = ?

A loan of €200 000 is repaid with equal monthly payments over 15 years at 6% per year, compounded monthly. How much interest I is paid in total over the life of the loan?

P = €200 0006% / 12N = 180
I = €
3.NPV & IRRMedium
years01234−60 00019 00020 00020 00023 000

An investment of €60 000 gives these cash flows at the end of years 1 to 4: €19 000, €20 000, €20 000, €23 000. The cost of capital is 9%. Find the net present value NPV.

r = 9%CF0 = −60 000CF1 = 19 000CF2 = 20 000CF3 = 20 000CF4 = 23 000
NPV = €
4.NPV & IRRHard
years0123−10 0004 0003 5005 000

A project costs €10 000 and returns €4 000, €3 500, €5 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −10 000CF1 = 4 000CF2 = 3 500CF3 = 5 000
IRR = %
5.Bond pricing & yieldMedium
years0123…12P€30€30€30 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 3% and matures in 12 years. Its yield to maturity changes from 5.5% to 7%. By how much ΔP does the bond price change?

F = €1 000coupon = 3%n = 12y1 = 5.5%y2 = 7%
ΔP = €
6.Bond pricing & yieldHard
years012345€1 065€60€60 + €1 000

A bond with a face value of €1 000 and an annual coupon of 6% matures in 5 years and sells for €1 065. Find its yield to maturity YTM in percent.

F = €1 000coupon = 6%n = 5P = €1 065
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€6 000 000re​ = 8%debt€1 000 000rd​ = 7%T = 30%

A company has equity worth €6 000 000 and debt worth €1 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 7% and the tax rate is 30%. Find the WACC.

E = €6 000 000D = €1 000 000re = 8%rd = 7%T = 30%
WACC = %
8.CAPM & WACCHard
rf​ = 3%, rm​ − rf​ = 6%D/E = 0.25equityβL​ = 1.1debtD/E = 0.75equityre​ = ?debtT = 25%

A company's shares have a beta of 1.1 at its current debt-to-equity ratio of 0.25. It plans to raise its debt-to-equity ratio to 0.75. The tax rate is 25%, the risk-free rate 3% and the market risk premium 6%. Find the cost of equity re after the change.

βL = 1.1D/E1 = 0.25D/E2 = 0.75T = 25%rf = 3%rm − rf = 6%
re = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 42, 11, 42, 28, 57.

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

A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is above 173?

μ = 170σ = 5x = 173
P =
11.ProbabilityMedium
red: 5blue: 7

A bag holds 5 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 7
P =
12.ProbabilityHard
xp(x)00.3520.240.3560.1μ

X has the distribution shown, and Y = −2X + 5. Find the standard deviation σY of Y.

x: 0, 2, 4, 6p(x): 0.35, 0.2, 0.35, 0.1Y = −2X + 5
σY =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9147

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9147. Find z to two decimals.

A = 0.9147
z =
14.Normal distributionHard
35455565758595pointstop 95%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points95%
x = points
15.Confidence intervalsMedium
n = 8, sd​ = 4.90LUd̄ = 8.8E = ?95%

The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 8.8 and standard deviation 4.9. Find the margin of error E of the 95% confidence interval for the mean difference.

n = 8d̄ = 8.8sd = 4.995%t(7) = 2.365
E = mmHg
16.Confidence intervalsHard
σ = 23, n = ?x̄ − Ex̄ + Ex̄E = 0.595%

The population standard deviation is 23. 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 = 23E = 0.595%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.96p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 20. A random sample of 100 observations has a mean of 46.08. Test H₀: μ = 50 against H₁: μ < 50. Find the p-value p.

μ₀ = 50σ = 20n = 100x̄ = 46.08H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 49 measurements has mean 203.4. The population standard deviation is 20. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 203.4σ = 20n = 4990%
U =
19.Linear regression & correlationMedium
xy2942062583410431250xy5102040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 181Σx² = 364Σy² = 6611Σxy = 1550
r =
20.Linear regression & correlationHard
xy148116871890201082212124125xy15202580100120x = 17

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 17.

n = 6x = 17
ŷ =

3Solutions

1.Time value of money

Answer: PV = 6319 €

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 − 0.7473)/0.06 = 4.212

Present value

PV = 1 500·4.212 = €6 318.55

Answer

PV = €6 318.55
2.Time value of money

Answer: I = 103788 €

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 = 0.005, N = 180
A = 200 000·0.005/(1 − (1 + 0.005)−180) = €1 687.71

Total interest

I = N·A − P = 180·1688 − 200 000 = €103 788.46

Answer

I = €103 788.46
3.NPV & IRR

Answer: NPV = 6002 €

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 = −60 000
CF1/(1 + 0.09)1 = 19 000/1.09 = 17 431.19
CF2/(1 + 0.09)2 = 20 000/1.188 = 16 833.60
CF3/(1 + 0.09)3 = 20 000/1.295 = 15 443.67
CF4/(1 + 0.09)4 = 23 000/1.412 = 16 293.78

Add them up

NPV = €6 002.24

Answer

NPV = €6 002.24

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

4.NPV & IRR

Answer: IRR = 11.55 %

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%) = 285.50
NPV(15%) = −587.66

Narrow down

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

IRR ≈ 10 + 5·285.50/(285.50 − (−587.66)) = 11.63%
exact: NPV(11.55%) = 0 → IRR = 11.55%

Answer

IRR = 11.55%
5.Bond pricing & yield

Answer: ΔP = −102.2 €

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 = €30.00, i = 0.055, N = 12
€30.00·(1 − 0.526)/0.055 + €1 000·0.526 = €258.56 + €525.98 = €784.54

Price at the new yield

C = €30.00, i = 0.07, N = 12
€30.00·(1 − 0.444)/0.07 + €1 000·0.444 = €238.28 + €444.01 = €682.29

Change

ΔP = 682.29 − 784.54 = −€102.24

Answer

ΔP = −€102.24
6.Bond pricing & yield

Answer: YTM = 4.519 %

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

€1 065 = €60·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵

Solve numerically

P(4%) = €1 089.04, P(5%) = €1 043.29
YTM = 4.519%

Answer

YTM = 4.519%
7.CAPM & WACC

Answer: WACC = 7.557 %

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 = 0.8571, D/V = 0.1429

WACC

WACC = 0.8571·8% + 0.1429·7%·(1 − 0.3) = 7.557%

Answer

WACC = 7.557%
8.CAPM & WACC

Answer: re = 11.68 %

Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.

Unlever the beta

βU = 1.1/(1 + 0.75·0.25) = 0.9263

Relever at the new D/E

βL = 0.9263·(1 + 0.75·0.75) = 1.447

Cost of equity (CAPM)

re = 3% + 1.447·6% = 11.68%

Answer

re = 11.68%
9.Descriptive statistics

Answer: s = 17.33

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 = 180/5 = 36

Squared deviations

Σ(x − x̄)² = 6² + (−25)² + 6² + (−8)² + 21² = 1202

Standard deviation

s = √(1202/4) = 17.33

Answer

s = 17.33
10.Descriptive statistics

Answer: P = 0.2743

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
11.Probability

Answer: P = 0.1515

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.

First and second draw

P = 5/12 · 4/11 = 20/132 = 0.1515

Answer

P = 0.1515
12.Probability

Answer: σY = 4.118

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.

Spread of X

μ = 2.4, Var(X) = 4.24, σX = 2.059

Linear change

σY = |−2|·2.059 = 4.118

Answer

σY = 4.118

The mean does move: E(Y) = −2·2.4 + 5 = 0.2.

13.Normal distribution

Answer: z = −1.37

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 − 0.9147 = 0.0853

Read the z-table backwards

Find 0.0853 inside the table and read off z.

z = −1.37

Answer

z = −1.37
14.Normal distribution

Answer: x = 48.6 points

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

Φ(z) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 65 − 1.64·10 = 48.6 points

Answer

x = 48.6 points
15.Confidence intervals

Answer: E = 4.097 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 = 2.365

Margin of error

E = 2.365·4.9/√8 = 4.097

Answer

E = 4.097 mmHg
16.Confidence intervals

Answer: n = 8129

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 = (1.96·23/0.5)² = 8129

Round up

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

n = 8129

Answer

n = 8129
17.Hypothesis tests & confidence intervals

Answer: p = 0.025

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 = 20/√100 = 2
z = (46.08 − 50)/2 = −1.96

Tail area

Φ(1.96) = 0.9750
1 − Φ(1.96) = 0.0250

p-value

p = 0.025

Answer

p = 0.025

p ≤ 0.05, so H₀ is rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: U = 208.1

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 = 1.645·20/√49 = 4.7

Interval limit

U = 203.4 + 4.7 = 208.1

Answer

U = 208.1
19.Linear regression & correlation

Answer: r = 0.9971

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 = 1550 − 42·181/6 = 283
Sxx = 364 − 42²/6 = 70
Syy = 6611 − 181²/6 = 1151

Correlation coefficient

r = 283/√(70·1151) = 0.9971

Answer

r = 0.9971
20.Linear regression & correlation

Answer: ŷ = 92.29

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 = 114/6 = 19
ȳ = 612/6 = 102

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 340

Regression line

b = 340/70 = 4.857
a = 102 − 4.857·19 = 9.714
ŷ = 9.714 + 4.857·x

Prediction

ŷ = 9.714 + 4.857·17 = 92.29

Answer

ŷ = 92.29

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