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…10PV = ?€40 000

You need €40 000 in 10 years. An account pays 5% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €40 0005% / 12n = 10
PV = €
2.Time value of moneyHard
months0123…204€500€500€500FV = ?

You save €500 at the end of every month for 17 years. The account pays 7.2% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €5007.2% / 12N = 204
FV = €
3.NPV & IRRMedium
years0123…9−200 00011 000+2%+2%…r = 8%, NPV = ?

A project costs €200 000 today. It generates €11 000 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 = €200 000CF1 = €11 000g = 2%r = 8%
NPV = €
4.NPV & IRRHard
years0123−50 00037 00033 00029 500

A project costs €50 000 and returns €37 000, €33 000, €29 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 37 000CF2 = 33 000CF3 = 29 500
IRR = %
5.Bond pricing & yieldMedium
years0123…12€1 051C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 12 years and pays annual coupons. It sells for €1 051 when the yield to maturity is 2.5%. Find the coupon rate c in percent.

F = €1 000n = 12P = €1 051YTM = 2.5%
c = %
6.Bond pricing & yieldHard
years01234567€1 120€60€60 + €1 000

A bond with a face value of €1 000 and an annual coupon of 6% matures in 7 years and sells for €1 120. Find its yield to maturity YTM in percent.

F = €1 000coupon = 6%n = 7P = €1 120
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€10 000 000re​ = 10%preferred shares€500 000rp​ = 7.5%debt€5 000 000rd​ = 3%T = 20%

A company is financed with ordinary shares worth €10 000 000, preferred shares worth €500 000 and debt worth €5 000 000. The cost of equity is 10%, the cost of preferred stock 7.5% and the pre-tax cost of debt 3%. The tax rate is 20%. Find the WACC.

E = €10 000 000Ps = €500 000D = €5 000 000re = 10%rp = 7.5%rd = 3%T = 20%
WACC = %
8.CAPM & WACCHard
WACC = ?equity5 000 000 × €42.50β = 1.3rf​ = 3.5%, rm​ = 9.5%debt40 000 × €1 000rd​ = 3.5%T = 28%

A company has 5 000 000 shares at €42.50 each and 40 000 bonds trading at €1 000 each. The pre-tax cost of debt is 3.5% and the tax rate is 28%. The share’s beta is 1.3, the risk-free rate 3.5% and the market return 9.5%. Find the WACC.

β = 1.3rf = 3.5%rm = 9.5%rd = 3.5%T = 28%
WACC = %
9.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 20, 58, 44, 52, 54, 22, 52.

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

Exam scores are normally distributed with mean 100 and standard deviation 20. Find the z-score z of a score of 120.

μ = 100σ = 20x = 120
z =
11.ProbabilityMedium
xp(x)00.1510.220.2530.340.1μ = 2

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

x: 0, 1, 2, 3, 4p(x): 0.15, 0.2, 0.25, 0.3, 0.1
Var(X) =
12.ProbabilityHard
n = 9, p = 0.3: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?

n = 9p = 0.3
P =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.6102

Drag z so that the shaded area between −z and z under the standard normal curve is 0.6102. Find z to two decimals.

A = 0.6102
z =
14.Normal distributionHard
μ = 32x = ?25%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 25% of them lie (the 25th percentile).

μ = 32σ = 6P = 25%
x = min
15.Confidence intervalsMedium
n = 8, s = 3L = ?Ux̄ = 64E99%

A sample of 8 items has mean 64 and standard deviation s = 3. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 8x̄ = 64s = 399%t(7) = 3.499
lower limit =
16.Confidence intervalsHard
σ = 39, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

The population standard deviation is 39. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 1.5?

σ = 39E = 1.599%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 193.4 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 193.4s = 15n = 49
t =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 1.5α = 0.01βσ = 6, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 1.5 units with a power of 90%. The standard deviation is 6. Find the minimum sample size n.

α = 0.01power = 90%δ = 1.5σ = 6zα = 2.326zβ = 1.282
n =
19.Linear regression & correlationMedium
xy363470575680785xy467080

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 25Σy = 373Σx² = 135Σy² = 28119Σxy = 1919
r =
20.Linear regression & correlationHard
xy2114216268361045xy5102040x = 3

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 = 3.

n = 5x = 3
ŷ =

3Solutions

1.Time value of money

Answer: PV = 24286 €

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 = 0.05/12 = 0.004167
N = 12·10 = 120

Present value

PV = FV/(1 + i)N = 40 000/1.647 = €24 286.44

Answer

PV = €24 286.44
2.Time value of money

Answer: FV = 199029 €

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 = 0.07200000000000001/12 = 0.006
N = 12·17 = 204

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 500·(3.388 − 1)/0.006 = €199 029.04

Answer

FV = €199 029.04

You paid in €102 000 yourself; the rest is interest.

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 = 11 000/(0.08 − 0.02) = 183 333.33

NPV

NPV = 183 333.33 − 200 000 = −€16 666.67

Answer

NPV = −€16 666.67

NPV is negative, so the project earns less than the required return.

4.NPV & IRR

Answer: IRR = 46.52 %

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(45%) = 889.34
NPV(50%) = −1 925.93

Narrow down

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

IRR ≈ 45 + 5·889.34/(889.34 − (−1 925.93)) = 46.58%
exact: NPV(46.52%) = 0 → IRR = 46.52%

Answer

IRR = 46.52%
5.Bond pricing & yield

Answer: c = 2.997 %

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 = (1051 − 743.56)/10.26 = €29.97

Coupon rate

c = 29.97/1000 = 2.997%

Answer

c = 2.997%
6.Bond pricing & yield

Answer: YTM = 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.

Equation for the yield

€1 120 = €60·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷

Solve numerically

P(4%) = €1 120.04, P(5%) = €1 057.86
YTM = 4.001%

Answer

YTM = 4.001%
7.CAPM & WACC

Answer: WACC = 7.468 %

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 = 10 000 000 + 500 000 + 5 000 000 = €15 500 000
E/V = 0.6452, Ps/V = 0.03226, D/V = 0.3226

WACC

WACC = 0.6452·10% + 0.03226·7.5% + 0.3226·3%·(1 − 0.2) = 7.468%

Answer

WACC = 7.468%
8.CAPM & WACC

Answer: WACC = 9.909 %

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 = 5 000 000·€42.50 = €212 500 000
D = 40 000·€1 000 = €40 000 000

Cost of equity (CAPM)

re = 3.5% + 1.3·6% = 11.3%

Weights

V = €252 500 000, E/V = 0.8416, D/V = 0.1584

WACC

WACC = 0.8416·11.3% + 0.1584·3.5%·(1 − 0.28) = 9.909%

Answer

WACC = 9.909%
9.Descriptive statistics

Answer: s = 15.7

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 = 302/7 = 43.14

Squared deviations

Σ(x − x̄)² = (−23.14)² + 14.86² + 0.8571² + 8.857² + 10.86² + (−21.14)² + 8.857² = 1479

Standard deviation

s = √(1479/6) = 15.7

Answer

s = 15.7
10.Descriptive statistics

Answer: z = 1

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 = (120 − 100)/20 = 1

Answer

z = 1
11.Probability

Answer: Var(X) = 1.5

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 + 1·0.2 + 2·0.25 + 3·0.3 + 4·0.1 = 2

Mean of the squares

E(X²) = 0·0.15 + 1·0.2 + 4·0.25 + 9·0.3 + 16·0.1 = 5.5

Variance

Var(X) = 5.5 − 2² = 1.5

Answer

Var(X) = 1.5
12.Probability

Answer: P = 0.196

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.

Zero and one failure

P(X = 0) = 0.79 = 0.04035
P(X = 1) = 9·0.3·0.78 = 0.1556

Add them

P = 0.04035 + 0.1556 = 0.196

Answer

P = 0.196
13.Normal distribution

Answer: z = 0.86

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 + 0.6102)/2 = 0.8051

Read the z-table backwards

Find 0.8051 inside the table and read off z.

z = 0.86

Answer

z = 0.86
14.Normal distribution

Answer: x = 27.95 min

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) = 0.25 → z = −0.6745

Back to x

x = 32 + −0.6745·6 = 27.95 min

Answer

x = 27.95 min
15.Confidence intervals

Answer: lower limit = 60.29

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 = 7, t = 3.499

Margin of error

E = 3.499·3/√8 = 3.711

Interval

64 ± 3.711 → [60.29, 67.71]

Answer

lower limit = 60.29
16.Confidence intervals

Answer: n = 4486

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 = (2.576·39/1.5)² = 4486

Round up

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

n = 4486

Answer

n = 4486
17.Hypothesis tests & confidence intervals

Answer: t = −3.08

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 = 15/√49 = 15/7 = 2.143

Test statistic

t = (193.4 − 200)/2.143 = −3.08

Answer

t = −3.08
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 209

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((2.326 + 1.282)·6/1.5)² = 208.3

Round up

n = 209

Answer

n = 209
19.Linear regression & correlation

Answer: r = 0.9973

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 = 1919 − 25·373/5 = 54
Sxx = 135 − 25²/5 = 10
Syy = 28119 − 373²/5 = 293.2

Correlation coefficient

r = 54/√(10·293.2) = 0.9973

Answer

r = 0.9973
20.Linear regression & correlation

Answer: ŷ = 15.35

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 = 139/5 = 27.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 166

Regression line

b = 166/40 = 4.15
a = 27.8 − 4.15·6 = 2.9
ŷ = 2.9 + 4.15·x

Prediction

ŷ = 2.9 + 4.15·3 = 15.35

Answer

ŷ = 15.35

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