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…14€1 000FV = ?

You invest €1 000 at 3% per year with continuous compounding. Find the value FV after 14 years.

PV = €1 000r = 3%t = 14
FV = €
2.Time value of moneyHard
months0123…228A?AA€100 000

You want €100 000 in 19 years. You save the same amount at the end of every month in an account paying 5.4% per year, compounded monthly. Find the monthly deposit A.

FV = €100 0005.4% / 12N = 228
A = €
3.NPV & IRRMedium
years0123−60 00022 00014 00020 000

An investment of €60 000 gives these cash flows at the end of years 1 to 3: €22 000, €14 000, €20 000. The cost of capital is 15%. Find the net present value NPV.

r = 15%CF0 = −60 000CF1 = 22 000CF2 = 14 000CF3 = 20 000
NPV = €
4.NPV & IRRHard
years012345−50 00011 000+5%+5%+5%+5%r = 10%, NPV = ?

A project costs €50 000 today. Its cash flow is €11 000 at the end of year 1 and grows by 5% per year until the last cash flow at the end of year 5. The cost of capital is 10%. Find the net present value NPV.

C0 = €50 000CF1 = €11 000g = 5%n = 5r = 10%
NPV = €
5.Bond pricing & yieldMedium
half-years0123…24€20€20€20 + €1 000P = ?

A bond with a face value of €1 000 pays a 4% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 5% per year. Find the price P of the bond.

F = €1 000coupon = 4%YTM = 5%n = 12
P = €
6.Bond pricing & yieldHard
years0123456€1 050€50€50 + €1 000

A bond with a face value of €1 000 and an annual coupon of 5% matures in 6 years and sells for €1 050. Find its yield to maturity YTM in percent.

F = €1 000coupon = 5%n = 6P = €1 050
YTM = %
7.CAPM & WACCMedium
WACC = 8.6%equity€5 000 000re​ = ?debt€5 000 000rd​ = 6%T = 30%

A company has equity worth €5 000 000 and debt worth €5 000 000. Its WACC is 8.6%, its pre-tax cost of debt is 6% and the tax rate is 30%. Find the cost of equity re.

E = €5 000 000D = €5 000 000WACC = 8.6%rd = 6%T = 30%
re = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €33.50β = 0.9rf​ = 3.5%, rm​ = 9.5%debt40 000 × €1 020rd​ = 7%T = 21%

A company has 2 500 000 shares at €33.50 each and 40 000 bonds trading at €1 020 each. The pre-tax cost of debt is 7% and the tax rate is 21%. The share’s beta is 0.9, the risk-free rate 3.5% and the market return 9.5%. Find the WACC.

β = 0.9rf = 3.5%rm = 9.5%rd = 7%T = 21%
WACC = %
9.Descriptive statisticsMedium
15202530354045

Find the sample standard deviation s of the data: 23, 17, 45, 11, 41, 22.

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

A quantity is normally distributed with mean 100 and standard deviation 10. What proportion P of values is above 80?

μ = 100σ = 10x = 80
P =
11.ProbabilityMedium
xp(x)00.1510.320.330.25μ = 1.65

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

x: 0, 1, 2, 3p(x): 0.15, 0.3, 0.3, 0.25
Var(X) =
12.ProbabilityHard
xp(x)10.3520.2530.2540.15μ

X has the distribution shown, and Y = −2X + 1. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.35, 0.25, 0.25, 0.15Y = −2X + 1
σY =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0170

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0170. Find z to two decimals.

A = 0.0170
z =
14.Normal distributionHard
μ = 32x = ?75%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 75% of them lie (the 75th percentile).

μ = 32σ = 6P = 75%
x = min
15.Confidence intervalsMedium
x̄E₁ = 4.499%, n = 50x̄E₂ = ?95%, n = 50

A 99% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 4.4. What is the margin of error E₂ of a 95% interval from the same data?

99%n₁ = 50E₁ = 4.495%
E₂ =
16.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0190%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.01?

p = 0.5E = 0.0190%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.262.26p = ?

The mean of a population is claimed to be 100; the population standard deviation is known to be 12. A random sample of 64 observations has a mean of 103.39. Test H₀: μ = 100 against H₁: μ ≠ 100. Find the p-value p.

μ₀ = 100σ = 12n = 64x̄ = 103.39H₁: μ ≠ μ₀
p =
18.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 49 measurements has mean 501. The population standard deviation is 4. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 501σ = 4n = 4999%
U =
19.Linear regression & correlationMedium
xy1869218824103271183013733157xy202530100150

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
xy123222324422519617xy24620x = 2.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 20) is made. Find its residual e = y − ŷ.

n = 6(2.5, 20)
e =

3Solutions

1.Time value of money

Answer: FV = 1522 €

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.

Exponent

r·t = 0.03·14 = 0.42
e0.42 = 1.522

Future value

FV = 1 000·1.522 = €1 521.96

Answer

FV = €1 521.96
2.Time value of money

Answer: A = 252.3 €

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 = 0.054000000000000006/12 = 0.0045
N = 12·19 = 228

Deposit

A = FV·i/((1 + i)N − 1) = 100 000·0.0045/(2.783 − 1) = €252.32

Answer

A = €252.32
3.NPV & IRR

Answer: NPV = −17133 €

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 + 0.15)1 = 22 000/1.15 = 19 130.43
CF2/(1 + 0.15)2 = 14 000/1.322 = 10 586.01
CF3/(1 + 0.15)3 = 20 000/1.521 = 13 150.32

Add them up

NPV = €−17 133.23

Answer

NPV = €−17 133.23

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

4.NPV & IRR

Answer: NPV = −4343 €

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 growing annuity

((1.05)/(1.1))5 = 0.7925
PV = 11 000/(0.1 − 0.05)·(1 − 0.7925) = 45 656.50

NPV

NPV = 45 656.50 − 50 000 = −€4 343.50

Answer

NPV = −€4 343.50

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

5.Bond pricing & yield

Answer: P = 910.6 €

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 = €20.00, i = 0.025, N = 24

Present value of the coupons

€20.00·(1 − 0.5529)/0.025 = €357.70

Present value of the face value

€1 000·0.5529 = €552.88

Price

P = €910.58

Answer

P = €910.58

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

6.Bond pricing & yield

Answer: YTM = 4.045 %

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

€1 050 = €50·(1 − (1 + y)⁻⁶)/y + €1 000/(1 + y)⁶

Solve numerically

P(4%) = €1 052.42, P(5%) = €1 000.00
YTM = 4.045%

Answer

YTM = 4.045%
7.CAPM & WACC

Answer: re = 13 %

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, E/V = 0.5, D/V = 0.5

Debt part of the WACC

0.5·6%·(1 − 0.3) = 2.1%

Cost of equity

re = (8.6% − 2.1%)/0.5 = 13%

Answer

re = 13%
8.CAPM & WACC

Answer: WACC = 7.796 %

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·€33.50 = €83 750 000
D = 40 000·€1 020 = €40 800 000

Cost of equity (CAPM)

re = 3.5% + 0.9·6% = 8.9%

Weights

V = €124 550 000, E/V = 0.6724, D/V = 0.3276

WACC

WACC = 0.6724·8.9% + 0.3276·7%·(1 − 0.21) = 7.796%

Answer

WACC = 7.796%
9.Descriptive statistics

Answer: s = 13.53

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 = 159/6 = 26.5

Squared deviations

Σ(x − x̄)² = (−3.5)² + (−9.5)² + 18.5² + (−15.5)² + 14.5² + (−4.5)² = 915.5

Standard deviation

s = √(915.5/5) = 13.53

Answer

s = 13.53
10.Descriptive statistics

Answer: P = 0.9772

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
11.Probability

Answer: Var(X) = 1.028

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 + 1·0.3 + 2·0.3 + 3·0.25 = 1.65

Mean of the squares

E(X²) = 0·0.15 + 1·0.3 + 4·0.3 + 9·0.25 = 3.75

Variance

Var(X) = 3.75 − 1.65² = 1.028

Answer

Var(X) = 1.028
12.Probability

Answer: σY = 2.154

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 = 2.2, Var(X) = 1.16, σX = 1.077

Linear change

σY = |−2|·1.077 = 2.154

Answer

σY = 2.154

The mean does move: E(Y) = −2·2.2 + 1 = −3.4.

13.Normal distribution

Answer: z = 2.12

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 − 0.0170 = 0.9830

Read the z-table backwards

Find 0.9830 inside the table and read off z.

z = 2.12

Answer

z = 2.12
14.Normal distribution

Answer: x = 36.05 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) = 0.75 → z = 0.6745

Back to x

x = 32 + 0.6745·6 = 36.05 min

Answer

x = 36.05 min
15.Confidence intervals

Answer: E₂ = 3.348

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.

Ratio of margins

E₂/E₁ = (1.96/2.576)·√(50/50) = 0.7609

New margin

E₂ = 4.4·0.7609 = 3.348

Answer

E₂ = 3.348
16.Confidence intervals

Answer: n = 6766

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.645²·0.5·0.5/0.01² = 6765

Round up

n = 6766

Answer

n = 6766
17.Hypothesis tests & confidence intervals

Answer: p = 0.02382

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 = 12/√64 = 1.5
z = (103.39 − 100)/1.5 = 2.26

Tail area

Φ(2.26) = 0.9881
1 − Φ(2.26) = 0.0119

p-value

p = 2·0.0119 = 0.02382

Answer

p = 0.02382

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

18.Hypothesis tests & confidence intervals

Answer: U = 502.5

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 = 2.576·4/√49 = 1.472

Interval limit

U = 501 + 1.472 = 502.5

Answer

U = 502.5
19.Linear regression & correlation

Answer: r² = 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.

Means

x̄ = 153/6 = 25.5
ȳ = 672/6 = 112

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 903
Syy = Σ(y − ȳ)² = 5192

Coefficient of determination

r² = 903²/(157.5·5192) = 0.9971

Answer

r² = 0.9971

About 99.71% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = −2.338

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 = 21/6 = 3.5
ȳ = 127/6 = 21.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −20.5

Regression line

b = −20.5/17.5 = −1.171
a = 21.17 − (−1.171)·3.5 = 25.27
ŷ = 25.27 − 1.171·x

Predicted value

ŷ = 25.27 + (−1.171)·2.5 = 22.34

Residual

e = 20 − 22.34 = −2.338

Answer

e = −2.338

The point lies below the line.

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