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…9PV = ?€20 000+1.5%+1.5%…r = 6.5%

A property is expected to earn net rent of €20 000 next year. The rent is expected to grow by 1.5% per year forever. The required return is 6.5% per year. Find the value PV of the property today.

C1 = €20 000g = 1.5%r = 6.5%
PV = €
2.Time value of moneyHard
months0123…84€15 000AAAA3% / 12, I = ?

A loan of €15 000 is repaid with equal monthly payments over 7 years at 3% per year, compounded monthly. How much interest I is paid in total over the life of the loan?

P = €15 0003% / 12N = 84
I = €
3.NPV & IRRMedium
years0123456−50 00019 00013 50020 50016 00011 50014 500

A project costs €50 000 today and returns €19 000, €13 500, €20 500, €16 000, €11 500, €14 500 at the end of years 1 to 6. The cost of capital is 8%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 8%CF0 = −50 000CF1 = 19 000CF2 = 13 500CF3 = 20 500CF4 = 16 000CF5 = 11 500CF6 = 14 500
DPB = years
4.NPV & IRRHard
years0123…8−50 0007 500+3%+3%+3%r = 10%, NPV = ?

A project costs €50 000 today. Its cash flow is €7 500 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 8. The cost of capital is 10%. Find the net present value NPV.

C0 = €50 000CF1 = €7 500g = 3%n = 8r = 10%
NPV = €
5.Bond pricing & yieldMedium
years01234567€1 120C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 7 years and pays annual coupons. It sells for €1 120 when the yield to maturity is 4%. Find the coupon rate c in percent.

F = €1 000n = 7P = €1 120YTM = 4%
c = %
6.Bond pricing & yieldHard
years012345€915€40€40 + €1 000

A bond with a face value of €1 000 and an annual coupon of 4% matures in 5 years and sells for €915. Find its yield to maturity YTM in percent.

F = €1 000coupon = 4%n = 5P = €915
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€4 000 000re​ = 13%debt€2 000 000rd​ = 4.5%T = 30%

A company has equity worth €4 000 000 and debt worth €2 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 4.5% and the tax rate is 30%. Find the WACC.

E = €4 000 000D = €2 000 000re = 13%rd = 4.5%T = 30%
WACC = %
8.CAPM & WACCHard
WACC = ?equity5 000 000 × €41.00D0​ = €0.90g = 4%debt€205 000 000rd​ = 4%T = 20%

A company has 5 000 000 shares trading at €41.00; the book value of its equity is €123 000 000. It just paid a dividend of €0.90 per share, and dividends grow by 4% per year. Its debt has a market value of €205 000 000 and a pre-tax cost of 4%. The tax rate is 20%. Find the WACC, using the dividend growth model for the cost of equity.

N = 5 000 000P0 = €41.00book equity = €123 000 000D0 = €0.90g = 4%D = €205 000 000rd = 4%T = 20%
WACC = %
9.Descriptive statisticsMedium
15202530354045

Find the sample standard deviation s of the data: 31, 44, 32, 48, 12, 19.

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

Exam scores are normally distributed with mean 70 and standard deviation 10. Find the z-score z of a score of 76.

μ = 70σ = 10x = 76
z =
11.ProbabilityMedium
P(at least one six) = ??1?2?3P(six) = 1/6 on each roll

A fair die is rolled 3 times. What is the probability P of getting at least one six?

n = 3
P =
12.ProbabilityHard
xp(x)00.3510.3520.1530.15μ

X has the distribution shown, and Y = 0.5X − 3. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.35, 0.35, 0.15, 0.15Y = 0.5X − 3
σY =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9297

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9297. Find z to two decimals.

A = 0.9297
z =
14.Normal distributionHard
μ = 65x = ?80%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 80% of them lie (the 80th percentile).

μ = 65σ = 10P = 80%
x = points
15.Confidence intervalsMedium
x̄E₁ = 5.190%, n = 64x̄E₂ = ?95%, n = 16

A 90% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 5.1. What is the margin of error E₂ of a 95% interval based on 16 observations?

90%n₁ = 64E₁ = 5.195%n₂ = 16
E₂ =
16.Confidence intervalsHard
σ = 38, n = ?x̄ − Ex̄ + Ex̄E = 395%

The population standard deviation is 38. 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 3?

σ = 38E = 395%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.821.82p = ?

The mean of a population is claimed to be 120; the population standard deviation is known to be 15. A random sample of 36 observations has a mean of 124.55. Test H₀: μ = 120 against H₁: μ ≠ 120. Find the p-value p.

μ₀ = 120σ = 15n = 36x̄ = 124.55H₁: μ ≠ μ₀
p =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.05βσ = 22, power 80%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 22. Find the minimum sample size n.

α = 0.05power = 80%δ = 4σ = 22zα = 1.645zβ = 0.842
n =
19.Linear regression & correlationMedium
xy1243315327399351140xy51025303540

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy123220319420520615xy2461520x = 4.5

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. Predict ŷ for x = 4.5.

n = 6x = 4.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 400000 €

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.

Growing perpetuity

PV = C1/(r − g) = 20 000/(0.065 − 0.015) = 20 000/0.05 = €400 000.00

Answer

PV = €400 000.00
2.Time value of money

Answer: I = 1649 €

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 = 0.0025, N = 84
A = 15 000·0.0025/(1 − (1 + 0.0025)−84) = €198.20

Total interest

I = N·A − P = 84·198.2 − 15 000 = €1 648.76

Answer

I = €1 648.76
3.NPV & IRR

Answer: DPB = 3.388 years

Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.

Discounted and cumulative cash flows

year 1: 19 000/1.08 = 17 592.59 → Σ = 17 592.59
year 2: 13 500/1.166 = 11 574.07 → Σ = 29 166.67
year 3: 20 500/1.26 = 16 273.56 → Σ = 45 440.23
year 4: 16 000/1.36 = 11 760.48 → Σ = 57 200.71

Fraction of the last year

DPB = 3 + 4 559.77/11 760.48 = 3.388

Answer

DPB = 3.388 years

The discounted payback is always longer than the simple payback (2.854 years).

4.NPV & IRR

Answer: NPV = −6174 €

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

((1.03)/(1.1))8 = 0.591
PV = 7 500/(0.1 − 0.03)·(1 − 0.591) = 43 825.97

NPV

NPV = 43 825.97 − 50 000 = −€6 174.03

Answer

NPV = −€6 174.03

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

5.Bond pricing & yield

Answer: c = 5.999 %

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 + 0.04)−7 = €759.92

Coupon

C = (1120 − 759.92)/6.002 = €59.99

Coupon rate

c = 59.99/1000 = 5.999%

Answer

c = 5.999%
6.Bond pricing & yield

Answer: YTM = 6.019 %

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

€915 = €40·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵

Solve numerically

P(6%) = €915.75, P(7%) = €876.99
YTM = 6.019%

Answer

YTM = 6.019%
7.CAPM & WACC

Answer: WACC = 9.717 %

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 = €6 000 000, E/V = 0.6667, D/V = 0.3333

WACC

WACC = 0.6667·13% + 0.3333·4.5%·(1 − 0.3) = 9.717%

Answer

WACC = 9.717%
8.CAPM & WACC

Answer: WACC = 4.741 %

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).

Cost of equity

D1 = 0.9·1.04 = €0.94
re = 0.936/41 + 0.04 = 6.283%

Market values

E = 5 000 000·€41.00 = €205 000 000
V = €410 000 000, E/V = 0.5, D/V = 0.5

WACC

WACC = 0.5·6.283% + 0.5·4%·(1 − 0.2) = 4.741%

Answer

WACC = 4.741%
9.Descriptive statistics

Answer: s = 13.89

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 = 186/6 = 31

Squared deviations

Σ(x − x̄)² = 0² + 13² + 1² + 17² + (−19)² + (−12)² = 964

Standard deviation

s = √(964/5) = 13.89

Answer

s = 13.89
10.Descriptive statistics

Answer: z = 0.6

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 = (76 − 70)/10 = 0.6

Answer

z = 0.6
11.Probability

Answer: P = 0.4213

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.

Complement

P(no six) = (5/6)3 = 0.5787
P = 1 − 0.5787 = 0.4213

Answer

P = 0.4213
12.Probability

Answer: σY = 0.522

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 = 1.1, Var(X) = 1.09, σX = 1.044

Linear change

σY = |0.5|·1.044 = 0.522

Answer

σY = 0.522

The mean does move: E(Y) = 0.5·1.1 − 3 = −2.45.

13.Normal distribution

Answer: z = 1.81

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 + 0.9297)/2 = 0.9649

Read the z-table backwards

Find 0.9649 inside the table and read off z.

z = 1.81

Answer

z = 1.81
14.Normal distribution

Answer: x = 73.42 points

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) = 0.8 → z = 0.8416

Back to x

x = 65 + 0.8416·10 = 73.42 points

Answer

x = 73.42 points
15.Confidence intervals

Answer: E₂ = 12.15

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 = (1.96/1.645)·√(64/16) = 2.383

New margin

E₂ = 5.1·2.383 = 12.15

Answer

E₂ = 12.15
16.Confidence intervals

Answer: n = 617

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 = (1.96·38/3)² = 616.4

Round up

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

n = 617

Answer

n = 617
17.Hypothesis tests & confidence intervals

Answer: p = 0.06876

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 = 15/√36 = 2.5
z = (124.55 − 120)/2.5 = 1.82

Tail area

Φ(1.82) = 0.9656
1 − Φ(1.82) = 0.0344

p-value

p = 2·0.0344 = 0.06876

Answer

p = 0.06876

p > 0.05, so H₀ is not rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: n = 188

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 = ((1.645 + 0.842)·22/4)² = 187.1

Round up

n = 188

Answer

n = 188
19.Linear regression & correlation

Answer: r² = 0.807

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 = 36/6 = 6
ȳ = 201/6 = 33.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 99
Syy = Σ(y − ȳ)² = 173.5

Coefficient of determination

r² = 99²/(70·173.5) = 0.807

Answer

r² = 0.807

About 80.7% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 18.39

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 = 117/6 = 19.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −19.5

Regression line

b = −19.5/17.5 = −1.114
a = 19.5 − (−1.114)·3.5 = 23.4
ŷ = 23.4 − 1.114·x

Prediction

ŷ = 23.4 + (−1.114)·4.5 = 18.39

Answer

ŷ = 18.39

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