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
years01234€1 000FV = ?

You invest €1 000 at a nominal rate of 3.6% per year, compounded quarterly. Find the value FV of the investment after 4 years.

PV = €1 0003.6% / 4n = 4
FV = €
2.Time value of moneyHard
months0123…N = ?€10 000€200€200€200€2003.6% / 12

You borrow €10 000 at 3.6% per year, compounded monthly, and repay €200 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.

P = €10 000A = €2003.6% / 12
N = payments
3.NPV & IRRMedium
years0123−30 00012 0008 00013 000

An investment of €30 000 gives these cash flows at the end of years 1 to 3: €12 000, €8 000, €13 000. The cost of capital is 15%. Find the net present value NPV.

r = 15%CF0 = −30 000CF1 = 12 000CF2 = 8 000CF3 = 13 000
NPV = €
4.NPV & IRRHard
years0123−40 00017 00024 50024 500

A project costs €40 000 and returns €17 000, €24 500, €24 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −40 000CF1 = 17 000CF2 = 24 500CF3 = 24 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…24€22.50€22.50€22.50 + €1 000P = ?

A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 4% per year. Find the price P of the bond.

F = €1 000coupon = 4.5%YTM = 4%n = 12
P = €
6.Bond pricing & yieldHard
years0123…8€1 110€40€40 + €1 000

A bond with a face value of €1 000 and an annual coupon of 4% matures in 8 years and sells for €1 110. Find its yield to maturity YTM in percent.

F = €1 000coupon = 4%n = 8P = €1 110
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€6 000 000re​ = 10%preferred shares€1 500 000rp​ = 6%debt€4 000 000rd​ = 3%T = 28%

A company is financed with ordinary shares worth €6 000 000, preferred shares worth €1 500 000 and debt worth €4 000 000. The cost of equity is 10%, the cost of preferred stock 6% and the pre-tax cost of debt 3%. The tax rate is 28%. Find the WACC.

E = €6 000 000Ps = €1 500 000D = €4 000 000re = 10%rp = 6%rd = 3%T = 28%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €29.50β = 1.3rf​ = 1%, rm​ = 7%debt5 000 × €1 020rd​ = 7%T = 30%

A company has 2 000 000 shares at €29.50 each and 5 000 bonds trading at €1 020 each. The pre-tax cost of debt is 7% and the tax rate is 30%. The share’s beta is 1.3, the risk-free rate 1% and the market return 7%. Find the WACC.

β = 1.3rf = 1%rm = 7%rd = 7%T = 30%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 21, 12, 51, 27, 56, 50, 34.

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

A quantity is normally distributed with mean 60 and standard deviation 8. What proportion P of values is below 63.2?

μ = 60σ = 8x = 63.2
P =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6P(six) = 1/6 on each roll

A fair die is rolled 6 times. What is the probability P of getting at least one six?

n = 6
P =
12.ProbabilityHard
n = 8, p = 0.6: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 8p = 0.6
P =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.3688

Drag z so that the shaded area between −z and z under the standard normal curve is 0.3688. Find z to two decimals.

A = 0.3688
z =
14.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
15.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 80 phones, mean 76.5 h, known standard deviation 4 h. Model 2: 50 phones, mean 68.5 h, known standard deviation 15 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 76.5 hσ₁ = 4 hn₂ = 50x̄₂ = 68.5 hσ₂ = 15 h90%
lower limit = h
16.Confidence intervalsHard
p ≈ 0.15, n = ?p̂ − Ep̂ + Ep̂E = 0.0490%

A pilot study suggests that about 0.15 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.04?

p ≈ 0.15E = 0.0490%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 204.8 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 204.8s = 15n = 81
t =
18.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 1.599%
n =
19.Linear regression & correlationMedium
xy16761910022112251252814031162xy15202530100150

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
xy126224319417515611xy24610152025x = 5.5

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 (5.5, 10) is made. Find its residual e = y − ŷ.

n = 6(5.5, 10)
e =

3Solutions

1.Time value of money

Answer: FV = 1154 €

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 = r/m = 0.036000000000000004/4 = 0.009
N = m·n = 4·4 = 16

Future value

FV = 1 000·(1 + 0.009)16 = 1 000·1.154 = €1 154.14

Answer

FV = €1 154.14
2.Time value of money

Answer: N = 55 payments

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.

Rearrange the annuity formula

i = 0.003
(1 + i)−N = 1 − 10 000·0.003/200 = 0.85

Solve with logarithms

N = −ln(0.85)/ln(1.003) = 54.25

Round up to whole payments

54 full payments leave a small balance, which the last, smaller payment clears.

N = 55

Answer

N = 55 payments
3.NPV & IRR

Answer: NPV = −4968 €

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 = −30 000
CF1/(1 + 0.15)1 = 12 000/1.15 = 10 434.78
CF2/(1 + 0.15)2 = 8 000/1.322 = 6 049.15
CF3/(1 + 0.15)3 = 13 000/1.521 = 8 547.71

Add them up

NPV = €−4 968.36

Answer

NPV = €−4 968.36

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

4.NPV & IRR

Answer: IRR = 27.87 %

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(25%) = 1 824.00
NPV(30%) = −1 274.47

Narrow down

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

IRR ≈ 25 + 5·1 824.00/(1 824.00 − (−1 274.47)) = 27.94%
exact: NPV(27.87%) = 0 → IRR = 27.87%

Answer

IRR = 27.87%
5.Bond pricing & yield

Answer: P = 1047 €

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 = €22.50, i = 0.02, N = 24

Present value of the coupons

€22.50·(1 − 0.6217)/0.02 = €425.56

Present value of the face value

€1 000·0.6217 = €621.72

Price

P = €1 047.28

Answer

P = €1 047.28

The yield is below the coupon rate, so the bond sells at a premium.

6.Bond pricing & yield

Answer: YTM = 2.468 %

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

€1 110 = €40·(1 − (1 + y)⁻⁸)/y + €1 000/(1 + y)⁸

Solve numerically

P(2%) = €1 146.51, P(3%) = €1 070.20
YTM = 2.468%

Answer

YTM = 2.468%
7.CAPM & WACC

Answer: WACC = 6.751 %

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 + 1 500 000 + 4 000 000 = €11 500 000
E/V = 0.5217, Ps/V = 0.1304, D/V = 0.3478

WACC

WACC = 0.5217·10% + 0.1304·6% + 0.3478·3%·(1 − 0.28) = 6.751%

Answer

WACC = 6.751%
8.CAPM & WACC

Answer: WACC = 8.49 %

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·€29.50 = €59 000 000
D = 5 000·€1 020 = €5 100 000

Cost of equity (CAPM)

re = 1% + 1.3·6% = 8.8%

Weights

V = €64 100 000, E/V = 0.9204, D/V = 0.07956

WACC

WACC = 0.9204·8.8% + 0.07956·7%·(1 − 0.3) = 8.49%

Answer

WACC = 8.49%
9.Descriptive statistics

Answer: s = 16.87

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 = 251/7 = 35.86

Squared deviations

Σ(x − x̄)² = (−14.86)² + (−23.86)² + 15.14² + (−8.857)² + 20.14² + 14.14² + (−1.857)² = 1707

Standard deviation

s = √(1707/6) = 16.87

Answer

s = 16.87
10.Descriptive statistics

Answer: P = 0.6554

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 0.6554

Answer

P = 0.6554
11.Probability

Answer: P = 0.6651

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)6 = 0.3349
P = 1 − 0.3349 = 0.6651

Answer

P = 0.6651
12.Probability

Answer: P = 0.00852

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) = 0.48 = 0.0006554
P(X = 1) = 8·0.6·0.47 = 0.007864

Add them

P = 0.0006554 + 0.007864 = 0.00852

Answer

P = 0.00852
13.Normal distribution

Answer: z = 0.48

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 + 0.3688)/2 = 0.6844

Read the z-table backwards

Find 0.6844 inside the table and read off z.

z = 0.48

Answer

z = 0.48
14.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
15.Confidence intervals

Answer: lower limit = 4.434 h

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.

Point estimate

x̄₁ − x̄₂ = 76.5 − 68.5 = 8

Standard error

SE = √(4²/80 + 15²/50) = 2.168

Margin of error

E = 1.645·2.168 = 3.566

Interval

8 ± 3.566 → [4.434, 11.57]

Answer

lower limit = 4.434 h

The interval does not contain 0, so the means differ significantly.

16.Confidence intervals

Answer: n = 216

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 = 1.645²·0.15·0.85/0.04² = 215.6

Round up

n = 216

Answer

n = 216
17.Hypothesis tests & confidence intervals

Answer: t = 2.88

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/√81 = 15/9 = 1.667

Test statistic

t = (204.8 − 200)/1.667 = 2.88

Answer

t = 2.88
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 295

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).

Solve for n

n = (2.576·10/1.5)² = 294.9

Round up

n = 295

Answer

n = 295
19.Linear regression & correlation

Answer: r² = 0.9876

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 = 141/6 = 23.5
ȳ = 715/6 = 119.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 844.5
Syy = Σ(y − ȳ)² = 4585

Coefficient of determination

r² = 844.5²/(157.5·4585) = 0.9876

Answer

r² = 0.9876

About 98.76% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = −2.724

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 = 112/6 = 18.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −52

Regression line

b = −52/17.5 = −2.971
a = 18.67 − (−2.971)·3.5 = 29.07
ŷ = 29.07 − 2.971·x

Predicted value

ŷ = 29.07 + (−2.971)·5.5 = 12.72

Residual

e = 10 − 12.72 = −2.724

Answer

e = −2.724

The point lies below the line.

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