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
years012€8 000FV = ?

You invest €8 000 at a nominal rate of 3% per year, compounded monthly. Find the value FV of the investment after 2 years.

PV = €8 0003% / 12n = 2
FV = €
2.Time value of moneyHard
months01…108…240€250 000AB = ?A3.6% / 12

A mortgage of €250 000 is repaid with equal monthly payments over 20 years at 3.6% per year, compounded monthly. How much is still owed just after the payment at the end of year 9? Find the outstanding balance B.

P = €250 0003.6% / 12N = 240k = 108
B = €
3.NPV & IRRMedium
years0123456−20 0005 0005 5006 0007 0004 5006 500

A project costs €20 000 today and returns €5 000, €5 500, €6 000, €7 000, €4 500, €6 500 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 = −20 000CF1 = 5 000CF2 = 5 500CF3 = 6 000CF4 = 7 000CF5 = 4 500CF6 = 6 500
DPB = years
4.NPV & IRRHard
years012−40 00022 50027 500

A project costs €40 000 and returns €22 500, €27 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −40 000CF1 = 22 500CF2 = 27 500
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 7% per year. Find the price P of the bond.

F = €1 000coupon = 4.5%YTM = 7%n = 12
P = €
6.Bond pricing & yieldHard
years012345€935€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 €935. Find its yield to maturity YTM in percent.

F = €1 000coupon = 4%n = 5P = €935
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€5 000 000re​ = 13%preferred shares€2 000 000rp​ = 6.5%debt€5 000 000rd​ = 3%T = 28%

A company is financed with ordinary shares worth €5 000 000, preferred shares worth €2 000 000 and debt worth €5 000 000. The cost of equity is 13%, the cost of preferred stock 6.5% and the pre-tax cost of debt 3%. The tax rate is 28%. Find the WACC.

E = €5 000 000Ps = €2 000 000D = €5 000 000re = 13%rp = 6.5%rd = 3%T = 28%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €29.50β = 1.5rf​ = 2%, rm​ = 9%debt20 000 × €1 050rd​ = 3%T = 25.8%

A company has 2 000 000 shares at €29.50 each and 20 000 bonds trading at €1 050 each. The pre-tax cost of debt is 3% and the tax rate is 25.8%. The share’s beta is 1.5, the risk-free rate 2% and the market return 9%. Find the WACC.

β = 1.5rf = 2%rm = 9%rd = 3%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
202530354045

Find the sample standard deviation s of the data: 41, 29, 19, 48, 27, 33.

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

Exam scores are normally distributed with mean 500 and standard deviation 60. Find the z-score z of a score of 470.

μ = 500σ = 60x = 470
z =
11.ProbabilityMedium
n = 7k = 5N = ?

In how many ways N can a committee of 5 people be chosen from a group of 7? The order does not matter.

n = 7k = 5
N =
12.ProbabilityHard
€p(x)−50.7550.2950.05

A game costs €5 to play. You win €10 with probability 0.2 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€10: p = 0.2€100: p = 0.05
E(G) = €
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9780

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9780. Find z to two decimals.

A = 0.9780
z =
14.Normal distributionHard
380420460500540580620hbottom 90%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h90%
x = h
15.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?95%

The battery life of two phone models is compared. Model 1: 50 phones, mean 81 h, known standard deviation 19 h. Model 2: 100 phones, mean 70.5 h, known standard deviation 13 h. Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 81 hσ₁ = 19 hn₂ = 100x̄₂ = 70.5 hσ₂ = 13 h95%
E = h
16.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.02595%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.025?

p = 0.5E = 0.02595%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 43.4 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 43.4s = 12n = 16
t =
18.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 195%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 195%
n =
19.Linear regression & correlationMedium
xy342439541643753xy46404550

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy274166228271035xy510102030x = 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. A new observation (3, 13) is made. Find its residual e = y − ŷ.

n = 5(3, 13)
e =

3Solutions

1.Time value of money

Answer: FV = 8494 €

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 = 0.03/12 = 0.0025
N = m·n = 12·2 = 24

Future value

FV = 8 000·(1 + 0.0025)24 = 8 000·1.062 = €8 494.06

Answer

FV = €8 494.06
2.Time value of money

Answer: B = 159245 €

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 = 0.003, N = 240
A = 250 000·0.003/(1 − (1 + 0.003)−240) = €1 462.78

Present value of the remaining payments

Whatever is still owed must be exactly paid off by the remaining payments.

N − k = 240 − 108 = 132
B = 1463·(1 − (1 + 0.003)−132)/0.003 = €159 245.02

Answer

B = €159 245.02
3.NPV & IRR

Answer: DPB = 4.244 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: 5 000/1.08 = 4 629.63 → Σ = 4 629.63
year 2: 5 500/1.166 = 4 715.36 → Σ = 9 344.99
year 3: 6 000/1.26 = 4 762.99 → Σ = 14 107.99
year 4: 7 000/1.36 = 5 145.21 → Σ = 19 253.20
year 5: 4 500/1.469 = 3 062.62 → Σ = 22 315.82

Fraction of the last year

DPB = 4 + 746.80/3 062.62 = 4.244

Answer

DPB = 4.244 years

The discounted payback is always longer than the simple payback (3.5 years).

4.NPV & IRR

Answer: IRR = 15.68 %

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(15%) = 359.17
NPV(20%) = −2 152.78

Narrow down

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

IRR ≈ 15 + 5·359.17/(359.17 − (−2 152.78)) = 15.71%
exact: NPV(15.68%) = 0 → IRR = 15.68%

Answer

IRR = 15.68%
5.Bond pricing & yield

Answer: P = 799.3 €

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 = 0.035, N = 24

Present value of the coupons

€22.50·(1 − 0.438)/0.035 = €361.31

Present value of the face value

€1 000·0.438 = €437.96

Price

P = €799.27

Answer

P = €799.27

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

6.Bond pricing & yield

Answer: YTM = 5.523 %

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

€935 = €40·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵

Solve numerically

P(5%) = €956.71, P(6%) = €915.75
YTM = 5.523%

Answer

YTM = 5.523%
7.CAPM & WACC

Answer: WACC = 7.4 %

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 = 5 000 000 + 2 000 000 + 5 000 000 = €12 000 000
E/V = 0.4167, Ps/V = 0.1667, D/V = 0.4167

WACC

WACC = 0.4167·13% + 0.1667·6.5% + 0.4167·3%·(1 − 0.28) = 7.4%

Answer

WACC = 7.4%
8.CAPM & WACC

Answer: WACC = 9.803 %

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 = 20 000·€1 050 = €21 000 000

Cost of equity (CAPM)

re = 2% + 1.5·7% = 12.5%

Weights

V = €80 000 000, E/V = 0.7375, D/V = 0.2625

WACC

WACC = 0.7375·12.5% + 0.2625·3%·(1 − 0.258) = 9.803%

Answer

WACC = 9.803%
9.Descriptive statistics

Answer: s = 10.36

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 = 197/6 = 32.83

Squared deviations

Σ(x − x̄)² = 8.167² + (−3.833)² + (−13.83)² + 15.17² + (−5.833)² + 0.1667² = 536.8

Standard deviation

s = √(536.8/5) = 10.36

Answer

s = 10.36
10.Descriptive statistics

Answer: z = −0.5

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 = (470 − 500)/60 = −0.5

Answer

z = −0.5
11.Probability

Answer: N = 21

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.

Combinations

C(7, 5) = (7·6·5·4·3)/(5·4·3·2·1) = 21

Answer

N = 21
12.Probability

Answer: E(G) = 2 €

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.

Net outcomes

G: 5 (p = 0.2), 95 (p = 0.05), −5 (p = 0.75)

Expected value

E(G) = 5·0.2 + 95·0.05 − 5·0.75 = 2 €

Answer

E(G) = 2 €

A positive value means you win money on average.

13.Normal distribution

Answer: z = 2.29

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 + 0.9780)/2 = 0.9890

Read the z-table backwards

Find 0.9890 inside the table and read off z.

z = 2.29

Answer

z = 2.29
14.Normal distribution

Answer: x = 551 h

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) = 0.9

z from the table

z ≈ 1.28

Back to the original units

x = 500 + 1.28·40 = 551 h

Answer

x = 551 h
15.Confidence intervals

Answer: E = 5.851 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 = 81 − 70.5 = 10.5

Standard error

SE = √(19²/50 + 13²/100) = 2.985

Margin of error

E = 1.96·2.985 = 5.851

Answer

E = 5.851 h

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

16.Confidence intervals

Answer: n = 1537

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 = 1.96²·0.5·0.5/0.025² = 1537

Round up

n = 1537

Answer

n = 1537
17.Hypothesis tests & confidence intervals

Answer: t = −2.2

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 = 12/√16 = 12/4 = 3

Test statistic

t = (43.4 − 50)/3 = −2.2

Answer

t = −2.2
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 554

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 = (1.96·12/1)² = 553.2

Round up

n = 554

Answer

n = 554
19.Linear regression & correlation

Answer: r = 0.7531

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 = 25/5 = 5
ȳ = 218/5 = 43.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 26
Syy = Σ(y − ȳ)² = 119.2

Correlation coefficient

r = 26/√(10·119.2) = 0.7531

Answer

r = 0.7531
20.Linear regression & correlation

Answer: e = 1.65

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 = 107/5 = 21.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 134

Regression line

b = 134/40 = 3.35
a = 21.4 − 3.35·6 = 1.3
ŷ = 1.3 + 3.35·x

Predicted value

ŷ = 1.3 + 3.35·3 = 11.35

Residual

e = 13 − 11.35 = 1.65

Answer

e = 1.65

The point lies above the line.

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