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
years0123PV = ?€15 000

You need €15 000 in 3 years. An account pays 4.8% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €15 0004.8% / 4n = 3
PV = €
2.Time value of moneyHard
months0123…N = ?€10 000€450€450€450€4503% / 12

You borrow €10 000 at 3% per year, compounded monthly, and repay €450 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 = €4503% / 12
N = payments
3.NPV & IRRMedium
years0123456−40 00016 00014 00015 00012 00013 00012 000

A project costs €40 000 today and returns €16 000, €14 000, €15 000, €12 000, €13 000, €12 000 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 = −40 000CF1 = 16 000CF2 = 14 000CF3 = 15 000CF4 = 12 000CF5 = 13 000CF6 = 12 000
DPB = years
4.NPV & IRRHard
years012−50 00036 50022 000

A project costs €50 000 and returns €36 500, €22 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 36 500CF2 = 22 000
IRR = %
5.Bond pricing & yieldMedium
years0123…10€1 154C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 10 years and pays annual coupons. It sells for €1 154 when the yield to maturity is 5%. Find the coupon rate c in percent.

F = €1 000n = 10P = €1 154YTM = 5%
c = %
6.Bond pricing & yieldHard
years0123…10€1 155€70€70 + €1 000

A bond with a face value of €1 000 and an annual coupon of 7% matures in 10 years and sells for €1 155. Find its yield to maturity YTM in percent.

F = €1 000coupon = 7%n = 10P = €1 155
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€12 000 000re​ = 11%debt€6 000 000rd​ = 4%T = 20%

A company has equity worth €12 000 000 and debt worth €6 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 4% and the tax rate is 20%. Find the WACC.

E = €12 000 000D = €6 000 000re = 11%rd = 4%T = 20%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €15.00β = 0.9rf​ = 3%, rm​ = 10%debt40 000 × €960rd​ = 4%T = 20%

A company has 2 000 000 shares at €15.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 4% and the tax rate is 20%. The share’s beta is 0.9, the risk-free rate 3% and the market return 10%. Find the WACC.

β = 0.9rf = 3%rm = 10%rd = 4%T = 20%
WACC = %
9.Descriptive statisticsMedium
15202530354045505560

Find the sample standard deviation s of the data: 60, 15, 54, 58, 31, 15.

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

A quantity is normally distributed with mean 60 and standard deviation 10. What proportion P of values is above 72.5?

μ = 60σ = 10x = 72.5
P =
11.ProbabilityMedium
red: 8blue: 9

A bag holds 8 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 9
P =
12.ProbabilityHard
n = 5, p = 0.7: P(X ≤ 1) = ?k012345

5 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?

n = 5p = 0.7
P =
13.Normal distributionMedium
6285.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 62 and 85.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 62x2 = 85.5
P =
14.Normal distributionHard
157164171178185192199cmtop 20%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 20% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm20%
x = cm
15.Confidence intervalsMedium
n = 8, s = 10LU = ?x̄ = 44.5E99%

A sample of 8 items has mean 44.5 and standard deviation s = 10. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 8x̄ = 44.5s = 1099%t(7) = 3.499
upper limit =
16.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0490%

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 0.04?

p = 0.5E = 0.0490%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 116.1 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 116.1s = 12n = 64
t =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.10 with a sample of 64; the standard deviation is known to be 20. In reality the mean is 56.5. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 56.5σ = 20n = 64α = 0.10
1 − β = %
19.Linear regression & correlationMedium
xy1541185521682471279630108xy1520253050100

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy18862091221062411726121xy2025100120x = 25

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (25, 117) is made. Find its residual e = y − ŷ.

n = 5(25, 117)
e =

3Solutions

1.Time value of money

Answer: PV = 12999 €

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 = 0.048/4 = 0.012
N = 4·3 = 12

Present value

PV = FV/(1 + i)N = 15 000/1.154 = €12 999.45

Answer

PV = €12 999.45
2.Time value of money

Answer: N = 23 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 = 0.0025
(1 + i)−N = 1 − 10 000·0.0025/450 = 0.9444

Solve with logarithms

N = −ln(0.9444)/ln(1.002) = 22.89

Round up to whole payments

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

N = 23

Answer

N = 23 payments
3.NPV & IRR

Answer: DPB = 3.145 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: 16 000/1.08 = 14 814.81 → Σ = 14 814.81
year 2: 14 000/1.166 = 12 002.74 → Σ = 26 817.56
year 3: 15 000/1.26 = 11 907.48 → Σ = 38 725.04
year 4: 12 000/1.36 = 8 820.36 → Σ = 47 545.40

Fraction of the last year

DPB = 3 + 1 274.96/8 820.36 = 3.145

Answer

DPB = 3.145 years

The discounted payback is always longer than the simple payback (2.667 years).

4.NPV & IRR

Answer: IRR = 12.21 %

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(10%) = 1 363.64
NPV(15%) = −1 625.71

Narrow down

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

IRR ≈ 10 + 5·1 363.64/(1 363.64 − (−1 625.71)) = 12.28%
exact: NPV(12.21%) = 0 → IRR = 12.21%

Answer

IRR = 12.21%
5.Bond pricing & yield

Answer: c = 6.994 %

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 + 0.05)−10 = €613.91

Coupon

C = (1154 − 613.91)/7.722 = €69.94

Coupon rate

c = 69.94/1000 = 6.994%

Answer

c = 6.994%
6.Bond pricing & yield

Answer: YTM = 4.993 %

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

€1 155 = €70·(1 − (1 + y)⁻¹⁰)/y + €1 000/(1 + y)¹⁰

Solve numerically

P(4%) = €1 243.33, P(5%) = €1 154.43
YTM = 4.993%

Answer

YTM = 4.993%
7.CAPM & WACC

Answer: WACC = 8.4 %

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 = €18 000 000, E/V = 0.6667, D/V = 0.3333

WACC

WACC = 0.6667·11% + 0.3333·4%·(1 − 0.2) = 8.4%

Answer

WACC = 8.4%
8.CAPM & WACC

Answer: WACC = 5.875 %

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·€15.00 = €30 000 000
D = 40 000·€960 = €38 400 000

Cost of equity (CAPM)

re = 3% + 0.9·7% = 9.3%

Weights

V = €68 400 000, E/V = 0.4386, D/V = 0.5614

WACC

WACC = 0.4386·9.3% + 0.5614·4%·(1 − 0.2) = 5.875%

Answer

WACC = 5.875%
9.Descriptive statistics

Answer: s = 21.18

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 = 233/6 = 38.83

Squared deviations

Σ(x − x̄)² = 21.17² + (−23.83)² + 15.17² + 19.17² + (−7.833)² + (−23.83)² = 2243

Standard deviation

s = √(2243/5) = 21.18

Answer

s = 21.18
10.Descriptive statistics

Answer: P = 0.1056

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
11.Probability

Answer: P = 0.2059

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.

First and second draw

P = 8/17 · 7/16 = 56/272 = 0.2059

Answer

P = 0.2059
12.Probability

Answer: P = 0.03078

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) = 0.35 = 0.00243
P(X = 1) = 5·0.7·0.34 = 0.02835

Add them

P = 0.00243 + 0.02835 = 0.03078

Answer

P = 0.03078
13.Normal distribution

Answer: P = 0.5977

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-scores

z₁ = (62 − 65)/10 = −0.3
z₂ = (85.5 − 65)/10 = 2.05

Area between

P = Φ(2.05) − Φ(−0.3) = 0.9798 − 0.3821 = 0.5977

Answer

P = 0.5977
14.Normal distribution

Answer: x = 183.9 cm

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) = 0.8

z from the table

z ≈ 0.84

Back to the original units

x = 178 + 0.84·7 = 183.9 cm

Answer

x = 183.9 cm
15.Confidence intervals

Answer: upper limit = 56.87

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 = 3.499·10/√8 = 12.37

Interval

44.5 ± 12.37 → [32.13, 56.87]

Answer

upper limit = 56.87
16.Confidence intervals

Answer: n = 423

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 = 1.645²·0.5·0.5/0.04² = 422.8

Round up

n = 423

Answer

n = 423
17.Hypothesis tests & confidence intervals

Answer: t = −2.6

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/√64 = 12/8 = 1.5

Test statistic

t = (116.1 − 120)/1.5 = −2.6

Answer

t = −2.6
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: 1 − β = 90.62 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 20/√64 = 2.5
x̄c = 50 + 1.282·2.5 = 53.2

z under the true mean

z = (53.2 − 56.5)/2.5 = −1.318

Power

power = P(Z > −1.318) = 90.62%
1 − β = 90.62%

Answer

1 − β = 90.62%
19.Linear regression & correlation

Answer: r = 0.9847

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 = 135/6 = 22.5
ȳ = 439/6 = 73.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 691.5
Syy = Σ(y − ȳ)² = 3131

Correlation coefficient

r = 691.5/√(157.5·3131) = 0.9847

Answer

r = 0.9847
20.Linear regression & correlation

Answer: e = −1.6

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 = 110/5 = 22
ȳ = 521/5 = 104.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 192

Regression line

b = 192/40 = 4.8
a = 104.2 − 4.8·22 = −1.4
ŷ = −1.4 + 4.8·x

Predicted value

ŷ = −1.4 + 4.8·25 = 118.6

Residual

e = 117 − 118.6 = −1.6

Answer

e = −1.6

The point lies below the line.

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