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
years01234567PV = ?€25 000

You need €25 000 in 7 years. An account pays 3.6% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €25 0003.6% / 12n = 7
PV = €
2.Time value of moneyHard
months0123…300€200 000A?AAA

A loan of €200 000 is repaid with equal monthly payments over 25 years. The interest rate is 3.6% per year, compounded monthly. Find the monthly payment A.

P = €200 0003.6% / 12N = 300
A = €
3.NPV & IRRMedium
years0123−75 00016 00037 00035 000

An investment of €75 000 gives these cash flows at the end of years 1 to 3: €16 000, €37 000, €35 000. The cost of capital is 5%. Find the net present value NPV.

r = 5%CF0 = −75 000CF1 = 16 000CF2 = 37 000CF3 = 35 000
NPV = €
4.NPV & IRRHard
years0123−50 00026 00035 00029 500

A project costs €50 000 and returns €26 000, €35 000, €29 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 26 000CF2 = 35 000CF3 = 29 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…20€30€30€30 + €1 000P = ?

A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 10 years. The yield to maturity is 3.5% per year. Find the price P of the bond.

F = €1 000coupon = 6%YTM = 3.5%n = 10
P = €
6.Bond pricing & yieldHard
half-years0123…16€929€10€10€10 + €1 000

A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 8 years. It sells for €929. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 2% (2×)n = 8P = €929
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€6 000 000re​ = 10%debt€4 000 000rd​ = 4%T = 25%

A company has equity worth €6 000 000 and debt worth €4 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 4% and the tax rate is 25%. Find the WACC.

E = €6 000 000D = €4 000 000re = 10%rd = 4%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity€15 000 000re​ = 9%bonds5 000 × €1 119coupon 6%, n = 10T = 20%

A company has equity with a market value of €15 000 000 and a cost of equity of 9%. Its debt consists of 5 000 bonds (face value €1 000, annual coupon 6%, 10 years to maturity) that trade at €1 119 each. The tax rate is 20%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.

E = €15 000 000re = 9%bonds: 5 000 × €1 119coupon = 6%n = 10T = 20%
WACC = %
9.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 22, 16, 25, 53, 35, 45.

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

Exam scores are normally distributed with mean 170 and standard deviation 5. Find the z-score z of a score of 180.

μ = 170σ = 5x = 180
z =
11.ProbabilityMedium
red: 8blue: 5

A bag holds 8 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 5
P =
12.ProbabilityHard
n = 5, p = 0.6: P(X ≤ 1) = ?k012345

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

n = 5p = 0.6
P =
13.Normal distributionMedium
1020 kg1041 kgT = X₁ + … + X₁₂?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 12 of them exceeds 1041 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 12> 1041 kg
P =
14.Normal distributionHard
157164171178185192199cmtop 60%

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 60% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm60%
x = cm
15.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E95%

The battery life of two phone models is compared. Model 1: 40 phones, mean 88 h, known standard deviation 9 h. Model 2: 30 phones, mean 88.5 h, known standard deviation 15 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 40x̄₁ = 88 hσ₁ = 9 hn₂ = 30x̄₂ = 88.5 hσ₂ = 15 h95%
lower limit = h
16.Confidence intervalsHard
σ = 22, n = ?x̄ − Ex̄ + Ex̄E = 399%

The population standard deviation is 22. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 3?

σ = 22E = 399%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 101.3 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 101.3s = 10n = 81
t =
18.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 81 measurements has mean 199.6. The population standard deviation is 8. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 199.6σ = 8n = 8199%
U =
19.Linear regression & correlationMedium
xy2114206288331043xy51010203040

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy164318502064227124832683xy2025406080ŷ = 57

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 57?

n = 6ŷ = 57
x =

3Solutions

1.Time value of money

Answer: PV = 19438 €

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 = 0.036000000000000004/12 = 0.003
N = 12·7 = 84

Present value

PV = FV/(1 + i)N = 25 000/1.286 = €19 438.45

Answer

PV = €19 438.45
2.Time value of money

Answer: A = 1012 €

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 payments

i = 0.036000000000000004/12 = 0.003
N = 12·25 = 300

Payment

The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.

(1 + i)−N = 0.4071
A = 200 000·0.003/(1 − 0.4071) = €1 012.01

Answer

A = €1 012.01
3.NPV & IRR

Answer: NPV = 4033 €

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 = −75 000
CF1/(1 + 0.05)1 = 16 000/1.05 = 15 238.10
CF2/(1 + 0.05)2 = 37 000/1.103 = 33 560.09
CF3/(1 + 0.05)3 = 35 000/1.158 = 30 234.32

Add them up

NPV = €4 032.50

Answer

NPV = €4 032.50

NPV is positive, so the project earns more than the required return.

4.NPV & IRR

Answer: IRR = 35.66 %

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(35%) = 453.69
NPV(40%) = −2 820.70

Narrow down

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

IRR ≈ 35 + 5·453.69/(453.69 − (−2 820.70)) = 35.69%
exact: NPV(35.66%) = 0 → IRR = 35.66%

Answer

IRR = 35.66%
5.Bond pricing & yield

Answer: P = 1209 €

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 = €30.00, i = 0.0175, N = 20

Present value of the coupons

€30.00·(1 − 0.7068)/0.0175 = €502.59

Present value of the face value

€1 000·0.7068 = €706.82

Price

P = €1 209.41

Answer

P = €1 209.41

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

6.Bond pricing & yield

Answer: YTM = 3.005 %

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 = €10.00, N = 16

Solve numerically

i = 1.503% per half-year

YTM

YTM = 2·1.503% = 3.005%

Answer

YTM = 3.005%
7.CAPM & WACC

Answer: WACC = 7.2 %

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 = 0.6, D/V = 0.4

WACC

WACC = 0.6·10% + 0.4·4%·(1 − 0.25) = 7.2%

Answer

WACC = 7.2%
8.CAPM & WACC

Answer: WACC = 7.532 %

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 debt (YTM)

1119 = 60·(1 − (1 + y)−10)/y + 1000/(1 + y)10
y = 4.496%

Market values

D = 5 000·€1 119 = €5 595 000
V = €20 595 000, E/V = 0.7283, D/V = 0.2717

WACC

WACC = 0.7283·9% + 0.2717·4.496%·(1 − 0.2) = 7.532%

Answer

WACC = 7.532%
9.Descriptive statistics

Answer: s = 14.29

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 = 196/6 = 32.67

Squared deviations

Σ(x − x̄)² = (−10.67)² + (−16.67)² + (−7.667)² + 20.33² + 2.333² + 12.33² = 1021

Standard deviation

s = √(1021/5) = 14.29

Answer

s = 14.29
10.Descriptive statistics

Answer: z = 2

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 = (180 − 170)/5 = 2

Answer

z = 2
11.Probability

Answer: P = 0.359

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 = 8/13 · 7/12 = 56/156 = 0.359

Answer

P = 0.359
12.Probability

Answer: P = 0.08704

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) = 0.45 = 0.01024
P(X = 1) = 5·0.6·0.44 = 0.0768

Add them

P = 0.01024 + 0.0768 = 0.08704

Answer

P = 0.08704
13.Normal distribution

Answer: P = 0.3085

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 total

μT = 1020 kg, σT = 12·√12 = 41.57 kg

Standardise

z = (1041 − 1020)/41.57 = 0.5

Probability

P = 1 − Φ(0.5) = 0.3085

Answer

P = 0.3085
14.Normal distribution

Answer: x = 176.3 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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 178 − 0.25·7 = 176.3 cm

Answer

x = 176.3 cm
15.Confidence intervals

Answer: lower limit = −6.549 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 = 88 − 88.5 = −0.5

Standard error

SE = √(9²/40 + 15²/30) = 3.086

Margin of error

E = 1.96·3.086 = 6.049

Interval

−0.5 ± 6.049 → [−6.549, 5.549]

Answer

lower limit = −6.549 h

The interval contains 0, so the data do not show a significant difference.

16.Confidence intervals

Answer: n = 357

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 = (2.576·22/3)² = 356.9

Round up

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

n = 357

Answer

n = 357
17.Hypothesis tests & confidence intervals

Answer: t = 1.17

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 = 10/√81 = 10/9 = 1.111

Test statistic

t = (101.3 − 100)/1.111 = 1.17

Answer

t = 1.17
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: U = 201.9

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 = 2.576·8/√81 = 2.29

Interval limit

U = 199.6 + 2.29 = 201.9

Answer

U = 201.9
19.Linear regression & correlation

Answer: r = 0.9957

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 = 135/5 = 27

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 154
Syy = Σ(y − ȳ)² = 598

Correlation coefficient

r = 154/√(40·598) = 0.9957

Answer

r = 0.9957
20.Linear regression & correlation

Answer: x = 19.02

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 = 126/6 = 21
ȳ = 394/6 = 65.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 306

Regression line

b = 306/70 = 4.371
a = 65.67 − 4.371·21 = −26.13
ŷ = −26.13 + 4.371·x

Solve for x

x = (57 − (−26.13))/4.371 = 19.02

Answer

x = 19.02

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