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 = ?€8 000+1.5%+1.5%…r = 6.5%

A property is expected to earn net rent of €8 000 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 = €8 000g = 1.5%r = 6.5%
PV = €
2.Time value of moneyHard
months0123…48€8 000A?AAA

A loan of €8 000 is repaid with equal monthly payments over 4 years. The interest rate is 4.2% per year, compounded monthly. Find the monthly payment A.

P = €8 0004.2% / 12N = 48
A = €
3.NPV & IRRMedium
years012345−30 0006 5006 5006 5006 50012 500

A machine costs €30 000 today. It generates €6 500 at the end of each year for 5 years and can then be sold for €6 000. The cost of capital is 12%. Find the net present value NPV.

C0 = €30 000CF = €6 500S = €6 000n = 5r = 12%
NPV = €
4.NPV & IRRHard
years0123−10 0007 0005 5005 500

A project costs €10 000 and returns €7 000, €5 500, €5 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −10 000CF1 = 7 000CF2 = 5 500CF3 = 5 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…10€10€10€10 + €1 000P = ?

A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 5 years. The yield to maturity is 3% per year. Find the price P of the bond.

F = €1 000coupon = 2%YTM = 3%n = 5
P = €
6.Bond pricing & yieldHard

A bond trades at €1 055 with a yield to maturity of 7% (annual compounding) and a Macaulay duration of 11.3 years. Use the modified duration to estimate the new price P if the yield rises by 25 basis points.

P0 = €1 055y = 7%D = 11.3Δy = +25 bp
P = €
7.CAPM & WACCMedium
WACC = ?equity€2 000 000re​ = 8%debt€4 000 000rd​ = 5%T = 25.8%

A company has equity worth €2 000 000 and debt worth €4 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 5% and the tax rate is 25.8%. Find the WACC.

E = €2 000 000D = €4 000 000re = 8%rd = 5%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €15.00β = 1.5rf​ = 2%, rm​ = 6%debt5 000 × €960rd​ = 7%T = 28%

A company has 2 000 000 shares at €15.00 each and 5 000 bonds trading at €960 each. The pre-tax cost of debt is 7% and the tax rate is 28%. The share’s beta is 1.5, the risk-free rate 2% and the market return 6%. Find the WACC.

β = 1.5rf = 2%rm = 6%rd = 7%T = 28%
WACC = %
9.Descriptive statisticsMedium
3035404550

Find the sample standard deviation s of the data: 44, 42, 52, 48, 26, 30.

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

A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is below 176?

μ = 170σ = 10x = 176
P =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7P(six) = 1/6 on each roll

A fair die is rolled 7 times. What is the probability P of getting at least one six?

n = 7
P =
12.ProbabilityHard
€p(x)−30.84120.15270.01

A game costs €3 to play. You win €15 with probability 0.15 and €30 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€15: p = 0.15€30: p = 0.01
E(G) = €
13.Normal distributionMedium
93.25127.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 93.25 and 127.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 93.25x2 = 127.75
P =
14.Normal distributionHard
35455565758595pointstop 80%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points80%
x = points
15.Confidence intervalsMedium
n = 25, s = 2L = ?Ux̄ = 54.5E99%

A sample of 25 items has mean 54.5 and standard deviation s = 2. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 25x̄ = 54.5s = 299%t(24) = 2.797
lower limit =
16.Confidence intervalsHard
μ₁ − μ₂, t(14) = 1.7610L = ?Ux̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 10 plots, mean yield 11 kg, standard deviation 7 kg. Plot group 2: 6 plots, mean 5 kg, standard deviation 8 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 10x̄₁ = 11s₁ = 7n₂ = 6x̄₂ = 5s₂ = 890%t(14) = 1.761
lower limit = kg
17.Hypothesis tests & confidence intervalsMedium
−2.0642.064t = ?df = 24

Two teaching methods are compared. Group 1: 9 students, mean score 21.5, standard deviation 11. Group 2: 17 students, mean score 14.9, standard deviation 6. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 9x̄₁ = 21.5s₁ = 11n₂ = 17x̄₂ = 14.9s₂ = 6
t =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 500 against H₁: μ < 500 is done at α = 0.01 with a sample of 64; the standard deviation is known to be 4. In reality the mean is 498.05. Find the power 1 − β of the test in percent.

μ₀ = 500μ = 498.05σ = 4n = 64α = 0.01
1 − β = %
19.Linear regression & correlationMedium
xy1783208923110261182913832160xy202530100150

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
xy1222163174105766xy2465101520ŷ = 12

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. For which value x does the line predict ŷ = 12?

n = 6ŷ = 12
x =

3Solutions

1.Time value of money

Answer: PV = 160000 €

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) = 8 000/(0.065 − 0.015) = 8 000/0.05 = €160 000.00

Answer

PV = €160 000.00
2.Time value of money

Answer: A = 181.3 €

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 = 0.042/12 = 0.0035
N = 12·4 = 48

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 = 0.8456
A = 8 000·0.0035/(1 − 0.8456) = €181.35

Answer

A = €181.35
3.NPV & IRR

Answer: NPV = −3164 €

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 yearly cash flows

6 500·3.605 = 23 431.05

Present value of the salvage value

6 000/1.762 = 3 404.56

NPV

NPV = −30 000 + 23 431.05 + 3 404.56 = −€3 164.39

Answer

NPV = −€3 164.39
4.NPV & IRR

Answer: IRR = 38.43 %

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%) = 438.45
NPV(40%) = −189.50

Narrow down

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

IRR ≈ 35 + 5·438.45/(438.45 − (−189.50)) = 38.49%
exact: NPV(38.43%) = 0 → IRR = 38.43%

Answer

IRR = 38.43%
5.Bond pricing & yield

Answer: P = 953.9 €

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, i = 0.015, N = 10

Present value of the coupons

€10.00·(1 − 0.8617)/0.015 = €92.22

Present value of the face value

€1 000·0.8617 = €861.67

Price

P = €953.89

Answer

P = €953.89

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

6.Bond pricing & yield

Answer: P = 1027 €

Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.

Modified duration

Dmod = 11.3/1.07 = 10.56

Estimated price change

%ΔP ≈ −10.56·(0.0025) = −2.64%
P ≈ 1055·(1 − 0.0264) = €1 027.15

Answer

P = €1 027.15

Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.

7.CAPM & WACC

Answer: WACC = 5.14 %

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 = 0.3333, D/V = 0.6667

WACC

WACC = 0.3333·8% + 0.6667·5%·(1 − 0.258) = 5.14%

Answer

WACC = 5.14%
8.CAPM & WACC

Answer: WACC = 7.592 %

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 = 5 000·€960 = €4 800 000

Cost of equity (CAPM)

re = 2% + 1.5·4% = 8%

Weights

V = €34 800 000, E/V = 0.8621, D/V = 0.1379

WACC

WACC = 0.8621·8% + 0.1379·7%·(1 − 0.28) = 7.592%

Answer

WACC = 7.592%
9.Descriptive statistics

Answer: s = 10.23

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 = 242/6 = 40.33

Squared deviations

Σ(x − x̄)² = 3.667² + 1.667² + 11.67² + 7.667² + (−14.33)² + (−10.33)² = 523.3

Standard deviation

s = √(523.3/5) = 10.23

Answer

s = 10.23
10.Descriptive statistics

Answer: P = 0.7257

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 0.7257

Answer

P = 0.7257
11.Probability

Answer: P = 0.7209

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)7 = 0.2791
P = 1 − 0.2791 = 0.7209

Answer

P = 0.7209
12.Probability

Answer: E(G) = −0.45 €

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: 12 (p = 0.15), 27 (p = 0.01), −3 (p = 0.84)

Expected value

E(G) = 12·0.15 + 27·0.01 − 3·0.84 = −0.45 €

Answer

E(G) = −0.45 €

A negative value means you lose money on average.

13.Normal distribution

Answer: P = 0.6415

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 = (93.25 − 100)/15 = −0.45
z₂ = (127.75 − 100)/15 = 1.85

Area between

P = Φ(1.85) − Φ(−0.45) = 0.9678 − 0.3264 = 0.6415

Answer

P = 0.6415
14.Normal distribution

Answer: x = 56.6 points

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) = 0.2

z from the table

z ≈ −0.84

Back to the original units

x = 65 − 0.84·10 = 56.6 points

Answer

x = 56.6 points
15.Confidence intervals

Answer: lower limit = 53.38

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 = 24, t = 2.797

Margin of error

E = 2.797·2/√25 = 1.119

Interval

54.5 ± 1.119 → [53.38, 55.62]

Answer

lower limit = 53.38
16.Confidence intervals

Answer: lower limit = −0.7046 kg

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.

Pooled standard deviation

sp² = (9·7² + 5·8²)/14 = 54.36
sp = 7.373

Margin of error

E = 1.761·7.373·√(1/10 + 1/6) = 6.705

Interval

6 ± 6.705 → [−0.7046, 12.7]

Answer

lower limit = −0.7046 kg
17.Hypothesis tests & confidence intervals

Answer: t = 1.996

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.

Pooled variance

sp² = (8·11² + 16·6²)/24 = 64.33
sp = 8.021

Standard error

SE = 8.021·√(1/9 + 1/17) = 3.306

Test statistic

t = (21.5 − 14.9)/3.306 = 1.996

Answer

t = 1.996

With df = 24, the two-sided 5% critical value is 2.064; |t| is smaller, so H₀ is not rejected.

18.Hypothesis tests & confidence intervals

Answer: 1 − β = 94.23 %

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 = 4/√64 = 0.5
x̄c = 500 − 2.326·0.5 = 498.8

z under the true mean

z = (498.8 − 498.05)/0.5 = 1.574

Power

power = P(Z < 1.574) = 94.23%
1 − β = 94.23%

Answer

1 − β = 94.23%
19.Linear regression & correlation

Answer: r = 0.9869

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 = 147/6 = 24.5
ȳ = 698/6 = 116.3

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 810
Syy = Σ(y − ȳ)² = 4277

Correlation coefficient

r = 810/√(157.5·4277) = 0.9869

Answer

r = 0.9869
20.Linear regression & correlation

Answer: x = 3.807

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 = 21/6 = 3.5
ȳ = 78/6 = 13

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −57

Regression line

b = −57/17.5 = −3.257
a = 13 − (−3.257)·3.5 = 24.4
ŷ = 24.4 − 3.257·x

Solve for x

x = (12 − 24.4)/(−3.257) = 3.807

Answer

x = 3.807

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