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
years0123456PV = ?€10 000

You need €10 000 in 6 years. An account pays 5% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €10 0005% / 4n = 6
PV = €
2.Time value of moneyHard
years0123…18€2 000€2 000€2 000FV = ?

You deposit €2 000 at the start of every year for 18 years, the first deposit today. The account pays 6% per year. How much is in the account at the end of year 18? Find FV.

A = €2 000i = 6%n = 18
FV = €
3.NPV & IRRMedium
years0123−75 00037 00029 00032 000

An investment of €75 000 gives these cash flows at the end of years 1 to 3: €37 000, €29 000, €32 000. The cost of capital is 8%. Find the net present value NPV.

r = 8%CF0 = −75 000CF1 = 37 000CF2 = 29 000CF3 = 32 000
NPV = €
4.NPV & IRRHard
years0123−25 0007 50013 50014 000

A project costs €25 000 and returns €7 500, €13 500, €14 000 at the end of years 1 to 3. Cash flows can be reinvested at the cost of capital of 12%. Find the modified internal rate of return MIRR in percent.

r = 12%CF0 = −25 000CF1 = 7 500CF2 = 13 500CF3 = 14 000
MIRR = %
5.Bond pricing & yieldMedium
half-years0123…8€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 4 years. The yield to maturity is 3% per year. Find the price P of the bond.

F = €1 000coupon = 4.5%YTM = 3%n = 4
P = €
6.Bond pricing & yieldHard

A bond trades at €920 with a yield to maturity of 4% (annual compounding) and a Macaulay duration of 7.2 years. Use the modified duration to estimate the new price P if the yield rises by 25 basis points.

P0 = €920y = 4%D = 7.2Δy = +25 bp
P = €
7.CAPM & WACCMedium
WACC = 8.09%equity€3 000 000re​ = ?debt€1 000 000rd​ = 3%T = 21%

A company has equity worth €3 000 000 and debt worth €1 000 000. Its WACC is 8.09%, its pre-tax cost of debt is 3% and the tax rate is 21%. Find the cost of equity re.

E = €3 000 000D = €1 000 000WACC = 8.09%rd = 3%T = 21%
re = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €40.50D0​ = €1.40g = 5%debt€40 500 000rd​ = 3%T = 25%

A company has 1 000 000 shares trading at €40.50; the book value of its equity is €12 200 000. It just paid a dividend of €1.40 per share, and dividends grow by 5% per year. Its debt has a market value of €40 500 000 and a pre-tax cost of 3%. The tax rate is 25%. Find the WACC, using the dividend growth model for the cost of equity.

N = 1 000 000P0 = €40.50book equity = €12 200 000D0 = €1.40g = 5%D = €40 500 000rd = 3%T = 25%
WACC = %
9.Descriptive statisticsMedium
15202530354045505560

Find the sample standard deviation s of the data: 36, 44, 12, 26, 30, 37, 60.

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

Exam scores are normally distributed with mean 170 and standard deviation 8. Find the z-score z of a score of 173.2.

μ = 170σ = 8x = 173.2
z =
11.ProbabilityMedium
red: 6blue: 8

A bag holds 6 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

6 / 8
P =
12.ProbabilityHard
€p(x)−100.83400.151900.02

A game costs €10 to play. You win €50 with probability 0.15 and €200 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€50: p = 0.15€200: p = 0.02
E(G) = €
13.Normal distributionMedium
640 g632.8 gT = X₁ + … + X₁₆?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 16 of them exceeds 632.8 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 16> 632.8 g
P =
14.Normal distributionHard
235240245250255260265gbottom 5%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 5% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g5%
x = g
15.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?90%

The battery life of two phone models is compared. Model 1: 50 phones, mean 41 h, known standard deviation 3 h. Model 2: 50 phones, mean 47 h, known standard deviation 18 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 41 hσ₁ = 3 hn₂ = 50x̄₂ = 47 hσ₂ = 18 h90%
E = h
16.Confidence intervalsHard
p ≈ 0.3, n = ?p̂ − Ep̂ + Ep̂E = 0.0290%

A pilot study suggests that about 0.3 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 0.02?

p ≈ 0.3E = 0.0290%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 48.5 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 48.5s = 10n = 36
t =
18.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 49 measurements has mean 51.3. The population standard deviation is 15. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 51.3σ = 15n = 4990%
U =
19.Linear regression & correlationMedium
xy18582069228024832699xy20256080100

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 110Σy = 389Σx² = 2460Σy² = 31215Σxy = 8750
r =
20.Linear regression & correlationHard
xy131225327425523xy242530x = 1.5

The table gives 5 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 (1.5, 34) is made. Find its residual e = y − ŷ.

n = 5(1.5, 34)
e =

3Solutions

1.Time value of money

Answer: PV = 7422 €

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 = 0.05/4 = 0.0125
N = 4·6 = 24

Present value

PV = FV/(1 + i)N = 10 000/1.347 = €7 421.97

Answer

PV = €7 421.97
2.Time value of money

Answer: FV = 65520 €

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.

Ordinary annuity factor

((1 + 0.06)18 − 1)/0.06 = 30.91

Annuity due

Each deposit is invested one year longer.

FV = 2 000·30.91·1.06 = €65 519.98

Answer

FV = €65 519.98
3.NPV & IRR

Answer: NPV = 9525 €

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 + 0.08)1 = 37 000/1.08 = 34 259.26
CF2/(1 + 0.08)2 = 29 000/1.166 = 24 862.83
CF3/(1 + 0.08)3 = 32 000/1.26 = 25 402.63

Add them up

NPV = €9 524.72

Answer

NPV = €9 524.72

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

4.NPV & IRR

Answer: MIRR = 15.51 %

Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.

Terminal value of the inflows

7 500·1.122 = 9 408.00
13 500·1.121 = 15 120.00
14 000·1.120 = 14 000.00
TV = 38 528.00

MIRR

MIRR = (38 528.00/25 000)1/3 − 1 = 15.51%

Answer

MIRR = 15.51%
5.Bond pricing & yield

Answer: P = 1056 €

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 = 0.015, N = 8

Present value of the coupons

€22.50·(1 − 0.8877)/0.015 = €168.43

Present value of the face value

€1 000·0.8877 = €887.71

Price

P = €1 056.14

Answer

P = €1 056.14

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

6.Bond pricing & yield

Answer: P = 904.1 €

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 = 7.2/1.04 = 6.923

Estimated price change

%ΔP ≈ −6.923·(0.0025) = −1.731%
P ≈ 920·(1 − 0.01731) = €904.08

Answer

P = €904.08

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

7.CAPM & WACC

Answer: re = 9.997 %

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 = €4 000 000, E/V = 0.75, D/V = 0.25

Debt part of the WACC

0.25·3%·(1 − 0.21) = 0.5925%

Cost of equity

re = (8.09% − 0.5925%)/0.75 = 9.997%

Answer

re = 9.997%
8.CAPM & WACC

Answer: WACC = 5.44 %

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 equity

D1 = 1.4·1.05 = €1.47
re = 1.47/40.5 + 0.05 = 8.63%

Market values

E = 1 000 000·€40.50 = €40 500 000
V = €81 000 000, E/V = 0.5, D/V = 0.5

WACC

WACC = 0.5·8.63% + 0.5·3%·(1 − 0.25) = 5.44%

Answer

WACC = 5.44%
9.Descriptive statistics

Answer: s = 14.98

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 = 245/7 = 35

Squared deviations

Σ(x − x̄)² = 1² + 9² + (−23)² + (−9)² + (−5)² + 2² + 25² = 1346

Standard deviation

s = √(1346/6) = 14.98

Answer

s = 14.98
10.Descriptive statistics

Answer: z = 0.4

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 = (173.2 − 170)/8 = 0.4

Answer

z = 0.4
11.Probability

Answer: P = 0.1648

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 = 6/14 · 5/13 = 30/182 = 0.1648

Answer

P = 0.1648
12.Probability

Answer: E(G) = 1.5 €

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: 40 (p = 0.15), 190 (p = 0.02), −10 (p = 0.83)

Expected value

E(G) = 40·0.15 + 190·0.02 − 10·0.83 = 1.5 €

Answer

E(G) = 1.5 €

A positive value means you win money on average.

13.Normal distribution

Answer: P = 0.9332

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 = 640 g, σT = 1.2·√16 = 4.8 g

Standardise

z = (632.8 − 640)/4.8 = −1.5

Probability

P = 1 − Φ(−1.5) = 0.9332

Answer

P = 0.9332
14.Normal distribution

Answer: x = 241.8 g

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) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 250 − 1.64·5 = 241.8 g

Answer

x = 241.8 g
15.Confidence intervals

Answer: E = 4.245 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 = 41 − 47 = −6

Standard error

SE = √(3²/50 + 18²/50) = 2.581

Margin of error

E = 1.645·2.581 = 4.245

Answer

E = 4.245 h

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

16.Confidence intervals

Answer: n = 1421

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 = 1.645²·0.3·0.7/0.02² = 1421

Round up

n = 1421

Answer

n = 1421
17.Hypothesis tests & confidence intervals

Answer: t = −0.9

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).

Standard error

SE = 10/√36 = 10/6 = 1.667

Test statistic

t = (48.5 − 50)/1.667 = −0.9

Answer

t = −0.9
t0.025; df = 35 = 2.03

|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: U = 54.82

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 = 1.645·15/√49 = 3.525

Interval limit

U = 51.3 + 3.525 = 54.82

Answer

U = 54.82
19.Linear regression & correlation

Answer: r = 0.9845

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.

Centred sums

Sxy = 8750 − 110·389/5 = 192
Sxx = 2460 − 110²/5 = 40
Syy = 31215 − 389²/5 = 950.8

Correlation coefficient

r = 192/√(40·950.8) = 0.9845

Answer

r = 0.9845
20.Linear regression & correlation

Answer: e = 5.4

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 = 15/5 = 3
ȳ = 131/5 = 26.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −16

Regression line

b = −16/10 = −1.6
a = 26.2 − (−1.6)·3 = 31
ŷ = 31 − 1.6·x

Predicted value

ŷ = 31 + (−1.6)·1.5 = 28.6

Residual

e = 34 − 28.6 = 5.4

Answer

e = 5.4

The point lies above the line.

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