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 = ?€20 000

You need €20 000 in 6 years. An account pays 2% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €20 0002% / 4n = 6
PV = €
2.Time value of moneyHard
years0123…29€3 000€3 000€3 000FV = ?

You deposit €3 000 at the start of every year for 29 years, the first deposit today. The account pays 5% per year. How much is in the account at the end of year 29? Find FV.

A = €3 000i = 5%n = 29
FV = €
3.NPV & IRRMedium
years0123−75 00030 00025 00024 000

An investment of €75 000 gives these cash flows at the end of years 1 to 3: €30 000, €25 000, €24 000. The cost of capital is 9%. Find the net present value NPV.

r = 9%CF0 = −75 000CF1 = 30 000CF2 = 25 000CF3 = 24 000
NPV = €
4.NPV & IRRHard
years0123−25 00013 00010 00013 500

A project costs €25 000 and returns €13 000, €10 000, €13 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −25 000CF1 = 13 000CF2 = 10 000CF3 = 13 500
IRR = %
5.Bond pricing & yieldMedium

You buy a bond with 8 years to maturity (face value €1 000, annual coupon 4.5%) when its yield to maturity is 6%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 5%. Find your holding period return HPR in percent.

F = €1 000coupon = 4.5%n = 8y0 = 6%y1 = 5%
HPR = %
6.Bond pricing & yieldHard

A bond trades at €950 with a yield to maturity of 6% (annual compounding) and a Macaulay duration of 11.1 years. Use the modified duration to estimate the new price P if the yield rises by 75 basis points.

P0 = €950y = 6%D = 11.1Δy = +75 bp
P = €
7.CAPM & WACCMedium
WACC = 6.73%equity€3 000 000re​ = ?debt€5 000 000rd​ = 4%T = 25.8%

A company has equity worth €3 000 000 and debt worth €5 000 000. Its WACC is 6.73%, its pre-tax cost of debt is 4% and the tax rate is 25.8%. Find the cost of equity re.

E = €3 000 000D = €5 000 000WACC = 6.73%rd = 4%T = 25.8%
re = %
8.CAPM & WACCHard
WACC = ?equity4 000 000 × €12.50β = 0.8rf​ = 1%, rm​ = 5.5%debt10 000 × €1 020rd​ = 4.5%T = 21%

A company has 4 000 000 shares at €12.50 each and 10 000 bonds trading at €1 020 each. The pre-tax cost of debt is 4.5% and the tax rate is 21%. The share’s beta is 0.8, the risk-free rate 1% and the market return 5.5%. Find the WACC.

β = 0.8rf = 1%rm = 5.5%rd = 4.5%T = 21%
WACC = %
9.Descriptive statisticsMedium
202530354045

Find the sample standard deviation s of the data: 45, 30, 30, 23, 16, 44.

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

A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 58?

μ = 70σ = 8x = 58
P =
11.ProbabilityMedium
red: 5blue: 7

A bag holds 5 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 7
P =
12.ProbabilityHard
€p(x)−50.7100.25450.05

A game costs €5 to play. You win €15 with probability 0.25 and €50 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€15: p = 0.25€50: p = 0.05
E(G) = €
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.8186

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8186. Find z to two decimals.

A = 0.8186
z =
14.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
15.Confidence intervalsMedium
n = 30, s = 9.5LU = ?x̄ = 61E95%

A sample of 30 items has mean 61 and standard deviation s = 9.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 30x̄ = 61s = 9.595%t(29) = 2.045
upper limit =
16.Confidence intervalsHard
n = 15Lσ = ?s = 3.895%

The diameters of 15 machined shafts have a sample standard deviation of 3.8 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.

n = 15s = 3.8 µmχ² = 5.629, 26.119
σ = µm
17.Hypothesis tests & confidence intervalsMedium
−2.1312.131t = ?df = 15

Two teaching methods are compared. Group 1: 8 students, mean score 42, standard deviation 10.5. Group 2: 9 students, mean score 47.6, standard deviation 10.5. 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 = 8x̄₁ = 42s₁ = 10.5n₂ = 9x̄₂ = 47.6s₂ = 10.5
t =
18.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 16 measurements has mean 505.4. The population standard deviation is 8. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 505.4σ = 8n = 1690%
L =
19.Linear regression & correlationMedium
xy126224324421521617xy2462025

The table gives 6 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 21Σy = 133Σx² = 91Σy² = 2999Σxy = 437
r =
20.Linear regression & correlationHard
xy241340452552658xy246405060x = 4.5

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (4.5, 53) is made. Find its residual e = y − ŷ.

n = 5(4.5, 53)
e =

3Solutions

1.Time value of money

Answer: PV = 17744 €

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 = 0.02/4 = 0.005
N = 4·6 = 24

Present value

PV = FV/(1 + i)N = 20 000/1.127 = €17 743.71

Answer

PV = €17 743.71
2.Time value of money

Answer: FV = 196317 €

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 + 0.05)29 − 1)/0.05 = 62.32

Annuity due

Each deposit is invested one year longer.

FV = 3 000·62.32·1.05 = €196 316.54

Answer

FV = €196 316.54
3.NPV & IRR

Answer: NPV = −7903 €

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 + 0.09)1 = 30 000/1.09 = 27 522.94
CF2/(1 + 0.09)2 = 25 000/1.188 = 21 042.00
CF3/(1 + 0.09)3 = 24 000/1.295 = 18 532.40

Add them up

NPV = €−7 902.66

Answer

NPV = €−7 902.66

NPV is negative, so the project earns less than the required return.

4.NPV & IRR

Answer: IRR = 21.5 %

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(20%) = 590.28
NPV(25%) = −1 288.00

Narrow down

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

IRR ≈ 20 + 5·590.28/(590.28 − (−1 288.00)) = 21.57%
exact: NPV(21.5%) = 0 → IRR = 21.5%

Answer

IRR = 21.5%
5.Bond pricing & yield

Answer: HPR = 12.04 %

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.

Purchase price

C = €45.00, i = 0.06, N = 8
€45.00·(1 − 0.6274)/0.06 + €1 000·0.6274 = €279.44 + €627.41 = €906.85

Selling price

C = €45.00, i = 0.05, N = 7
€45.00·(1 − 0.7107)/0.05 + €1 000·0.7107 = €260.39 + €710.68 = €971.07

Holding period return

HPR = (45 + 971.07 − 906.85)/906.85 = 12.04%

Answer

HPR = 12.04%
6.Bond pricing & yield

Answer: P = 875.4 €

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 = 11.1/1.06 = 10.47

Estimated price change

%ΔP ≈ −10.47·(0.0075) = −7.854%
P ≈ 950·(1 − 0.07854) = €875.39

Answer

P = €875.39

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

7.CAPM & WACC

Answer: re = 13 %

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 = €8 000 000, E/V = 0.375, D/V = 0.625

Debt part of the WACC

0.625·4%·(1 − 0.258) = 1.855%

Cost of equity

re = (6.73% − 1.855%)/0.375 = 13%

Answer

re = 13%
8.CAPM & WACC

Answer: WACC = 4.423 %

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 = 4 000 000·€12.50 = €50 000 000
D = 10 000·€1 020 = €10 200 000

Cost of equity (CAPM)

re = 1% + 0.8·4.5% = 4.6%

Weights

V = €60 200 000, E/V = 0.8306, D/V = 0.1694

WACC

WACC = 0.8306·4.6% + 0.1694·4.5%·(1 − 0.21) = 4.423%

Answer

WACC = 4.423%
9.Descriptive statistics

Answer: s = 11.45

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 = 188/6 = 31.33

Squared deviations

Σ(x − x̄)² = 13.67² + (−1.333)² + (−1.333)² + (−8.333)² + (−15.33)² + 12.67² = 655.3

Standard deviation

s = √(655.3/5) = 11.45

Answer

s = 11.45
10.Descriptive statistics

Answer: P = 0.06681

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 0.06681

Answer

P = 0.06681
11.Probability

Answer: P = 0.1515

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 = 5/12 · 4/11 = 20/132 = 0.1515

Answer

P = 0.1515
12.Probability

Answer: E(G) = 1.25 €

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: 10 (p = 0.25), 45 (p = 0.05), −5 (p = 0.7)

Expected value

E(G) = 10·0.25 + 45·0.05 − 5·0.7 = 1.25 €

Answer

E(G) = 1.25 €

A positive value means you win money on average.

13.Normal distribution

Answer: z = −0.91

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 − 0.8186 = 0.1814

Read the z-table backwards

Find 0.1814 inside the table and read off z.

z = −0.91

Answer

z = −0.91
14.Normal distribution

Answer: P = 0.8133

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 difference

μ = 25.02 − 25 = 0.02 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.02)/0.02236 = −0.89

Probability

P = 1 − Φ(−0.89) = 0.8133

Answer

P = 0.8133
15.Confidence intervals

Answer: upper limit = 64.55

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 = 29, t = 2.045

Margin of error

E = 2.045·9.5/√30 = 3.547

Interval

61 ± 3.547 → [57.45, 64.55]

Answer

upper limit = 64.55
16.Confidence intervals

Answer: σ = 5.993 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 14·3.8² = 202.2

Limits for σ²

202.2/26.119 = 7.74
202.2/5.629 = 35.91

Limit for σ

σ = √35.91 = 5.993 µm

Answer

σ = 5.993 µm

95% interval for σ: [2.782, 5.993] µm.

17.Hypothesis tests & confidence intervals

Answer: t = −1.098

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 = (7·10.5² + 8·10.5²)/15 = 110.3
sp = 10.5

Standard error

SE = 10.5·√(1/8 + 1/9) = 5.102

Test statistic

t = (42 − 47.6)/5.102 = −1.098

Answer

t = −1.098

With df = 15, the two-sided 5% critical value is 2.131; |t| is smaller, so H₀ is not rejected.

18.Hypothesis tests & confidence intervals

Answer: L = 502.1

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 = 1.645·8/√16 = 3.29

Interval limit

L = 505.4 − 3.29 = 502.1

Answer

L = 502.1
19.Linear regression & correlation

Answer: r = −0.9555

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 = 437 − 21·133/6 = −28.5
Sxx = 91 − 21²/6 = 17.5
Syy = 2999 − 133²/6 = 50.83

Correlation coefficient

r = −28.5/√(17.5·50.83) = −0.9555

Answer

r = −0.9555
20.Linear regression & correlation

Answer: e = 2.1

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 = 20/5 = 4
ȳ = 243/5 = 48.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 46

Regression line

b = 46/10 = 4.6
a = 48.6 − 4.6·4 = 30.2
ŷ = 30.2 + 4.6·x

Predicted value

ŷ = 30.2 + 4.6·4.5 = 50.9

Residual

e = 53 − 50.9 = 2.1

Answer

e = 2.1

The point lies above the line.

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