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…10€20 000FV = ?

You invest €20 000 at a nominal rate of 6% per year, compounded monthly. Find the value FV of the investment after 10 years.

PV = €20 0006% / 12n = 10
FV = €
2.Time value of moneyHard
months0123…60€15 000A?AAA

A loan of €15 000 is repaid with equal monthly payments over 5 years. The interest rate is 5.4% per year, compounded monthly. Find the monthly payment A.

P = €15 0005.4% / 12N = 60
A = €
3.NPV & IRRMedium
years0123456−20 0005 5005 5005 5005 5005 5005 500

A project costs €20 000 today and returns €5 500 at the end of each year for 6 years. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 12%CF0 = −20 000CF1 = 5 500CF2 = 5 500CF3 = 5 500CF4 = 5 500CF5 = 5 500CF6 = 5 500
DPB = years
4.NPV & IRRHard
years012−25 00017 50015 500

A project costs €25 000 and returns €17 500, €15 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −25 000CF1 = 17 500CF2 = 15 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…24€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 12 years. The yield to maturity is 3% per year. Find the price P of the bond.

F = €1 000coupon = 2%YTM = 3%n = 12
P = €
6.Bond pricing & yieldHard
years012345€1 060€80€80 + €1 000

A bond with a face value of €1 000 and an annual coupon of 8% matures in 5 years and sells for €1 060. Find its yield to maturity YTM in percent.

F = €1 000coupon = 8%n = 5P = €1 060
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€6 000 000re​ = 14%preferred shares€1 000 000rp​ = 7.5%debt€2 000 000rd​ = 6%T = 25.8%

A company is financed with ordinary shares worth €6 000 000, preferred shares worth €1 000 000 and debt worth €2 000 000. The cost of equity is 14%, the cost of preferred stock 7.5% and the pre-tax cost of debt 6%. The tax rate is 25.8%. Find the WACC.

E = €6 000 000Ps = €1 000 000D = €2 000 000re = 14%rp = 7.5%rd = 6%T = 25.8%
WACC = %
8.CAPM & WACCHard
rf​ = 1%, rm​ − rf​ = 6%D/E = 0.2equityβL​ = 1.4debtD/E = 1.5equityre​ = ?debtT = 25%

A company's shares have a beta of 1.4 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 1.5. The tax rate is 25%, the risk-free rate 1% and the market risk premium 6%. Find the cost of equity re after the change.

βL = 1.4D/E1 = 0.2D/E2 = 1.5T = 25%rf = 1%rm − rf = 6%
re = %
9.Descriptive statisticsMedium
4045505560

Find the sample standard deviation s of the data: 44, 37, 46, 60, 48.

n = 5
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is above 155?

μ = 170σ = 15x = 155
P =
11.ProbabilityMedium
P(at least one six) = ??1?2P(six) = 1/6 on each roll

A fair die is rolled 2 times. What is the probability P of getting at least one six?

n = 2
P =
12.ProbabilityHard
n = 12, p = 0.2: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 12p = 0.2
P =
13.Normal distributionMedium
105 s110.1 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 110.1 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 110.1 s
P =
14.Normal distributionHard
235240245250255260265gtop 20%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 20% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g20%
x = g
15.Confidence intervalsMedium
n = 12, s = 10LUx̄ = 68E = ?95%

A sample of 12 items has mean 68 and standard deviation s = 10. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 12x̄ = 68s = 1095%t(11) = 2.201
E =
16.Confidence intervalsHard
σ = 26, n = ?x̄ − Ex̄ + Ex̄E = 195%

The population standard deviation is 26. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1?

σ = 26E = 195%
n =
17.Hypothesis tests & confidence intervalsMedium
1.1p = ?

The mean of a population is claimed to be 120; the population standard deviation is known to be 24. A random sample of 36 observations has a mean of 124.4. Test H₀: μ = 120 against H₁: μ > 120. Find the p-value p.

μ₀ = 120σ = 24n = 36x̄ = 124.4H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
σ = 5, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 5. Find the smallest sample size n.

σ = 5E = 0.590%
n =
19.Linear regression & correlationMedium
xy14361756206423632681xy152025406080

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy2524566648791089xy5106080x = 5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (5, 58) is made. Find its residual e = y − ŷ.

n = 5(5, 58)
e =

3Solutions

1.Time value of money

Answer: FV = 36388 €

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 = r/m = 0.06/12 = 0.005
N = m·n = 12·10 = 120

Future value

FV = 20 000·(1 + 0.005)120 = 20 000·1.819 = €36 387.93

Answer

FV = €36 387.93
2.Time value of money

Answer: A = 285.8 €

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 = 0.054000000000000006/12 = 0.0045
N = 12·5 = 60

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 = 0.7638
A = 15 000·0.0045/(1 − 0.7638) = €285.83

Answer

A = €285.83
3.NPV & IRR

Answer: DPB = 5.062 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: 5 500/1.12 = 4 910.71 → Σ = 4 910.71
year 2: 5 500/1.254 = 4 384.57 → Σ = 9 295.28
year 3: 5 500/1.405 = 3 914.79 → Σ = 13 210.07
year 4: 5 500/1.574 = 3 495.35 → Σ = 16 705.42
year 5: 5 500/1.762 = 3 120.85 → Σ = 19 826.27
year 6: 5 500/1.974 = 2 786.47 → Σ = 22 612.74

Fraction of the last year

DPB = 5 + 173.73/2 786.47 = 5.062

Answer

DPB = 5.062 years

The discounted payback is always longer than the simple payback (3.636 years).

4.NPV & IRR

Answer: IRR = 21.17 %

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%) = 347.22
NPV(25%) = −1 080.00

Narrow down

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

IRR ≈ 20 + 5·347.22/(347.22 − (−1 080.00)) = 21.22%
exact: NPV(21.17%) = 0 → IRR = 21.17%

Answer

IRR = 21.17%
5.Bond pricing & yield

Answer: P = 899.8 €

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 = 24

Present value of the coupons

€10.00·(1 − 0.6995)/0.015 = €200.30

Present value of the face value

€1 000·0.6995 = €699.54

Price

P = €899.85

Answer

P = €899.85

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

6.Bond pricing & yield

Answer: YTM = 6.554 %

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

€1 060 = €80·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵

Solve numerically

P(6%) = €1 084.25, P(7%) = €1 041.00
YTM = 6.554%

Answer

YTM = 6.554%
7.CAPM & WACC

Answer: WACC = 11.16 %

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 + 1 000 000 + 2 000 000 = €9 000 000
E/V = 0.6667, Ps/V = 0.1111, D/V = 0.2222

WACC

WACC = 0.6667·14% + 0.1111·7.5% + 0.2222·6%·(1 − 0.258) = 11.16%

Answer

WACC = 11.16%
8.CAPM & WACC

Answer: re = 16.52 %

Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.

Unlever the beta

βU = 1.4/(1 + 0.75·0.2) = 1.217

Relever at the new D/E

βL = 1.217·(1 + 0.75·1.5) = 2.587

Cost of equity (CAPM)

re = 1% + 2.587·6% = 16.52%

Answer

re = 16.52%
9.Descriptive statistics

Answer: s = 8.367

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 = 235/5 = 47

Squared deviations

Σ(x − x̄)² = (−3)² + (−10)² + (−1)² + 13² + 1² = 280

Standard deviation

s = √(280/4) = 8.367

Answer

s = 8.367
10.Descriptive statistics

Answer: P = 0.8413

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

Standard normal table

Φ(−1) = 0.1587
P = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
11.Probability

Answer: P = 0.3056

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)2 = 0.6944
P = 1 − 0.6944 = 0.3056

Answer

P = 0.3056
12.Probability

Answer: P = 0.2749

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) = 0.812 = 0.06872
P(X = 1) = 12·0.2·0.811 = 0.2062

Add them

P = 0.06872 + 0.2062 = 0.2749

Answer

P = 0.2749
13.Normal distribution

Answer: P = 0.2119

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 sum

μ = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (110.1 − 105)/6.403 = 0.8

Upper tail

P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119

Answer

P = 0.2119
14.Normal distribution

Answer: x = 254.2 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) = 0.8

z from the table

z ≈ 0.84

Back to the original units

x = 250 + 0.84·5 = 254.2 g

Answer

x = 254.2 g
15.Confidence intervals

Answer: E = 6.354

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 = 11, t = 2.201

Margin of error

E = 2.201·10/√12 = 6.354

Answer

E = 6.354
16.Confidence intervals

Answer: n = 2597

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 = (1.96·26/1)² = 2597

Round up

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

n = 2597

Answer

n = 2597
17.Hypothesis tests & confidence intervals

Answer: p = 0.1357

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.

Test statistic

SE = σ/√n = 24/√36 = 4
z = (124.4 − 120)/4 = 1.1

Tail area

Φ(1.1) = 0.8643
1 − Φ(1.1) = 0.1357

p-value

p = 0.1357

Answer

p = 0.1357

p > 0.05, so H₀ is not rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: n = 271

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).

Solve for n

n = (1.645·5/0.5)² = 270.6

Round up

n = 271

Answer

n = 271
19.Linear regression & correlation

Answer: r² = 0.8893

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 = 100/5 = 20
ȳ = 300/5 = 60

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 291
Syy = Σ(y − ȳ)² = 1058

Coefficient of determination

r² = 291²/(90·1058) = 0.8893

Answer

r² = 0.8893

About 88.93% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = −5.15

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 = 340/5 = 68

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 194

Regression line

b = 194/40 = 4.85
a = 68 − 4.85·6 = 38.9
ŷ = 38.9 + 4.85·x

Predicted value

ŷ = 38.9 + 4.85·5 = 63.15

Residual

e = 58 − 63.15 = −5.15

Answer

e = −5.15

The point lies below the line.

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