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 = ?€50 000

You need €50 000 in 6 years. An account pays 3% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €50 0003% / 4n = 6
PV = €
2.Time value of moneyHard
months0123…192€500€500€500FV = ?

You save €500 at the end of every month for 16 years. The account pays 5.4% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €5005.4% / 12N = 192
FV = €
3.NPV & IRRMedium
years0123…9−250 00027 000+1%+1%…r = 10%, NPV = ?

A project costs €250 000 today. It generates €27 000 at the end of the first year, and the cash flow then grows by 1% per year forever. The cost of capital is 10%. Find the net present value NPV.

C0 = €250 000CF1 = €27 000g = 1%r = 10%
NPV = €
4.NPV & IRRHard
years0123−40 00023 00028 00027 000

A project costs €40 000 and returns €23 000, €28 000, €27 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −40 000CF1 = 23 000CF2 = 28 000CF3 = 27 000
IRR = %
5.Bond pricing & yieldMedium

You buy a bond with 7 years to maturity (face value €1 000, annual coupon 4.5%) when its yield to maturity is 2.5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 4%. Find your holding period return HPR in percent.

F = €1 000coupon = 4.5%n = 7y0 = 2.5%y1 = 4%
HPR = %
6.Bond pricing & yieldHard
years0123…10€1 155€70€70 + €1 000

A bond with a face value of €1 000 and an annual coupon of 7% matures in 10 years and sells for €1 155. Find its yield to maturity YTM in percent.

F = €1 000coupon = 7%n = 10P = €1 155
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€10 000 000re​ = 13%debt€2 000 000rd​ = 6%T = 20%

A company has equity worth €10 000 000 and debt worth €2 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 6% and the tax rate is 20%. Find the WACC.

E = €10 000 000D = €2 000 000re = 13%rd = 6%T = 20%
WACC = %
8.CAPM & WACCHard
rf​ = 4%, rm​ − rf​ = 5.5%D/E = 0.5equityβL​ = 1.1debtD/E = 1.5equityre​ = ?debtT = 30%

A company's shares have a beta of 1.1 at its current debt-to-equity ratio of 0.5. It plans to raise its debt-to-equity ratio to 1.5. The tax rate is 30%, the risk-free rate 4% and the market risk premium 5.5%. Find the cost of equity re after the change.

βL = 1.1D/E1 = 0.5D/E2 = 1.5T = 30%rf = 4%rm − rf = 5.5%
re = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 57, 29, 13, 34, 41.

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

A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is above 76?

μ = 70σ = 15x = 76
P =
11.ProbabilityMedium
xp(x)00.210.1520.0530.340.3μ = 2.35

For the distribution shown, find the variance Var(X) of X.

x: 0, 1, 2, 3, 4p(x): 0.2, 0.15, 0.05, 0.3, 0.3
Var(X) =
12.ProbabilityHard
xp(x)00.220.3540.360.15μ

X has the distribution shown, and Y = 3X + 4. Find the standard deviation σY of Y.

x: 0, 2, 4, 6p(x): 0.2, 0.35, 0.3, 0.15Y = 3X + 4
σY =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0594

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0594. Find z to two decimals.

A = 0.0594
z =
14.Normal distributionHard
380420460500540580620hbottom 20%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 20% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h20%
x = h
15.Confidence intervalsMedium
n = 8, sd​ = 4.10LU = ?d̄ = 8.7E95%

The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 8.7 and standard deviation 4.1. Find the upper limit of the 95% confidence interval for the mean difference.

n = 8d̄ = 8.7sd = 4.195%t(7) = 2.365
upper limit = mmHg
16.Confidence intervalsHard
σ = 28, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

The population standard deviation is 28. 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 1.5?

σ = 28E = 1.599%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.122.12p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 12.5. A random sample of 25 observations has a mean of 55.3. Test H₀: μ = 50 against H₁: μ ≠ 50. Find the p-value p.

μ₀ = 50σ = 12.5n = 25x̄ = 55.3H₁: μ ≠ μ₀
p =
18.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 49 measurements has mean 120.1. The population standard deviation is 5. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 120.1σ = 5n = 4999%
L =
19.Linear regression & correlationMedium
xy284961581710251225xy5101020

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy127226322420518616xy2462025x = 4.5

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. Predict ŷ for x = 4.5.

n = 6x = 4.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 41792 €

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 = 0.03/4 = 0.0075
N = 4·6 = 24

Present value

PV = FV/(1 + i)N = 50 000/1.196 = €41 791.57

Answer

PV = €41 791.57
2.Time value of money

Answer: FV = 152004 €

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 deposits

i = 0.054000000000000006/12 = 0.0045
N = 12·16 = 192

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 500·(2.368 − 1)/0.0045 = €152 004.23

Answer

FV = €152 004.23

You paid in €96 000 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = 50000 €

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.

Value of the cash flows

PV = 27 000/(0.1 − 0.01) = 300 000.00

NPV

NPV = 300 000.00 − 250 000 = €50 000.00

Answer

NPV = €50 000.00

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

4.NPV & IRR

Answer: IRR = 41.05 %

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(40%) = 553.94
NPV(45%) = −1 964.00

Narrow down

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

IRR ≈ 40 + 5·553.94/(553.94 − (−1 964.00)) = 41.1%
exact: NPV(41.05%) = 0 → IRR = 41.05%

Answer

IRR = 41.05%
5.Bond pricing & yield

Answer: HPR = −4.949 %

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 = 0.025, N = 7
€45.00·(1 − 0.8413)/0.025 + €1 000·0.8413 = €285.72 + €841.27 = €1 126.99

Selling price

C = €45.00, i = 0.04, N = 6
€45.00·(1 − 0.7903)/0.04 + €1 000·0.7903 = €235.90 + €790.31 = €1 026.21

Holding period return

HPR = (45 + 1 026.21 − 1 126.99)/1 126.99 = −4.949%

Answer

HPR = −4.949%
6.Bond pricing & yield

Answer: YTM = 4.993 %

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

€1 155 = €70·(1 − (1 + y)⁻¹⁰)/y + €1 000/(1 + y)¹⁰

Solve numerically

P(4%) = €1 243.33, P(5%) = €1 154.43
YTM = 4.993%

Answer

YTM = 4.993%
7.CAPM & WACC

Answer: WACC = 11.63 %

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 = €12 000 000, E/V = 0.8333, D/V = 0.1667

WACC

WACC = 0.8333·13% + 0.1667·6%·(1 − 0.2) = 11.63%

Answer

WACC = 11.63%
8.CAPM & WACC

Answer: re = 13.19 %

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 = 1.1/(1 + 0.7·0.5) = 0.8148

Relever at the new D/E

βL = 0.8148·(1 + 0.7·1.5) = 1.67

Cost of equity (CAPM)

re = 4% + 1.67·5.5% = 13.19%

Answer

re = 13.19%
9.Descriptive statistics

Answer: s = 16.13

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 = 174/5 = 34.8

Squared deviations

Σ(x − x̄)² = 22.2² + (−5.8)² + (−21.8)² + (−0.8)² + 6.2² = 1041

Standard deviation

s = √(1041/4) = 16.13

Answer

s = 16.13
10.Descriptive statistics

Answer: P = 0.3446

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446

Answer

P = 0.3446
11.Probability

Answer: Var(X) = 2.328

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.

Mean

μ = 0·0.2 + 1·0.15 + 2·0.05 + 3·0.3 + 4·0.3 = 2.35

Mean of the squares

E(X²) = 0·0.2 + 1·0.15 + 4·0.05 + 9·0.3 + 16·0.3 = 7.85

Variance

Var(X) = 7.85 − 2.35² = 2.328

Answer

Var(X) = 2.328
12.Probability

Answer: σY = 5.817

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.

Spread of X

μ = 2.8, Var(X) = 3.76, σX = 1.939

Linear change

σY = |3|·1.939 = 5.817

Answer

σY = 5.817

The mean does move: E(Y) = 3·2.8 + 4 = 12.4.

13.Normal distribution

Answer: z = 1.56

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 − 0.0594 = 0.9406

Read the z-table backwards

Find 0.9406 inside the table and read off z.

z = 1.56

Answer

z = 1.56
14.Normal distribution

Answer: x = 466 h

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 = 500 − 0.84·40 = 466 h

Answer

x = 466 h
15.Confidence intervals

Answer: upper limit = 12.13 mmHg

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.

Critical value

df = 7, t = 2.365

Margin of error

E = 2.365·4.1/√8 = 3.428

Interval

8.7 ± 3.428 → [5.272, 12.13]

Answer

upper limit = 12.13 mmHg
16.Confidence intervals

Answer: n = 2313

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 = (2.576·28/1.5)² = 2312

Round up

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

n = 2313

Answer

n = 2313
17.Hypothesis tests & confidence intervals

Answer: p = 0.03401

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 = 12.5/√25 = 2.5
z = (55.3 − 50)/2.5 = 2.12

Tail area

Φ(2.12) = 0.9830
1 − Φ(2.12) = 0.0170

p-value

p = 2·0.0170 = 0.03401

Answer

p = 0.03401

p ≤ 0.05, so H₀ is rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: L = 118.3

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 = 2.576·5/√49 = 1.84

Interval limit

L = 120.1 − 1.84 = 118.3

Answer

L = 118.3
19.Linear regression & correlation

Answer: r² = 0.945

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 = 42/6 = 7
ȳ = 99/6 = 16.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 135
Syy = Σ(y − ȳ)² = 275.5

Coefficient of determination

r² = 135²/(70·275.5) = 0.945

Answer

r² = 0.945

About 94.5% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 19.19

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 = 129/6 = 21.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −40.5

Regression line

b = −40.5/17.5 = −2.314
a = 21.5 − (−2.314)·3.5 = 29.6
ŷ = 29.6 − 2.314·x

Prediction

ŷ = 29.6 + (−2.314)·4.5 = 19.19

Answer

ŷ = 19.19

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