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…8PV = ?€5 000

You need €5 000 in 8 years. An account pays 6% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €5 0006% / 12n = 8
PV = €
2.Time value of moneyHard
months0123…72€20 000A?AAA

A loan of €20 000 is repaid with equal monthly payments over 6 years. The interest rate is 5.4% per year, compounded monthly. Find the monthly payment A.

P = €20 0005.4% / 12N = 72
A = €
3.NPV & IRRMedium
years012345−20 0006 0006 0006 0006 0006 000

A project costs €20 000 today and returns €6 000 at the end of each year for 5 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 = 6 000CF2 = 6 000CF3 = 6 000CF4 = 6 000CF5 = 6 000
DPB = years
4.NPV & IRRHard
years0123…9−150 00016 500+3%+3%+3%r = 8%, NPV = ?

A project costs €150 000 today. Its cash flow is €16 500 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 9. The cost of capital is 8%. Find the net present value NPV.

C0 = €150 000CF1 = €16 500g = 3%n = 9r = 8%
NPV = €
5.Bond pricing & yieldMedium
half-years0123…16€40€40€40 + €1 000P = ?

A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 8 years. The yield to maturity is 7% per year. Find the price P of the bond.

F = €1 000coupon = 8%YTM = 7%n = 8
P = €
6.Bond pricing & yieldHard
years01234€1 070€70€70 + €1 000

A bond with a face value of €1 000 and an annual coupon of 7% matures in 4 years and sells for €1 070. Find its yield to maturity YTM in percent.

F = €1 000coupon = 7%n = 4P = €1 070
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€10 000 000re​ = 9%preferred shares€2 000 000rp​ = 6.5%debt€4 000 000rd​ = 6%T = 25%

A company is financed with ordinary shares worth €10 000 000, preferred shares worth €2 000 000 and debt worth €4 000 000. The cost of equity is 9%, the cost of preferred stock 6.5% and the pre-tax cost of debt 6%. The tax rate is 25%. Find the WACC.

E = €10 000 000Ps = €2 000 000D = €4 000 000re = 9%rp = 6.5%rd = 6%T = 25%
WACC = %
8.CAPM & WACCHard
rf​ = 1%, rm​ − rf​ = 4%no debtequityβU​ = 1.1D/E = 0.6equityre​ = ?debtT = 25.8%

A company's business has an unlevered (asset) beta of 1.1. It plans a debt-to-equity ratio of 0.6. The tax rate is 25.8%, the risk-free rate is 1% and the market risk premium is 4%. Find the cost of equity re at the new capital structure.

βU = 1.1D/E = 0.6T = 25.8%rf = 1%rm − rf = 4%
re = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 51, 50, 51, 13, 25.

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

A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is below 70?

μ = 50σ = 10x = 70
P =
11.ProbabilityMedium
xp(x)100.05200.35300.35400.25μ = 28

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

x: 10, 20, 30, 40p(x): 0.05, 0.35, 0.35, 0.25
Var(X) =
12.ProbabilityHard
xp(x)10.220.130.3540.150.25μ

X has the distribution shown, and Y = 0.5X − 3. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.2, 0.1, 0.35, 0.1, 0.25Y = 0.5X − 3
σY =
13.Normal distributionMedium
320 g325.1 gT = X₁ + … + X₈?

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 8 of them is below 325.1 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 8< 325.1 g
P =
14.Normal distributionHard
μ = 500x = ?99%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 99% of them lie (the 99th percentile).

μ = 500σ = 40P = 99%
x = h
15.Confidence intervalsMedium
n = 12, s = 13.5L = ?Ux̄ = 29E90%

A sample of 12 items has mean 29 and standard deviation s = 13.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 12x̄ = 29s = 13.590%t(11) = 1.796
lower limit =
16.Confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 195%

The population standard deviation is 10. 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 = 10E = 195%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.2622.262t = ?df = 9

10 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 0.76 and the standard deviation of the differences is 1. Find the paired test statistic t for H₀: μd = 0.

n = 10d̄ = 0.76sd = 1
t =
18.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 16 measurements has mean 494.2. The population standard deviation is 8. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 494.2σ = 8n = 1690%
L =
19.Linear regression & correlationMedium
xy3585747819851193xy5106080

The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy24045566087410751287xy510406080ŷ = 83

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 83?

n = 6ŷ = 83
x =

3Solutions

1.Time value of money

Answer: PV = 3098 €

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 = 0.06/12 = 0.005
N = 12·8 = 96

Present value

PV = FV/(1 + i)N = 5 000/1.614 = €3 097.62

Answer

PV = €3 097.62
2.Time value of money

Answer: A = 325.8 €

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 = 12·6 = 72

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 = 0.7238
A = 20 000·0.0045/(1 − 0.7238) = €325.82

Answer

A = €325.82
3.NPV & IRR

Answer: DPB = 4.522 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: 6 000/1.12 = 5 357.14 → Σ = 5 357.14
year 2: 6 000/1.254 = 4 783.16 → Σ = 10 140.31
year 3: 6 000/1.405 = 4 270.68 → Σ = 14 410.99
year 4: 6 000/1.574 = 3 813.11 → Σ = 18 224.10
year 5: 6 000/1.762 = 3 404.56 → Σ = 21 628.66

Fraction of the last year

DPB = 4 + 1 775.90/3 404.56 = 4.522

Answer

DPB = 4.522 years

The discounted payback is always longer than the simple payback (3.333 years).

4.NPV & IRR

Answer: NPV = −35395 €

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 growing annuity

((1.03)/(1.08))9 = 0.6527
PV = 16 500/(0.08 − 0.03)·(1 − 0.6527) = 114 605.23

NPV

NPV = 114 605.23 − 150 000 = −€35 394.77

Answer

NPV = −€35 394.77

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

5.Bond pricing & yield

Answer: P = 1060 €

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 = €40.00, i = 0.035, N = 16

Present value of the coupons

€40.00·(1 − 0.5767)/0.035 = €483.76

Present value of the face value

€1 000·0.5767 = €576.71

Price

P = €1 060.47

Answer

P = €1 060.47

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

6.Bond pricing & yield

Answer: YTM = 5.025 %

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

€1 070 = €70·(1 − (1 + y)⁻⁴)/y + €1 000/(1 + y)⁴

Solve numerically

P(5%) = €1 070.92, P(6%) = €1 034.65
YTM = 5.025%

Answer

YTM = 5.025%
7.CAPM & WACC

Answer: WACC = 7.563 %

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 = 10 000 000 + 2 000 000 + 4 000 000 = €16 000 000
E/V = 0.625, Ps/V = 0.125, D/V = 0.25

WACC

WACC = 0.625·9% + 0.125·6.5% + 0.25·6%·(1 − 0.25) = 7.563%

Answer

WACC = 7.563%
8.CAPM & WACC

Answer: re = 7.359 %

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.

Levered beta

βL = 1.1·(1 + 0.742·0.6) = 1.59

Cost of equity (CAPM)

re = 1% + 1.59·4% = 7.359%

Answer

re = 7.359%
9.Descriptive statistics

Answer: s = 17.86

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 = 190/5 = 38

Squared deviations

Σ(x − x̄)² = 13² + 12² + 13² + (−25)² + (−13)² = 1276

Standard deviation

s = √(1276/4) = 17.86

Answer

s = 17.86
10.Descriptive statistics

Answer: P = 0.9772

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 0.9772

Answer

P = 0.9772
11.Probability

Answer: Var(X) = 76

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 = 10·0.05 + 20·0.35 + 30·0.35 + 40·0.25 = 28

Mean of the squares

E(X²) = 100·0.05 + 400·0.35 + 900·0.35 + 1600·0.25 = 860

Variance

Var(X) = 860 − 28² = 76

Answer

Var(X) = 76
12.Probability

Answer: σY = 0.7053

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 = 3.1, Var(X) = 1.99, σX = 1.411

Linear change

σY = |0.5|·1.411 = 0.7053

Answer

σY = 0.7053

The mean does move: E(Y) = 0.5·3.1 − 3 = −1.45.

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 = 320 g, σT = 1.2·√8 = 3.394 g

Standardise

z = (325.1 − 320)/3.394 = 1.5

Probability

P = Φ(1.5) = 0.9332

Answer

P = 0.9332
14.Normal distribution

Answer: x = 593.1 h

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 for the percentile

Φ(z) = 0.99 → z = 2.326

Back to x

x = 500 + 2.326·40 = 593.05 h

Answer

x = 593.05 h
15.Confidence intervals

Answer: lower limit = 22

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 = 1.796

Margin of error

E = 1.796·13.5/√12 = 6.999

Interval

29 ± 6.999 → [22, 36]

Answer

lower limit = 22
16.Confidence intervals

Answer: n = 385

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 = (1.96·10/1)² = 384.2

Round up

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

n = 385

Answer

n = 385
17.Hypothesis tests & confidence intervals

Answer: t = 2.403

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 of the mean difference

SE = 1/√10 = 0.3162

Test statistic

t = 0.76/0.3162 = 2.403

Answer

t = 2.403

With df = 9, the two-sided 5% critical value is 2.262; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: L = 490.9

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).

Margin of error

E = 1.645·8/√16 = 3.29

Interval limit

L = 494.2 − 3.29 = 490.9

Answer

L = 490.9
19.Linear regression & correlation

Answer: r = 0.969

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 = 35/5 = 7
ȳ = 391/5 = 78.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 162
Syy = Σ(y − ȳ)² = 698.8

Correlation coefficient

r = 162/√(40·698.8) = 0.969

Answer

r = 0.969
20.Linear regression & correlation

Answer: x = 11.04

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 = 391/6 = 65.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 309

Regression line

b = 309/70 = 4.414
a = 65.17 − 4.414·7 = 34.27
ŷ = 34.27 + 4.414·x

Solve for x

x = (83 − 34.27)/4.414 = 11.04

Answer

x = 11.04

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