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…16PV = ?€500€500€500€500

An annuity pays €500 at the end of each year for 16 years. The interest rate is 8% per year. Find the present value PV of the annuity.

A = €500i = 8%n = 16
PV = €
2.Time value of moneyHard
months0123…N = ?€8 000€375€375€375€3757.2% / 12

You borrow €8 000 at 7.2% per year, compounded monthly, and repay €375 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.

P = €8 000A = €3757.2% / 12
N = payments
3.NPV & IRRMedium
years012345−50 00016 50012 00010 50014 00014 000

A project costs €50 000 today and returns €16 500, €12 000, €10 500, €14 000, €14 000 at the end of years 1 to 5. The cost of capital is 6%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 6%CF0 = −50 000CF1 = 16 500CF2 = 12 000CF3 = 10 500CF4 = 14 000CF5 = 14 000
DPB = years
4.NPV & IRRHard
years0123−40 00028 50021 50026 000

A project costs €40 000 and returns €28 500, €21 500, €26 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −40 000CF1 = 28 500CF2 = 21 500CF3 = 26 000
IRR = %
5.Bond pricing & yieldMedium
half-years0123…8€30€30€30 + €1 000P = ?

A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 4 years. The yield to maturity is 2.5% per year. Find the price P of the bond.

F = €1 000coupon = 6%YTM = 2.5%n = 4
P = €
6.Bond pricing & yieldHard

A bond trades at €985 with a yield to maturity of 7% (annual compounding) and a Macaulay duration of 10.8 years. Use the modified duration to estimate the new price P if the yield rises by 25 basis points.

P0 = €985y = 7%D = 10.8Δy = +25 bp
P = €
7.CAPM & WACCMedium
WACC = 5.78%equity€5 000 000re​ = ?debt€4 000 000rd​ = 4%T = 25%

A company has equity worth €5 000 000 and debt worth €4 000 000. Its WACC is 5.78%, its pre-tax cost of debt is 4% and the tax rate is 25%. Find the cost of equity re.

E = €5 000 000D = €4 000 000WACC = 5.78%rd = 4%T = 25%
re = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €39.50β = 0.9rf​ = 3%, rm​ = 7.5%debt5 000 × €1 000rd​ = 7%T = 25.8%

A company has 1 000 000 shares at €39.50 each and 5 000 bonds trading at €1 000 each. The pre-tax cost of debt is 7% and the tax rate is 25.8%. The share’s beta is 0.9, the risk-free rate 3% and the market return 7.5%. Find the WACC.

β = 0.9rf = 3%rm = 7.5%rd = 7%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 24, 33, 52, 19, 37, 17, 11.

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

A quantity is normally distributed with mean 500 and standard deviation 25. What proportion P of values is above 470?

μ = 500σ = 25x = 470
P =
11.ProbabilityMedium
P(at least one six) = ??1?2?3P(six) = 1/6 on each roll

A fair die is rolled 3 times. What is the probability P of getting at least one six?

n = 3
P =
12.ProbabilityHard
xp(x)00.310.3520.35μ

X has the distribution shown, and Y = 0.5X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 0.5X + 1
σY =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.3616

Drag z so that the shaded area between −z and z under the standard normal curve is 0.3616. Find z to two decimals.

A = 0.3616
z =
14.Normal distributionHard
557085100115130145pointstop 40%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points40%
x = points
15.Confidence intervalsMedium
n = 8, s = 12.5LU = ?x̄ = 55.5E90%

A sample of 8 items has mean 55.5 and standard deviation s = 12.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 8x̄ = 55.5s = 12.590%t(7) = 1.895
upper limit =
16.Confidence intervalsHard
σ = 32, n = ?x̄ − Ex̄ + Ex̄E = 495%

The population standard deviation is 32. 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 4?

σ = 32E = 495%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.84p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 40. A random sample of 100 observations has a mean of 42.64. Test H₀: μ = 50 against H₁: μ < 50. Find the p-value p.

μ₀ = 50σ = 40n = 100x̄ = 42.64H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 390%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 15. Find the smallest sample size n.

σ = 15E = 390%
n =
19.Linear regression & correlationMedium
xy352447558662760xy46505560

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 25Σy = 279Σx² = 135Σy² = 15721Σxy = 1426
r =
20.Linear regression & correlationHard
xy2504486588691071xy510506070ŷ = 69

The table gives 5 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 = 69?

n = 5ŷ = 69
x =

3Solutions

1.Time value of money

Answer: PV = 4426 €

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.

Annuity factor

(1 − (1 + i)−n)/i = (1 − 0.2919)/0.08 = 8.851

Present value

PV = 500·8.851 = €4 425.68

Answer

PV = €4 425.68
2.Time value of money

Answer: N = 23 payments

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.

Rearrange the annuity formula

i = 0.006
(1 + i)−N = 1 − 8 000·0.006/375 = 0.872

Solve with logarithms

N = −ln(0.872)/ln(1.006) = 22.9

Round up to whole payments

22 full payments leave a small balance, which the last, smaller payment clears.

N = 23

Answer

N = 23 payments
3.NPV & IRR

Answer: DPB = 4.368 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: 16 500/1.06 = 15 566.04 → Σ = 15 566.04
year 2: 12 000/1.124 = 10 679.96 → Σ = 26 246.00
year 3: 10 500/1.191 = 8 816.00 → Σ = 35 062.00
year 4: 14 000/1.262 = 11 089.31 → Σ = 46 151.31
year 5: 14 000/1.338 = 10 461.61 → Σ = 56 612.92

Fraction of the last year

DPB = 4 + 3 848.69/10 461.61 = 4.368

Answer

DPB = 4.368 years

The discounted payback is always longer than the simple payback (3.786 years).

4.NPV & IRR

Answer: IRR = 41.62 %

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%) = 801.75
NPV(45%) = −1 590.47

Narrow down

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

IRR ≈ 40 + 5·801.75/(801.75 − (−1 590.47)) = 41.68%
exact: NPV(41.62%) = 0 → IRR = 41.62%

Answer

IRR = 41.62%
5.Bond pricing & yield

Answer: P = 1132 €

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 = €30.00, i = 0.0125, N = 8

Present value of the coupons

€30.00·(1 − 0.9054)/0.0125 = €227.04

Present value of the face value

€1 000·0.9054 = €905.40

Price

P = €1 132.44

Answer

P = €1 132.44

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

6.Bond pricing & yield

Answer: P = 960.1 €

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 = 10.8/1.07 = 10.09

Estimated price change

%ΔP ≈ −10.09·(0.0025) = −2.523%
P ≈ 985·(1 − 0.02523) = €960.14

Answer

P = €960.14

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

7.CAPM & WACC

Answer: re = 8.004 %

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 = €9 000 000, E/V = 0.5556, D/V = 0.4444

Debt part of the WACC

0.4444·4%·(1 − 0.25) = 1.333%

Cost of equity

re = (5.78% − 1.333%)/0.5556 = 8.004%

Answer

re = 8.004%
8.CAPM & WACC

Answer: WACC = 6.841 %

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 = 1 000 000·€39.50 = €39 500 000
D = 5 000·€1 000 = €5 000 000

Cost of equity (CAPM)

re = 3% + 0.9·4.5% = 7.05%

Weights

V = €44 500 000, E/V = 0.8876, D/V = 0.1124

WACC

WACC = 0.8876·7.05% + 0.1124·7%·(1 − 0.258) = 6.841%

Answer

WACC = 6.841%
9.Descriptive statistics

Answer: s = 14.07

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 = 193/7 = 27.57

Squared deviations

Σ(x − x̄)² = (−3.571)² + 5.429² + 24.43² + (−8.571)² + 9.429² + (−10.57)² + (−16.57)² = 1188

Standard deviation

s = √(1188/6) = 14.07

Answer

s = 14.07
10.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
11.Probability

Answer: P = 0.4213

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)3 = 0.5787
P = 1 − 0.5787 = 0.4213

Answer

P = 0.4213
12.Probability

Answer: σY = 0.4023

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 = 1.05, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |0.5|·0.8047 = 0.4023

Answer

σY = 0.4023

The mean does move: E(Y) = 0.5·1.05 + 1 = 1.525.

13.Normal distribution

Answer: z = 0.47

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 + 0.3616)/2 = 0.6808

Read the z-table backwards

Find 0.6808 inside the table and read off z.

z = 0.47

Answer

z = 0.47
14.Normal distribution

Answer: x = 103.8 points

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) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 100 + 0.25·15 = 103.8 points

Answer

x = 103.8 points
15.Confidence intervals

Answer: upper limit = 63.87

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 = 7, t = 1.895

Margin of error

E = 1.895·12.5/√8 = 8.375

Interval

55.5 ± 8.375 → [47.13, 63.87]

Answer

upper limit = 63.87
16.Confidence intervals

Answer: n = 246

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 = (1.96·32/4)² = 245.9

Round up

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

n = 246

Answer

n = 246
17.Hypothesis tests & confidence intervals

Answer: p = 0.03288

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 = 40/√100 = 4
z = (42.64 − 50)/4 = −1.84

Tail area

Φ(1.84) = 0.9671
1 − Φ(1.84) = 0.0329

p-value

p = 0.03288

Answer

p = 0.03288

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

18.Hypothesis tests & confidence intervals

Answer: n = 68

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 = (1.645·15/3)² = 67.65

Round up

n = 68

Answer

n = 68
19.Linear regression & correlation

Answer: r = 0.793

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 = 1426 − 25·279/5 = 31
Sxx = 135 − 25²/5 = 10
Syy = 15721 − 279²/5 = 152.8

Correlation coefficient

r = 31/√(10·152.8) = 0.793

Answer

r = 0.793
20.Linear regression & correlation

Answer: x = 9.111

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 = 296/5 = 59.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 126

Regression line

b = 126/40 = 3.15
a = 59.2 − 3.15·6 = 40.3
ŷ = 40.3 + 3.15·x

Solve for x

x = (69 − 40.3)/3.15 = 9.111

Answer

x = 9.111

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