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
years01234PV = ?€5 000€5 000€5 000€5 000

An annuity pays €5 000 at the end of each year for 4 years. The interest rate is 7% per year. Find the present value PV of the annuity.

A = €5 000i = 7%n = 4
PV = €
2.Time value of moneyHard
months0123…300€200 000AAAA3.6% / 12, I = ?

A loan of €200 000 is repaid with equal monthly payments over 25 years at 3.6% per year, compounded monthly. How much interest I is paid in total over the life of the loan?

P = €200 0003.6% / 12N = 300
I = €
3.NPV & IRRMedium
years0123456−50 00018 00018 00018 00018 00018 00018 000

A project costs €50 000 today and returns €18 000 at the end of each year for 6 years. The cost of capital is 8%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 8%CF0 = −50 000CF1 = 18 000CF2 = 18 000CF3 = 18 000CF4 = 18 000CF5 = 18 000CF6 = 18 000
DPB = years
4.NPV & IRRHard
years0123−20 0007 5007 5007 000

A project costs €20 000 and returns €7 500, €7 500, €7 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 7 500CF2 = 7 500CF3 = 7 000
IRR = %
5.Bond pricing & yieldMedium
years0123…9€1 114C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 9 years and pays annual coupons. It sells for €1 114 when the yield to maturity is 3.5%. Find the coupon rate c in percent.

F = €1 000n = 9P = €1 114YTM = 3.5%
c = %
6.Bond pricing & yieldHard
years01234567€1 055€70€70 + €1 000

A bond with a face value of €1 000 and an annual coupon of 7% matures in 7 years and sells for €1 055. Find its yield to maturity YTM in percent.

F = €1 000coupon = 7%n = 7P = €1 055
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€8 000 000re​ = 11%debt€5 000 000rd​ = 4%T = 28%

A company has equity worth €8 000 000 and debt worth €5 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 4% and the tax rate is 28%. Find the WACC.

E = €8 000 000D = €5 000 000re = 11%rd = 4%T = 28%
WACC = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €50.50β = 0.6rf​ = 3%, rm​ = 9.5%debt40 000 × €980rd​ = 4.5%T = 25%

A company has 1 000 000 shares at €50.50 each and 40 000 bonds trading at €980 each. The pre-tax cost of debt is 4.5% and the tax rate is 25%. The share’s beta is 0.6, the risk-free rate 3% and the market return 9.5%. Find the WACC.

β = 0.6rf = 3%rm = 9.5%rd = 4.5%T = 25%
WACC = %
9.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 44, 21, 43, 50, 22, 21, 18.

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

A quantity is normally distributed with mean 60 and standard deviation 20. What proportion P of values is above 44?

μ = 60σ = 20x = 44
P =
11.ProbabilityMedium
n = 10k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 10? The order does not matter.

n = 10k = 2
N =
12.ProbabilityHard
n = 9, p = 0.4: P(X = 1) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 1 correct answers?

n = 9p = 0.4k = 1
P =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.4778

Drag z so that the shaded area between −z and z under the standard normal curve is 0.4778. Find z to two decimals.

A = 0.4778
z =
14.Normal distributionHard
μ = 32x = ?90%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 90% of them lie (the 90th percentile).

μ = 32σ = 6P = 90%
x = min
15.Confidence intervalsMedium
n = 30, s = 14.5L = ?Ux̄ = 99.5E99%

A sample of 30 items has mean 99.5 and standard deviation s = 14.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 30x̄ = 99.5s = 14.599%t(29) = 2.756
lower limit =
16.Confidence intervalsHard
σ = 26, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 26. 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 4?

σ = 26E = 499%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

Two teaching methods are compared. Group 1: 13 students, mean score 33, standard deviation 6.5. Group 2: 8 students, mean score 38.4, standard deviation 6. 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 = 13x̄₁ = 33s₁ = 6.5n₂ = 8x̄₂ = 38.4s₂ = 6
t =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ < 50 is done at α = 0.01 with a sample of 36; the standard deviation is known to be 12. In reality the mean is 43.6. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 43.6σ = 12n = 36α = 0.01
1 − β = %
19.Linear regression & correlationMedium
xy2842062883410431249xy5102040

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
xy136247347450557xy244050x = 4.5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.

n = 5x = 4.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 16936 €

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 − 0.7629)/0.07 = 3.387

Present value

PV = 5 000·3.387 = €16 936.06

Answer

PV = €16 936.06
2.Time value of money

Answer: I = 103602 €

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 payment

i = 0.003, N = 300
A = 200 000·0.003/(1 − (1 + 0.003)−300) = €1 012.01

Total interest

I = N·A − P = 300·1012 − 200 000 = €103 601.63

Answer

I = €103 601.63
3.NPV & IRR

Answer: DPB = 3.273 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: 18 000/1.08 = 16 666.67 → Σ = 16 666.67
year 2: 18 000/1.166 = 15 432.10 → Σ = 32 098.77
year 3: 18 000/1.26 = 14 288.98 → Σ = 46 387.75
year 4: 18 000/1.36 = 13 230.54 → Σ = 59 618.28

Fraction of the last year

DPB = 3 + 3 612.25/13 230.54 = 3.273

Answer

DPB = 3.273 years

The discounted payback is always longer than the simple payback (2.778 years).

4.NPV & IRR

Answer: IRR = 4.98 %

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(0%) = 2 000.00
NPV(5%) = −7.56

Narrow down

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

IRR ≈ 0 + 5·2 000.00/(2 000.00 − (−7.56)) = 4.981%
exact: NPV(4.98%) = 0 → IRR = 4.98%

Answer

IRR = 4.98%
5.Bond pricing & yield

Answer: c = 4.998 %

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.

Present value of the face value

1000·(1 + 0.035)−9 = €733.73

Coupon

C = (1114 − 733.73)/7.608 = €49.98

Coupon rate

c = 49.98/1000 = 4.998%

Answer

c = 4.998%
6.Bond pricing & yield

Answer: YTM = 6.014 %

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

€1 055 = €70·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷

Solve numerically

P(6%) = €1 055.82, P(7%) = €1 000.00
YTM = 6.014%

Answer

YTM = 6.014%
7.CAPM & WACC

Answer: WACC = 7.877 %

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 = €13 000 000, E/V = 0.6154, D/V = 0.3846

WACC

WACC = 0.6154·11% + 0.3846·4%·(1 − 0.28) = 7.877%

Answer

WACC = 7.877%
8.CAPM & WACC

Answer: WACC = 5.36 %

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·€50.50 = €50 500 000
D = 40 000·€980 = €39 200 000

Cost of equity (CAPM)

re = 3% + 0.6·6.5% = 6.9%

Weights

V = €89 700 000, E/V = 0.563, D/V = 0.437

WACC

WACC = 0.563·6.9% + 0.437·4.5%·(1 − 0.25) = 5.36%

Answer

WACC = 5.36%
9.Descriptive statistics

Answer: s = 13.68

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 = 219/7 = 31.29

Squared deviations

Σ(x − x̄)² = 12.71² + (−10.29)² + 11.71² + 18.71² + (−9.286)² + (−10.29)² + (−13.29)² = 1123

Standard deviation

s = √(1123/6) = 13.68

Answer

s = 13.68
10.Descriptive statistics

Answer: P = 0.7881

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
11.Probability

Answer: N = 45

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.

Combinations

C(10, 2) = (10·9)/(2·1) = 45

Answer

N = 45
12.Probability

Answer: P = 0.06047

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.

Binomial formula

P(X = 1) = 9·0.41·0.68 = 9·0.4·0.0168 = 0.06047

Answer

P = 0.06047
13.Normal distribution

Answer: z = 0.64

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 + 0.4778)/2 = 0.7389

Read the z-table backwards

Find 0.7389 inside the table and read off z.

z = 0.64

Answer

z = 0.64
14.Normal distribution

Answer: x = 39.69 min

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) = 0.9 → z = 1.282

Back to x

x = 32 + 1.282·6 = 39.69 min

Answer

x = 39.69 min
15.Confidence intervals

Answer: lower limit = 92.2

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 = 2.756

Margin of error

E = 2.756·14.5/√30 = 7.296

Interval

99.5 ± 7.296 → [92.2, 106.8]

Answer

lower limit = 92.2
16.Confidence intervals

Answer: n = 281

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 = (2.576·26/4)² = 280.4

Round up

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

n = 281

Answer

n = 281
17.Hypothesis tests & confidence intervals

Answer: t = −1.901

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 = (12·6.5² + 7·6²)/19 = 39.95
sp = 6.32

Standard error

SE = 6.32·√(1/13 + 1/8) = 2.84

Test statistic

t = (33 − 38.4)/2.84 = −1.901

Answer

t = −1.901

With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.

18.Hypothesis tests & confidence intervals

Answer: 1 − β = 80.89 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 12/√36 = 2
x̄c = 50 − 2.326·2 = 45.35

z under the true mean

z = (45.35 − 43.6)/2 = 0.874

Power

power = P(Z < 0.874) = 80.89%
1 − β = 80.89%

Answer

1 − β = 80.89%
19.Linear regression & correlation

Answer: r² = 0.9882

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 = 182/6 = 30.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 280
Syy = Σ(y − ȳ)² = 1133

Coefficient of determination

r² = 280²/(70·1133) = 0.9882

Answer

r² = 0.9882

About 98.82% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 54.15

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 = 15/5 = 3
ȳ = 237/5 = 47.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 45

Regression line

b = 45/10 = 4.5
a = 47.4 − 4.5·3 = 33.9
ŷ = 33.9 + 4.5·x

Prediction

ŷ = 33.9 + 4.5·4.5 = 54.15

Answer

ŷ = 54.15

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