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…9€2 000FV = ?

You invest €2 000 at a nominal rate of 4.8% per year, compounded monthly. Find the value FV of the investment after 9 years.

PV = €2 0004.8% / 12n = 9
FV = €
2.Time value of moneyHard
months0123…108A?AA€75 000

You want €75 000 in 9 years. You save the same amount at the end of every month in an account paying 7.2% per year, compounded monthly. Find the monthly deposit A.

FV = €75 0007.2% / 12N = 108
A = €
3.NPV & IRRMedium
years0123456−60 00016 50016 50016 50016 50016 50016 500

A project costs €60 000 today and returns €16 500 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 = −60 000CF1 = 16 500CF2 = 16 500CF3 = 16 500CF4 = 16 500CF5 = 16 500CF6 = 16 500
DPB = years
4.NPV & IRRHard
years01234−40 00016 00021 00016 50014 000

A project costs €40 000 and returns €16 000, €21 000, €16 500, €14 000 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 8%. Find the modified internal rate of return MIRR in percent.

r = 8%CF0 = −40 000CF1 = 16 000CF2 = 21 000CF3 = 16 500CF4 = 14 000
MIRR = %
5.Bond pricing & yieldMedium
half-years0123…14€25€25€25 + €1 000P = ?

A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 8% per year. Find the price P of the bond.

F = €1 000coupon = 5%YTM = 8%n = 7
P = €
6.Bond pricing & yieldHard
years0123456P€40€40€40 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 4% and matures in 6 years. The yield to maturity is 3%. Find the Macaulay duration D in years.

F = €1 000coupon = 4%n = 6YTM = 3%
D = years
7.CAPM & WACCMedium
WACC = ?equity€3 000 000re​ = 14%debt€5 000 000rd​ = 3%T = 25%

A company has equity worth €3 000 000 and debt worth €5 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 3% and the tax rate is 25%. Find the WACC.

E = €3 000 000D = €5 000 000re = 14%rd = 3%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €32.00β = 1.3rf​ = 1.5%, rm​ = 5.5%debt10 000 × €1 000rd​ = 4%T = 21%

A company has 2 000 000 shares at €32.00 each and 10 000 bonds trading at €1 000 each. The pre-tax cost of debt is 4% and the tax rate is 21%. The share’s beta is 1.3, the risk-free rate 1.5% and the market return 5.5%. Find the WACC.

β = 1.3rf = 1.5%rm = 5.5%rd = 4%T = 21%
WACC = %
9.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 56, 22, 54, 36, 56.

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

Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 80.

μ = 70σ = 5x = 80
z =
11.ProbabilityMedium
xp(x)10.1520.330.340.1550.1μ = 2.75

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

x: 1, 2, 3, 4, 5p(x): 0.15, 0.3, 0.3, 0.15, 0.1
Var(X) =
12.ProbabilityHard
n = 7, p = 0.6: P(X ≤ 1) = ?k01234567

7 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 7p = 0.6
P =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.6579

Drag z so that the shaded area between −z and z under the standard normal curve is 0.6579. Find z to two decimals.

A = 0.6579
z =
14.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
15.Confidence intervalsMedium
n = 30, s = 11.5L = ?Ux̄ = 66.5E99%

A sample of 30 items has mean 66.5 and standard deviation s = 11.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 30x̄ = 66.5s = 11.599%t(29) = 2.756
lower limit =
16.Confidence intervalsHard
σ = 36, n = ?x̄ − Ex̄ + Ex̄E = 299%

The population standard deviation is 36. 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 2?

σ = 36E = 299%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 122.9 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 122.9s = 5n = 25
t =
18.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 1.595%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 1.595%
n =
19.Linear regression & correlationMedium
xy225326427537633xy246253035

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 20Σy = 148Σx² = 90Σy² = 4488Σxy = 619
r =
20.Linear regression & correlationHard
xy358461562667775875xy4686070x = 3.5

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.5.

n = 6x = 3.5
ŷ =

3Solutions

1.Time value of money

Answer: FV = 3078 €

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 = 0.048/12 = 0.004
N = m·n = 12·9 = 108

Future value

FV = 2 000·(1 + 0.004)108 = 2 000·1.539 = €3 078.02

Answer

FV = €3 078.02
2.Time value of money

Answer: A = 495.6 €

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 = 0.07200000000000001/12 = 0.006
N = 12·9 = 108

Deposit

A = FV·i/((1 + i)N − 1) = 75 000·0.006/(1.908 − 1) = €495.59

Answer

A = €495.59
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: 16 500/1.12 = 14 732.14 → Σ = 14 732.14
year 2: 16 500/1.254 = 13 153.70 → Σ = 27 885.84
year 3: 16 500/1.405 = 11 744.37 → Σ = 39 630.22
year 4: 16 500/1.574 = 10 486.05 → Σ = 50 116.26
year 5: 16 500/1.762 = 9 362.54 → Σ = 59 478.81
year 6: 16 500/1.974 = 8 359.41 → Σ = 67 838.22

Fraction of the last year

DPB = 5 + 521.19/8 359.41 = 5.062

Answer

DPB = 5.062 years

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

4.NPV & IRR

Answer: MIRR = 17.59 %

Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.

Terminal value of the inflows

16 000·1.083 = 20 155.39
21 000·1.082 = 24 494.40
16 500·1.081 = 17 820.00
14 000·1.080 = 14 000.00
TV = 76 469.79

MIRR

MIRR = (76 469.79/40 000)1/4 − 1 = 17.59%

Answer

MIRR = 17.59%
5.Bond pricing & yield

Answer: P = 841.6 €

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 = €25.00, i = 0.04, N = 14

Present value of the coupons

€25.00·(1 − 0.5775)/0.04 = €264.08

Present value of the face value

€1 000·0.5775 = €577.48

Price

P = €841.55

Answer

P = €841.55

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

6.Bond pricing & yield

Answer: D = 5.468 years

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.

Present value of each cash flow

t = 1: 40/1.03 = 38.83, t·PV = 38.83
t = 2: 40/1.061 = 37.70, t·PV = 75.41
t = 3: 40/1.093 = 36.61, t·PV = 109.82
t = 4: 40/1.126 = 35.54, t·PV = 142.16
t = 5: 40/1.159 = 34.50, t·PV = 172.52
t = 6: 1040/1.194 = 870.98, t·PV = 5 225.90

Duration

P = 1 054.17
D = 5 764.64/1 054.17 = 5.468

Answer

D = 5.468 years
7.CAPM & WACC

Answer: WACC = 6.656 %

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 = €8 000 000, E/V = 0.375, D/V = 0.625

WACC

WACC = 0.375·14% + 0.625·3%·(1 − 0.25) = 6.656%

Answer

WACC = 6.656%
8.CAPM & WACC

Answer: WACC = 6.222 %

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 = 2 000 000·€32.00 = €64 000 000
D = 10 000·€1 000 = €10 000 000

Cost of equity (CAPM)

re = 1.5% + 1.3·4% = 6.7%

Weights

V = €74 000 000, E/V = 0.8649, D/V = 0.1351

WACC

WACC = 0.8649·6.7% + 0.1351·4%·(1 − 0.21) = 6.222%

Answer

WACC = 6.222%
9.Descriptive statistics

Answer: s = 15.27

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 = 224/5 = 44.8

Squared deviations

Σ(x − x̄)² = 11.2² + (−22.8)² + 9.2² + (−8.8)² + 11.2² = 932.8

Standard deviation

s = √(932.8/4) = 15.27

Answer

s = 15.27
10.Descriptive statistics

Answer: z = 2

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 = (80 − 70)/5 = 2

Answer

z = 2
11.Probability

Answer: Var(X) = 1.387

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 = 1·0.15 + 2·0.3 + 3·0.3 + 4·0.15 + 5·0.1 = 2.75

Mean of the squares

E(X²) = 1·0.15 + 4·0.3 + 9·0.3 + 16·0.15 + 25·0.1 = 8.95

Variance

Var(X) = 8.95 − 2.75² = 1.387

Answer

Var(X) = 1.387
12.Probability

Answer: P = 0.01884

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) = 0.47 = 0.001638
P(X = 1) = 7·0.6·0.46 = 0.0172

Add them

P = 0.001638 + 0.0172 = 0.01884

Answer

P = 0.01884
13.Normal distribution

Answer: z = 0.95

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 + 0.6579)/2 = 0.8289

Read the z-table backwards

Find 0.8289 inside the table and read off z.

z = 0.95

Answer

z = 0.95
14.Normal distribution

Answer: P = 0.7389

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 difference

μ = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
15.Confidence intervals

Answer: lower limit = 60.71

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 = 2.756·11.5/√30 = 5.787

Interval

66.5 ± 5.787 → [60.71, 72.29]

Answer

lower limit = 60.71
16.Confidence intervals

Answer: n = 2150

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 = (2.576·36/2)² = 2150

Round up

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

n = 2150

Answer

n = 2150
17.Hypothesis tests & confidence intervals

Answer: t = 2.9

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).

Standard error

SE = 5/√25 = 5/5 = 1

Test statistic

t = (122.9 − 120)/1 = 2.9

Answer

t = 2.9
t0.025; df = 24 = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 171

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 = (1.96·10/1.5)² = 170.7

Round up

n = 171

Answer

n = 171
19.Linear regression & correlation

Answer: r = 0.8246

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 = 619 − 20·148/5 = 27
Sxx = 90 − 20²/5 = 10
Syy = 4488 − 148²/5 = 107.2

Correlation coefficient

r = 27/√(10·107.2) = 0.8246

Answer

r = 0.8246
20.Linear regression & correlation

Answer: ŷ = 58.79

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 = 33/6 = 5.5
ȳ = 398/6 = 66.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 66

Regression line

b = 66/17.5 = 3.771
a = 66.33 − 3.771·5.5 = 45.59
ŷ = 45.59 + 3.771·x

Prediction

ŷ = 45.59 + 3.771·3.5 = 58.79

Answer

ŷ = 58.79

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