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…9PV = ?€20 000+4%+4%…r = 9%

A property is expected to earn net rent of €20 000 next year. The rent is expected to grow by 4% per year forever. The required return is 9% per year. Find the value PV of the property today.

C1 = €20 000g = 4%r = 9%
PV = €
2.Time value of moneyHard
months0123…132€50€50€50FV = ?

You save €50 at the end of every month for 11 years. The account pays 6% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €506% / 12N = 132
FV = €
3.NPV & IRRMedium
years01234−45 00014 00017 00012 00011 000

An investment of €45 000 gives these cash flows at the end of years 1 to 4: €14 000, €17 000, €12 000, €11 000. The cost of capital is 9%. Find the net present value NPV.

r = 9%CF0 = −45 000CF1 = 14 000CF2 = 17 000CF3 = 12 000CF4 = 11 000
NPV = €
4.NPV & IRRHard
years01234−10 0005 0004 0003 5004 500

A project costs €10 000 and returns €5 000, €4 000, €3 500, €4 500 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 = −10 000CF1 = 5 000CF2 = 4 000CF3 = 3 500CF4 = 4 500
MIRR = %
5.Bond pricing & yieldMedium
years0123€899C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 3 years and pays annual coupons. It sells for €899 when the yield to maturity is 9%. Find the coupon rate c in percent.

F = €1 000n = 3P = €899YTM = 9%
c = %
6.Bond pricing & yieldHard
years0123…9€1 120€40€40 + €1 000

A bond with a face value of €1 000 and an annual coupon of 4% matures in 9 years and sells for €1 120. Find its yield to maturity YTM in percent.

F = €1 000coupon = 4%n = 9P = €1 120
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€4 000 000re​ = 8%debt€1 000 000rd​ = 5%T = 25%

A company has equity worth €4 000 000 and debt worth €1 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 5% and the tax rate is 25%. Find the WACC.

E = €4 000 000D = €1 000 000re = 8%rd = 5%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €56.50D0​ = €2.70g = 4%debt€56 500 000rd​ = 3%T = 25.8%

A company has 2 000 000 shares trading at €56.50; the book value of its equity is €67 800 000. It just paid a dividend of €2.70 per share, and dividends grow by 4% per year. Its debt has a market value of €56 500 000 and a pre-tax cost of 3%. The tax rate is 25.8%. Find the WACC, using the dividend growth model for the cost of equity.

N = 2 000 000P0 = €56.50book equity = €67 800 000D0 = €2.70g = 4%D = €56 500 000rd = 3%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
202530354045

Find the sample standard deviation s of the data: 28, 18, 26, 24, 45.

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

Exam scores are normally distributed with mean 100 and standard deviation 5. Find the z-score z of a score of 96.

μ = 100σ = 5x = 96
z =
11.ProbabilityMedium
n = 6k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 6? The order does not matter.

n = 6k = 3
N =
12.ProbabilityHard
€p(x)−50.83200.15450.02

A game costs €5 to play. You win €25 with probability 0.15 and €50 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€25: p = 0.15€50: p = 0.02
E(G) = €
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9788

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9788. Find z to two decimals.

A = 0.9788
z =
14.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
15.Confidence intervalsMedium
n = 7, s = 8.5L = ?Ux̄ = 72.5E95%

A sample of 7 items has mean 72.5 and standard deviation s = 8.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 7x̄ = 72.5s = 8.595%t(6) = 2.447
lower limit =
16.Confidence intervalsHard
n = 16Lσ = ?s = 5.895%

The diameters of 16 machined shafts have a sample standard deviation of 5.8 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.

n = 16s = 5.8 µmχ² = 6.262, 27.488
σ = µm
17.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 197 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 197s = 8n = 49
t =
18.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 16 measurements has mean 103.4. The population standard deviation is 5. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 103.4σ = 5n = 1699%
L =
19.Linear regression & correlationMedium
xy3385457489481160xy510405060

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy2104156278331040xy51010203040x = 5

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.

n = 5x = 5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 400000 €

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.

Growing perpetuity

PV = C1/(r − g) = 20 000/(0.09 − 0.04) = 20 000/0.05 = €400 000.00

Answer

PV = €400 000.00
2.Time value of money

Answer: FV = 9316 €

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 = 0.06/12 = 0.005
N = 12·11 = 132

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 50·(1.932 − 1)/0.005 = €9 316.13

Answer

FV = €9 316.13

You paid in €6 600 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = −788.5 €

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 each cash flow

CF0 = −45 000
CF1/(1 + 0.09)1 = 14 000/1.09 = 12 844.04
CF2/(1 + 0.09)2 = 17 000/1.188 = 14 308.56
CF3/(1 + 0.09)3 = 12 000/1.295 = 9 266.20
CF4/(1 + 0.09)4 = 11 000/1.412 = 7 792.68

Add them up

NPV = €−788.52

Answer

NPV = €−788.52

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

4.NPV & IRR

Answer: MIRR = 17.78 %

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

5 000·1.083 = 6 298.56
4 000·1.082 = 4 665.60
3 500·1.081 = 3 780.00
4 500·1.080 = 4 500.00
TV = 19 244.16

MIRR

MIRR = (19 244.16/10 000)1/4 − 1 = 17.78%

Answer

MIRR = 17.78%
5.Bond pricing & yield

Answer: c = 5.01 %

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 + 0.09)−3 = €772.18

Coupon

C = (899 − 772.18)/2.531 = €50.10

Coupon rate

c = 50.1/1000 = 5.01%

Answer

c = 5.01%
6.Bond pricing & yield

Answer: YTM = 2.495 %

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

€1 120 = €40·(1 − (1 + y)⁻⁹)/y + €1 000/(1 + y)⁹

Solve numerically

P(2%) = €1 163.24, P(3%) = €1 077.86
YTM = 2.495%

Answer

YTM = 2.495%
7.CAPM & WACC

Answer: WACC = 7.15 %

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 = €5 000 000, E/V = 0.8, D/V = 0.2

WACC

WACC = 0.8·8% + 0.2·5%·(1 − 0.25) = 7.15%

Answer

WACC = 7.15%
8.CAPM & WACC

Answer: WACC = 6.722 %

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).

Cost of equity

D1 = 2.7·1.04 = €2.81
re = 2.808/56.5 + 0.04 = 8.97%

Market values

E = 2 000 000·€56.50 = €113 000 000
V = €169 500 000, E/V = 0.6667, D/V = 0.3333

WACC

WACC = 0.6667·8.97% + 0.3333·3%·(1 − 0.258) = 6.722%

Answer

WACC = 6.722%
9.Descriptive statistics

Answer: s = 10.11

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 = 141/5 = 28.2

Squared deviations

Σ(x − x̄)² = (−0.2)² + (−10.2)² + (−2.2)² + (−4.2)² + 16.8² = 408.8

Standard deviation

s = √(408.8/4) = 10.11

Answer

s = 10.11
10.Descriptive statistics

Answer: z = −0.8

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 = (96 − 100)/5 = −0.8

Answer

z = −0.8
11.Probability

Answer: N = 20

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(6, 3) = (6·5·4)/(3·2·1) = 20

Answer

N = 20
12.Probability

Answer: E(G) = −0.25 €

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.

Net outcomes

G: 20 (p = 0.15), 45 (p = 0.02), −5 (p = 0.83)

Expected value

E(G) = 20·0.15 + 45·0.02 − 5·0.83 = −0.25 €

Answer

E(G) = −0.25 €

A negative value means you lose money on average.

13.Normal distribution

Answer: z = −2.03

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 − 0.9788 = 0.0212

Read the z-table backwards

Find 0.0212 inside the table and read off z.

z = −2.03

Answer

z = −2.03
14.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
15.Confidence intervals

Answer: lower limit = 64.64

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 = 6, t = 2.447

Margin of error

E = 2.447·8.5/√7 = 7.861

Interval

72.5 ± 7.861 → [64.64, 80.36]

Answer

lower limit = 64.64
16.Confidence intervals

Answer: σ = 8.977 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 15·5.8² = 504.6

Limits for σ²

504.6/27.488 = 18.36
504.6/6.262 = 80.58

Limit for σ

σ = √80.58 = 8.977 µm

Answer

σ = 8.977 µm

95% interval for σ: [4.285, 8.977] µm.

17.Hypothesis tests & confidence intervals

Answer: t = −2.625

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 = 8/√49 = 8/7 = 1.143

Test statistic

t = (197 − 200)/1.143 = −2.625

Answer

t = −2.625
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: L = 100.2

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/√16 = 3.22

Interval limit

L = 103.4 − 3.22 = 100.2

Answer

L = 100.2
19.Linear regression & correlation

Answer: r² = 0.8738

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 = 239/5 = 47.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 94
Syy = Σ(y − ȳ)² = 252.8

Coefficient of determination

r² = 94²/(40·252.8) = 0.8738

Answer

r² = 0.8738

About 87.38% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 21.1

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 = 125/5 = 25

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 156

Regression line

b = 156/40 = 3.9
a = 25 − 3.9·6 = 1.6
ŷ = 1.6 + 3.9·x

Prediction

ŷ = 1.6 + 3.9·5 = 21.1

Answer

ŷ = 21.1

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