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…10PV = ?€5 000

You need €5 000 in 10 years. An account pays 4% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €5 0004% / 4n = 10
PV = €
2.Time value of moneyHard
months0123…60€12 000A?AAA

A loan of €12 000 is repaid with equal monthly payments over 5 years. The interest rate is 4.8% per year, compounded monthly. Find the monthly payment A.

P = €12 0004.8% / 12N = 60
A = €
3.NPV & IRRMedium
years01234−60 00017 00013 00013 00029 000

An investment of €60 000 gives these cash flows at the end of years 1 to 4: €17 000, €13 000, €13 000, €29 000. The cost of capital is 5%. Find the net present value NPV.

r = 5%CF0 = −60 000CF1 = 17 000CF2 = 13 000CF3 = 13 000CF4 = 29 000
NPV = €
4.NPV & IRRHard
years0123−20 00010 50010 5008 000

A project costs €20 000 and returns €10 500, €10 500, €8 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 10 500CF2 = 10 500CF3 = 8 000
IRR = %
5.Bond pricing & yieldMedium
half-years0123…20€20€20€20 + €1 000P = ?

A bond with a face value of €1 000 pays a 4% annual coupon in two half-yearly payments and matures in 10 years. The yield to maturity is 5% per year. Find the price P of the bond.

F = €1 000coupon = 4%YTM = 5%n = 10
P = €
6.Bond pricing & yieldHard
years012345€1 163€70€70€70 + €1 070

A bond with 17 years to maturity (face value €1 000, annual coupon 7%) sells for €1 163. The issuer can call it in 5 years at a call price of €1 070. Find the yield to call YTC in percent.

F = €1 000coupon = 7%P = €1 163call price = €1 070call in 5 years
YTC = %
7.CAPM & WACCMedium
WACC = 10.97%equity€8 000 000re​ = ?debt€3 000 000rd​ = 4%T = 28%

A company has equity worth €8 000 000 and debt worth €3 000 000. Its WACC is 10.97%, its pre-tax cost of debt is 4% and the tax rate is 28%. Find the cost of equity re.

E = €8 000 000D = €3 000 000WACC = 10.97%rd = 4%T = 28%
re = %
8.CAPM & WACCHard
rf​ = 2.5%, rm​ − rf​ = 4.5%D/E = 0.4equityβL​ = 1.2debtD/E = 0.6equityre​ = ?debtT = 25.8%

A company's shares have a beta of 1.2 at its current debt-to-equity ratio of 0.4. It plans to raise its debt-to-equity ratio to 0.6. The tax rate is 25.8%, the risk-free rate 2.5% and the market risk premium 4.5%. Find the cost of equity re after the change.

βL = 1.2D/E1 = 0.4D/E2 = 0.6T = 25.8%rf = 2.5%rm − rf = 4.5%
re = %
9.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 43, 56, 22, 33, 22, 41, 32.

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

Exam scores are normally distributed with mean 500 and standard deviation 50. Find the z-score z of a score of 450.

μ = 500σ = 50x = 450
z =
11.ProbabilityMedium
xp(x)100.15200.3300.35400.2μ = 26

For the distribution shown, find the standard deviation σ of X.

x: 10, 20, 30, 40p(x): 0.15, 0.3, 0.35, 0.2
σ =
12.ProbabilityHard
xp(x)00.310.3520.35μ

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

x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 3X − 3
σY =
13.Normal distributionMedium
105 s101.8 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 101.8 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 101.8 s
P =
14.Normal distributionHard
557085100115130145pointsbottom 5%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 5% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points5%
x = points
15.Confidence intervalsMedium
n = 16, sd​ = 10L = ?Ud̄ = 8E99%

The blood pressure of 16 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 8 and standard deviation 1. Find the lower limit of the 99% confidence interval for the mean difference.

n = 16d̄ = 8sd = 199%t(15) = 2.947
lower limit = mmHg
16.Confidence intervalsHard
σ = 36, n = ?x̄ − Ex̄ + Ex̄E = 290%

The population standard deviation is 36. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 2?

σ = 36E = 290%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 202.9 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 202.9s = 12n = 49
t =
18.Hypothesis tests & confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 195%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 15. Find the smallest sample size n.

σ = 15E = 195%
n =
19.Linear regression & correlationMedium
xy25476108101015xy51051015

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy251356456560669774xy246506070x = 2.5

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.

n = 6x = 2.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 3358 €

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 = 0.04/4 = 0.01
N = 4·10 = 40

Present value

PV = FV/(1 + i)N = 5 000/1.489 = €3 358.27

Answer

PV = €3 358.27
2.Time value of money

Answer: A = 225.4 €

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 = 0.048/12 = 0.004
N = 12·5 = 60

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 = 0.787
A = 12 000·0.004/(1 − 0.787) = €225.36

Answer

A = €225.36
3.NPV & IRR

Answer: NPV = 3070 €

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 = −60 000
CF1/(1 + 0.05)1 = 17 000/1.05 = 16 190.48
CF2/(1 + 0.05)2 = 13 000/1.103 = 11 791.38
CF3/(1 + 0.05)3 = 13 000/1.158 = 11 229.89
CF4/(1 + 0.05)4 = 29 000/1.216 = 23 858.37

Add them up

NPV = €3 070.12

Answer

NPV = €3 070.12

NPV is positive, so the project earns more than the required return.

4.NPV & IRR

Answer: IRR = 22.23 %

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(20%) = 671.30
NPV(25%) = −784.00

Narrow down

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

IRR ≈ 20 + 5·671.30/(671.30 − (−784.00)) = 22.31%
exact: NPV(22.23%) = 0 → IRR = 22.23%

Answer

IRR = 22.23%
5.Bond pricing & yield

Answer: P = 922.1 €

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 = €20.00, i = 0.025, N = 20

Present value of the coupons

€20.00·(1 − 0.6103)/0.025 = €311.78

Present value of the face value

€1 000·0.6103 = €610.27

Price

P = €922.05

Answer

P = €922.05

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

6.Bond pricing & yield

Answer: YTC = 4.559 %

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

1163 = 70·(1 − (1 + y)−5)/y + 1070/(1 + y)5

Solve numerically

y = 4%: €1 191.09
y = 5%: €1 141.44
YTC = 4.559%

Answer

YTC = 4.559%

The yield to maturity is 5.5%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.

7.CAPM & WACC

Answer: re = 14 %

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 = €11 000 000, E/V = 0.7273, D/V = 0.2727

Debt part of the WACC

0.2727·4%·(1 − 0.28) = 0.7855%

Cost of equity

re = (10.97% − 0.7855%)/0.7273 = 14%

Answer

re = 14%
8.CAPM & WACC

Answer: re = 8.518 %

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.

Unlever the beta

βU = 1.2/(1 + 0.742·0.4) = 0.9254

Relever at the new D/E

βL = 0.9254·(1 + 0.742·0.6) = 1.337

Cost of equity (CAPM)

re = 2.5% + 1.337·4.5% = 8.518%

Answer

re = 8.518%
9.Descriptive statistics

Answer: s = 12.18

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 = 249/7 = 35.57

Squared deviations

Σ(x − x̄)² = 7.429² + 20.43² + (−13.57)² + (−2.571)² + (−13.57)² + 5.429² + (−3.571)² = 889.7

Standard deviation

s = √(889.7/6) = 12.18

Answer

s = 12.18
10.Descriptive statistics

Answer: z = −1

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 = (450 − 500)/50 = −1

Answer

z = −1
11.Probability

Answer: σ = 9.695

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 = 10·0.15 + 20·0.3 + 30·0.35 + 40·0.2 = 26

Mean of the squares

E(X²) = 100·0.15 + 400·0.3 + 900·0.35 + 1600·0.2 = 770

Variance

Var(X) = 770 − 26² = 94
σ = √94 = 9.695

Answer

σ = 9.695
12.Probability

Answer: σY = 2.414

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 = |3|·0.8047 = 2.414

Answer

σY = 2.414

The mean does move: E(Y) = 3·1.05 − 3 = 0.15.

13.Normal distribution

Answer: P = 0.6915

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 sum

μ = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (101.8 − 105)/6.403 = −0.5

Upper tail

P = 1 − Φ(−0.5) = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
14.Normal distribution

Answer: x = 75.4 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) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 100 − 1.64·15 = 75.4 points

Answer

x = 75.4 points
15.Confidence intervals

Answer: lower limit = 7.263 mmHg

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Critical value

df = 15, t = 2.947

Margin of error

E = 2.947·1/√16 = 0.7368

Interval

8 ± 0.7368 → [7.263, 8.737]

Answer

lower limit = 7.263 mmHg
16.Confidence intervals

Answer: n = 877

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 = (1.645·36/2)² = 876.8

Round up

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

n = 877

Answer

n = 877
17.Hypothesis tests & confidence intervals

Answer: t = 1.692

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 = 12/√49 = 12/7 = 1.714

Test statistic

t = (202.9 − 200)/1.714 = 1.692

Answer

t = 1.692
t0.025; df = 48 = 2.011

|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 865

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 = (1.96·15/1)² = 864.4

Round up

n = 865

Answer

n = 865
19.Linear regression & correlation

Answer: r = 0.9617

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 = 47/5 = 9.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 46
Syy = Σ(y − ȳ)² = 57.2

Correlation coefficient

r = 46/√(40·57.2) = 0.9617

Answer

r = 0.9617
20.Linear regression & correlation

Answer: ŷ = 51.97

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 = 27/6 = 4.5
ȳ = 366/6 = 61

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 79

Regression line

b = 79/17.5 = 4.514
a = 61 − 4.514·4.5 = 40.69
ŷ = 40.69 + 4.514·x

Prediction

ŷ = 40.69 + 4.514·2.5 = 51.97

Answer

ŷ = 51.97

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