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…8PV = ?€5 000

You need €5 000 in 8 years. An account pays 4% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €5 0004% / 12n = 8
PV = €
2.Time value of moneyHard
months0123…60€200€200€200FV = ?

You save €200 at the end of every month for 5 years. The account pays 3.6% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €2003.6% / 12N = 60
FV = €
3.NPV & IRRMedium
years01234−20 0007 0006 0005 0008 000

An investment of €20 000 gives these cash flows at the end of years 1 to 4: €7 000, €6 000, €5 000, €8 000. The cost of capital is 10%. Find the net present value NPV.

r = 10%CF0 = −20 000CF1 = 7 000CF2 = 6 000CF3 = 5 000CF4 = 8 000
NPV = €
4.NPV & IRRHard
years0123…8−40 00010 50010 50010 50010 500

An investment of €40 000 returns €10 500 at the end of each year for 8 years. Find its internal rate of return IRR in percent.

C0 = €40 000CF = €10 500n = 8
IRR = %
5.Bond pricing & yieldMedium
years0123…11€1 166C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 11 years and pays annual coupons. It sells for €1 166 when the yield to maturity is 5%. Find the coupon rate c in percent.

F = €1 000n = 11P = €1 166YTM = 5%
c = %
6.Bond pricing & yieldHard
years0123456€1 258€60€60€60 + €1 060

A bond with 13 years to maturity (face value €1 000, annual coupon 6%) sells for €1 258. The issuer can call it in 6 years at a call price of €1 060. Find the yield to call YTC in percent.

F = €1 000coupon = 6%P = €1 258call price = €1 060call in 6 years
YTC = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€12 000 000re​ = 14%preferred shares€1 000 000rp​ = 7%debt€4 000 000rd​ = 6%T = 20%

A company is financed with ordinary shares worth €12 000 000, preferred shares worth €1 000 000 and debt worth €4 000 000. The cost of equity is 14%, the cost of preferred stock 7% and the pre-tax cost of debt 6%. The tax rate is 20%. Find the WACC.

E = €12 000 000Ps = €1 000 000D = €4 000 000re = 14%rp = 7%rd = 6%T = 20%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €51.00β = 0.8rf​ = 2.5%, rm​ = 6.5%debt10 000 × €1 000rd​ = 6%T = 20%

A company has 2 000 000 shares at €51.00 each and 10 000 bonds trading at €1 000 each. The pre-tax cost of debt is 6% and the tax rate is 20%. The share’s beta is 0.8, the risk-free rate 2.5% and the market return 6.5%. Find the WACC.

β = 0.8rf = 2.5%rm = 6.5%rd = 6%T = 20%
WACC = %
9.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 45, 58, 40, 27, 25.

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

Exam scores are normally distributed with mean 170 and standard deviation 10. Find the z-score z of a score of 174.

μ = 170σ = 10x = 174
z =
11.ProbabilityMedium
xp(x)10.3520.2530.340.1μ = 2.15

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

x: 1, 2, 3, 4p(x): 0.35, 0.25, 0.3, 0.1
Var(X) =
12.ProbabilityHard
€p(x)−10.7320.25190.02

A game costs €1 to play. You win €3 with probability 0.25 and €20 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€3: p = 0.25€20: p = 0.02
E(G) = €
13.Normal distributionMedium
850 kg869 kgT = X₁ + … + X₁₀?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 10 of them is below 869 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 10< 869 kg
P =
14.Normal distributionHard
μ = 65x = ?10%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 10% of them lie (the 10th percentile).

μ = 65σ = 10P = 10%
x = points
15.Confidence intervalsMedium
n = 30, s = 8LU = ?x̄ = 95E99%

A sample of 30 items has mean 95 and standard deviation s = 8. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 30x̄ = 95s = 899%t(29) = 2.756
upper limit =
16.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0290%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.02?

p = 0.5E = 0.0290%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 78.64 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 78.64s = 15n = 49
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 88 of 250 patients on treatment A recover, and 50 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 88n₁ = 250x₂ = 50n₂ = 100
z =
19.Linear regression & correlationMedium
xy337548766979118813102xy51050100

The table gives 6 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 = 6
r² =
20.Linear regression & correlationHard
xy121221320415514615xy2461520ŷ = 17

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 17?

n = 6ŷ = 17
x =

3Solutions

1.Time value of money

Answer: PV = 3633 €

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 = 0.04/12 = 0.003333
N = 12·8 = 96

Present value

PV = FV/(1 + i)N = 5 000/1.376 = €3 632.68

Answer

PV = €3 632.68
2.Time value of money

Answer: FV = 13126 €

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 = 0.036000000000000004/12 = 0.003
N = 12·5 = 60

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 200·(1.197 − 1)/0.003 = €13 126.32

Answer

FV = €13 126.32

You paid in €12 000 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = 543 €

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 = −20 000
CF1/(1 + 0.1)1 = 7 000/1.1 = 6 363.64
CF2/(1 + 0.1)2 = 6 000/1.21 = 4 958.68
CF3/(1 + 0.1)3 = 5 000/1.331 = 3 756.57
CF4/(1 + 0.1)4 = 8 000/1.464 = 5 464.11

Add them up

NPV = €543.00

Answer

NPV = €543.00

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

4.NPV & IRR

Answer: IRR = 20.24 %

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.

Required annuity factor

(1 − (1 + i)−8)/i = 40 000/10 500 = 3.81

Solve for the rate

The annuity factor falls as the rate rises; bracket the rate and narrow down.

i = 20%: 3.837
i = 21%: 3.726
IRR = 20.24%

Answer

IRR = 20.24%
5.Bond pricing & yield

Answer: c = 6.998 %

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 + 0.05)−11 = €584.68

Coupon

C = (1166 − 584.68)/8.306 = €69.98

Coupon rate

c = 69.98/1000 = 6.998%

Answer

c = 6.998%
6.Bond pricing & yield

Answer: YTC = 2.293 %

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

1258 = 60·(1 − (1 + y)−6)/y + 1060/(1 + y)6

Solve numerically

y = 2%: €1 277.34
y = 3%: €1 212.76
YTC = 2.293%

Answer

YTC = 2.293%

The yield to maturity is 3.496%. 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: WACC = 11.42 %

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 = 12 000 000 + 1 000 000 + 4 000 000 = €17 000 000
E/V = 0.7059, Ps/V = 0.05882, D/V = 0.2353

WACC

WACC = 0.7059·14% + 0.05882·7% + 0.2353·6%·(1 − 0.2) = 11.42%

Answer

WACC = 11.42%
8.CAPM & WACC

Answer: WACC = 5.62 %

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·€51.00 = €102 000 000
D = 10 000·€1 000 = €10 000 000

Cost of equity (CAPM)

re = 2.5% + 0.8·4% = 5.7%

Weights

V = €112 000 000, E/V = 0.9107, D/V = 0.08929

WACC

WACC = 0.9107·5.7% + 0.08929·6%·(1 − 0.2) = 5.62%

Answer

WACC = 5.62%
9.Descriptive statistics

Answer: s = 13.58

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 = 195/5 = 39

Squared deviations

Σ(x − x̄)² = 6² + 19² + 1² + (−12)² + (−14)² = 738

Standard deviation

s = √(738/4) = 13.58

Answer

s = 13.58
10.Descriptive statistics

Answer: z = 0.4

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 = (174 − 170)/10 = 0.4

Answer

z = 0.4
11.Probability

Answer: Var(X) = 1.028

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 = 1·0.35 + 2·0.25 + 3·0.3 + 4·0.1 = 2.15

Mean of the squares

E(X²) = 1·0.35 + 4·0.25 + 9·0.3 + 16·0.1 = 5.65

Variance

Var(X) = 5.65 − 2.15² = 1.028

Answer

Var(X) = 1.028
12.Probability

Answer: E(G) = 0.15 €

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: 2 (p = 0.25), 19 (p = 0.02), −1 (p = 0.73)

Expected value

E(G) = 2·0.25 + 19·0.02 − 1·0.73 = 0.15 €

Answer

E(G) = 0.15 €

A positive value means you win money on average.

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 total

μT = 850 kg, σT = 12·√10 = 37.95 kg

Standardise

z = (869 − 850)/37.95 = 0.5

Probability

P = Φ(0.5) = 0.6915

Answer

P = 0.6915
14.Normal distribution

Answer: x = 52.18 points

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) = 0.1 → z = −1.282

Back to x

x = 65 + −1.282·10 = 52.18 points

Answer

x = 52.18 points
15.Confidence intervals

Answer: upper limit = 99.03

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 = 2.756·8/√30 = 4.025

Interval

95 ± 4.025 → [90.97, 99.03]

Answer

upper limit = 99.03
16.Confidence intervals

Answer: n = 1692

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.645²·0.5·0.5/0.02² = 1691

Round up

n = 1692

Answer

n = 1692
17.Hypothesis tests & confidence intervals

Answer: t = 1.699

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 = 15/√49 = 15/7 = 2.143

Test statistic

t = (78.64 − 75)/2.143 = 1.699

Answer

t = 1.699
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: z = −2.56

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.

Sample proportions

p̂₁ = 88/250 = 0.352
p̂₂ = 50/100 = 0.5
p̂ = 138/350 = 0.3943

Standard error under H₀

SE = √(0.3943·0.6057·(1/250 + 1/100)) = 0.05782

Test statistic

z = (0.352 − 0.5)/0.05782 = −2.56

Answer

z = −2.56

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

19.Linear regression & correlation

Answer: r² = 0.9929

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 = 48/6 = 8
ȳ = 420/6 = 70

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 458
Syy = Σ(y − ȳ)² = 3018

Coefficient of determination

r² = 458²/(70·3018) = 0.9929

Answer

r² = 0.9929

About 99.29% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: x = 3.917

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 = 21/6 = 3.5
ȳ = 106/6 = 17.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −28

Regression line

b = −28/17.5 = −1.6
a = 17.67 − (−1.6)·3.5 = 23.27
ŷ = 23.27 − 1.6·x

Solve for x

x = (17 − 23.27)/(−1.6) = 3.917

Answer

x = 3.917

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