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 = ?€40 000

You need €40 000 in 10 years. An account pays 4.8% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €40 0004.8% / 4n = 10
PV = €
2.Time value of moneyHard
months0123…72€200€200€200FV = ?

You save €200 at the end of every month for 6 years. The account pays 7.2% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €2007.2% / 12N = 72
FV = €
3.NPV & IRRMedium
years0123…9−500 00044 000+3%+3%…r = 11%, NPV = ?

A project costs €500 000 today. It generates €44 000 at the end of the first year, and the cash flow then grows by 3% per year forever. The cost of capital is 11%. Find the net present value NPV.

C0 = €500 000CF1 = €44 000g = 3%r = 11%
NPV = €
4.NPV & IRRHard
years012−20 00015 00014 000

A project costs €20 000 and returns €15 000, €14 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 15 000CF2 = 14 000
IRR = %
5.Bond pricing & yieldMedium
years0123…14P€50€50€50 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 14 years. Its yield to maturity changes from 2.5% to 1%. By how much ΔP does the bond price change?

F = €1 000coupon = 5%n = 14y1 = 2.5%y2 = 1%
ΔP = €
6.Bond pricing & yieldHard
years01234€1 128€60€60€60 + €1 060

A bond with 11 years to maturity (face value €1 000, annual coupon 6%) sells for €1 128. The issuer can call it in 4 years at a call price of €1 060. Find the yield to call YTC in percent.

F = €1 000coupon = 6%P = €1 128call price = €1 060call in 4 years
YTC = %
7.CAPM & WACCMedium
WACC = 7.55%equity€3 000 000re​ = ?debt€2 000 000rd​ = 3%T = 21%

A company has equity worth €3 000 000 and debt worth €2 000 000. Its WACC is 7.55%, its pre-tax cost of debt is 3% and the tax rate is 21%. Find the cost of equity re.

E = €3 000 000D = €2 000 000WACC = 7.55%rd = 3%T = 21%
re = %
8.CAPM & WACCHard
WACC = ?equity€12 000 000re​ = 11%bonds8 000 × €1 088coupon 5%, n = 11T = 28%

A company has equity with a market value of €12 000 000 and a cost of equity of 11%. Its debt consists of 8 000 bonds (face value €1 000, annual coupon 5%, 11 years to maturity) that trade at €1 088 each. The tax rate is 28%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.

E = €12 000 000re = 11%bonds: 8 000 × €1 088coupon = 5%n = 11T = 28%
WACC = %
9.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 38, 22, 59, 37, 48, 18.

n = 6
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is above 45?

μ = 60σ = 15x = 45
P =
11.ProbabilityMedium
xp(x)10.220.130.240.3550.15μ = 3.15

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

x: 1, 2, 3, 4, 5p(x): 0.2, 0.1, 0.2, 0.35, 0.15
σ =
12.ProbabilityHard
xp(x)00.310.3520.35μ

X has the distribution shown, and Y = −2X + 20. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = −2X + 20
σY =
13.Normal distributionMedium
200 g196 gT = X₁ + … + X₅?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 5 of them exceeds 196 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 5> 196 g
P =
14.Normal distributionHard
557085100115130145pointstop 20%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 20% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points20%
x = points
15.Confidence intervalsMedium
n = 12, s = 18.5LU = ?x̄ = 107E90%

A sample of 12 items has mean 107 and standard deviation s = 18.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 12x̄ = 107s = 18.590%t(11) = 1.796
upper limit =
16.Confidence intervalsHard
σ = 19, n = ?x̄ − Ex̄ + Ex̄E = 1.590%

The population standard deviation is 19. 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 1.5?

σ = 19E = 1.590%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.121.12p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 5. A random sample of 25 observations has a mean of 48.88. Test H₀: μ = 50 against H₁: μ ≠ 50. Find the p-value p.

μ₀ = 50σ = 5n = 25x̄ = 48.88H₁: μ ≠ μ₀
p =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 13, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 90%. The standard deviation is 13. Find the minimum sample size n.

α = 0.01power = 90%δ = 4σ = 13zα = 2.326zβ = 1.282
n =
19.Linear regression & correlationMedium
xy1263405467589731183xy510406080

The table gives 6 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 = 6Σx = 36Σy = 326Σx² = 286Σy² = 19974Σxy = 2352
r =
20.Linear regression & correlationHard
xy17571957216123822579xy2025607080ŷ = 71

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 71?

n = 5ŷ = 71
x =

3Solutions

1.Time value of money

Answer: PV = 24822 €

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 = 0.048/4 = 0.012
N = 4·10 = 40

Present value

PV = FV/(1 + i)N = 40 000/1.611 = €24 822.16

Answer

PV = €24 822.16
2.Time value of money

Answer: FV = 17945 €

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 = 12·6 = 72

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 200·(1.538 − 1)/0.006 = €17 944.94

Answer

FV = €17 944.94

You paid in €14 400 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = 50000 €

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.

Value of the cash flows

PV = 44 000/(0.11 − 0.03) = 550 000.00

NPV

NPV = 550 000.00 − 500 000 = €50 000.00

Answer

NPV = €50 000.00

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

4.NPV & IRR

Answer: IRR = 29.19 %

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(25%) = 960.00
NPV(30%) = −177.51

Narrow down

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

IRR ≈ 25 + 5·960.00/(960.00 − (−177.51)) = 29.22%
exact: NPV(29.19%) = 0 → IRR = 29.19%

Answer

IRR = 29.19%
5.Bond pricing & yield

Answer: ΔP = 227.9 €

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.

Price at the old yield

C = €50.00, i = 0.025, N = 14
€50.00·(1 − 0.7077)/0.025 + €1 000·0.7077 = €584.55 + €707.73 = €1 292.27

Price at the new yield

C = €50.00, i = 0.01, N = 14
€50.00·(1 − 0.87)/0.01 + €1 000·0.87 = €650.19 + €869.96 = €1 520.15

Change

ΔP = 1 520.15 − 1 292.27 = €227.88

Answer

ΔP = €227.88
6.Bond pricing & yield

Answer: YTC = 3.897 %

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

1128 = 60·(1 − (1 + y)−4)/y + 1060/(1 + y)4

Solve numerically

y = 3%: €1 164.82
y = 4%: €1 123.89
YTC = 3.897%

Answer

YTC = 3.897%

The yield to maturity is 4.499%. 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 = 11 %

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 = 0.6, D/V = 0.4

Debt part of the WACC

0.4·3%·(1 − 0.21) = 0.948%

Cost of equity

re = (7.55% − 0.948%)/0.6 = 11%

Answer

re = 11%
8.CAPM & WACC

Answer: WACC = 7.585 %

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 debt (YTM)

1088 = 50·(1 − (1 + y)−11)/y + 1000/(1 + y)11
y = 3.996%

Market values

D = 8 000·€1 088 = €8 704 000
V = €20 704 000, E/V = 0.5796, D/V = 0.4204

WACC

WACC = 0.5796·11% + 0.4204·3.996%·(1 − 0.28) = 7.585%

Answer

WACC = 7.585%
9.Descriptive statistics

Answer: s = 15.44

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 = 222/6 = 37

Squared deviations

Σ(x − x̄)² = 1² + (−15)² + 22² + 0² + 11² + (−19)² = 1192

Standard deviation

s = √(1192/5) = 15.44

Answer

s = 15.44
10.Descriptive statistics

Answer: P = 0.8413

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 = −1

Standard normal table

Φ(−1) = 0.1587
P = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
11.Probability

Answer: σ = 1.352

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 = 1·0.2 + 2·0.1 + 3·0.2 + 4·0.35 + 5·0.15 = 3.15

Mean of the squares

E(X²) = 1·0.2 + 4·0.1 + 9·0.2 + 16·0.35 + 25·0.15 = 11.75

Variance

Var(X) = 11.75 − 3.15² = 1.828
σ = √1.828 = 1.352

Answer

σ = 1.352
12.Probability

Answer: σY = 1.609

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 = |−2|·0.8047 = 1.609

Answer

σY = 1.609

The mean does move: E(Y) = −2·1.05 + 20 = 17.9.

13.Normal distribution

Answer: P = 0.9319

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 = 200 g, σT = 1.2·√5 = 2.683 g

Standardise

z = (196 − 200)/2.683 = −1.49

Probability

P = 1 − Φ(−1.49) = 0.9319

Answer

P = 0.9319
14.Normal distribution

Answer: x = 112.6 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) = 0.8

z from the table

z ≈ 0.84

Back to the original units

x = 100 + 0.84·15 = 112.6 points

Answer

x = 112.6 points
15.Confidence intervals

Answer: upper limit = 116.6

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 = 11, t = 1.796

Margin of error

E = 1.796·18.5/√12 = 9.592

Interval

107 ± 9.592 → [97.41, 116.6]

Answer

upper limit = 116.6
16.Confidence intervals

Answer: n = 435

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 = (1.645·19/1.5)² = 434.2

Round up

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

n = 435

Answer

n = 435
17.Hypothesis tests & confidence intervals

Answer: p = 0.2627

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.

Test statistic

SE = σ/√n = 5/√25 = 1
z = (48.88 − 50)/1 = −1.12

Tail area

Φ(1.12) = 0.8686
1 − Φ(1.12) = 0.1314

p-value

p = 2·0.1314 = 0.2627

Answer

p = 0.2627

p > 0.05, so H₀ is not rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: n = 138

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((2.326 + 1.282)·13/4)² = 137.5

Round up

n = 138

Answer

n = 138
19.Linear regression & correlation

Answer: r = 0.9953

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 = 2352 − 36·326/6 = 396
Sxx = 286 − 36²/6 = 70
Syy = 19974 − 326²/6 = 2261

Correlation coefficient

r = 396/√(70·2261) = 0.9953

Answer

r = 0.9953
20.Linear regression & correlation

Answer: x = 22.1

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 = 105/5 = 21
ȳ = 336/5 = 67.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 138

Regression line

b = 138/40 = 3.45
a = 67.2 − 3.45·21 = −5.25
ŷ = −5.25 + 3.45·x

Solve for x

x = (71 − (−5.25))/3.45 = 22.1

Answer

x = 22.1

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