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
years012345PV = ?€25 000

You need €25 000 in 5 years. An account pays 5% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €25 0005% / 4n = 5
PV = €
2.Time value of moneyHard
months0123…72€8 000A?AAA

A loan of €8 000 is repaid with equal monthly payments over 6 years. The interest rate is 4.8% per year, compounded monthly. Find the monthly payment A.

P = €8 0004.8% / 12N = 72
A = €
3.NPV & IRRMedium
years012345−20 0007 5008 0008 0006 0004 000

A project costs €20 000 today and returns €7 500, €8 000, €8 000, €6 000, €4 000 at the end of years 1 to 5. 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 = −20 000CF1 = 7 500CF2 = 8 000CF3 = 8 000CF4 = 6 000CF5 = 4 000
DPB = years
4.NPV & IRRHard
years0123−20 00014 5007 00012 500

A project costs €20 000 and returns €14 500, €7 000, €12 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −20 000CF1 = 14 500CF2 = 7 000CF3 = 12 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…8€40€40€40 + €1 000P = ?

A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 4 years. The yield to maturity is 3% per year. Find the price P of the bond.

F = €1 000coupon = 8%YTM = 3%n = 4
P = €
6.Bond pricing & yieldHard
years012345€1 108€60€60€60 + €1 060

A bond with 16 years to maturity (face value €1 000, annual coupon 6%) sells for €1 108. The issuer can call it in 5 years at a call price of €1 060. Find the yield to call YTC in percent.

F = €1 000coupon = 6%P = €1 108call price = €1 060call in 5 years
YTC = %
7.CAPM & WACCMedium
WACC = ?equity€5 000 000re​ = 8%debt€1 000 000rd​ = 4%T = 25.8%

A company has equity worth €5 000 000 and debt worth €1 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 4% and the tax rate is 25.8%. Find the WACC.

E = €5 000 000D = €1 000 000re = 8%rd = 4%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €42.50D0​ = €0.80g = 5%debt€51 000 000rd​ = 6%T = 25%

A company has 2 000 000 shares trading at €42.50; the book value of its equity is €51 000 000. It just paid a dividend of €0.80 per share, and dividends grow by 5% per year. Its debt has a market value of €51 000 000 and a pre-tax cost of 6%. The tax rate is 25%. Find the WACC, using the dividend growth model for the cost of equity.

N = 2 000 000P0 = €42.50book equity = €51 000 000D0 = €0.80g = 5%D = €51 000 000rd = 6%T = 25%
WACC = %
9.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 33, 29, 44, 59, 44, 36, 25.

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

A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is above 158?

μ = 170σ = 10x = 158
P =
11.ProbabilityMedium
xp(x)100.15200.35300.3400.2μ = 25.5

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

x: 10, 20, 30, 40p(x): 0.15, 0.35, 0.3, 0.2
Var(X) =
12.ProbabilityHard
xp(x)100.35200.05300.25400.35μ

X has the distribution shown, and Y = 3X + 10. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.35, 0.05, 0.25, 0.35Y = 3X + 10
σY =
13.Normal distributionMedium
27.535P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 27.5 and 35 min? Give P as a decimal.

μ = 32σ = 6x1 = 27.5x2 = 35
P =
14.Normal distributionHard
157164171178185192199cmtop 5%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 5% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm5%
x = cm
15.Confidence intervalsMedium
n = 10, sd​ = 6.90L = ?Ud̄ = 8.3E99%

The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 8.3 and standard deviation 6.9. Find the lower limit of the 99% confidence interval for the mean difference.

n = 10d̄ = 8.3sd = 6.999%t(9) = 3.25
lower limit = mmHg
16.Confidence intervalsHard
n = 15Lσ = ?s = 2.495%

The diameters of 15 machined shafts have a sample standard deviation of 2.4 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.

n = 15s = 2.4 µmχ² = 5.629, 26.119
σ = µm
17.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 122.3 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 122.3s = 12n = 81
t =
18.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 16 measurements has mean 122. The population standard deviation is 8. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 122σ = 8n = 1699%
U =
19.Linear regression & correlationMedium
xy33954975496411681370xy51040506070

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy246353460563668777xy24650607080x = 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. A new observation (2.5, 54) is made. Find its residual e = y − ŷ.

n = 6(2.5, 54)
e =

3Solutions

1.Time value of money

Answer: PV = 19500 €

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 = 0.05/4 = 0.0125
N = 4·5 = 20

Present value

PV = FV/(1 + i)N = 25 000/1.282 = €19 500.21

Answer

PV = €19 500.21
2.Time value of money

Answer: A = 128.1 €

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 = 12·6 = 72

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 = 0.7502
A = 8 000·0.004/(1 − 0.7502) = €128.10

Answer

A = €128.10
3.NPV & IRR

Answer: DPB = 3.323 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: 7 500/1.12 = 6 696.43 → Σ = 6 696.43
year 2: 8 000/1.254 = 6 377.55 → Σ = 13 073.98
year 3: 8 000/1.405 = 5 694.24 → Σ = 18 768.22
year 4: 6 000/1.574 = 3 813.11 → Σ = 22 581.33

Fraction of the last year

DPB = 3 + 1 231.78/3 813.11 = 3.323

Answer

DPB = 3.323 years

The discounted payback is always longer than the simple payback (2.563 years).

4.NPV & IRR

Answer: IRR = 33.67 %

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(30%) = 985.43
NPV(35%) = −337.86

Narrow down

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

IRR ≈ 30 + 5·985.43/(985.43 − (−337.86)) = 33.72%
exact: NPV(33.67%) = 0 → IRR = 33.67%

Answer

IRR = 33.67%
5.Bond pricing & yield

Answer: P = 1187 €

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 = €40.00, i = 0.015, N = 8

Present value of the coupons

€40.00·(1 − 0.8877)/0.015 = €299.44

Present value of the face value

€1 000·0.8877 = €887.71

Price

P = €1 187.15

Answer

P = €1 187.15

The yield is below the coupon rate, so the bond sells at a premium.

6.Bond pricing & yield

Answer: YTC = 4.625 %

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

1108 = 60·(1 − (1 + y)−5)/y + 1060/(1 + y)5

Solve numerically

y = 4%: €1 138.35
y = 5%: €1 090.31
YTC = 4.625%

Answer

YTC = 4.625%

The yield to maturity is 5.003%. 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 = 7.161 %

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 = €6 000 000, E/V = 0.8333, D/V = 0.1667

WACC

WACC = 0.8333·8% + 0.1667·4%·(1 − 0.258) = 7.161%

Answer

WACC = 7.161%
8.CAPM & WACC

Answer: WACC = 6.048 %

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 = 0.8·1.05 = €0.84
re = 0.84/42.5 + 0.05 = 6.976%

Market values

E = 2 000 000·€42.50 = €85 000 000
V = €136 000 000, E/V = 0.625, D/V = 0.375

WACC

WACC = 0.625·6.976% + 0.375·6%·(1 − 0.25) = 6.048%

Answer

WACC = 6.048%
9.Descriptive statistics

Answer: s = 11.47

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 = 270/7 = 38.57

Squared deviations

Σ(x − x̄)² = (−5.571)² + (−9.571)² + 5.429² + 20.43² + 5.429² + (−2.571)² + (−13.57)² = 789.7

Standard deviation

s = √(789.7/6) = 11.47

Answer

s = 11.47
10.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
11.Probability

Answer: Var(X) = 94.75

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 + 20·0.35 + 30·0.3 + 40·0.2 = 25.5

Mean of the squares

E(X²) = 100·0.15 + 400·0.35 + 900·0.3 + 1600·0.2 = 745

Variance

Var(X) = 745 − 25.5² = 94.75

Answer

Var(X) = 94.75
12.Probability

Answer: σY = 38.42

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 = 26, Var(X) = 164, σX = 12.81

Linear change

σY = |3|·12.81 = 38.42

Answer

σY = 38.42

The mean does move: E(Y) = 3·26 + 10 = 88.

13.Normal distribution

Answer: P = 0.4648

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-scores

z₁ = (27.5 − 32)/6 = −0.75
z₂ = (35 − 32)/6 = 0.5

Area between

P = Φ(0.5) − Φ(−0.75) = 0.6915 − 0.2266 = 0.4648

Answer

P = 0.4648
14.Normal distribution

Answer: x = 189.5 cm

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) = 0.95

z from the table

z ≈ 1.64

Back to the original units

x = 178 + 1.64·7 = 189.5 cm

Answer

x = 189.5 cm
15.Confidence intervals

Answer: lower limit = 1.209 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 = 9, t = 3.25

Margin of error

E = 3.25·6.9/√10 = 7.091

Interval

8.3 ± 7.091 → [1.209, 15.39]

Answer

lower limit = 1.209 mmHg
16.Confidence intervals

Answer: σ = 3.785 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 = 14·2.4² = 80.64

Limits for σ²

80.64/26.119 = 3.087
80.64/5.629 = 14.33

Limit for σ

σ = √14.33 = 3.785 µm

Answer

σ = 3.785 µm

95% interval for σ: [1.757, 3.785] µm.

17.Hypothesis tests & confidence intervals

Answer: t = 1.725

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/√81 = 12/9 = 1.333

Test statistic

t = (122.3 − 120)/1.333 = 1.725

Answer

t = 1.725
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: U = 127.2

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 = 2.576·8/√16 = 5.152

Interval limit

U = 122 + 5.152 = 127.2

Answer

U = 127.2
19.Linear regression & correlation

Answer: r = 0.9785

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 = 344/6 = 57.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 222
Syy = Σ(y − ȳ)² = 735.3

Correlation coefficient

r = 222/√(70·735.3) = 0.9785

Answer

r = 0.9785
20.Linear regression & correlation

Answer: e = 4.433

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 = 367/6 = 61.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 101.5

Regression line

b = 101.5/17.5 = 5.8
a = 61.17 − 5.8·4.5 = 35.07
ŷ = 35.07 + 5.8·x

Predicted value

ŷ = 35.07 + 5.8·2.5 = 49.57

Residual

e = 54 − 49.57 = 4.433

Answer

e = 4.433

The point lies above the line.

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