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
years01234PV = ?€50 000

You need €50 000 in 4 years. An account pays 3.6% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €50 0003.6% / 12n = 4
PV = €
2.Time value of moneyHard
months01…48…360€300 000AB = ?A7.2% / 12

A mortgage of €300 000 is repaid with equal monthly payments over 30 years at 7.2% per year, compounded monthly. How much is still owed just after the payment at the end of year 4? Find the outstanding balance B.

P = €300 0007.2% / 12N = 360k = 48
B = €
3.NPV & IRRMedium
years01234−45 00011 00018 00011 00019 000

An investment of €45 000 gives these cash flows at the end of years 1 to 4: €11 000, €18 000, €11 000, €19 000. The cost of capital is 15%. Find the net present value NPV.

r = 15%CF0 = −45 000CF1 = 11 000CF2 = 18 000CF3 = 11 000CF4 = 19 000
NPV = €
4.NPV & IRRHard
years01234−20 0007 0006 50011 50011 500

A project costs €20 000 and returns €7 000, €6 500, €11 500, €11 500 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 10%. Find the modified internal rate of return MIRR in percent.

r = 10%CF0 = −20 000CF1 = 7 000CF2 = 6 500CF3 = 11 500CF4 = 11 500
MIRR = %
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 3.5% per year. Find the price P of the bond.

F = €1 000coupon = 8%YTM = 3.5%n = 4
P = €
6.Bond pricing & yieldHard
years012345€1 190€50€50€50 + €1 050

A bond with 17 years to maturity (face value €1 000, annual coupon 5%) sells for €1 190. The issuer can call it in 5 years at a call price of €1 050. Find the yield to call YTC in percent.

F = €1 000coupon = 5%P = €1 190call price = €1 050call in 5 years
YTC = %
7.CAPM & WACCMedium
WACC = 8.79%equity€6 000 000re​ = ?debt€3 000 000rd​ = 3%T = 21%

A company has equity worth €6 000 000 and debt worth €3 000 000. Its WACC is 8.79%, its pre-tax cost of debt is 3% and the tax rate is 21%. Find the cost of equity re.

E = €6 000 000D = €3 000 000WACC = 8.79%rd = 3%T = 21%
re = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €28.50β = 1.1rf​ = 3.5%, rm​ = 10.5%debt20 000 × €1 000rd​ = 5%T = 20%

A company has 2 500 000 shares at €28.50 each and 20 000 bonds trading at €1 000 each. The pre-tax cost of debt is 5% and the tax rate is 20%. The share’s beta is 1.1, the risk-free rate 3.5% and the market return 10.5%. Find the WACC.

β = 1.1rf = 3.5%rm = 10.5%rd = 5%T = 20%
WACC = %
9.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 21, 19, 51, 53, 42.

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

Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 60.

μ = 70σ = 20x = 60
z =
11.ProbabilityMedium
xp(x)100.25200.2300.2400.35μ = 26.5

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

x: 10, 20, 30, 40p(x): 0.25, 0.2, 0.2, 0.35
σ =
12.ProbabilityHard
n = 8, p = 0.2: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 8p = 0.2
P =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.6255

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.6255. Find z to two decimals.

A = 0.6255
z =
14.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
15.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?99%

The battery life of two phone models is compared. Model 1: 50 phones, mean 123 h, known standard deviation 18 h. Model 2: 60 phones, mean 133 h, known standard deviation 18 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 123 hσ₁ = 18 hn₂ = 60x̄₂ = 133 hσ₂ = 18 h99%
E = h
16.Confidence intervalsHard
σ = 35, n = ?x̄ − Ex̄ + Ex̄E = 399%

The population standard deviation is 35. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 3?

σ = 35E = 399%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 76.05 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 76.05s = 4n = 64
t =
18.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 49 measurements has mean 53.4. The population standard deviation is 10. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 53.4σ = 10n = 4999%
L =
19.Linear regression & correlationMedium
xy1575188521108241242714930165xy15202530100150

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy23456881410181217xy51051015x = 3

The table gives 6 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 = 3.

n = 6x = 3
ŷ =

3Solutions

1.Time value of money

Answer: PV = 43304 €

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 = 0.036000000000000004/12 = 0.003
N = 12·4 = 48

Present value

PV = FV/(1 + i)N = 50 000/1.155 = €43 303.72

Answer

PV = €43 303.72
2.Time value of money

Answer: B = 286897 €

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 payment

i = 0.006, N = 360
A = 300 000·0.006/(1 − (1 + 0.006)−360) = €2 036.36

Present value of the remaining payments

Whatever is still owed must be exactly paid off by the remaining payments.

N − k = 360 − 48 = 312
B = 2036·(1 − (1 + 0.006)−312)/0.006 = €286 897.13

Answer

B = €286 897.13
3.NPV & IRR

Answer: NPV = −3728 €

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 + 0.15)1 = 11 000/1.15 = 9 565.22
CF2/(1 + 0.15)2 = 18 000/1.322 = 13 610.59
CF3/(1 + 0.15)3 = 11 000/1.521 = 7 232.68
CF4/(1 + 0.15)4 = 19 000/1.749 = 10 863.31

Add them up

NPV = €−3 728.21

Answer

NPV = €−3 728.21

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

4.NPV & IRR

Answer: MIRR = 19.9 %

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

7 000·1.13 = 9 317.00
6 500·1.12 = 7 865.00
11 500·1.11 = 12 650.00
11 500·1.10 = 11 500.00
TV = 41 332.00

MIRR

MIRR = (41 332.00/20 000)1/4 − 1 = 19.9%

Answer

MIRR = 19.9%
5.Bond pricing & yield

Answer: P = 1167 €

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 = 0.0175, N = 8

Present value of the coupons

€40.00·(1 − 0.8704)/0.0175 = €296.20

Present value of the face value

€1 000·0.8704 = €870.41

Price

P = €1 166.61

Answer

P = €1 166.61

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

6.Bond pricing & yield

Answer: YTC = 1.938 %

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

1190 = 50·(1 − (1 + y)−5)/y + 1050/(1 + y)5

Solve numerically

y = 1%: €1 241.71
y = 2%: €1 186.69
YTC = 1.938%

Answer

YTC = 1.938%

The yield to maturity is 3.498%. 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 = 12 %

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 = €9 000 000, E/V = 0.6667, D/V = 0.3333

Debt part of the WACC

0.3333·3%·(1 − 0.21) = 0.79%

Cost of equity

re = (8.79% − 0.79%)/0.6667 = 12%

Answer

re = 12%
8.CAPM & WACC

Answer: WACC = 9.622 %

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 = 2 500 000·€28.50 = €71 250 000
D = 20 000·€1 000 = €20 000 000

Cost of equity (CAPM)

re = 3.5% + 1.1·7% = 11.2%

Weights

V = €91 250 000, E/V = 0.7808, D/V = 0.2192

WACC

WACC = 0.7808·11.2% + 0.2192·5%·(1 − 0.2) = 9.622%

Answer

WACC = 9.622%
9.Descriptive statistics

Answer: s = 16.25

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 = 186/5 = 37.2

Squared deviations

Σ(x − x̄)² = (−16.2)² + (−18.2)² + 13.8² + 15.8² + 4.8² = 1057

Standard deviation

s = √(1057/4) = 16.25

Answer

s = 16.25
10.Descriptive statistics

Answer: z = −0.5

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 = (60 − 70)/20 = −0.5

Answer

z = −0.5
11.Probability

Answer: σ = 11.95

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 = 10·0.25 + 20·0.2 + 30·0.2 + 40·0.35 = 26.5

Mean of the squares

E(X²) = 100·0.25 + 400·0.2 + 900·0.2 + 1600·0.35 = 845

Variance

Var(X) = 845 − 26.5² = 142.8
σ = √142.8 = 11.95

Answer

σ = 11.95
12.Probability

Answer: P = 0.5033

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.

Zero and one failure

P(X = 0) = 0.88 = 0.1678
P(X = 1) = 8·0.2·0.87 = 0.3355

Add them

P = 0.1678 + 0.3355 = 0.5033

Answer

P = 0.5033
13.Normal distribution

Answer: z = −0.32

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 − 0.6255 = 0.3745

Read the z-table backwards

Find 0.3745 inside the table and read off z.

z = −0.32

Answer

z = −0.32
14.Normal distribution

Answer: P = 0.9515

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 = 25.03 − 25 = 0.03 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.03)/0.01803 = −1.66

Probability

P = 1 − Φ(−1.66) = 0.9515

Answer

P = 0.9515
15.Confidence intervals

Answer: E = 8.879 h

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.

Point estimate

x̄₁ − x̄₂ = 123 − 133 = −10

Standard error

SE = √(18²/50 + 18²/60) = 3.447

Margin of error

E = 2.576·3.447 = 8.879

Answer

E = 8.879 h

The interval does not contain 0, so the means differ significantly.

16.Confidence intervals

Answer: n = 904

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 = (2.576·35/3)² = 903.2

Round up

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

n = 904

Answer

n = 904
17.Hypothesis tests & confidence intervals

Answer: t = 2.1

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 = 4/√64 = 4/8 = 0.5

Test statistic

t = (76.05 − 75)/0.5 = 2.1

Answer

t = 2.1
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: L = 49.72

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 = 2.576·10/√49 = 3.68

Interval limit

L = 53.4 − 3.68 = 49.72

Answer

L = 49.72
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.

Means

x̄ = 135/6 = 22.5
ȳ = 706/6 = 117.7

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 987
Syy = Σ(y − ȳ)² = 6243

Correlation coefficient

r = 987/√(157.5·6243) = 0.9953

Answer

r = 0.9953
20.Linear regression & correlation

Answer: ŷ = 4.262

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 = 42/6 = 7
ȳ = 65/6 = 10.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 115

Regression line

b = 115/70 = 1.643
a = 10.83 − 1.643·7 = −0.6667
ŷ = −0.6667 + 1.643·x

Prediction

ŷ = −0.6667 + 1.643·3 = 4.262

Answer

ŷ = 4.262

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