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 6% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €40 0006% / 4n = 10
PV = €
2.Time value of moneyHard
months01…48…300€200 000AB = ?A4.2% / 12

A mortgage of €200 000 is repaid with equal monthly payments over 25 years at 4.2% 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 = €200 0004.2% / 12N = 300k = 48
B = €
3.NPV & IRRMedium
years0123−60 00016 00026 00014 000

An investment of €60 000 gives these cash flows at the end of years 1 to 3: €16 000, €26 000, €14 000. The cost of capital is 9%. Find the net present value NPV.

r = 9%CF0 = −60 000CF1 = 16 000CF2 = 26 000CF3 = 14 000
NPV = €
4.NPV & IRRHard
r (%)NPVABr* = ?0

Project A costs €30 000 and returns €17 000, €10 500, €7 000 at the end of years 1 to 3. Project B costs €35 000 and returns €9 000, €14 000, €18 000. Find the crossover rate r* at which both projects have the same NPV, in percent.

A: −30 000, 17 000, 10 500, 7 000B: −35 000, 9 000, 14 000, 18 000
r* = %
5.Bond pricing & yieldMedium
years0123456P€20€20€20 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 2% and matures in 6 years. Its yield to maturity changes from 4% to 5%. By how much ΔP does the bond price change?

F = €1 000coupon = 2%n = 6y1 = 4%y2 = 5%
ΔP = €
6.Bond pricing & yieldHard
years012345€960€60€60 + €1 000

A bond with a face value of €1 000 and an annual coupon of 6% matures in 5 years and sells for €960. Find its yield to maturity YTM in percent.

F = €1 000coupon = 6%n = 5P = €960
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€3 000 000re​ = 8%debt€2 000 000rd​ = 4.5%T = 25.8%

A company has equity worth €3 000 000 and debt worth €2 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 4.5% and the tax rate is 25.8%. Find the WACC.

E = €3 000 000D = €2 000 000re = 8%rd = 4.5%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €38.00β = 1.1rf​ = 3.5%, rm​ = 7.5%debt20 000 × €960rd​ = 3%T = 25%

A company has 2 500 000 shares at €38.00 each and 20 000 bonds trading at €960 each. The pre-tax cost of debt is 3% and the tax rate is 25%. The share’s beta is 1.1, the risk-free rate 3.5% and the market return 7.5%. Find the WACC.

β = 1.1rf = 3.5%rm = 7.5%rd = 3%T = 25%
WACC = %
9.Descriptive statisticsMedium
202530354045505560

Find the sample standard deviation s of the data: 60, 30, 43, 21, 17, 24, 39.

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

A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is below 452?

μ = 500σ = 40x = 452
P =
11.ProbabilityMedium
n = 8k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 8? The order does not matter.

n = 8k = 2
N =
12.ProbabilityHard
xp(x)00.3520.2540.260.2μ

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

x: 0, 2, 4, 6p(x): 0.35, 0.25, 0.2, 0.2Y = 2X + 20
σY =
13.Normal distributionMedium
225 kg224.3 kgT = X₁ + … + X₉?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 9 of them exceeds 224.3 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 9> 224.3 kg
P =
14.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
n = 8, sd​ = 1.30L = ?Ud̄ = −7.5E99%

The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −7.5 and standard deviation 1.3. Find the lower limit of the 99% confidence interval for the mean difference.

n = 8d̄ = −7.5sd = 1.399%t(7) = 3.499
lower limit = mmHg
16.Confidence intervalsHard
n = 21σ = ?Us = 195%

The diameters of 21 machined shafts have a sample standard deviation of 1 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.

n = 21s = 1 µmχ² = 9.591, 34.17
σ = µm
17.Hypothesis tests & confidence intervalsMedium
−2.0642.064t = ?df = 24

Two teaching methods are compared. Group 1: 12 students, mean score 26, standard deviation 11. Group 2: 14 students, mean score 35.8, standard deviation 3. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 12x̄₁ = 26s₁ = 11n₂ = 14x̄₂ = 35.8s₂ = 3
t =
18.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 25 measurements has mean 199.7. The population standard deviation is 15. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 199.7σ = 15n = 2595%
L =
19.Linear regression & correlationMedium
xy350457563665771xy46506070

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy349447559662764xy465060ŷ = 49

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 49?

n = 5ŷ = 49
x =

3Solutions

1.Time value of money

Answer: PV = 22050 €

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 = 0.06/4 = 0.015
N = 4·10 = 40

Present value

PV = FV/(1 + i)N = 40 000/1.814 = €22 050.49

Answer

PV = €22 050.49
2.Time value of money

Answer: B = 180286 €

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 = 0.0035, N = 300
A = 200 000·0.0035/(1 − (1 + 0.0035)−300) = €1 077.88

Present value of the remaining payments

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

N − k = 300 − 48 = 252
B = 1078·(1 − (1 + 0.0035)−252)/0.0035 = €180 286.33

Answer

B = €180 286.33
3.NPV & IRR

Answer: NPV = −12627 €

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 + 0.09)1 = 16 000/1.09 = 14 678.90
CF2/(1 + 0.09)2 = 26 000/1.188 = 21 883.68
CF3/(1 + 0.09)3 = 14 000/1.295 = 10 810.57

Add them up

NPV = €−12 626.85

Answer

NPV = €−12 626.85

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

4.NPV & IRR

Answer: r* = 5.224 %

Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.

Difference in cash flows

ΔCF0 = −35 000 − (−30 000) = −5 000
ΔCF1 = 9 000 − 17 000 = −8 000
ΔCF2 = 14 000 − 10 500 = 3 500
ΔCF3 = 18 000 − 7 000 = 11 000

IRR of the difference

NPVΔ(5%) = 57.77
NPVΔ(6%) = −196.37
r* = 5.224%

Answer

r* = 5.224%

Check: at 5.224% both NPVs are €1 647.78.

5.Bond pricing & yield

Answer: ΔP = −47.43 €

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 = €20.00, i = 0.04, N = 6
€20.00·(1 − 0.7903)/0.04 + €1 000·0.7903 = €104.84 + €790.31 = €895.16

Price at the new yield

C = €20.00, i = 0.05, N = 6
€20.00·(1 − 0.7462)/0.05 + €1 000·0.7462 = €101.51 + €746.22 = €847.73

Change

ΔP = 847.73 − 895.16 = −€47.43

Answer

ΔP = −€47.43
6.Bond pricing & yield

Answer: YTM = 6.975 %

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

€960 = €60·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵

Solve numerically

P(6%) = €1 000.00, P(7%) = €959.00
YTM = 6.975%

Answer

YTM = 6.975%
7.CAPM & WACC

Answer: WACC = 6.136 %

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

WACC

WACC = 0.6·8% + 0.4·4.5%·(1 − 0.258) = 6.136%

Answer

WACC = 6.136%
8.CAPM & WACC

Answer: WACC = 6.95 %

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·€38.00 = €95 000 000
D = 20 000·€960 = €19 200 000

Cost of equity (CAPM)

re = 3.5% + 1.1·4% = 7.9%

Weights

V = €114 200 000, E/V = 0.8319, D/V = 0.1681

WACC

WACC = 0.8319·7.9% + 0.1681·3%·(1 − 0.25) = 6.95%

Answer

WACC = 6.95%
9.Descriptive statistics

Answer: s = 15.02

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 = 234/7 = 33.43

Squared deviations

Σ(x − x̄)² = 26.57² + (−3.429)² + 9.571² + (−12.43)² + (−16.43)² + (−9.429)² + 5.571² = 1354

Standard deviation

s = √(1354/6) = 15.02

Answer

s = 15.02
10.Descriptive statistics

Answer: P = 0.1151

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 = 0.1151

Answer

P = 0.1151
11.Probability

Answer: N = 28

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.

Combinations

C(8, 2) = (8·7)/(2·1) = 28

Answer

N = 28
12.Probability

Answer: σY = 4.539

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 = 2.5, Var(X) = 5.15, σX = 2.269

Linear change

σY = |2|·2.269 = 4.539

Answer

σY = 4.539

The mean does move: E(Y) = 2·2.5 + 20 = 25.

13.Normal distribution

Answer: P = 0.6808

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 = 225 kg, σT = 0.5·√9 = 1.5 kg

Standardise

z = (224.3 − 225)/1.5 = −0.47

Probability

P = 1 − Φ(−0.47) = 0.6808

Answer

P = 0.6808
14.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
15.Confidence intervals

Answer: lower limit = −9.108 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 = 7, t = 3.499

Margin of error

E = 3.499·1.3/√8 = 1.608

Interval

−7.5 ± 1.608 → [−9.108, −5.892]

Answer

lower limit = −9.108 mmHg
16.Confidence intervals

Answer: σ = 0.7651 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 = 20·1² = 20

Limits for σ²

20/34.17 = 0.5853
20/9.591 = 2.085

Limit for σ

σ = √0.5853 = 0.7651 µm

Answer

σ = 0.7651 µm

95% interval for σ: [0.7651, 1.444] µm.

17.Hypothesis tests & confidence intervals

Answer: t = −3.207

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.

Pooled variance

sp² = (11·11² + 13·3²)/24 = 60.33
sp = 7.767

Standard error

SE = 7.767·√(1/12 + 1/14) = 3.056

Test statistic

t = (26 − 35.8)/3.056 = −3.207

Answer

t = −3.207

With df = 24, the two-sided 5% critical value is 2.064; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: L = 193.8

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 = 1.96·15/√25 = 5.88

Interval limit

L = 199.7 − 5.88 = 193.8

Answer

L = 193.8
19.Linear regression & correlation

Answer: r = 0.9867

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 = 25/5 = 5
ȳ = 306/5 = 61.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 50
Syy = Σ(y − ȳ)² = 256.8

Correlation coefficient

r = 50/√(10·256.8) = 0.9867

Answer

r = 0.9867
20.Linear regression & correlation

Answer: x = 3.4

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 = 25/5 = 5
ȳ = 281/5 = 56.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 45

Regression line

b = 45/10 = 4.5
a = 56.2 − 4.5·5 = 33.7
ŷ = 33.7 + 4.5·x

Solve for x

x = (49 − 33.7)/4.5 = 3.4

Answer

x = 3.4

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