Chapter II - Capital Investment

Capital Investment Risk and Uncertainty

• Risk can be defined as uncertainty

• In capital Expenditure ( Capex),• risk is variability likely to occur between • Estimated and future / actual return

• SD = Standard Deviation =• average magnitude of deviation from

expected value• SD2 = 2 = variance

Mode

• Definition: Mode is the

• most frequently occurring value • in a frequency distribution.

Example: To find the mode of 11,3,5,11,7,3,11,17,11,25

Mode• Step 1:Arrange the numbers in ascending order.

3,3,5,7,11,11,11,11,17,25

Step 2: In the above distribution Number 11 occurs 4 times, Number 3 occurs 2 times, Number 5 occurs 1 times, Number 7 occurs 1 times.

• Number 17 occurs 1 times.• Number 25 occurs 1 times.•

So the number with most occurrances ( four occurances) is 11 • and is the Mode of this distribution.

Mode = 11

Measures of risk

• 1) Range• Rg= ( Rh- Rl)

• Example

• CAT score of 6 students were

• 55, 66,45,77,88,99

• Range = 45 to 99

Range

• Definition : Range is the difference • between the highest and the lowest values• in a frequency distribution.

Example: To find the range in 3,4,5,7,3,9,11

Range

• Step 1:Arrange the numbers in ascending order. 3,3,4,5,7,911

Step 2: In the above distribution The largest number is11 The smallest value is 3 Range =( largest number - smallest number)

• Range = 11-3 = 8

Range

•Example :

• six mid-cap mutual funds, • five-year annual returns are • +10.1, % +7.7%, +5.0, +12.3%, +12.2% and +10.9%.

Range = Maximum – Minimum• = (+12.3%) - (+5.0%) = 7.3%

Mean absolute deviation

•

MAD• The mean absolute deviation (MAD) • is the mean of the absolute deviations • of a set of data about the data's mean. • For a sample size , the mean deviation is

defined by

where is the mean of the distribution.

MAD• Example:

six mid-cap mutual funds, • five-year annual returns are +10.1, +7.7%, +5.0, +12.3%, +12.2% and +10.9%.•

Mean absolute deviation starts by finding the mean:• (10.1% + 7.7% + 5.0% + 12.3% + 12.2% + 10.9%)/6 = 9.7%. =

Each of the six observations deviate from the 9.7%; the absolute deviation ignores +/-.

1st: 10.1 - 9.7 = 0.4 2nd: 7.7 - 9.7 = 2.0

3rd: 5.0 - 9.7 = 4.7 4th: 12.3 - 9.7 = 2.6

• 5th: 12.2 - 9.7 = 2.56th: 10.9 - 9.7 = 1.2

• Next, the absolute deviations are summed and divided by 6:• (0.4 + 2.0 + 4.7 + 2.6 + 2.5 + 1.2)/6 = 13.4/6 = 2.233333, or rounded, 2.2

Standard Deviation

• SD 2 = 2= variance = pi * ( Xi – X’)2

• Where

• pi = probability of occurance

• Xi = occurance of ith item

• X’ = mean of distribution

Variance• Variance (σ2)

• is a measure of dispersion • that in practice can be easier to apply than mean absolute deviation• because it removes +/- signs by squaring the deviations.

• five-year annual returns are +10.1, +7.7%, +5.0, +12.3%, +12.2% and +10.9%.

six deviations. • To compute variance, we take the square of each deviation, • add the terms together and • divide by the number of observations.

VarianceObservation Value Deviation from

mean +9.7%Square of Deviation

1 +10.1% 0.4 0.16

2 +7.7% 2.0 4.0

3 +5.0% 4.7 22.09

4 +12.3% 2.6 6.76

5 +12.2% 2.5 6.25

6 +10.9% 1.2 1.44

Variance

• Variance = • (0.16 + 4.0 + 22.09 + 6.76 + 6.25 + 1.44)/6 =

6.7833.• Variance is not in the same units as the

underlying data.• In this case, it's expressed as 6.7833%

squared -

Co-efficient of variation

• CV = / X’

• Where

• X’ = arithmatic mean of variable

Distributions

• Probability distributions:

• Discrete probability distributions…… the outcomes can be separated

• Continuous probability distributions…..• The outcomes can NOT be separated

Binomial Distribution

• Definition :• Probability distribution function• Is a function• Which serves as a tool• To determine ALL the probabilities• Of ALL the outcomes of a• Given experiment

Binomial Distribution

• Is a discrete, probability distribution function having the parameters as n and p

• And helps to find out the probabilities • When the outcomes are independent to each

other • Or the outcomes are independently defined.

Example of binomial distribution function

• Objective type question paper• Q 1

– A)…..blah….blah… blah– B) …..blah….blah… blah– C) …..blah….blah… blah– D) …..blah….blah… blah

Q2)A) …. More …..blah….blah… blahB) ….some more …..blah….blah… blahC) …. Still more …..blah….blah… blahD)….. Really … more and more …..blah….blah… blah

• P(s) = probability of success in a question = 0.25 = ¼ PER question• P(f) = probability of failure = 1-0.25 = 0.75 = ¾ per question• nCr = n! / ( n-r)! * r!• = n(n-1)(n-2)….(n-r-1) / 1*2*3….r• P(sssff) = p(s) * p(s) * p(s) * p(f)*p(f)• Because p(s f) = p(s) * p(f)• Where • = intersection• S and f are success and failure… independent events• = 0.25 * 0.25 *0.25*0.75*0.75• = (0.25)3* (0.75)2

Example

• A tyre wholesaler has 500 Super Brand Tyres in stock AND that

• 50 of them are slightly damaged.

• If a retailer buys 10 of these tyres from the wholesaler,

• what are the probability that the retailer receives EXACTLY 8 good tyres ?

• Given :• n = number of tyres bought = 10• r = number of successes = 8• p = probability of success of each item = 450/500• So, p(r=8) = 10C8 * (450/500)8 * (1-450/500)10-8

• = (10 * 9/1*2) * ((9/10)8)* ((1/10)2)

• = 0.194• The probability that the retailer receives EXACTLY 8 good

tyres is 0.194•

Poisson Distribution

• The poisson distribution resembles the binomial distribution

• if the probability of an event is very small.• The poisson distribution is an appropriate

model for count data

Poisson vs Binomial

• Poisson pdf is nothing but• binomial pdf where n is large and p is small

• If n >= 20, it is called a large number of experiments

• If p < 0.05 then p is small

Examples

• mortality of infants in a city,• the number of misprints in a book, • the number of bacteria on a plate,• and the number of activations of a geiger counter. • The poisson distribution was derived by the french

mathematician Poisson in 1837, • and the first application was the description of • the number of death by horse kicking in the

prussian army

The poisson distribution

• This is sometimes also known as the law of rare events,

• is a mathematical rule • that assigns probabilities to the number

occurances. • The probability density function of a Poisson

variable is given by

The Poisson distribution

• applies when: • (1) the event is something that can be counted in whole

numbers;• (2) occurrences are independent, so that one occurrence

neither diminishes nor increases the chance of another; • (3) the average frequency of occurrence for the time period in

question is known; and• (4) it is possible to count how many events have occurred, such

as the number of times a firefly lights up in my garden in a given 5 seconds, some evening,

• but meaningless to ask how many such events have not occurred.

• This last point sums up the contrast with the Binomial situation, • where the probability of each of two mutually exclusive events

(p and q) is known.

• Poisson distribution is a discrete variable distribution’ and you cannot have a decimal answer:

• For example, you can’t have 1.9 children, or 3.7 computers, or 2.3 cars

• The Poisson Distribution, so to speak, is the Binomial Distribution Without Q.

The classic Poisson example• is the data set of von Bortkiewicz (1898), for the chance of a

Prussian cavalryman being killed by the kick of a horse. • Ten army corps were observed over 20 years, • giving a total of 200 observations of one corps for a one year

period. • The period or module of observation is thus one year. • The total deaths from horse kicks were 122, • and the average number of deaths per year per corps was thus

122/200 = 0.61. • This is a rate of less than 1. • It is also obvious that it is meaningless to ask how many times per

year a cavalryman was not killed by the kick of a horse.

Poisson Distribution• In any given year, we expect to observe, • not exactly 0.61 deaths in one corps• (that is not possible; deaths occur in modules of 1), • but sometimes none, sometimes one, • occasionally two, • perhaps once in a while three, and (we might intuitively expect) • very rarely any more. • Here, then, is the classic Poisson situation:• a rare event, • whose average rate is small,• with observations made over many small intervals of time.

Example - Poisson

• 200 passengers have made reservations for a flight.

• If the probability that a passenger who has a reservation will not show up is 0.01, what is the probability that exactly three passengers will not show up ?

• For binomial distributions,• P(r) = nCr * pr * (1-p)n-r

• Probability of exactly r successes• • In case of Poisson distributions,• P( r) = ( e- * r) / r!• E= exponential value is taken as 2.718• Here, n = 200 which is greater than 20 so it satisfies condition (1)• P = 0.01 which is less than < 0.05 , so it satisfies condition (2)• = n * p = 200 * 0.01 = 2• So,• P(r=3) = e-2 * (2)3 / 3!• = 0.1804• The probability that exactly 3 passengers will NOT show up is 0.1804

Normal distribution

• All normal distributions are symmetric and • have bell-shaped density curves • with a single peak.• To speak specifically of any normal distribution, • two quantities have to be specified: • the mean , where the peak of the density occurs, • and the standard deviation , • which indicates the spread or girth of the bell

curve

Normal pdf

• Is a continuous pdf• Is symmetric with respect to its mean• Eg.,• Accident at a spot vis-à-vis accident at a

stretch of road• If area is not equal to 1, it is NOT a pd curve

• Normal pdf has two papameters• = mean • = standard deviation• For continuous pdf, you have to find by area, NOT by

height• Because probability at a particular point or place is

ZERO• In all standard normal variates or form, the mean and

sigma are expected to be 0 and 1 respectively• For ALL normal pdf, the shape of the curve is same

The 68-95-99.7% Rule

• Empirical Rule.• 68%of the observations fall within 1 standard

deviation of the mean, that is, • between µ - δ and µ + δ

• 95%of the observations fall within 2 standard deviations of the mean, that is,

• between µ - 2δ and µ + 2δ

• 99.7%of the observations fall within 3 standard deviations of the mean, that is,

• between µ - 3δ and µ + 3δ

• Thus, for a normal distribution, almost all values lie within 3 standard deviations of the mean.

Risk adjusted discount rate

• - assumption :• Investors expect a higher rate of return on

riskier projects

• rp= rf+p (rm-rf)• Where• rp=return of the portfolio

• rf = return of risk free asset

• p = beta of the portfolio

• Eg.,

Project A -20000 8000 8000 6000

Project B -20000 10000 12000 6000

Rf = 5 %rp(A) = 5 % rp(B) = 10 %

? NPV of the projects

Project A -20000 8000/1.1 8000/1.12 6000/1.13

Project B -20000 10000/1.1 12000/1.12 6000/1.13

NPV of Project A = (-1618)NPV of Project B = 1780

Since NPV of Project B is positive, select Project B

Risk adjusted discount rate rA= 10 % rB=15 %

Merits

• Simple and easy to calculate and understand• Takes into a/c investor’s risk averse attitude

Demerits

• Discount rate not objective• Wrong factor is adjusted ( discount rate

instead of cash flows)• Assumes increasing risk over time• Assumes investors are risk averse

Certainty Equivalent Co-Efficient

• CEC = correction factor• CEC = riskless cash flow / risky cash flow

Item Time period 1 T 2 T 3

Project A 0.9 0.8 0.6

Project B 0.8 0.7 0.5

CF A -20000 8000 8000 6000

CF B -20000 10000 12000 6000

• Rf = 5 %• Certain cash flows

Item Project

CF 0 CF 1 CF 2 CF 3

CFc Proj A -20000 8000*0.9= 7200 8000*0.8 = 6400 6000*0.6 = 3600

CFc ProjB -20000 10000*0.8 = 8000 12000*0.7 = 8400 6000*0.5= 3000

• DCF =• ProjA = -20000 + ( 7200/1.05) + ( 6400/1.052 )

+(3600/1.053 )• = -20000 + 6854 + 5804 + 3108• NPVA = - 4234

• ProjB = -20000 + ( 8000 / 1.05) + (8400/ 1.052 )+ (3000/1.053 )

• = -20000 + 7616 + 7618 + 2592• NPV = - 2174

Conclusion

• Since NPV of ProjA is greater than NPV of Project B

• Select Project A

Sensitivity Analysis

• More than one cash flow estimates in a year• Takes into consideration variability of return• Three scenarios• Optimistic• Pessimistic• Most likely

Cash flow estimates Scenario Cash

FlowCF 0 CF 1 CF 2 CF 3 …… CF 15 PVIFA

@10 %, 15

years

Pessimistic

CFA -20000 3000 3000 3000 …… 3000 7.606

Most likely

CFA -20000 4000 4000 4000 …… 4000 7.606

Optimistic

CFA -20000 5000 5000 5000 ……. 5000 7.606

Ke = 10 % = 0.10

DCF Project A

• Project A

Item CF 0 CF 1 to CF 15

PVIFA @10

%, 15 years

PV NPV

Pessimistic

-20000 3000 7.606 22818 2818

Most Likely

-20000 4000 7.606 30414 10414

Optimistic

-20000 5000 7.606 38030 18030

Cash flow estimates Scenario Cash

FlowCF 0 CF 1 CF 2 CF 3 …… CF 15 PVIFA

@10 %, 15

years

Pessimistic

CFB -20000 0 0 0 …… 0 7.606

Most likely

CFB -20000 4000 4000 4000 …… 4000 7.606

Optimistic

CFB -20000 8000 8000 8000 ……. 8000 7.606

Ke = 10 % = 0.10

DCF Project B

• Project B

Item CF 0 CF 1 to CF 15

PVIFA @10

%, 15 years

PV NPV

Pessimistic

-20000 0 7.606 0 0

Most Likely

-20000 4000 7.606 30414 10414

Optimistic

-20000 8000 7.606 60848 40848

Conclusions

• Project B is more profitable• And MORE RISKY than Project A

• So, select according to your risk appetite

• Issues : • Probability estimates of cash flows under different

conditions is NOT known

Problem

• Same # as above

Probability Pi

Project A Project B

Pessimistic 0.2 0.2

Most likely 0.6 0.4

Optimistic 0.2 0.4

Estimated CF Project A Project B

Pessimistic 3000*0.2= 600 0 * 0.2 = 0

Most likely 4000*0.6=2400 4000 * 0.4 = 1600

Optimistic 5000*0.2 = 1000 4000*0.4 = 3200

Expected CF 4000 4800

Probability = Likelihood of an event happening

Project AScenario CF 0 Expected CF

1 to 15PVIFA10 %, 15 years

Expected NPV

Pessimistic -20000 600 7.606 4562

Most Likely -20000 2400 7.606 18254

Optimistic -20000 1000 7.606 7606

Total 30422

Project BScenario CF 0 Expected CF

1 to 15PVIFA10 %, 15 years

Expected NPV

Pessimistic -20000 0 7.606 0

Most Likely -20000 1600 7.606 12088

Optimistic -20000 3200 7.606 24338

TOTAL 36426

Recommendations:

Project B has higher EMV than Project A … However, Project B is more Risky.

Mathematical Analysis• PCCF :Perfectly correlated cash flows

• Behaviour of cash flows in ALL periods is same• n * SD of cash flow is same• In other words,• If actual cash flow in time t1 is x times to expected value,• Then,• CF in other years will also be x times to expected value• i = standard deviation of CF in year I• n• (NPV) = (i’ / ( 1+rf)i

• i =1

n• NPV’ = ( Xi’ / ( 1+rf)i - CF0 • i =1

• = ( CFi’ / ( 1+rf)i – I• Where• NPV’ = mean or average NPV• i = time horizon• rf = risk free rate of return ( discount rate) … we are trying to

isolate risk of the project• Xi’ = mean or average cash flow in year i

Example

• Given : • CF0 = I = 20000

• rf = 0.06 or 6 %

• Year P 0.5 P 0.5 xi

’ I’

Year 1 13000 7000 x1’ = 10000 1

’ = 3000

Year 2 8000 4000 x1’ = 6000 1

’ = 2000

Year 3 12000 4000 x1’ = 8000 1

’ = 4000

Year 4 8400 3600 x1’ = 6000 1

’ = 2400

n• NPV’ = ( Xi’ / ( 1+rf)i - CF0 • i =1

• = -20000 + {(10000 / 1.06) + (6000/1.062) + ( 8000/1.063) + (6000/1.064)}

• = -20000 + 9432 + 5338 + 6716 + 4752• = -20000 + 26238 = + 6238

• n• (NPV) = (i’ / ( 1+rf)i

• i =1

• = (3000 / 1.06) + (2000/1.062) + (4000/1.063) + (2400/1.064)

• = 2830 + 1778 + 3358 + 1900• = 9866

Moderately Correlated Cash Flows

• Joint probability • p(xi to xn) = p( x1) *p ( x2/x1) * p(x3/x1,x2)….. *

p(xn/x1… to…. Xn-1)

• Example• CF0 = I = (-100000)

• Rf = 6 % = 0.06

Joint Probability CaseYear 1 Year 2 Year 2 Year 3 Year 3 CF JP

(p(1,2,3))

CF1 P1 CF2 JP2/1 CF3 JP3/2,1

30000 0.5 30000 0.8 35000 0.6 1 0.24

0.8 40000 0.4 2 0.16

40000 0.2 45000 0.5 3 0.05

0.2 50000 0.5 4 0.21

50000 0.5 50000 0.6 60000 0.7 5 0.21

70000 0.3 6 0.09

60000 0.4 75000 0.8 7 0.16

90000 0.2 8 0.04

Cash FlowsCFx Year 1 Year 2 Year 3

CF1 30000 30000 35000

CF2 30000 30000 40000

CF3 30000 40000 45000

CF4 30000 40000 50000

CF5 50000 50000 60000

CF6 50000 50000 70000

CF7 50000 60000 75000

CF8 50000 60000 90000

CF CF0 CF1 CF2 CF3 NPV JP

1 -100000 30000/1.06 30000/1.062 35000/1.063 -15612 0.24

2 -100000 30000/1.06 30000/1.062 40000/1.063 -11414 0.16

3 -100000 30000/1.06 40000/1.062 45000/1.063 1684 0.05

4 -100000 30000/1.06 40000/1.062 45000/1.063 5882 0.05

5 -100000 50000/1.06 50000/1.062 60000/1.063 42046 0.21

6 -100000 50000/1.06 50000/1.062 70000/1.063 50442 0.09

7 -100000 50000/1.06 60000/1.062 75000/1.063 63540 0.16

8 -100000 50000/1.06 60000/1.062 90000/1.063 76134 0.04

Decision Tree approach(cash flows from previous joint probability case)

Year 3

Year 1 Year 2

• 30000(0.8) • 30000(0.5) 45000(0.5) 3 0.05 1684

• 40000(0.2) 50000(0.5) 4 0.05 5882

• 50000(0.5) 50000(0.6) 60000(0.7) 5 0.21 42046

• 70000(0.3) 6 0.06 50442

• 60000(0.4)

35000(0.6) 1) JP 0.24 -15612

40000(0.4) 2 0.16 -11414

75000(0.8) 7 0.16 63540

90000(0.2) 8 0.04 76134

CF JP NPV

analysis

• The joint probability of the project having lowest NPV of -15612 is 0.24

• The joint probability of the project having highest NPV of 76134is 0.04

• The expected NPV of the project is +212702• Decision:• select

Uncorrelated cash flows

• Cash flows of various period are independent• In other words, no relationship between the

CF from one period to another

• NPV ‘ = -I + i = 1 to n (xi’ / ( 1+rf)i

• 2(NPV) = i = 1 to n (i2/ ( 1+rf)2i

• Where • NPV ‘ = expected NPV• xi’ = expected CF in year i

• rf = risk free rate of return

• I = CF0

• (NPV) = Standard Deviation of NPV• I = Standard deviation of Cash Flow for year i

• • Reason for rf : We are trying to separate time

value of money AND risk factor• NPV ‘ is computed using risk adjusted

discount rate and then viewed along with (NPV), it would result in double counting of risk factor

Example

Year CF Pi CF Pi CF Pi Xi’

Year 1 6000 0.3 10000 0.4 14000 0.3 10000

Year 2 4000 0.2 8000 0.6 12000 0.2 8000

Year 3 6000 0.3 10000 0.4 14000 0.3 10000

CF0 = -20000, rf = 0.06 ,

=-20000+10000/1.06+8000/1.062+10000/1.063 = +4948

• I2 = pi(xi-x1’) = {[0.3*(6000-10000)2] +

{[0.4*(10000-10000)2+{[0.3*(14000-10000)2]}• I

2 = 9600000

• 22 = pi(x2-x2’) = {[0.2*(4000-8000)2] +

{[0.6*(8000-8000)2+{[0.2*(12000-8000)2]}• 2

2 = 6400000

• 32 = pi(x2-x2’) = {[0.3*(6000-10000)2] +

{[0.4*(10000-10000)2+{[0.3*(14000-10000)2]}• 3

2 = 9600000

• 2 = 9600000/1.062

• +6400000/1.064

• +9600000/1.066

• =22356679

• (NPV) = 4728

Issues:

• Investment proposals differ in risk• Subjective probability distribution• Non-availability of objective evidence for

defining probability distributions• High degree of subjectivity

Practical Issues

• Conservative estimation of revenues• Safety margin in expenditure• Flexibilty in investment yardsticks• Like different post-tax rate of return, pay back period etc.,• Use of certainty index for scarce inputs• Judgement based on three point estimates – most likely,

optimistic, pessimistic• Subjective evaluation• Question remains : what is the probability that NPV of a

project are normally distributed ?

Case Study Airbus A3xx

• ?• ? Project economics• ? Break-even demand analysis• ? required rate of return• ? Realized price per plane• ? Capacity to produce• ? Market for VLA• ? Boeing’s competitive response• ? What happened ?

Sampa Video ,Inc

Case - Sampa Video

• To explore the concept of valuation• When the project can support debt capacity• What is the tax shield by debt• To find and understand• WACC• Adjusted PV• Capital cash flow• ? Vale of the firm with • all equity• 50 % debt• 25 % debt-to market value

Case-Sampa Video-Assignment Questions

• 1) What is the value of the firm assuming that the firm was entirely equity financed ? What are the annual projected cash flows ? What discount rate is appropriate ?

• 2) Value the project using the Adjusted Present Value (APV) approach assuming the firm maintains a constant 25 % debt-to-market ratio, in perpetuity

• 3)Value the firm using the APV approach assuming the firm raises $750 000 of debt to partially fund the project and keeps the level of debt constant in perpetuity

Assignment Questions - continued

• 4) What are the end-of-year debt balances implied by the 25% target debt-to-value ration ?

• 5) using the debt balances from the above, use the Capital Cash flow (CCF) approach to value the project.

• 6) How do the values from the APV,WACC and CCF approaches compare ? How do the assumptions about the financial policy differ across the three approaches?

• 7) Given the assumptions behind APV,WACC and CCF, when is one method more appropriate or easier to implement than the others ?

Case-Sampa Videos-All Equity Valuation of the Project

• Assumptions : initial investment of 1,500,000 USD is made immediately

• No NWC • 5% growth in free cash flow as in perpetuity after

2005• Market risk premium = long run excess return of USA

equities = 7.2 %• Unlevered cost of equity = expected return of assets =

(5%+ 1.5*7.2) = 15.8 %• Value of unlevered firm = 1.228 M USD

Calculations

• Interest tax shield = ITS =debt*tax rate*cost of debt/debt weightage = $750000 *0.40*6.8%/6.8% = $300000

• WACC = (1-t) *kd*D/V) + (ke*E/V)• Where V = Value of firm = debt + equity• Capitalisation ratio for debt = 25 %• Capitalisation ratio for equity = 75 %• Cost of debt = kd= 6.8 %

• Beta of the firm’s assets = (weighted average beta of debt and equity)

• a = (D/D+E)* d + (E/D+E) * e

• Levered beta = e = 1.92• Levered return on equity = 18.8%

• WACC = (0.25*6.8) + (0.75 * 18.8) = 15.12 %• • Required rate of return on levered equity

using M&M proposition :• Re = Ra + (D/E) * (Ra-Rd) = 18.8 %

• Value of the project with no debt = NPV = $1,228,500

• Value of the project with 50% debt $750000 = NPV = $1,528,500

• Value of the project with 25 % D/V forever = NPV = $1,470,000

Free cash flows Item Source 2002E 2003E 2004E 2005E 2006e

Ebiat (A) From Ex 2

-12000 81000 201000 339000 495000

Depreciation

(b) From Ex 3

200000 225000 250000 275000 300000

Capex © from ex 4

300000 300000 300000 300000 300000

Change in NWC

(d) From Ex 5

0 0 0 0 0

Free CF (d) = a+b-c-d

-112000 6000 151000 314000 495000

Expected ReturnsItem Sourc Value

Asset beta (a) From Ex 2 1.5

Risk free rate (b) From Ex 2 5%

Market risk premium © from Ex 2 7.2

Asset return (d) = (b)+(a)*c 15.8

Debt beta (e) From Ex 2 0.25

Debt % (f) From case 25%

Debt return (g) = (b)+(f)*© 6.8%

Debt beta contribution (h) = (e)* f 0.06

Equity beta (i) = (a) –(h) /j) 1.92

Equity % J=1-f 75%

Equity return K=(b) + i*c 18.8%

Equity beta contribution (l) = i*j 1.44

Asset beta (m) = (h) + (l) = (a) 1.5

Tax rate (n) From Ex 2 40%

WACC (o) = (1-n) *f*g+j*k 15.1

All equity ValuationItem Source 2002 E 2003 2004E 2005E 2006E Terminal

ValueFree CF -112000 6000 151000 314000 495000 4821500

Discount Rate

15.8% 15.8% 15.8% 15.8% 15.8% 15.8%

Discount factor

PV of $1 0.864 0.746 0.644 0.556 0.480 0.480

PV (d)=a*c -96700 4500 97200 174600 237700 2311100

Total PV of FCF

2728500

Less initial investment

1500000

Net PV +1228500

ALL Equity – Terminal Value

• = 2006E (estimated) cash flow ,• Grown by 5 %• And discounted at R-g• Where R is the appropriate discount rate (15.8%)• And g is the growth rate (5%)• = 495*1.05/(0.158-0.050)• = 495*1.05/0.108• =4812.5 * 1000 $

WACC valuation with a target debt-to-value ratio of 25 %

Item Source 2002E 2003E 2004E 2005E 2006E Terminal Value

Free CF -112000 6000 151000 314000 495000 5135900

Discount Rate

15.1% 15.1% 15.1% 15.1% 15.1%

Discount Factor

C=PV of $1

0.869 0.755 0.655 0.569 0.495 0.495

PV D=a*c -97300 4500 99000 178800 244800 2540200

Total PV of FCF

E=sum of d

2970000

Less initial investment

F from case

1500000

NPV G=e-f + 1470000

Terminal value with a target debt-value ratio of 25%

• Equal to 2006E (estimated) cash flow• Grown by 5%• And discounted by R-g• Where R is the appropriate discount rate(15.1%)• And g is the growth rate (5%)• =495*1.05/(0.1512 -0.050) • = 495 * 1.05 /(0.1012)• =5135.9 ( including rounding-off error)

Capital Cash flow valuation with a target debt-to-valuation ratio of 25%

Item Source 2002E 2003E 2004E 2005E 2006E Terminal Value

PV of FCF

case 2970000 3531000 4058000 4521600 4891300

Debt at 25% of value

B=25% of a

742500 882800 101470 130400 1222800

Debt rate

C from ex 3

6.80% 6.80% 6.80% 6.80% 6.80%

Tax rate D from Ex 3

40% 40% 40% 40% 40%

Expected interest tax shield

E=b*c*d 20200 24000 27600 30700 33300

Free tax cash flow

F -112000 6000 151000 314000 495000

Expected interest tax shield

G=e 20200 24000 27600 30700 33300

Capital cash flow

H=g+f -91800 30000 178600 344700 528300 5135900

Discount rate

B 15.8% 15.8% 15.8% 15.8% 15.8%

Discount factor

C=PV of $1

0.864 0.746 0.644 0.556 0.480 0.480

PV D=a*c -79300 22400 115000 191700 253700 2466400

Total PV of FCF

E=sum of d

2970000

Initial Investment

F from case

1500000

NPV G=e-f 1470000

Terminal cash flows with a target debt-to value ratio of 25%

• Equal to 2006E (estimated) cash flows• Grown by 5%• And discounted at R-r• Where R is the appropriate discount rate• And g is the growth rate @5%• = free cash flow + expected interest tax shield =

495+33.3 = 528.3 *1.05/(0.158-0.05)• =528.3*1.05/0.108• = 5135.9*1000 $

Normal Distribution

• A continuous probability distribution function (pdf)

• Which is systematic with respect to mean ()• 2 parameters– Mean () of population or x’ of sample distribution– SD of population or of sampleYou have to convert the normal distribution to

standard normal form

Pi

Xi

=0 Z

• Z=(x-) /

• Or ( Xi -x )/ • Where • z = standard normal variant or standard difference• Xi = normal variant or random variable• In normal distribution, • = 1 and x = 0

• For continuous pdf, you have to calculate by area , NOT by height

• Because, probability (p ) at a particular point or place is zero

• A range has to be given for x• In ALL standard normal variates or form, the mean

() or standard deviation (SD) are expected to be equal to 0 and 1 respectively

• In ANY normal distribution, we need parameters () and