chapter ii - capital investment
TRANSCRIPT
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