Download 14) Probabilistic Cash Flow Analysis

Lecture No. 39
Chapter 12
Contemporary Engineering Economics
Copyright, © 2010
Contemporary Engineering Economics, 5th edition, © 2010
Probability Concepts for Investment
Decisions
 Random variable: variable that
can have more than one possible
value
 Discrete random variables:
random variables that take on
only isolated (countable) values
 Continuous random variables:
random variables that can have
any value in a certain interval
 Probability distribution: the
assessment of probability for each
random event
Contemporary Engineering Economics, 5th edition, © 2010
Types of Probability Distribution
 Continuous Probability
Distribution
Distribution
 Triangular distribution
 Uniform distribution
 Normal distribution
 Discrete Probability
Distribution
 Cumulative Probability
 Discrete
j
F (x)  P( X  x)   p j
j 1
 Continuous
 f(x)dx
Contemporary Engineering Economics, 5th edition, © 2010
Useful Continuous Probability Distributions in
Cash Flow Analysis
(a) Triangular Distribution
(b) Uniform Distribution
L: minimum value
Mo: mode (most-likely)
H: maximum value
Contemporary Engineering Economics, 5th edition, © 2010
Discrete Distribution -Probability Distributions for Unit
Demand (X) and Unit Price (Y) for BMC’s Project
Product Demand (X)
Unit Sale Price (Y)
Units (x)
P(X = x)
Unit price (y)
P(Y = y)
1,600
0.20
$48
0.30
2,000
0.60
50
0.50
2,400
0.20
53
0.20
Contemporary Engineering Economics, 5th edition, © 2010
Cumulative Probability Distribution for X
Unit Demand
(x)
Probability
P(X = x)
1,600
0.2
2,000
0.6
2,400
0.2
F (x)  P( X  x)  0.2, x  1,600
0.8, x  2,000
1.0, x  2,400
Contemporary Engineering Economics, 5th edition, © 2010
Probability and Cumulative Probability
Distributions for Random Variable X and Y
Unit Demand (X)
Unit Price (Y)
Contemporary Engineering Economics, 5th edition, © 2010
Measure of Expectation
 Discrete case
Event
j
E[ X ]     ( p j ) x j
j 1
 Continuous case
E[X] =  xf(x)dx
Return
(%)
Probability
Weighted
1
6%
0.40 2.4%
2
9%
0.30 2.7%
3
18%
0.30 5.4%
Expected Return (μ) 10.5%
Contemporary Engineering Economics, 5th edition, © 2010
Measure of Variation
Formula:
Variance Calculation:
μ = 10.5%
 j
2
discrete case
  (x j   ) (p j ),
j

1

Var  X    X2   H
 (x   )2 f (x)dx , continuous case


L
or
Event
Probability
Deviation
Squared
Weighted
Deviation
1
0.40
(6 – 10.5)2
8.10
2
0.30
(9 – 10.5)2
0.68
3
0.30
(18 – 10.5)2
16.88
Var  X   E  X 2   (E  X )2
Variance (σ2) = 25.66
σ = 5.07%
 x  Var  X 
Contemporary Engineering Economics, 5th edition, © 2010
Example 12.5 Calculation of Mean & Variance
Xj
Pj
Xj(Pj)
(Xj-E[X])
(Xj-E[X])2 (Pj)
1,600
0.20
320
(-400)2
32,000
2,000
0.60
1,200
0
0
2,400
0.20
480
(400)2
32,000
E[X] = 2,000
Var[X] = 64,000
  252.98
Yj
Pj
Yj(Pj)
[Yj-E[Y]]2
(Yj-E[Y])2 (Pj)
$48
0.30
$14.40
(-2)2
1.20
50
0.50
25.00
(0)
0
53
0.20
10.60
(3)2
1.80
E[Y] = 50.00
Var[Y] = 3.00
  1.73
Contemporary Engineering Economics, 5th edition, © 2010
Joint and Conditional Probabilities
P( x, y)  P( X  x Y  y) P(Y  y)
P( x , y )  P ( x ) P ( y )
P( x , y )  P(1,600,$48)
 P(x  1,600 y  $48)P(y  $48)
 (0.10)(0.30)
 0.03
Contemporary Engineering Economics, 5th edition, © 2010
Assessments of Conditional and Joint
Probabilities
Unit Price Y
$48
50
53
Marginal
Probability
0.30
0.50
0.20
Conditional
Unit Sales X
Conditional
Probability
Joint
Probability
1,600
0.10
0.03
2,000
0.40
0.12
2,400
0.50
0.15
1,600
0.10
0.05
2,000
0.64
0.32
2,400
0.26
0.13
1,600
0.50
0.10
2,000
0.40
0.08
2,400
0.10
0.02
Contemporary Engineering Economics, 5th edition, © 2010
Marginal Distribution for X
Xj
P( x )   P( x , y )
y
1,600
P(1,600, $48) + P(1,600, $50) + P(1,600, $53) = 0.18
2,000
P(2,000, $48) + P(2,000, $50) + P(2,000, $53) = 0.52
2,400
P(2,400, $48) + P(2,400, $50) + P(2,400, $53) = 0.30
Contemporary Engineering Economics, 5th edition, © 2010
Covariance and Coefficient of Correlation
Cov( X , Y )   xy
 E ( X  E[ X ])(Y  E[Y ])
 E ( XY )  E ( X ) E (Y )
  xy x y
 xy 
Cov( X , Y )
 x y
Contemporary Engineering Economics, 5th edition, © 2010
Calculating the Correlation Coefficient
between X and Y
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Meanings of Coefficient of Correlation
 Case 1: 0 <ρXY < 1
 Positively correlated – When X increases in value, there is
a tendency that Y also increases in value. When ρXY = 1, it
is known as a perfect positive correlation.
 Case 2: ρXY = 0
 No correlation between X and Y. If X and Y are statistically
independent each other, ρXY = 0.
 Case 3: -1 < ρXY < 0
 Negatively correlated – When X increases in value, there is
a tendency that Y will decrease in value. When ρXY =-1, it is
known as a perfect negative correlation.
Contemporary Engineering Economics, 5th edition, © 2010
1.
2.
3.
4.
How to develop a probability distribution of NPW
How to calculate the mean and variance of NPW
How to aggregate risks over time
How to compare mutually exclusive risky alternatives
Contemporary Engineering Economics, 5th edition, © 2010
Example 12.6 Probability
Distribution of an NPW
Step 1:
Item
Express After-Tax Cash
Flow as a Function of
Unknown Unit Demand (X)
and Unit Price (Y).
0
1
2
3
4
5
X(1-0.4)Y
0.6XY
0.6XY
0.6XY
0.6XY
0.6XY
0.4 (dep)
7,145
12,245
8,745
6,245
2,230
-9X
-9X
-9X
-9X
-9X
-6,000
-6,000
-6,000
-6,000
-6,000
0.6X(Y-15)
+1,145
0.6X(Y-15)
+6,245
0.6X(Y-15)
+2,745
0.6X(Y-15)
+245
0.6X(Y-15)
+33,617
Cash inflow:
Net salvage
Cash outflow:
Investment
-125,000
-X(1-0.4)($15)
-(10.4)($10,000)
Net Cash Flow
-125,000
Contemporary Engineering Economics, 5th edition, © 2010
Step 2:
 Develop an NPW
Function Based on After-Tax
Project Cash Flows.
 Cash Inflow:
PW(15%) = 0.6XY(P/A, 15%, 5) + $44,490
= 2.0113XY + $44,490
 Cash Outflow:
PW(15%) = $125,000 + (9X + $6,000)(P/A, 15%, 5)
= 30.1694X + $145,113.
 Net Cash Flow:
PW(15%) = 2.0113X(Y - $15) - $100,623
Contemporary Engineering Economics, 5th edition, © 2010
Step 3:
Event No.
x
y
P[x,y]
Cumulative
Joint
Probability
NPW
1
1,600
$48.00
0.06
0.06
$5,574
2
1,600
50.00
0.10
0.16
12,010
 Sample Calculation:
3
1,600
53.00
0.04
0.20
21,664
X = 1,600
4
2,000
48.00
0.18
0.38
32,123
 Y = $48
5
2,000
50.00
0.30
0.68
40,168
 PW(15%) =
6
2,000
53.00
0.12
0.80
52,236
7
2,400
48.00
0.06
0.86
58,672
8
2,400
50.00
0.10
0.96
68,326
9
2,400
53.00
0.04
1.00
82,808
 Calculate the
NPW for
Each Event with PW(15%) =
2.0113X(Y - $15) - $100,623
2.0113(1,600)(48 – 15) $100,623
= $5,574
Contemporary Engineering Economics, 5th edition, © 2010
Step 4:
 Plot the
NPW Probability
Distribution Assuming X and
Y are Independent
Contemporary Engineering Economics, 5th edition, © 2010
Step 5:
 Calculation of the
Mean of the NPW
Distribution.
Event
No.
x
y
P[x,y]
Cumulative
Joint
Probability
1
1,600
$48.00
0.06
0.06
$5,574
$334
2
1,600
50.00
0.10
0.16
12,010
1,201
3
1,600
53.00
0.04
0.20
21,664
867
4
2,000
48.00
0.18
0.38
32,123
5,782
5
2,000
50.00
0.30
0.68
40,168
12,050
6
2,000
53.00
0.12
0.80
52,236
6,268
7
2,400
48.00
0.06
0.86
58,672
3,520
8
2,400
50.00
0.10
0.96
68,326
6,833
9
2,400
53.00
0.04
1.00
82,808
3,312
NPW
E[NPW] =
Contemporary Engineering Economics, 5th edition, © 2010
Weighted
NPW
$40,168
Step 6:
 Calculation
of the
Variance of the NPW
Distribution.
Event
No.
x
y
P[x,y]
NPW
Weighted
(NPW-E[NPW])
1
1,600
$48.00
0.06
$5,574
1,196,769,744
$71,806,185
2
1,600
50.00
0.10
12,010
792,884,227
79,228,423
3
1,600
53.00
0.04
21,664
342,396,536
13,695,861
4
2,000
48.00
0.18
32,123
64,725,243
11,650,544
5
2,000
50.00
0.30
40,168
0
0
6
2,000
53.00
0.12
52,236
145,631,797
17,475,816
7
2,400
48.00
0.06
58,672
342,396,536
20,543,792
8
2,400
50.00
0.10
68,326
792,884,227
79,288,423
9
2,400
53.00
0.04
82,808
1,818,132,077
72,725,283
Var[NPW] =
366,474,326
(NPW-E[NPW])2
σ = $19,144
Contemporary Engineering Economics, 5th edition, © 2010
Aggregating Risk Over Time
 Approach: Determine the
mean and variance of
cash flows in each period,
and then aggregate the
risk over the project life
in terms of NPW.
0
1
2
3
4
5
E[NPW]
Var[NPW]
NPW
Contemporary Engineering Economics, 5th edition, © 2010
Case 1: Independent Random Cash Flows
E[ PW (i )] 
N

n 0
Var[ PW (i )] 
E ( An )
(1  i ) n
N

n 0
Var( An )
(1  i ) 2 n
where
i = a risk-free discount rate,
An = net cash flow in period n,
E[An ] = expected net cash flow in period n
Var[An ] = variance of the net cash flow in period n
E[PW(i)] = expected net present worth of the project
Var[PW(i)] = variance of the net present worth of the project
Contemporary Engineering Economics, 5th edition, © 2010
Case 2: Dependent Cash Flows
N
E (An )
E[PW(i)]  
n
n  0 (1  i )
Contemporary Engineering Economics, 5th edition, © 2010
Example 12.7
Aggregation of Risk
Over Time
 Net Cash Flow Statement Using the Generalized Cash Flow
Approach
0
Contemporary Engineering Economics, 5th edition, © 2010
1
2
3
Case 1: Independent Cash Flows
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Case 2: Dependent Cash Flows
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Normal Distribution
Assumption
 The
distribution of a sum
of a large number of
independent variables is
approximately normal –
Central-Limit-Theorem.
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NPW Distribution with ±3σ
Contemporary Engineering Economics, 5th edition, © 2010
Expected Return/Risk Trade-off
Probability (%)
Investment A
Investment B
-30
-20
-10
0
10
20
30
Return (%)
Contemporary Engineering Economics, 5th edition, © 2010
40
50
Example 12.8 Comparing
Risky Mutually Exclusive
Projects
 Green Engineering has
developed a prototype
conversion unit that allows
a motorist to switch from
gasoline to compressed
natural gas.
 Four models with
different NPW distributions
at MARR = 10%.
 Find the best model to
market.
Event (NPW)
(unit: thousands)
Probabilities
Model 1
Model 2
Model 3
Model 4
1,000
0.35
0.10
0.40
0.20
1,500
0
0.45
0
0.40
2,000
0.40
0
0.25
0
2,500
0
0.35
0
0.30
3,000
0.20
0
0.20
0
3,500
0
0
0
0
4,000
0.05
0
0.15
0
4,500
0
0.10
0
0.10
Contemporary Engineering Economics, 5th edition, © 2010
Comparison Rule
If EA > EB and VA  VB, select
A.
If EA = EB and VA  VB, select
A.
If EA < EB and VA  VB, select
B.
If EA > EB and VA > VB, Not
clear.
Model Type
E (NPW)
Var (NPW)
Model 1
$1,950
747,500
Model 2
2,100
915,000
Model 3
2,100
1,190,000
Model 4
2,000
1,000,000
Model 2 vs. Model 3
Model 2 vs. Model 4
Model 2 vs. Model 1
Contemporary Engineering Economics, 5th edition, © 2010
Model 2 >>> Model 3
Model 2 >>> Model 4
Can’t decide
Mean-Variance Chart Showing Project
Dominance
Contemporary Engineering Economics, 5th edition, © 2010
Summary
 Project risk—the possibility that an investment project
will not meet our minimum return requirements for
acceptability.
 Our real task is not to try to find “risk-free” projects—
they don’t exist in real life. The challenge is to decide
what level of risk we are willing to assume and then,
having decided on your risk tolerance, to understand
the implications of that choice.
 Three of the most basic tools for assessing project risk
are (1) sensitivity analysis, (2) break-even analysis, and
(3) scenario analysis.
Contemporary Engineering Economics, 5th edition, © 2010

Sensitivity, break-even, and scenario analyses are
reasonably simple to apply, but also somewhat
simplistic and imprecise in cases where we must
deal with multifaceted project uncertainty.

Probability concepts allow us to further refine the
analysis of project risk by assigning numerical values
to the likelihood that project variables will have
certain values.

The end goal of a probabilistic analysis of project
variables is to produce an NPW distribution.
Contemporary Engineering Economics, 5th edition, © 2010
From the NPW distribution, we can extract such
useful information as the expected NPW value, the
extent to which other NPW values vary from , or
are clustered around the expected value,
(variance), and the best- and worst-case NPWs.
All other things being equal, if the expected
returns are approximately the same, choose the
portfolio with the lowest expected risk (variance).
Contemporary Engineering Economics, 5th edition, © 2010