Decision Making.ppt - Sihombing15's (Haery Sihombing) economy studies. Decision making is fraught...
Transcript of Decision Making.ppt - Sihombing15's (Haery Sihombing) economy studies. Decision making is fraught...
4/18/2011
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Engineering Economy
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
Chapter : DECISION MAKING
The objective of this chapter is to discuss and illustrate several probabilistic methods that are useful in analyzing risk and
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
y guncertainty associated with
engineering economy studies.
Decision making is fraught with risk and uncertainty.
• Decisions under risk are those where the decision maker can estimate probabilities of occurrence of particular outcomes.
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
p• Decisions under uncertainty are those
where estimates of probabilities of the several unknown future states cannot be estimated.
Four major sources of uncertainty are present in engineering economy studies
• Possible inaccuracy of cash-flow estimates• The type of business involved in relation to
the future health of the economy
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
the future health of the economy• The type of physical plant and equipment
involved• The length of the study period used in the
analysis
Factors such as revenues, costs, salvage values, etc., can often be considered
random variables.
For discrete random variables X, the probability X takes on any particular value
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xi is
where
Some other properties of discrete random variables.
Probability mass function
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Cumulative distribution function
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For continuous random variables…
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The probability that X takes on any particular value is 0.
The cumulative distribution function (CDF) is
which leads to
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
which leads to
The expected value (mean, central moment), E(X), and variance (measure of dispersion), V(X), of a random variable
X, are
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
Some properties of the mean and variance.
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
Acme manufacturing has installed a much-needed new CNC machine. The initial investment in this
machine is $180,000 and annual expenses are $12,000. The life of the machine is expected to be 5
years, with a $20,000 market value at that time. Acme’s MARR is 10%. Possible revenues follow
the probabilities given below
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the probabilities given below.
Revenue Probability$35,000 0.1$44,000 0.3$50,000 0.4
$60,000 0.2
What are the expected value and variance of Acme’s revenue?
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What are the expected value and variance of Acme’s revenue?
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Probability tree diagrams can display prospective cash flows, and their respective probabilities that occur in each time period.
End-of-Year0 1 2
0.4 $1,100
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0.5
0.5
0.6
$700
$825
$800
$1,000
$9000.3
0.4
0.3
0.3
0.7
$625
$700
$750-$2,000
Continuous random variables present special challenges, and special opportunities.
• Two frequently used assumptions are– cash-flow amounts are distributed according to a
normal distribution, and– cash flows are statistically independent.
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
Thus, if
then
Apply these concepts to cash flows over time to fine the expected PW, and SD of PW, for the expected values and standard
deviations in the table below. (Use i=8%.)
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End of Year,k
Expected Value of Net Cash Flow, Fk
SD of Net Cash Flow, Fk
0 -$10,000 $01 $4,000 $4002 $4,000 $6003 $5,000 $650
For the expected PW.
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For V(PW) and SD(PW)
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With our estimates of cash flow variables, and using the normal distribution, we can find the
probability of events about the random variable occurring. For instance, in the
previous example, what is the probability that th PW f th h fl i iti ? R ll
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the PW of the cash flows is positive? Recall
The standard normal (mean=0 and standard deviation=0) variable, Z, is
defined as
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For our problem, since our random variable is present worth,
The probability is found by looking up a value in the standard normal table
(Appendix E).
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So
Another way to handle uncertainty is to use Monte Carlo simulation.
• Based on the probability of different outcomes for each random variable, a particular value is randomly generated.
• The numbers generated for all random variables constitute an instance or realization reflecting a particular outcome
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an instance or realization reflecting a particular outcome.• Hundreds or thousands of these instances are generated,
and these are examined to assist in decision making.• A caution: these will yield long term, average results, and
you will be able to see the variation over time. However, your decision may be a one-time decision, so don’t expect the “average” outcome to be your outcome.
Remember Acme Manufacturing and the new CNC machine. The revenues and associated probabilities are given below, and also now the expenses have been given probabilities.
Revenue Probability Expenses Probability
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
Revenue Probability$35,000 0.1$44,000 0.3$50,000 0.4
$60,000 0.2
Expenses Probability$6,000 0.1$8,000 0.4$11,000 0.3
$14,000 0.2
Using the RAND() function in Excel, and following the probabilities on the previous
slide, we generated 1000 revenue and expense values (each using a separate random number
for independence), and found the resulting PW for Acme. We found the following useful
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
ginformation (we could discuss a lot more).
Not a good deal!1. The average PW was -$20,6702. The number of positive PW values, out of the
1000 simulated, was 216.
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Monte Carlo simulation is very flexible.• The example used a discrete distribution, but there is a way
to use any discrete or continuous distribution. If Excel is used, it has several special functions that generate random variates (e.g., NORMINV to generate normal random variates and BETAINV for beta random variates.)Th l k h f f
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• There are many ways to look at the performance of an alternative using Monte Carlo simulation (we examined only two in the previous example). Graphs can be especially valuable.
• Generate lots of data, through many trials. When average values converge to a fairly constant amount, you probably have enough data.
Decision trees can be helpful in examining sequential decision problems
with outcomes that vary over time.• Break down large problems into a series of smaller
problems.
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• Provide objective analysis that explicitly considers the risk and effect of the future.
• A decision tree is built from a series of nodes, where decisions are made (square symbols) or chance outcomes are noted (circle symbols), and branches, which specify outcomes.
Revisiting Acme Manufacturing and their CNC machine decision.
Acme already has a machine they can use that is adequate for their needs. However, they wish to make a decision about their purchase of the new machine. The time horizon is 4 years, and they know that they won’t replace their existing machine if there is only one years of the
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
their existing machine if there is only one years of the time horizon remaining. So, they will make a decision today, and at the end of the first and second years regarding the new machine. We assume no chance elements, only a deterministic decision tree. The tree on the following slide depicts the situation. Cash inflows and durations are above the arrows, and capital investments below the arrows.
Acme’s decision tree.
Old: 30k/yr1 yr
Old: 25k/yr1 yr
Old: 20k/yr2 yr
-15k -20k -30k 0 1 2
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New: 55k/yr4 yr
New: 70k/yr3 yr
New: 70k/yr2 yr
Analyze decision trees from the last decision, backward to the first.
Decision Point Alternative Monetary Outcome Choice
2Old $20k(2)-$30k = $10k
Old
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New $70k(2)-$180k = -$40k
1Old $10k+$25k(1)-$20k = $15k
NewNew $70k(3)-$180k = $30k
0Old $30k+$30k(1)-$15k = $45k
OldNew $55k(4)-$180k = $40k
Acme should keep their current machine for one more year.
• The table reveals that at decision point 2, if Acme still has the old machine, they should keep it.
• At decision point 1, purchasing the new machine provides greater return than keeping the old one.
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
• At decision point 0 (today), it is more advantageous to keep the old (current) machine for one more year, given that the best decision at decision point 1 is to get the new machine.
• It would be appropriate to include the time value of money, so cash flows should be discounted to the present and the analysis performed again.
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Adding probabilities to decision trees.• Most decisions also include chance outcomes, so we use
chance nodes.• All alternatives emanating from either a decision or chance
node must be mutually exclusive (no more than one may be selected) and exhaustive (contain all possible outcomes).
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• The probabilities on the branches from a chance node must sum to one (like probability tree diagrams).
• The value assigned to a chance node is the expected value of the possible outcomes along each of the branches leaving the node.
Mitselfik, Inc. believes new scheduling software (at a cost of $150,000) will allow them to better manage
product flow and therefore increase sales. The projection of increased annual sales (for the next 5 years), and the associated probabilities, are below. The following slide shows the decision tree, and
resulting PW at a MARR of 12%
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
resulting PW at a MARR of 12%.
Increased annual sales Probability$75,000 0.35$60,000 0.45
$40,000 0.15$30,000 0.05
New
Sales increase
75,000 $120,360
$68,992
PW Probability
0.45
0.15
0.35
40,000
60,000 $66,288
$5 808
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
software
0.05 30,000
$0
$0Current software
-$5,808
-$41,856 $68,992
Mitselfik should purchase the software; the expected PW of the investment is
$68,992.• The PW of each annual sales increase amount is
given in the far right of the tree.• The expected value of the annual sales increase is
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Engineering Economy, Fourteenth EditionBy William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
The expected value of the annual sales increase is $218,992 (the sum of the probabilities times the respective PW).
• Subtracting the initial cost yields a net PW of $68,992, which is superior to “do nothing” (which is eliminated, signified by the double lines on that decision branch).
How much would we pay to have perfect information about the future?
• Perhaps with additional information we might have a better estimate of sales, or exact knowledge of sales (“perfect” information).
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• The cost of reducing the uncertainty must be balanced against the value.
• Perfect information is not obtainable, so the expected value of perfect information (EVPI) is an upper limit on what we would consider spending.
• EVPI = the value of the decision based on perfect information minus the value without the information.
Mitselfik could make the right decision with perfect information.
Decision with perfect information
Increased sales Probability Decision Outcome
Prior decision
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y$75,000 0.35 Purchase $120,360 $120,360
$60,000 0.45 Purchase $66,288 $66,288
$40,000 0.15 No purchase $0 -$5,808
$30,000 0.05 No purchase $0 -$41,856
Expected Value $71,956 $68,992
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How much should Mitselfik pay for perfect information?
• If Mitselfik had perfect information they would decide not to purchase the software if the increase in sales were $30,000 or $40 000
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$40,000.• EVPI = $71,956 - $68,992 = $2,964.• It is possible to find the expected value of
any additional information that is not “perfect.” This is discussed in detail in the text.
Decision trees can be used to assist in analyzing real options.
• Real options, similar to financial call options, allow decision makers to invest capital now or postpone all or part of the
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p p p pinvestment until later.
• When a firm makes an irreversible capital investment that could be postponed, it exercises its call option, which has value by virtue of the flexibility it gives the firm.
A good example of postponable investment is a plant addition.
Consider Mitselfik, Inc., which in addition to purchasing software needs to expand the facility. It can complete the entire expansion
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y p pnow at a cost of $7 million, leading to anticipated net cash flows (after tax) of $1.2 million for the next ten years. At an after-tax MARR of 12%, is this an attractive investment?
This does not look attractive for Mitselfik.
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What if demand should change, rising higher than originally anticipated? Perhaps Mitselfik, Inc. could be prepared and have an option available that would allow them to respond to this increased demand.
Assume that demand could balloon to $3.5 million (or, it could go to zero). The original expansion could handle sales of $1.5 million, and Mitselfik could acquire additional space (an option) to handle the additional increased demand of $2 million at a cost of $4 million If this additional demand did not
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cost of $4 million. If this additional demand did not materialize, the original expansion could be sold for $1 million. What is the best decision for Mitselfik?
This is modeled as a decision tree on the next slide.
Mitselfik’s decision tree
Buildinge
$9.20mil
-$3.79mil
$1.48mil
-$6.55mil Sales
$2.1mil
Add
Continue
Abandon
Add
PW
$9.20mil
-$0.22mil
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gxpansion
-$10.6mil
-$0.22mil
-$6.11mil with option
Negligible sales
Sales $1.2mil
Add
Continue
Continue
Abandon
Abandon
-$7.00mil
-$6.11mil
Expected value = $0.96mil if all outcomes are equally likely.
-$6.11mil
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Mitselfik should strongly consider investing since the option to add capacity
can provide a positive return.
• Using the decision tree model to assess the option reveals that the expected return given the best
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reveals that the expected return, given the best decisions along the way, is $960,000.
• However, losses could be large, and there is a 2/3 chance of a loss (if all outcomes are equally likely). Issues like this are covered in Chapter 14.
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Decision-Making Tools
Decision-Making Tools
PowerPoint presentation to accompany PowerPoint presentation to accompany
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o e o t p ese tat o to acco pa yo e o t p ese tat o to acco pa yHeizer and Render Heizer and Render Operations Management, 10e Operations Management, 10e Principles of Operations Management, 8ePrinciples of Operations Management, 8e
PowerPoint slides by Jeff Heyl
OutlineOutline
The Decision Process in OperationsFundamentals of Decision Making
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Decision Tables
Outline Outline –– ContinuedContinued
Types of Decision-Making Environments
Decision Making Under Uncertainty
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Decision Making Under RiskDecision Making Under CertaintyExpected Value of Perfect Information (EVPI)
Outline Outline –– ContinuedContinued
Decision TreesA More Complex Decision TreeUsing Decision Trees in Ethical
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gDecision MakingThe Poker Decision Problem
Learning ObjectivesLearning ObjectivesWhen you complete this module you When you complete this module you should be able to:should be able to:
1. Create a simple decision tree2 B ild d i i t bl
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2. Build a decision table3. Explain when to use each of the three
types of decision-making environments
4. Calculate an expected monetary value (EMV)
Learning ObjectivesLearning ObjectivesWhen you complete this module you When you complete this module you should be able to:should be able to:
5. Compute the expected value of perfect information (EVPI)
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perfect information (EVPI)6. Evaluate the nodes in a decision tree7. Create a decision tree with sequential
decisions
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Decision to Go All InDecision to Go All In
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The Decision Process in The Decision Process in OperationsOperations
1. Clearly define the problems and the factors that influence it
2. Develop specific and measurable objectives
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objectives3. Develop a model4. Evaluate each alternative solution5. Select the best alternative6. Implement the decision and set a
timetable for completion
Fundamentals of Fundamentals of Decision MakingDecision Making
1. Terms:a. Alternative – a course of action or
strategy that may be chosen by the
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strategy that may be chosen by the decision maker
b. State of nature – an occurrence or a situation over which the decision maker has little or no control
Fundamentals of Fundamentals of Decision MakingDecision Making
2. Symbols used in a decision tree:a. – decision node from which one
of several alternatives may be
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of several alternatives may be selected
b. – a state-of-nature node out of which one state of nature will occur
Decision Tree ExampleDecision Tree Example
Favorable market
Unfavorable market
A decision node A state of nature node
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Favorable market
Unfavorable market
Construct small plant
Figure A.1
Decision Table ExampleDecision Table Example
State of NatureAlternatives Favorable Market Unfavorable Market
Construct large plant $200,000 –$180,000
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Table A.1
Construct small plant $100,000 –$ 20,000Do nothing $ 0 $ 0
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DecisionDecision--Making Making EnvironmentsEnvironments
Decision making under uncertaintyComplete uncertainty as to which state of nature may occur
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Decision making under riskSeveral states of nature may occurEach has a probability of occurring
Decision making under certaintyState of nature is known
UncertaintyUncertainty1. Maximax
Find the alternative that maximizes the maximum outcome for every alternative
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Pick the outcome with the maximum numberHighest possible gainThis is viewed as an optimistic approach
UncertaintyUncertainty2. Maximin
Find the alternative that maximizes the minimum outcome for every alternative
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Pick the outcome with the minimum numberLeast possible lossThis is viewed as a pessimistic approach
UncertaintyUncertainty
3. Equally likelyFind the alternative with the highest average outcome
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Pick the outcome with the maximum numberAssumes each state of nature is equally likely to occur
Uncertainty ExampleUncertainty ExampleStates of Nature
Favorable Unfavorable Maximum Minimum RowAlternatives Market Market in Row in Row AverageConstruct
large plant $200,000 -$180,000 $200,000 -$180,000 $10,000Construct
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1. Maximax choice is to construct a large plant2. Maximin choice is to do nothing3. Equally likely choice is to construct a small plant
Maximax Maximin Equally likely
Constructsmall plant $100,000 -$20,000 $100,000 -$20,000 $40,000
Do nothing $0 $0 $0 $0 $0
RiskRisk
Each possible state of nature has an assumed probabilityStates of nature are mutually exclusive
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Probabilities must sum to 1Determine the expected monetary value (EMV) for each alternative
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Expected Monetary ValueExpected Monetary Value
EMV (Alternative i) = (Payoff of 1st state of nature) x (Probability of 1st
state of nature)
+ (Payoff of 2nd state of
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+ (Payoff of 2nd state of nature) x (Probability of 2nd
state of nature)
+…+ (Payoff of last state of nature) x (Probability of last state of nature)
EMV ExampleEMV ExampleTable A.3
States of NatureFavorable Unfavorable
Alternatives Market MarketConstruct large plant (A1) $200,000 -$180,000Construct small plant (A2) $100,000 -$20,000
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1. EMV(A1) = (.5)($200,000) + (.5)(-$180,000) = $10,0002. EMV(A2) = (.5)($100,000) + (.5)(-$20,000) = $40,0003. EMV(A3) = (.5)($0) + (.5)($0) = $0
p ( 2) $ , $ ,Do nothing (A3) $0 $0Probabilities .50 .50
EMV ExampleEMV ExampleTable A.3
States of NatureFavorable Unfavorable
Alternatives Market MarketConstruct large plant (A1) $200,000 -$180,000Construct small plant (A2) $100,000 -$20,000
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1. EMV(A1) = (.5)($200,000) + (.5)(-$180,000) = $10,0002. EMV(A2) = (.5)($100,000) + (.5)(-$20,000) = $40,0003. EMV(A3) = (.5)($0) + (.5)($0) = $0
Best Option
p ( 2) $ , $ ,Do nothing (A3) $0 $0Probabilities .50 .50
CertaintyCertainty
Is the cost of perfect information worth it?Determine the expected value of
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Determine the expected value of perfect information (EVPI)
Expected Value of Expected Value of Perfect InformationPerfect Information
EVPI is the difference between the payoff under certainty and the payoff under risk
EVPI = –Expected value
with perfect Maximum EMV
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pinformation EMV
Expected value with perfect information (EVwPI)
= (Best outcome or consequence for 1st state of nature) x (Probability of 1st state of nature)
+ Best outcome for 2nd state of nature) x (Probability of 2nd state of nature)
+ … + Best outcome for last state of nature) x (Probability of last state of nature)
EVPI ExampleEVPI Example
1. The best outcome for the state of nature “favorable market” is “build a large facility” with a payoff of $200,000. The best outcome for “unfavorable” is “do
$
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nothing” with a payoff of $0.
Expected value with perfect information
= ($200,000)(.50) + ($0)(.50) = $100,000
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EVPI ExampleEVPI Example
2. The maximum EMV is $40,000, which is the expected outcome without perfect information. Thus:
M i
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= $100,000 – $40,000 = $60,000
EVPI = EVwPI – Maximum EMV
The most the company should pay for perfect information is $60,000
Decision TreesDecision Trees
Information in decision tables can be displayed as decision treesA decision tree is a graphic display of the decision process that indicates decision
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alternatives, states of nature and their respective probabilities, and payoffs for each combination of decision alternative and state of natureAppropriate for showing sequential decisions
Decision TreesDecision Trees
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Decision TreesDecision Trees1. Define the problem2. Structure or draw the decision tree3. Assign probabilities to the states of
nature
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4. Estimate payoffs for each possible combination of decision alternatives and states of nature
5. Solve the problem by working backward through the tree computing the EMV for each state-of-nature node
Decision Tree ExampleDecision Tree Example= (.5)($200,000) + (.5)(-$180,000)EMV for node 1
= $10,000
Payoffs
$200,000Favorable market (.5)
Unfavorable market (.5)1
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EMV for node 2= $40,000 = (.5)($100,000) + (.5)(-$20,000)
-$180,000
$100,000
-$20,000
$0
Construct small plant
Unfavorable market (.5)
Favorable market (.5)
Unfavorable market (.5)2
Figure A.2
Complex Complex Decision Decision
Tree Tree ExampleExample
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Figure A.3
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Complex ExampleComplex Example
1. Given favorable survey results
EMV(2) = (.78)($190,000) + (.22)(-$190,000) = $106,400EMV(3) = (.78)($90,000) + (.22)(-$30,000) = $63,600
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The EMV for no plant = -$10,000 so, if the survey results are favorable, build the large plant
Complex ExampleComplex Example
2. Given negative survey results
EMV(4) = (.27)($190,000) + (.73)(-$190,000) = -$87,400EMV(5) = (.27)($90,000) + (.73)(-$30,000) = $2,400
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The EMV for no plant = -$10,000 so, if the survey results are negative, build the small plant
Complex ExampleComplex Example3. Compute the expected value of the
market survey
EMV(1) = (.45)($106,400) + (.55)($2,400) = $49,200
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The EMV for no plant = $0 so, given no survey, build the small plant
4. If the market survey is not conducted
EMV(6) = (.5)($200,000) + (.5)(-$180,000) = $10,000EMV(7) = (.5)($100,000) + (.5)(-$20,000) = $40,000
Decision Trees in Ethical Decision Trees in Ethical Decision MakingDecision Making
Maximize shareholder value and behave ethically
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Technique can be applied to any action a company contemplates
Yes
No
Decision Trees in Ethical Decision Trees in Ethical Decision MakingDecision Making
Yes
Is it ethical? (Weigh the affect on employees, customers, suppliers,
community verses shareholder benefit)
D ti
Do it
Don’t do it
Action outcome
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No
Yes
NoIs it ethical not to take
action? (Weigh the harm to shareholders
verses benefits to other stakeholders)No
YesDoes action
maximize company returns?
Is action legal?
Figure A.4Don’t do it
Don’t do it
Do it, but notify appropriate parties
do it
The Poker Design ProcessThe Poker Design ProcessIf T. J. folds,
EMV = (.80)($99,000)= $79,200
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If T. J. calls,EMV = .20[(.45)($853,000) - Phillips’ bet of $422,000]
= .20[$383,850 - $422,000]= .20[-$38,150] = -$7,630
Overall EMV = $79,200 - $7,630 = $71,750
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All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying,
recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America.
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Make or Buy DecisionMake or Buy Decision
EVALUATING ALTERNATIVE
Example:1• Ayam Galing restaurant is experiencing a boom in business.
The owner expects to serve 80,000 meals this year. Although the kitchen is operating at 100 % capacity, the dining room can handle 105,000 dinners per year. Forecasted demand for the next years is 90,000 meals for next year, followed by a 10,000-meal increase in each of the succeeding years. One alternative is to expand both the kitchen and the dining room now, b i i th i it t 130 000 l Thbringing their capacity up to 130,000 meals per year. The initial would be RM 200,000, made at the end of this years (year 0). The average meal is priced at RM 10, and the before-tax profit margin is 20%. The 20% figure was arrived at by determining that, for each RM 10 meal, RM 6 covers variable costs and RM 2 goes toward fixed costs (other than depreciation). The remaining RM2 goes to pretax profit.
• What are the pretax cash flows from this project for the next five years compared to those of the base case of doing nothing?
• Solution:Recall that the base case of doing nothing results in losing all potential sales beyond 80,000. With the new capacity, the cash flow would equal the extra meals served by having a 130,000-meal capacity, multiplied by a profit of RM 2 per meal. In year 0, the only cash flow is –RM200,000 or the initial investment. In year 1, the 90,000-meal demand will be completely satisfied by the expanded capacity, so the incremental cash flow is (90,000 – 80,000)
Example:1
by the expanded capacity, so the incremental cash flow is (90,000 80,000) (RM 2) = RM 20,000. For subsequent years, the figures are as follows:
Year 2 :Demand = 100,000; Cash flow = (100,000 – 80,000) RM2 = RM40,000Year 3: Demand = 110,000; Cash flow = (110,000 – 80,000) RM2 = RM60,000Year 4 :Demand = 120,000; Cash flow = (120,000 – 80,000) RM2 = RM80,000Year 5 :Demand = 130,000; Cash flow = (130,000 – 80,000) RM2 = RM100,000
• If the capacity were smaller than the expected demand in any year, we would subtract the base case capacity from the new capacity (rather than the demand).
• NPV = -200,000 + [(20,000/1.1)] + [40,000/(1.1)2]+ [60,000/(1.1)3]+ [80,000/(1.1)4]+ [100,000/(1.1)5]
Example:1
= -200,000 +18,181.82 + 33,057.85 + 45,078.89 + 54,641.07 + 62,092.13
= RM 13,051.76
• An operations manager's staff has compiled the information below for four manufacturing alternatives (E, F, G, and H) that vary by production technology and the capacity of the machinery. All choices enable the same level of total production and have the same lifetime. The four states of nature represent four levels of consumer acceptance of the firm's products. Values in the table are net present value of future profits in millions of dollars. Forecasts indicate that there is a 0.1 probability of acceptance level 1, 0.2 chance of acceptance level 2, 0.4 chance of acceptance level 3, and 0.3 change of acceptance level 4.
St t f t
Example:2
States of nature
1 2 3 4 Alternative E 50 50 70 60 Alternative F 30 50 80 130Alternative G 70 80 70 60Alternative H -140 -10 150 220
• Using the criterion of expected monetary value, which production alternative should be chosen?
• The expected values are:
• E = .1*50 + .2*50 + .4*70 + .3*60 = 5 + 10 + 28 + 18= 61
• F = .1*30 + .2*50 + .4*80 + .3*130 = 3 + 10 + 32 + 39 = 84
Example:2
= 84• G = .1*70 + .2*80 + .4*70 + .3*60 = 7 + 16 + 28 + 18
= 69• H = .1 *-140 + .2*-10 + .4*150 + .3*220 = -10 -2 + 60 + 66
= 110
The highest of these occurs with production alternative H.
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• A toy manufacturer makes stuffed kittens and puppies which have relatively lifelike motions. There are three different mechanisms which can be installed in these "pets." These toys will sell for the same price regardless of the mechanism installed, but each mechanism has its own variable cost and setup cost. Profit, therefore, is dependent upon the choice of mechanism and upon the level of demand The manufacturer has in hand a
Example:3
level of demand. The manufacturer has in hand a forecast of demand that suggests a 0.2 probability of light demand, a 0.45 probability of moderate demand, and a probability of 0.35 of heavy demand. Payoffs for each mechanism-demand combination appear in the table as follows:
Example:3Demand Wind-up action Pneumatic
actionElectronic
actionLight $250,000 $90,000 -$100,000
Moderate 400,000 440,000 400,000
Heavy 650 000 740 000 780 000Heavy 650,000 740,000 780,000
• Construct the appropriate decision tree to analyze this problem. Use standard symbols for the tree. Analyze the tree to select the optimal decision for the manufacturer.
0.2Light demand
250000250,000 250000
0.45Wind-up Moderate demand
4000000 457500 400,000 400000
0.35Heavy demand
650000650,000 650000
0.2Light demand
9000090,000 90000
0.45Pneumatic Moderate demand
Example:3
eu at c ode ate de a d2 440000
475000 0 475000 440,000 440000
0.35Heavy demand
740000740,000 740000
0.2Light demand
-100000-100,000 -100000
0.45Electronic Moderate demand
4000000 433000 400,000 400000
0.35Heavy demand
780000780,000 780000The best choice is Pneumatic, $475,000.
• Earl Shell owns his own Sno-Cone business and lives 30 miles from a beach resort. The sale of Sno-Cones is highly dependent upon his location and upon the weather. At the resort, he will profit $120 per day in fair weather, $10 per day in bad weather. At home, he will profit $60 in fair weather $35 in bad weather Assume
Example:4
profit $60 in fair weather, $35 in bad weather. Assume that on any particular day, the weather service suggests a 40% chance of foul weather.
• a. Construct Earl's decision tree.• b. What decision is recommended by the expected value
criterion?
• Resort has a higher EMV ($76) than HomeExample:4
• A toy manufacturer has three different mechanisms that can be installed in a doll that it sells. The different mechanisms have three different setup costs (overheads) and variable costs and, therefore, the profit from the dolls is d d t th l f l Th
Example:5
dependent on the volume of sales. The anticipated payoffs are as follows.
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Light Demand Moderate Demand Heavy Demand
Probability 0.25 0.45 0.3Wind-up action $325,000 $190,000 $170,000 Pneumatic action $300,000 $420,000 $400,000
Example:5
Electrical action -$400,000 $240,000 $800,000
a. What is the EMV of each decision alternative?b. Which action should be selected?c. What is the expected value under certainty?d. What is the expected value of perfect information?
• (a) Wind-up =.25*$325,000 + .45*$190,000 + .3*$170,000 = $217,750Pneumatic =.25*$300,000 + .45*$420,000 + .3*$400,000 = $384,000 Electrical =.25*(-$400,000) + .45*$240,000 + .3*$800,000 =
$248,000.
Example:5
• (b) Pneumatic has the best EMV at $384,000.
• (c) Expected value under certainty is .25*$325,000 + .45* $420,000 + .3* $800,000 = $510,250
• (d) EVPI =$510,250 - $384,000 = $126,250.