PowerPoint Presentation

							Decision Analysis and
   Decision Tree


      Dr Wang ShouQing
School of Building & Real Estate
National University of Singapore
        Decision Analysis/Making
 Many instances when a decision has to be
  made without exact knowledge to problem.
 Decision Analysis: study of the rational
  factors associated & the techniques used
  to attain consistent & acceptable results.
       Techniques themselves cannot minimize the
        uncertainty, nor guarantee a precise
        knowledge about outcomes.
 Decision Making (3 steps): finding occasions;
  possible actions; then choosing an action.
 Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
Strategic/Operational Decision
 Strategic Decision:
       Is of long-term importance.
       Is fundamental to the main objectives of the
        decision-maker, eg a developer deciding to
        invest a property project.
 Operational Decision:
       Day-to-day operations of an organization, eg
        a contractor reordering raw material once
        stock levels had fallen to a certain level.
Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
            Structure of Decision
 Objectives:
       Profitability, market share, level of costs etc.
 Courses of Action (Strategies):
   Alternative courses of action must be
    available to the decision-maker.
   If this does not hold, then no decision to be
    made although a problem is faced.
 Pay-off Matrix:
       Values of possible outcomes associated with
        each course of action.
Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
                       Decision Tree
 One of the tools for decision-making that
  has remained simple & effective.
 Involves structuring/evaluating of decision
  problems by presenting their 4 basics:
     Determine the possible actions which can be
      pursued but only one selected.
     Outline the events/outcomes of each action.

     Calculate value/pay-off of different actions.

     Choose the criterion upon which the
      alternative actions can be judged.
Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
             Decision Tree (con’t)

 By above structuring/evaluating, the
  decision tree is able to incorporate the
  answers to the questions :
       What is the objective?
       What are the alternatives of action?
       What is the basis for comparison?
       What are the possible outcomes of the
        alternatives?

Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
 Example 1 – Problem & Pay-Off
A property company is considering a commercial
project which will take 2 years to complete. During
which, the economy which is on the verge of a
recovery can either improve or face a collapse.
Question:           Table 1: Pay-Off Matrix
To contract            Outcome Economy        Economy
the project                     Improved(b1) Collapse(b2)
entirely alone      Action
or enter into a
                Alone (a1)                    2,000,000               -1,000,000
JV in order to
spread risks? JV (a2)                         1,400,000                -200,000
  Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
           Example 1 – Decision Tree
 Initial        Action                  Outcome                  Probability           Pay off
decision                                                      (state of nature)
 point
                                                                       0.6           2,000,000
                                 Economy improved
                      800,000
           Alone                                                       0.4           -1,000,000
                                         Collapse
 800,000
                      760,000 Economy improved                         0.6           1,400,000

           Joint Venture
                                         Collapse                      0.4           -200,000

                                Sequential and logical order

  Time nearest the present                                            The most distant future

      Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
     Decision Tree Components
 Action nodes        : points when an option
  to be chosen from a few alternatives.
 Event nodes        : points from which the
  different possible outcomes are accruing.
 Pay-offs: the results of different courses
  of action. Usually expressed in currency.
 Probabilities: the likelihood of the future
  outcome happening. Ranges from 0
  (impossibility) to 1 (certainty).
 Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
Action Node vs Outcome Node
 An action, even if it is to be taken in the
  future, is under the control of the
  decision-maker.
 An outcome is usually a state of nature
  and is beyond the complete control of the
  decision-maker. It is usually associated
  with and represented in probabilities.
       e.g. the reactions of a competitor, the
        response of a large number of consumers or
        the state of the economy etc.
Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
    Decision Criteria - Pessimism
    Maximin/Minimax Cost Rule
     (Criterion of Pessimism)
Decision-maker holds a                 Table 2: Minimax Cost
pessimistic view of life &
assumes that whatever                     Outcome         Economy Minimum
                                                         Collapse(b2) Loss
action he takes, nature will
arrange the worst possible             Action
outcome. Hence, in the                 Alone (a1) -1,000,000
example, consider b2 and
                                       JV (a2)     -200,000 -200,000
choose a2. Snag: "do
                                       Do nothing      0       0
nothing" will be best.

   Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
      Decision Criteria - Optimism
    Maximax Rule (Criterion of Optimism)

Decision-maker holds Table 3: Maximax Pay-Off
a optimism view (risk-
                       Outcome Economy     Maximum
lover) & assumes the          Improved(b1) Pay-Off
best outcome (b1)                           (Profit)
what ever the action. Action
                                    Alone (a1)           2,000,000            2,000,000
In the example, he will
                        JV (a2)                          1,400,000
hence plum for a1.


    Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
           Decision Criteria - EMV
 Expected Monetary Value (EMV) Approach
   To determine by the law of averages which
    action has the max. monetary expectation.
   Two important ingredients: the probabilities
    assessed & the pay-offs determined.
   For Example 1, decision should be a1 (alone):
         EMV(a1)=0.6(2000000)+0.4(1000000)= 800000
         EMV(a2)=0.6(1400000)+0.4(200000)=760000
      The EMV approach is a systematic way of
       calculating the max. expectation which will lead
       to consistent decision-making all the time.
 Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
            Decision Criteria - EOL
 Definition of Regret (Opportunity Loss)
     The deference between the minimum cost (or
      maximum pay-off) under that outcome and the
      cost (or pay-off) resulting from the action taken
      & outcome combination.
 Expected Opportunity Loss (EOL) Approach
     An alternative to the EMV approach is to look at
      the minimum EOL (sum of all regrets, weighted
      by the respective probabilities).
  Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
     Decision Criteria – EOL (con’t)
     EOL represents cost-pay-off differential
       between the action which should have
       been taken & the action which was taken.
For Example 1,         Table 4: Minimax Regret
if don’t
                         Outcome           Economy     Economy Minimum
consider the                             Improved(b1) Collapse(b2) Regret
probabilities,                               (0.6)        (0.4)
the decision is        Action
a2 which has a         Alone (a1)               0                 800,000
lower regret.          JV (a2)               600,000                 0               600,000

     Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
  Decision Criteria – EOL (con’t)
 However, EOL which combines the regret
 & the probabilities of the outcomes:
     EOL(a1) = 0  0.6 + 800,000  0.4 = 320,000
     EOL(a2) = 600,000  0.6 + 0  0.4 = 360,000
 The decision is reversed with the
 introduction of probabilities.
 A decision-maker will opt for a1 which is the
 same decision as using EMV approach.

  Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
                Roll-Back Concept
 In a decision tree, a decision is made by
  analyzing the pay-offs to a criterion.
 The process of moving from the “right” of
  the decision tree where the pay-offs are
  analyzed, along the outcome paths where
  the probabilities are incorporated, and
  from the outcome node to the decision
  node along the decision path, is called
  the roll-back concept, or backward
  induction.

Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
             Example 2 - Problem
 A contractor considers a work valued at 15,000.
 If good weather, he can carry out the work
  himself at a cost of 10,000, but if the weather
  were bad it could cost him 20,000.
 He knows that a subcontractor will complete the
  work for 12,000, irrespective of the weather.
 Based on past weather data, the probabilities
  are 0.7 for good weather & 0.3 for bad weather.
 The contractor has then to decide whether to
  undertake the work himself or to subcontract.
Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
            Example 2 – Solution
 Draw decision point with two alternative actions
 Draw chance event & two outcomes
 Calculate profit of each possible outcome
 Calculate: EMV(direct)=0.7x5000+0.3x(-5000)=2000
             EMV(sublet)=1.0x3000=3000
 Decision: to sub-contract the work




Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
             Example 3 - Problem
 A company is considering two alternative
  plans for a high-rise building complex.
 Plan I calls for a 70-storey apartment
  building & a 40-storey office building.
 Plan II envisages the construction of a
  single, 100-storey building with 45 storeys
  for offices and 55 storeys for apartments.
 The estimated costs and life-time returns
  for the two plans are in following table.
Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
Example 3 – Solution
             Plan I
     Cost       Probability
     100             0.6
      95             0.3
      90             0.1
    Return      Probability
     300             0.5
     250             0.4
     200             0.1
             Plan II
     Cost       Probability
     150           0.7
     120           0.2
     100           0.1
    Return      Probability
     450           0.2
     350           0.4
     250           0.3
     200           0.1
Basic Structure of Decision Tree
 The single-stage, static problem provides a
  model for basic structure of a decision tree.
 The decision is mainly confined to a single
  problem with one objective, eg Example 1-3.
 The action node is confined mainly to the
  various possible actions representing business
  strategies, policies & other managerial areas.
 The event node shows the different chance
  outcomes to the chosen action.
 The paths of these outcomes determine the
  ultimate pay-offs.
 Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
        Extended Decision Tree
 In real world, decision problems often consist of
  a series of decisions spread over a period of
  time before the entire problem is solved.
       For example, the decision whether to acquire further
        information in order to attain a more definite picture
        of how the outcome will be.
 Such a problem is a sequential and information
  acquisition decision & to be solved with EDT.
 The issue is not so much whether the
  information acquired will be useful, but whether
  the cost and time spent in obtaining the
  information is worthwhile.
Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
 Expected Value of Perfect Info
 EVPI = EMVUC  initial EMV
  EMVUC is EMV under certainty (or with perfect information - PI).
 PI is rarely obtainable but as a theoretical concept
  helps to set upper limit to the amount that might
  be worth spending on acquisition of information.
 Example 4 – game of tossing a coin: The stake of
  each game is $2. A win will get $5, and a loss get nothing.
      EMV without PI = 0.5 x (5-2) + 0.5 x (-2) = $0.5
      EMV for not playing = 0
  Suppose a crystal-ball gazer is able to predict accurately:
      EMV (perfect Information) = 100% x (5-2) = $3
  $3-$0.5=$2.5 is max fee (EVPI) one would pay for the info.
  Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
  Represented in Decision Tree
 Whether to acquire the information or not:
    Has one possible initial decision.
    Concerns the purchasing of info of a fixed
     quantity with defined quality at a pre-said price.
 If information should be acquired, then:
    Which source & what level of info to obtain.
    The decision is of a more complex structure.
    The event nodes resulting from a decision to
      acquire info will show messages from the info
      source which will in turn influence the outcome.
  Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
              Sample Information
 Decision situations are almost invariably
  analyzed in an uncertain environment.
 To find the true state, additional info may be
  beneficial to the decision-making.
 Info may be acquired to achieve a partial
  solution because it is not practical to obtain all
  (perfect) info which will result in absolute
  certainty of the outcomes .
 The info can be of the form of sample surveys
  of a market, experiments, tests or further
  research in a particular field.
Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
 Prior vs Posterior Probabilities
 The extra info is then combined with the original or
  prior probabilities (assessed with existing info) to
  form revised or posterior probabilities which may be
  less uncertain.
 The intention
  is to use the
  info to weight
  the prior
  probabilities
  resulting in
  posterior
  probabilities.
  Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
Expected Value of Sample Info

 EVSI = EMV (with info) – EMV (w/o info)

  cf. EVPI = EMVUC – initial EMV
 One would spend X (cost of info) only on
  obtaining info provided that EVSI > X.
 Expected Net Gain from Sample Info

         ENGSI = EVSI – X

Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
                      Bayes Theorm
 If events Ei (i = 1, 2 …r) are exclusive, and an
  event F can occur only if one of Ei happens,
  then the probability that Ej happens when F is
  know to have occurred is: P( Ej / F )  rP( Ej )  P( F / Ej )
  where:                                
                                        i 1
                                                                     P ( Ei )  P ( F / Ei )

   P(Ei) is the prior probability of event Ei;

   P(F/Ei) is the conditional probability that
    event F occurs given that Ei has occurred;
   P(Ei/F) is the posterior probability of event Ei
    given that event F has occurred.
Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
     Example 5 - Based on Example 1
  The company consults a firm of economists to advise
  on the level of current economic activity which will, in
  turn, indicate the future state of economy. Their
  relationships (conditional probabilities) shown below.
Probabilities Current Level of
of Economy Economy Activity
Improved or
Collapse       High     Low
Improved(b1)           0.8          0.2
Collapse(b2)           0.3          0.7
Conditional Probabilities: p(H|b1)=0.8, p(L|b1)=0.2, p(H|b2)=0.3, p(L|b2)=0.7
      Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
Example 5 - Solution
              Prior Probabilities:
              p(b1)=0.6, p(b2)=0.4
              Posterior Probabilities:
              p(b1|H)=[p(b1).p(H|b1)]/
                      [p(b1).p(H|b1+
                       p(b2).p(H|b2)]
                     =0.6x0.8/(0.6x0.8
                      +0.4x0.3)=0.8
              p(b2|H)=0.4x0.3/(0.6x0.8
                      +0.4x0.3)=0.2
              p(b1|L)=0.6x0.2/(0.6x0.2
                      +0.4x0.7)=0.3
              p(b2|L)=0.4x0.7/(0.6x0.2
                      +0.4x0.7)=0.7
              p(H)=p(H|b1)+p(H|b2)=
                  =0.6x0.8+0.4x0.3=0.6
              p(L)=p(L|b1)+p(L|b2)=0.4
              EVSI=952–800=152 (103)
              ENGSI=EVSI–(info fee)
                             Example 6
Suppose: a) TV tubes are made by manufacturers M1
& M2; b) Events E1 & E2 represent the production of
defective or effective TV tubes; c) One TV tube selected
at random is defective.
Question: What is the probability that it came from M1?
Solution: It is to find the posterior probability which is
represented by P(M1|E1). According to Bayes' theorem:
P(M1|E1)=P(M1)P(E1|M1)/[P(M1)P(E1|M1)+P(M2)P(E1|M2)]
Where P(E1|M1) is the prior probability, i.e. the
probability of a defective tube given it is manufactured
by M1. P(M1) is the probability that a tube chosen at
random came from M1.
  Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
                    Example 6 (con’t)
If 40% of the tubes are
produced by M1 & 60%
by M2, and 1% & 2%
respectively of the tubes
produced by M1 & M2
turn out to be defective,
ie P(M1)=0.4, P(M2)=0.6,
P(E1|M1)=0.01,
P(E1|M2)=0.02, then:
P(M1|E1) = 0.4 x 0.01 /
    (0.4 x 0.01+0.6 x 0.02)
    = 0.25.
  Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
              Example 7 - Problem
Patients showing a given set of symptoms are
thought to have a 0.70 probability of having
contracted a type of kidney disease. A further test
for this disease has found that 60% of patients with
the disease give a positive response to this test.
Unfortunately 10% of patients who do not have the
disease also give a positive test.
Suppose a patient showing the initial set of
symptoms gives a positive result to the new test.
What is the probability that he is suffering from this
type of kidney disease?
 Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore
                Example 7 - Solution
Let T1 be the event of positive
response to the test & T2 the
negative response; D1 be the
event of having the disease &
D2 without the disease.
We have p(D1)=0.70,
p(D2)=0.3, p(T1|D1)=0.60,
p(T1|D2)=0.10
Hence p(D1|T1) =
p(D1).p(T1|D1) / [p(D1).
p(T1|D1) + p(D2).p(T1|D2)]
=0.70x0.60 / (0.70x0.60 +
0.30x0.10) = 0.93
   Dr Wang ShouQing, School of Building & Real Estate, National University of Singapore