BUS 202 Quantitative Analysis II

Solutions to Assignment 5

  • 1.
    •  
    a)Laplace     b)MaxMin    c)MaxMax    d)MinRegret 
      23.5           12           35           21 
      22.5           18           27           15 
      25             22           28           13 
      26.5           20           33           15 
        
     
    Regret Table: 
                               Max 
        0    3    3    21       21 
        8    0    8    15       15  
       13    0    3     5       13 
       15    0    0     0       15
    Laplace criterion chooses decision 4; MaxMin decision 3; MaxMax decision 1; and MinRegret decision 3. [Any questions?] 
     
  • 2.
    • a)Laplace     b)MaxMin    c)MaxMax    d)MinRegret 
      6.67            5            8           2 
      6.00            6            6           3 
      4.33            1            9           7 
      
        
     
    Regret Table: 
                          Max 
        1    2    0       2 
        0    3    2       3  
        3    0    7       7 
     
    Laplace criterion chooses decision 1; MaxMin decision 2; MaxMax decision 3; and MinRegret decision 1. [Any questions?]

  • 3.
    • [go to top] [Any questions?]
  • 5.
    • [go to top] [Any questions?]
  • 6.

    • a) 
     
    Small:       (0 + 1000 + 2000 + 3000)/4  =  $1,500 
    Medium:     (-1000 + 0 + 3000 + 6000)/4  =  $2,000 
    Large:    (-3000 -1000 + 4000 + 8000)/4  =  $2,000
    Medium and Large tie for best expected dollar return

     b)Utilities of the payoffs from the graph: 

     
    Decision   Cold     Cool     Warm    Hot        E(U) 
    Small       .66     .73      .78     .83        .75 
    Medium      .55     .66      .83     .93        .7425 
    Large       .14     .55      .87     .98        .635
    Small has the maximum expected utility. Working backwards with the utility graph (finding the $ value corresponding to the expected utilities of the alternatives) we find that "Small" has a certainity equivalent of about $1,800; "Medium" of about $1700 and "Large" less than $0.
    [go to top] [Any questions?]
  • 8.

    • a) Equivalent lottery for a net cash flow of $0 is obtained from p*50,000 + (1-p)*(-20,000)= $0 solving for p gives p = .28.
    Equivalent lottery for a net cash flow of $20,000 is obtained from p*50,000 + (1-p)*(-20,000)= $20,000 solving for p gives p = .57.
    b) .
    For me, the lottery of winning $50,000 or loosing $20,00  if the probability of winning is .40 is barely worth a sure $0. Thus my utility of a sure $0 is 0.4 ( > .28).

     Likewise the lottery of winning $50,000 or loosing $20,000 is just about as desirable as a sure $20,000 if the winning probability is .75, thus my utility for a sure $20,000 is .75 (> .57).

     Therefore, my utility function for money is given in red .

    c)I am risk averse since my utility function is concave. To accept a lottery (with risk) I demand a premium in the form of higher winning probability (40% as opposed to 28% in the case of $0; and 75% as opposed to 57% in the case of $20,000). If I were risk indifferent my utility function would have been the blue line for which the utility of the lottery with expected value $0 and the sure $0 has the same utility.
    [go to top] [Any questions?]


  • 10.

    • a) 
                       .25              .75 
    Decision          Rainy            Sunny            E(V) 
      GO            -15,000           10,000          $3,750 
      CANCEL         -1,000           -1,000         $-1,000
    Thus GO is the best decision

     b) Expected Value (EV) with perfect information:
    .25 *(-1,000) + .75 *(10,000) = $7,250
    Hence: Expected Value of Perfect Information (EVPI) = 7,250 - 3,750 = 3,500
    [go to top] [Any questions?]

  • 11.
    • States of nature:
    Sunny: S
    Rainy: R
    Result of report (Event):
    Predict Sunny: PS
    Predict Rainy: PR
     
    Priors:  P(S) = .75          P(R) = .25
    Reliability of the weather report: (conditional probabilities)

          P(PS|S) = .80       P(PS|R) = .10
          P(PR|S) = .20       P(PR|R) = .90


    Joint Probabilities  =  Priors * Conditionals
    Marginals
    P(PS and S) = .60 
    P(PS and R) = .025 
    P(PS) = .625 
    P(PR and S) =.15 P(PR and R)  = .225  P(PR) = .375 
    Posteriors: (Joints/marginals) 

    P(S|PS) = .96   P(R|PS) = .04   
    P(S|PR) = .40   P(R|PR) = .60
    The decision tree:
    Expected value of sample information (EVSI) is the difference between expected dollar value with the report which is $5,250 and expected dollar value without the report which is $3,750.
    Hence EVSI= $5,250 - $3750 = $1500
    Here is another way to arrive at the same results:
    If the report predicts Sunny (PS) use probalities of .96 and .04 respectively for sunny and rainy conditions. In which case the best alternative is to go with the show with EV of $9,000.
    Whereas if the report indicates Rainy conditions (PR) use probalities of .40 and .60 respectively for sunny and rainy conditions. In which case the better alternative is to cancel with EV of $-1,000.
    There is .625 chance the report will predict sunny in which case we will go with show and make $9,000 while there is a probability of .375 that the report will predict rain in which case we will cancel and suffere a loss of $1,000. Thus the Overall EV:
    .625 * 9,000 + .375 * (-1,000) = $5,250 (this is the Expected value with the report)
    Therefore EVSI = EV (with the report) - EV(without the report) = 5,250 - 3,750 = $1,500.

    [go to top] [Any questions?]