A decision maker faces a risky gamble in which she may obtain one of five outcomes. Label the outcomes A, B, C, D, and E. A is the most preferred, and E is least preferred. She has made the following three assessments. - PowerPoint PPT Presentation

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A decision maker faces a risky gamble in which she may obtain one of five outcomes. Label the outcomes A, B, C, D, and E. A is the most preferred, and E is least preferred. She has made the following three assessments.

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Label the outcomes A, B, C, D, and E. A is the most preferred, and E is least ... She is indifferent between having C for sure or a lottery in which she wins A ... – PowerPoint PPT presentation

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Title: A decision maker faces a risky gamble in which she may obtain one of five outcomes. Label the outcomes A, B, C, D, and E. A is the most preferred, and E is least preferred. She has made the following three assessments.


1
  • 13.21
  • A decision maker faces a risky gamble in which
    she may obtain one of five outcomes. Label the
    outcomes A, B, C, D, and E. A is the most
    preferred, and E is least preferred. She has made
    the following three assessments.
  • She is indifferent between having C for sure or
    a lottery in which she wins A with probability
    0.5 or E with probability 0.5
  • She is indifferent between having B for sure or
    lottery in which she wins A with probability 0.4
    or C with probability 0.6
  • She is indifferent between these two lotteries
  • A 50 chance at B and a 50 chance at D
  • A 50 chance at A and a 50 chance at E
  • What are U(A), U(B), U(C), U(D), and U(E)?

2
First, because A and E are best and worst
outcomes, respectively, we can set U(A) 1, and
U(E) 0.
Then, based on the indifference assessment, we
can set up equations
First assessment U(C) 0.5U(A) 0.5U(E)
0.5(1) 0.5(0) 0.5
Second assessment U(B) 0.4U(A) 0.6U(C)
0.4(1) 0.6(0.5) 0.7
Third assessment 0.5U(B) 0.5U(D) 0.5U(A)
0.5U(E) ? 0.5(0.7) 0.5U(D) 0.5(1) 0.5(0)
?U(D) 0.30
3
  • 15.14
  • A friend of yours is in a market for a new
    computer. Four different machines are under
    consideration. The four computers are essentially
    the same, but they vary in price and reliability.
    The least expensive model is also the least
    reliable, the most expensive is the most
    reliable, and the other two are in between.
  • The computers are described as follows
  • Price 998.95 Expected number of days in the
    shop per year 4
  • Price 1300.00 Expected number of days in the
    shop per year 2
  • Price 1350.00 Expected number of days in the
    shop per year 2.5
  • Price 1750.00 Expected number of days in the
    shop per year 0.5
  • The computer will be an important part of your
    friends livelihood for the next two years. The
    magnitude of the losses are uncertain but are
    estimated to be approximately 180 per day that
    the computer is down.
  • Can you give your friend any advice without doing
    calculations?
  • Use the information given to determine weights kP
    and kR. What assumptions are you making?
  • Calculate overall utilities for the computers.
    What do you conclude?
  • Sketch three indifference curves that reflect
    your friends trade-off between reliability and
    price.

4
Let P Price of computers () R Reliability
(the number of days in the shop per year)
a. Compared to machine B, machine C has a higher
price yet more number of days in the shop per
year. Therefore, machine C is dominated by B and
thus can be eliminated from the analysis
b. Because A is the best on price but the worst
on reliability, UP (A) 1 and UR (A) 0 Because
D is the worst on price but the best on
reliability, UP (D) 0 and UR (D) 1
UP (B) (1300-1750)/(998.95-1750) 0.60 and UR
(B) (2-4)/(0.5-4) 0.57
Utility Alternative Alternative Alternative
Utility A B D
UP 1 0.6 0
UR 0 0.57 1
5
An extra day of reliability is worth 180, and
also assume proportionality for the utilities,
then Machine A, with price 998.95 and expected
down time of 4 days per year would be equivalent
to hypothetical Machine E for (998.951801178.95
) and expected down time of (4-13) days per
year.
Because A is the best on price but the worst on
reliability, U(A) kP(1) kR (0) kP
UP (E) (1178.95-1750)/(998.95-1750) 0.76 and
UR (E) (3-4)/(0.5-4) 0.29
U(E) kP (0.76) kR (0.29)
U(A) U(E) ?0.76kP 0.29kR kP (Eq. 1)
kP kR 1 (Eq. 2)
Solving Equations 1 and 2 for kP and kR , we can
get kP 0.547 and kR 0.453
c. U(A) 0.547(1) 0.453(0) 0.547
U(B) 0.547(0.60) 0.453(0.57) 0.586
U(D) 0.547(0) 0.453(1) 0.453
6
d.
Utility ?
Days in shop/per year
0.547
0.586
0.453
A
E
B
D
Price ()
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