Operational Research Paper

David R. Gulick

NUID: 300046769

IMSE 828

Assignment 3 Part II

p214, #2 Customers buy cars from three auto companies. Given the company from which a customer last bought a car, the probability that she will buy her next car from each company is as follows:

P = | 0.80  0.10  0.10 |

        | 0.05  0.85  0.10 |

        | 0.10  0.20  0.70 |

a) If someone currently owns a company 1 car, what is the probability that at least one of the next two cars she buys will be a company 1 car?

b) At present, it costs company 1 an average of $5,000 to produce a car, and the average price a customer pays for one is $8,000. Company 1 is considering instituting a five-year warranty. It estimates that this will increase the cost per car by $300, but a market research survey indicates that the probabilities will change as follows:

P = | 0.85  0.10  0.05 |

                    | 0.10  0.80  0.10 |

                     | 0.15  0.10  0.75 |

Should company 1 institute the five-year warranty?

a) .10

.10

P = | 0.80  0.10  0.10 |

b) CO2

CO2

CO1

CO1

        | 0.05  0.85  0.10 |

.05

.05

.850

.850

.80

.80

        | 0.10  0.20  0.70 |

.20

.20

.10

.10

.10

.10

.10

.10

CO3

CO3

.70

.70

.80

.80

.80

.80

CO1

CO1

.05

.05

.10

.10

CO1

CO1

CO2

CO2

CO1

CO1

CO3

CO3

.10

.10

.10

.10

Pr(at least 1 car of the next 2 will be CO1) = (0.80)*(0.80) + (0.10)*(0.05) + (0.10)*(0.10)

Pr(at least 1 car of the next 2 will be CO1) = 0.64 + 0.005 + 0.01

Pr(at least 1 car of the next 2 will be CO1) = 0.655

c) For the non-warranty, I get based on

Π  * P = [ Π0  Π1  Π2 ] | 0.80  0.10  0.10 |

                                           | 0.05  0.85  0.10 |

                                           | 0.10  0.20  0.70 |

Π0 = 0.80Π0 + 0.05Π1 + 0.10Π2

Π1 = 0.10Π0 + 0.85Π1 + 0.20Π2

Π2 = 0.10Π0 + 0.10Π1 + 0.70Π2

Π0 +...