Mat 540 Assignment 1

1. In Excel, use a suitable method for generating the number of days needed to repair the copier, when it is out of service, according to the discrete distribution shown.

Assumptions: The number of days to repair a printer is random. rV is the random value.

0<rV<0.2, then 1 day

0.2<rV<0.65, then 2 days

0.65<rV<0.90, then 3 days

0.9<rV<1, then 4 days

2. In Excel, use a suitable method for simulating the interval between successive breakdowns, according to the continuous distribution shown.

The probability distribution of the rV will vary between zero and six weeks. The probability will increase as the time approaches six weeks.

Function:

F(x)=x/18, for 0x6, where x = weeks between copier breakdowns.

Distribution function:

F(x)=x2/36 for 0x6

If r1 is between 0 and 1 then:

F(x)=x2/36≥x=6*√r1

3. In Excel, use a suitable method for simulating the lost revenue for each day the copier is out of service.

The number of copies sold per day is between 2000 and 8000. The cost per copy is $.10. Therefore r3 is a number between 2000 and 8000. The lost business on a specific day would be r3*repair time. The lost revenue would be equal to ($0.1*)(r3*repair time).

4. Put all of this together to simulate the lost revenue due to copier breakdowns over 1 year to answer the question asked in the case study.

Repair Distribution | | | | | |

P(x) | Cumulative | Repair Time | | | | | |

0.2 | 0 | 1 | | | | | |

0.45 | 0.2 | 2 | | | | | |

0.25 | 0.65 | 3 | | | | | |

0.1 | 0.9 | 4 | | | | | |

| | | | | | | |

Breakdown | Random r1 | Time Between Breaking | Random r2 | Repair Time | Random r3 | Lost Revenue | Cumulative Time |

1 | 0.638013 | 4.79254120791 | 0.01957 | 2 | 5072 | $ 1,014.40 | 2.65758223 |

2 | 0.762761 | 5.24017026627 | 0.365292 | 3 | 7359 | $ 2,207.70 | 4.12805291 |

3 | 0.47049 | 4.11553544494 | 0.673794 | 1 | 6090 | $ 609.00 | 2.14428382 |

4 |...