Jet Copies Paper

1. Days-to-repair: If you assume that the number of days needed to repair a copier is random, you can generate a random number denoted r2 between 0 and 1:

0 < random value < 0.2, then it takes 1 day

0.2 < random value < 0.65, then it takes 2 days

0.65 < random value < 0.90, then it takes 3 days

0.9 < random value < 1, then it takes 4 days

2. Intervals between successive breakdowns: The probability distribution of the random variable varies between the times of 0 to 6 weeks, with the probability increasing as time goes on. This can be approximated by the function

F(x) = x/18, for 0≤x≤6, where x= weeks between machine breakdowns

Therefore the distribution function is:

F(x) = x²/36 for 0≤x≤6

If we set this equal to another random number r1 that is between 0 and 1 then

r1 = x²/36 => x=6*sqrt (r1)

3. Lost revenue: Since the number of copies sold per day is a uniform probability distribution between 2000 to 8000 copies, r3 is a random number between 2000 and 8000. To get the amount of business lost on a particular day is r3*repair time, and the lost revenue is then equal to 0.1*r3*repair time, since they charge $0.10 per copy.

4. These are the results after running the simulation in Excel:

1.Repair Distribution | | | | | | |

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

0.20 | 0.00 | 1.00 | | | | | |

0.45 | 0.20 | 2.00 | | | | | |

0.25 | 0.65 | 3.00 | | | | | |

0.10 | 0.90 | 4.00 | | | | | |

1.00 | | | | | | | |

| | | | | | | |

Breakdown | Random | Time between | Random | Repair | Random | Lost | Cumulative |

| r1 | breaking (weeks) | r2 | Time | r3 | Revenue | Time |

1 | 0.47638934 | 4.141257817 | 0.694977 | 3 | 5009 | $1,502.70 | 4.141 |

2 | 0.45265588 | 4.036782342 | 0.339443 | 2 | 2918 | $583.60 | 8.178 |

3 | 0.49535053 | 4.222868568 | 0.411788 | 2 | 3207 | $641.40 | 12.401 |

4 | 0.16026329 | 2.401973849 | 0.647771 | 2 | 7658 | $1,531.60 | 14.803 |

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