Week 3 Case Problem

1. If the number of days needed to repair the copier are random. Jet can create random numbers (R2) between 0 and 1. See Below

Repair Distribution | | |

P(x) | Cumlative Distribution | Repair time in days | |

0.20 | 0.00 | 1 | 0-19 |

0.45 | 0.20 | 2 | 20-64 |

0.25 | 0.65 | 3 | 65-89 |

0.10 | 0.90 | 4 | 90-100 |

1.00

2. The probability distribution for random variables is 0-6 weeks. X equals the number of weeks between breakdowns. if a random number is between 0 and 1 the weeks between breakdowns can be computed as below.

Weeks between breakdown=6(square root of R1)

3. To calculate the lost revenue jet has decided to use a uniform probability distribution between 2000-8000 copies sold be day. To generate the (R3) random number between 2000-8000 I used the excel equation =Rand (8000-2000)+2000.

Total revenue lost is computed in the following equation

( .10 Price per copy)*R3*Repair time in days.

4. The Total amount of revenue lost is about $11592.60. However, more than one simulation would have to be conducted to find the average of total revenue lost. After running several simulations they would then take the average of all total revenues lost to find a more accurate amount of the revenue lost due to breakdowns.

5. To compute the revenue lost, Jet needs to do the following (A) find out the number of days needed to repair the copier. Then multiply it by the (B) amount of customers they may receive in a day finally multiply the (C) number of copies the produced.

The time between breakdowns can tell you how often occur. Any breakdown that occurs after 52 weeks will not be counted because there is only 52 weeks in 1 year.

Lost revenue=A*B*C

6. According to this simulation Jet may not feel a need to by an extra copier since the total revenue loss is only $11592.60, which is only $407.40 less then their cut off of $12,000. However, because of the limitations presented in this problem I would...