# Assignment #1: JET Copies Case Problem

Read the "JET Copies" Case Problem on pages 678-679 of the text. Using simulation estimate the loss of revenue due to copier breakdown for one year, as follows:

- 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.
- In Excel, use a suitable method for simulating the interval between successive breakdowns, according to the continuous distribution shown.
- In Excel, use a suitable method for simulating the lost revenue for each day the copier is out of service.
- 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.
- In a word processing program, write a brief description/explanation of how you implemented each component of the model. Write 1-2 paragraphs for each component of the model (days-to-repair; interval between breakdowns; lost revenue; putting it together).
- Answer the question posed in the case study. How confident are you that this answer is a good one? What are the limits of the study? Write at least one paragraph.

There are two deliverables for this Case Problem, the Excel spreadsheet and the written description/explanation.

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## MAT 540 Assignment for you is attached (as discussed)

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MAT 540 Assignment xxx xxx is attached xxx discussed)

file1.xlsx preview (474 words)

# xxxxxx

xxxxxxxxxxxxxxxxxxxxxxxxxxxContinuous Probability | ||||||||||

#1- xxx average number of xxxx needed to repair the xxxxxx when xx xxxxxxxxxx | #2- The average xxxxxx xx weeks between xxxxxxxxxxx | |||||||||

Probability | Cumulative Probability | Repair Time-days | Repair Time-days | Random #'s | xxxxxx Time-days | Average xxxxxx Time xxxxxxx xxxxxxxxxxxxxxxx | 3.84 | xxxxxxxxxx | xx xxxxxx xxx xxx xxxxxx xxxxxxxxxx | Time xxxxxxx break downs xxxxxxx |

xxxx | xxxx | x | 1 | xxxxxxxxxxxx | x | x | xxxxxxxxx | xxxxxxxxxxxx | ||

0.45 | 0.20 | 2 | 2 | xxxxxxxxxxxx | 2 | 2 | 0.118795422 | xxxxxxxxxxxx | ||

0.25 | xxxx | 3 | x | xxxxxxxxxxxx | x | 3 | 0.1300556639 | xxxxxxxxxxxx | ||

0.10 | xxxx | x | x | xxxxxxxxxxxx | 3 | x | xxxxxxxxxxxx | xxxxxxxxxxxx | ||

xxxx | x | 0.7023251249 | x | x | xxxxxxxxxxxx | xxxxxxxxxxxx | ||||

xxxxxxx xxxxxx |

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file2.doc preview (472 words)

**Assignment #1: JET xxxxxx Case Problem**

**Do xxx xxxxxx xxxx work as your xxxx xxxxxx to xxxxxx xxxxxxxxxxx**

xx

**The xxxxxxx xxxxxx xx days xxxxxx xx xxxxxx the xxxxxxx Continuous xxxxxxxxxxx Distributions
**

I did xxx xxxxxxxxxx probability for the four xxxxxx xxxx xxxxx xxxx I named the xxxxxxxxxx probability xxx repair time xxxx’ xxxxxxx x decided xx expand my xxxxxx time days xx xxx to xxx a xxxxxx xxxxxxx xxx I used the xxxxxxx =rand() to xxx my random numbers. I froze my random xxxxxxxx xx convert xx random xxxxxxx to xxxxxx time days, I used xxx xxxxxxx =vlookup(F6,vlookup,2) and average the xxx repair xxxx – xxxxx

Equals= xxxxx

**The average xxxxxx xx xxxxx between xxxxxxxxxxx**

I xxxx xxx repair time xxxxx then x xxxxx () to get xx random xxxxxxxx The random xxxxxxx xxxx xxxxxxx xx xxx my time xxxxxxx xxxxxxxxxx x used the formula =6*sqrt(m7) xxx continuous xxxxxxxxxxx xxxxxxxx and xxxxxxxx x

Equals= xxxx

**Lost xxxxxxx Due to xxxxxx xxxxx Out of xxxxxxxx**

The xxxxxxx estimated

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## UPDATED 100% correct answer A+ TUTORIAL GUARANTEED PERFECT PLAGIARISM FREE WORK

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# repair xxxx simulation

xxxxxxxxcum. Prob. | repair time | probability | simulated xxxx | uniform |

x | x | xxx | 2 | xxxxxxxxxxxxxxxxxxx |

xxx | 2 | 0.45 | ||

xxxx | x | 0.25 | ||

0.9 | 4 | xxx |

# xxxxxxxxx interval xxxxxxxxxx

xsimulated xxxx | uniform |

33 | xxxxxxxxxxxxxxxxxxx |

# xxxxx loss xxx xxx

xrate | xxxx sold | loss |

$0.10 | 5734 | $573.40 |

# simulation for 1 year

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxx | breakdown xx day xxx | xxxxx number of xxxxxxxxxx | xxxxxx time | modified repair xxxx | sales lost(in xx | xxxx xxxxx | xxxxxx time | |||

xxx x | day 2 | day x | day 4 | |||||||

26 | 26 | 12 | 2 | x | 508.70000000000005 | xxxxxxxxxxxxxxxxxx | x | x | 0 | 1 |

xx | xx | x | 4 | xxxxx | xxxxxxxxxxxxxxxxxx | xxxxx | xxxxx | xxx | x | |

xx | xxx | 2 | x | xxxxx | 562.9 | 0 | 0 | xxxx | 3 | |

xx | xxx | x | x | xxxxxxxxxxxxxxxxxx | xxx | x | 0 | 0.9 | 4 | |

10 | xxx | 2 | x | 532.4 | 335.1 | 0 | x | |||

29 | xxx | 1 | 1 | xxx | 0 | x | 0 | |||

xx | 208 | 3 | 3 | 243.70000000000002 | xxxxxxxxxxxxxxxxxx | xxxxx | 0 | |||

30 | 238 | 3 | x | 303 | xxxxx | xxxxx | x | |||

37 | xxx | 2 | x | 381.6 | xxxxx | x | 0 | |||

38 | 313 | x | x | 632.40000000000009 | xxxxxxxxxxxxxxxxxx | 0 | 0 | |||

xx | xxx | x | x | xxxxx | 652.70000000000005 | 249.70000000000002 | 510.8 | |||

xx | xxx | 3 | x | 798.90000000000009 | 376.8 | 0 | 0 | |||

25 | 388 | |||||||||

Estimated total xxxx | ||||||||||

15780.3 | ||||||||||

file2.docx preview (567 words)

1,2,3,4. xxx the xxxxxxxx xxxxx file.

5. xxxxxxxxxx xxx number xx xxxx to xxxxxx xx the first problem. xx is a discrete probability distribution. xxx xx generated xxxxxx xxxxxxx xxxx it xx the following xxxx xx xxxxx xxxxxxxxxx xxxxxxxxx type cumulative probabilities xxx then xx generated xxx xxxxxxx xxxxx random variable xxxxx xxxxxxxxx xxx generated uniform xxx is in [0,.2) xx return 1, if xx xx in [.2,.65) xxxx xx xxxxxx x xxx so on. Thus xx xxxxxxxxx the number of xxxx to xxxxxxx

xxxxxxxxxxxcum. Prob. | xxxxxx time | xxxxxxxxxxx |

x | x | 0.2 |

0.2 | x | xxxx |

0.65 | 3 | xxxx |

0.9 | x | xxx |

For xxxxxxxxxx the xxxxxxxx xxxxxxx successive xxxxxxxxxx the p.d.f. is , xxx x xx xxxxxx The xxxxxxxxxxxx function is for x xx [0,6].

We simulate xxx distribution xx generating one uniform xxxxx xxx xxxxxxx xx xx the inverse xx xxx xxxxxxxxxxxx function.

The inverse function is x

xxxxxxx we round xxx the xxxxxxxxx no. of xxxxx to xxxxxxx xxx xx xxxxx (as xxxx x xxxxxxxxxx distribution no. of

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## Jet Copies solution

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# xxx xxxxxx xxxxxxxxxx solution.xlsx

# xxxxxx xxxx xxxxxxxxxx

xxxxxxxxxxxxxxxxxxxxxxxxxxxxcum. xxxxx | repair xxxx | xxxxxxxxxxx | simulated xxxx | xxxxxxx |

x | x | xxx | x | 0.24366721125313728 |

xxx | x | 0.45 | ||

xxxx | 3 | 0.25 | ||

0.9 | 4 | 0.1 |

# xxxxxxxxx xxxxxxxx xxxxxxxxxx

xxxxsimulated xxxx | uniform |

33 | xxxxxxxxxxxxxxxxxxx |

# sales xxxx per xxx

xxxxrate | xxxx xxxx | xxxx |

xxxxx | xxxx | xxxxxxx |

# xxxxxxxxxx xxx 1 year

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxbreakdown xxxxxxxxxxxxxxxxxxx | breakdown xx day no. | xxxxx number xx breakdowns | Repair time | xxxxxxxx repair xxxx | xxxxx lost(in $) | cum. Prob. | xxxxxx time | |||

day x | xxx x | day 3 | xxx x | |||||||

26 | 26 | xx | 2 | 2 | 508.70000000000005 | 323.20000000000005 | x | 0 | 0 | 1 |

xx | 66 | x | x | xxxxx | 645.30000000000007 | 786.7 | 601.6 | 0.2 | x | |

xx | xxx | 2 | 2 | xxxxx | xxxxx | x | x | 0.65 | x | |

24 | 131 | 2 | x | 572.70000000000005 | 678 | 0 | 0 | xxx | 4 | |

xx | xxx | x | 2 | 532.4 | xxxxx | 0 | x | |||

29 | 170 | x | x | 482 | x | 0 | x | |||

xx | 208 | 3 | x | 243.70000000000002 | xxxxxxxxxxxxxxxxxx | xxxxx | 0 | |||

xx | xxx | x | x | xxx | 533.9 | 451.6 | 0 | |||

xx | 275 | x | x | xxxxx | 763.1 | 0 | 0 | |||

xx | xxx | 2 | x | xxxxxxxxxxxxxxxxxx | xxxxxxxxxxxxxxxxxx | 0 | 0 | |||

20 | 333 | 4 | x | xxxxx | 652.70000000000005 | 249.70000000000002 | xxxxx | |||

30 | xxx | 3 | x | xxxxxxxxxxxxxxxxxx | xxxxx | 0 | x | |||

xx | xxx | |||||||||

Estimated total loss | ||||||||||

xxxxxxx | ||||||||||

# jet copies xxxxxxxxxxxx

xxxxxxxx See xxx attached excel xxxxx

xx xxxxxxxxxx the xxxxxx of days to xxxxxx is the xxxxx problem. xx is a discrete xxxxxxxxxxx distribution. xxx xx generated random numbers xxxx xx xx the following way. xx first calculated less-than type cumulative

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