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# Mat540

Submitted By noraraeb
Words 530
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MAT 540
Assign 3
Nora Bennett

Q5.
Using the discrete distribution chart shown, I created an Excel chart featuring the Probability of each repair time in days and the cumulative probability of each, so I could use the =LOOKUP formula to create the time between breakdowns in Q2.
This shows that if the random number falls: between 0 and 0.2 it will take 1 day to repair between 0.2 and 0.65 it will take 2 days to repair between 0.65 and 0.90 it will take 3 days to repair and between 0.90 and 1 it will take 4 days to repair.

Using the continuous distribution chart shown, I first set up a table with 52 options for random numbers. I then used the formula =RAND() in Column B to simulate 52 random numbers to then be used for the continuous distribution formula to find: Time Between Breakdowns: | Chart: 6/.33 = 18.18 | f(x)= x/18, 0 ≤ X ≤ 6 | f(x)= x²/36 = x = 6√RN1 | |
I then used this formula in Column C to find the time between breakdowns for the copy machine. I averaged this column for use in answering question 4, and also divided 52 weeks by this average to help answer question 4.

To simulate the lost revenue for each day the copier is out of service, I generated a new column of random numbers, RN2, (Column D) to use to find the repair time in days. In column E, I used the formula =LOOKUP that I set up in question one, to input the amount of days of repair for the random numbers I set up in column D. I averaged these numbers at the bottom of the column for use in answering question 4.
In column F I set up a final column of random numbers (RN3), between the range of 2000 and 8000 using the =RANDBETWEEN(2000,8000) formula. This column shows how much money is lost when the copier is down. In column G, I used the formula = Repair time in Days * RN3 * 0.10
To find the amount of money lost each day if the copier was broken down. I averaged this column into cell G66 to help find the answer for question 4 as well.

For Question 4 I used the averages I created to find: Takes an Average of 3.88 weeks for the machine to breakdown or 13.41 times a year | This takes an average of 1.42 days to get fixed | This costs an Average of \$681.05 revenue loss per day | Meaning JET loses about (681.05*1.42)*13.41 or | \$ 12,968.69 | in one year | | | | |

Q6. To answer the question in the case study, yes, James, Ernie, and Terri should invest the \$8,000 into a back-up copier. The limits of the study could include more data than 52 points, or multiple simulations, and could then provide a more solid answer as to exactly how much money they would be losing. With that being said, I feel confident in my findings that a loss of greater than \$12,000 would take place each year.

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