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JET Copies Case Problem

Venetia Carter

Professor Donald Ray West
Quantitative Methods
MAT 540
January 18, 2011

1. The number of days needed to repair the copier when it is out of service , according to the discrete distribution shown : Repair time (days) | Probability | Contribution to average repair time | 1 | 0.20 | 0.20 | 2 | 0.45 | 0.90 | 3 | 0.25 | 0.75 | 4 | 0.10 | 0.40 | Total (average days) | 1.00 | 2.25 |

2. Number of weeks between breakdowns simplified # of weeks | Cumulative chance of breakdowns | Simplified | 0 | 0* 0/18 *1/2 | 0 | 1 | 1*1/18*1/2 | 1/36 | 2 | 2*2/18*1/2 | 4/36 | 3 | 3*3/18*1/2 | 9/36 | 4 | 4*4/18*1/2 | 16/36 | 5 | 5*5/18*1/2 | 25/36 | 6 | 6*6/18*1/2 | 36/36 |

The chance of a breakdown within x weeks after a previous breakdown is x^2/36. To find the average value of x we set x^2/36=1/2 to get x =3 *sqrt (2) =4.243. This gives us 4.243 weeks between breakdowns.

3. Revenue lost per day the copier is out of service # of copies sold per day Price of each copy Revenue per day Low estimateHigh estimate | 20008000 | $0.10$0.10 | $200$800 | Average 5000 $0.10 $ 500

4. Lost revenue due to copier breakdowns over 1 year (average)

Weeks per year | 52 | Weeks between breakdowns | 4.243 | Breakdowns per year | 12.255 | Days lost per breakdown | 2.25 | Revenue lost per day | $500 | Revenue lost per breakdown | $1,125 |
Revenue lost per year $13,787

5. Write a brief explanation or description of how you implemented each component of the model.
To find the average time it takes to repair the copier, I looked at the given table. The average time will be the possible times (1-4 days) weighted by the probabilities of each. So I multiplied each number of days by the probability of that number of days, and added the results to get 2.25 days.

The average time between breakdowns was a little trickier. The area under the diagonal line in the given graph, between two points on the x-axis, represents the probability of the copier breaking down within the times indicated by those two points. So, for example, the area under the whole line from x = 0 to x = 1 is the area of a triangle with sides 6 and 1/3, which is 6 * 1/3 * 1/2 = 1. This is what we expect because we are told in the problem that the copier will certainly break down within six weeks.

We can use the same logic to find the probability that the copier will break down between x = 0 and any z. The line shown goes through (0,0) and (6, 1/3), so it is given by y = x/18. The probability is then the area of a triangle with sides z and z/18, or z * z/18 * 1/2 = z^2/36.

We now want to find the number of weeks where we have an equal probability of the copier breaking down before or after this time—this will be the average time between breakdowns. So we need z such that z^2/36 = 1/2. Solving gives z = 4.243 weeks.

We assumed a uniform distribution for the number of copies sold per day, so the average number of copies sold will be the average of the high and low estimates, which is 5000 copies per day. At 10 cents each, we expect revenue of $500 per day, and this amount will be lost while the copier is broken.

Putting everything together, we find the average number of breakdowns per year to be 12.255 (dividing 52 weeks/year by 4.243 weeks/breakdown), the revenue lost per breakdown to be $1,125 (days lost/breakdown * revenue lost/day), and finally the revenue lost per year due to breakdowns to be $13,787 (breakdowns/year * revenue lost/breakdown).

6. Since the revenue lost per year due to breakdowns is $13,787, which is greater than $12,000, they should purchase a backup copier. However, I can’t put high confidence in this answer because the study has serious limitations. Every step of the process used best guesses based on what staff members suggested, and not a rigorous analysis of other similar machines.

The repair times and probabilities seem reasonable, and I could expect that data gathered from the college of business to be fairly accurate. But the graph for the probability of breakdowns is at best a rough approximation. A copier won’t suddenly stop working after 6 weeks simply because it’s reached the six-week mark—yet this is what our model implies. And just because the probability of a breakdown increases as time goes by, we don’t know whether the graph should be a line or some other shape.

The estimate of the number of copies sold is seemingly arbitrary, and the students even admit they have no way to guess the distribution among these possible values. Perhaps the students could start their business before deciding on whether to purchase a backup copier. This would give them more reliable information about repair time, the frequency of breakdowns, and number of copies sold.

Reference: * Taylor, B. M. (2010). Introduction to management science (10th Ed.). Upper Saddle River, NJ: Pearson/Prentice Hall

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