472 WordsFeb 26, 20132 Pages

Running Head: JET Copies Case Problem – Solution
JET Copies Case Problem – Solution
JET Copies Case Problem – Solution
5. The average time will be the possible times (1-4 days) weighted by the probabilities of each. So multiply each number of days by the probability of that number of days, and added the results to get 2.25 days. 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. 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 the region of a triangle with sides z and z/18, or z * z/18 * 1/2 = z^2/36.
The equal probability of the copier breaking down before or after is the average time between breakdowns. So z^2/36 = 1/2. Solving gives z = 4.243 weeks. The uniform distribution for the number of copies sold per day, standard number of copies sold will be the average of the high and low estimates, which is 5000 copies per day. At 10 cents each, the expected revenue of $500 per day, and the amount will be lost while the copier is broken. The standard number of breakdowns per year to be 12.255 (dividing 52 weeks/year by 4.243 weeks/breakdown), the profit lost per breakdown to be $1,125 (days lost/breakdown * revenue lost/day), and finally the profit lost per year due to breakdowns to be $13,787 (breakdowns/year * profit lost/breakdown).
6. The profits lost per year due to breakdowns is $13,787, which is bigger than $12,000, they should procure a backup copier. A high

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