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: Jet Copies Case Problem.

Submitted By honey1985
Words 477
Pages 2
1. Number of days needed to repair the copier.

Machine Probability of Cumulative Random number repair time repair time P(y) probability range, r2
1 0.2 0.2 0.00 - 0.20
2 0.45 0.65 0.21- 0.65
3 0.25 0.9 0.66 - 0.90
4 0.1 1 0.91 - 1

2. Intervals between successive breakdowns: The probability distribution of the random variable varies between the times of 0 to 6 weeks, with the probability increasing as time goes on. This can be approximated by the function
F(x) = x/18, for 0”x”6, where x= weeks between machine breakdowns
Therefore the distribution function is:
F(x) = x²/36 for 0”x”6
Lost revenue: Since the number of copies sold per day is a uniform probability distribution between 2000 to 8000 copies, r3 is a random number between 2000 and 8000. To get the amount of business lost on a particular day is r1, r2 and r3 are random number picked, as well as repair time, and the lost revenue is then calculated by their charging price of \$0.10 per copy.

Simulation Random Times between Cumulative S. Random Repair S. Random Z= 6r + 2 Lost Revenue
Breakdowns r1 bkdowns X (weeks) time r2 time (y) r3 Z Copies # \$
1 45 4.02 4.02 61 2 19; 65 9.04 9040 904
2 90 5.7 9.72 11 1 51 5.06 5060 506
3 84 5.5 15.22 19 1 17 3.02 3020 302
4 17 2.47 17.69 58 3 63; 85; 37 17.1 17100 1710
5 74 5.16 22.85 30 2 89; 76 6.56 6560 656
6 94 5.82 28.67 95 4 71; 34; 11; 27 16.58 16580 1658
7 7 1.58 30.25 38 2 10; 59 8.14 8140 814
8 15 2.32 32.57 24 2 87; 08 9.7 9700 970
9 4 1.2 33.77 68 3 08; 89; 42 14.34 1434 43.4
10 31 3.34 37.11 41 2 79; 79 13.48 13480 1348
11 7 1.58 38.69 35 2 97; 26 11.38 11380 1138
12 99 5.97 44.66 18 1 6 5.6 5600 560
15 97 5.9 50.56 71 3 87; 39; 28 15.24 15240 1524
16 73 5.12 55.68 59 2 97; 69 13.96 13960 1396
17 13 2.16 57.84 32 2 39; 28 8.02 8020 802
18 3 1.04 58.88 17 1 97 7.82 7820 782
19 62 4.72 63.6 69 3 69; 33; 87 17.34 17340 1734 20 47 4.11 67.71 40 2 99; 93 15.52 15510 1551 sum: 18.398,4

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