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:

1. 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.

2. In Excel, use a suitable method for simulating the interval between successive

breakdowns, according to the continuous distribution shown.

3. In Excel, use a suitable method for simulating the lost revenue for each day the copier is

out of service.

4. 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.

5. 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).

6. 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.

1

ASSIGNMENT 1

1. Days-to-repair: If you assume that the number of days needed to repair a copier is random,

you can generate a random number denoted r2 between 0 and 1:

0 < random value < 0.2, then it takes 1 day

0.2 < random value < 0.65, then it takes 2 days

0.65 < random value < 0.90, then it takes 3 days

0.9 < random value < 1, then it takes 4 days

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

If we set this equal to another random number r1 that is between 0 and 1 then

r1 = x²/36 => x=6*sqrt (r1)

3. 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 r3*repair time, and the lost revenue is then equal to

0.1*r3*repair time, since they charge $0.10 per copy.

4. These are the results after running the simulation in Excel:

2

ASSIGNMENT 1

1.Repair Distribution

3

ASSIGNMENT 1

P(x) Cumulative Repair Time0.20 0.00 1.000.45 0.20 2.000.25 0.65 3.000.10 0.90 4.001.00

Breakdown Random Time between Random Repair Random Lost Cumulative

r1 breaking (weeks) r2 Time r3 Revenue Time

1 0.47638934 4.141257817 0.694977 3 5009 $1,502.70 4.1412 0.45265588 4.036782342 0.339443 2 2918 $583.60 8.1783 0.49535053 4.222868568 0.411788 2 3207 $641.40 12.4014 0.16026329 2.401973849 0.647771 2 7658 $1,531.60 14.8035 0.36734912 3.63655994 0.941637 4 6530 $2,612.00 18.4396 0.332808 3.461370813 0.397173 2 5150 $1,030.00 21.9017 0.73259303 5.135498912 0.897752 3 7645 $2,293.50 27.0368 0.73802422 5.154500149 0.686301 3 2586 $775.80 32.1919 0.60207711 4.655617678 0.367898 2 2469 $493.80 36.846

10 0.88683273 5.650307803 0.876709 3 4270 $1,281.00 42.49711 0.63896082 4.796101505 0.821057 3 6272 $1,881.60 47.29312 0.76560163 5.24991989 0.188747 1 7834 $783.40 52.54313 0.99158349 5.97469712 0.040863 1 3075 $307.50 58.51714 0.24882376 2.992934237 0.248753 2 2886 $577.20 61.51015 0.07945108 1.691224026 0.293336 2 4485 $897.00 63.20216 0.08830862 1.783006021 0.378375 2 3067 $613.40 64.985

Total Revenue Lost = $17,805.50

The amount of revenue lost is approximately US$17,805.50. This isn't entirely accurate since

you would have to run a large number of trials and take the average to find exactly what you lost.

Also, the amount of time doesn't always add up to 1 year, so you need to account for that by only

adding up to the revenue that doesn't surpass 1 year.

5. The amount of revenue lost can be found by looking at how many days it takes to repair, and

then multiplying that number by the amount of customers you may have received on a given day

multiplied by the amount of copies they produced. The time between breakdowns will tell you

4

ASSIGNMENT 1

how often this occurs, and the sum in weeks is not to exceed 52 (since there are 52 weeks in a

year).

6. Buying an extra copier would probably be a good choice, since the amount of revenue lost

almost doubles (US$17,805.50 vs. US$8,000.00) the cost of buying an extra copier. I feel

confident with my answer, although there are some limitations to it. As was mentioned before,

the sum of the weeks will not always add up to 1 years worth, so that needs to be taken into

account. Also, this simulation must be run several times to find the average amount lost, since

that would reflect a more accurate result of revenue lost per year.

References

5

ASSIGNMENT 1

Taylor, B. M. (2010). Introduction to management science (11th ed.). Upper Saddle River, NJ:

Pearson/Prentice Hall.

6