# Math540 week 3 assignment, chapter 14,

This is an unformatted preview. Please download the rooted muniment for the first format.

MATH540 Week 3 Assignment, Chapter 14, Jet Copies, Set up

Provided by Professor Aungst, Supplemental Instruction

Information from Jet Copies Case Study:

- Students bought an \$18,000 copier to set-on-foot their own observation concern.

- Wanted to donation a smaller copier for \$8,000 as back-up

- Created a airs to deem the sum of enrichment that would be past if they did not feel a

backup

- Duration betwixt breakdowns is 0 weeks to 6 weeks (see affectlihood power on page 679, and

provided succeeding in this set up

- Plain aftercited affectlihood dispensation of restore durations:

Repair Duration (days)

Probability

1

0.20

2

0.45

3

0.25

4

0.10

- Estimated they would vend betwixt 2,000 and 8,000 copies per day at 10 cents (0.10) per observation

- Used a consistent affectlihood dispensation betwixt 2,000 and 8,000 to deem how multifarious copies

they would vend per day

- If waste of enrichment due to means downduration during 1 year is greater than or resembling to \$12,000,

then they should donation the back-up copier

- Decided to persuade a manual airs of this course for 1 year to see if the example was

working correctly

- Our assignment is to achieve this manual airs for JET copies and mention the

waste of enrichment for 1 year.

Here’s some precursive Set Up information:

The affectlihood power for duration betwixt restores, f(x), is,

f(x) = x/18, 0 <= x <= 6

and, r = x^2/36

x2 = 36r

x = 6*sqrt of r (use this formula in the shaft you mention as duration betwixt restores)

You could enunciate the cumulative dispensation and wild sum ranges for the dispensation of

restore durations for allusion if you would affect that for allusion.

Repair Time

Repair Time

y (days)

1

2

3

P(y)

0.2

0.45

0.25

Cumulative

Probability

RN Ranges

4

0.10

The affectlihood power for daily ask-for is enunciateed by determining the direct power

for the consistent dispensation, which is,

f(z) = 1 / b – a which resemblings 1/6

Letting F(z) = r in the Integrated Function, and solving for z we get: z = 6r + 2 (this is the

formula for copies past)

There are diverse ways to set up the Monte Carlo airs in Excel using the formulas we

learned in Chapter 14 … namely Wild Sum Generation (which is =RAND) and

VLOOKUP which allows us to “apex back” to a affectlihood board and insinuate a affectlihood based

on that Wild Sum and the Likelihood associated delay it in the board.

Most students set-on-foot delay enunciateing the affectlihood board for Restore duration to succeeding be used as the

VLOOKUP Board for Restore Duration affectlihood.

P(x)

Cumulative

Repair

Time

The Airs itself would be for 52 weeks (which would be when the cumulative “time

betwixt breakdowns” reached 52 weeks). You could commence delay a Wild Sum (r1) which

would be divers by shaft 2, the Duration Betwixt Breakdown (in weeks) formula of 6*square

root of r1

You could then sum those variables in a cumulative roll in shaft 3 (so you could report when the

airs reached 52 weeks).

In shaft 4 you could beget another wild sum (say, r2) to count the shaft 5

Repair duration in y days.

That r2 could be used in a shaft 5 for Restore Duration in y days which could be countd by

using the =VLOOKUP power which would recite that r2 to probabilities in the Restore Time

likelihood board firstly set up.

You susceptibility then set up some wild sum shafts and remainder shafts for restores importation 1 day,

2 days, 3 days and 4 days.

At some apex, you would deficiency to illustration out how to count copies past in a day in thousands

and that would probably comprise the formula z = 6r + 2