Question

please post solutions using R a manufacturing facility has a repair shop with two repairmen who...

please post solutions using R

a manufacturing facility has a repair shop with two repairmen who repair failed machines on a first-fail-first-serve basis. They work together on the machine if there is one machine down (the repair still takes the same amount of time), and otherwise, each works on a separate machine. Thus, if there are more than two machines down, new failures simply wait for their turn to be repaired. Assume that machine failures arrive in a combined failure stream, so that we do not need to track machine identity. More specifically, assume that the times between machine failures are equally likely between 10 and 20 hours, and that repair times are equally likely between 5 and 55 hours for each machine.

To manually simulate the manufacturing facility, use the approach of items (a) and (b) of Exercise 1 utilizing the following sequence of uniform random numbers (between 0 and 1), U = {0.2, 0.5, 0.9, 0.7, 0.8, 0.1, 0.5, 0.2, 0.7, 0.4, 0.3} to generate times to failure and repair times. Simulate repair completions, and calculate the following statistics:

• Fraction of time a machine is down

• Average number of down machines waiting to be repaired

• Fraction of time each repairman is busy (repairman utilization)

• Fraction of time the repair facility is idle

• Average time a failed machine waits until its repair starts

• Throughput of the repair facility (number of repair completions per hour)

Note: These manual procedures will simulate the system for a short period of time. The tedium involved in simulating the system should make you realize that you need a computer program for long simulations or of even those of moderate complexity.

Homework Answers

Answer #1

Generating 11 Random Numbers from 10 to 20 Using R,

Use the below syntax,

#machine failures are equally likely between 10 and 20 hours

> x<-runif(11, 10,20)
> x
[1] 13.65028 15.97950 19.61011 18.82964 19.80892 18.47822 17.64867 12.03148 16.45946 10.74006 10.54702

> #repair times are equally likely between 5 and 55 hours
> y<-runif(11, 5,55)
> y
[1] 27.385872 27.236466 51.656502 10.440647 35.598574 44.447306 19.832874 35.021982 33.827839 9.460412 36.841597

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