Assignment3_Ans_2015.pdf

Assignment #3 Announced on September 30 Due by October 9 (5PM via Sakai; There will be a penalty for late submissions) This assignment is composed of 6 problems. You need to submit Arena models (.doe) and a word (or PDF) document containing your answers into e-Learning. Alternately, if you solve the problems by hand, you will need to scan (in PDF file) your answers and submit it into Sakai. Note: you must WRITE OUT YOUR CALCULATIONS for full credit!!! 1. Problem 1 (21 points; 7 points for each question) Real data have been collected for a) inter-arrival time for high type b) inter-arrival time for low type, and c) processing time, and they are shown blow. Using the input analyzer in Arena, find out the best-fit distribution along with associated parameters for a) inter-arrival time for high type b) inter-arrival time for low type, and c) processing time. In your answering sheet, explain why you have chosen a specific distribution for each of them. If one of your answers happens to be a Normal distribution, it should be noted that Normal distribution is not bounded in both sides. a) Data for inter-arrival time for high type (unit: minute) 10.3 10.9 12 11.5 15.2 14 8.9 7.8 11.3 9.5 16.5 17.5 5.5 8.6 9.5 12 13.7 12.1

13.6 17.8 11.6

10.3 12.1 4.5

b) Data for inter-arrival time for low type (unit: minute) 5.3 1.2 7.8 2.6 5 6.2 6.9 8.3 3.7 6.1 2.9 5.9 5.4 3.9 4.8 7.2 6.8 3.9

5.5 6.3 5.1

3.4 5.3 5.1

c) Processing time (unit: minute) 7.2 6.9 10.3 8.9 8.3 9.4 8.8 10.9 8.4 6.9 10.5 6.4

9.2 8.8 7.5

8.7 12.5 6.7

4.5 11.5 5.9

11.2 8.9 7.8

Since we have 24 data, taking number of intervals between “5” and “10” is reasonable. The analysis for the corresponding situations is below: 1) Data for inter-arrival time for high type (unit: minute) 10.3 10.9 12 11.5 15.2 14 8.9 7.8 11.3 9.5 16.5 17.5 5.5 8.6 9.5 12 13.7 12.1

13.6 17.8 11.6

10.3 12.1 4.5

(An example of multiple solutions) I chose to utilize 7 intervals for this data set as it was sufficiently large enough to spread out the data and not contain unpopulated bins. I next conducted a “Fit-All” test and the analyzer yielded a Triangular distribution described in the left-hand box as the best fit. Given a small Mean-Square Error and P-values greater than 0.15 (for both the Chi-Square and K-S tests), I agree that a Triangular Distribution is the most suitable distribution to model our data.

(Expression of the fitted theoretical distribution: 4 points) (Description of the goodness of fit test: 3 points)

2) Data for inter-arrival time for low type (unit: minute) 5.3 1.2 7.8 2.6 5 6.2 6.9 8.3 3.7 6.1 2.9 5.9 5.4 3.9 4.8 7.2 6.8 3.9

5.5 6.3 5.1

3.4 5.3 5.1

(An example of multiple solutions) I chose to utilize 10 intervals for this data set as it was sufficiently large enough to spread out the data and not contain unpopulated bins. I next conducted a “Fit-All” test and the analyzer yielded a Triangular distribution described in the lefthand box as the best fit. Given a small Mean-Square Error and P-values greater than 0.15 (for both the Chi-Square and K-S tests), I agree that a Triangular Distribution is the most suitable distribution to model our data. For this set of data I also considered 9 intervals, for which the analyzer proposed a Normal Distribution as the best fit. I summarily discarded this option in favor of the Triangular Distribution at 10 intervals for a couple reasons. First the meansquare error was less for Triangular Distribution at 10 intervals than it was for a Normal Distribution at 9 intervals. Second, the Triangular Distribution yielded P-Values greater than 0.15 for both tests while the Normal Distribution yielded a P-Value less than 0.005 for the Chi-Square test. (Expression of the fitted theoretical distribution: 4 points) (Description of the goodness of fit test: 3 points)

3) Processing time (unit: minute) 7.2 6.9 10.3 8.9 8.3 9.4 8.8 10.9 8.4 6.9 10.5 6.4

4.5 11.5 5.9

11.2 8.9 7.8

9.2 8.8 7.5

8.7 12.5 6.7

(An example of multiple solutions) I chose to utilize 6 intervals for this data set as it was sufficiently large enough to spread out the data and not contain unpopulated bins. When compared against the options of using 9, 8, 7, and 5 intervals, I concluded that the best fit for all cases was a Triangular Distribution. I next concluded that the choice of 6 intervals yields the greatest P-Values for the Chi-Square and K-S tests and had the smallest Mean-Square Error.

(Expression of the fitted theoretical distribution: 4 points) (Description of the goodness of fit test: 3 points)

2. Problem 2 (9 points; 3 points for each line)

Let’s consider Process A taking 30 minutes to service a customer. We would like to schedule one hour break of Process A at 1:30 pm but Customer 1 enters Process A at 1:15 pm. Note that the capacity of Process A is one and becomes zero during the break time. Fill in the blank in the table shown below:

Schedule rule

Scheduled break start

Break start

Wait Ignore Preempt

1:30pm 1:30pm 1:30pm

1:45pm 1:45pm 1:30pm

Processing start (Customer 1) 1:15 pm 1:15 pm 1:15 pm

Scheduled break end 2:30 pm 2:30 pm 2:30 pm

Break end

Processing end (Customer 1) 2:45 pm 1:45pm 2:30 pm 1:45pm 2:30 pm 2:45 pm

3. Problem 3 (15 points) Let’s suppose we performed data collection of the time that customers spend at a local post office. Twenty (20) data have been collected, and they are shown in the following table. 0.1

0.3

0.9

1.1

1.2

1.4

1.8

1.9

2.3

2.5

2.8

3.1

3.2

3.2

3.4

3.5

3.6

3.7

3.8

3.9

Using the “Chi-square test”, determine whether the collected data are independent samples from this fitted distribution or not at level  = 0.05. Let’s assume the number of interval (k) used in the “Chisquare test” is 4. Let’s also assume that the range of the values is 0 to 4. Note: 23, 0.05 is 7.81. Note that the chi-square statistics is defined by: χ2 = ∑𝑘𝑗=1 (

2

(𝑁𝑗 −𝐸𝑗 ) 𝐸𝑗

), where Nj = number of Xi’s in the jth interval,

and Ej = expected number of Xi’s in the jth interval if the PDF was correct. A CDF of hypothesized distribution (i.e., a theoretical distribution function): 0 𝑖𝑓 𝑥 < 0 1⁄16 𝑖𝑓 0 ≤ 𝑥 < 1 4⁄16 𝑖𝑓 1 ≤ 𝑥 < 2 𝐹̂ (𝑥) = 9⁄16 𝑖𝑓 2 ≤ 𝑥 < 3 16⁄16 𝑖𝑓 3 ≤ 𝑥 < 4 { 16⁄16 𝑖𝑓 4 ≤ 𝑥 a) Fill in the table below Range Interval j Nj Ej (Nj-Ej)2/Ej 2

[0,1) 1 3 20*1/16=1.25 2.45 4.564

[1,2) 2 5 20*3/16=3.75 .4167

[2,3) 3 3 20*5/16=6.25 1.69

[3,4) 4 9 20*7/16=8.75 .0071

Ej = # of data × area of interval j in pdf b) Determined whether we need to reject H0 or not and explain why you have made that decision. H0= The data are consistent with a specified distribution at level α=0.05. Ha= The data are not consistent with a specified distribution at level α=0.05. χ2 = 4.564 < χ23,0.05 = 7.81. Thus, we cannot reject H0 (accept the distribution).

4. Problem 4 (15 points) In Problem 3, suppose that we conduct the “Kolmogorove-Smirnov (K-S) test” to find the best-fitted distribution for the collected data. Determine whether the collected data are independent samples from this fitted distribution or not at level  = 0.05. Note: D20 at level  = 0.05 is 0.294. Note that the K-S statistic is defined by: 𝐷 = Sup𝑥 (𝐹(𝑥) − 𝐹̂ (𝑥)), where 𝐹(𝑥) = the empirical CDF based on the data set in Problem 2 and 𝐹̂ (𝑥)= a CDF of hypothesized distribution (i.e., a theoretical distribution function).

a) F(x) 0 𝑖𝑓 𝑥 < 0 3⁄20 𝑖𝑓 0 ≤ 𝑥 < 1 8⁄20 𝑖𝑓 1 ≤ 𝑥 < 2 𝐹(𝑥) = 11⁄20 𝑖𝑓 2 ≤ 𝑥 < 3 20⁄20 𝑖𝑓 3 ≤ 𝑥 < 4 { 20⁄20 𝑖𝑓 4 ≤ 𝑥 b) 𝐷 = Sup𝑥 (𝐹(𝑥) − 𝐹̂ (𝑥)) |0.25-0.4|=0.15 where 1 ≤ 𝑥 < 2.

c) Determined whether we need to reject H0 or not and explain why you have made that decision. H0= The data are consistent with a specified distribution at level α=0.05. Ha= The data are not consistent with a specified distribution at level α=0.05. D= 0.15 < D20 = 0.294. Thus. we cannot reject H0 (accept the distribution).

5. Problem 5 (Exercise 3-9 in Textbook; 20 points) - In Model 3-1, suppose that instead of having a single source of parts, there are three sources of arrival, one for each of three different kinds of parts that arrive: Blue (as before), Green, and Red. For each color of arriving part, interarrival times are exponentially distributed with a mean of 16 minutes. Run the simulation for 480 minutes and compute the same performance measures as for Model 3-1. Once the parts are in the system, they retain their correct color (for the animation) but are not differentiated for collection of statistics on time in queue, queue length, or utilization (that is, they’re lumped together for purposes of processing and statistics collection on these output performance measures); however, collect statistics separately by part color for total time in system. Processing times at the drilling center are the same as in Model 3-1 and are the same regardless of the color of the part. Note: you must submit your simulation model (.doe) for full credit!!!

Q1: Add a snapshot of your model. (An example of multiple solutions)

Q2: Summarize results of your model.  Total time in system



Average

Minimum Value

Maximum Value

Blue

6.8337

2.0110

19.0057

Green

7.1205

2.2155

16.8567

Red

5.7569

1.9194

14.2642

The statistics on time in queue, queue length, and utilization

Average

Minimum Value

Maximum Value

Time in queue

3.0077

0.00

15.6421

Queue length

0.5514

0.00

4.0000

Utilization

0.6474

0.00

1.0000

6. Problem 6 (Exercise 4-4 in Textbook; 20 points) - Two different part types arrive at a facility for processing. Parts of Type 1 arrive with interarrival times following a lognormal distribution with a log mean of 11.5 hours and log standard deviation of 2.0 hours (note that these values are the mean and standard deviation of this lognormal random variable itself); the first arrival is at time 0. Theses arriving parts wait in a queue designated for only Part type 1’s until a (human) operator is available to process them (there’s only one such human operator in the facility) and the processing times follow a triangular distribution with parameters 5, 6, and 8 hours. Parts of type 2 arrive with interarrival times following an exponential distribution with mean of 15.1 hours; the first arrival is at time 0. These parts wait in a second queue (designated for Part Type 2’s only) until the same lonely (human) operator is available to process them; processing times follow a triangular distribution with parameters 3, 7, and 8 hours. After being processed by the human operator, all parts are sent for processing to an automatic machine not requiring a human operator, which has processing times distributed as triangular with parameters 4, 6, and 8 hours for both part types; all parts share the same first-come, first-served queue for this automatic machine. Completed parts exit the system. Assume that the times for all part transfers are negligible. Run the simulation for 5,000 hours to determine the average total time in system (sometimes called cycle time) for all parts (lumped together regardless of type), and the average number of items in the queues designated for the arriving parts. Animate your model, including use of different pictures for the different part types, and use resources that look different for busy vs. idle. Note: you must submit your simulation model (.doe) for full credit!!! Q1: Add a snapshot of your model (use a “Print Screen” key). - The snapshot when the human operator is idle. (An example of multiple solutions)

- The snapshot when the human operator is busy. (An example of multiple solutions)

Q2: Summarize results of your model.  Total time in system

Part1 Part2 

Average

Half width

31.2090 32.9864

5.03176 5.24000

Minimum Value 9.5027 10.1114

Maximum Value 81.5985 83.1269

Number of items in the queue

Process name

Average

Automatic 0.3413 Machine Process for part 1.4894 1 Process for part2 1.2489

Minimum Value

Maximum Value

0.00

3.0000

0.00

7.0000

0.00

7.0000