Brenda SaldanaHomework Assignment #2

IE 429 – Simulation

Parisay’s comments are in red.Consider the following system. Create an Arena model.  Create a Word file to contain the followings:

a) Problem statement as stated hereb) Draw logical model using software (past it into your Word file)c) Copy of statistical output (past it into your Word file)d) Create a summary table of major performance measurese) Plot number in line during simulation for Station 2 and Station 5.f)  Write a report for a manager, using a simple language; explain the performance of this system.

Problem Statement:

  Funny parts called AB arrive to a section of our system and go through different stations for some processes and then leave the system. Interarrvial time is distributed as expo(5 hr).  The process at Station 1requires 1 worker and it will take expo(4 hr). The process at Station 2 requires 1 doctor and 2 nurse and it will take expo(5 hr). The process at Station 3 requires 1 bed and it will take expo(13 hr). The process at Station 4 requires 3 of tool X and 2 of tool Y and it will take expo(10 hr). The process at Station 5 requires 2 carts and it will take expo(5 hr). We have one of Station 1, one of Station 2, 3 identical Station 3, 3 parallel Station 4, and one of Station 5. Simulate for 10,000 hrs.

Solution:

Logical Model:

Have Have Have Have Have

Simulate 10,000 hours

Station 43 Tool X2 Tool Y

Expo (10)

ArriveFunny parts

ABExpo (5)

Station 22 nurses1 doctorExpo (5)

Station 31 bed

Expo (13)

Station 11 workerExpo (4)

RestStation 52 carts

Expo (5)

Statistical Output :

Summary Table:

Acceptable range for utilization: Human: 70% – 85% Machine: 70% - 90%

PerformanceParameters

Funny Parts

Station 1

Station 2

Station 3

Station 4

Station 5

Worker Nurses Doctor BedTool

XTool

YCarts

Utilization 78% 99% 99% 84% 64% 64% 98%

Number Scheduled

1 2 1 3 9 6 2

Number used .78 1.9 .99 2.5 5.8 3.8 1.9

Average Wait Time

307.1 14.2 103.4 14 3.3 173.2

Maximum Wait Time

684.8 93 270 95.4 44.8 445.6

Average Number in

Line68.7 2.8 20.6 2.7 .66 34.5

Maximum Number in

Line142 19 53 26 11 89

Total Time 750.66

Number in 2002

Number out 1887

Plot Number in Line:

Report to Manager:

A simulation of the process was made in order to better understand how funny parts AB efficiently go through the 5 different stations and if all the stations are being utilized to their specifications. The situation was simulated over a span of 10,000 hours. In this time Funny parts AB were processed through stations 1 through 5. {This comment is not important for manager. Instead you can write about throughput. A total of 2002 parts entered the system, while only 1887 completed the process. This is an indication that there are parts that are still waiting to be processed during this time. You can say the throughput is 1887/10000}

Based on our summary table above the average waiting times for Station 2 and Station 5 are the highest compared to the others, respectively they are 103.4 and 173.4 hours. This indicates that most of the parts are being held here for a long time that could be cause for a bottleneck in the system. It should be noted that in station 2 and 5 the maximum number of parts in line are respectively 53 and 89, which is a high indication of a possible bottleneck in these stations. This is represented in the graphs above. It is to my recommendation that theses Stations be reviewed and improved so that waiting times are decreased.

In addition, because Stations 2 and 5 have the highest average waiting times, utilization in these stations are high as well. For station 2, the utilization for nurses and doctor is both 90%. For station 5, utilization of the carts is 98%. An acceptable range of utilization is noted above for humans between 70% and 85% and for machines between 70% and 90%. According to the acceptable range of utilization, the

nurses, doctor and carts are being over overworked causing safety and maintenance concerns. It is recommended that the company hire more nurses and another doctor for station 2 and invest in more carts for station 5 to bring the utilization to the acceptable range.

Below is a good summary table and report by Mr. Lossio.

Report to Manager: Actual Resources summary table:

Station Resource Needed Quantity Needed/Operation

Actual Capacity

1 Worker 1 1

2 Nurse 2 2

Doctor 1 1

3 Bed 1 3

4 Tool X 3 9

Tool Y 2 6

5 Carts 2 2

In the

hospital

in

average, a patient tends to spend 307.11 hours in line. Also, in average there are 68.75 patients in the hospital either being

Performance Measurements Calculated Values Acceptable ValuesAverage service time 36.84 hr

Average Waiting time in line 307.11 hr Less or equal than 36.84 Average number of Patient in the Hospital 68.75

Maximum number of patients in the Hospital 142Average waiting time in Station 1 14.23 hr

Average waiting time in Station 2 103.48hr Less or equal than 36.84Average waiting time in Station 3 14.05hrAverage waiting time in Station 4 3.33 hr

Average waiting time in Station 5 173.24 hr Less or equal than 36.84Utilization of Worker 78.84% Between 75% to 85%

Utilization of Doctor and Nurses 99.51% Between 75% to 85%Utilization of Bed 84.81% 85% to 90%

Utilization of tool X and Y 64.48% 85% to 90%Utilization of Carts 98.40% 85% to 90 %

Average patient in line in Station 1 2.85Average patient in line in Station 2 20.63 10 in line and one being

servedAverage patient in line in Station 3 2.79Average patient in line in Station 4 0.66

Average patient in line in Station 5 34.58 10 in line

served or waiting in line; therefore for layout purposes there should be a consideration of the occupancy of 68.75 in

average and a maximum of 142 patients. In additional, in average a patient is being served for 36.84 hours which if it is

compared to the average waiting time, a patient waits more than double its service time in the hospital.

Based on our analysis, it shows that there are 115 patients being entered in the hospital and not being served.

There is excessive utilization of doctors and nurses of 99.51%; as a consequence, it shows that in station 2, patients wait

in average of 103.48 hr and there is a long line of 20.63 patients in average. This means that if there is an emergency,

chances there will no time. A room for improvement is needed in this station. Considering an option of hiring one more

doctor and 2 more nurses will reduce the waiting time about a 50% as well the number of patients waiting in line.

In station 5, Carts are being utilized at 98.40% in average; as a result, there is an average of 173.24 hr waiting time for this

station as well there is an average of 34.58 patients waiting in line. The acceptable value is that at least there are 10 in line

which means that there is a need of purchasing more carts. Right now, there are two carts in the hospital, and in order to

reduce the waiting time in line and the number of patients waiting in line there should be at least 3, but more analysis

should be done in order to reach the acceptable values.

Also there is a low utilization of 64.48% of the tool X and Y, and this reflects the low average waiting time in line of 3.33

hr, and the average number of patients in line of 0.66

Part 2

Problem Statement:

Assume parts arrive at a facility for service. After service they leave the facility. We would like to simulate the system, using Arena, for 30000 unit of time and find the average waiting time, average number in queue, maximum number in queue, and server utilization from the statistical output. Create a model for the following scenarios. (one hour is unit of time)

Interarrival times follow an exponential distribution with a mean of 8 units of time.

a) Service times are exponentially distributed with a mean of 5 units of time b) Service times are normally distributed with a mean of 5 units of time and a standard deviation of 1

units of time. c) Service times are uniformly distributed with a min of 1 units of time and a max of 9 units of time. d) Service times are distributed as triangular distribution with a 2, 6, and 7 units of time for mix, mode, and max, respectively.

Print statistical outputs. Create a summary table where columns indicate each scenario and rows are:

1) Specifications (interarrival and service times distributions) for each model 2) Average waiting time 3) Average number in line4) Maximum number in line 5) Utilization of server

Analyze this summary table and write (type) your findings and conclusions.

Solution:

Logical Model

A)

Simulate 30,000 hours

B)

Simulate 30,000 hours

C)

Simulate 30,000 hours

D)

ArriveParts

Expo (8)

ServiceExpo (5)

Leave

ArriveParts

Expo (8)

Leave

LeaveServiceTria (2,6,7)

ArriveParts

Expo (8)

ServiceUnif (1, 9)

LeaveServiceNorm (5,1)

ArriveParts

Expo (8)

Simulate 30,000 hours

Statistical Output:

A) EXPO (5)

B) NORM (5,1)

C) UNIF (1,9)

D) TRIA (2,6,7)

Summary Table:

Acceptable Utilization Range: human 70% - 85%

Analysis and Conclusion:

By Mr. Lossio: The utilization of the server for all four scenarios is similar. Scenario B and C have a utilization of 63.7 %, scenario D has a 63.8% utilization, and scenario A with 64.26%, and this is due to the fact that average service time is the same for all service distribution (5). {This is an important understanding of the behavior of the queuing systems.}In scenario A, there is much greater waiting time in line as opposed to the rest of the scenarios, and this is due to the larger variation (std dev) of its exponential service distribution compared to the other distributions. {This is an important understanding of the behavior of the queuing systems. We can use this concept for improvement of a system!}For the same reason, scenario “A” has an average of 1.19 parts waiting in line, with 19 parts as maximum in line. This difference is due to the randomness {You need to be specific on how randomness affected the case.} of the system, but it only occurs two times in 30000 hours length simulation.

With a normal service time distribution of N (5, 1), the waiting time is the least time of 4.37 from all four scenarios. The average of parts in line is .56, but the maximum number in line is 7 which are due to smaller variation.

Part 3

Performance Parameters

Exponential Normal Uniform Triangular

Input infoInterarrival

time distExp(8hr) Exp(8hr) Exp(8hr) Exp(8hr)

Service time dist

Exponential (5) Normal (5, 1) Uniform (1, 9) Triangular (2, 6, 7)

Mean service time

5 5 5 5

Std. dev service time

5 1 2.31 1.08

Output infoAverage

waiting time9.3 4.3 5.7 4.5

Average number in line

1.19 .55 .72 .58

Maximum number in line

19 7 13 8

Utilization of server

64.2% 63.7% 63.7% 63.8%

Problem Statement:

For each one of the following distributions draw (a rough graph) a probability distribution function, write down the number of their parameters and what is the name of each parameter, mention what is its rang, and calculate its mean and standard deviation. (Use Appendix D in your textbook)

Normal(7,1); Expo (6); Uniform(2,8); Beta(3,5); Erlang(2,7); Gama(5,2); and Triangular(2,7,10)

Dist Normal Exponential Uniform Beta Erlang Gama Triangular

Number of Parameters

2 1 2 2 2 2 3

Name of each

parameter

Norm(

Mean (Mu) =

Standard Deviation =

Expo(

Mean =

Unif(a,b)Minimum = aMaximum = b

Beta (

Shape parameters

Beta = Alpha

=

Erlang (

Beta = Number

of exponential random variables =

Gama (

Shape parameter (Alpha) =

Scale parameter (Beta)=

Tria(a,m,b)Minimum = a

Mode = mMaximum = b

Range ( [0, [a,b] [0,1] [0, [0, [a,b]

EquationsMean =

SD =

Mean =

SD =

Mean =

SD =

Mean =

SD Mean =

SD =

Mean =

SD =

Mean =

SD =

CalculateNorm(7,1) Mean = 7

SD = 1

Expo(6)Mean = 6

SD = 6

Unif(2,8)Mean = 5SD = 1.73

Beta (3,5)

 

= .375 SD =

= .161

Erlang(2,7)Mean = (7*2) = 14

SD =

Gama(5,2)Mean = (2*5) =

10

SD =  7.07

Tria(2,7,10)

Mean = = 6.3

SD=

= 2

Graph

Part 4

Problem Statement:

Finish your quiz of 1-12-2010 as below:

a) How do we draw a histogram? Explain using an example. Data Sample: 3, 6, 10, 13, 14 n = 5

Number of classes: k = = = 2.23 = 2

Interval:

b) Draw a uniform distribution and a normal distribution. What are their parameters?

Parameters: Parameters:

Unif(a,b) Norm(

c) What is the difference between continuous random variable and discrete random variable? Explain and provide an example.

Continuous R.V – describe infinite possibilitiesEx: Time between arrival of parts

Discrete R.V. – describe only limited (finite) possibilities

Uniform Distribution Normal

Distribution

Ex: Flipping a coin, is either heads or tail

d) What is the relationship between histogram and probability function?

By: Mr. Lossio: There is a close relationship between histograms and probability function because probability is calculated based on number of observation in one cell divided to total number. Therefore the cells of probability function is proportionate to the cells of histogram. Therefore, changing histogram parameters

will lead to represent different probability distributions models

e) Draw a triangular distribution and an exponential distribution. What are their parameters?

Parameters: Parameters:

Expo( Tria(a,m,b)

f) What is the difference between density function and cumulative function? Explain and provide example.

By Mr. Lossio: The probability distribution function (pdf) for a continuous random variable is a function that assigns a probability to each value of the random variable. The probability that the random variable X assumes for any specific value xi is the value of the pdf for xi and is denoted Px(x=xi)=f(xi).

The cumulative probability distribution functions for a random variable X is defined as the probability that the random variable is less than or equal to a specific value x. and is denoted Px(x<=xi)=F(xi).

Exponential Distribution

Triangular Distribution