Problem Formulation

Introduction:

In systems engineering there are many tools that are useful in the improvement of industry and in many life aspects. One of the major tools is the simulation. It gives the imitation of the operation of a real world process. Simulation is used in many contexts, such as simulation of technology for performance optimization, safety engineering, testing, training, education, and video games.Furthermore, it can be used for scientific modelling of natural systems or human systems in order to gain insight into their functioning. In this project, our system is about a full gas station with a their facilities (gas pumpers, supermarket, car washing and ATM machine) .the concept of theour simulation is used to simulate a real gas station, improve the system and set some alternatives that may improve the system in the future.

Problem Formulation:

In this project we simulated a real system of gas station located in AL-Doha neighbourhood near KFUPM. The station works 7 days in the week with 18 hours daily. The gas station has:

A) Two kinds of gas (octane 95 and octane 91), and there are 2 workers for each kind.

B) Supermarket with one worker.

C) Carwash with oneworkers.

D) One ATM machine.

We noticed that there are 10 types of vehicles entering the gas station, and we differentiated between these types based on their sequences. The following details will explain what their sequences are:

Type 1 follows Sequence 1 / ATM / 91 octane / Supermarket
Type 2 follows Sequence 2 / ATM / 95 octane / Supermarket
Type 3 follows Sequence 3 / ATM / 91 octane
Type 4 follows Sequence 4 / ATM / 95 octane
Type 5 follows Sequence 5 / 91 octane
Type 6 follows Sequence 6 / 95 octane
Type 7 follows Sequence 7 / ATM / Carwash
Type 8 follows Sequence 8 / Carwash
Type 9 (Trucks) follows Sequence 9 / Diesel
Type 10 follows Sequence 10 / Supermarket

Data collection and analysis:

We have collected our data during last month. Each member of our group gathered one third of the data. We have attached our data in excel file.

Model construction and validation:

In the following figures, we have specified what is the appropriate distribution for each process and we validate each process by a goodness of fit test ( chi-square and Kolmogorov tests ).

(Small cars - Octane-91)

Type of distribution and analysis / graph
Distribution:Beta
Expression:1 + 3.85 * BETA(1.53, 2.59)
Square Error:0.042479
Chi Square Test
Number of intervals= 6
Degrees of freedom = 3
Test Statistic = 100
Corresponding p-value< 0.005
Kolmogorov-Smirnov Test
Test Statistic= 0.111
Corresponding p-value< 0.01
Data Summary
Number of Data Points= 500
Min Data Value = 1.02
Max Data Value = 4.5
Sample Mean = 2.43
Sample StdDev = 0.831
Histogram Summary
Histogram Range = 1 to 4.85
Number of Intervals= 7 /

(Large cars - Octane-95)

Type of distribution and analysis / graph
Distribution:Gamma
Expression: 1 + GAMM(0.647, 3.37)
Square Error:0.000432
Chi Square Test
Number of intervals= 6
Degrees of freedom = 3
Test Statistic = 3.05
Corresponding p-value= 0.401
Kolmogorov-Smirnov Test
Test Statistic= 0.12
Corresponding p-value< 0.01
Data Summary
Number of Data Points= 500
Min Data Value = 1.19
Max Data Value = 5.54
Sample Mean = 3.18
Sample StdDev = 1.06
Histogram Summary
Histogram Range = 1 to 5.98
Number of Intervals= 6 /

(Tracks - Diesel)

Type of distribution and analysis / graph
Distribution:Weibull
Expression:3 + WEIB(2.34, 2.02)
Square Error:0.001135
Chi Square Test
Number of intervals= 6
Degrees of freedom = 3
Test Statistic = 3.28
Corresponding p-value= 0.369
Kolmogorov-Smirnov Test
Test Statistic= 0.122
Corresponding p-value< 0.01
Data Summary
Number of Data Points= 500
Min Data Value = 3.03
Max Data Value = 7.52
Sample Mean = 5.09
Sample StdDev = 1.05
Histogram Summary
Histogram Range = 3 to 7.97
Number of Intervals= 6 /

(Supermarket)

Type of distribution and analysis / graph
Distribution:Normal
Expression:NORM(2.85, 0.823)
Square Error:0.018561
Chi Square Test
Number of intervals= 7
Degrees of freedom = 4
Test Statistic = 82
Corresponding p-value< 0.005
Kolmogorov-Smirnov Test
Test Statistic= 0.119
Corresponding p-value< 0.01
Data Summary
Number of Data Points= 500
Min Data Value = 1.05
Max Data Value = 4.55
Sample Mean = 2.85
Sample StdDev = 0.824
Histogram Summary
Histogram Range = 1 to 4.9
Number of Intervals= 7 /

(ATM)

Type of distribution and analysis / graph
Distribution Summary
Distribution:Normal
Expression:NORM(2.79, 0.792)
Square Error:0.021620
Chi Square Test
Number of intervals= 8
Degrees of freedom = 5
Test Statistic = 79.6
Corresponding p-value< 0.005
Kolmogorov-Smirnov Test
Test Statistic= 0.107
Corresponding p-value< 0.01
Data Summary
Number of Data Points= 500
Min Data Value = 1.12
Max Data Value = 4.43
Sample Mean = 2.79
Sample StdDev = 0.793
Histogram Summary
Histogram Range = 1 to 4.77
Number of Intervals= 8 /

(Carwash)

Type of distribution and analysis / graph
Distribution Summary
Distribution:Weibull
Expression:8 + WEIB(4.58, 2.29)
Square Error:0.000473
Chi Square Test
Number of intervals= 8
Degrees of freedom = 5
Test Statistic = 2.72
Corresponding p-value= 0.743
Kolmogorov-Smirnov Test
Test Statistic= 0.0704
Corresponding p-value= 0.0146
Data Summary
Number of Data Points= 500
Min Data Value = 8.09
Max Data Value = 16.4
Sample Mean = 12.1
Sample StdDev = 1.85
Histogram Summary
Histogram Range = 8 to 17
Number of Intervals= 8 /

The Simulation Model:

Based on the data we have collected, we built a model with arena software to visualize the real system

And inside the station services there are:

Model Verification and Validation:

For verification:

we have checked our simulation model with our lab instructor MR. Adel Fadhel.
Flow diagram for our model actions are:

1-Vehicle arrival event:

The vehicle comes to the server and finds it either:

-Idle server:

The vehicle begins the service immediately

-Busy server:

-The vehicle enters the queue

In our model each type will have a sequence so, there will be a difference in the flow chart. We did it for the first sequence only.

HINT: It is not possible for the server to be idle while the queue is nonempty.

2-Vehicle departure event:

-Case 1:If there are no vehicles waiting in the queue, then the server status changes to idle.

-Case 2: If there is one or more vehicle waiting in the queue, then the next vehicle begins service.

Hint: In both cases the number of vehicles in the system decreases by one vehicle.

We have examined the model output for reasonableness under a variety of setting of the input parameters

For validation:

For the validation of the model to be possible, we will construct a theory to compare with real system data on the assumption that the two data are collected in the same time (from the model, from the real system).

Theory:

We will compare the average process time, which we got from the simulation (Ŷ) and we will compare it with (μi) by using the t-test with α=0.05, n= 5 replication

Hypothesis Testing:

If H0 is not rejected, then, there is no reason to consider the model invalid

If H0 is rejected, the current version of the model is rejected, and we need to improve the model

1-Validation of the mean process time for 91 pumps:

Let the true mean of the system.2.43 min. Y is the process time for 91.

H0: E(Y)=

From Arena, using 5 replications, average processing time, E(Y)= 2.236min.

S=0.831 min. Therefore, t0==0.522<t.025,4= 2.7764. Therefore H0 is accepted.

2- Validation of the mean process time of the 95 pumps:

Let the true mean of the system. 3.18 min. Y is the process time for 95.

H0: E(Y)=

From Arena, using 5 replications, average processing time, E(Y)=3.296 min.

S=1.05 min. Therefore, t0==0.244<t.025,4=2.7764. Therefore H0 is accepted.

3- Validation of the mean process time for the Supermarket:

Let the true mean of the system. 2.85 min. Y is the process times of supermarket.

H0: E(Y)=

From Arena, using 5 replications, average processing time, E(Y) = 3.7 min.

S=0.824 min. Therefore, t0==2.306<t.025,4 =2.7764. Therefore H0 is accepted.

4-Validation of the mean process time for ATM :

Letthe true mean of the system. 2.79 min. Y is the process times of ATM.

H0: E(Y)=

From Arena, using 5 replications, average processing time, E(Y) = 3.21min.

From Excel, S=0.793 min. Therefore, t0==1.18<t.025,4 =2.7764. Therefore H0 is accepted.

5- Validation of the mean process time for Carwash:

Let the true mean of the system. 12.1 min. Y is the process times of Carwash.

H0: E(Y)=

From Arena, using 5 replications, average processing time, E(Y)=13.3min.

S=1.85 min. Therefore, t0= =1.18<t.025,4=2.7764. Therefore H0 is accepted.

Output Analysis:

1-Number of all customers in 7 days, 18 hours per day

2-Average total time spent in the system for each vehicle type,

As clarified before, each type follows a certain sequence

3-Average waiting time in queue for each service in the gas station

4-Instantaneous and scheduled utilization for the workers and the ATM machine

5-Total number of customers seized by the workers and the ATM machine

Conclusion and recommendation:

In this project we applied what we have learned in the course by using a simulation program “ ARENA ‘’ in order to simulate and improve a real gas station system with their facilities (91&95 Pumps, Supermarket, ATM and Carwash). We began with identifying the system and the observed vehicles sequences so we gathered their data and we tried to be unbiased as much as possible. We noticed that there is a large waiting time in the carwashingqueue,It is 32.59 minutes so we recommend to increase the number of workers in the carwash into 2 workers. If there are two workers the waiting time in the queue 1.269 minute