OIE-501 Simulation Balancing Process Capacity
docx
keyboard_arrow_up
School
Worcester Polytechnic Institute *
*We aren’t endorsed by this school
Course
500
Subject
Industrial Engineering
Date
Feb 20, 2024
Type
docx
Pages
2
Uploaded by SuperMetal13209
Introduction
As a car operations manager, I am given a budget of $3 million dollars, and I must decide on how to allocate it. I have to allocate the $3 million between vacuum, machine wash, and hand dry stations, to maximize efficiency and improve profitability. It costs $675 per day to run the car wash, at the beginning of every day the profit is -$675, each car wash yields revenue of $10 with it costing $5 per car. There are two challenges in this simulation: 1] No Variation
2] Variation
1] Challenge 1: No Variation
In this type of challenge, the capacity of each station remains constant. To, achieve maximum efficiency and profitability, the capacity of each station should be the same. Let x cars/hr be the capacity of each station. Initially, we know that:
Machine
Initial Capacity
Time required for each car
Vacuum and Trash Control
36 cars/hr
1.67 minutes
Machine Wash
30 cars/hr
2 minutes
Hand dry
24 cars/hr
2.5 minutes
The $3 million would be used to increase the capacity of each station to x.
Change in capacity for vacuum and trash control = (x-36) cars/hr
Change in capacity for Machine Wash = (x-30)/hr
Change in capacity for Hand dry = (x-24)/hr
In the simulation we are given information about the cost of the station operation for the number of cars.
Vacuum Station - $1 million for 1.5 cars/hr
Hence, cost for (x-36) cars/hr = (x-36)/1.5 million
Machine Wash - $1 million for 15 cars/hr
Hence, cost for (x-30) cars/hr = (x-30)/15 million
Hand dry - $1 million for 9 cars/hr
Hence, cost for (x-24) cars/hr = (x-24)/9 million
The total cost for the upgrade is $3 million.
(x-36)/1.5 + (x-30)/15 + (x-24)/9 = 3
Solving, for the value of x, we get x=37.5
When x=37.5, (x-36)/1.5 = (37.5-36)/1.5 = 1.5/1.5 = $1 million
When x=37.5, (x-30)/15 = (37.5-30)/15 = 7.5/15 = $0.5 million
When x=37.5, (x-24)/9 = (37.5-24)/9 = 13.5/9 = $1.5 million
Hence, the optimal solution is, Vacuum Station - $1 million, Machine wash = $0.5 million, Hand dry - $1.5 million.
Putting the above values in the simulator, we get the profit as $825.
Simulation results:
Challenge 2:
For Challenge 2, I started with substituting values obtained from Challenge 1.
Profit obtained was $655 but the hand dry capacity was not utilized completely. After multiple trials I realized that the capacity of each station varies. Through several trials I saw that the fluctuations lead to longer cycle time than the one in Challenge 1. The variables create differences in process performance and in turn affect the overall profits. Hence, if the processes are longer the profitability will decrease. In Challenge one, I had a profit of $825 a day, $103.13 an hour, and 300
cars serviced, while in challenge 2, the most I was able to achieve was $685, $85.63 an hour, and 272 cars serviced.
Below values give us the profit of $685.
Vacuum and Trash removal - $1.3 million
Machine wash - $0.6 million
Hand Dry and Windows - $1.1 million
Conclusion
In conclusion the key takeaways from this simulation are capacity, washing station allocation and utilization. In both challenges we learned that the more capacity we used the greater profit we gained overall. We also learned that decreasing the allocation of the was station and having it the least funded station in both challenges is what yielded the best results and highest profits. Lastly, we saw that the higher our utilization rate for each station, the higher our capacity could be and overall, we would be able to maximize our profits for the day.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help