Manufacturing capacity planning simply comes down to proper asset allocation and utilization. Make sure the company’s machines are never idle, run smoothly and rarely encounter any down time and everything should take care of itself. Unfortunately, the basis for capacity planning relies upon linear & lean manufacturing principles that are often more impressive in theory, than they are in application.
Lean manufacturing may tout the benefits of optimizing manufacturing resources, but to ensure a manufacturer never encounters down time is often next to impossible. Therefore, when it comes to manufacturing capacity planning, linear programming is more representative of the ideal situation, than the reality that manufacturers face. However, what can be learned by adopting linear & lean manufacturing principles?
First, before venturing down the road of the ideal situation, it’s important to note that to increase production throughput means to 1) eliminate down time, or 2) to reduce cycle times. In essence, either will suffice and ultimately to do one means to accomplish the other. Keep in mind that linear programming is the best case scenario and doesn’t account for those common everyday issues all manufacturers face.
In order for this to resonate with you, the reader, we’ll first look at the calculation behind linear programming, and then explain why this “best case” scenario can only be applied once the company has eliminated all causes of lost time.
There is value to linear programming and several companies use advanced manufacturing software programs to plan production. However, by and large, this approach is only useful once the company has first eliminated its down time and maximized its productivity rates.
To learn about properly laying out and design work cells on your shop floor, please read: Manufacturing Work Cell Optimization: Design, Layout and Analysis
Manufacturing Capacity Planning – Linear Programming’s Ideal World
Let’s assume that a company manufacturers two products. To make this example as simple as possible, we’ll limit the number of variables and work operations. For simplicity, we’ll call these two products widget “X” & widget “Y”. The company makes a profit of $300 on widget “X” and $400 on Widget “Y”. They have also broken down their work processes into two operations: CNC machining & assembly. In addition, they have properly identified their manufacturing capacity for each widget based on running two shifts, five days a week for a full month. They have 1400 hours available to make widget X, and 1000 hours available to make widget Y.
The two operations:
- CNC Machining
- Assembly
Manufacturing capacity in hours based on one month’s production:
- Widget X = 1400 hours
- Widget Y = 1000 hours
Profit derived from each widget:
- Widget X = $300
- Widget Y = $400
The following table shows each operation for each product line, with a summary of the manufacturing capacity for each and the corresponding profit.
The questions that must be answered include the following:
- How much should the company make of Widget X?
- How much should the company make of Widget Y?
- How much overall profit will the company make with both product lines?
The process to make this analysis complete involves using the linear programming calculation.
- Widget X = 2(x) + 1(y) <= 1400
- Widget Y = 1(x) + 3(y) <= 1000
In our example, we’ll isolate the Y value first.
We do this by taking the first equation above (Widget X equation). I’ve highlighted it red to show it later in the process.
- 2(x) + 1(y) <= 1400
To isolate the Y value, we move the 2(x) to the opposite side of the equation.
- 1(y) <= 1400 – 2(x) this portion of the equation states that 1(y) is equal to or less than 1400 – 2(x)
Now, we revert to the second equation (Widget Y equation) and switch in the 1(y) value above. This process is highlighted above & below in bold BLUE.
Widget Y = 1(x) + 3(y) <= 1000
- 1(x) + 3{1400-2(x)} <= 1000
- 1(x) + 4200 – 6 (x) <= 1000
- - 5 (x) <= - 4200 - 1000
- - 5(x) = -3200
- X = 640
Now, Having the X value, we can now determine our Y value. We’ll do this by taking the first equation we started with {2(x) + 1(y) <= 1400}
- 2(x) + 1(y) <- 1400
- 2 (640) + 1(y) <= 1400
- 1280 + 1(y) <= 1400
- 1(y) <= 1400 – 1280
- Y = 120
Testing the above values is very straightforward and will be done by keeping the current equation {2(x) + 1(y) <= 1400} and plugging in the values for X & Y.
- 2(640) + 1(120) <= 1400
- 1280 + 120 <= 1400
- 1400 = 1400 – this is correct!
At the end of the day, the company will maximize manufacturing capacity by manufacturing 640 units of Widget “X” and 120 units of Widget “Y”. Finally, the company’s profit totals for each product line will be the following:
- Widget X = 640 Units * $300 = $192,000.00
- Widget Y = 120 Units * $400 = $48,000.00
The video above provides a more in-depth approach to increasing production throughput. It is taken from the post: Simplifying Lean Manufacturing: Work Cell Output, Cycle Time Variances & Production Volumes
When Manufacturing Capacity Planning Works
As mentioned, the above must always be considered a “best case” scenario. In this case the company doesn’t encounter any down time. In addition, we’ve only looked at two work operations and every manufacturer has more than just two. However, the best manufacturing software programs are able to calculate multiple variables and multiple work operations. Regardless, our example showed the company was able to run non-stop without interruption. Now, most manufacturers know this simply isn’t possible, UNLESS, the company improves its manufacturing productivity rates by eliminating its down time and or, reducing cycle times.
Succeeding in manufacturing capacity planning does involve theory, but the most successful manufacturers know that to run an efficient manufacturing facility takes constant effort, and the willingness to tackle the biggest problems. For linear programming to be successful, the company would first determine its productivity rate, and then set about plans to eliminate its down time and improve its cycle times. Only then can these approaches be useful.
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