On Running Changeovers

Women Relay RaceChangeover times and their reduction are popular topics in lean manufacturing. In this post I would like to introduce the idea of running changeovers for production lines. The idea behind it is simple, and probably many of you do it already. Nevertheless, I have found little info on it online. I also would like to go into more detail on the benefits of a running changeover in comparison to the alternatives.

A “Non-Running” Changeover

Parallel Changeovers

For a production-line changeover, the “easier” approach is to do all changeovers while the line is stopped. The sequence of steps is shown below and also animated in the image below.

  1. Empty the line of all products (i.e., the ramp down).
  2. Change over all machines.
  3. Fill the line with products again (i.e., the ramp up).
Parallel change over while line is stopped
Parallel changeover while line is stopped

The animation above changes all processes over to the new product at the same time. Unless the changeovers are fully automatic or nearly so, this may be a manpower problem. You would need the people to change over all machines at the same time!

Sequential Changeovers

More realistically, you may not have the manpower to change all machines simultaneously. Rather, you change the machines one by one as shown in the animation below.

Sequential change over while line is stopped
Sequential changeover while line is stopped

Depending on your manpower or your level of automation, you can also imagine situations that are a mix of the two above. The changeovers could overlap without being fully parallel. For example, an oven could heat up on its own as part of the changeover while the mechanic is working on the changeover of the next process.

A Running Changeover

The improvement idea should be pretty obvious by now. You simply do the changeovers one machine after another while the line is running. This is called a running changeover and is shown in the animation below.

Sequential change over while line running
Sequential changeover while line running

Functional Requirements

While this running changeover looks pretty neat, it is not suitable for all situations.

First of all, it makes sense only in a flow line (or flow shop). There has to be a clear sequence of processes that the parts follow. In comparison, in a job shop there is no clear sequence, and hence the changeover cannot “run” along the line.

A running changeover may also require the parts to stay in a process longer than normal. In other words, if the changeover takes longer than a process cycle, the parts in the previous process have to wait until the changeover is completed. This is often no problem for steps such as milling or assembly. It is, however, a problem for any kind of heat treatment (for example). If a changeover makes your bread wait in the oven three times the normal time, then you will not get nice bread but a burnt loaf of ugliness. Therefore, the parts must be able to wait in the process longer than normal, or the changeover process must be reliably (!) faster than a cycle time.

Overall, there is a higher demand on the changeover timing – which is actually not a bad thing! During a “normal” non-running change over, it usually makes no difference if you start the line ten minutes earlier or later. The waiting parts during a running changeover, however, do create a sense of urgency. Workers are then somewhat less likely to dawdle and may complete the changeovers faster.

Performance Comparison

Parallel Changeover while Line Is Stopped

Let’s first look at the changeover where all processes are changed at the same time while the entire line is stopped. The total time lost consists of two parts: 1) The longest changeover time of the processes, and 2) the ramp-up time to fill the line until the first new part leaves the system again.  For clarification I have here the animation again from above.

Parallel change over while line is stopped
Parallel changeover while line is stopped

Mathematically speaking, if we have n processes, all with a cycle time CTi and a changeover time COi, then the total delay time T would be as follows (not including fine-tuning and adjusting while the line is running):

\[ T = Max(CO_1, … CO_i, …CO_n) +  \sum_{i=1}^{n} CT_i \]

Sequential Changeover while Line Is Stopped

Of course, this differs when the changeover times are not parallel, but sequential. Here is the animation again for reference.

Sequential change over while line is stopped
Sequential changeover while line is stopped

Mathematically speaking, this would be the sum of all changeover times COi plus the sum of all cycle times CTi for the ramp up.

\[ T = \sum_{i=1}^{n} CO_i +  \sum_{i=1}^{n} CT_i \]

Running Changeover

Finally, we look at the running changeover, where all processes are changed in sequence while the line is running.

Sequential change over while line running
Sequential change over while line running

Mathematically speaking, the total time between the last part and the first new part T is a bit more complex. It also depends on whether the line is a pulsed or continuously moving line, or a more flexible unstructured pacing system where the pieces move individually.

If the line is a pulsed or continuously moving line, then you lose at least one cycle time CTPulse due to one “empty slot” on the pulse line being used for the changeover. If all  changeover times COi are faster than the CTPulse, you would not lose any additional time. Only if a changeover time COi is slower than the slowest cycle time CTPulse does this extend the “pulse” and slow down the line even more. In this case the additional time T is calculated as follows.

\[ T = CT_{Pulse} + \sum_{i=1}^{n} Max(CO_i – CT_{Pulse}; 0) \]

If the line has an unstructured pacing, the flow of the material through the system is very similar, except that the pulse speed is now the largest cycle time CTMax.

\[CT_{Max} = Max(CT_1, … CT_i, …CT_n) \] \[ T = CT_{Max} + \sum_{i=1}^{n} Max(CO_i – CT_{Max};0) \]

Please be aware that these equations are an approximation, and may also depend on buffer sizes and random fluctuations.

Overall, at best you may lose only one cycle time CTMax or CTPulse for the changeover. At worst you may lose one cycle plus the sum of all changeover times in excess of this one cycle. If your changeover times are very long, the total duration may even exceed the duration of a parallel changeover.

Summary

Depending on your situation, any of the above changeover approaches for production lines may be the best.

A sequential changeover while the line is stopped may be best if  1) you cannot do all changeovers at the same time (i.e., not enough manpower or not a fully automatic changeover), and 2) the changeover durations would cause unacceptable waiting times for the products in a running changeover. To improve this situation, it may be an idea to solve the issues that prevent you from a running changeover.

A parallel changeover while the line is stopped may be the best if you 1) can do all changeovers at the same time (either with manpower available or automatic changeover), and 2) this would be faster than a running changeover. To improve this situation, you may simply reduce changeover times, and possibly check if a running changeover may be better afterward.

A running changeover while the line is running may be the best if 1) additional delay from a longer changeover time would not cause problems for the products or processes or there is no additional delay; and 2) it is faster than the parallel changeover. To improve this situation, reduce changeover times, ideally starting with the longest changeover times that exceed your cycle time.

I hope this article was interesting to you, and I hope I did not dig too deep into small details, although these small details can often make a big difference in lean. Now, go out, run your changeovers, and organize your industry!

PS: The highly influential blog and website Curious Cat by John Hunter maintains a list of Curious Cat Top Management Improvement Blogs. This ranking is based on different metrics like MOZ page authority and page rank, traffic rank, number of subscribers, and so on. I am immensely thrilled that AllAboutLean.com ranks number 4 out of 48 on  this list, right after such long established and respected blogs like the Curious Cat, the Deming Institute Blog, and Mark Grabham’s Lean Blog. Many thanks to John for the compliments 🙂

3 thoughts on “On Running Changeovers”

  1. Please help me clarify more detailed the meaning of “CT Pulse” vs “CT Max” on the post

  2. Hi Nguyen, if the changeover time at every station of a pulsed line is shorter than the pulse time, the changeover can be done while the line is running, and instead of one part there is one changeover going through the line. If the changeover takes longer than a pulse, however, the rest of the line has to wait for the changeover.

  3. Thanks Christoph for your explanation

    I have one more concern related to this formula:
    T=CTMax+∑i=1nMax(COi–CTMax;0)

    Let me clarify and help me point out if my thinking is correct or not:
    – At each station have each C/O time –> So, we take C/O time of each station minus C/T Max, and compare with 0
    – Then we will sum all of value that is higher than 0 and plus C/T Max to get the delay time T, is that correct?

    For example:
    – We have 4 stations (1,2,3,4) and C/T of each station is 1,2,3,4 (sec) respectively
    –> C/T Max = 4 sec
    – C/O of each station (1,2,3,4) is 5,10,15,20 sec respectively
    – C/O1 (5 sec) – C/T Max (4 sec) = 1
    C/O2 (10 sec) – C/T Max (4 sec) = 6
    C/O3 (15 sec) – C/T Max (4 sec) = 11
    C/O4 (20 sec) – C/T Max (4 sec) = 16
    ==> T = 4 + (1+6+11+16) = 38 sec
    Is the above result following your formula with C/T Max is correct or not? I’m really appreciated that you would help me for this understanding correctly

    Thanks a lot
    Best Regards

Leave a Comment

Cookie Consent with Real Cookie Banner