I've been relating my observations about operator utilization, a rather controversial and complex subject, and I will continue to do so for the next few months. If you would like to comment, please e-mail me and explain how your own experiences compare with what I say. I’ll relate the responses I get in an upcoming column.
As stated last month, knowing your machine and operator costs is key to making wise operator-utilization decisions—at least from a cost standpoint. At this point, I’ll assume you know what your company pays per hour for the use of your CNC machines and your CNC operators.
There are certain ideal applications when it makes sense to use one operator to run two or more machines. Consider, for example, a company that dedicates its CNC machines to running the same part, day in and day out. Some automotive companies fit into this category. Here are four things that perfect applications have in common:
For an ideal application, one operator can truly maintain two machines as efficiently as two operators would. This means, of course, that the cost of the second operator can be eliminated with zero to minimal impact on the output of the two machines.
Ideal applications don’t just happen. They must be engineered. In most, a tool-life management system is required to keep all tool maintenance internal to the machining cycle. That means that one machine will never sit idle waiting for the operator to change inserts on the other. In like fashion, automatic loading devices ensure that a machine will never wait for part loading.
Even for ideal applications, there is a limit to how much money a company can expect to save by having one operator run two machines. A simple formula can be used to determine this savings:
Savings % = (1-(O1+M1+M2) / (O1+M1+O2+M2)) * 100
Where:
O1, O2—cost of operator/s per hour.
M1—cost of machine one per hour.
M2—cost of machine two per hour.
Machine A cost/hour: |
Machine B cost/hour: |
Operator cost/hour: |
Savings: |
$20.00 |
$20.00 | $10.00 | 16.00% |
$30.00 | $30.00 | $15.00 | 16.67% |
$40.00 | $40.00 | $18.00 | 15.52% |
$50.00 | $50.00 | $22.00 | 15.28% |
$60.00 | $60.00 | $25.00 | 14.71% |
$70.00 | $70.00 | $28.00 | 14.29% |
For the sake of this formula, we are assuming that if two operators are used, they are paid the same.
For example, say that operator cost is $25 per hour (each), and machine cost is $40 per hour (each). In this case, savings will be 19.231 percent. In a cost-related context, an 8-hour shift will cost $1,040 ($135 per hour) if a separate operator is used to run each machine. It will cost only $840 ($105 per hour) if one operator runs both machines. The savings ($200) is 19.231 percent of $1,040. Again, this formula assumes that one machine is never sitting idle waiting for the operator to do something on the other. It is the best-case scenario.
When it comes to costs and cost savings, notice that only the machine and operator costs are involved. Also, note that the savings comes exclusively from the relationship between operator costs and machine costs. The higher your operator costs, the more the savings; the higher the machine costs, the lower the savings.
Even if you don’t have an ideal application, this is still a very important formula. It will tell you the very best you can hope for, based on your machine and operator costs. You should apply it before making your operator-utilization decision. The chart at the top of the page shows a few examples that stress the use of the formula.
Notice that as the difference between machine cost and operator cost grows, the percentage of savings shrinks. Again, the entire savings you can expect is based upon this difference. With a very expensive machine run by a very low cost operator, you can’t expect to gain much.
For example, if machines A and B both cost $100 per hour, and operator cost is $10 per hour, the savings is only 5.55 percent.
In applications that are less than ideal, of course, one machine will—at times—be sitting idle waiting for the operator to do something at the other machine. The term I use to describe this condition is interference. Interference, which will reduce the amount of saving you can achieve by having one person operate two CNC machines, will be the topic for next month’s column.