3.2 Total evaluation function
The overall evaluation function of the tool flow should consider both the utilization of the tool and the high production efficiency of the FMS. thereforeG j =lF (j) +LT ij
| In the middle | G j ---- total evaluation function of tool j |
| l ---- adjustment factor, l ≥ 1 |
In this online control, the G j of each tool is calculated for each round, and the tool with the smallest G j is selected. All types of sister knives can be selected by the above methods to complete the control of all tool flows.
4 Examples and conclusionsThe following examples are optimized using the mean control method. Each example gives a sequence of requirements for a number of parts for a type of sister knife. The length of the sequence is the number of times the part is required for that type of tool, and each demand has a period of use. Each tool has a service life. In the example, the total part is used for the tool slightly longer than the total life of the tool.
Mean Control Tool Flow Optimization Example and Results
| Case number | 1 | 2 | 3 | 4 | 5 |
| Number of tools | 3 | 3 | 19 | 30 | 30 |
| Maximum tool life (min) | 40 | 85 | 91 | 91 | 440 |
| Minimum tool life (min) | 25 | twenty one | 20 | 20 | twenty four |
| Average tool life (min) | 33.667 | 51 | 51.15 | 56.4 | 120 |
| Tool life mean square error | 120 | 2072 | 9924 | 15471 | 219807 |
| Total tool life (min) | 101 | 153 | 972 | 1692 | 3602 |
| Number of parts required for the tool (times) | 10 | 10 | 46 | 76 | 200 |
| Maximum usage time (min) | 13 | twenty three | 40 | 43 | 37 |
| Minimum usage time (min) | 8 | 11 | 10 | 10 | 10 |
| Average usage time (min) | 10.4 | 16.6 | 20.24 | 20.842 | 20.16 |
| Time-averaged variance | 0.6517 | 1.5 | 0.9769 | 0.7761 | 0.4865 |
| Tool maximum remaining life (min) | 5 | 4 | 13 | 15 | 13 |
| Tool minimum remaining life (min) | 0 | 2 | 0 | 0 | 0 |
| Tool average remaining life (min) | 3 | 3.33 | 4.632 | 4.8 | 4.2667 |
| Total remaining life (min) | 9 | 10 | 88 | 144 | 128 |
| Calculation time (s) | 19 | 20 | 32 | 36 | 36 |
| Total tool life utilization | 0.911 | 0.935 | 0.905 | 0.915 | 0.9645 |
This calculation uses the Pentium P155 computer, the memory is 16M, and it is programmed with Visual Basic 5.0. Since the total tool requirement usage time is greater than the total tool life, the tool requirements follow a predetermined sequence, and the remaining life of each tool in the example after calculation and dispatch can no longer complete the next requirement of the tool. That is, the tool has done its best at this time.
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