Cycle and reaction times
5.6 Sample calculations
CPU 31xC and CPU 31x, Technical Data
Manual, 01/2006 Edition, A5E00105475-06
5-23
Calculating the cycle time
The cycle time for the example results from the following times:
• User program execution time:
approx. 5 ms x CPU-specific factor 1.10 = approx. 5.5 ms
• Process image transfer time
Process image of inputs: 100 μs + 8 Byte x 37 μs = approx. 0.4 ms
Process image of outputs: 100 μs + 8 Byte x 37 μs = approx. 0.4 ms
• Operating system runtime at scan cycle checkpoint:
Approx. 0.5 ms
Cycle time = 5.5 ms + 0.4 ms + 0.4 ms + 0.5 ms = 6.8 ms.
Calculation of the actual cycle time
• There is no active communication.
• There is no interrupt handling.
Hence, the physical cycle time is 6 ms.
Calculating the longest response time
Longest response time:
6.8 ms x 2 = 13.6 ms.
• I/O delay can be neglected.
• Neither PROFIBUS DP, nor PROFINET IO are being used, so you do not have to make
allowances for any DP cycle times on PROFIBUS DP or for PROFINET IO update times.
• There is no interrupt handling.
5.6.2 Sample of response time calculation
Design
You have configured an S7300 and equipped it with the following modules in two racks:
• a CPU 314C-2
Configuring the cycle load as a result of communication: 40 %
• 4 digital input modules SM 321; DI 32 x 24 VDC (4 bytes each in the PI)
• 3 digital output modules SM 322; DO 16 x 24 VDC/0.5 A (2 bytes each in the PI)
• 2 analog input modules SM 331; AI 8 x 12-bit (not in the PI)
• 2 analog output modules SM 332; AO 4 x 12 bit (not in the PI)
To Next Page
Loading...
Need help?
General
