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Precautions in Using Socket Services Section 6-9
Note 4: When the CPU Unit cycle time is less than 2 ms, A = 27, and when it
is 2 ms or more, A = 53
Note 1. The values obtained from the above equations are guidelines for the trans-
mission delay time when one socket in the Ethernet Unit is used only. The
execution time required for the user program is not included.
2. The communications time for the remote nodes depends on the device be-
ing used. For remote nodes that are not Ethernet Units, calculate the com-
munications time according to the device's operation manual.
3. The actual operating environment can cause transmission delays larger
than those calculated with the methods given here. Among the causes of
longer delays are the following: traffic on the network, window sizes at net-
work nodes, traffic through the Ethernet Unit (e.g., simultaneous servicing
of multiple sockets and socket service communications, etc.), and the sys-
tem configuration.
4. The above values are guidelines when the default (4%) for the uniform pe-
ripheral servicing time in the PLC System Setup is used.
5. By increasing the value of the uniform peripheral servicing time, the maxi-
mum transmission delay time for socket services can be shorter.
Example: When using TCP socket services between two PLCs (CS-H CPU
Units) by executing the CMND(490) instruction to send/receive 512 bytes in
both directions the guideline for the maximum transmission delay time can be
calculated according to the following conditions as shown in the table below.
CPU Unit cycle time (local node) = 5 ms
CPU Unit cycle time (remote node) = 10 ms
Item Calculation
Reception processing
time (local node)
5
× 2 + 512 × 0.001 + 53 = 63.512 ms ≈ 63.5 ms
Transmission processing
time (local node)
5
× 2 + 512 × 0.001 + 53 = 63.512 ms ≈ 63.5 ms
Transmission processing
time (remote node)
10
× 2 + 512 × 0.001 + 53 = 73.512 ms ≈ 73.5 ms
Reception processing
time (remote node)
10
× 2 + 512 × 0.001 + 53 = 73.512 ms ≈ 73.5 ms
Maximum transmission
delay
63.5 + 63.5 + 73.5 + 73.5 = 274 ms