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Transmission delay calculation example
The following shows a calculation example for the transmission delay time for single network system
The system configuration and its settings are as shown below.
Item
CPU module on sending side, receiving side
Total station numbers per one network
Total number of points of link devices
Refresh cycle
Transient transmission
Station-based block data assurance
CC-Link IE Controller Network module
Error station
■Link refresh time (β
The calculation formula is as follows:
Time
Link refresh time (ms)
The values of each variable are as follows:
• KM1: Constant= 1.0
• KM2: Constant = 0.36 × 10
• LB, LW: Total number of points of link devices = 1024
• LX, LY: Total number of points of link devices = 0
Link refresh time can be found by putting values of each variable in the following formula:
• β
, β
= 1.0 + 0.36 × 10
T
R
■Transmission delay time
When the station-based block data assurance setting is enabled and LT is longer than LS (LT > LS), the calculation formula is
as follows:
Calculated
Transmission delay time (ms)
value
(LT + β
Normal value
Maximum
(LT + β
value
The values of each variable are as follows:
• LT, LR: Refresh cycle = 100 (ms)
: Link refresh time 1.39 (ms)
• β
, β
T
R
• LS: Link scan time 1.25 (ms)
Transmission delay time (normal and maximum value) can be found by putting values of each variable in the following
formula:
• Normal value = (100+1.39) × 1.5 + 1.25 × 0.5 + (100+1.39) × 0.5 = 203.405 (ms)
• Maximum value = (100+1.39) × 2 + 1.25 × 1 + 100 + 1.39 = 305.420 (ms)
5 ACCESS FUNCTION USING NETWORK MODULE
102
5.5 Processing Time
, β
)
T
R
Formula
β
, β
= KM1 + KM2 × {(LB + LX + LY) ÷ 16 + LW}
T
R
-3
-3
× {((1024 + 0 + 0) ÷ 16) + 1024} 1.39 (ms)
) × 1.5 + LS × 0.5 + (LR + β
) × 0.5
T
R
) × 2 + LS × 1 + LR + β
T
R
Description
C Controller module
Two stations (one control station and one normal station)
LB/LW
1024 points
LX/LY
0 points
SB/SW
0 points
100 ms
Disabled
Enabled
Main base unit
Disabled
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