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5 DATA LINK PROCESSING TIME
MELSEC-QnA
(b) For writing
MB = [SL1 + LS × 2 + {(No. of writing points + 16)/16}
1
× LS + α
+ SL
2 + LS × 3 + β + constant bps] × N [ms]
SL
1 : Send end local station sequence scan time
SL
2: Receive end local station sequence scan time
LS : Link scan time (Refer to section 5.2)
α : Send end internal processing time
No. of writing points 1 to 120 points 121 to 240 points 241 to 360 points 361 to 480 points
α LS LS × 2 LS × 3 LS × 4
β : Receive end internal processing time
LS
[Constant bps]
Transmission speed 156kbps 625kbps 2.5Mbps 5Mbps 10Mbps
constant bps LS LS × 2 LS × 4 LS × 6 LS × 7
N : No. of execution instructions simultaneously
1 : Raise the decimals to a unit
(Example) When send end local station sequence scan time is 10ms, receive end local
station sequence scan time is 10ms, link scan time is 5ms, no. of writing
points is 20 words, transmission speed is 10Mbps, no. of execution
instructions simultaneously is 1
MB = [SL
1 + LS × 2 + {(No. of writing points + 16)/16}
1
× LS + α
+ SL
2 + LS × 3 + β + constant bps] × N
= [10 + 5 × 2 + {(20 + 16)/16}
1
× 5 + 5 + 10 + 5 × 3 + 5 + 5 × 7] × 1
= [10 + 5 × 2 + 3 × 5 + 5 + 10 + 5 × 3 + 5 + 5 × 7] × 1
= 105 [ms]