
2. SEQUENCE PROGRAM
B–61863E/15
PMC SEQUENCE PROGRAM
58
(3) Processing time calculation example (for PMC–SB)
(a) 1st level sequence
Basic instruction: 100 steps
Functional instruction:
CTR: 2 times,
COMPB: 2 times
CTR execution time constant: 26
COMPB execution time constant: 24
END.1 execution time constant: 206
HT={100+(26 2+24 2+206) 10} 0.15 =474 µsec
(b) 2nd level sequence
Basic instruction: 6,000 steps
Functional instruction:
TMR: 35 times,
DECB: 25 times,
ROTB: 2 times
TMR execution time constant: 23
DECB execution time constant: 20
ROTB execution time constant: 33
END.2 execution time constant: 32
LT={6,000+(23 35+20 25+33 2+32) 10} 0.15=3004.5msec
(c) Determination of the number of divisions (n)
3004.5 µsec
n=
1250µsec – 474 µsec
+1 = 4.87
(d) Processing time calculation
Sequence processing time=4 (number of division) 8 msec=32
msec
NOTE
For the PMC–SB/SC, see the execution time constant of
each function instruction in Table 5 (b) in Section I–5, ”PMC
FUNCTION INSTRUCTIONS.”