Summary

You found the bottleneck with an instrument, subordinated the release to it and saw that only it is worth speeding up. All the new electronics fits in a few lines:

  • probe — CD4017: CLOCK = S, outputs O0…O4 = P1…P5, RESET = P6 OR R — the «one» goes round the five machines;
  • multiplexer — SB = (P1 · B1) + (P2 · B2) + … + (P5 · B5): the BUSY of the chosen machine (CD4081 and CD4072);
  • idle meter — two CD4029: CLOCK = CKR, CI = SB — a second is added only while the machine stands; reset PE = R OR S;
  • divider — CD4013: D1 = QN1, CL1 = REL — HALF rises once every 2 seconds;
  • rhythm — CD4019: RELS = REL · NOT SLOW + HALF · SLOW.
FL-01 panelCK · RUN · R · REL · CKRSbuttonCD4017P1…P5AND · ORCD4081 · CD4072 → SB2 × CD4029idleCIdisplay ICKRCD4013HALFRELCD4019SLOWsourceRELS
The kit’s new electronics on top of the FL-01 panel: the probe and idle meter (bottom) and the divider with the rhythm switch (middle).

What's next

  • Make machine 5 slow instead of machine 3. Find it with the instrument without looking at the machines.
  • Give Herbie a die 1…6. His idle time stops being zero: sometimes he waits. Why is that dangerous for the whole factory?
  • Release at Herbie's rhythm, but give him a small buffer (cap = 1). Is one part in the queue enough to keep him busy?
  • Make two machines slow — 2 and 4. Which of them is the real bottleneck?