# Fixed Repeating Schedule – Product Family Turnover Rate

Here’s another example for “sizing the wheel” for a mixed model fixed repeating schedule.

Given:

1. Changeover Time = 100 minutes
2. Available Production Time = 2 shifts * 7 hrs/shift * 60 min/hr = 840 min/day

and

 Product Daily Demand (pcs) Cycle Time (min.) Load (min.) A 80 5 400 B 40 4 160 C 20 4 80 D 10 6 60 700

Now some math:

1. Total Load = 700 min.
2. Time Available for Changeovers = 840 min/day – 700 min/day = 140 min/day
3. Changeover Time for group = 100 min * 4 products = 400 min/family
4. Changeovers per day = 140 min/day / 400 min/family = 0.35 group/day or a Replenishment Time of 2.85 days/family

So the fixed repeating wheel will turn once in 2.86 days.  Production run sizes as follows:

 Product Daily Demand * Replen Time Cycle Time (min) Load (min) A 229 5 1145 B 114 4 456 C 57 4 228 D 29 6 174 2003

Plus 4 changeovers of 100 min each = 2403 min = 2.86 days.

When it comes time to run product A, run 2.86 days worth. Got it?

Would be nice if we could run just one piece.  But until we can make the 100 minute changeover go away we’re stuck running a batch of some size.