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How many machines can one operator tend? Enter load/unload, automatic and walking times and see who waits, how many parts per hour come out and which arrangement gives the lowest cost per part.
Pre-sizing and estimating tool: results do not replace design, calculations or reports signed by a licensed professional, nor verification against the applicable codes and standards.
Two steady-state cycles, in minutes: what the operator and each machine are doing at each moment.
Bars: operator and machine cost per part. Line: parts per hour of the group.
Share of the cycle idle, in %. Left of N₀ the operator waits; right of it, the machines.
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It is how many machines one operator can tend at the same time. While one machine runs on its own, the operator loads and unloads the others. The starting point is N₀ = (l + m) ÷ (l + w): the number of machines with which neither the operator nor the machines wait. It is also called multiple machine tending or synchronous servicing.
l is load and unload, when operator and machine are busy together and the machine is stopped. m is the time the machine runs on its own after starting. w is operator work that does not stop the machine: walking to the next one, gauging, deburring. Time each one at the station several times; m comes from the machine display or the CNC program.
With l = 1.2 min, m = 4.5 min and w = 0.4 min, N₀ = 5.7 ÷ 1.6 = 3.56. With 3 lathes the cycle is 5.7 min, the operator waits 0.9 min per cycle and the group makes 31.6 parts/h at $7.38 per part. With 4 lathes the cycle rises to 6.4 min, each lathe stands idle 0.7 min and the part costs $7.95. With labor at $38/h and the machine at $65/h, 3 lathes per operator is cheapest; for 12 lathes, that is 4 operators.
Below N₀ the operator waits; above it, the machines do. The better choice depends on whose hour is more expensive. In the example, 4 machines would only beat 3 if the operator cost more than 2.33 times the machine. With expensive labor and fully depreciated machines the upper option usually wins; with costly machines, the lower one.
Use "Different machines" mode and enter l, m and w for each one in walking order. The steady-state cycle is the larger of the operator’s loop (sum of l + w) and the l + m of the slowest machine. In the 4-machine example the loop takes 6.2 min and the machining center needs 7.5 min: the operator waits 1.3 min per cycle and the drill press stands idle 5.4 min.
Operator cost is the loaded cost per productive hour (wages, payroll taxes, benefits), which you can work out on the operator labor cost page. Machine cost is the hourly rate without labor: depreciation, interest, power, maintenance, floor space and tooling, as on the machine hour rate page. Do not count the operator twice.
It assumes fixed times, machines always tended in the same order, one part per machine per cycle and no random interference (breakdowns, tool changes, scrap). With unpredictable stops, real machine waiting is higher than calculated. Fatigue allowances are not included. Loading several machines requires guarding at the point of operation (OSHA 29 CFR 1910.212) and an ergonomic review of the task.
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