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Man-machine ratio calculator

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.

Pre-sizingRevised 11/02/2026
Times per machine

Times in minutes (45 s = 0.75 min). l stops the machine; w is operator work while the machine runs.

Hourly costs
Arrangement

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—Machines per operator
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Man-machine chart (multiple activity chart)

Two steady-state cycles, in minutes: what the operator and each machine are doing at each moment.

Cost per part and output by number of machines

Bars: operator and machine cost per part. Line: parts per hour of the group.

Tap the chart to see the values.

Who waits: operator or machines

Share of the cycle idle, in %. Left of N₀ the operator waits; right of it, the machines.

Tap the chart to see the values.

Comparison by number of machines

Next step

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Method, assumptions and references

Equations used

N₀ = (l + m) ÷ (l + w) (l = load/unload; m = automatic; w = walk and inspection)
n ≤ N₀: the operator waits; cycle = l + m; cost per part = (C_op + n·C_mach)·(l + m) ÷ n
n > N₀: the machines wait; cycle = n·(l + w); cost per part = (C_op + n·C_mach)·(l + w)
Operator idle = cycle − n·(l + w); each machine idle = cycle − (l + m); parts/h = 60·n ÷ cycle (min)
Break-even between n₁ = ⌊N₀⌋ and n₂ = ⌈N₀⌉: C_op ÷ C_mach = [n₂·(l + w) − n₁·a] ÷ [a − (l + w)], with a = (l + m) ÷ n₁
Different machines, fixed order: steady-state cycle = max[Σ (lᵢ + wᵢ); maxᵢ (lᵢ + mᵢ)]; checked by discrete simulation of the loop

Assumptions and limits

  • Deterministic times; the operator always tends the machines in the same order; one part per machine per cycle; the machine is stopped during load and unload.
  • Random interference (breakdowns, tool changes, scrap) and fatigue allowances are not included: with frequent stops, real machine waiting rises and the ideal number of machines falls.
  • Cost per part covers only operator and machines (machine rate without labor); material, per-part tooling and overhead are excluded.
  • In the chart, the cycle starts when the pace-setting machine is served; in steady state, the starting point does not change the cycle or the waits.

Standards and technical documents

  • Freivalds, A.; Niebel, B. W. Niebel’s Methods, Standards, and Work Design, 13th ed., McGraw-Hill, 2014 (worker-machine relationships: synchronous servicing)
  • Groover, M. P. Work Systems and the Methods, Measurement, and Management of Work, Pearson Prentice Hall, 2007 (worker-machine systems)
  • Kanawaty, G. (ed.). Introduction to Work Study, 4th ed., International Labour Office, Geneva, 1992 (multiple activity charts)
  • Barnes, R. M. Motion and Time Study: Design and Measurement of Work, 7th ed., Wiley, 1980 (man and machine charts)

U.S. regulations

  • OSHA 29 CFR 1910.212: general requirements for all machines (point-of-operation guarding)
  • OSHA General Duty Clause, Section 5(a)(1) of the OSH Act, for ergonomic hazards

Frequently asked questions

What is the man-machine ratio?

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.

What are l, m and w and how do I measure them?

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.

How do I read the example result?

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.

Why compare the floor and ceiling of N₀ instead of rounding?

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.

What if the machines are different?

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.

Where do I get the hourly costs?

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.

What are the limits?

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.