Illustration of a semicircular utilization gauge with low, medium and high bands and a pink needle

How Equipment Utilization Changes Automation Payback

Low, medium and high utilization scenarios for one $60,000 automation case, with a sensitivity check and a break-even hours formula.

You can buy the same machine at the same price and get very different results. In the example below it pays back in 16 months at one plant and in nearly seven years at another. The difference is how many hours a year it actually produces.

We run one $60,000 investment (the same one as in the payback, ROI and TCO article) through three utilization scenarios, test which input matters most, and work backward to the hours you need for a given payback target.

The model

Simple payback assumes a steady annual benefit. Here the benefit is tied to productive hours, because most savings (labor not paid, material not wasted, orders filled in-house) only occur while the machine is producing.

B = (net cash benefit per productive hour x productive hours per year)
    - fixed annual cash costs

Payback (years) = I / B, only when B > 0
Payback (months) = payback in years x 12

Net cash benefit per productive hour is the cash saved or earned per hour of running, after the variable costs of running. Fixed annual cash costs are paid whether or not the machine runs, a service contract for example. When B is zero or negative, these inputs give no positive payback.

Inputs and scenarios

Worked example. Replace each input with a quote or a measured figure from the plant before using the result.

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Input Value
Installed investment (I) $60,000
Net cash benefit per productive hour $12
Fixed annual cash costs $3,000

The three scenarios differ only in productive hours per year. A single 8-hour shift over about 250 working days is 2,000 hours if the machine runs the whole time, so 1,000 hours is roughly half a shift, 2,000 is one shift and 4,000 is two. Real plants lose hours to changeovers and stoppages, so treat that mapping as rough.

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Scenario Productive hours per year Net benefit (hours x $12) Fixed costs B Payback (years) Payback (months)
Low 1,000 $12,000 $3,000 $9,000 6.67 about 80
Medium 2,000 $24,000 $3,000 $21,000 2.86 about 34
High 4,000 $48,000 $3,000 $45,000 1.33 16

The arithmetic: 60,000 / 9,000 = 6.67 years (x 12 = 80 months); 60,000 / 21,000 = 2.86 years (x 12 = 34.3 months); 60,000 / 45,000 = 1.33 years (x 12 = 16 months).

Bar chart of payback months by productive hours per year: about 80 months at 1,000 hours, about 34 at 2,000 and 16 at 4,000
Payback falls from about 80 months to 16 as productive hours rise from 1,000 to 4,000. Numbers from the worked example. Illustration: WonksAnonymous.

Why the relationship is not linear

Doubling hours from 1,000 to 2,000 doesn’t simply halve the payback. B rises from $9,000 to $21,000, a factor of 2.33, because the $3,000 fixed cost is paid regardless. Payback falls from 80 to 34 months. Doubling again to 4,000 hours lifts B to $45,000 (a factor of 2.14) and payback falls to 16 months.

Fixed costs also set a floor. At $12 per hour and $3,000 fixed, the machine has to run 3,000 / 12 = 250 hours a year just to cover them. Below that, B is negative and there is no payback.

At low hours, fixed costs eat a large share of the gross benefit: 3,000 of 12,000, or 25%, in the low case, against 6.25% in the high case. That is why the low scenario looks worse than a straight-line reading of hours would suggest.

What pushes productive hours down

Several things pull productive hours below what the shift schedule suggests.

Order volume comes first. If orders fill half a shift, the machine runs half a shift, because capacity doesn’t create demand. When volume is too low to keep a machine busy, compare contract packing versus owning a line. Shifts matter too: a machine staffed for one shift can’t reach the two-shift scenario without added labor cost, and that cost belongs in B.

Changeovers take minutes or hours out of every product switch, and frequent small batches multiply the loss, as shown in small batches and changeover cost. Stoppages cut hours and can leave orders late; see measuring the cost of unplanned production downtime.

Finally, a machine that waits for material or for the next station isn’t productive, however fast its rated speed. See why faster machines do not always increase line output, and cost per good pack for counting good output instead of nameplate speed.

Sensitivity check

Start from the medium scenario (2,000 hours, payback 2.86 years, about 34 months) and change one input at a time.

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Change New B New I Payback (years) Payback (months) Change vs. baseline
Baseline $21,000 $60,000 2.86 about 34 none
Net benefit per hour 25% lower ($9) $15,000 $60,000 4.00 48 about 14 months longer
Installed cost 20% higher ($72,000) $21,000 $72,000 3.43 about 41 about 7 months longer

The arithmetic: at $9 per hour, B = 9 x 2,000 - 3,000 = 15,000, and 60,000 / 15,000 = 4.00 years (48 months). At 20% higher cost, I = 60,000 x 1.2 = 72,000, and 72,000 / 21,000 = 3.43 years (41.1 months).

In this example, a 25% shortfall in hourly benefit hurts about twice as much as a 20% cost overrun. So verify the hourly benefit and the productive hours first. They drive B, and B is the denominator, so an error in either moves payback further than the same-sized error in installed cost, which you can check against a quote and a scope list. See the true installed cost of packaging automation for that check.

Horizontal bar chart of payback months in the medium scenario: baseline about 34, hourly benefit 25% lower 48, installed cost 20% higher about 41
A 25% shortfall in hourly benefit lengthens payback more (to 48 months) than a 20% cost overrun (about 41). Numbers from the worked example. Illustration: WonksAnonymous.

A 25% drop in hours has a similar effect to a 25% drop in hourly benefit, since the two multiply, so check both. After installation, compare these hours with measured ones in a 90-day review.

Break-even utilization

You can also run the model backward. Given a target payback, how many productive hours do you need?

Hours needed = (I / target years + fixed annual costs) / net benefit per hour

Take a 3-year target with the same inputs. I / target years = 60,000 / 3 = 20,000 per year. Add fixed costs: 20,000 + 3,000 = 23,000. Divide by $12: 23,000 / 12 = 1,917 productive hours per year (rounded up).

To check: 1,917 x 12 = 23,004; minus 3,000 = 20,004; 60,000 / 20,004 = 3.00 years. A 2-year target would need (30,000 + 3,000) / 12 = 2,750 hours. If your realistic hours fall short of what the formula gives, the target isn’t met under these inputs.

Estimate your own utilization

  • Use order history, not capacity. Count the hours that produced paid output last year, not the hours the machine could have run.
  • Subtract changeovers and downtime, using logged records where they exist. If they don’t, estimate conservatively and say it’s an estimate.
  • Separate committed orders from hoped-for orders. Run the model on committed volume first and show the hoped-for volume as an upside case.
  • Confirm the cash benefit holds at that volume. If the saving is labor, check that wages actually stop. See labor savings versus redeployment.

When current volume is well below break-even hours, the model may say the machine isn’t justified yet, or that a smaller or semi-automatic solution fits better. That is a legitimate result.

Keep reading

Assumptions and limits

  • Net benefit per productive hour is held constant. In practice it shifts with product mix, scrap rates and order size.
  • Taxes, financing, the time value of money, ramp-up, residual value and useful life are ignored, so this is a simple payback, not NPV or IRR.
  • The shift-to-hours mapping assumes about 250 working days of 8 hours.
  • Replace every input with figures from the plant’s own records.
  • The model is for comparing options and is not professional financial or engineering advice.