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Manufacturing · Chapter 10 of 10 · Intermediate

Automation, CNC, and Economics

Processes make parts; economics decides whether to. This closing chapter weighs automation against volume and finds the cutting speed that minimises cost, the judgment that turns a process into a business.

01

Readiness check

This capstone ties processes to cost. Tick only what you can do closed-notes.

  • Build a unit-cost model (fixed plus variable).
  • Find a break-even quantity (Chapter 1).
  • Use Taylor's tool-life equation (Chapter 6).
  • Compute a payback as investment over annual saving.
  • Recognise a curve with a minimum.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsReview the cost model in Chapter 1 and tool life in Chapter 6.
3 or more weak itemsRevisit both before continuing.
02

The core idea

Automation pays off above a break-even volume, and within a process there is an operating point, such as the cutting speed, that minimises total cost rather than time.

payback = investment / annual savingcost/part = machining + toolingeconomic speed minimises total cost

Automation (CNC, robots, flexible systems) trades high capital cost for low per-part cost, so it wins above a break-even quantity, just like tooling in Chapter 1. Within an operation, pushing faster cuts machining time but, through Taylor's law, shortens tool life and raises tooling cost. Total cost per part is therefore a U-shaped curve in cutting speed, with a minimum at the economic speed, which is slower than the maximum-rate speed.

The skill works when: you justify automation by volume and payback, and pick the operating point that minimises total cost, not time.
The skill breaks down when: automation is bought below its break-even, or speed is maximised for rate while tooling cost is ignored.
The concept. Automation matches production volume and variety: manual shops for low-volume, high-variety work; programmable CNC for batches; dedicated fixed automation for high-volume, low-variety runs.
03

The skills, taught in order

This chapter turns processes into decisions. Five skills cover automation types, CNC, integrated manufacturing, cost justification, and economic optimization.

10.1 Types of automation

Automation is chosen by production volume and product variety.

TypeVolumeVarietyExample
Manual / job shoplowhighprototypes, one-offs
Programmable (CNC)low to mediummediumbatch machining
Flexible (FMS)mediummediumpart families in a cell
Fixed / hardvery highlowtransfer lines

10.2 Computer numerical control

CNC drives a machine along programmed paths (G-code) on two to five axes, with interpolation for curves. It gives programmable flexibility, the same machine makes a new part by loading a new program, which is why CNC dominates low-to-medium-volume precision work.

10.3 Integrated and flexible manufacturing

Robots handle and assemble; flexible manufacturing systems link CNC machines, automated handling, and control so a family of parts flows with little changeover. Computer-integrated manufacturing (CIM) ties design, planning, and production into one data system.

10.4 Manufacturing cost and automation payback

The unit-cost model from Chapter 1 extends to capital: automation adds investment but lowers per-part cost. It is justified above a break-even volume, and its payback period is the investment divided by the annual saving it produces.

10.5 Economic optimization

Within an operation, total cost per part is machining cost plus tooling and tool-change cost. Higher speed cuts the first but, via Taylor's VTn = C, raises the second. The sum is U-shaped, so the economic cutting speed that minimises cost is below the speed that minimises time.

Engineering connection: these decisions, automate or not, how fast to run, set the real cost of every part and close the loop opened by the process-selection logic of Chapter 1.

04

Worked example 1: justifying automation

An automated cell costs 150 000 to install and cuts the per-part cost from 9 (manual) to 1.50. Annual volume is 30 000 parts. Find the annual saving, the payback period, and the break-even volume.

Figure 1. The automated cell starts with a large fixed cost but a shallow per-part slope; above the break-even volume its total cost falls below the manual line.
  1. ProblemFind the annual saving, payback, and break-even volume for the cell in Figure 1.
  2. Given / findInvestment 150 000, manual 9/part, automated 1.50/part, volume 30 000/yr. Find saving, payback, N*.
  3. AssumptionsPer-part savings are steady; ignore the time value of money for a first estimate.
  4. ModelAnnual saving is the per-part saving times volume; payback is investment over saving; break-even sets the automated total equal to the manual total.
  5. Equationssaving = (cmanual − cauto)·volume payback = investment/saving N* = investment/(cmanual − cauto)
  6. SolveSaving = (9 − 1.50) × 30 000 = 7.5 × 30 000 = 225 000/yr. Payback = 150 000/225 000 = 0.67 yr (about 8 months). N* = 150 000/7.5 = 20 000 parts.
  7. CheckThe annual volume (30 000) exceeds the break-even (20 000), so the cell pays for itself within the year, consistent with the eight-month payback. Below 20 000 parts, manual would be cheaper.
  8. ConclusionAutomation is a volume bet: a fast payback at 30 000 parts a year, but a loss at low volume. The same break-even logic as process and tooling selection, now applied to capital.
Result. Saving 225 000/yr, payback ≈ 8 months, break-even 20 000 parts.
05

Worked example 2: the economic cutting speed

A turning job runs at a machine rate of 1/min, with a 2 min tool change and a 5 tool cost. The tool follows Taylor's VTn = C (n = 0.25, C = 400). Compare the cost per part at 150 m/min (4 min machining) and 250 m/min (2.4 min machining).

Figure 2. Machining cost falls with speed while tooling cost rises through Taylor's law. Their sum is U-shaped, with a minimum at the economic speed, below the fastest speed.
  1. ProblemFind the cost per part at the two speeds in Figure 2 and say which is cheaper.
  2. Given / findCm = 1/min, tchange = 2 min, Ctool = 5, n = 0.25, C = 400. At V = 150, tm = 4 min; at V = 250, tm = 2.4 min. Find cost/part.
  3. AssumptionsOne tool change per worn tool; cost is machining plus tool-related cost spread over the tool's parts.
  4. ModelTool life from Taylor; cost per part is machining cost plus the tool-change and tool cost times the fraction of a tool life used per part (tm/T).
  5. EquationsT = (C/V)1/n cost = Cmtm + (Cmtchange + Ctool)(tm/T)
  6. SolveAt 150: T = (400/150)4 = 50.6 min, cost = 1×4 + (2 + 5)(4/50.6) = 4.00 + 0.55 = 4.55. At 250: T = (400/250)4 = 6.6 min, cost = 1×2.4 + 7(2.4/6.6) = 2.40 + 2.56 = 4.96.
  7. CheckThe faster speed machines in less time (2.4 vs 4 min) yet costs more, because its short tool life multiplies tool-change and tool cost. The cheaper speed (150) is the slower one, confirming the U-curve.
  8. ConclusionThe fastest speed is not the cheapest. The economic speed minimises the sum of time and tooling cost, the optimization that unites Taylor's law (Chapter 6) with the cost model (Chapter 1) and closes the course.
Result. 4.55/part at 150 m/min versus 4.96 at 250 m/min: the slower speed is cheaper.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Automate everythingIdle expensive cell at low volume"Is the volume above break-even?"Automation pays only above its break-even quantity.
Fastest speed is cheapestTool cost dominates, cost rises"Did I add tooling cost?"Total cost is U-shaped; use the economic speed.
Ignoring payback timeCapital approved with no horizon"How long until it pays back?"Compare payback against the planning horizon.
CNC for any volumeDedicated mass part run on CNC"Volume high and variety low?"Very high volume favours fixed automation over programmable CNC.
07

Practice ladder

Level 1 · Direct skill

A robot costs 90 000 and saves 5 per part. At 25 000 parts a year, find the payback period.

Show answer

Annual saving = 5 × 25 000 = 125 000. Payback = 90 000/125 000 = 0.72 yr, about 9 months. Fast enough to justify in most planning horizons.

Level 2 · Mixed concept

For the Worked Example 2 job, is 200 m/min likely cheaper than 150 m/min? Reason qualitatively.

Show answer

At 200, machining time drops (about 3 min) but tool life falls to (400/200)4 = 16 min, so tooling cost is 7 × 3/16 = 1.31, giving 3.00 + 1.31 = 4.31. That is below both 4.55 and 4.96, so the economic speed lies near 200, between the two tested speeds.

Level 3 · Independent problem

Two processes: manual (no fixed cost, 12/part) and a CNC cell (40 000 fixed, 4/part). Find the break-even volume and recommend a process for an order of 4000 parts.

Show answer

N* = 40 000/(12 − 4) = 5000 parts. At 4000 (below 5000), manual is cheaper; the CNC cell only pays off above 5000. Volume decides, exactly as in Chapter 1.

Level 4 · Transfer to real engineering

Pick a real production scenario and decide its automation level and a sensible operating speed. Justify both with a break-even or payback and a cost-versus-speed argument.

What good work looks like

An automation choice tied to volume and payback, and an operating point justified as minimising total cost rather than time.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that my automation case clears its break-even volume."
"Give me five operations; I will say whether faster or slower is cheaper."
"Should I automate?" Building the payback and break-even yourself is the skill.
"What speed is cheapest?" Balancing machining and tooling cost is the point.

Portfolio task

Make one make-or-automate decision and one speed decision for a real part: justify the automation with payback and break-even, and pick a speed near the cost minimum.

Must include: a break-even volume, a payback period, and a total-cost comparison at two speeds.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. How is automation matched to production?

By volume and variety: manual for low volume, CNC for batches, fixed automation for high volume.

2. What does CNC provide?

Programmable flexibility: a new part by loading a new program on the same machine.

3. Write the automation payback.

payback = investment / annual saving.

4. Why is total machining cost U-shaped in speed?

Machining cost falls with speed, but tool cost rises (Taylor), so their sum has a minimum.

5. Is the economic speed faster or slower than the maximum-rate speed?

Slower; the cost minimum lies below the time minimum.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the payback and the two-speed cost from a blank page.
+3 daysOne automation and one economic-speed problem.
+7 daysTie cost back to process selection in Chapter 1.
+30 daysReview the whole course from the Manufacturing hub.
10

Textbook mapping

ItemMapping
Primary sourceKalpakjian and Schmid, Manufacturing Engineering and Technology, Chapters 37 to 40 (automation, CNC, CIM, economics)
Cross-referenceGroover, Automation, Production Systems, and CIM · DeGarmo, automation chapters
Core topics10.1 Automation types · 10.2 CNC · 10.3 CIM and FMS · 10.4 Cost and payback · 10.5 Economic optimization
Engineering connectionThe cost of every part, and the make-or-automate decision behind production.
Read nextYou have completed the Manufacturing course. Return to the course hub or continue to Machine Elements.