Tool-Life Economics: Is a 25% Speed Increase Really Cheaper per Piece?
A small CNC turning shop machines a component requiring 160 m of cutting travel per piece. The current programme runs at 40 m/min, at which speed the inserts last 60 min per edge. The shop's tool-life data follow Taylor's relation VT^n = C with n = 0.125. Machine and operator cost is ₹5 per minute, a tool change takes 1.5 min, and each reground edge costs ₹30. The production manager proposes a 25% speed increase to 50 m/min, arguing that output per shift will rise by a quarter at almost no extra cost. The costing sheet, however, still treats cost per piece as the governing figure, because the customer's price is fixed per piece. A premium grade insert with a higher C value is available but costs 40% more per edge.
1. Evaluate the Taylor constant C and the tool life expected at 50 m/min.
Read the worked answer
Taylor's equation: VT^n = C, with n = 0.125, so 1/n = 8.
C = 40 × 60^0.125. Now 60^0.125 = e^(0.125 × ln 60) = e^(0.125 × 4.0943) = e^0.5118 = 1.668. Hence C ≈ 66.7.
At V = 50 m/min: T = (C/V)^(1/n) = (66.72/50)^8 = (1.3345)^8 ln 1.3345 = 0.2887, × 8 = 2.3094, e^2.3094 = 10.07
T ≈ 10 min per edge, against 60 min at 40 m/min.
The small exponent is the point: a 25% rise in speed cuts tool life by a factor of about six. Candidates often read n = 0.125 as "tool life barely changes", but it appears in the exponent 1/n = 8, which is why modest speed changes cause dramatic tool consumption.
2. Compute the cost per piece at 40 m/min and at 50 m/min and decide whether the proposal should be accepted.
Read the worked answer
Cutting time per piece = 160/V. Tools consumed per piece = cutting time ÷ tool life. Cost of tool change per piece = tools/piece × 1.5 min × ₹5/min; tool cost = tools/piece × ₹30.
At 40 m/min: cutting time = 160/40 = 4.0 min → ₹20.00. Tools/piece = 4.0/60 = 0.0667. Change cost = 0.0667 × 1.5 × 5 = ₹0.50. Tool cost = 0.0667 × 30 = ₹2.00. Total ₹22.50 per piece.
At 50 m/min: cutting time = 160/50 = 3.2 min → ₹16.00. Tools/piece = 3.2/10.07 = 0.318. Change cost = 0.318 × 1.5 × 5 = ₹2.38. Tool cost = 0.318 × 30 = ₹9.53. Total ₹27.91 per piece.
The proposal raises cost per piece by about ₹5.41, roughly 24%, even though machining time falls by 20%. With a fixed selling price, every rupee saved on cycle time is more than lost on tools and tool changes. The production manager is measuring output, not profit - the correct decision is to reject the speed increase on the existing grade.
3. Find the tool life that would make 50 m/min cost-neutral, and state what it implies for the premium insert.
Read the worked answer
Set the 50 m/min cost equal to ₹22.50, letting T be the unknown tool life. Tools per piece = 3.2/T, and each tool costs 1.5 × 5 + 30 = ₹37.50 in change time plus edge cost.
16.00 + (3.2/T) × 37.50 = 22.50 (3.2/T) × 37.50 = 6.50 120/T = 6.50 → T = 18.5 min
So the edge must last 18.5 min at 50 m/min instead of 10.07 min. Convert to a Taylor constant: C_required = 50 × 18.5^0.125 = 50 × e^(0.125 × 2.918) = 50 × 1.440 ≈ 72.0
That is about 8% above the current C = 66.7. So a premium grade is justified only if its C is at least about 72 - not merely "better". Then check the price: the premium edge costs 40% more, i.e. ₹42 instead of ₹30. Recompute at 50 m/min if T = 18.5 min: tool cost per piece = (3.2/18.5) × 42 = ₹7.26; change cost = (3.2/18.5) × 1.5 × 5 = ₹1.30; machining ₹16.00 → ₹24.56, still above ₹22.50. The premium grade pays only if it also lifts T well beyond 18.5 min. Buying a better grade is a calculation, not a slogan.