Cutting Speed Selection on a Late Order
A job shop must turn 400 pieces of 1045 cold-drawn bar, Ø80 mm × 300 mm long, in a single pass at 2.5 mm depth of cut and 0.3 mm/rev on a CNC lathe. Taylor constants for the coated-carbide insert at this feed are n = 0.25 and C = 250 (V in m/min, T in min). The machine-plus-operator rate is P600/h, indexing to a fresh edge costs P60 in insert consumption, a tool change takes 1.5 min, and loading is 2 min/piece. The Ra 1.6 µm finish requirement is satisfied at any speed below 200 m/min. The customer has pulled the delivery date in by two days and the lathe is already booked to 85% of its available hours that week.
1. Calculate the cutting speed for minimum cost per piece and the corresponding tool life.
Read the worked answer
Convert the tool cost to machine-time equivalent: P60 per edge ÷ P10/min = 6 min.
Optimum life for minimum cost: Topt = (1/n − 1)(t_change + tool cost in time) = (4 − 1)(1.5 + 6) = 3 × 7.5 = 22.5 min
Vopt = C/T^n = 250/22.5^0.25 ln 22.5 = 3.1135 → × 0.25 = 0.7784 → e^0.7784 = 2.178 Vopt = 250/2.178 = 114.8 m/min
Check: at this speed N = 1000(114.8)/(π × 80) = 457 rpm, so cutting time = 300/(0.3 × 457) = 2.19 min, i.e. about 10 pieces per edge. That is a sensible shop-floor picture; a speed giving a 2-minute tool life would be an obvious error signal.
2. Calculate the speed for maximum production rate and compare the cost per piece of the two policies.
Read the worked answer
Maximum production rate (ignore tool cost, count only the 1.5 min change): Topt = (1/n − 1) × t_change = 3 × 1.5 = 4.5 min V = 250/4.5^0.25 = 250/1.4565 = 171.6 m/min → N = 683 rpm → t = 300/(0.3 × 683) = 1.464 min
Cost per piece = machining time × P10/min + (machining time/T) × (P60 + 1.5 × P10)
Minimum-cost policy: 2.189 × 10 + (2.189/22.5)(75) = 21.89 + 7.30 = P29.19/piece Maximum-production policy: 1.464 × 10 + (1.464/4.5)(75) = 14.64 + 24.40 = P39.04/piece
Difference: P9.85/piece, or P3 940 over the 400 pieces. The 2 min handling is identical in both cases and drops out of the comparison. Note the finish limit (200 m/min) is not binding.
3. Which speed do you run, given the compressed delivery date? Show the arithmetic that drives the decision.
Read the worked answer
Time saved by the high-speed policy: (2.189 − 1.464) × 400 = 290 min ≈ 4.8 machine-hours. Cost of buying those 4.8 hours: P3 940, i.e. about P820 per hour of lathe time.
Decision: run the minimum-cost speed (114.8 m/min) unless the lathe is the binding constraint and the marginal value of a lathe-hour exceeds about P820 - for example if the 85% booking means overtime, subcontracting or a late-delivery penalty above that rate. Do not simply default to maximum production speed; on this job it buys less than five hours at a 35% cost penalty.
Two practical checks before committing: (a) confirm the 85% booking really leaves no slack - 15% of a week is often several hours; (b) if a speed change is made, verify chip control and surface finish at the higher speed, since the Ra limit was quoted only up to 200 m/min and was not measured at 171.6 m/min.