Three Assembly Lines: Scrap Count Is Not the Performance Measure
A plant runs three parallel assembly lines producing the same fitting. All lines are scheduled for 450 min per shift. Line A had 45 min of breakdown, an ideal cycle time of 2.0 min/unit, produced 170 units and scrapped 8. Line B had 20 min of breakdown, an ideal cycle time of 1.5 min/unit, produced 240 units and scrapped 5. Line C had 60 min of breakdown, an ideal cycle time of 2.5 min/unit, produced 140 units and scrapped 2. The operations manager compares defect counts and concludes that Line A is the best performer because its scrap rate is 'only 8 units'. A consultant proposes spending Rs 5 lakh on a quick-change die for Line C, claiming it will cut breakdown time by 40 min per shift. An alternative is to spend Rs 3 lakh on a tooling upgrade on Line B that would reduce its ideal cycle time by 0.1 min/unit. The plant runs two shifts per day, 300 days per year, and each good unit contributes Rs 40 in margin.
1. Compute availability, performance, quality and OEE for each line, and rank the lines.
Read the worked answer
Use OEE = availability × performance × quality, where performance = (ideal cycle time × output)/run time.
Line A: Run time = 450 − 45 = 405 min. Availability = 405/450 = 0.900. Performance = (2.0 × 170)/405 = 340/405 = 0.8395. Quality = (170 − 8)/170 = 162/170 = 0.9529. OEE = 0.900 × 0.8395 × 0.9529 = 0.720 (72.0%).
Line B: Run time = 430 min. Availability = 430/450 = 0.9556. Performance = (1.5 × 240)/430 = 360/430 = 0.8372. Quality = 235/240 = 0.9792. OEE = 0.9556 × 0.8372 × 0.9792 = 0.783 (78.3%).
Line C: Run time = 390 min. Availability = 390/450 = 0.8667. Performance = (2.5 × 140)/390 = 350/390 = 0.8974. Quality = 138/140 = 0.9857. OEE = 0.8667 × 0.8974 × 0.9857 = 0.767 (76.7%).
Rank: B (78.3%), C (76.7%), A (72.0%). Line A is the weakest overall, not the best.
2. Critique the manager's use of absolute scrap count and explain a better interpretation.
Read the worked answer
The manager compares 8, 5 and 2 scraps without normalizing by output. Line A produced only 170 units, so its scrap rate is 8/170 = 4.7%. Line B is 5/240 = 2.1%, and Line C is 2/140 = 1.4%. Line A has the worst quality rate, not the best. Absolute scrap count rewards low production volume and hides the loss.
A better comparison uses quality rate (good units/output), scrap per good unit, or cost of poor quality, and combines it with availability and performance in OEE. On OEE, Line A is last at 72.0%. Its availability is 90%, which is second, but its performance is 83.95% and quality is 95.29%, the lowest quality of the three. The manager should also check whether the scrap on A is concentrated at startup, because that changes the corrective action.
3. Evaluate the two investments using annual incremental good units and simple payback, and recommend one. State any assumption.
Read the worked answer
Assume the performance ratio and quality rate remain unchanged when run time or ideal cycle changes.
Line C quick-change die: Run time rises from 390 to 430 min. Output at the same units per run minute = 140 × (430/390) = 154.4 units. Good units = 154.4 × 0.9857 = 152.2, versus 138.0 now. Gain = 14.2 good units/shift. Annual gain = 14.2 × 2 shifts × 300 days = 8,520 units. Margin = 8,520 × 40 = Rs 3,40,800/year. Payback = 5,00,000/3,40,800 = 1.47 years.
Line B tooling upgrade: Ideal cycle falls from 1.5 to 1.4 min/unit, a 7.14% speed gain. Output rises to 240 × (1.5/1.4) = 257.1 units. Good units = 257.1 × 0.9792 = 251.8, versus 235.0 now. Gain = 16.8 good units/shift. Annual gain = 16.8 × 600 = 10,080 units. Margin = 10,080 × 40 = Rs 4,03,200/year. Payback = 3,00,000/4,03,200 = 0.74 years.
Line B has the shorter payback and higher annual benefit, so it is the better first investment. Confirm that the added output can be sold and that the 0.1 min/unit speed gain is achievable without quality loss.