GB560 Unit 4 Cycle Time Analysis Example

3.34 Days Elapsed, 11.4 Minutes Worked: A Unit 4 Cycle Time Analysis of 120 Returns

[Student Name]

Purdue University Global

GB560: Designing, Improving, and Implementing Processes

Unit 4 Cycle Time Analysis

[Instructor Name]

[Date]

The seller, its returns center and every time measurement are composite and illustrative; the methods and sources cited are real.

Data and Method

Sample. 120 returns, every return scanned at the dock during sampled hours over two weeks, covering both shifts and every weekday.

Elapsed time. From the dock scan timestamp to the refund posting timestamp, both taken from system records.

Touch time. Timed at each step from the Unit 3 map, with loop steps added for returns that went backward.

Histogram

The 120 elapsed times are skewed to the right: most returns finish in two to five days, and a long tail stretches past a week.

Under 1 day: 3. 1 to 2 days: 14. 2 to 3 days: 34. 3 to 4 days: 24. 4 to 5 days: 15. 5 to 7 days: 18. 7 to 10 days: 8. 10 to 15 days: 4.

Summary statistics: median 3.34 days; mean 4.10; standard deviation 2.34; 90th percentile 7.05; longest 14.9.

A handful of very slow returns pulls the mean up while leaving the median almost untouched. Reporting the median, the mean and the 90th percentile together tells a reader how the process behaves for a typical shopper and for an unlucky one. Measured against the refund-timing requirement set in Unit 2, the process is far short: about 42 percent finish within three calendar days, and even measured in business days the share is far below 95 percent.

Process Cycle Efficiency

Touch time on the standard path is 11.4 minutes, and 12.4 minutes on average once loop steps are included. Dividing 11.4 minutes by the median elapsed time of 3.34 days, measured in clock hours, gives a process cycle efficiency of about 0.24 percent. More than 99.7 percent of the time a typical return spends in the process, nobody is working on it.

The returns are not slow because people are slow; they are slow because they wait, and the waiting is built into the process.

Very low efficiency figures are common in administrative work, which is why the ratio is useful: it shows that cutting touch time, the focus of the dock board's returns per person-hour, can barely move elapsed time.

Segment Table

Each wait is ranked by its share of mean elapsed time, 4.10 days or about 98.4 hours.

Bulk cartons before unpacking: about 35.0 hours, 35.6 percent.

Overnight refund-keying queue: about 21.1 hours, 21.4 percent.

Wait for the 16:00 export: about 13.1 hours, 13.3 percent.

Grading queue: about 9.5 hours, 9.7 percent.

All other waits, including weekends and loop delays: about 19.7 hours, 20.0 percent.

The two batch points identified in Unit 3, the export and the overnight keying, together account for about 34.7 percent of elapsed time, more than a day per return. Neither involves any work.

Path Comparison

No usable authorization: 14 returns took a median of 8.94 days, because they waited in the problem cage for an agent to identify the order.

All others: 106 returns took a median of 3.21 days.

The 14 returns are under 12 percent of the sample but account for much of the long tail. They lift the mean far more than the median, which is why the mean alone would exaggerate how slow the typical return is.

First Pass Yield

Of the 120 returns, 96 passed through without a loop, a first pass yield of 80.0 percent. The other 24 went backward at least once: 14 to the problem cage, 7 for a regrade and 3 for a refund correction after the shopper called.

How Much Data Is Enough

One hundred twenty returns over two weeks is enough to show the shape of the distribution and the main paths, and to cover every weekday on both shifts. It is not enough to estimate the 90th percentile precisely: a handful of different returns in the tail could move it by half a day. The analysis therefore reports the median as its main figure and treats the 90th percentile as an indication rather than a target to the hour. A second sample after any redesign should use the same method so the comparison is fair.

Work in Process

Little (1961) proved that, in a stable system, items in process equal throughput times the time each spends inside. At about 207 returns a calendar day and a mean of 4.10 days, the center holds roughly 850 returns in process at any moment. Hopp and Spearman (2011) note that cutting elapsed time cuts work in process in proportion, so halving elapsed time would free roughly 425 boxes' worth of floor space and the cash tied up in their refunds.

What the Analysis Shows

Three conclusions carry into the redesign. The two batch points are the largest removable delay. The bulk carton wait is the largest single delay and depends on how the dock and bench crews hand work over. And the returns without authorization, though few, need their own path, because they account for much of the tail. Rother and Shook (1999) recommend designing a future state around the biggest waits first, and this analysis says which they are.

References

Hopp, W. J., & Spearman, M. L. (2011). Factory physics (3rd ed.). Waveland Press.

Little, J. D. C. (1961). A proof for the queuing formula: L = λW. Operations Research, 9(3), 383-387. https://doi.org/10.1287/opre.9.3.383

Rother, M., & Shook, J. (1999). Learning to see: Value stream mapping to add value and eliminate muda. Lean Enterprise Institute.