Shrinkage Allowance on a School Shirt Order
A small uniform unit in Hubballi has won an order for 300 cotton-poplin school shirts. The buyer specifies a finished shirt length of 72 cm, chest 92 cm and sleeve 58 cm. The unit cut the first 40 shirts from a pattern that includes a 1.5 cm hem allowance but no shrinkage allowance. A washed sample now measures 69.5 cm long and 90.5 cm at chest; the fabric supplier's test report states warp shrinkage 3.5% and weft shrinkage 1.5%. The cutting master argues that the 2.5 cm length loss is within the buyer's tolerance, but the buyer has rejected the washed sample and threatened to cancel. The remaining fabric is limited, and re-cutting all 300 panels would miss the school-opening deadline. The unit has 12 operators, one cutting table and no in-house washing or compressive shrinking range.
1. Identify the likely causes of the rejection and explain why the cutting master's tolerance argument is weak. Use the measured data.
Read the worked answer
The pattern was cut without a shrinkage allowance. The washed sample is 72 − 69.5 = 2.5 cm short, which is 2.5/72 × 100 = 3.47%, almost exactly the reported 3.5% warp shrinkage. The chest loss is 92 − 90.5 = 1.5 cm, or 1.63%, close to the 1.5% weft figure. The 1.5 cm hem allowance cannot compensate because it is already part of the 73.5 cm cut length; after shrinkage the cut length becomes about 73.5 × 0.965 = 70.93 cm, and after turning the hem the finished length is about 69.4 cm. The cutting master is treating a predictable fibre-relaxation shrinkage as if it were random cutting error. A 2.5 cm shortfall is 3.5% of the specified length and is outside a normal woven-shirt tolerance of about ±1 cm. The correct conclusion is that the pattern, not the operator, is the source of the nonconformity.
2. Calculate the revised cut length for a 72 cm finished shirt with a 1.5 cm hem allowance and 3.5% warp shrinkage. Decide whether the 40 already-cut panels can be salvaged.
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Required cut length = (finished length + hem allowance) ÷ (1 − shrinkage rate). Here, (72 + 1.5) ÷ (1 − 0.035) = 73.5 ÷ 0.965 = 76.17 cm, so the pattern should be revised to about 76.2 cm before cutting. The existing panels were cut at about 73.5 cm. After 3.5% shrinkage they are only about 70.9 cm before hemming, so they cannot produce a 72 cm finished length. Adding a false hem or contrast band would need about 2.7 cm of matching fabric and buyer approval; otherwise the 40 panels should be segregated and re-cut. They may be usable as a shorter size only if the buyer accepts a revised measurement chart. The chest pattern should also be reviewed: to finish at 92 cm after 1.5% weft shrinkage, the cut chest dimension must be increased by roughly 1.4 cm plus normal seam allowance.
3. Recommend a production control plan for the remaining 260 shirts so that the same failure does not repeat.
Read the worked answer
First, stop cutting and test a one-metre sample from the remaining fabric through a representative wash, because shrinkage can vary by lot. Second, either buy pre-shrunk poplin or pre-shrink the available fabric before cutting. Third, revise the pattern with the measured warp and weft allowances and issue a new cutter's marker. Fourth, cut and sew a first-off sample, wash it, measure length and chest, and obtain buyer approval before bulk cutting. Fifth, add in-process checks after cutting, after joining the yoke and before hemming; record measurements on a simple chart. Sixth, keep the 40 short panels physically segregated so they cannot be mixed with the corrected lot. Finally, tell the buyer the revised delivery plan in writing and ask for acceptance of the corrected measurement method rather than hiding the shrinkage loss.