example of "abnormal" rework? Be specific
Rework : As the name suggests RE + Work = repeating the work
Abnormal rework means the situation where we have to repeat the entire process due to the circumstances that are not normal . for example repeating the process of painting an iron rod is normal rework if standard requires double coating.Similarly : inspection used to be done as a part of process after completion of work is an example of normal rework.
Now , lets say there is a fire in the warehouse and goods tends to be burnt .Now to check the position of goods saved inspection of goods is done again to ascertain whether they match the standards or not is an example of abnormal rework . Because this inspection was not the part of the process but is the requirement of the situation .
2nd example : In a chair manufacturing company , process is to first make a chair ,then sharpen its edges and then painting and then inspection is done . paint to done is 21st shade of red ( hypothetical) but operator who has to choose the color made a mistake n choose 20th shade of red . Now customer rejected to accept the chairs because this is not of the color they demanded.
Now company have to paint it again with 21st shade of red . This is an abnormal rework because it was not a part of usual process of company ad cause cost addition to company.
Data Table 1. Specific gravity of urine Standard urine Abnormal urine Real urine (optional) 1. What is the significance of the specific gravity of the abnormal urine? Edit & C
Problem 5.4: Process Cost example with Normal Loss, Abnormal Loss, Abnormal Galli The following information is available about process B. 20000 units were transferred to process B @ Rs. 4 per unit. Normal loss in the production process is estimated @ 10% of process input. Output of process B is transferred to finished stock account. Amount (Rs. '000) Particulars Labor Cost 1000 Materials 4000 Proportionate share of factory overhead 700 You are required to prepare the process cost account for...
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Rework Example 10.3.1 by replacing фп(x) by the conventional Legendre polynomial. Pn (x): [Pn (x)' dx = 2n + 1 Using Eqs. (10.47a), and (10.49a), construct Po, P1(x), and P2(x).
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