Shifting the process mean to minimize surplus components and unacceptable products in selective assembly

We consider a product assembled from two components such that its quality characteristic is the clearance between the mating components. Random assembly of mating components may lead to an unacceptably large number of rejected products, i.e. those which do not satisfy a given clearance specification. In such a situation, selective assembly should be effective in reducing the rejection rate. Most previous studies have focused on equal width and equal probability partitioning schemes in selective assembly. When there is a large difference between the variances of the two component dimensions, equal width partitioning will result in a large number of surplus components and equal probability partitioning will result in some rejected products. Some authors have proposed a method of manufacturing the component with smaller variance at two (or more) shifted means to cope with this difficulty. This paper deals with the problem of determining the optimal mean shift for both equal width and equal probability partitioning schemes. Some numerical results are given which show that using the optimal mean shift can considerably reduce the number of surplus components in equal width partitioning and may enable us to ensure that all products satisfy the clearance specification in equal probability partitioning. We also show some advantages and disadvantages of equal width and equal probability partitioning schemes.