The other B3:7/x bits on the rung determine if the next broken bottle is large AND if the box already contains "145" units or more of glass. The attached picture shows a way to compute the broken glass, then use only 2 comparisons to always fill the scrap box to 93.5% or 100% full. You must multiply your Small and Large multipliers (1 and 1.5) by 10 to come up with an integer value for the broken glass. Because your LogixPro (and the Allen Bradley SLC 5/02 that it simulates) does not have Floating-Point decimal numbers, you cannot divide by 0.6666 and you cannot multiply by 1.5. You cannot just add them together because the small is only 2/3 the quantity of the large (or Large = 1.5 times the Small). So first you create a value ("BROKEN GLASS" in N7:2) that is composed of both the Small and Large Broken bottles. As in many programming applications, the key is to recognize that you need a simple way to compare the total of 2 unequal items. For your "Botle Boxing" program, your Rung 009 with 11 rung branches and 22 comparisons is way too complicated, but still it does not maximize the amount of broken glass going into each scrap box.
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