0
|
2217 Views
|
5 Replies
|
4 Total Likes
View groups...
Share
GROUPS:

# Find sub-item costs

Posted 9 years ago
 Hi, Sorry if this is a mundane question for you all but I need some help that should be simple. I have already imported my data, it has 18 columns, the first being the the name of the product, and the last being the final cost of 1 product. The rest of the 16 columns are amounts of sub-items that make up the product. Not all 16 sub-items are used for every product. I do not have sub-item costs, whole or partial. I think I need to analyze all 84 products and based upon that data I think I could retroactively find the cost of each sub-item. The problem is, I don't know where to go from here. Can anyone help?
5 Replies
Sort By:
Posted 9 years ago
Posted 9 years ago
 Here is a possible start. Drop the first column since that's just a part's name. And take the final column out but save it as an 84 element list. This is the vector of prices of your products: call it pProducts (for product prices).Now you are left with a 84 by 16 matrix (84 rows, 16 columns). Call it qSubItems (q for quantity).Lets say that the price per unit quantity of each sub item is a 16 element vector, call it pSubitems.So, in effect, you have a matrix equation of the form qSubItems.pSubitems=pProductsGiven that this is potentially 84 linear equations in 16 unknowns, it is likely that it is overdetermined. And it may not have a consistent solution because the measurements may not be completely accurate, or the quantities may not be exact--they may be rounded, or the final product prices may have been rounded or modified in some way.So, a very practical approach might be to take 16 of these equations randomly and see if there is sufficient information in them (i.e., that the submatrix from qSubItems has non-zeor determinant) to solve for the pSubitems vector.
Posted 9 years ago
 I think Bill and I almost tied for this ... ;-)
Posted 9 years ago
 I really appreciate the help and direction on where to start. Thank you!
Posted 9 years ago
 Better tied than died. But tie-dyed would have been more colorful..