Bom calculates the final loss rate of finished products directly to the item

Assume that production from B is a and the loss rate of each process is C1, C2, and C3 respectively. Production from D to B goes through one process, the job loss rate is C4. What is the loss rate of materials created in Bom?

My understanding is as follows:
Output of a = input of B (1-c1) (1-c2) (1-c3)
So the input of B = the output of a/(1-c1) (1-c2) (1-c3)
Therefore, when creating a Bom, the loss rate of Material B is: 1/(1-c1) (1-c2) (1-c3)-1
The aboveAlgorithmYes
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Assume that the composition consumption is 1.
Output of a = input of B * output rate
Output of a = input of B (1-c1) (1-c2) (1-c3)
=> Input of B = output of A/[(1-c1) (1-c2) (1-c3)]
Because
B's input = (A's output + B's input * Loss Rate) rather than B's input = A's output (1 + Loss Rate)
=> Input of B = output of A/(1-Loss Rate)
So 1-Loss Rate = (1-c1) (1-c2) (1-c3)
Therefore, the loss rate is 1-(1-c1) (1-c2) (1-c3)

I agree with this algorithm.

There is a multi-order relationship in the BOM. You need an algorithm to know the loss rate of the component directly from the finished product.
I found a post online and thought it was good.
Http://www.itpub.net/viewthread.php? Tid = 706926 & extra = & page = 1