calculating repair requirements based on scrap rate

[dmac13]

Hi im wondering if anyone can help?

i am trying to calculate the amount of items i would need to send for repair based on my requirement and a scrap rate.

e.g. i need refurbished 10 parts, the part in question has a 50% scrap rate when sent on repair, therefore i would need to send 20 non-refurbished parts to produce 10 refurbished parts.

how can i get to the answer of 20 from qty 10 and a scrap rate of 50%?

 

[HallsofIvy]

Hi im wondering if anyone can help?

i am trying to calculate the amount of items i would need to send for repair based on my requirement and a scrap rate.

e.g. i need refurbished 10 parts, the part in question has a 50% scrap rate when sent on repair, therefore i would need to send 20 non-refurbished parts to produce 10 refurbished parts.

how can i get to the answer of 20 from qty 10 and a scrap rate of 50%?

]
If 50%= 1/2 will be scrapped then 1- 1/2= 1/2 will be return "refurbished". So if x is the number sent to be refurbished, in order to get 10 back, x must satisfy (1/2)x= 10.

 

[dmac13]

]
If 50%= 1/2 will be scrapped then 1- 1/2= 1/2 will be return "refurbished". So if x is the number sent to be refurbished, in order to get 10 back, x must satisfy (1/2)x= 10.

The scrap rate can differ from part to part so I'm looking formula I can use in excel .

Thanks for your feedback

 

[HallsofIvy]

The scrap rate can differ from part to part so I'm looking formula I can use in excel .

Thanks for your feedback

If the "scrap rate" is "r" then you will get back 1- r of them. If you want to get back "B" of them then the number you send, x, must satisfy \(\displaystyle (1-r)x= B\) so \(\displaystyle x= \frac{B}{1- r}\).

In the example with r= 1/2 and B= 10, that is \(\displaystyle x= \frac{10}{1- 1/2}= \frac{10}{1/2}= 10(\frac{2}{1})=20\).

 

[dmac13]

If the "scrap rate" is "r" then you will get back 1- r of them. If you want to get back "B" of them then the number you send, x, must satisfy \(\displaystyle (1-r)x= B\) so \(\displaystyle x= \frac{B}{1- r}\).

In the example with r= 1/2 and B= 10, that is \(\displaystyle x= \frac{10}{1- 1/2}= \frac{10}{1/2}= 10(\frac{2}{1})=20\).

Simple when you kno how thanks for your help


Scrap rate = x
Requirement = y

Repair qty = y / (1-x)