Weighted cost per result?

[Mackensie]

Hello,
In advertising my business, I have two products, one $997 and the other $497. Purchasing online advertising, I spend $2500 and end up getting 4 $997 sales and 5 $497 sales. How can I determine what a weighted cost per result is? If I calculate it separately (2500/5 and 2500/4), they don't take into account that some of that overall money went to get the other product's sales.

 

[blamocur]

Are you assuming that the advertising costs are proportional to the number of sales and to the price of the product? I.e., do you feel that you have to spend twice as much to advertise the $997 product compared to the $497 product? And do you feel you need to spend twice as much to make 4 sales as opposed to 2 sales?

 

[BigBeachBanana]

Let N be the number of units sold then:
[math]\text{Cost per Unit} = \frac{\text{2500}}{\text{E(N)}}[/math]where E(N) is the expected number of units sold
The question is how to calculate the expected number of units?
Using Law of Total Expectation: [math]E(N)=E[E[N|Cost]]= \\E[N|Cost=997]*Pr(Cost=997) + E[N|Cost=497]*Pr(Cost=497)=\\4*Pr(Cost=997)+5*Pr(Cost=497)[/math]That leaves the question of how many of the $997-product you have in inventory compared to the $497 product?

 

[BigBeachBanana]

[EDIT] Wrong question asked from the previous post:
"That leaves the question of how many of the $997-product you have in inventory compared to the $497 product?"

[math]\text{Let N be the number of units sold then:}\\ \text{Cost per Unit} = \frac{\text{2500}}{\text{E(N)}}\\ \text{Using the Law of Total Expectation:}\\ E(N)=E[E[N|Cost]]= \\E[N|Cost=997]*Pr(Cost=997) + E[N|Cost=497]*Pr(Cost=497)=\\4*Pr(Cost=997)+5*Pr(Cost=497)[/math]Correct question: This leads to the question of how likely is it to see an ad for the $997-product vs. the $497?