I have n products. $(y_1, y_2, ...y_n)$ is my inventory vector. This should last for a certain period.
Assume $m$ transactions will be made in this period.
Each transaction will have only one of these $n$ items (they are substitutional products). The probabilities of purchase of each product are known, $p_1,p_2,p_3,...,p_n.$ They add up to 1.
Revenue given by each product is known $r_1,r_2,...r_n$.
OBJECTIVE
Maximize :- $R_m(y_1, y_2, ...,y_n)$ which is the revenue made in this period, given $m$ purchases are made and given an inventory vector.
Decision variable is obviously the inventory vector.
CONSTRAINT
$y_1+y_2+...+y_n \leq C$ [capacity constraint]
How do you approach this problem? It's more tricky than your usual lp problem which is what I'm familiar with. I have done a bit of searching but didn't across similar problems.
Expected revenue for one transaction is just sumproduct of probablities and profits. So we have $m$ transactions and just multiplying by $m$ is wrong since we don't know if the products still remain in inventory by the time we reach $m-th$ transaction (quantity is what we have to decide).
If in each transaction only one unit is sold, I guess your objective function is something like
$$ \sum_j^m \sum_i^n r_i.p_i·1_{ij},$$
where $1_{ij}$ is an indicator function, that takes value $1$ if after $j$ transactions there is still inventory left for product $i$ and $0$ otherwise. If initially you have $y_i$ units of this particular product, after $j$ transactions you can expect to have $y_i-j·p_i$ units left (because the number of units sold has a binomial distribution). Therefore $1_{ij}=1_{y_i-j·p_i>0},$ the indicator function will take value $1$ if $1y_i-j·p_i>0,$ and $0$ otherwise.