On Nov 18, 3:24 pm, "danhey...@yahoo.com" <danhey...@yahoo.com> wrote: > On Nov 17, 8:32 pm, junoexpress <mtbrenne...@gmail.com> wrote: > > > > > > > > > > > On Nov 17, 5:22 pm, "danhey...@yahoo.com" <danhey...@yahoo.com> wrote: > > > > Since you discard all end of year inventory, B cannot exceed A_n. > > > In this problem you don't discard all end of year inventory. The place > > where you get extra M&Ms when you need them is like a very big > > reservoir that has some fixed capacity C. In this problem, we are > > essentially assuming that C is large enough that it can cover the > > deficit from all the years. We will let c(i) denote the number of M&Ms > > in the reservoir in the ith year. The goal is to find the maximum > > value of B such that c(n) = C. You only discard M&Ms when the excess, > > i.e. A_i - B) + c(i-1) > C (which is the excess that the reservoir > > cannot contain). > > > Matt > > This is what you wrote in your original message. > > If I have a deficit of M&Ms (i.e. F_i < 0), I add this to any deficit > I already have. > If I have a surplus of M&Ms in any yr, I cover any deficit I have, and > then throw out the excess. > > This doesn't seem to agree with what you just wrote. By inventory, I > mean the amount of M&Ms at the end of the year, not the vast supply > used to cover deficits. If you want to include a reservoir of C M&Ms, > change the recursion to > > X_(i+1)=min[C,X_i + F_i] > > with X_0=C.
The only thing that is thrown away are any M&Ms that when added to the current reservoir of M&Ms, causes it to exceed C. So I agree with your condition X_(i+1)=min[C,X_i + F_i], where X(i) denotes the number in the reservoir in year i. Sorry for any confusion caused on my part.
You do have some upper bounds on B. A(n), as you note is one. The mean of the A(n) is another.