Using recursion to organize office productivity

Question

I am working on a project for a campus consulting firm with regards to optimizing a client's office productivity. On average, a phone call between employees lasts about 5 minutes. An average phone call with a client lasts 10 minutes. I am trying to develop a closed summation with regards to recursion that shows the possible ways of organizing the distribution of calls within a $5x$ minute time period.

This equation should show the number of ways it is possible to organize phone calls in a given $5x$ minute time period.

For example, when $x=1$, this summation should evaluate to 1, because our only option is to fit in a 5 minute employee phone call.

And when $x = 2$, the summation should return 2, since we can either split this time into 2 employee phone calls or 1 client call.

I am assuming that my base case will be when $x=1$, as we only have one option as to how we can use this time, an employee call.

I am wondering as to how I should proceed, as I run into a bit of trouble when I have significantly more time. The difficulty arises with, say we have $x = 3$, so 15 minutes. Say I assign 2 employee calls for the first 10 minutes. How can I use this equation to correctly consider how much time I have used, and how much I have left?

Any help with this would be greatly appreciated!

Answer 1

Let's call the number of ways we can use $5x$ minutes $f_x$. From your statements earlier, we have $f_1 = 1$ and $f_2 = 2$. From the way that we've defined $f_x$, we also have $f_0 = 1$, since trivially, there is only one way to use $0$ minutes: by doing nothing.

Consider $y, z \in \mathbb{N}$ such that $x = y + z$. Let's try to think about how we can express $f_x$ in terms of $f_y$ and $f_z$. The first $5y$ minutes can be used $f_y$ different ways, and the next $5z$ minutes can be used $f_z$ ways, but that is missing some ways we can use the $5x$ minutes. Specifically, what if we do some configuration of calls with the first $5(y-1)$ minutes, then do a client call taking 10 minutes, then do some configuration of calls with the last $5(z-1)$ minutes? In other words, when $5y$ minutes have passed, we can either be inbetween calls, or in the middle of a client call. That means we have

$$f_x = f_yf_z + f_{y-1}f_{z-1}$$

since those two situations are disjoint. I played around with this for a few minutes and used it to figure out $f_x$ for a few values until it looked familiar, then I realized to try $y = x-1$ and $z = 1$, getting us

$$f_x = f_{x-1}f_1 + f_{x-2}f_0 = f_{x-1} + f_{x-2}.$$

Look familiar?