I'm studying the trade-off between speed and quality. We need to finish a fixed number of jobs, let's say $n$ jobs. We receive a reward for finishing a job and this reward is decreasing and convex in time. All jobs are the same and share the same reward function. At each job, we can work at a normal rate and get a normal reward; or we can work fast and get a discounted reward, but this allows us to get to the other jobs faster. Our objective is to maximize the total reward for all $n$ jobs. I formulated the problem as follows:
Let the reward function $f(x)$ be decreasing and convex in $x$, $0\leq f(x)\leq 1$ and $x\geq 0$. When you work normally, it takes you $μ$ units of time to finish a job. When you speed up, it takes you $μ'$ units of time to finish a job. $μ$ and $μ'$ are fixed. Let $g_i(x)$ be the maximum expected reward from job i to job n when you start job $i$ at time $x$, $i∈(1,2,...,n)$. Let the discount in reward when you speed up be $α$, $0<α<1$. For a fixed $\alpha$, we can construct $g_i(x)$ recursively as: \begin{equation}g_i(x)=\max (f(x)+g_{i+1}(x+\mu), \alpha f(x)+g_{i+1}(x+\mu')) \end{equation} where $\mu>\mu'>0$ and $i\in(1,2,...,n-1)$.
Whether we should speed up for a job depends on the comparison between the fixed $\alpha$ and $1-\frac{g_{i+1}(x+μ′)−g_{i+1}(x+μ)}{f(x)}$. We speed up if $\alpha$ is greater. I've been trying to understand how $1-\frac{g_{i+1}(x+μ′)−g_{i+1}(x+μ)}{f(x)}$ changes with respect to $i$ and $x$ without giving $f(x)$ an explicit form. What are some sufficient conditions so that $1-\frac{g_{i+1}(x+μ′)−g_{i+1}(x+μ)}{f(x)}$ is monotonically increasing or decreasing in $i$ or $x$?
With the current formulation, I find it very difficult to obtain any useful analytical results, is there a better way to formulate this problem?