I am not a math student, merely a biology student trying my hand at behavioral modeling, so I apologize in advance if the terminology or notation I use isn't appropriate for the field.
Set up
Individual ants within a colony of $N$ ants can be in one of two states: it can perform an action or it does not.
The decision to start a task is based on the current need for the task, $S$ (which can vary between $0$ and $1$). The higher $S$, the more likely an ant will start performing an action.
Every timestep, $S$ increases by the constant $C$ and decreases every time an ant performs the task. The probability that an ant will start a task is $S^2/(S^2+.25)$. The probability that an ant will continue doing a task given that it is already doing it is $D$.
The current value of $S$ is given by equation $(1)$:
$S_{i+1} = S_i + C - \frac{2C}{N}n_i \tag{1}$
If $t$ timesteps pass, then equation $(2)$ gives the value of $S$:
$$S_t= S_i + Ct - \frac{2C}{N}n_i - \frac{2C}{N}n_{i+1} - \frac{2C}{N}n_{i+2} -\ldots-\frac{2C}{N}n_t\tag{2}$$
The number of active ants updates with equation $(3)$:
$$n_{i+1} = (N-n_i)\frac{S^2}{S^2+.25} + n_iD \tag3$$
When $D = .5$, $C = .1$, I know that eventually, this system of equations will eventually converge on the point $S = .5$, $n = N/2$ (see attached figure of vector field).
Question
My question is for a given $S_\text{initial}$ and $n_\text{initial}$ set of values, at what value of $t$ will the system converge at this point?
Vector field of the model
The vertical line is the point at which $S$ stops changing, and the curve is the line at which $n$ stops changing, thus their intersection point is the equilibrium point.
You're looking at the nonlinear recurrence $(S(i+1), n(i+1)) = f(S(i),n(i))$ where $$ f(S,n) = \left(\frac{1}{10} + S - \frac{n}{5N}, \; \frac{(N-n) S^2}{S^2+1/4} + \frac{n}{2} \right) $$ There is essentially no possibility of closed-form solutions. However, what we can do is linearize around the fixed point $S=1/2, n=N/2$. The Jacobian matrix there is $$ \pmatrix{1 & - 1/(5N)\cr N/2 & 0\cr} $$ which has eigenvalues $\lambda_{\pm} = 1/2 \pm \sqrt{15}/10 \approx 0.8872983346$ and $0.1127016654$. Since these are both $< 1$ in absolute value, the fixed point is attractive. An eigenvector for the "slow" eigenvalue $ \lambda_+ = 1/2 + \sqrt{15}/10$ is $$ \pmatrix{1 \cr 5 N \lambda_-}$$ Thus for almost all initial states, for large $i$ we should have (for some $c$ depending on the initial state) $$ S(i) \approx 1/2 + c \lambda_+^i,\; n(i) \approx N/2 + 5 c N \lambda_- \lambda_+^i$$
I don't think there's a good way to determine $c$ for a given initial state except by numerical simulation. Thus I can't really answer your (revised) question, but I hope this is of some help.