How many time steps does it take for iterative model to reach equilibrium? 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.
 A: 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.
